diff --git "a/data_all_eng_slimpj/shuffled/split2/finalzzfxnm" "b/data_all_eng_slimpj/shuffled/split2/finalzzfxnm" new file mode 100644--- /dev/null +++ "b/data_all_eng_slimpj/shuffled/split2/finalzzfxnm" @@ -0,0 +1,5 @@ +{"text":"\\section{Introduction}\n\nRecently quantum field theories on a noncommutative space-time have received\na lot of attention, see for example \\cite{Obzor} and references there. The\nnoncommutative $d+1$ space-time can be realized by the coordinate operators $%\n\\hat{q}^{\\mu },$ $\\mu =0,1,...,d\\,,$ satisfying%\n\\begin{equation}\n\\left[ \\hat{q}^{\\mu },\\hat{q}^{\\nu }\\right] =i\\theta ^{\\mu \\nu }\\,,\n\\label{a.1}\n\\end{equation}%\nwhere, in the general case, the noncommutativity parameters enter in the\ntheory via an antisymmetric matrix $\\theta ^{\\mu \\nu \\,}$. Obviously, many\nof principle problems related to the noncommutativity can be examined\nalready in the noncommutative quantum mechanics (QM). Some of articles in\nthis direction consider generalization of the well-known QM problems\n(harmonic oscillator \\cite{Smalagic1}, Landau problem \\cite{Nair}, Hydrogen\natom spectrum \\cite{Chaichian1}, a particle in the Aharonov-Bohm field \\cite%\n{Chaichian}, and a system in a central potential \\cite{Gamboa}) for the\nnoncommutative case, trying to extract possible observable differences with\nthe commutative case. In this connection path integral representations in\nnonrelativistic QM were studied \\cite{Mangano}-\\cite{Tan} and calculated for\nsimple cases of the harmonic oscillator \\cite{Acatrinei} and a free particle \n\\cite{Tan}.\n\nOne ought to say that classical actions of field theories on a\nnoncommutative space-time can be written as some modified classical actions\nalready on the commutative space-time (introducing a star product). Then the\nquantization of such modified actions (let us call them $\\theta$-modified\nactions in what follows) reproduces both space-time noncommutativity and\nusual quantum mechanical features of the corresponding field theory.\nConsidering QM of one particle (or a system of $N$ particles) with\nnoncommutative coordinates, one can ask the question how to construct a $%\n\\theta$-modified classical action (with already commuting coordinates) for\nthe system. As in the case of field theory, such $\\theta$-modified classical\nactions in course of a quantization must reproduce both noncommutativity of\ncoordinates and usual QM features of the corresponding finite-dimensional\nphysical system. For nonrelativistic QM, the latter problem was solved by\nDeriglazov in \\cite{Deriglazov}. In the present article we discuss the\nproblem of constructing $\\theta$-modified actions for relativistic QM. We\nconstruct $\\theta$-modified actions for relativistic spinless and spinning\nparticles. The key idea is to extract $\\theta$-modified actions of the\nrelativistic particles from path integral representations of the\ncorresponding noncommtative field theory propagators. We consider $\\theta$%\n-modified Klein-Gordon and Dirac equations with external backgrounds for the\ncausal propagators. Then, using technics developed in \\cite{FG,Gitman} for\nusual commutative case, we construct for them path-integral representations.\nEffective actions in such path-integral representations, we treat as $\\theta$%\n-modified actions of the relativistic particles. To confirm this\ninterpretation, we quantize canonically these actions. Thus, we obtain the\nabove mentioned $\\theta$-modified Klein-Gordon and Dirac equations. The $%\n\\theta$-modified action of the relativistic spinning particle is a\ngeneralization of the Berezin-Marinov pseudoclassical action \\cite{Berezin}\nfor the noncommutative case. One ought to say that effects of the\nnoncommutativity appear to be essential only due to the external background.\nFinally, we consider a noncommutative $d$-dimensional nonrelativistic QM\nwith no restrictions on the noncommutativity parameters $\\theta^{\\mu\\nu\\,}$\nand formally arbitrary Hamiltonian. We construct a path integral\nrepresentation for the corresponding propagation function and demonstrate\nthat the effective action in our path-integral representation is just $%\n\\theta $-modified action for nonrelativistic QM proposed in \\cite{Deriglazov}%\n.\n\n\\section{Path integral representations for particle propagators in\nnoncommutative field theory}\n\n\\subsection{Spinless case}\n\nIn field theories the effect of the noncommutativity of the space-time can\nbe realized by substitution of usual function product by the Weil-Moyal star\nproduct \n\\begin{align}\nf\\left( x\\right) \\ast g\\left( x\\right) & =f\\left( x\\right) \\exp\\left\\{ \\frac{%\ni}{2}\\overleftarrow{\\partial}_{\\mu}\\theta^{\\mu\\nu }\\overrightarrow{%\n\\partial_{\\nu}}\\right\\} g\\left( x\\right) \\notag \\\\\n& =f\\left( x^{\\mu}+\\frac{i}{2}\\theta^{\\mu\\nu}\\partial_{\\nu}\\right) g\\left(\nx\\right) \\,, \\label{13}\n\\end{align}\nwhere $\\ f\\left( x\\right) $ and $g\\left( x\\right) $ are two arbitrary\ninfinitely differentiable functions of the commutative variables $x^{\\mu}$\nand the second line in (\\ref{13}) holds if perturbation in $\\theta$ are\npossible (see \\cite{Chaichian}). The latter is presumably since the effect\nof noncommutativity should be small.\n\nThe action of a noncommutative field theory of a scalar field $\\Phi$ that\ninteracts with an external electromagnetic field $A_{\\mu}\\left( x\\right) $\nreads%\n\\begin{equation}\nS_{\\mathrm{scal-field}}^{\\theta}=\\int d^{D}x\\left[ \\left( P_{\\mu}\\ast\n\\Phi\\right) \\ast\\left( P^{\\mu}\\ast\\bar{\\Phi}\\right) +m^{2}\\Phi\\bar{\\Phi }%\n\\right] \\,,\\;P_{\\mu}=i\\partial_{\\mu}-gA_{\\mu}\\left( x\\right) \\,. \\label{15}\n\\end{equation}\nThe corresponding Euler-Lagrange equation%\n\\begin{equation}\n\\frac{\\delta S_{\\mathrm{scal-field}}^{\\theta}}{\\delta\\bar{\\Phi}}=0\\Rightarrow%\n\\left[ P_{\\mu}\\ast P^{\\mu}-m^{2}\\right] \\ast\\Phi=0\\,, \\label{16}\n\\end{equation}\nbeing rewritten by the help of (\\ref{13}) takes the form%\n\\begin{align}\n& \\left( \\tilde{P}^{2}-m^{2}\\right) \\Phi=0~,\\ \\ \\tilde{P}^{2}=\\tilde {P}%\n_{\\mu}\\tilde{P}^{\\mu}, \\label{16a} \\\\\n& \\tilde{P}_{\\mu}=i\\partial_{\\mu}-gA_{\\mu}\\left( x^{\\mu}+\\frac{i}{2}%\n\\theta^{\\mu\\nu}\\partial_{\\nu}\\right) \\,, \\label{16b}\n\\end{align}\nand is an analog of the Klein-Gordon equation for noncommutative case. The\npropagator in the noncommutative scalar field theory is the causal Green\nfunction\\ $D^{c}\\left( x,y\\right) $ of the equation (\\ref{16a}), \n\\begin{equation}\n\\left( \\tilde{P}^{2}-m^{2}\\right) D^{c}\\left( x,y\\right) =-\\delta\\left(\nx-y\\right) ~. \\label{17}\n\\end{equation}\n\nFrom this point on, we are going to follow the way elaborated in \\cite{FG}\nto construct a path integral representation for the propagator: We consider $%\nD^{c}\\left( x,y\\right) $ as a matrix element of an operator $\\hat{D}^{c}$ in\na Hilbert space%\n\\begin{equation}\nD^{c}\\left( x,y\\right) =\\left\\langle x\\right| \\hat{D}^{c}\\left|\ny\\right\\rangle \\,. \\label{18}\n\\end{equation}\nHere $\\left| x\\right\\rangle $ are eigenvectors of some self-adjoint and\nmutually commuting operators $\\hat{x}^{\\mu},$ \n\\begin{equation}\n\\hat{x}^{\\mu}=\\hat{q}^{\\mu}+\\frac{1}{2\\hbar}\\theta^{\\mu\\nu}\\hat{p}_{\\nu},\n\\label{cc}\n\\end{equation}\nwhere operators $\\hat{q}^{\\mu}$ obey the commutation relations (\\ref{a.1})\nand $\\hat{p}_{\\mu}$ are momentum operators conjugated to $\\hat{x}^{\\mu},$%\n\\begin{align}\n& \\left[ \\hat{x}^{\\mu},\\hat{p}_{\\nu}\\right] =i\\hbar\\delta_{\\nu}^{\\mu }\\,,\\;%\n\\left[ \\hat{x}^{\\mu},\\hat{x}^{\\nu}\\right] =\\left[ \\hat{p}_{\\mu},\\hat{p}_{\\nu}%\n\\right] =0\\,, \\notag \\\\\n& \\hat{x}^{\\mu}\\left| x\\right\\rangle =x^{\\mu}\\left| x\\right\\rangle\n\\,,\\;=\\delta^{D}\\left( x-y\\right) \\,,\\;\\int|x>,\\;\\;a,b=1,2,\\ldots,2^{d}\\,,\n\\label{b6}\n\\end{equation}\nwhere the spinor indices $a,b$ are written here explicitly for clarity and\nwill be omitted hereafter. The equation (\\ref{b5}) implies $\\hat{S}%\n^{c}=\\left( \\Pi_{n}\\Gamma^{n}\\right) ^{-1},$ where $\\Pi_{\\mu}$ are defined\nin (\\ref{20}), and $\\Pi_{D}=-m$. Using a generalization of the Schwinger\nproper-time representation, proposed in \\cite{FG}, we write the Green\nfunction (\\ref{b6}) in the form \n\\begin{align}\n& \\tilde{G}^{c}=\\tilde{G}^{c}(x_{out},x_{in})=\\int_{0}^{\\infty}\\,d\\lambda\n\\int\\langle x_{\\mathrm{out}}|e^{-i\\hat{\\mathcal{H}}(\\lambda,\\chi )}|x_{%\n\\mathrm{in}}\\rangle d\\chi\\,, \\label{b10} \\\\\n& \\hat{\\mathcal{H}}(\\lambda,\\chi)=\\lambda\\left( m^{2}-\\Pi^{2}+\\frac{ig}{2}%\nF_{\\mu\\nu}\\Gamma^{\\mu}\\Gamma^{\\nu}\\right) +\\Pi_{n}\\Gamma^{n}\\,\\chi\\;. \\notag\n\\end{align}\n\nSimilar to \\cite{FG}, we present the matrix element entering in the\nexpression (\\ref{b10}) by means of a Hamiltonian path integral%\n\\begin{align}\n& \\tilde{G}^{c}=\\exp\\left( i\\Gamma^{n}\\frac{\\partial_{l}}{\\partial\n\\varepsilon^{n}}\\right) \\int_{0}^{\\infty}\\,d\\lambda_{0}\\int\nd\\chi_{0}\\int_{\\lambda_{0}}D\\lambda\\int_{\\chi_{0}}D\\chi%\n\\int_{x_{in}}^{x_{out}}Dx\\int Dp\\int D\\pi\\int D\\nu \\label{b15a} \\\\\n& \\times\\int_{\\psi(0)+\\psi(1)=\\varepsilon}\\mathcal{D}\\psi\\exp\\left\\{\ni\\int_{0}^{1}\\left[ \\lambda\\left( \\mathcal{P}^{2}-m^{2}+2igF_{\\mu\\nu}^{\\ast\n}\\psi^{\\mu}\\psi^{\\nu}\\right) +2i\\mathcal{P}_{n}\\psi^{n}\\chi\\right. \\right. \n\\notag \\\\\n& -i\\psi_{n}\\dot{\\psi}^{n}+\\left. \\left. p_{\\mu}\\dot{x}^{\\mu}+\\pi \\dot{%\n\\lambda}+\\nu\\dot{\\chi}\\right] d\\tau+\\left. \\psi_{n}(1)\\psi ^{n}(0)\\right\\}\n\\right\\vert _{\\varepsilon=0}\\,. \\notag\n\\end{align}\nHere $\\varepsilon^{n}$ are odd variables, anticommuting with the $\\Gamma $%\n-matrices,%\n\\begin{equation}\n\\mathcal{P}_{\\mu}=-p_{\\mu}-gA_{\\mu}\\left( x^{\\mu}-\\frac{1}{2\\hbar}\\theta\n^{\\mu\\nu}p_{\\nu}\\right) ,\\;\\mathcal{P}_{D}=-m,\\;F_{\\mu\\nu}^{\\ast}=F_{\\mu\\nu\n}^{\\ast}\\left( x^{\\mu}-\\frac{1}{2\\hbar}\\theta^{\\mu\\nu}p_{\\nu}\\right) , \n\\notag\n\\end{equation}\nthe function $F_{\\mu\\nu}^{\\ast}\\left( q\\right) $ is defined in (\\ref{20a}),\\\nand the integration goes over even trajectories $x\\left( \\tau\\right)\n,~p\\left( \\tau\\right) ,~\\lambda\\left( \\tau\\right) ,$ $\\pi\\left( \\tau\\right)\n, $ and odd trajectories $\\psi_{n}(\\tau),$ $\\chi (\\tau),\\;\\nu(\\tau),$\nparametrized by some invariant parameter $\\tau\\in\\left[ 0,1\\right] $ and\nobeying the boundary conditions $\\,x(0)=x_{\\mathrm{in}}$, $x(1)=x_{\\mathrm{%\nout}}$, $\\lambda(0)=\\lambda_{0}$, $\\chi(0)=\\chi_{0}$.\n\nPerforming the change of variables (\\ref{24})$\\ $in (\\ref{b15a}), we obtain\nanother representaion for $\\tilde{G}^{c},$%\n\\begin{align}\n& \\tilde{G}^{c}=\\exp \\left( i\\Gamma ^{n}\\frac{\\partial _{l}}{\\partial\n\\varepsilon ^{n}}\\right) \\int_{0}^{\\infty }\\,d\\lambda _{0}\\int d\\chi\n_{0}\\int_{\\lambda _{0}}D\\lambda \\int_{\\chi _{0}}D\\chi \\overset{\\infty }{%\n\\underset{-\\infty }{\\int }}Dp\\overset{x_{out}-\\theta p\/2\\hbar }{\\underset{%\nx_{in}-\\theta p\/2\\hbar }{\\int }}Dq\\int D\\pi \\int D\\nu \\label{b16} \\\\\n& \\times \\int_{\\psi (0)+\\psi (1)=\\varepsilon }\\mathcal{D}\\psi \\left. \\exp\n\\left\\{ i\\left[ S_{\\mathrm{spin-part}}^{\\theta }+S_{\\mathrm{GF}}\\right]\n+\\psi _{n}(1)\\psi ^{n}(0)\\right\\} \\right\\vert _{\\varepsilon =0}\\,, \\notag\n\\end{align}%\nwhere \n\\begin{subequations}\n\\begin{align}\n& S_{\\mathrm{spin-part}}^{\\theta }=\\int_{0}^{1}\\left[ \\lambda \\left( \\left(\np_{\\mu }+gA_{\\mu }\\right) ^{2}-m^{2}+2igF_{\\mu \\nu }^{\\ast }\\psi ^{\\mu }\\psi\n^{\\nu }\\right) +2i\\left( p_{\\mu }+gA_{\\mu }\\left( q\\right) \\right) \\psi\n^{\\mu }\\chi \\right. \\notag \\\\\n& \\left. -2im\\psi ^{D}\\chi -i\\psi _{n}\\dot{\\psi}^{n}+p_{\\mu }\\dot{q}^{\\mu }+%\n\\frac{1}{2\\hbar }\\dot{p}_{\\mu }\\theta ^{\\mu \\nu }p_{\\nu }\\right] d\\tau \\;,\n\\label{BM} \\\\\n& S_{\\mathrm{GF}}=\\int_{0}^{1}\\left( \\pi \\dot{\\lambda}+\\nu \\dot{\\chi}\\right)\nd\\,\\tau . \\label{GF}\n\\end{align}\n\nNote that in the work \\cite{BB} was made an attempt to construct the path\nintegral representation of Green function of noncommutative Dirac equation.\nHowever the consideration was perturbative in $\\theta $ (taken into account\nonly the first order perturbation). As a consequence the authors did not\nobtain the corresponding action (\\ref{BM}), moreover the essential term $%\n\\dot{p}_{\\mu }\\theta ^{\\mu \\nu }p_{\\nu }\/2\\hbar $ was missed.\n\n\\section{Pseudoclassical action of spinning particle in noncommutative space\ntime}\n\nSimilar to the spinless case, the exponent in the integrand (\\ref{b16}) can\nbe considered as an effective and non-degenerate Hamiltonian action of a\nspinning particle in the noncommutative space time. It consists of two\nprincipal parts. The first one $S_{\\mathrm{GF}}$ with derivatives of $%\n\\lambda $ and $\\chi$ can be treated as a gauge fixing term, which\ncorresponds to gauge conditions $\\dot{\\lambda}=\\dot{\\chi}=0$. The rest part $%\nS_{\\mathrm{spin-part}}^{\\theta}$ can be treated as a gauge invariant action\nof a spinning particle in the noncommutative space time. The action $S_{%\n\\mathrm{spin-part}}^{\\theta}$ is a $\\theta$-modification of the Hamiltonian\nform of the Berezin-Marinov action \\cite{Berezin}. It will be studied and\nquantized below to justify such an interpretation.\n\nOne can easily verify that $S_{\\mathrm{spin-part}}^{\\theta}$ is\nreparametrization invariant. Explicit form of supersymmetry transformations,\nwhich generalize ones for the Berezin-Marinov action, is not so easily to\nderive. Their presence will be proved in an indirect way. Namely, we are\ngoing to prove the existence of two primary first-class constraints in the\ncorresponding Hamiltonian formulation.\n\nLet us consider $S_{\\mathrm{spin-part}}^{\\theta}$ as a Lagrangian action\nwith generalized coordinates $Q_{A}=\\left( q^{\\mu},p_{\\mu}\\right) $, $%\nA=(\\zeta,\\mu),$ $\\zeta=1,2,\\;Q_{1\\mu}=q^{\\mu},\\;Q_{2\\mu}=p_{\\mu}$ ; $\\chi,$ $%\n\\psi,$ and $\\lambda,$ and let us perform a Hamiltonization of such an\naction. To this end, we introduce the canonical momenta $P$ conjugate to the\ngeneralized coordinates as follows: \n\\end{subequations}\n\\begin{align}\n& P_{Q_{A}}=\\frac{\\partial L}{\\partial\\dot{Q}^{A}}=J_{A}\\left( Q\\right)\n\\,,\\;J_{1\\mu}=p_{\\mu}\\,,\\ \\ J_{2\\mu}=\\frac{1}{2\\hbar}\\theta^{\\mu\\nu}p_{\\nu\n}\\,, \\notag \\\\\n& P_{\\lambda}=\\frac{\\partial L}{\\partial\\dot{\\lambda}}=0,\\;P_{\\chi }=\\frac{%\n\\partial_{r}L}{\\partial\\dot{\\chi}}=0,\\;P_{n}=\\frac{\\partial_{r}L}{\\partial%\n\\dot{\\psi}^{n}}=-i\\psi_{n}\\,. \\label{d1}\n\\end{align}\n\\break It follows from equations (\\ref{d1}) that there exist primary\nconstraints $\\Phi^{(1)}=0$, \n\\begin{equation}\n\\Phi_{l}^{(1)}=\\left\\{ \n\\begin{array}{l}\n\\Phi_{1A}^{(1)}=P_{A}-J_{A}\\left( Q\\right) ~, \\\\ \n\\Phi_{2}^{(1)}=P_{\\lambda}\\,\\,\\,,\\ \\ \\Phi_{3}^{\\left( 1\\right) }=P_{\\chi }~,\n\\\\ \n\\Phi_{4n}^{(1)}=P_{n}+i\\psi_{n}\\,\\,\\,.%\n\\end{array}\n\\right. \\label{d2}\n\\end{equation}\nThe Poisson brackets of primary constraints are%\n\\begin{align}\n& \\{\\Phi_{1A}^{\\left( 1\\right) },\\Phi_{1B}^{\\left( 1\\right)\n}\\}=\\Omega_{AB}=\\left( \n\\begin{array}{cc}\n\\mathbf{0} & \\mathbb{I} \\\\ \n-\\mathbb{I} & \\mathbf{\\theta}\/\\hbar%\n\\end{array}\n\\right) \\,,\\;\\left\\{ \\Phi_{4n}^{(1)},\\Phi_{4m}^{(1)}\\right\\} =2i\\eta _{nm}~,\n\\\\\n& \\{\\Phi_{1A}^{\\left( 1\\right) },\\Phi_{4n}^{\\left( 1\\right)\n}\\}=\\{\\Phi_{1A}^{\\left( 1\\right) },\\Phi_{2,3}^{\\left( 1\\right)\n}\\}=\\{\\Phi_{4n}^{\\left( 1\\right) },\\Phi_{2,3}^{\\left( 1\\right) }\\}=0~. \n\\notag\n\\end{align}\nwhere $\\mathbf{\\theta}=\\theta^{\\mu\\nu\\,},$\\ $\\mathbb{I}$ is a $D\\times D$\nunit matrix, and $\\mathbf{0}$ denotes an $D\\times D$ zero matrix. Note that $%\n\\det\\Omega_{AB}=1$, and \n\\begin{equation*}\n\\omega^{AB}=\\Omega_{AB}^{-1}=\\left( \n\\begin{array}{cc}\n\\mathbf{\\theta}\/\\hbar & -\\mathbb{I} \\\\ \n\\mathbb{I} & \\mathbf{0}%\n\\end{array}\n\\right) ~.\n\\end{equation*}\n\nNow we construct the total Hamiltonian $H^{(1)}$, according to the standard\nprocedure \\cite{GTbook}. Thus, we obtain: \n\\begin{align}\n& H^{(1)}=H+\\Lambda_{l}\\Phi_{l}^{(1)}, \\notag \\\\\n& H=-\\lambda\\left[ \\left( p_{\\mu}+gA_{\\mu}\\right) ^{2}-m^{2}+2igF_{\\mu\\nu\n}^{\\ast}\\left( q\\right) \\psi^{\\mu}\\psi^{\\nu}\\right] +2i\\chi\\left( \\left(\np_{\\mu}+gA_{\\mu}\\right) \\psi^{\\mu}-m\\psi^{D}\\,\\right) \\,. \\label{d3}\n\\end{align}\nwhere $\\Lambda_{l}$\\ .... The consistency conditions $\\dot{\\Phi}%\n_{1A,4n}^{(1)}=\\left\\{ {\\Phi}_{1A,4n}^{(1)},H^{(1)}\\right\\} =0$ for the\nprimary constraints $\\Phi_{1A}^{(1)}$ and $\\Phi_{4n}^{(1)}$ allow us to fix\nthe Lagrange multipliers $\\lambda^{1A}$ and $\\lambda^{4n}$. The consistency\nconditions for the constraints $\\Phi_{2,3}^{(1)}$ imply secondary\nconstraints $\\Phi_{1,2}^{(2)}=0$, \n\\begin{align}\n& \\Phi_{1}^{(2)}=\\left( p_{\\mu}+gA_{\\mu}\\right) \\psi^{\\mu}-m\\psi ^{D}=0\\,,\n\\label{d4} \\\\\n& \\Phi_{2}^{(2)}=\\left( p_{\\mu}+gA_{\\mu}\\right) ^{2}-m^{2}+2igF_{\\mu\\nu\n}^{\\ast}\\psi^{\\mu}\\psi^{\\nu}=0\\,. \\label{d5}\n\\end{align}\nThus, the Hamiltonian $H$ appears to be proportional to constraints, as\nalways in the case of a repara\\-metrization invariant theory, \n\\begin{equation*}\nH=2i\\chi\\Phi_{1}^{(2)}-\\lambda\\Phi_{2}^{(2)}.\n\\end{equation*}\nNo more secondary constraints arise from the Dirac procedure, and the\nLagrange multipliers $\\lambda^{2}$ and $\\lambda^{3}$ remain undetermined, in\nperfect correspondence with the fact that the number of gauge\ntransformations parameters equals two for the theory in question.\n\nOne can go over from the initial set of constraints $\\left(\n\\Phi^{(1)},\\Phi^{(2)}\\right) $ to the equivalent one $\\left(\n\\Phi^{(1)},T\\right) ,$ where: \n\\begin{equation}\nT=\\Phi^{(2)}+\\frac{\\partial\\Phi^{\\left( 2\\right) }}{\\partial q^{A}}%\n\\omega^{AB}\\Phi_{1B}^{(1)}+\\frac{i}{2}\\frac{\\partial_{r}\\Phi}{\\partial\\psi\n^{n}}^{(2)}\\Phi_{4n}^{(1)}\\,. \\label{d6}\n\\end{equation}\nThe new set of constraints can be explicitly divided in a set of first-class\nconstraints, which is $\\left( \\Phi_{2,3}^{(1)},T\\right) $ and in a set of\nsecond-class constraints, which is $\\left(\n\\Phi_{1A}^{(1)},\\Phi_{4n}^{(1)}\\right) $.\n\nNow we consider an operator quantization. To this end we perform a partial\ngauge fixing, imposing gauge conditions $\\Phi_{1,2}^{\\mathrm{G}}=0$ to the\nprimary first-class constraints $\\Phi_{1,2}^{(1)}\\,$, \n\\begin{equation}\n\\Phi_{1}^{\\mathrm{G}}=\\chi=0,\\,\\,\\,\\,\\,\\,\\Phi_{2}^{\\mathrm{G}}=\\lambda=1\/m\\,.\n\\label{d7}\n\\end{equation}\nOne can check that the consistency conditions for the gauge conditions (\\ref%\n{d7}) lead to fixing the Lagrange multipliers $\\lambda_{2}$ and $\\lambda_{3}$%\n. Thus, on this stage we reduced our Hamiltonian theory to one with the\nfirst-class constraints $T$ and second-class ones $\\varphi=\\left(\n\\Phi^{(1)},\\Phi^{\\mathrm{G}}\\right) $. Then, we apply the so called Dirac\nmethod for systems with first-class constraints \\cite{Dirac64}, which, being\ngeneralized to the presence of second-class constraints, can be formulated\nas follow: the commutation relations between operators are calculated\naccording to the Dirac brackets with respect to the second-class constraints\nonly; second-class constraints as operators equal zero; first-class\nconstraints as operators are not zero, but, are considered in sense of\nrestrictions on state vectors. All the operator equations have to be\nrealized in a Hilbert space.\n\nThe subset of the second-class constraints $\\left( \\Phi_{2,3}^{(1)},\\Phi^{%\n\\mathrm{G}}\\right) $ has a special form \\cite{GTbook}, so that one can use\nit for eliminating of the variables $\\lambda,P_{\\lambda},\\chi,P_{\\chi}$,\nfrom the consideration, then, for the rest of the variables $q,p,\\psi^{n}$,\nthe Dirac brackets with respect to the constraints $\\varphi$ reduce to ones\nwith respect to the constraints $\\Phi_{1A}^{(1)}$ and $\\Phi_{4n}^{(1)}$ only\nand can be easy calculated, \n\\begin{equation*}\n\\left\\{ Q^{A},Q^{B}\\right\\}\n_{D(\\Phi^{(1)})}=\\omega^{AB}\\,,\\,\\,\\,\\,\\,\\,\\,\\left\\{\n\\psi^{n},\\psi^{m}\\right\\} _{D(\\Phi^{(1)})}=\\frac{i}{2}\\eta^{nm}\\,,\n\\end{equation*}\nwhile all other Dirac brackets vanish. Thus, the commutation relations for\nthe operators $\\hat{q}^{\\mu},\\hat{p}_{\\mu},\\hat{\\psi}^{n}$, which correspond\nto the variables $q^{\\mu},p_{\\mu},\\psi^{n}$ respectively, are \n\\begin{align}\n& \\left[ \\hat{q}^{\\mu},\\hat{p}_{\\nu}\\right] _{-}=i\\hbar\\omega^{\\mu,D+\\nu\n}=i\\hbar\\delta_{\\nu}^{\\mu}\\,,\\;\\left[ \\hat{q}^{\\mu},\\hat{q}^{\\nu}\\right]\n=i\\hbar\\omega^{\\mu\\nu}=i\\theta^{\\mu\\nu},\\;\\left[ \\hat{p}_{\\mu},\\hat{p}_{\\nu }%\n\\right] =0, \\notag \\\\\n& \\left[ \\hat{\\psi}^{m},\\hat{\\psi}^{n}\\right] _{+}=i\\left\\{\n\\psi^{m},\\psi^{n}\\right\\} _{D(\\Phi^{(1)})}=-\\frac{1}{2}\\eta^{mn}\\,.\n\\label{d8}\n\\end{align}\nBesides, the following operator equations hold: \n\\begin{equation}\n\\hat{\\Phi}_{1A}^{(1)}=\\hat{P}_{A}-J_{A}\\left( \\hat{Q}\\right) ,\\;\\hat{\\Phi }%\n_{4n}^{(1)}=\\hat{P}_{n}+i\\hat{\\psi}_{n}=0. \\label{d9}\n\\end{equation}\nTaking that into account, one can construct a realization of the commutation\nrelations (\\ref{d8}) in a Hilbert space whose elements $\\Psi$ are $2^{d}$%\n-component columns dependent only on $x$, such that \n\\begin{equation}\n\\hat{q}^{\\mu}=\\left( x^{\\mu}+\\frac{i}{2}\\theta^{\\mu\\nu}\\partial_{\\nu}\\right) \n\\mathbf{I}\\,,\\;\\;\\hat{p}_{\\mu}=-i\\partial_{\\mu}\\mathbf{I}\\,,\\;\\;\\hat{\\psi}%\n^{n}=\\frac{i}{2}\\Gamma^{n}\\,, \\label{c13}\n\\end{equation}\nwhere $\\mathbf{I}$ is $2^{d}\\times2^{d}$ unit matrix, and $\\Gamma^{n}$, are\ngamma-matrices (\\ref{b4}). The first-class constraints $\\hat{T}$ as\noperators have to annihilate physical vectors; in virtue of (\\ref{d9}) and (%\n\\ref{d6}) that implies the equations: \n\\begin{equation}\n\\hat{\\Phi}_{1}^{(2)}\\Psi=0\\,,\\;\\;\\hat{\\Phi}_{2}^{(2)}\\Psi=0\\,, \\label{d10}\n\\end{equation}\nwhere $\\hat{\\Phi}_{1,2}^{(2)}$ are operators, which correspond to\nconstraints (\\ref{d4}), (\\ref{d5}). Taking into account the realizations of\nthe commutation relations (\\ref{d8}), one easily can see that the first\nequation (\\ref{d10}) takes the form of the $\\theta$-modified Dirac equation, \n\\begin{equation}\n\\left( \\tilde{P}_{\\mu}\\tilde{\\gamma}^{\\mu}-m\\gamma^{D+1}\\right)\n\\Psi=0\\Longleftrightarrow\\left( P_{\\mu}\\gamma^{\\mu}+m\\right) \\ast\\Psi=0~,\n\\label{d11}\n\\end{equation}\n\nSince $\\hat{\\Phi}_{2}^{(2)}=\\left( \\hat{\\Phi}_{1}^{(2)}\\right) ^{2}$, the\nsecond equation (\\ref{d10}) is a consequence of the first one.\n\nThus, we have constructed a $\\theta$-modification of the Berezin-Marinov\naction (\\ref{BM}) which, being quantized, leads to a quantum theory based on\nthe $\\theta$-modified Dirac equation.\n\nNote that space-time non-commutativity $\\left[ \\hat{q}^{0},\\hat{q}^{i}\\right]\n=i\\theta^{0i}$ can be obtained also from the canonical quantization of the\nconventional Lagrangian action of a relativistic spinless particle by\nimposing a special gauge condition $\\Phi_{gf}=x^{0}+\\theta\n^{0i}p_{i}-\\tau=0, $ see \\cite{PS}.\n\n\\section{Path integral in nonrelativistic quantum mechanics on a\nnoncommutative space}\n\nIn this section, we construct a path integral representation for the\npropagation function (a symbol of the evolution operator) in nonrelativistic\nQM on a noncommutative space. We compare our result with some previous\nconstructions and use it to extract a $\\theta$-modified first-order\nclassical Hamiltonian action for such a system.\n\nWe consider a $d$-dimensional nonrelativistic QM with basic canonical\noperators of coordinates $\\hat{q}^{k}$ and momenta $\\hat{p}_{j},$ $%\nk,j=1,...,d$ that obey the following commutation relations \n\\begin{equation}\n\\left[ \\hat{q}^{k},\\hat{q}^{j}\\right] =i\\theta^{kj}\\,,\\;\\left[ \\hat{q}^{k},%\n\\hat{p}_{j}\\right] =i\\hbar\\delta_{j}^{k}\\,,\\;\\left[ \\hat{p}_{k},\\hat {p}_{j}%\n\\right] =0\\,. \\label{1}\n\\end{equation}\nIt is supposed that the nonzeroth commutation relations for the coordinate\noperators in (\\ref{1}) have emerged from the noncommutative properties of\nthe position space. The time evolution of the system under consideration is\ngoverned by a self-adjoint Hamiltoniam $\\hat{H}.$ We believe that behind\nsuch a QM there exist a classical theory with a $\\theta$-modified action\n(which we are going to restore in what follows), such that a quantization of\nthis action leads to the QM.\n\nIn conventional nonrelativistic QM, one constructs a path integral\nrepresentations for matrix elements (in a coordinate representation) of the\nevolution operator $\\hat{U}\\left( t,t^{\\prime}\\right) .$ In the QM under\nconsideration, we also start with such an operator. It obeys the Schr\\\"{o}%\ndinger equation and for time independent $\\hat{H}$ (which we consider for\nsimplicity in what follows) has the form%\n\\begin{equation}\n\\hat{U}\\left( t^{\\prime},t\\right) =\\exp\\left\\{ -\\frac{i}{\\hbar }\\hat{H}%\n\\left( t^{\\prime}-t\\right) \\right\\} \\,. \\label{2}\n\\end{equation}\n\nSince the coordinate operators $\\hat{q}$ do not commute, they do not posses\na common complete set of eigenvectors. Therefore, there is no $q$-coordinate\nrepresentation and one cannot speak about matrix elements of the evolution\noperator in such a representation. Consequently, one cannot define a\nprobability amplitude of a transition between two points in the position\nspace. Nevertheless, one can consider another types of matrix elements of\nthe evolution operator that are probability amplitudes (evolution functions)\nand can be represented via path integrals. Below, we consider two types of\nsuch matrix elements, \n\\begin{equation}\nG_{p}=\\left\\langle p^{out}\\right| \\hat{U}\\left( t_{out},t_{in}\\right) \\left|\np^{in}\\right\\rangle \\;\\mathrm{and\\;}G_{x}=\\left\\langle x_{out}\\right| \\hat{U}%\n\\left( t_{out},t_{in}\\right) \\left| x_{in}\\right\\rangle \\,. \\label{2a}\n\\end{equation}\nIn (\\ref{2a}) $\\left| p\\right\\rangle $ is a complete set of eigenvectors of\ncommuting operators $\\hat{p},$%\n\\begin{align}\n& \\hat{p}_{j}\\left| p\\right\\rangle =p_{j}\\left| p\\right\\rangle\n\\,,\\;=\\delta\\left( p-p^{\\prime}\\right) \\,,\\;\\int\n|p>=\\frac{1}{\\left( 2\\pi\\hbar\\right) ^{d\/2}}\\exp\\left\\{ -\\frac{i}{\\hbar}%\np_{i}x^{i}\\right\\} \\,,\\;=i\\hbar\\frac{\\partial }{%\n\\partial p}\\,, \\label{3}\n\\end{align}\nand $\\left| x\\right\\rangle $ is a complete set of eigenvectors of some\ncommuting and canonically conjugated to $\\hat{p}$ operators $\\hat{x}^{k}.$\nWe chose these operators as follows\\footnote{%\nFor the first time the commuting operators $\\hat{x}^{k}$ were introduced in \n\\cite{Chaichian1}.}:%\n\\begin{align}\n& \\hat{x}^{k}=\\hat{q}^{k}+\\frac{\\theta^{kj}\\hat{p}_{j}}{2\\hbar}\\,,\\;\\left[ \n\\hat{x}^{k},\\hat{x}^{j}\\right] =0\\,,\\;\\left[ \\hat{x}^{k},\\hat{p}_{j}\\right]\n=i\\hbar\\delta_{j}^{k}\\,, \\notag \\\\\n& \\hat{x}^{\\mu}\\left| x\\right\\rangle =x^{\\mu}\\left| x\\right\\rangle\n\\,,\\;=\\delta^{D}\\left( x-y\\right) \\,,\\;\\int|x>\\,, \\label{6}\n\\end{equation}\nwhere $p^{\\left( 0\\right) }=p^{\\left( in\\right) }$, $p^{\\left( N\\right)\n}=p^{\\left( out\\right) }$, and $p^{\\left( k\\right) }=(p_{i}^{\\left( k\\right)\n}).$ Bearing in mind the limiting process $N\\rightarrow\\infty$ or $\\Delta\nt\\rightarrow0$ and using the completeness relation (\\ref{4a}) for the\neigenvectors $\\left\\vert x\\right\\rangle $, one can approximately calculate\nthe matrix element from (\\ref{6}),%\n\\begin{equation}\n\\thickapprox\\int dx_{(k)}\\,,\n\\label{7}\n\\end{equation}\nwhere $x_{(k)}=\\left( x_{(k)}^{i}\\right) $ and $dx_{(k)}=%\n\\prod_{i}dx_{(k)}^{i}.$ A result of this calculation can be expressed in\nterms of a classical Hamiltonian $H$, however, in general case, it will\ndepend on the choice of the correspondance rule between the classical\nfunction and quantum operator. For our calculations we choose the Weyl\nordering. In this case the matrix element (\\ref{7}) will take the form%\n\\begin{equation*}\n\\int\\frac{dx_{(k)}}{\\left( 2\\pi\\hbar\\right) ^{d}}\\exp\\left\\{ \\frac{i}{\\hbar}%\n\\left[ -x_{(k)}^{i}\\frac{p_{i}^{\\left( k\\right) }-p_{i}^{\\left( k-1\\right) }%\n}{\\Delta t}-H\\left( x_{\\left( k\\right) }-\\frac{\\mathbf{\\theta }p^{\\left(\nk\\right) \\prime}}{2\\hbar},p^{\\left( k\\right) \\prime}\\right) \\right] \\Delta\nt+O\\left( \\Delta t^{2}\\right) \\right\\} ,\n\\end{equation*}\nwhere $p^{\\left( k\\right) \\prime}=\\frac{p^{\\left( k\\right) }+p^{\\left(\nk-1\\right) }}{2},$ and $H\\left( x-\\frac{\\mathbf{\\theta}p}{2\\hbar},p\\right) $\nis the Weyl symbol of the operator $\\hat{H}.$ Using the above formula and\ntaking the limit $N\\rightarrow\\infty$ in the integral (\\ref{6}), we get for $%\nG_{p}$ the following path integral representation:%\n\\begin{equation}\nG_{p}=\\overset{p^{\\left( out\\right) }}{\\underset{p^{\\left( in\\right) }}{\\int}%\n}Dp\\int Dx\\exp\\left\\{ \\frac{i}{\\hbar}\\int dt\\left[ -x_{j}\\dot{p}^{j}-H\\left(\nx-\\frac{\\mathbf{\\theta}p}{2\\hbar},p\\right) \\right] \\right\\} \\,. \\label{8}\n\\end{equation}\n\nIn the same manner, one can construct a path integral representation for the\nevolution function $G_{x},$ which, is%\n\\begin{equation}\nG_{x}=\\int Dp\\int_{x_{(in)}}^{x_{(out)}}Dx\\exp\\left\\{ \\frac{i}{\\hbar}\\int dt%\n\\left[ p_{j}\\dot{x}^{j}-H\\left( x-\\frac{\\mathbf{\\theta}p}{2\\hbar },p\\right) %\n\\right] \\right\\} \\,. \\label{9}\n\\end{equation}\n\nLet us pass to the integration over trajectories $q=x-\\frac{\\mathbf{\\theta}p%\n}{2\\hbar}$ in path integrals (\\ref{8}) and (\\ref{9}). Then we get%\n\\begin{align}\n& G_{x}=\\int Dp\\int_{x_{(in)}-\\theta p\/2\\hbar}^{x_{\\left( out\\right)\n}-\\theta p\/2\\hbar}Dq\\exp\\left\\{ \\frac{i}{\\hbar}S_{\\mathrm{nonrel}}^{\\theta\n}\\right\\} \\,, \\label{12} \\\\\n& G_{p}=\\overset{p_{out}}{\\underset{p_{in}}{\\int}}Dp\\int Dq\\exp\\left\\{ \\frac{%\ni}{\\hbar}\\tilde{S}_{\\mathrm{nonrel}}^{\\theta}\\right\\} \\,, \\label{14}\n\\end{align}\nwhere%\n\\begin{align}\nS_{\\mathrm{nonrel}}^{\\theta} & =\\int dt\\left[ p_{j}\\dot{q}^{j}-H\\left(\np,q\\right) +\\dot{p}_{j}\\theta^{ji}p_{i}\/2\\hbar\\right] , \\label{11a} \\\\\n\\tilde{S}_{\\mathrm{nonrel}}^{\\theta} & =\\int dt\\left[ -q_{j}\\dot{p}%\n^{j}-H\\left( p,q\\right) -p_{j}\\theta^{ji}\\dot{p}_{i}\/2\\hbar\\right] .\n\\label{12a}\n\\end{align}\n\nOne ought to stress that the actions $S_{\\mathrm{nonrel}}^{\\theta}$ and $%\n\\tilde{S}_{\\mathrm{nonrel}}^{\\theta}$ differ by a total time derivative.\n\nThe path-integral (\\ref{12}) is a generalization of the result obtained in \n\\cite{Acatrinei} for arbitrary nonrelativistic system and without any\nrestrictions on the matrix $\\mathbf{\\theta}$. One ought to say that path\nintegrals on noncommutative plane for matrix elements of the evolution\noperator in coherent state representations were studied in \\cite{Smailagic}\nand \\cite{Tan}. They have specific forms which is difficult to compare with\nour results.\n\nIn the convetional ''commutative'' nonsingular QM the action $S_{\\mathrm{%\nnonrel}}^{\\theta}$ (at $\\theta=0)$ is just the Hamiltonian action of the\nclassical system under consideration. The canonical quantization of this\naction reproduces the initial QM of the system. In the noncommutative case\nthis action is modified by a new term $\\dot{p}_{k}\\theta^{kj}p_{j}\/2\\hbar$.\nOne can treat the action (\\ref{11a}) as a the $\\theta$-modified Hamiltonian\naction of the classical system under consideration (see the Introduction).\nThis interpretation can be justified by the canonical quantization of the\naction, see \\cite{Deriglazov}.\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction}\nThe noisy wiretap channel was first studied by Wyner \\cite{wyner}, in which a\nlegitimate transmitter (Alice) wishes to send a message to a legitimate receiver (Bob), and hide it from an eavesdropper (Eve). Wyner proved that Alice can send positive secure rate using channel coding. He derived capacity-equivocation region for the degraded\nwiretap channel. Later, Csiszar and Korner found capacity-equivocation the region for the\ngeneral wiretap channel \\cite{csi}, which was\nextended to the Gaussian wiretap channel by Leung-Yan-\nCheong and Hellman \\cite{leu}.\n\nA significant amount of work was carried thereafter to study the information theoretic physical layer security for different network models. The relay assisted wiretap channel was studied in \\cite{secop}. The secure degrees of freedom (SDoF) region of multiple access channel (MAC) was presented in \\cite{sennur_mac}. The SDoF is the the pre-log of the secrecy capacity region in the high-SNR regime. \nUsing MIMO systems for securing the message was an intuitive extension due to the spatial gain provided by multiple antennas. The MIMO wiretap channel was studied in \\cite{mimo_wire,mimo_secure,mimo_note,mimo_hass,yener_coop,yener_mimo,yener_mac,mimo_shafie,mimo_confidential} and the secrecy capacity was identified in \\cite{mimo_hass}.\n \nAll the aforementioned work assumes availability of either partial or complete channel state information (CSI) at the transmitter. Given that the eavesdropper is passive, it is of high importance to study the case where the channel state information is completely unknown to the legitimate users. The authors in\\cite{yener_mimo,yener_mac} study the secrecy capacity and SDoF for different MIMO channels when the eavesdropper channel is arbitrarily varying and known to the eavesdropper only. In~\\cite{yener_mimo} the authors established the SDoF region for a scenario comprising a single eavesdropper assuming a square full-rank legitimate channel matrix (i.e. has zero null space). Meanwhile, in \\cite{yener_mac} the authors addressed the SDoF of the MAC assuming a single eavesdropper where all the transmitted symbols are intended for the legitimate receiver (no exploitation of jamming).\n\nMeanwhile, the idea of cooperative jamming was proposed in \\cite{yener_coop},\n where some of the users transmit independent and identically distributed (i.i.d.)\nGaussian noise towards the eavesdropper to improve the sum secrecy rate. Cooperative jamming was used for deriving the SDoF for different networks. In \\cite{sennur_mac}, cooperative jamming was used to jam the eavesdropper and proved that the K-user MAC with single antenna nodes can achieve $\\frac{K(K-1)}{K(K-1)+1}$ SDoF.\n\nIn this paper, we study the MIMO MAC with unknown eavesdroppers' CSI at the legitimate transmitters and receiver. We provide a new upperbound for the achievable SDoF and determine the exact sum SDoF by providing an achievable scheme. We show that our scheme is optimal and that the achievable bound and the new upperbound are tight. Our scheme requires neither cooperation between the legitimate transmitters nor cooperation between any legitimate transmitter and the legitimate receiver, and precoding matrices and the jamming vectors are determined in distributed manner.\n\nThe major contributions of our work as compared to existing literature can be summarized as follows:\n\\begin{itemize}\n \\item We present the sum SDoF of the multiple access channel with unknown fading eavesdroppers channels.\n\\item For the case of known eavesdropper channels with constant or time varying channels, we show that it has the same sum SDoF as the previous case, closing an open problem since the best known achievable region was presented in \\cite{khan}.\n\\item our work incorporates the more general scenario of multiple eavesdroppers.\n \\item we assume no restrictions on the relation between the number of antennas at transmitters and the receiver. \n \\item We study the more comprehensive case where all the eigenvalues of the legitimate channel has non-zero values\\footnote{The cases where some of the eigenvalues are equal to zero represent special degraded cases of the more general non-zero eigenvalues case, where the SDoF decreases for every zero eigenvalue till it collapses to the trivial case of zero SDoF for all-zero eigenvalues.}.\n \\item we deduce a new upperbound on the SDoF.\n \\item we provide a scheme that achieves the new upperbound. For the special case of a single eavesdropper, our proposed scheme achieves a sum SDoF superior to those reported in the literature. \n\\end{itemize}\n\nThe paper is organized as follows. Section~\\ref{sec:model} defines the system model and the secrecy constraints. The main results are presented in Section~\\ref{sec:results}. In Section~\\ref{sec:bound}, the new upperbound is derived and the achievable scheme is presented in Section~\\ref{sec:scheme}. The paper is concluded in Section~\\ref{sec:conclusion}. We use the following notation, $\\vec{a}$ for vectors, $\\vec{A}$ for matrices, $\\vec{A}^{\\dagger}$ for the hermitian transpose of $\\vec{A}$, $[A]^+$ for the $\\max{A,0}$ and $\\mathrm{Null}(\\vec{A})$ to define the nullspace of $\\vec{A}$. \n\n\n\n\\section{System model}\n\\label{sec:model}\n\nWe consider a communication system of two transmitters and a single receiver in vicinity of an unknown number of passive eavesdroppers. Transmitters one and two are equipped with $M_1$ and $M_2$ antennas, respectively. The legitimate receiver is equipped with $N$ antennas, while the $j$th eavesdropper is equipped with $N_{Ej}\\leq N_E$ antennas. Let $\\vec{x}_i$ denote the $M_i \\times 1$ vector of Gaussian coded symbols to be transmitted by transmitter $i$, where $i\\in\\{1,2\\}$. We can\nwrite the received signal at the legitimate receiver at time (sample) $k$ as\n\\begin{equation}\\label{Received_signal}\n\\vec{Y}(k)=\\sum_{i=1}^2\\vec{H}_i \\vec{V}_i\\vec{x}_i(k)+ \\vec{n}(k)\n\\end{equation}\nand the received signal at the $j$th eavesdropper as\n\\begin{equation}\\label{Received_signal}\n\\vec{Z}_{j}(k)=\\sum_{i=1}^2\\vec{G}_{i,j}(k) \\vec{V}_i\\vec{x}_i(k)+ \\vec{n}_{Ej}(k),\n\\end{equation}\n\n\\noindent where $\\vec{H}_i$ is the $N \\times M_i$ matrix containing\nthe channel coefficients from transmitter $i$ to the receiver, $\\vec{G}_{i,j}(k)$ is the $N_{Ej} \\times M_i$ matrix containing the i.i.d time varying channel coefficients from transmitter $i$ to the eavesdropper $j$ drawn from a continuous distribution with mean $\\eta$ and variance $\\sigma_e^2$, $\\vec{V}_i$ is the precoding unitary matrix (i.e. $\\vec{V}_i\\vec{V}_i^\\dagger = \\vec{I}$) at transmitter $i$, $\\vec{n}(k)$ and $\\vec{n}_{Ej}(k)$ are the $N\\times 1$ and the $N_{Ej}\\times 1$ additive white Gaussian\nnoise vectors with zero mean and variance $\\sigma^2$ at the legitimate receiver and the $j$th eavesdropper, respectively. We assume that the transmitters know the maximum number of antennas any eavesdropper can possess; namely, $N_E$, but they do not know any of the eavesdroppers' channels $\\vec{G}_{i,j}(k)$. We assume that $N_E< M$, where $M=M_1+M_2$. \n\nWe define the $M_i \\times 1$ channel input from legitimate transmitter $i$ as\n\\begin{equation}\n\\vec{X}_i(k)= \\vec{V}_i \\vec{x}_i(k).\n\\end{equation}\n\n\n\\begin{figure}\n \\begin{center}\n\\hspace{-4mm} \\includegraphics[width=.50\\textwidth]{sys_macsecrecy.png}\\vspace{-2mm}\n\\caption{System model}\n \\label{sys}\n \\end{center}\n \\end{figure}\n\n\nEach transmitter $i$ intends to send a message $W_i$ over $n$ channel uses (samples) to the legitimate receiver simultaneously while preventing the eavesdroppers from decoding its message. The encoding occurs under a constrained power given by\n\\begin{equation}\n\\sum_i^2 \\text{E}\\left\\{\\vec{X}_i\\vec{X}^{\\dagger}\\right\\} \\leq P\n\\end{equation}\n\nExpanding the notations over $n$ channel extensions we have $\\vec{H}_i^n = \\vec{H}_i(1), \\vec{H}_i(2), \\ldots, \\vec{H}_i(n)$, $\\vec{G}_{i,j}^{n} = \\vec{G}_{i,j}(1), \\vec{G}_{i,j}(2), \\ldots, \\vec{G}_{i,j}(n)$ and similarly the time extended channel input, $\\vec{X}_i^n$, time extended channel output at legitimate receiver, $\\vec{Y}^n$ and time extended channel output at eavesdropper $j$, $\\vec{Z}_{j}^n$ as well as noise at legitimate receiver, $\\vec{n}^n$ and noise at eavesdroppers, $\\vec{n}_{Ej}^n$.\n\nAt each transmitter, the message $W_i$ is uniformly and independently chosen from a set of possible secret messages for transmitter $i$, $\\mathcal{W}_i = \\{1,2, \\ldots, 2^{nR_i}\\}$. The rate for message $W_i$ is $R_i \\triangleq \\frac{1}{n} \\log\\left|\\mathcal{W}_i\\right|$, where $|\\cdot|$ denotes the cardinality of the set. Transmitter $i$ uses a stochastic encoding function $f_i: W_i \\longrightarrow \\vec{X}_i^n$ to map the secret message into a transmitted symbol. The receiver has a decoding function $\\phi: \\vec{Y}^n \\longrightarrow (\\hat{W}_1,\\hat{W}_2)$, where $\\hat{W}_i$ is an estimate of $W_i$.\n\\begin{definition}\nA secure rate tuple $(R_1, R_2)$ is said to be achievable if for any $\\epsilon > 0$ there exist $n$-length codes such that the legitimate receiver can decode the messages reliably, i.e.,\n\\begin{equation}\n\\text{Pr}\\{(W_1,W_2) \\neq (\\hat{W}_1, \\hat{W}_2)\\} \\leq \\epsilon\n\\end{equation}\nand the messages are kept information-theoretically secure against the eavesdroppers, i.e.,\n\\begin{equation}\\label{eqn:cond}\nH(W_1, W_2|\\vec{Z}_{j}^{n})\\geq H(W_1, W_2)-\\epsilon \\\\\n\\end{equation}\n\\begin{equation}\nH(W_i|\\vec{Z}_{j}^{n})\\geq H(W_i) - \\epsilon \\forall i =1,2,\n\\end{equation}\n\\noindent where $H(\\cdot)$ is the Entropy function and~\\eqref{eqn:cond} implies the secrecy for any subset $\\mathbb{S} \\subset \\{1,2\\}$ of messages including individual messages~\\cite{sennur_mac}.\n\\end{definition}\n\\begin{definition}\nThe sum SDoF is defined as\n\\begin{equation}\nD_s = \\lim_{P\\rightarrow \\infty} \\sup{\\sum_i \\frac{R_i}{\\frac{1}{2}\\log P}},\n\\end{equation}\n\\noindent where the supremum is over all achievable secrecy rate tuples $(R_1, R_2)$, $D_s = d_1 + d_2$, and $d_{1}$ and $d_{2}$ are the secure DoF of transmitters one and two, respectively. \n\\end{definition}\n\n\n\\section{Main Results}\n\\label{sec:results}\n\n\\begin{theorem}\nThe sum SDoF of the two user MAC channel is\n\\small\n\\begin{equation}\nD_s = \\begin{cases}\n M-N_E & \\mathscr{C}_1\\\\\n \\frac{1}{2}\\max(M_1,N)+\\frac{1}{2}\\max(M_2,N)-\\frac{1}{2}N_E & \\mathscr{C}_2\\\\\n N & \\mathscr{C}_3 \\\\\n\\end{cases}\n\\end{equation}\n\\noindent where the conditions $\\mathscr{C}_1$, $\\mathscr{C}_2$ and $\\mathscr{C}_3$ are given by:\n\\begin{eqnarray}\n\\nonumber &&\\mathscr{C}_1: M \\leq N\\\\ \n\\nonumber &&or\\;M_1N \\text{ and } N_E \\geq 2(M-N)\\\\\n\\nonumber &&or\\;M_1>N, M_2N, M_2N, M_2\\geq N \\text{ and } N_E \\geq M-2N \\\\\n\\nonumber &&\\\\\n\\nonumber &&\\mathscr{C}_3: N_E < [M_1-N]^++[M_2-N]^+\n\\end{eqnarray}\n\\end{theorem}\n\n\\begin{proof}\nTo prove the theorem, we deduce the converse in Section~\\ref{sec:bound} and provide the achievable scheme in Section~\\ref{sec:scheme}.\n\\end{proof}\n\\normalsize\n\n\n\\section{Converse}\\label{sec:bound}\nWithout loss of generality, we consider the case of only one eavesdropper with $N_E$ antennas (maximum possible number of antennas in any eavesdropper). The SDoF of the single eavesdropper scenario is certainly an upperbound for the multiple eavesdroppers case\\footnote{Increasing the number of eavesdroppers can only reduce the SDoF of the legitimate users.}. Accordingly, we omit the eavesdropper subscript for simplicity of notation. Starting from the sum rates and following an approach similar to~\\cite{sennur_mac}, we have:\n\\begin{eqnarray}\nn\\sum_{i=1}^2 R_i & \\leq & I(W_1,W_2;\\vec{Y}^n)- I(W_1,W_2;\\vec{Z}^n)+nc_1\\\\ \\label{secrecy}\n\\nonumber & \\leq & I(W_1,W_2;\\vec{Y}^n,\\vec{Z}^n)- I(W_1,W_2;\\vec{Z}^n)+nc_1\\\\\n\\nonumber & = & I(W_1,W_2;\\vec{Y}^n|\\vec{Z}^n)+nc_1\\\\\n\\nonumber & \\leq & I(\\vec{X}_1^n,\\vec{X}_2^n;\\vec{Y}^n|\\vec{Z}^n)+nc_1\\\\\n\\nonumber & = & h(\\vec{Y}^n|\\vec{Z}^n)-h(\\vec{Y}^n|\\vec{Z}^n,\\vec{X}_1^n,\\vec{X}_2^n)+nc_1\\\\\n\\nonumber & = & h(\\vec{Y}^n|\\vec{Z}^n)-h(\\vec{n}^n|\\vec{Z}^n,\\vec{X}_1^n,\\vec{X}_2^n)+nc_1\\\\\n\\nonumber & \\leq & h(\\vec{Y}^n|\\vec{Z}^n)+nc_2\\\\\n\\nonumber & = & h(\\vec{Y}^n,\\vec{Z}^n)-h(\\vec{Z}^n)+nc_2\\\\\n\\nonumber & = & h({\\vec{X}}_1^n,{\\vec{X}}_2^n,\\vec{Y}^n,\\vec{Z}^n)-h({\\vec{X}}_1^n,{\\vec{X}}_2^n|\\vec{Y}^n,\\vec{Z}^n)\\\\\n& & -h(\\vec{Z}^n)+nc_2\\\\ \\label{add}\n\\nonumber &\\leq & h({\\vec{X}}_1^n,{\\vec{X}}_2^n,\\vec{Y}^n,\\vec{Z}^n)-h(\\vec{Z}^n)+nc_2\\\\\n\\nonumber &= & h({\\vec{X}}_1^n,{\\vec{X}}_2^n)+h(\\vec{Y}^n,\\vec{Z}^n|{\\vec{X}}_1^n,{\\vec{X}}_2^n)\\\\\n\\nonumber & & -h(\\vec{Z}^n)+nc_2\\\\\n\\nonumber & \\leq & h({\\vec{X}}_1^n,{\\vec{X}}_2^n)-h(\\vec{Z}^n)+nc_3\\\\ \\label{22}\n\\nonumber & = & \\sum_{m=1}^{M_1}h({x}_{1m}^n)+\\sum_{m=1}^{M_2}h({x}_{2m}^n)-h(\\vec{Z}^n)+nc_3\\\\ \n\\label{up}\n\\end{eqnarray}\n\n\n\n\nlet $\\vec{\\bar{\\vec{x}}}$ be a concatenating vector of $\\vec{x}_1$ and $\\vec{x}_2$, $\\bar{\\vec{x}}={[{\\vec{x}_1}^{\\mathsmaller T} \\hspace{2mm}{\\vec{x}_2}^{\\mathsmaller T}]}^{\\mathsmaller T}$.let $B$ be a permutation matrix when multiplied by $\\vec{Z}$ results in a vector $\\bar{\\vec{Z}}$ with the $m$th element ($m \\in \\{1, 2, \\ldots, M_1+M_2\\}$) depending on $\\bar{x}_m$ and $e_m$, where $e_{m}$ are constants depending on $B$ and $G_{i \\; : \\; i\\in \\{1,2\\}}$.\n\\begin{eqnarray}\n\\bar{Z}=BZ=\n\\begin{bmatrix}\ne_1 \\bar{x}_1\\\\\ne_2 \\bar{x}_2\\\\\n\\vdots\\\\\ne_{N_E-1} \\bar{x}_{N_E-1}\\\\\ne_{N_E} \\bar{x}_{N_E}+..+e_{M_1+M_2} \\bar{x}_{M_1+M_2}\n\\end{bmatrix}\n +\\vec{B} \\vec{n}_E\n\\end{eqnarray}\nand, \n\\begin{eqnarray}\nh(\\bar{\\vec{Z}})=h(\\vec{Z})+log|B| \\label{perm}\n\\end{eqnarray}\nSubstituting (\\ref{perm}) into (\\ref{up}),\n\n\\begin{eqnarray}\n\\nonumber n\\sum_{i=1}^2 R_i & \\leq & n\\sum_{m=1}^{M_1}h({x}_{1m})+n\\sum_{m=1}^{M_2}h({x}_{2m})-nh(\\bar{\\vec{Z}})\\\\\n\\nonumber & + & \\log|B|+nc_3 \\\\\n\\nonumber \\sum_{i=1}^2 R_i & \\leq & \\sum_{m=1}^{M_1+M_2}h(\\bar{x}_m)-\\sum^{N_E-1}_{m=1}h(e_m\\bar{x}_m)\\\\\n\\nonumber &-&h\\left(\\sum_{m=1}^{M_1+M_2-N_E+1}e_{m+N_E-1}\\bar{x}_{m+N_E-1}\\right)\\\\\n\\nonumber &+& N_E h(n)+c_4\\\\ \n\\nonumber & \\leq & (M_1+M_2-N_E+1) \\log P\\\\\n\\nonumber & - & \\log \\left(\\left|\\left|[e_{N_E} \\ldots e_{M_1+M_2}]\\right|\\right|^2 P\\right) +nc_5\n\\end{eqnarray}\n\nHence, \n\\begin{equation}\nD_s \\leq M_1+M_2-N_E,\n\\label{eqn:15}\n\\end{equation}\n\n\\noindent where \\eqref{secrecy} comes from the perfect secrecy constraint~\\cite{khan, sennur_mac}, $x_{im}$ is the $m$th element of $\\vec{x}_i$. All $c_{i \\; : \\; i\\in\\{1,2, \\ldots 5\\}}$ are constants independent of $P$. \nGiven that the receiver has only $N$ antennas, then\n\\begin{equation}\\label{2}\nD_s \\leq N\n\\end{equation}\n\\\\\n\nLet $W_{e1}$ and $W_{e2}$ be the messages sent by transmitter one and transmitter two, respectively, which the eavesdropper can decode. Let $d_e^1$ and $d_e^2$ be the maximum degrees of freedom for messages sent by transmitter one and two, respectively, which can be decoded by the eavesdropper. \n\nSuppose that we can set all coefficients of $\\vec{G}_{1}$ to zero to hide all information sent by transmitter one from the eavesdropper, so the resulting channel becomes identical to the Z channel of Figure~\\ref{zch}(b). However, setting $\\vec{G}_{1}$ coefficients to zero cannot decrease the performance of the coding scheme. Therefore, the DoF tuple for the modified channel is upperbounded by\n\\begin{equation}\\label{z1}\nd_{1}+d_{2}+d_{e}^2 \\leq \\text{max}(M_2, N)\\\\\n\\end{equation}\n\\\\\n\\\\\nSimilarly, using the modified Z channel in Figure~\\ref{zch}(a),\n\\hspace{5mm}\n\n\\begin{equation}\\label{z2}\nd_{1}+d_{2}+d_{e}^1\\leq \\text{max}(M_1,N)\\\\\n\\end{equation} \nCombining (\\ref{z1}) and (\\ref{z2}), we have\n\\begin{equation}\n2(d_1+d_2)+d_{e}^1+d_{e}^2\\leq \\text{max}(M_1,N) +\\text{max}(M_2, N).\\\\\n\\end{equation}\nMoreover, since the eavesdropper has $N_E < M$ antennas then \n\\begin{equation} \nd_{e}^1+d_{e}^2=N_E\n\\end{equation}\nthen,\n\\begin{equation}\\label{c3}\nD_s\\leq \\frac{\\text{max}(M_1,N) +\\text{max}(M_2, N)-N_E}{2}\n\\end{equation} \n\\begin{figure} \\label{zch}\n \\begin{center}\n\\hspace{-4mm} \\includegraphics[width=.35\\textwidth]{zch.png}\\vspace{-2mm}\n\\caption{DoF-equivalent Z channel for two transmitter MAC channel with an eavesdropper}\n\\label{zch} \n\\end{center} \\vspace{-6mm}\n \\end{figure}\nFrom (\\ref{eqn:15}),(\\ref{2}) and (\\ref{c3}), we have\n\n\\begin{equation}\nD_s\\leq \\text{min}\\left(\\frac{\\text{max}(M_1,N) +\\text{max}(M_2, N)-N_E}{2}, M-N_E,N\\right)\\\\\n\\end{equation} \n\n\\vspace{1mm}\n\n\n\\section{Achievable scheme}\n\\label{sec:scheme}\nFor securing the legitimate messages, the transmitters send $N_E$ jamming signal vector $\\vec{r}=[\\vec{r}_1 \\text{ } \\vec{r}_2]^T$ with random symbols using $\\vec{V}^J_{1}$ and $\\vec{V}^J_{2}$ as jamming precoders\\footnote{For the special case $N_E=1$, only one user sends a single jamming symbol.}. Hence, the transmitted coded signal can be broken into legitimate signal, $\\vec{s}_i$, and jamming signal, $\\vec{r}_i$, such that \n$$\\vec{x}_i = \\left[\\begin{array}{c} \\vec{s}_i\\\\ \\vec{r}_i \\end{array}\\right], i \\in\\{1,2\\}.$$\nAccordingly, the precoder, $\\vec{V}_i$ can be also broken into legitimate, $\\vec{V}^L_i$, and jamming, $\\vec{V}^J_i$ precoders such that\n$$\\vec{V}_i = \\left[\\begin{array}{cc} \\vec{V}^L_i & \\vec{V}^J_i \\end{array}\\right] i \\in\\{1,2\\}.$$\n \nChoosing $\\vec{V}^J$ to be the unitary matrix, the jamming power becomes $P^J=\\text{E}\\{\\text{tr}(\\vec{r}_i\\vec{r}_i^{\\dagger})\\} = \\alpha P$, where $\\alpha$ is a constant controlled by the transmitter.\n\n\\begin{proposition}\nThe jamming signal, $\\vec{r}$, overwhelms the eavesdropper signal space, and the eavesdropper ends up decoding zero DoF of the legitimate messages.\\\\\n\\end{proposition}\n\n\\begin{proof}\n\\begin{eqnarray}\n\\nonumber n R_e && \\leq I(\\vec{Z}^n; \\vec{s}_1^n, \\vec{s}_2^n)\\\\\n\\nonumber && = h(\\vec{Z}^n) -h(\\vec{Z}^n|\\vec{s}_1^n, \\vec{s}_2^n)\\\\\n\\nonumber R_e && = h (\\vec{Z}) - h (\\vec{G}_1(\\vec{s}_1+\\vec{r}_1) +\\vec{G}_2(\\vec{s}_2+\\vec{r}_2) \\\\\n\\nonumber && + \\vec{n}_E | \\vec{s}_1, \\vec{s}_2)\\\\\n\\nonumber && = h (\\vec{Z}) - h (\\vec{G}_1 \\vec{s}_1^J+\\vec{G}_2\\vec{s}_2^J + \\vec{n}_E)\\\\\n\\nonumber && \\leq\\frac{1}{2}\\log \\frac{\\left| \\vec{I} \\sigma^2+(\\sum_{i=1}^2\\text{E}\\{\\vec{G}_i\\vec{V}^L_i \\vec{s}_i\\vec{s}^{\\dagger}_i\\vec{V}^{L\\dagger}_i\\vec{G}_i^{\\dagger})\\} \\right|}{\\left| \\vec{I} \\sigma^2+\\sum_{i=1}^2 \\text{E}\\{(\\vec{G}_i\\vec{V}^J_i\\vec{r}^J_i\\vec{r}^{J\\dagger}_i\\vec{V}^{J\\dagger}_i\\vec{G}_i^{\\dagger}) \\}\\right|}\\\\\n&&\\leq C \\label{con}\n\\end{eqnarray}\n\\noindent where $C$ is a constant that does not depend on $P$ and known to the transmitter. Therefor, \n\\begin{equation}\n\\nonumber \\lim_{P\\longrightarrow \\infty} \\frac{R_e(P)}{\\frac{1}{2} \\log_2 P} \\leq \\lim_{P\\longrightarrow \\infty} \\frac{C}{\\frac{1}{2} \\log_2 P} = 0,\n\\end{equation}\n\\end{proof}\n\n\\begin{remark}\nThe constant eavesdropper rate comes from the fact that $P^J$ is controlled by the transmitter. Hence, setting $P^J=rP$ for some constant $r$ we guarantee a constant SNR at the eavesdropper and a constant rate independent of $P$. For the case of the constant known eavesdropper channel, the constant $C$ is known for the transmitter. While for the case that the transmitter does not know the eavesdropper channel, it knows the expectation in \\ref{con}. \n\\end{remark}\n\n\\bigskip\n\nThe transmitters use the rate difference to transmit perfectly secure messages using a stochastic encoder similar to the one described in~\\cite{khan} according to the eavesdropper's rate, $C$, in worst case scenario. Using the results in~\\cite{khan}, for some postprocessing matrix $\\vec{U}$ projecting the received signal into a jamming free space, the achievable secrecy sum rate can be lowerbounded by\n\n \\begin{eqnarray}\n\\nonumber \\sum_{i=1}^2 R_i \\hspace{-7mm}&& \\geq \\frac{1}{2} \\log \\left| \\vec{I}+ \\sum_{i=1}^2 (\\vec{U}\\vec{H}_i\\vec{V}_i^L\\vec{s}_i\\vec{s}^{\\dagger}_i\\vec{V}^{L\\dagger}_i\\vec{H}_i^{\\dagger}\\vec{U}^{\\dagger})\\right| -R_e\\\\ \n&& \\geq \\frac{1}{2} \\log \\left| \\vec{I}+ \\sum_{i=1}^2 (\\vec{U}\\vec{H}_i\\vec{V}_i^L\\vec{s}_i\\vec{s}^{\\dagger}_i\\vec{V}^{L\\dagger}_i\\vec{H}_i^{\\dagger}\\vec{U}^{\\dagger}) \\right| - C\n\\label{secrate}\n\\end{eqnarray}\n\nAs $R_e$ is upperbound by a constant for all values of $\\vec{G}_i$ and $P$, a positive secrecy rate, which is monotonically increasing with $P$, is achieved. Computing the secrecy degrees of freedom boils down to calculating the degrees of freedom for the first term in the right hand side of \\eqref{secrate}, which represents the receiver DoF after jamming is applied.\n \nWith the eavesdropper completely blocked, it remains to show how the jamming signal directions are designed to achieve the maximum possible secure DoF. First, we study the secure DoF for $M \\leq N$, then go for $M > N$ with different regions of the relations between $(M_1, M_2, N, N_E)$.\n\n\\subsection{Achievability for $M \\leq N$}\n\nFor this region, transmitters one and two send the jamming signals using precoders $\\vec{V}^J_{1}$ and $\\vec{V}^J_{2}$, with dimensions $J_1$ and $J_2$, respectively, such that $J_1+J_2=N_E$. \n\n\\subsubsection*{Random jamming}\nThe jamming precoders and symbols are randomly chosen. We call this method random jamming. The receiver zero-forces the jamming signal using the post processing-matrix $\\vec{U}$ as in~\\eqref{zero}. Accordingly, $M-N_E$ secure DoF can be sent. Since $N \\geq M $, the receiver can decode $M-N_E$ DoF after zero-forcing the jamming signal.\\\\\n\n\\begin{equation}\\label{zero}\n\\vec{U}= [\\vec{I} - \\vec{a}\\vec{a}^{-1}]\n\\end{equation}\n where\n\\begin{equation}\n\\vec{a}= \\vec{H}_1 \\vec{V}_1^J+\\vec{H}_2 \\vec{V}_2^J\n\\end{equation}\n\n\n\\subsection{Achievability for $M>N$}\nFor this region we use three methods for jamming, aligned jamming, nullspace jamming and random jamming. As random jamming was described above, the other two will be explained in the following.\n\n\\subsubsection*{Aligned jamming}\nThe jamming signals of both transmitters are aligned at the legitimate receiver signal space. \nLet $\\mathcal{I}$ be the jamming space at the receiver. Each transmitter aligns a part or the whole of its jamming signal into this jamming space. The total signal space of transmitter one and transmiter two occupies \\emph{only} $M_1$ and $M_2$ dimensions, receptively, at the receiver. These two spaces are distinct if $M_1N \\hspace{1mm} \\text{ and } \\hspace{1mm}N_E< M_1-N+[M_2-N]^+ $:}\n\nIn this region, nullspace jamming is used alone. Transmitter one sends $J_1=\\text{min}(N_E, M_1-N)$ dimensional jamming signal and transmitter two sends $J_2=N_E-J_1$ dimensional jamming signal in the null spaces of the legitimate receiver channels.\nThis leaves the receiver with $N$ jamming free dimensions to decode the $N$ SDoF transmitted. Consequently, the upperbound $N$ is achieved.\\\\\n\n\\emph{Case $M_1>N, M_2N, M_2N, M_2N, M_2N, M_2\\geq N \\text{ and } N_E \\geq M-2N$:}\n\nIn this region, two jamming methods are used. Each transmitter jamming signal is divided into two parts of sizes \n\\begin{eqnarray}\n\\nonumber J_{1,i}=M_i-N, \\;\\;\\;\\;\\ \\forall \\; i\\in\\{1,2\\}\\\\\n\\nonumber J_2= \\frac{N_E-\\sum_{i\\in\\{1,2\\}}J_{1,i}}{2}.\n\\end{eqnarray}\nBoth transmitters use nullspace jamming for their first part and aligned jamming for the second part. The jamming occupies $J_s=J_2$ dimensions at the receiver.\n\n\\begin{lemma}\nFor the proposed scheme under case $M_1>N, M_2\\geq N \\text{ and } N_E \\geq M-2N$, the achievable secure degrees of freedom is\n\\begin{equation}\nd_1+d_2 \\leq \\frac{\\text{max}(M_1,N)+\\text{max}(M_2,N)-N_E}{2}\n\\end{equation}\n\\end{lemma}\n\n\\begin{proof}\n\\begin{eqnarray}\n\\nonumber J_s &=& \\frac{N_E-(M_1-N+M_2-N)}{2}\\\\\n\\nonumber N-J_s &=& N- \\frac{N_E-[M-2N]}{2}\\\\\n\\nonumber N-J_s &=&\\frac{2N-N_E+M-2N}{2}\\\\\n\\nonumber N-J_s &=& \\frac{M-N_E}{2}.\n\\end{eqnarray}\nThus, \n\\begin{eqnarray}\nd_1+d_2 \\leq \\frac{M-N_E}{2},\n\\end{eqnarray}\nwhich can be rewritten as\n\\begin{equation}\n\\nonumber d_1+d_2 \\leq \\frac{\\text{max}(M_1,N)+\\text{max}(M_2,N)-N_E}{2}\n\\end{equation}\n\\end{proof}\n\n\n \n \n \n \n %\n \n \n\\section{Conclusion}\n\\label{sec:conclusion}\nWe studied the two-transmitter Gaussian multiple access wiretap channel with multiple antennas at the transmitters, legitimate receivers and eavesdroppers. Generalizing new upperbound was established and a new achievable scheme was provided. We used the new optimal scheme to derive the sum secure DoF of the channel. We showed that the our scheme meets the upperbound or all $M_1, M_2, N_E$ combinations. We showed that Cooperative Jamming is SDoF optimal even without the eavesdropper CSI available at the transmitters by showing that jamming signal independent of the eavsdropper channel and only dependes on the signal transmitted power make the eavsdropper decoded DoF. Finally we showed that if any eavsdropper has more antennas that the sum of the transmiting antennas or the receiving antennas the SDoF is zero.\n\n\n ","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction}\n\\label{intro}\nThe {\\it discriminator} of a sequence ${\\bf a}=\\{a_n\\}_{n\\ge 1}$\nof distinct integers is the sequence given by\n$$\n{\\mathcal D}_{\\bf a}(n)=\\min\\{m: a_0,\\ldots,a_{n-1}~{\\text{\\rm are~pairwise~distinct~modulo}}~m\\}.\n$$\nIn other words,\n${\\mathcal D}_{\\bf a}(n)$\nis the smallest\ninteger $m$ that allows one to discriminate (tell\napart) the integers $a_0,\\ldots, a_{n-1}$ on reducing modulo $m$.\n\nNote that since $a_0,\\ldots,a_{n-1}$ are $n$ distinct residue classes modulo ${\\mathcal D}_{\\bf a}(n)$ it follows that ${\\mathcal D}_{\\bf a}(n)\\ge n$.\nOn the other hand obviously\n$${\\mathcal D}_{\\bf a}(n)\\le \\max\\{a_0,\\ldots,a_{n-1}\\}\n-\\min\\{a_0,\\ldots,a_{n-1}\\}.$$\nPut\n$$\n{\\mathcal D}_{\\bf a}=\\{{\\mathcal D}_{\\bf a}(n): n\\ge 1\\}.\n$$\nThe main problem is to give an easy description or characterization of\nthe discriminator (in many cases such a characterization does not seem to exist). The discriminator was named and introduced by Arnold, Benkoski and McCabe in \\cite{A}. \nThey considered the sequence $\\bf u$ with terms $u_j = j^2$.\nMeanwhile the case where $u_j=f(j)$ with $f$ a polynomial has been well-studied,\nsee, for example, \\cite{B,M,MM,Z}. The most general result in this direction is due to Zieve \\cite{Z}, who improved\non an earlier result by Moree \\cite{M}.\n\nIn this paper we study the discriminator problem for\nLucas sequences (for a basic\naccount of Lucas sequences see, for \nexample, Ribenboim\n\\cite[2.IV]{Ri}).\nOur main results are\nTheorem \\ref{main} ($k=1$) and Theorem\n\\ref{main2} ($k>2$). Taken together with Theorem \\ref{twee} ($k=2$) \nthey evaluate the\ndiscriminator for the infinite family of\nsecond-order recurrences \\eqref{basicfamily} with for each $k$ at most finitely\nmany not covered values.\n\\par All members in the family \\eqref{basicfamily} have a characteristic\nequation that is irreducible over\nthe rationals. Very recently,\nCiolan and Moree \\cite{CM} determined the\ndiscriminator for another infinite family,\nthis time with all members having a reducible characteristic\nequation.\nFor every prime $q\\ge 7$ they computed the discriminator of the sequence $$u_q(j)=\\frac{3^j-q(-1)^{j+(q-1)\/2}}{4},~j=1,2,3,\\ldots$$\nthat was first considered in this context\nby Jerzy Browkin.\nThe case\n$q=5$ was earlier dealt with by\nMoree and Zumalac\\'arregui\n\\cite{PA}, who showed that,\nfor this value of $q,$ the smallest positive integer $m$\ndiscriminating $u_q(1),\\ldots,u_q(n)$ modulo $m$\nequals $\\min\\{2^e,5^f\\},$ where $e$ is the smallest integer\nsuch that $2^e\\ge n$ and $f$ is the smallest integer\nsuch that $5^f\\ge 5n\/4$.\n\\par Despite structural similarities\nbetween the present paper and\n\\cite{CM} (for example the\nindex of appearance $z$ in the present paper plays the\nsame role as the period $\\rho$\nin \\cite{CM}), there are also many differences.\nFor example, Ciolan and Moree have to work\nmuch harder to exclude small prime numbers as discriminator values.\nThis is related to the sequence of good\ndiscriminator candidate\nvalues in that case being much sparser, namely\nbeing $O(\\log x)$ for the values $\\le x$,\nversus $\\gg \\log^2 x$.\nIn our\ncase one has to work with elements and ideals in\nquadratic number fields, whereas in \\cite{CM}\nin the proof of the main result the realm of the rationals is never left.\n\n\\par Let $k\\ge 1$. For\n$n\\ge 0$ consider the sequence $\\{U_n(k)\\}_{n\\ge 0}$\nuniquely determined by\n\\begin{equation}\n\\label{basicfamily}\nU_{n+2}(k)\n=(4k+2)U_{n+1}(k)-U_n(k),~U_0(k)=0,~U_1(k)=1.\n\\end{equation}\nFor $k=1$, the sequence $\\{U_n(1)\\}_{n\\ge 0}$ is\n$$\n0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214,\\ldots.\n$$\nThis is A001109 in OEIS.\nOn noting that\n$$U_{n+2}(k)-U_{n+1}(k)=4kU_{n+1}(k)\n+U_{n+1}(k)-U_n(k)\\ge 1,$$\none sees that the sequence $U_n(k)$\nconsists of strictly increasing\nnon-negative numbers. Therefore we\ncan consider ${\\mathcal D}_{U(k)}$,\nwhich for notational convenience we\ndenote by ${\\mathcal D}_{k}$.\n\\par In May 2016, Jeffrey Shallit,\nwho was the first to consider\n${\\mathcal D}_k$, wrote the third\nauthor that numerical evidence suggests\nthat ${\\mathcal D}_{1}(n)$\nis the smallest number of the form\n$250\\cdot 2^i$ or $2^i$ greater than\nor equal to $n$, but\nthat he was reluctant to conjecture such a weird thing.\nMore extensive numerical experiments show that\nif we compute ${\\mathcal D}_1(n)$ for all $n\\le 2^{10}$,\nthen they are powers of $2$ except for $n\\in [129,150]$, and other similar instances such as\n$$\nn\\in [2^a+1,2^{a-6}\\cdot 75],\\text{\\rm ~for~which~}{\\mathcal D}_1(n)=2^{a-6}\\cdot 125\\text{~and~}a\\in \\{7,8,9\\}.\n$$\nThus the situation is more weird than Shallit expected\nand this is confirmed by\nTheorem \\ref{main}.\n\nAs usual, by $\\{x\\}$\nthe\nfractional part of the real number $x$ is denoted. Note that $\\{x\\}=x-\\lfloor x\\rfloor$.\n\\begin{theorem}\n\\label{main}\nLet $v_n$ be the smallest power of two\nsuch that\n$v_n\\ge n$. Let $w_n$ be the smallest integer\nof the form $2^a5^b$ satisfying $2^a5^b\\ge 5n\/3$\nwith $a,b\\ge 1$.\nThen\n$$\n{\\mathcal D}_{1}(n)=\\min\\{v_n,w_n\\}.\n$$\nLet\n$$\n{\\mathcal M}=\\left\\{m\\ge 1: \\left\\{m \\frac{\\log 5}{\\log 2}\\right\\}\\ge 1-\\frac{\\log(6\/5)}{\\log 2}\\right\\}\n=\\{3,6,9,12,15,18,21,\\ldots\\}.\n$$\nWe have\n$$\\{{\\mathcal D}_1(2),{\\mathcal D}_1(3),{\\mathcal D}_1(4),\n\\ldots\\}=\\{2^a5^b:a\\ge 1,~b\\in {\\mathcal M}\\cup \\{0\\}\\}.$$\n\\end{theorem}\n\\noindent A straightforward application of\nWeyl's criterion (cf. the proof of\n\\cite[Proposition 1]{PA} or \\cite[Proposition 1]{CM}) gives\n$$\\lim_{x\\rightarrow \\infty}\\frac{\\#\\{m\\in {\\mathcal M}:m\\le x\\}}{x}=\\frac{\\log(6\/5)}{\\log 2}\n=0.263034\\ldots.$$\n\\par In contrast to the case $k=1$, the\ncase $k=2$ turns out to be especially easy.\n\\begin{theorem}\n\\label{twee}\nLet $e\\ge 0$ be the smallest integer such that\n$2^e\\ge n$ and $f\\ge 1$ the smallest integer such that\n$3\\cdot 2^f\\ge n$. Then\n${\\mathcal D}_2(n)=\\min\\{2^e,3\\cdot 2^f\\}$.\n\\end{theorem}\n\\indent Our second main result shows\nthat the behavior of the discriminator \n${\\mathcal D}_k$ \nwith $k>2$ is very different from that of\n${\\mathcal D}_1$.\n\\begin{theorem}\\setcounter{thm}{1}\n\\label{main2}\nPut\n$${\\mathcal A}_k=\\begin{cases}\n\\{m~{\\text{\\rm odd}}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k\\}\\text{~if~}\nk\\not\\equiv 6\\pmod*{9};\\cr\n\\{m~{\\text{\\rm odd}},~9\\nmid m:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k\\}\\text{~if~}\nk\\equiv 6\\pmod*{9},\n\\end{cases}\n$$\nand $${\\mathcal B}_k\n=\\begin{cases}\n\\{m~{\\text{\\rm even}}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k(k+1)\\}\n\\text{~if~}\nk\\not\\equiv 2\\pmod*{9};\\cr\n\\{m~{\\text{\\rm even},~9\\nmid m}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k(k+1)\\}\n\\text{~if~}\nk\\equiv 2\\pmod*{9}.\n\\end{cases}\n$$\nLet $k>2$. We have\n\\begin{equation*}\n{\\mathcal D}_{k}(n)\\le \\min\\{m\\ge n:\nm\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}\\},\n\\end{equation*}\nwith equality if the interval \n$[n,3n\/2)$ contains an integer \n$m\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}$ and with at most \nfinitely many $n$ for which \nstrict inequality holds.\nFurthermore, we have ${\\mathcal D}_{k}(n)=n$ if and only \nif $n\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}$.\n\\end{theorem}\n\\begin{remark}\nThe condition on the interval $[n,3n\/2)$ is sufficient, but not always\nnecessary. The proof also works for $k=2$ in which\ncase the interval becomes $[n,5n\/3)$. However,\nwe prefer to give a short proof from scratch of\nTheorem \\ref{twee} (in Section \\ref{generalkintro}).\n\\end{remark}\nTheorems \\ref{twee} and \\ref{main2} \ntaken together have\nthe following corollary.\n\\begin{corollary}\nFor $k>1$ there is a finite set ${\\mathcal F}_k$ such that\n\\begin{equation}\n\\label{ABF}\n{\\mathcal D}_{k}=\n{\\mathcal A}_k\\cup\n{\\mathcal B}_k\\cup {\\mathcal F}_k.\n\\end{equation}\n\\end{corollary}\nNote that\n${\\mathcal A}_1=\\{1\\},~\n{\\mathcal B}_1=\\{2^e:e\\ge 1\\}$\nand that by Theorem \\ref{main} identity\n\\eqref{ABF} holds true with\n${\\mathcal F}_1=\\{2^a\\cdot 5^m: a\\ge 1\n\\text{~and~} m\\in {\\mathcal M}\\}.$\nIn particular, ${\\mathcal F}_1$ is not\nfinite. In\ncontrast to this, Theorem \\ref{twee} says that\n${\\mathcal F}_2$ is empty and Theorem \\ref{main2} says that ${\\mathcal F}_k$ is \nfinite for $k>1$.\nIn part II \\cite{CLM} the problem\nof explicitly computing ${\\mathcal F}_k$ is considered.\n\\par Despite the progress made in this paper, for most second order\nrecurrences (and the Fibonacci numbers belong to\nthis class), the discriminator remains\nquite mysterious, even conjecturally. Thus in this\npaper we only reveal the tip of an iceberg.\n\\section{Preliminaries}\n\\label{sec:2}\n\nWe start with some considerations about $U(k)$ valid for any $k\\ge 1$. The characteristic equation of this recurrence is\n$$\nx^2-(4k+2)x+1=0.\n$$\nIts roots are $(\\alpha(k),\\alpha(k)^{-1})$, where\n$$\n\\alpha(k)=2k+1+2{\\sqrt{k(k+1)}}.\n$$\nIts discriminant is $$\\Delta(k)=\\Big(\\alpha(k)-\\frac{1}{\\alpha(k)}\\Big)^2=16k(k+1).$$\nWe have $\\alpha(k)=\\beta(k)^2$, where $\\beta(k)={\\sqrt{k+1}}+{\\sqrt{k}}$. Thus,\n$$\nU_n(k)=\\frac{\\alpha(k)^{n}-\\alpha(k)^{-n}}{\\alpha(k)-\\alpha(k)^{-1}}=\\frac{\\beta(k)^{2n}-\\beta(k)^{-2n}}{\\beta(k)^2-\\beta(k)^{-2}}\n$$\nis both the Lucas sequence having roots $(\\alpha(k),\\alpha(k)^{-1})$, as well as the sequence of even indexed members of the Lehmer sequence having roots $(\\beta(k),\\beta(k)^{-1})$\n(cf. Bilu and\nHanrot \\cite {BH} or Ribenboim \\cite[pp. 69-74]{Ri}). \n\nFirst we study the congruence $U_i(k)\\equiv U_j(k)\\pmod*{m}$ in case $m$ is an arbitrary integer. By the Chinese Remainder Theorem, it suffices to study this congruence only in the case where $m$ is a prime power.\nIn this section we will only deal with the easiest case\nwhere $m$ is a power of two.\n\\begin{lemma}\n\\label{poweroftwo}\nIf $U_i(k)\\equiv U_j(k)\\pmod*{2^{a}}$, then $i\\equiv j\\pmod*{2^{a}}$.\n\\end{lemma}\n\n\\begin{proof} This is clear for $a=0$. When $a=1$, we have $U_0(k)=0,~U_1(k)=1$ and $U_{n+2}(k)\\equiv -U_n(k)\\pmod*{2}$. Thus, $U_{n+2}(k)\\equiv U_{n}(k)\\pmod*{2}$. This shows that $U_n(k)\\equiv n\\pmod*{2}$ for all $n\\ge 0$.\nTherefore $U_i(k)\\equiv U_j(k)\\pmod*{2}$ implies that $i\\equiv j\\pmod*{2}$, which is what we wanted.\nWe now proceed by induction on $a$. Assume\nthat $a>1$ and that the lemma has been proved for $a-1$. Let $i\\le j$ be such that $U_i(k)\\equiv U_j(k)\\pmod*{2^{a}}$. In particular, $U_i(k)\\equiv U_j(k)\\pmod* 2$\nand so $i\\equiv j\\pmod*{2}$.\nIt is easy to check that putting $V_n(k)$ for the sequence given by $V_0(k)=2,~V_1(k)=4k+2$,\nwe have\n$$\nU_j(k)-U_i(k)=U_{(j-i)\/2}(k)V_{(j+i)\/2}(k).\n$$\nThe sequence $\\{V_n(k)\\}_{n\\ge 0}$ satisfies the same recurrence as $\\{U_n(k)\\}_{n\\ge 0}$, namely $$V_{n+2}(k)=(4k+2)V_{n+1}(k)-V_n(k).$$\nNote that $V_n(k)=\\alpha(k)^n+\\alpha(k)^{-n}$.\nFurther, by induction on $n$ using the fact that $2\\| V_0(k)$ and $2\\| V_1(k)$ and the recurrence for $V(k)$, we conclude that if $2\\| V_n(k)$ and $2\\| V_{n+1}(k)$, then\n$$V_{n+2}(k)=(4k+2)V_{n+1}(k)-V_n(k)\\equiv -V_n(k)\\equiv 2\\pmod*{4},$$\nso $2\\| V_{n+2}(k)$. Hence, since $2^{a}\\mid U_i(k)-U_j(k)=U_{(i-j)\/2}(k)V_{(i+j)\/2}(k)$, and $2\\| V_{(i+j)\/2}(k)$, we get that $2^{a-1}\\mid U_{(i-j)\/2}(k)$. Thus,\n$U_{(i-j)\/2}(k)\\equiv U_0(k)\\pmod*{2^{a-1}}$ and by the induction hypothesis we get that $(i-j)\/2\\equiv 0\\pmod*{2^{a-1}}$. Thus, $i\\equiv j\\pmod*{2^{a}}$ and the induction is complete.\n\\end{proof}\n\\begin{corollary}\nWe have ${\\mathcal D}_{k}(n)\\le \\min\\{2^e:2^e\\ge n\\}$.\n\\end{corollary}\n\n\\section{Index of appearance}\nWe now need to study the congruence $U_i(k)\\equiv U_j(k)\\pmod*{p^{b}}$ for odd primes $p$ and integers $b\\ge 1$. We start with the easy case when $j=0$. Given $m$, the smallest $n\\ge 1$ such that $U_n(k)\\equiv \n0\\pmod*{m}$ exists, cf. \\cite{BH}, and is called\nthe {\\it index of appearance of $m$ in $U(k)$} and\nis denoted by $z(m)$.\n(For notational convenience we suppress\nthe dependence of $z(m)$ on $k$.)\nThe following result is well-known, cf. Bilu and Hanrot \\cite{BH}. We write $\\nu_p(m)$ for the exponent of the prime $p$ in the factorization of the positive integer $m$. For an odd prime $p$ we write $(\\frac{\\bullet}{p})$ for the Legendre symbol with respect to $p$.\n\\begin{lemma}\n\\label{lem:flauw}\nThe index of appearance $z$ of the\nsequence $U(k)$ has the\nfollowing properties.\n\\begin{itemize}\n\\item[{\\rm i)}] If $p\\mid \\Delta(k)$, then $z(p)=p$.\n\\item[{\\rm ii)}] If $p\\nmid \\Delta(k)$, then $z(p)\\mid p-e$, where $e=(\\frac{\\Delta(k)}{p})$.\n\\item[{\\rm iii)}] Let $c=\\nu_p(U_{z(p)}(k))$. Then $z(p^{b})=p^{\\max\\{b-c,0\\}} z(p).$\n\\item[{\\rm iv)}] If $p|U_m(k)$, then\n$z(p)|m$.\n\\item[{\\rm v)}] If $n=m_1\\cdots m_s$ with $m_1,\\ldots,m_s$ pairwise coprime, then\n$$z(m_1\\cdots m_s)={\\text{\\rm lcm}}[z(m_1),\\ldots,z(m_s)].$$\n\n\\end{itemize}\n\\end{lemma}\nPart i says that $z(p^b)=p^b$ in\ncase $p\\mid \\Delta(k)$ and\n$b\\ge 1$. The\nnext result describes what happens\nfor arbitrary $b$ and $p>2$.\n\\begin{lemma}\n\\label{speciaal}\nAssume that $p>2$ is such that $p\\mid \\Delta(k)$. Let $z(p^b)$ be the index of appearance\nof $p^b$ in the sequence\n$U(k)$.\n\\begin{itemize}\n\\item[i)] If $p>3$, then $\\nu_p(U_p)=1$. In particular, $z(p^b)=p^b$ holds for all $b\\ge 1$.\n\\item[ii)] If $p=3$, then\n$$\nU_3=16k(k+1)+3.\n$$\nIn particular, $\\nu_3(U_3)=c>1$ exactly when $k\\equiv 2,6\\pmod* 9$. In these cases, $z(p^b)=p^{\\max\\{b-c,0\\}}$.\nHence, $z(p^b)\\mid p^{b-1}$ for all $b\\ge 2$.\n\\end{itemize}\n\\end{lemma}\n\\begin{proof}\nRecall that $\\Delta(k)=16k(k+1)$.\nPart i is known. As\nfor ii, we compute\n$$\nU_3=\\frac{\\alpha^3-\\alpha^{-3}}{\\alpha-\\alpha^{-1}}=\\alpha^2+1+\\alpha^{-2}=16k(k+1)+3.\n$$\nSince\nby assumption $3\\mid 16k(k+1)$, it follows that either $3\\mid k$ or $3\\mid (k+1)$. In the first case, $k=3k_0$ and\n$$\nU_3=3(16k_0(3k_0+1)+1).\n$$\nThe number in parenthesis is congruent to $16k_0+1\\pmod*{3}$, which is a multiple of $3$ exactly when $k_0\\equiv 2\\pmod*{3}$; hence, $k\\equiv 6\\pmod*{9}$. In the second case, $k+1=3k_1$, so\n$$\nU_3=3(16k_1(3k_1-1)+1)\n$$\nand the number in parenthesis is congruent to $-16k_1+1\\pmod*{3}$ which is a multiple of $3$ exactly when $k_1\\equiv 1\\pmod*3$, so \n$k\\equiv 2\\pmod*{9}$.\n\\end{proof}\n\\subsection{Index of appearance in case $k=1$}\nFor notational convenience we ignore\nwhere appropriate the index $k=1$ in $U(k),~\\alpha(k),~\u00ca\\beta(k)$ and so we only write $U,~\u00ca\\alpha,~\\beta$.\nWe have $\\Delta(1)=8$ and the relevant\nquadratic field is ${\\mathbb K}={\\mathbb Q}[{\\sqrt{2}}]$,\nwhich has ${\\mathbb Z}[{\\sqrt{2}}]$\nas its ring of integers.\nIf $\\gamma,\\delta\\in \\mathbb Z[\\sqrt{2}]$,\nthen we write $\\gamma \\equiv \\delta \n\\pmod*{p}$\nif and only if $(\\gamma-\\delta)\/p\\in \\mathbb Z[\\sqrt{2}]$. If $\\rho=a+b\\sqrt{2}\\in {\\mathbb K}$\nwith $a$ and $b$ rational numbers, then\nthe norm $N_{\\mathbb K}(\\rho)=\\rho \\cdot {\\overline \\rho}=a^2-2b^2,$ where ${\\overline \\rho}$ is the conjugate\nof $\\rho$ obtained by sending $\\sqrt{2}$ to $-\\sqrt{2}$.\n\\par For odd $p$, $z(p)$ is\na divisor of either $p-1$\nor $p+1$ by Lemma \\ref{lem:flauw} ii. The next lemma shows\nthat even more is true.\n\\begin{lemma}\n\\label{z(pb)}\nLet $k=1$ and $p$ be an odd prime.\nThen\n$$z(p^b)\\mid p^{b-1}\\Big(p-\\Big(\\frac{2}{p}\\Big)\\Big)\/2.$$\n\\end{lemma}\n\\begin{proof}\\hfil\\break\n\\noindent i) The case $e=(\\frac{2}{p})=1$.\n\\medskip\n\\par \\noindent Then $2^{(p-1)\/2}\\equiv \n1\\pmod* p$. We have\n$$\n\\beta^p=(1+{\\sqrt{2}})^p\\equiv 1+2^{p\/2}\\equiv 1+{\\sqrt{2}} \\cdot 2^{(p-1)\/2} \\equiv\\beta\\pmod*{p}.\n$$\nHere we used Euler's theorem that $2^{(p-1)\/2}\\equiv e\\pmod* p$. Since $\\beta$ is a unit, we\ninfer from $\\beta^{p}\\equiv \\beta\\pmod* p$ that\n$\\beta^{p-1}\\equiv 1\\pmod* p$. Thus,\n$$\n\\alpha^{(p-1)\/2}=(\\beta^2)^{(p-1)\/2}=\\beta^{p-1}\\equiv 1\\pmod*{p}.\n$$\nThe same congruence holds for $\\alpha$ replaced by $\\alpha^{-1}$. Hence, subtracting the\ntwo congruences we get that $\\alpha^{(p-1)\/2}-\\alpha^{-(p-1)\/2}\\equiv \n0\\pmod* p$. Thus, $p$ divides the\ndifference\n$\\alpha^{(p-1)\/2}-\\alpha^{-(p-1)\/2}$.\nOn noting that\n$$N_{\\mathbb K}(\\alpha^{(p-1)\/2}-\\alpha^{-(p-1)\/2})=32U_{(p-1)\/2}^2,$$ we infer that\n$p\\mid U_{(p-1)\/2}$.\nThus, $z(p)\\mid (p-1)\/2$, therefore $z(p^b)$ divides $p^{b-1}(p-1)\/2$ by Lemma \\ref{lem:flauw} iii.\n\\medskip \\hfil\\break\n\\noindent ii) The case $e=(\\frac{2}{p})=-1$. \\hfil\\break\n\\medskip\n\\noindent Then $2^{(p-1)\/2}\\equiv -1\\pmod* p$.\nNow we have\n$$\n\\beta^p=(1+{\\sqrt{2}})^p\\equiv 1+2^{p\/2}\\equiv 1+{\\sqrt{2}} \\cdot 2^{(p-1)\/2} \\equiv -\\beta^{-1}\\pmod*{p}.\n$$\nThus, $\\beta^{p+1}\\equiv -1\\pmod* p$. In particular,\n\\begin{equation}\n\\label{eq:minuseen}\n\\alpha^{(p+1)\/2}=(\\beta^2)^{(p+1)\/2}=\\beta^{p+1}\\equiv -1\\pmod*{p}.\n\\end{equation}\nThe same congruence holds for $\\alpha$ replaced by $\\alpha^{-1}$. Subtracting the\ntwo congruences, we get that $\\alpha^{(p+1)\/2}-\\alpha^{-(p+1)\/2}\\equiv 0\\pmod* p$.\nNoting that $$N_{\\mathbb K}(\\alpha^{(p+1)\/2}-\\alpha^{-(p+1)\/2})=32U_{(p+1)\/2}^2,$$ we obtain that\n$p\\mid U_{(p+1)\/2}$.\nWe have,\nin particular, $z(p)\\mid (p+1)\/2$ and\nhence $z(p^b)\\mid p^{b-1}(p+1)\/2$\nby Lemma \\ref{lem:flauw} iii.\n\\end{proof}\n\nLet us recall the following well-known result.\n\n\\begin{lemma}\n\\label{zpb}\nLet $p$ be odd such that $e=(\\frac{2}{p})=-1$ and\nlet $b\\ge 1$ be an integer. Then $z(p^b)$ is the minimal $m\\ge 1$ such that $\\alpha^m\\equiv \\pm 1\\pmod*{p^b}$.\n\\end{lemma}\n\n\\begin{proof} Assume that $m\\ge 1$ is such that $\\alpha^m\\equiv \\varepsilon \\pmod*{p^b}$ for some $\\varepsilon\\in \\{1,-1\\}$. Then $\\alpha^{-m}\\equiv \\varepsilon \\pmod*{p^b}$. Subtracting both congruences we get that $p^b$ divides $\\alpha^m-\\alpha^{-m}$. Computing\nnorms\nwe see that\n$p^{2b}\\mid N_{\\mathbb K}(\\alpha^m-\\alpha^{-m})$,\nand so $p^{2b}\\mid 32U_m^2,$\nand therefore $p^b|U_m$, showing that $z(p^b)|m$.\nNext assume that $p^b\\mid U_m$ for\nsome $m\\ge 1$. Then $\\alpha^m\\equiv \\alpha^{-m}\\pmod*{p^b}$, so $\\alpha^{2m}\\equiv 1\\pmod*{p^b}$. Thus, $p^b\\mid (\\alpha^m-1)(\\alpha^m+1)$.\nThe assumption on $e$ implies that $p$ is inert in ${\\mathbb Z}[{\\sqrt{2}}]$. Since, moreover,\n$p$ cannot divide both $\\alpha^m-1$ and $\\alpha^m+1$,\nit follows that $p^b$ must divide\neither $\\alpha^m-1$ or $\\alpha^m+1$. \n\\end{proof}\n\n\\section{Structure of the discriminator ${\\mathcal D}_1$}\n\nNow we are ready to restrict the number of\nvalues the discriminator can assume.\n\n\\begin{lemma}\n\\label{restrict}\nLet $m={\\mathcal D}_1(n)$ for some $n>1$. Then\n\\begin{itemize}\n\\item[{\\rm i)}] $m$ has at most one odd prime\ndivisor.\n\\item[{\\rm ii)}] If $m$ is divisible by\nexactly one odd prime $p$, then $e=(\\frac{2}{p})=-1$ and $z(p)=(p+1)\/2$.\n\\item[{\\rm iii)}]\nIf $m$ is not a power of $2$, then $m$ can\nbe written as $2^ap^{b}$ with $a,b\\ge 1$ and $p\\equiv 5\\pmod* 8$. \n\\end{itemize}\n\\end{lemma}\n\\begin{proof}\nAssume that ${\\mathcal D}_1(n)=m$ and write it as\n$$\nm=2^a p_1^{b_1}\\cdots p_r^{b_r},\n$$\nwhere the $p_i$ are distinct\nodd primes.\nAssume first that $r\\ge 2$. Then $n\\le z(m)$ (otherwise if $z(m)p_1+p_2+1$. Assuming $3\\le p_11$, which is obviously satisfied. \\par\nSince $z(m)1$, then ${\\mathcal D}_{1}(n)$ is even.\n\\end{lemma}\n\n\\begin{proof}\nAssume that ${\\mathcal D}_1(n)=m$ is odd. By the previous lemma, it follows that $m=p_1^{b_1}$, where $({2\\over p_1})=-1$ and $z(p_1)=(p_1+1)\/2$. Further, in this\nsituation \\eqref{eq:minuseen} applies and\nwe have\n$$\n\\alpha^{(p_1+1)\/2}=-1+p_1\\gamma\n$$\nfor some algebraic integer $\\gamma\\in {\\mathbb Z}[{\\sqrt{2}}]$. By induction on $m\\ge 0$ one establishes that\n$$\n\\alpha^{p_1^m(p_1+1)\/2}\\equiv\n-1\\pmod*{p_1^{m+1}}.\n$$\nLet\n$$\ni=\\left\\lfloor \\frac {p_1^{b_1-1}(p_1+1)}{4}\\right\\rfloor-1\\quad {\\text{\\rm and}}\\quad j=\\frac{p_1^{b_1-1}(p_1+1)}{2}-\\left(\\left\\lfloor \\frac{p_1^{b_1-1}(p_1+1)}{4}\\right\\rfloor-1\\right).\n$$\nSince $b_1\\ge 1$ and $p_1\\ge 3$, we have that $i\\ge 0$. Further,\n$$\nj\\ge \\frac{p_1^{b_1-1}(p_1+1)}{2}-\\frac{p_1^{b_1-1}(p_1+1)}{4}+1=\\frac{p_1^{b_1-1}(p_1+1)}{4}+1\\ge i+2,\n$$\nand\n\\begin{equation}\n\\label{afschatting}\nj \\le \\frac{p_1^{b_1-1}(p_1+1)}{2}-\\left(\\frac{p_1^{b_1-1}(p_1+1)}{4}-\\frac{3}{4}\\right)+1= \\frac{p_1^{b_1-1}(p_1+1)}{4}+\\frac{7}{4}.\n\\end{equation}\nSince $i+j=p_1^{b_1-1}(p_1+1)\/2$, we have\n$\n\\alpha^{i+j}\\equiv -1\\pmod*{p_1^{b_1}}.\n$\nThus,\n$$\n\\alpha^j\\equiv -\\alpha^{-i}\\pmod*{p_1^{b_1}},\n$$\nand also\n$$\n\\alpha^{-j}\\equiv -\\alpha^i\\pmod*{p_1^{b_1}}.\n$$\nTaking the difference of the latter\ntwo congruences we get that\n$$\n(\\alpha^j-\\alpha^{-j})-(\\alpha^i-\\alpha^{-i})\\equiv 0\\pmod*{p_1^{b_1}}.\n$$\nThus, taking norms and using the fact that $p_1$ is inert in ${\\mathbb K}$ and\nso has norm $p_1^2$, we get\n$\np_1^{2b_1}\\mid N_{\\mathbb K}((\\alpha^j-\\alpha^{-j})-(\\alpha^i-\\alpha^{-i}))$, that is\n$$p_1^{2b_1}\\mid32(U_{j}-U_i)^2,\n$$\ngiving\n$$\nU_j\\equiv U_i\\pmod*{p_1^{b_1}}.\n$$\nSince $i7+p_1^{b_1}+p_1^{b_1-1}$, which is equivalent to $p_1^{b_1-1}(p_1-1)>7$. This holds whenever $p_1^{b_1}\\ge 11$. Thus, only the cases $m=p_1^{b_1}\\le 9$ need to be checked, so $n<9$. We check in this range\nand we get no odd discriminant. Thus, indeed $n1$, then $p_1\\| U_{z(p_1)}$.\n\\end{lemma}\n\n\\begin{proof}\nThis is trivial. Indeed, if $b_1>1$ and $p_1^2\\mid U_{z(p_1)}$, then $z(p_1^{b_1})\\mid p_1^{b_1-2}(p_1+1)\/2$\nby Lemma \\ref{lem:flauw} iii. Thus, in this case\n$$\nz(m)\\mid {\\text{\\rm lcm}}[2^a, p_1^{b_1-2}(p_1+1)\/2]\\mid 2^{a-1}p_1^{b_1-2} (p_1+1),$$\nSince $2^{a-1}p_1^{b_1-2} (p_1+1)=m (p_1+1)\/(2p_1^2)1$, some power of $p_1$ may divide the first factor and some power of $p_1$ can divide the second factor.\nWe investigate each of these options.\\medskip\n\\hfil\\break\n\\noindent i) $p_1^{b_1}\\mid (\\alpha^i-\\alpha^j)$.\n\\medskip \\hfil\\break\n\\noindent Then $\\alpha^{i-j}\\equiv \n1\\pmod*{p_1^{b_1}}$.\nSince $i$ and $j$ are of the same parity, it follows that\n$\\alpha^{(i-j)\/2}\\equiv \\pm 1\\pmod*{p_1^{b_1}}$.\nBy Lemma \\ref{zpb} we then infer that\n$z(p_1^{b_1})|(i-j)\/2$.\nBy Lemma \\ref{z(pb)} we have\n$z(p_1^{b_1})|p_1^{b_1-1}(p_1+1)\/2$.\nSince by assumption $p_1\\equiv 5\\pmod* 8$ it\nfollows that $z(p_1^{b_1})$ is odd\nand so divides $i-j$.\nSince $i-j$ is also divisible by $2^a=z(2^a)$, it is\ndivisible by ${\\text{\\rm lcm}}[z(2^a), z(p_1^{b_1})]=z(m)$.\\medskip \\hfil\\break\n\\noindent ii) $p_1^{b_1}$ does not divide $\\alpha^i-\\alpha^j$.\n\\medskip\n\\hfil\\break\n\\noindent We want to show that this case does not occur. If it does, then $p_1$ divides\n\\begin{equation}\n\\label{eq:7}\n\\alpha^i+\\alpha^j-4{\\sqrt{2}}\\lambda.\n\\end{equation}\nAssume first that $p_1\\mid \\lambda$. Then $p_1\\mid U_i$ and $p_1\\mid U_j$ so both $i$ and $j$ are divisible by the\nodd number $z(p_1)=(p_1+1)\/2$. Also, $i\\equiv j\\pmod*{2}$. Since\n$i=z(p_1)i_1$ and $j=z(p_1)j_1$, where\n$i_1\\equiv j_1\\pmod*{2}$ and $\\alpha^{z(p_1)}\\equiv -1\\pmod*{p_1}$, it follows that $\\alpha^i$ and $\\alpha^j$ are both congruent either to $1$ (if $i_1$ and $j_1$ are even) or to $-1$ (if $i_1$ and $j_1$ are odd) modulo $p_1$.\nThus, modulo $p_1$ the expression \\eqref{eq:7} is in fact congruent to $\\pm 2$ modulo $p_1$, which is certainly not zero. Thus, $\\lambda\\ne 0$. Then\n\\begin{equation}\n\\label{eq:een}\n\\alpha^i+\\alpha^j\\equiv 4{\\sqrt{2}}\\lambda\n\\pmod*{p_1}.\n\\end{equation}\nThe prime $p_1$ is inert so we can conjugate the above relation to get\n\\begin{equation}\n\\label{eq:twee}\n\\alpha^{-i}+\\alpha^{-j}\\equiv -4{\\sqrt{2}}\\lambda\\pmod*{p_1}.\n\\end{equation}\nMultiplying the second congruence by $\\alpha^{i+j}$ and subtracting\n\\eqref{eq:twee} from \\eqref{eq:een}, we get $4{\\sqrt{2}} \\lambda(\\alpha^{i+j}+1)\\equiv 0\\pmod* {p_1}$. Thus, $\\alpha^{i+j}\\equiv -1\\mod {p_1}$. But the smallest $k$ such that $\\alpha^k\\equiv -1\\pmod* {p_1}$ is $k=z(p_1)=(p_1+1)\/2$ which is odd. Hence, $i+j$ is an odd multiple of $z(p_1)$, therefore an odd number itself, which is a contradiction since $i\\equiv j\\pmod* 2$. Thus, this case does not appear.\nThis implies that $i\\equiv j\\pmod* {z(m)}$ if $U_i\\equiv U_j\\pmod* m$.\\medskip\n\\hfil\\break\n\\indent Conversely, assume $i>j$ and $i\\equiv j\\pmod* {z(m)}$. We need to show that $U_i\\equiv U_j\\pmod* m$. Since $i\\equiv j\\pmod* {2^a}$, it follows that $i-j$ is even and hence $U_i-U_j=U_{(i-j)\/2}V_{(i+j)\/2}$. Since $2^{a-1}\\mid (i-j)\/2$, we get, by iteratively applying the formula $U_{2n}=U_nV_n$, that\n$$\nU_i-U_j=U_{(i-j)\/2^{a}} V_{(i-j)\/2^{a}} V_{(i-j)\/2^{a-1}}\\cdots V_{(i-j)\/4} V_{(i+j)\/2}.\n$$\nIn the right--hand side we have $a$ factors from the $V$ sequence and each of\nthem is a multiple of $2$. Hence, $2^a\\mid (U_i-U_j)$. As for the divisibility by $p_1^{b_1}$, note that since $z(p_1^{b_1})\\mid (i-j)$ and $i-j$ is even, it follows that\n$$i-j=p_1^{b_1-1}(p_1+1)\\ell,$$ for some positive integer $\\ell$. Since $\\alpha^{p_1^{b_1-1}(p_1+1)\/2}\\equiv -1\\pmod* {p_1^{b_1}}$, it follows that $\\alpha^{i-j}\\equiv 1\\pmod* {p_1^{b_1}}$. The same holds if we replace $\\alpha$ by $\\alpha^{-1}$. Thus,\n$$\n\\alpha^i\\equiv \\alpha^j \\alpha^{i-j}\\equiv\n\\alpha^{j}\\pmod*{p_1^{b_1}},\n$$\nand the same congruence holds if $\\alpha$\nis replaced by $\\alpha^{-1}$. Subtracting\nthese two congruences we get $p_1^{b_1}\\mid ((\\alpha^i-\\alpha^{-i})-(\\alpha^j-\\alpha^{-j}))$. Computing norms in ${\\mathbb K}$ and using the fact that $p_1$ is inert, we get\n$p_1^{2b_1}\\mid N_{\\mathbb K}((\\alpha^i-\\alpha^{-i})-(\\alpha^j-\\alpha^{-j}))$\nand so $$p_1^{2b_1}\\mid 32(U_i-U_j)^2.$$\nThus, $U_i\\equiv U_j\\pmod* {p_1^{b_1}}$. Hence, $U_i\\equiv U_j\\pmod* m$.\n\\end{proof}\n\\section{The end of the proof or why $5$ and not $37$?}\nWe need a few more results before we are prepared well enough to\nestablish Theorem \\ref{main}.\n\\begin{lemma}\n\\label{BP}\nFor $n\\ge 2^{24}\\cdot 5^3$ the interval\n$[5n\/3,37n\/19)$ contains a number of\nthe form\n$2^a\\cdot 5^{b}$ with $a\\ge 1$ and $b\\ge 0$.\n\\end{lemma}\n\\begin{proof}\nIt is enough to show that there exists an\nstrictly increasing sequence of integers\n$\\{m_i\\}_{i=1}^{\\infty}$ of the form\n$m_i=2^{a_i+1}\\cdot 5^{b_i}$ with\n$a_1=23$ and $b_1=3$, $a_i,b_i\\ge 0$, having\nthe property\nthat\n$$1<\\frac{m_{i+1}}{m_i}<\\frac{111}{95}.$$\nSince both\n$2^7\/5^3$ and $5^{10}\/2^{23}$ \nare in $(1,111\/95)$, the idea is to use\nthe substitutions $5^3\\rightarrow 2^7$ and $2^{23}\\rightarrow 5^{10}$ to\nproduce a strictly increasing sequence starting from $m_1$. Note that we\ncan at each stage make one of these substitutions as otherwise\nwe have reached a number dividing $2\\cdot 2^{22}\\cdot 5^{2}5$, $a,b\\ge 1$.\nIf $m\\ge \\frac{37}{19}\\cdot 2^{24}\\cdot 5^{3}$, then\n$m$ is not a discriminator value.\n\\end{corollary}\n\\begin{proof}\nSuppose that $\\mathcal{D}_1(n)=m$, then we must have\n$$z(m)=2^a\\cdot p^{b-1}(p+1)\/(2k)\\ge 19m\/37\\ge n,$$ that\nis $m\\ge 37n\/19$. By Lemma \\ref{BP} in the\ninterval $[5n\/3,37n\/19)$ there is an integer\nof the form $m=2^c\\cdot 5^{d}$ with $c\\ge 1$. This integer\ndiscriminates the first $n$ terms of the sequence and is smaller\nthan $m$. This contradicts the definition of the discriminator.\n\\end{proof}\nThus we see that in some sense there is an\nabundance of the numbers of the form\n$m=2^{a}\\cdot 5^b$ that are in addition fairly\nregularly distributed. Since they discriminate\nthe first $n$ terms provided that $m\\ge 5n\/3$, rather\nthan the weaker $m\\ge 2np\/(p+1)$ for $p>5$, they\nremain as values, whereas numbers of the form\n$m=2^{a}\\cdot p^b$ with $p>5$ do not.\n\\begin{lemma}\n\\label{2a5b}\nIf $n>1,$ then ${\\mathcal D}_1(n)=2^a\\cdot 5^b$ for\nsome $a\\ge 1$ and\n$b\\ge 0$.\n\\end{lemma}\n\\begin{proof}\nBy Lemma \\ref{iseven} we have $a\\ge 1$. If $m={\\mathcal D}_1(n)\\ne 2^a$\nfor some $a\\ge 1$, then by Lemma \\ref{restrict} iii it is\nof the form $m=2^a\\cdot p_1^{b_1}$ for some $p_1\\equiv 5\\pmod*{8}$.\nLet assume for\nthe sake of contradiction that we have discriminators of the form $m=2^ap_1^{b_1}$ for some odd $p_1> 5$. Then $z(p_1^{b_1})=p_1^{b_1-1}(p_1+1)\/2$. Let $A$ be minimal with $p_1^{b_1}<2^{A+8}$. Then $A\\ge 2$ by our calculation\nbecause we did not find any such $p_1$ on calculating ${\\mathcal D}_1(n)$ for $n\\le 2^{10}$ (cf. Section \\ref{intro}). Then\n$$\n2^{A+7}50$, $2^{A+8}>p_1^{b_1}$ and that $p_1^{b_1}+p_1^{b_1-1}>2^{A+8}$,\nwe obtain\n$$\n0<2^{A+8}-p_1^{b_1}2^{A+8}\/50,\n$$\nit follows that $p_1^{b_1}>2^{A+1} 5^3$. We now claim that $p_1^{b_1-1}(p_1+1)\/2<2^{A+1} \\cdot 3\\cdot 5^2$. Indeed, for that it suffices that\n$$\n2^{A+8} \\left(\\frac{p_1+1}{2p_1}\\right)<2^{A+1}\\cdot 3\\cdot 5^2,\n$$\nso $2p_1\/(p_1+1)>128\/75$, which is equivalent to $22p_1>128$, which is true for $p_1>50$.\n\nSince $2^a\\cdot p_1^{b_1}$ discriminates the first\n$2^a\\cdot p_1^{b_1-1}(p_1+1)\/2$ terms of the sequence (but not more)\nand the same integers are discriminated by the smaller number\n$2^{a+A+1}\\cdot 5^3$, the number $2^a\\cdot p_1^{b_1}$ is not a\ndiscriminator value.\n\nSo, it remains to check primes $p_1<50$. Since $p_1\\equiv 5\\pmod* 8$, we just need to check $p_1\\in \\{13,29,37\\}$. Fortunately, $13$ is a Wiefrich prime for $\\alpha$ in that\n$$\n(3+2{\\sqrt{2}})^7\\equiv -1\\pmod*{13^2},\n$$\nso we cannot use powers $13^{b_1}$ with $b_1>1$, while for $b_1=1$ the interval $[z(p_1),p_1]=[7,13]$ contains $8$ which is a power of $2$. For $29$, we have that $z(29)=5$ (instead of $(29+1)\/2$), so\n$29$ is not good either.\n\nIt remains to deal with $p=37$. We will show\nthat for $k_i=2\\cdot 37^i$ and $1\\le i\\le 5$,\nthere is a power of the form $2^{e_i}1$. By Lemma \\ref{2a5b} it then follows that\neither $m=2^a$ for some $a\\ge 1$ or\n$m=2^a\\cdot 5^b$ with $a,b\\ge 1$.\nOn invoking Lemma \\ref{twocases} we infer\nthat the first assertion holds true.\n\\par It remains to determine the image of\nthe discriminator ${\\mathcal D}_1$.\nLet us suppose that $m=2^a\\cdot 5^b$ with $a,b\\ge 1$\noccurs as value. Let $\\alpha$ be the unique integer\nsuch that $2^{\\alpha}<2^a\\cdot 5^b<2^{\\alpha+1}$.\nBy Lemma \\ref{twocases} it now follows that we must have\n$z(m)>2^{\\alpha}$, that is\n$2^a\\cdot 5^{b-1}\\cdot3>2^{\\alpha}$. It follows that $m$ occurs\nas value iff\n\\begin{equation}\n\\label{ongelijk}\n\\frac{5}{3}\\cdot 2^{\\alpha}<2^a\\cdot 5^b<2^{\\alpha+1}.\n\\end{equation}\n(Indeed, under these\nconditions we have ${\\mathcal D}_1(n)=2^a\\cdot 5^b$ for\n$n\\in [2^a+1,2^a\\cdot 5^{b-1}\\cdot3]$.) Inequality \\eqref{ongelijk}\ncan be rewritten as $5\/6<2^{a-\\alpha-1}<1$ and, after\ntaking logarithms, is seen to have\na solution iff $b\\in {\\mathcal M}$.\nIf it has a solution, then we must have\n$\\alpha-a=\\lfloor b\\log 5\/\\log 2\\rfloor$. In\nparticular for each $a\\ge 1$ and $b\\in {\\mathcal M}$,\nthe number $2^a\\cdot 5^b$ occurs as value.\n\\end{proof}\n\n\\section{General $k$}\n\\subsection{Introduction}\n\\label{generalkintro}\nWhat is happening for $k>1$? It turns out that the situation is quite different. \n\\par For $k=2$ we have the following result.\n\\begin{theorem}\n\\label{tweee}\nLet $e\\ge 0$ be the smallest integer such that\n$2^e\\ge n$ and $f\\ge 1$ the smallest integer such that\n$3\\cdot 2^f\\ge n$. Then\n${\\mathcal D}_2(n)=\\min\\{2^e,3\\cdot 2^f\\}$.\n\\end{theorem}\n\\begin{proof}\nWe have that if\n$z(m)=m$, then $m|3\\cdot 2^a$ for some $a\\ge 0$.\nFor the other integers $m$ we have $z(m)\\le 3m\/5$ (actually\neven $z(m)\\le 7m\/13$).\nIt follows that if $m$ discriminates the first $n$ values\nof the sequence $U(2)$, then we must have\n$m\\ge 5n\/3$.\nIt is easy to check that for every $n\\ge 2$\nthere is a power of two or a number of the form\n$3\\cdot 2^a$ in the interval $[n,5n\/3)$.\nAs ${\\mathcal D}_2(1)=1$ we are done.\n\\end{proof}\n\nFor the\nconvenience of the reader we recall the\ntheorem from the introduction which deals\nwith the case $k>2$.\n\\begin{theorem}\n\\label{thm:2}\nPut\n$${\\mathcal A}_k=\\begin{cases}\n\\{m~{\\text{\\rm odd}}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k\\}\\text{~if~}\nk\\not\\equiv 6\\pmod*{9};\\cr\n\\{m~{\\text{\\rm odd}},~9\\nmid m:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k\\}\\text{~if~}\nk\\equiv 6\\pmod*{9},\n\\end{cases}\n$$\nand $${\\mathcal B}_k\n=\\begin{cases}\n\\{m~{\\text{\\rm even}}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k(k+1)\\}\n\\text{~if~}\nk\\not\\equiv 2\\pmod*{9};\\cr\n\\{m~{\\text{\\rm even},~9\\nmid m}:\\text{\\rm if~}p\\mid m,~{\\text{\\rm then}}~p\\mid k(k+1)\\}\n\\text{~if~}\nk\\equiv 2\\pmod*{9}.\n\\end{cases}\n$$\nLet $k>2$. We have\n\\begin{equation}\n\\label{dkinequal2}\n{\\mathcal D}_{k}(n)\\le \\min\\{m\\ge n:\nm\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}\\},\n\\end{equation}\nwith equality if the interval $[n,3n\/2)$ contains an integer \n$m\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}$.\nThere are at most \nfinitely many $n$ for which in \\eqref{dkinequal2} strict inequality holds.\nFurthermore, we have \n\\begin{equation}\n\\label{Dkn=n}\n{\\mathcal D}_{k}(n)=n \n\\iff n\\in {\\mathcal A}_{k}\\cup {\\mathcal B}_{k}.\n\\end{equation}\n\\end{theorem}\n\nIn our proof of this result the rank of appearance\nplays a crucial role. Its most \nimportant properties are summarized in Lemma \\ref{appearance}.\n\\subsection{The index of appearance}\n\\subsubsection{The case where $p\\mid k(k+1)$}\nThe index of appearance for\nprimes $p$ dividing $k(k+1)$ is determined in\nLemma \\ref{speciaal} for $p>2$.\nBy Lemma \\ref{poweroftwo} we have\n$z(2^b)=2^b$. In general $z(p^b)=p^b$ for\nthese primes, but for a prime which we call {\\it special} a\ncomplication can arise giving rise to\n$z(p^b)\\mid p^{b-1}$ for $b\\ge 2$.\n\\begin{defi}\nA prime $p$ is said to be special if\n$p|k(k+1)$ and $p^2|U_p$.\n\\end{defi}\nThe special feature of a special prime\n$p$ is that $p^b$ with $b\\ge 2$ cannot\ndivide a discriminator value.\nRecall that $z(p^a)=p^{\\max\\{a-c,0\\}} z(p)$, where $c=\\nu_p(U_{z(p)})$ by Lemma \\ref{lem:flauw}.\n\\begin{lemma}\n\\label{drienogmaals}\nLet $p\\ge 3$ be an odd prime.\nIf $z(p^{b})|p^{b-1}$, then $m=p^bm_1$\nwith $p\\nmid m_1$ is not a discriminator\nvalue.\n\\end{lemma}\n\\begin{proof}\nTaking $i=0$ and $j=p^{b-1}z(m_1)$ we have\n$U_i\\equiv U_j\\equiv 0 \\pmod*{m}.$\nIt follows that $n\\le p^{b-1} z(m_1)\\le m\/p$ so any power of 2 in $[m\/3,m)$ (and such a\npower exists) is a better discriminator than $m$.\n\\end{proof}\nBy Lemma \\ref{speciaal} only 3 can be special.\n\\subsubsection{The case where $p\\nmid k(k+1)$}\nLet us now look at odd prime numbers $p$ such that $p\\nmid k(k+1)$. These come in two flavors according to the sign of \n\\begin{equation}\n\\label{ep}\ne_p=\\Big(\\frac{k(k+1)}{p}\\Big).\n\\end{equation}\nSuppose that $e_p=1$. Then\neither\n$$\n\\left(\\frac{k}{p}\\right)=\\left(\\frac{k+1}{p}\\right)=1\\quad {\\text{\\rm or}}\\quad \\left(\\frac{k}{p}\\right)=\\left(\\frac{k+1}{p}\\right)=-1.\n$$\nIn the first case,\n\\begin{eqnarray*}\n\\beta^p & = & ({\\sqrt{k+1}}+{\\sqrt{k}})^p\\equiv {\\sqrt{k+1}} (k+1)^{(p-1)\/2}+{\\sqrt{k}} k^{(p-1)\/2}\\\\\n& \\equiv & {\\sqrt{k+1}}+{\\sqrt{k}}\\equiv \\beta\\pmod*{p}.\n\\end{eqnarray*}\nIn the second case, a similar calculation shows that $\\beta^p\\equiv -\\beta$. Thus, $\\beta^{p-1}\\equiv \\pm 1\\pmod* p$ and since $\\alpha=\\beta^2$, we get that $\\alpha^{(p-1)\/2}=\\beta^{p-1}\\equiv \\pm 1\\pmod* p$. Since the\nlast congruence implies that $\\alpha^{-(p-1)\/2}\\equiv \\pm 1\\pmod* p$ we obtain\non subtracting these two congruences that\n$p\\mid U_{(p-1)\/2}$. Thus, $z(p)\\mid (p-1)\/2$.\nIn case $e_p=-1$, a similar calculation shows that $\\beta^p\\equiv \\pm \\beta^{-1}\\pmod* p$, so $\\beta^{p+1}\\equiv \\pm 1\\pmod p$. This shows that $\\alpha^{(p+1)\/2}\\equiv \\pm 1\\pmod* p$, which leads to $z(p)\\mid (p+1)\/2$. \nThere is one more observation which is useful here. Assume that $e_p=-1$, which implies\nthat $z(p)\\mid (p+1)\/2$. Suppose that $p\\equiv 3\\pmod* 4$. Then $(p+1)\/2$ is even. Assume further that\n$$\n\\left(\\frac{k+1}{p}\\right)=1\\quad{\\text{\\rm ~~and ~~}}\\quad\\left(\\frac{k}{p}\\right)=-1.\n$$\nIn this case, by the above arguments, we have that $\\beta^{p}\\equiv \\beta^{-1}\\pmod* p$, so $\\beta^{p+1}\\equiv 1\\pmod* p$. This gives $\\alpha^{(p+1)\/2}\\equiv 1\\pmod* p$. Since $(p+1)\/2$ is even we conclude that\n$$\np\\mid(\\alpha^{(p+1)\/4}-1)(\\alpha^{(p+1)\/4}+1).\n$$\nSince $p$ is inert in ${\\mathbb K}$, we get that $\\alpha^{(p+1)\/4}\\equiv \\pm 1\\pmod* p$, which later leads to $p\\mid U_{(p+1)\/4}$, Hence, $z(p)\\mid (p+1)\/4$ in this case.\n\n\\subsubsection{General $m$}\n\\begin{lemma}\n\\label{appearance} Let $k\\ge 1$.\nWe have $z(m)=m$ if and only if\n$$\n\\begin{cases}\nm\\in {\\mathcal P}(k(k+1)),~9\\nmid m;\\cr\nm\\in {\\mathcal P}(k(k+1)),~9\\mid m,~\n\\text{and~}3{~is~not~special.}\n\\end{cases}\n$$\nFor the remaining integers $m$ we have\n$$z(m)\\le \\alpha_k m,$$\nwith\n\\begin{equation}\n\\label{alphadef}\n\\alpha_k:=\\lim\\sup_{m\\rightarrow \\infty}\\Big\\{\\frac{z_k(m)}{m}:z_k(m)2$, then $k(k+1)$ has an odd prime factor\nthat is not special.\n\\end{lemma}\n\\begin{proof}\nIf $k(k+1)$ only has an odd prime factor that is special, then\nit must be $3$ and $k\\equiv 2,6\\pmod*{9}$. It follows that for\nsuch a $k$ there are $a,b$ for which the Diophantine equation\n\\begin{equation}\n\\label{Diophantine}\nk(k+1)=2^a\\cdot 3^b,\n\\end{equation}\nhas a solution. However, this is easily shown to \nbe impossible for $k>2$.\n\\end{proof}\nIt is slightly more challenging to\nfind {\\tt all} solutions $k\\ge 1$ of \\eqref{Diophantine}.\nIn that case one is led\nto the Diophantine equation\n$$2^a-3^b\\equiv \\pm 1,$$\nwhich was already solved centuries\nago by Levi ben Gerson (alias Leo\nHebraeus), who lived in Spain from\n1288 to 1344, cf. Ribenboim \\cite[p. 5]{Rib}.\nIt has the solutions $(a,b)=(1,0),(0,1),(2,1)$\nand $(a,b)=(3,2)$, corresponding to,\nrespectively, $k=1,2,3$ and $k=8$.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\\subsection{Bertrand's Postulate for S-units}\nBefore we embark on the proof of our main result we\nmake a small excursion in Diophantine approximation.\n\\begin{lemma}\n\\label{eventuallythere}\nLet $\\alpha>1$ be a real number and \n$p$ be an arbitrary odd prime. Then there \nexists a real number $x(\\alpha)$ such\nthat for every $n\\ge x(\\alpha)$ the interval $[n,n\\alpha)$ contains\nan even integer of the form\n$2^a\\cdot p^b$.\n\\end{lemma}\n\\begin{proof}\nAlong the lines of the proof \nof Lemma \\ref{BP}. If $\\beta$\nis irrational, then the sequence of integers\n$\\{m\\beta\\}_{m=1}^{\\infty}$ is uniformly distributed.\nThis allows one to find quotients\n$2^c\/p^d$ and $p^r\/2^s$ that are in the interval\n$(1,\\alpha)$. Then proceed as in the proof\nof Lemma \\ref{BP}.\n\\end{proof}\nThe result also holds for S-units of the \nform $\\prod_{i=1}^s p_i^{b_i}$ with $p_1<\\ldots < p_s$ primes\nand $s\\ge 2$. \n\\par In \\cite{CLM} we consider\nthe {Bertrand's Postulate for S-units in \ngreater detail.\n\n\n\\subsection{Proof of the main result for\ngeneral $k$}\nFinally we are in the position to prove our main\nresult for $k>1$.\n\\begin{proof}[Proof of Theorem \\ref{thm:2}] Let $k>2$.\n\\hfil\\break\n\\indent First case: $m\\in {\\mathcal A}_k\\cup {\\mathcal B}_k.$ (Note that \n$z(m)=m$ for these $m$.)\\hfil\\break\nBy Lemma \\ref{lem:iui} we infer that the inequality \\eqref{dkinequal2}\nholds true and moreover the equivalence \n\\eqref{Dkn=n}. The ``$\\Leftarrow$\" implication\nin \\eqref{Dkn=n} yields\n${\\mathcal A}_k\\cup {\\mathcal B}_k\n\\subseteq {\\mathcal D}_k$.\n\\par Second case: $z(m)=m$ and $m\\not \\in\n{\\mathcal A}_k\\cup {\\mathcal B}_k$.\\hfil\\break\nIn this case $m$ has a \nodd prime divisor $p$ \nthat also divides $k+1$. \nNow write $m=p^a\\cdot m_1$ with $p\\nmid m_1$ and $m_1$ odd. Note\nthat $z(p^a)=p^a$.\nConsider $i=(p^a-1)m_1\/2$ and $j=(p^a+1)m_1\/2$. Then $i\\not\\equiv j\\pmod* 2$ and \n$p^{a}\\mid (i+j)$. Thus, $U_i\\equiv U_j\\pmod* {p^{a}}$\nby Lemma \\ref{tweedeelgevallen}.\nSince\n$m_1\\mid i$ and $m_1\\mid j$ and\n$m_1$ is composed of primes dividing\n$\\Delta(k)$, it follows that $U_i\\equiv U_j\\equiv 0\\pmod* {m_1}$. This shows that\nif $m$ discriminates the \nnumbers $U_0(k),\\ldots,U_{n-1}(k),$ then\n$$\nn\\le \\left(\\frac{p^{a}+1}{2}\\right)m_1.\n$$\nThe interval $[(p^a+1)\/2,p^{a})$ contains a power of $2$, say $2^b$. Then \n$2^bm$ is a better discriminator than $p^{a}m_1=m$. \nThus if $z(m)=m$ and $m\\not \\in\n{\\mathcal A}_k\\cup {\\mathcal B}_k,$ then $m$ is not a discriminator value.\n\\par Third case: $z(m)2$, by \nLemma \\ref{dioflauw} there exists\na non-special\nodd prime $p$ dividing $k(k+1)$ \nand hence if $a,b\\ge 0,$ then $2^{1+a}\\cdot p^b\n\\in {\\mathcal A}_k\\cup {\\mathcal B}_k.$ It now\nfollows by Lemma \\ref{eventuallythere} that for every $n$ large enough the interval $[n,3n\/2)$ \ncontains an element from ${\\mathcal A}_k\\cup {\\mathcal B}_k$ and so \nthere are at most \nfinitely many $n$ for which in \\eqref{dkinequal2} strict inequality holds.\n\\end{proof}\n\\subsection{The set ${\\mathcal F}_k$}\nAs was remarked in the introduction a consequence of Theorems \\ref{twee}\nand \\ref{main2} is that\nfor $k>1$ there is a finite set ${\\mathcal F}_k$ such that\n$$\n{\\mathcal D}_{k}=\n{\\mathcal A}_k\\cup\n{\\mathcal B}_k\\cup {\\mathcal F}_k.\n$$\nThe set ${\\mathcal F}_k$ \nis not a figment\nof our proof of this result, as the\nfollowing result shows.\n\\begin{lemma}\nThere are infinitely many $k$ for the \nfinite set ${\\mathcal F}_k$ is non-empty.\nIt can have a cardinality larger than\nany given bound.\n\\end{lemma}\n\\begin{proof}\nLet $N$ be large and $k\\equiv 1\\pmod* {N!}$. Then $U(k)\\pmod* m$ is the same as $U(1)\\pmod* m$ for all\n$m\\le N$. In particular, if $N>2\\cdot 5^{m_s}$, where $m_s$ is the\n$s^\\text{th}$ element of the set ${\\mathcal M}$, then certainly ${\\mathcal D}_1\\cap [1,N]$ will contain the numbers $2\\cdot 5^{m_i}$ for $i=1,\\ldots,s$, and $5\\nmid k(k+1)$ (in fact,\n$k\\equiv 1\\pmod*{5}$, so $5\\nmid k(k+1)$), therefore all such numbers\nare in the set ${\\mathcal F}_k$ for such values of $k$.\n\\end{proof}\nThus it is illusory\nto want to describe ${\\mathcal F}_k$ \ncompletely for every $k\\ge 1$. Nevertheless, \nin part II \\cite{CLM} we will explore\nhow far we can get in this respect.\n\\section{Analogy with the polynomial\ndiscriminator}\nIn our situation for $k\\ge 1$ on the one hand there\nare enough $m$ with $z(m)=m$ and \n${\\mathcal D}_k(m)=m$, on the other hand for the\nremaining $m$ either $z(m)=m$ and $m$ is not a discriminator\nvalue or we have $z(m)\\le \\alpha_km$ \nwith $\\alpha_k<1,$ a constant not depending on \n$m$. Thus the distribution of\n$\\{z(m)\/m:m\\ge 1\\}$ shows a gap directly below 1 (namely\n$(\\alpha_k,1)$).\n\\par For polynomial discriminators the analogue of $z(p)$ is $V(p)$, the number of values\nassumed by the polynomial modulo $p$. If on\nthe one hand there are enough integers $m$ such that $f$ permutes $\\mathbb Z\/m\\mathbb Z$, and on the other hand $V(p)\/p$ with $V(p) 1$.\n\nUnder the hypothesis that $e > 1$, Snyder has proved that $|G| \\le (2e)!$ (see \\cite{Snyder}). He also showed that if $e = 2$, then $|G| \\le 8$ and if $e = 3$, then $|G| \\le 54$, and in both of these cases, there exist examples of these orders; hence, the bounds given are best possible for $e = 2$ and $3$.\n\nIsaacs has shown that $|G| \\le B e^6$ for some universal constant $B$ and in many cases that $|G| \\le e^6 + e^4$ (see \\cite{large}). Finally, Durfee and Jensen have proved in \\cite{DuJe} (without using the classification of nonabelian simple groups) that $|G| \\le e^6 - e^4$. When $G$ is solvable and either $e$ is a prime or $e$ is divisible by at least two distinct primes, they prove that $|G| \\le e^4 - e^3$. Hence, the only time it is possible that $G$ is solvable and $|G| > e^4 - e^3$ is when $e$ is a prime power that is not prime. Notice that when $e = 2$ and $e = 3$, the expression $e^4 - e^3$ yields the bound found by Snyder.\n\nIsaacs also shows that there exists a (solvable) group $G$ for every prime power $q$ of order $q^3 (q-1)$ where $d = q (q-1)$. It is easy to compute that $e = q$, so $d = e^2 - e$ and $|G| = e^4 - e^3$. On the other hand, there are no known groups $G$ where $|G| > e^4 - e^3$, so it seems likely that $|G| \\le e^4 - e^3$ is the correct bound. In this paper, we prove this bound for groups with a nontrivial, abelian subgroup.\n\n\\begin{thm} \\label{thm2}\nLet $G$ be a group with a nontrivial, abelian normal subgroup, and let $d$ and $e$ be defined as above. If $e > 1$, then $d \\le e^2 - e$ and $|G| \\le e^4 - e^3$.\n\\end{thm}\n\nSince all solvable groups have a nontrivial, abelian normal subgroup, we obtain the bound for solvable groups.\n\n\\begin{cor}\nLet $G$ be a solvable group, and let $d$ and $e$ be defined as\nabove. If $e > 1$, then $|G| \\le e^4 - e^3$.\n\\end{cor}\n\nWe also are able to make use of the results of Durfee and Jensen to improve the bound for all groups.\n\n\\begin{cor} \\label{cor1}\nLet $G$ be a group with $d$ and $e$ defined as above. If $e > 1$, then $d < e^2$ and $|G| < e^4 + e^3$.\n\\end{cor}\n\nAt this time, it is still an open question as to whether there exists any groups where $e^4 - e^3 < |G| < e^4 + e^3$. We do not have any examples of such groups, but at this time, we have not been able to prove that such a group cannot exist. By Theorem \\ref{thm2}, we know that if such a group does exist, then all the normal subgroups of $G$ must be nonabelian. An important subcase, which we believe is still open is whether the bound $|G| \\le e^4 - e^3$ can be proved when $G$ is a simple group.\n\nLike \\cite{DuJe}, \\cite{large}, and \\cite{Snyder}, we consider groups first studied by Gagola. Gagola studied groups that have an irreducible character that vanish on all but two conjugacy classes. He proved that if $G$ has a character $\\chi \\in \\irr G$ so that $\\chi$ vanishes on all but two conjugacy classes of $G$. We will say that $\\chi$ is a {\\it Gagola character}. Gagola proved that such Gagola characters are unique. Furthermore, he proved that $G$ has a unique minimal normal subgroup $N$. He proved that $N$ is an elementary abelian $p$-group for some prime $p$. We say $(G,N)$ is a {\\it $p$-Gagola pair} if $G$ has a Gagola character and $N$ is the unique minimal normal subgroup of $G$ and is a $p$-group. The main result of this paper is the following theorem.\n\n\\begin{thm} \\label{thm1}\nLet $(G,N)$ be a $p$-Gagola pair some prime $p$. If $P$ is a Sylow $p$-subgroup of $G$, then $d \\le e^2 - e$ and $|N|^2 \\le |P:N| = |G:N|_p$.\n\\end{thm}\n\nThe proof of this result splits into three cases: when $p$ is odd. when $p = 2$ and $G$ is solvable, and when $p = 2$ and $G$ is nonsolvable. We have to work the hardest to prove the result when $p = 2$ and $G$ is solvable. In particular, when $p = 2$ and $G$ is solvable, we need to study in detail Suzuki $2$-groups, and we compute the full automorphism group of a Suzuki $2$-group, which to our knowledge has not been published before.\n\nDurfee and Jensen prove that if $G$ has a nontrivial, abelian normal subgroup, then $G$ has a normal subgroup $N$ so that $(G,N)$ is a $p$-Gagola pair for some prime $p$. Thus, if there exists a group $G$ with $e^4 - e^3 > |G|$, then $d > e^2 - e$ and all the nontrivial, normal subgroups of $G$ are nonabelian.\n\nWe would like to particularly thank David Gluck and Stephen Gagola, Jr. for a number of useful conversations while we were preparing this this paper.\n\n\n\n\n\n\n\\section{Gagola pairs}\n\nWe first use the results in \\cite{DuJe} to reduce the initial\nquestion to a question regarding Gagola pairs. Since $|G| = d\n(d+e)$, to prove $|G| \\le e^4 - e^3$ it suffices to prove that $d \\le e^2 - e$ and to prove $|G| \\le e^4 + e^3$, it suffices to prove that $d \\le e^2$. To prove this, we need to introduce some terminology from \\cite{DuJe}. If $\\chi, \\psi \\in \\irr G$, then we say $\\chi$ {\\it dominates} $\\psi$ if $\\psi \\chi = \\psi (1) \\chi$. Note that since solvable groups have a nontrivial, abelian normal subgroup, this next lemma applies when $G$ is solvable.\n\n\\begin{lemma} \\label{gagola 1}\nIf $G$ has a nontrivial, abelian normal subgroup and if $e > 1$ and $d \\ge e^2 - e$ where $d$ and $e$ are defined as above, then $G$ has a Gagola character $\\chi$ with $\\chi (1) = d$.\n\\end{lemma}\n\n\\begin{proof}\nWe may apply Theorem 5.2 of \\cite{DuJe} to see that $G$ has a\ncharacter $\\chi \\in \\irr G$ which dominates all the other\nirreducible characters of $G$ and by Lemma 2.1 of \\cite{DuJe}, $\\chi (1) = d$. By Lemma 4.2 of \\cite{DuJe}, we know that $G$ has a minimal normal subgroup $N$ such that $\\chi$ vanishes off of $N$ and $G$ acts transitively on $N \\setminus \\{ 1 \\}$. This implies that $\\chi$ vanishes on all but two conjugacy classes of $G$, and so, $\\chi$ is a Gagola character.\n\\end{proof}\n\nWhen we have a Gagola pair, we can compute $d$ and $e$ in terms of a Sylow subgroup of $G$.\n\n\\begin{lemma} \\label{gagola 2}\nLet $(G,N)$ be a $p$-Gagola pair, and let $P$ be a Sylow\n$p$-subgroup of $G$. If $\\chi$ is the associated Gagola character and $d = \\chi(1)$, then $e^2 = |P:N|$ and $d = e(|N| - 1)$.\n\\end{lemma}\n\n\\begin{proof}\nLet $\\chi \\in \\irr G$ be the Gagola character for $G$. Let $\\lambda \\in \\irr N$ be a constituent of $\\chi_N$. By Lemma 2.2 of \\cite{Gagola}, we may choose $\\lambda$ so that $P$ is the stabilizer of $\\lambda$, and it is shown in that lemma that $\\lambda$ is fully-ramified with respect to $P\/N$. Let $b$ be the integer so that $b^2 = |P:N|$. We have $\\chi (1) = b |G:P|$. By Corollary 2.3 of \\cite{Gagola}, we know that $P$ is a point stabilizer in the action of $G$ on $N \\setminus \\{ 1 \\}$, and from Lemma 2.1 of \\cite{Gagola}, we know that $G$ acts transitively on $N \\setminus \\{ 1 \\}$. By the Fundamental Counting Principle, we have $|G:P| = |N| - 1$, and so, $\\chi (1) = b(|N| - 1)$.\n\nWe now need to show that $b = e$. We have\n$$\n|G| = |G:P| |P| = (|N| - 1)|P:N||N| = d(d+e).\n$$\nDividing by $d$, we obtain\n$$\nd + e = \\frac {(|N| - 1)|P:N||N|}{b(|N|-1)} = \\frac {|P:N|}b |N| = b|N|,\n$$\nsince $|P:N| = b^2$. Thus, $e = b|N| - b (|N| - 1) = b$, as\ndesired.\n\\end{proof}\n\nWe next prove that the first conclusion of Theorem \\ref{thm1} is a consequence of the second conclusion.\n\n\\begin{corollary} \\label{gagola 3}\nLet $(G,N)$ be a $p$-Gagola pair, and let $P$ be a Sylow\n$p$-subgroup of $G$. If $|P:N| \\ge |N|^2$, then $d \\le e^2 - e$.\n\\end{corollary}\n\n\\begin{proof}\nBy Lemma \\ref{gagola 2}, we have $d = e (|N| - 1)$ and $e^2 = |P:N|$. Assuming $|N|^2 \\le |P:N|$ implies that $|N|^2 \\le e^2$, and so, $|N| \\le e$. This yields $d \\le e(e - 1) = e^2 - e$, as desired.\n\\end{proof}\n\nWe now prove that Theorem \\ref{thm2} is a corollary to Theorem \\ref{thm1}.\n\n\\begin{proof}[Proof of Theorem \\ref{thm2}]\nBy Lemma \\ref{gagola 1}, we know since $G$ has a nontrivial, abelian normal subgroup that it has a Gagola character. Hence, there is a normal $p$-subgroup $N$ so that $(G,N)$ is a $p$-Gagola pair. By Theorem \\ref{thm2}, we know that $|G:N|_p \\ge |N|^2$, and applying Corollary \\ref{gagola 3}, we obtain $d \\le e^2 - e$. We then obtain $|G| \\le (e^2 - e)((e^2 - e) + e) = (e^2 - e)e^2 = e^4 - e^3$.\n\\end{proof}\n\nWith this, we can use the results of Durfee and Jensen to prove Corollary \\ref{cor1}.\n\n\\begin{proof}[Proof of Corollary \\ref{cor1}]\nSuppose $d \\ge e^2$. In Corollary 3.4 of \\cite{DuJe}, Durfee and Jensen prove that there is a character $\\chi \\in \\irr G$ that dominates all the other characters in $\\irr G$. We then use Lemma 4.1 of \\cite{DuJe} to see that $G$ has a normal subgroup $N$ so that $(G,N)$ is a Gagola pair. We then apply Theorem \\ref{thm1} and Corollary \\ref{gagola 3} to see that $d \\le e^2 - e$. Since $e > 1$, this contradicts $d \\ge e^2$. Therefore, we conclude that $d < e^2$ and $|G| < e^2 (e^2 + e) = e^4 + e^3$.\n\\end{proof}\n\nTherefore, for the rest of this paper, we will be concerned with proving the second conclusion of Theorem \\ref{thm1} that if $(G,N)$ is a $p$-Gagola pair, then $|G:N|_p \\ge |N|^2$.\n\nIt turns out that Gagola pairs are examples of a more general\nconstruction that has been studied with some depth. Let $G$ be a group with a normal subgroup $N$ with $1 < N < G$. We say that $(G,N)$ is a Camina pair if for every $g \\in G \\setminus N$, the conjugacy class of $g$ contains $gN$. Camina pairs have been studied in a number of different places. There are a number of different conditions that are equivalent to being Camina pairs. For example, $(G,N)$ is a Camina pair if and only if $|C_G (g)| = |C_{G\/N} (gN)|$ for all $g \\in G \\setminus N$. In addition $(G,N)$ is a Camina pair if and only if for every every pair of elements $g \\in G$ and $x \\in N$, there exists an element $y \\in G$ so that $[g,y] = x$. Also, $(G,N)$ is a Camina pair if and only if there is a unique character in $\\irr {G \\mid \\nu}$ for each nonprincipal character $\\nu \\in \\irr N$. Based on this last condition, it follows that if $(G,N)$ is a Gagola pair, then $(G,N)$ is a Camina pair.\n\n\n\nWe start with a well-known result regarding Camina pairs where $G$ is a $p$-group and $G\/N$ is abelian.\n\n\\begin{lemma}\\label{abel pair}\nIf $(G,N)$ is a Camina pair where $G$ is a $p$-group for some prime $p$ and $G\/N$ is abelian, then $|G:N| \\ge |N|^2$.\n\\end{lemma}\n\n\\begin{proof}\nSince $G\/N$ is abelian, we have $G' \\le N$. On the other hand, because $(G,N)$ is a Camina pair, we know that every element in $N$ lies in $G'$, so $G' = N$. Under this hypothesis, it is proved in Theorem 3.2 of \\cite{MacD1} that $|N|^2 \\le |G:N|$.\n\\end{proof}\n\n\n\n\n\n\n\n\n\n\n\n\n\nThis next result should be compared with Theorem 6.2 of \\cite{Gagola}. In particular, we generalize a result regarding Gagola pairs to Camina pairs.\n\n\\begin{lemma} \\label{comm}\nLet $(G,N)$ be a Camina pair with $N$ a $p$-group for some prime $p$, and let $H$ be a $p'$-subgroup of $G$ such that ${\\bf O}_p (G)H$ is normal in $G$. If $G$ is not a Frobenius group with Frobenius kernel $N$, then $N < [O_p (G), H]$.\n\\end{lemma}\n\n\\begin{proof}\nWe know that $H$ acts coprimely on ${\\bf O}_p (G)$, so\n$$\n{\\bf O}_p (G) = [{\\bf O}_p (G),H] C_{{\\bf O}_p (G)} (H).\n$$\nWe know from\n\\cite{ChMc} that $NH$ is a Frobenius group so $N \\cap N_G (H) = 1$. Since ${\\bf O}_p (G) H$ is normal in $G$, we may use\nthe Frattini argument to see that $G = {\\bf O}_p (G) N_G (H) = [{\\bf\nO}_p (G),H] N_G (H)$. If $N = [{\\bf O}_p (G),H]$, then $N$ is\ncomplemented in $G$ by $N_G (H)$. By Proposition 3.2 of \\cite{ChMc}, this\nimplies that $G$ is a Frobenius group with Frobenius kernel $N$, a\ncontradiction.\n\\end{proof}\n\nThe following lemma considers the case when we have a ``large'' normal abelian subgroup.\n\n\\begin{lemma} \\label{abel}\nLet $(G,N)$ be a $p$-Gagola pair. Suppose $M$ is a normal abelian $p$-subgroup of $G$ with $N < M$ and $P$ is a Sylow $p$-subgroup of $G$. Then $|P:M| \\ge |M:N|$. If in addition, $|N| \\le |M:N|$, then $|N|^2 \\le |P:N|$.\n\\end{lemma}\n\n\\begin{proof}\nLet $\\lambda \\in \\irr N$ with $\\lambda \\ne 1_N$. By \\cite{Gagola}, we know that $\\lambda$ is fully ramified with respect to $P\/N$. It follows that $P\/M$ must act transitively on $\\irr {M \\mid \\lambda}$. We conclude that $|P:M| \\ge |\\irr {M \\mid \\lambda}| = |M:N|.$ If $|M:N| \\ge |N|$, then $|P:N| \\ge |N|^2$, and the result is proved.\n\\end{proof}\n\n\\section{Frobenius complements}\n\nWe start with presenting two general lemmas regarding solvable Frobenius complements. We would not be surprised if these results were known. A group $G$ is said to be a {\\it Z-group} if all of its Sylow subgroups are cyclic. It is known that if $G$ is a Z-group then $G\/G'$ and $G'$ are cyclic groups of coprime order (see Satz IV.2.11 of \\cite{hup} or Theorem 5.16 of \\cite{isa}). This first extends to Z-groups that are Frobenius complements, a result that is well-known for Frobenius complements of odd order, and as we will see the proof is essentially identical. (See Satz V.8.18(b) of \\cite{hup} or Theorem 6.19 of \\cite{isa}.)\n\n\\begin{lemma} \\label{Z-group}\nLet $G$ be a Frobenius complement that is a Z-group. If $p$ is a prime that divides $|G|$, then $G$ has a unique subgroup of order $p$.\n\\end{lemma}\n\n\\begin{proof}\nWe know that $p$ divides only one of $|G'|$ or $|G:G'|$. If $p$ divides $|G'|$, then since $G'$ is cyclic, we know $G'$ has a unique subgroup of order $p$. Since $p$ does not divide $|G:G'|$ and $G'$ is normal in $G$, it follows that this is the unique subgroup of $G$ having order $p$.\n\nSuppose that $p$ divides $|G:G'|$. We know that $G\/G'$ is cyclic, so there is a unique subgroup $X\/G'$ having order $p$. Observe that $X$ contains all the subgroups of $G$ having order $p$, so it suffices to prove that $X$ has a unique subgroup of order $p$. Let $P$ be a subgroup of $X$ having order $p$. Let $Q$ be any subgroup of $G'$ having prime order, say $|Q| = q$. Then $PQ$ is a subgroup of order $pq$, and by either Satz V.8.15 (b) of \\cite{hup} or Theorem 6.9 of \\cite{isa}, we see that $PQ$ is cyclic. Hence, $Q$ centralizes $P$. It follows that every subgroup of $G'$ of prime order centralizes $P$, and thus, $C_{G'} (P)$ contains every subgroup of $G'$ having prime order. Since $G'$ is abelian and has order coprime to $p$, we may use Fitting's theorem to see that $G' = [G',P] \\times C_{G'} (P)$. Since $C_{G'} (P)$ contains all the subgroups of prime order, we obtain $G' = C_{G'} (P)$. Now, $X = G' \\times P$, and this proves the result.\n\\end{proof}\n\nThis next result uses a Theorem due to Zassenhaus.\n\n\\begin{theorem} \\label{solv Frob}\nLet $G$ be a solvable Frobenius complement. If $p$ is a prime divisor of $|G|$ and $P$ is a subgroup of $G$ of order $p$, then either:\n\\begin{enumerate}\n\\item $P$ is normal in $G$; or\n\\item $p = 3$, $9$ does not divide $|G|$, and a Sylow $2$-subgroup of $G$ is quaternion of order $8$.\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\nIf $p = 2$, then the result is true by either Satz V.8.18 (a) of \\cite{hup} or Theorem 6.3 of \\cite{isa}. Also, if $|G|$ is odd, then we have seen that the result is true. Therefore, we may assume that $|G|$ is even and $p$ is odd. By a theorem of Zassenhaus (Theorem 18.2 of \\cite{perm}), $G$ has a normal subgroup $N$ so that $N$ is a Z-group and $G\/N$ is isomorphic to a subgroup of $S_4$. Thus, if $p > 3$ or $9$ divides $|G|$, then $p$ divides $|N|$. Since subgroups of Frobenius complements are Frobenius complements, we may apply Lemma \\ref{Z-group} to see that $P$ is characteristic in $N$, and so $P$ is characteristic in $G$.\n\nThus, we may assume that $p = 3$ and $9$ does not divide $|G|$. Let $H$ be a Hall $2$-complement of $G$ containing $P$. Thus, $H$ is a Frobenius complement with odd order. Let $Q$ be a Sylow $2$-subgroup of $G$ so that $QP$ is a subgroup of $G$. Since $P$ is a subgroup of $H$ with prime order, we know that $P$ is normal in $H$. If $P$ is central in $PQ$, then $P$ is normal in $G$ since $G = HQ$. Thus, we may assume that $P$ is not central in $PQ$, and so, the center of $PQ$ is a $2$-group. By Lemma 18.3 (iii) of \\cite{perm}, we know that either $P$ is normal in $PQ$ and hence $G$, or the Fitting subgroup of $PQ$ is isomorphic to the quaternions. This implies that $Q$ is quaternion of order $8$.\n\\end{proof}\n\n\n\\section{Semi-linear groups}\n\nThis first result of this section is suggested by Theorem 2.4 of \\cite{Gagola} and Theorem B of \\cite{Kuisch}. We would like to be able to replace the hypothesis that $(G,N)$ is a Gagola pair with the hypothesis that $(G,N)$ is a Camina pair with $N$ an elementary abelian $p$-group (that is minimal normal in $G$).\n\nIf $V$ is an elementary abelian $p$-group, then we define $\\Gamma (V)$ to be the semi-linear group as defined in Chapter 2 of \\cite{MaWo}. We can view $V$ as the additive group of some finite field $F$. In particular, $\\Gamma = \\Gamma (V)$ is the semi-direct product of $S$ acting on $\\Gamma_o (V)$ where $\\Gamma_o (V)$ is isomorphic to the multiplicative group of $F$ and $S$ is isomorphic to the Galois group of $F$ over $Z_p$. If $|V| = p^n$, then $|\\Gamma_o (V)| = p^n - 1$ and $|S| = n$. It is not difficult to show that $C_{\\Gamma} (\\Gamma_o (V)) = \\Gamma_o (V)$.\n\nIf $p$ is a prime, then recall that the group $G$ is {\\it $p$-closed} if $G$ has a normal Sylow $p$-subgroup.\n\n\\begin{lemma} \\label{aff}\nLet $(G,N)$ be a $p$-Gagola pair where $G$ is solvable. If $G$ is not $p$-closed, then either\n\\begin{enumerate}\n\\item $G\/{\\bf O}_p (G) \\le \\Gamma (N)$ and $|N| = p^{ap}$ for some positive integer $a$, and $G\/{\\bf O}_p (G)$ has a normal $p$-complement, or\n\\item $p = 3$, $|N| = 9$, and $G\/{\\bf O}_3 (G)$ is isomorphic to ${\\rm SL}_2 (3)$.\n\\end{enumerate}\n\\end{lemma}\n\n\\begin{proof}\nBy Corollary 2.3 of \\cite{Gagola}, we know that $G\/{\\bf O}_p (G)$ acts on $N$ and that $C_{G\/{\\bf O}_p (G)} (x)$ is a Sylow $p$-subgroup of $G\/{\\bf O}_p (G)$ for every $x \\in N \\setminus \\{ 1 \\}$. This satisfies the hypotheses of Lemma 1 of \\cite{bound}. The conclusion of Lemma 1 of \\cite{bound} is that either $|N| = 9$ and $G\/{\\bf O}_3 (G)$ is isomorphic to either ${\\rm SL}_2 (3)$ or ${\\rm GL}_2 (3)$, or $G\/{\\bf O}_p (G) \\le \\Gamma (N)$. If the first conclusion holds, then we have our conclusion (2) since $|G:{\\bf O}_p (G)| = |N| - 1 = 48$. Thus, we may assume that $G\/{\\bf O}_p (G) \\le \\Gamma (N)$. Using the notation from \\cite{MaWo}, we know that $\\Gamma (N)$ has a normal cyclic subgroup $\\Gamma_o (N)$ whose order is $|N| - 1$. Because $N$ is a $p$-group, we have $|N| = p^n$ for some positive integer $n$. Since $p$ divides $|G:{\\bf O}_p (G)|$ and $p$ does not divide $|\\Gamma_o (N)| = p^n - 1$, we conclude that $p$ divides $|\\Gamma (N):\\Gamma_o (N)| = n$, and thus, $n = ap$ for some positive integer $a$ as desired. Since $\\Gamma (N)\/\\Gamma_o (N)$ is cyclic, we see that $G\/{\\bf O}_p (G)$ has a normal $p$-complement.\n\\end{proof}\n\nThis next result is a number theoretic condition. It is probably known. It is related to the number theoretic results proved in Lemma 2.4 of \\cite{nor}.\n\n\\begin{lemma}\\label{nor cond}\nLet $p$ be a prime and $n$ a positive integer. Suppose that $q$ is a prime so that $n = q^a m$ where $a$ is a positive integer and $m$ is not divisible by $q$. If $q^2$ divides $p^m - 1$, then $(p^n - 1)_q = q^a (p^m - 1)_q$.\n\\end{lemma}\n\n\\begin{proof}\nObserve that\n$$\np^n - 1 = \\prod_{i=1}^a \\left( \\frac {p^{m q^i} - 1}{p^{m q^{i-1}} - 1} \\right) (p^m - 1).\n$$\nThus, it suffices to prove that $\\displaystyle \\left( \\frac {p^{m q^i} - 1}{p^{m q^{i-1}} - 1} \\right)_q = q$ for all $i$. By Lemma 2.4 of \\cite{nor}, we know that $q$ divides $({p^{m q^i} - 1})\/({p^{m q^{i-1}} - 1})$ and that ${\\rm gcd} (p^{m q^{i-1}} -1,({p^{m q^i} - 1})\/({p^{m q^{i-1}} - 1}))$ divides $q$. Since $q^2$ divides $p^m - 1$ which divides $p^{m q^{i-1}} - 1$, it follows that $\\displaystyle \\left( \\frac {p^{m q^i} - 1}{p^{m q^{i-1}} - 1} \\right)_q = q$.\n\\end{proof}\n\nWe now apply this to obtain a result about subgroups of semi-linear groups.\n\n\\begin{lemma} \\label{prime pow}\nLet $V$ be an elementary abelian group of order $2^n$. Let $H$ be a subgroup of $\\Gamma (V)$ of order $2^n - 1$ that acts transitively on $V \\setminus \\{ 1 \\}$. If $d$ is a nontrivial prime power that divides $n$, then $H \\cap \\Gamma_o (V)$ contains an element of order $2^d - 1$.\n\\end{lemma}\n\n\\begin{proof}\nSince $\\Gamma_o (V)$ is cyclic, it suffices to show that $H \\cap \\Gamma_o (V)$ contains an element of order $(2^d - 1)_r$ for every prime $r$ that divides $2^d - 1$. Fix a prime divisor $r$ of $2^d - 1$. Suppose $(2^d - 1)_r = r$. We know that $H$ has a normal subgroup of order $r$. Thus, this subgroup will centralize $H \\cap \\Gamma_o (V)$. We know that the centralizer of $\\Gamma_o (V)$ in $\\Gamma (V)$ is $\\Gamma_o (V)$, so $H \\cap \\Gamma_o (V)$ will contain the subgroup of order $r$. Thus, we may assume that $(2^d - 1)_r \\ge r^2$.\n\nLet $p$ be the prime so that $d$ is a power of $p$. We know that $2^d \\cong 1 ~({\\rm mod}~r)$, so the order of $2$ modulo $r$ must be a power of $p$. This implies that $p$ divides $r - 1$, and so, $r$ does not divide $d$. If we write $n = r^a m$ where $a$ is a positive integer and $r$ does not divide $m$, then we conclude that $2^d - 1$ will divide $2^m - 1$. In particular, $r^2$ divides $2^m - 1$.\n\nLet $R$ be a Sylow $r$-subgroup of $\\Gamma (V)$ so that $R \\cap H$ is a Sylow $r$-subgroup of $H$. We know that $|R| = (2^n - 1)_r n_r$. Observe that $R \\cap \\Gamma_o (V)$ is a Sylow $r$-subgroup of $\\Gamma_o (V)$, so $|R \\cap \\Gamma_o (V)| = (2^n - 1)_r$. It follows that $|R:R \\cap \\Gamma_o (V)| = n_r = r^a$. Since $H$ is a Frobenius complement, we know that $R \\cap H$ is cyclic. Let $x$ be a generator for $R \\cap H$. Since $R \\cap \\Gamma_o (V)$ is normal in $R$, we see that $x^{r^a} \\in R \\cap \\Gamma_o (V) \\le H \\cap \\Gamma_o (V)$. The order of $x^{r^a}$ will be $(2^n - 1)_r\/r^a$, and by Lemma \\ref{nor cond}, this equals $(2^m - 1)_r$. Since this is divisible by $(2^d - 1)_r$, this yields the desired conclusion.\n\\end{proof}\n\n\\section{$p$ odd} \\label{thm1sec}\n\nWe now prove Theorem \\ref{thm1} when $p$ is odd.\n\nWe begin with a result that applies both when $p$ is odd and when $p = 2$.\n\n\\begin{lemma} \\label{|N|p^2}\nLet $(G,N)$ be a $p$-Gagola pair. If $|N| \\le p^2$, then $|G:N|_p \\ge |N|^2$.\n\\end{lemma}\n\n\\begin{proof}\nIn \\cite{Gagola}, it is proved that $N < {\\bf O}_p (G)$, $N \\le Z ({\\bf O}_p (G))$, and $|G:N|_p$ is a square. Consider an element $x \\in {\\bf O}_p (G)$, and let $D (x) = \\{ g \\in G | [x,g] \\in N \\}$. We know that $D(x)\/N = C_{G\/N} (xN)$, so $|D(x):N| = |C_G (x)|$. This implies that $|D(x):C_G(x)| = |N|$. Notice that $\\langle N,x \\rangle \\le C_G (x)$. Since $xN$ will have nontrivial $p$-power order. This implies that $|G:N|_p > |N|$.\n\nIf $|N| = p$, then $|G:N|_p \\ge p^2 = |N|^2$. If $|N| = p^2$, then $|G:N|_p \\ge p^3$. Since we know that $|G:N|_p$ is square, this implies that $|G:N| \\ge p^4 = |N|^2$. This proves the lemma.\n\\end{proof}\n\nLet $p$ be a prime and $a$ a positive integer. We say that $q$ is a {\\it Zsigmondy} prime of $p^a - 1$ if $q$ divides $p^a - 1$ and $q$ does not divide $p^b - 1$ for every positive integers $b < a$. The Zsigmondy prime theorem states that a Zsigmondy primes exists for primes $p$ and positive integers $a$ except when $p$ is a Mersenne prime (i.e., $p + 1$ is a power of $2$) and $a = 2$ and when $p = 2$ and $n = 6$. (See\nTheorem IX.8.3 of \\cite{HBII}.)\n\nLet $p$ be a prime greater than $3$. We know that $p^2$ is congruent to $1$ modulo $3$, so $3$ divides $p^2 - 1$. This implies the following fact. If $p$ is prime, $n$ is a positive integer, and $3$ is a Zsigmondy prime divisor of $p^n - 1$, then $n \\le 2$.\n\nWe begin with a lemma about $p$-groups of nilpotence class $3$.\n\n\\begin{lemma} \\label{class3}\nLet $p$ be an odd prime, and suppose $P$ is a $p$-group of nilpotence class $3$. Suppose that $[P',P] \\le Z \\le Z(P) \\cap P'$. If $P$ has an automorphism of order $2$ that inverts all the elements in $P\/P'$ and in $Z$ and centralizes the elements in $P'\/Z$, then for every element $x \\in P \\setminus P'$, there is an element $y \\in xP'$ so that $C_{P\/Z} (y Z) = C_P (y)P'\/Z$.\n\\end{lemma}\n\n\\begin{proof}\nLet $\\sigma$ be the given automorphism of order $2$. For each element $g \\in P$, define $D (g)\/Z = C_{P\/Z} (gZ)$. Since $P'\/Z$ is central in $P\/Z$, we see that if $h \\in gP'$, then $D(h) = D (g)$.\n\nSuppose $r \\in P \\setminus P'$. Define $R = \\langle P',r \\rangle$. Since $P'\/Z$ is central, $R\/Z$ is abelian. We know that $\\sigma$ inverts $rP'$, so $\\sigma$ acts on $R$. Notice that $C_{R\/Z} (\\sigma) = P'\/Z$, so by Fitting's lemma, $R\/Z = P'\/Z \\times [R,\\sigma]\/Z$. Thus, the coset $rP'$ contains an element $s \\in [R,\\sigma]$. Notice that $\\langle s, Z \\rangle$ is abelian, and $\\sigma$ acts Frobeniusly on both $\\langle s, Z \\rangle\/Z$ and $Z$, so $\\sigma$ acts Frobeniusly on $\\langle s, Z \\rangle$. We conclude that $\\sigma$ inverts $s$.\n\nNow, consider $x \\in P \\setminus P'$. By the previous paragraph, we can find an element $y \\in xP'$ so that $y^\\sigma = y^{-1}$. Observe that $C_P (y) \\le D (y)$ and $P' \\le D (y)$, so $C_P (y) P' \\le D (y)$.\n\nConsider $d \\in D(y)$. We need to show that $d \\in C_P (y) P'$. If $d \\in P'$, then this holds, so we may assume $d \\not\\in P'$. By the previous paragraph, we can find $c \\in dP'$ so that $c^\\sigma = c^{-1}$. We know that $c \\in D (y)$, so $[c,y] \\in Z$. It follows that $[c,y]^\\sigma = [c^\\sigma,y^\\sigma] = [c^{-1},y^{-1}] = [c,y]$. Since $\\sigma$ inverts all the elements of $Z$, this implies that $[c,y] = 1$, and so, $c \\in C_P (y)$. We now have $dP' = cP' \\le C_P (y) P'$, and we conclude that $d \\in C_P (y) P'$. This proves that $D (y) = C_P (y)P'$, and this proves the result.\n\\end{proof}\n\nThis next theorem proves Theorem \\ref{thm1} under the hypothesis that $p$ is an odd prime. For nonsolvable groups, we apply a result that is found in \\cite{Gagola} which relied on the classification of finite simple groups. Thus, the nonsolvable case of this result also relies on the classification of finite simple groups. On the other hand, if one assumes solvability, then the classification is not needed.\n\n\\begin{theorem} If $(G,N)$ is a $p$-Gagola pair where $p$ is odd, then $|N|^2 \\le |G:N|_p$.\n\\end{theorem}\n\n\\begin{proof}\nBy Lemma \\ref{|N|p^2}, we may assume that $|N| = p^a$ where $a \\ge 3$. We claim that it suffices to find a normal $p$-subgroup $M$ and a subgroup $Y$ of order $2$ so that $[M',M] \\le N \\le M'$, $|M:M'| \\ge |N|$, and $Y$ acts Frobeniusly on $M\/M'$. (We know that $Y$ must act Frobeniusly on $N$.) If $M' = N$, then since $C_N (Y) = 1$, we conclude that $C_M (Y) = 1$. It follows that $Y$ acts Frobeniusly on $M$. We conclude that $M$ is abelian which is a contradiction. Thus, we may assume that $N < M'$. Since $[M',M] \\le N$ and $N$ is central in $M$ (see page 367 of \\cite{Gagola}), this implies that $M$ has nilpotence class $3$.\n\nWe know that $C_M (Y) \\le M'$ since $Y$ acts Frobeniusly on $M\/M'$. Since $M'\/N$ is central in $M\/N$, we see that $C_M (Y) N$ is normal in $M$. Notice that $Y$ will act Frobeniusly on $M\/(C_M (Y) N)$, so $M\/(C_M (Y) N)$ is abelian, and thus, $M' = C_M (Y) N$. It follows that $Y$ centralizes $M'\/N$. Hence, we may apply Lemma \\ref{class3}. By that result, we can find an element $x \\in M \\setminus M'$ so that $C_{M\/N} (xN) = C_M (x) M'\/N$. Let $D (x)\/N = C_{G\/N} (xN)$, and this implies that $D (x) \\cap M = C_M (x) M'$.\n\nWe now compute\n$$\n|M D(x):C_G (x) \\cap M'| = |M D(x):M| |M:M'| |M': C_G (x) \\cap M'|.\n$$\nWe know that $|MD (x):M| = |D(x): D(x) \\cap M| = |D(x): C_M (x)M'|$ and $|M':C_G (x) \\cap M'| = | C_M (x) M':C_M (x)|$. We have\n$$\n|D(x): C_M (x)M'| | C_M (x) M':C_M (x)| = |D (x):C_M (x)|.\n$$\nSince $|D (x): C_M (x)| = |D (x): C_G (x) \\cap M|$ is divisible by $|D(x):C_G (x)| = |N|$ and $|M:M'| \\ge |N|$, this yields the desired conclusion. We now work to find such subgroups $M$ and $Y$.\n\nSuppose first that $G$ is solvable. Since $a \\ge 3$, we know that $|N| - 1 = p^a - 1$ has a Zsigmondy prime divisor $q$. As we noted earlier, $q \\ne 3$. Let $H$ be a Hall $p$-complement of $G$. We know from Lemma 4.3 in \\cite{ChMc} that $NH$ is a $2$-transitive Frobenius group. Since $a \\ge 3$, we may apply Huppert's theorem (Proposition 19.10 of \\cite{perm} or Theorem 6.9 of \\cite{MaWo}) to see that $H$ is isomorphic to a subgroup of $\\Gamma (N)$, and so, $H$ has a normal subgroup $K$ of order $q$. We claim that ${\\bf O}_p (G) K$ is normal in $G$. If $G$ is $p$-closed, then this is immediate. If $G$ is not $p$-closed, then the existence of a Zsigmondy prime implies that we are in case 2 of Lemma \\ref{aff}. In particular, $G\/{\\bf O}_p (G) \\le \\Gamma (N)$, and thus, ${\\bf O}_p (G)K$ is normal in $G$. Hence, we may use Lemma \\ref{comm} to see that $N < [{\\bf O}_p (G), K] = M$. Notice that $K$ acts Frobeniusly on $M\/M'$. If $M' = 1$, then $K$ acts Frobeniusly on $M\/N$, and since $|K| = q$ is a Zsigmondy prime divisor of $p^a - 1$, we have $|M:N| \\ge |N|$. And by Lemma \\ref{abel}, we have the result. Thus, we may assume that $M' > 1$, and since $N$ is the unique minimal normal subgroup of $G$, this implies that $N \\le M'$. This implies that ${\\bf O}_p (G) = M C_{{\\bf O}_p (G)} (K)$, and by the Frattini argument, $G = {\\bf O}_p (G) N_G (K) = M N_G (K)$. If $M\/M'$ is not irreducible under the action of $K$, then $|M:M'| \\ge p^{2n}$, and the result follows. Thus, we may assume that $M\/M'$ is irreducible under the action of $K$.\n\n\nSince $p$ is odd, $2$ divides $|H|$. Let $Y \\le H$ be a subgroup of order $2$. We know that $Y$ is normal in $H$ by Theorem \\ref{solv Frob}. In particular, $K$ centralizes $Y$. Hence, $N_G (K)$ normalizes $C_M (Y)$. Since $M\/M'$ is abelian, $M$ normalizes $C_M (Y) M'$, and so, $C_M (Y) M'$ is normal in $G$. If $M = C_M (Y) M' = C_M (Y) \\Phi (M)$, then Frattini's theorem implies $M = C_M (Y)$, and this is a contradiction since $Y$ acts Frobeniusly on $N$. Thus, $C_M (Y) M' < M$. Since $M\/M'$ is irreducible under the action of $K$, we have $C_M (Y) \\le M'$. We deduce that $Y$ acts Frobeniusly on $M\/M'$.\n\n\nIf $K$ acts nontrivially on $M'\/N$, then we will have $|M':N| \\ge |N|$, and the result will follow since $|M:N| \\ge |N|^2$. Thus, we may assume that $K$ centralizes $M'\/N$. It follows that $[M',K] \\le N$. We then have $[M',K,M] \\le [N,M] = 1$ since $N \\le Z (M)$, and $[M,M',K] \\le [M',K] = N$. By the three subgroups lemma, this implies that $[K,M,M'] \\le N$, and since $M = [K,M]$, we conclude that $[M,M'] \\le N$. We now have the desired subgroups $M$ and $Y$, and the result follows in this case.\n\n\n\n\n\n\nFinally, we may assume $G$ is not solvable and $|N| = p^a$ where $a > 2$. In this case, we may apply Theorem 5.6 of \\cite{Gagola} to see that $G\/{\\bf O}_p (G)$ either has ${\\rm SL} (2,q)$ where $q$ is a power of $p$ as normal subgroup whose quotient is a $p$-group, has ${\\rm SL} (2,5)$ as a normal subgroup of index $2$ that is a nonsplit extension, or is isomorphic to ${\\rm SL} (2,13)$. In all of these cases, $G\/{\\bf O}_p (G)$ has a unique minimal normal subgroup $L\/{\\bf O}_p (G)$ which is of order $2$. Taking $Y$ to be a subgroup of order $2$ in $L$, we have $L = {\\bf O}_p (G) Y$ is normal in $G$.\n\nAgain, we may apply Lemma \\ref{comm} to find $M$ so that $N < M = [{\\bf O}_p (G), Y]$. It is not difficult to see that $Y$ acts Frobeniusly on $M\/M'$. Let $C\/N = C_{G\/N}{M\/N}$. Since ${\\bf O}_p (G)\/N$ is a $p$-group, we know that $Z({\\bf O}_p (G)\/N) \\cap M\/N > 1$, and since $M\/N$ is a chief factor for $G$, it follows that $M\/N \\le Z ({\\bf O}_p (G)\/N)$. This implies that ${\\bf O}_p (G) \\le C$. We see that $C \\cap L= {\\bf O}_p (G)$. Since $L\/{\\bf O}_p (G)$ is the unique minimal normal subgroup of $G\/{\\bf O}_p (G)$, we conclude that $C = {\\bf O}_p (G)$.\n\nLet $K$ be a subgroup of $G$ of order $q$. Since $K$ is not contained in $C$, we see that $K$ acts nontrivially on $M\/N$. If $M' = 1$, then it acts Frobeniusly on $M$, and this implies that $|M:N| \\ge |N|$, we obtain the result by Lemma \\ref{abel}. Thus, we may assume that $M' > 1$. By the uniqueness of $N$, this implies $M' \\ge N$. Since $K$ acts nontrivially on $M\/N$, it acts nontrivially on $M\/M'$. It follows that $|M:M'| \\ge |N|$. If $K$ acts nontrivially on $M'\/N$, then $|M':N| \\ge |N|$, and the result holds. Thus, we may assume that $K$ centralizes $M'\/N$. This implies that the centralizer of $M'\/N$ in $G\/{\\bf O}_p (G)$ is nontrivial. By the uniqueness of $L\/{\\bf O}_p (G)$, this implies that $L$ centralizes $M'\/N$, and so $Y$ centralizes $M'\/N$. We have $[M',Y,M] \\le [N,M] \\le N$. Also, $[M,M',Y] \\le [M',Y] \\le N$. By the Three Subgroups Lemma (see Lemma 4.9 of \\cite{isa}), $[Y,M,M'] \\le N$, and since $[M,Y] = M$, we obtain $[M,M'] \\le N$. We now have subgroups $M$ and $Y$ as desired. This completes the proof.\n\\end{proof}\n\n\n\n\n\n\n\n\\section{$p=2$ preliminary results}\n\nWe now left with the case that $N$ is a $2$-group. We break up the case when $p = 2$ into two subcases: $G$ is solvable and $G$ is not solvable. In this section, we include some results that are common to both cases.\n\n\n\\begin{lemma} \\label{five}\nLet $K$ have normal subgroup $N < M$ so that $N$ is an elementary abelian $2$-group, $M\/N$ is cyclic of odd order, $|K:M| = 2$, $C_{N}(M\/N) = 1$, and $K\/N$ is a dihedral group. Then $K \\setminus M$ contains an involution.\n\\end{lemma}\n\n\\begin{proof}\nLet $C$ be a Hall $2$-complement of $M$. Since $C$ is a Hall subgroup of $M$, we can apply the Frattini argument to see that $K = M N_G(C) = NC N_G (C) = N N_G (C)$. This implies that $2$ divides $|N_G (C)|$. Let $A = N_G (C) \\cap N$. Then $[A,C] \\le C$ and $[A,C] \\le [N,C] \\le N$, so $[A,C] \\le C \\cap N = 1$. This implies that $[A,C] = 1$. We deduce that $[A,M] = [A,NC] \\le [A,N][A,C] = [A,N] \\le N$, and we conclude that $A \\le C_{N}(M\/N) = 1$. Thus, we have $N_G (C) \\cap N = 1$. Let $T$ be a Sylow $2$-subgroup of $N_G (C)$. It follows that $T > 1$ and $T \\cap N = 1$. Thus, $T$ contains an involution which is not in $N$. Since $N$ is the Sylow $2$-subgroup of $M$, this implies that $T$ contains an involution not in $M$.\n\\end{proof}\n\nThe following fact is essentially the heart of the proof when $p = 2$ of the proof of Theorem 5.1 of \\cite{large}.\n\n\\begin{lemma} \\label{six}\nLet $(G,N)$ be a $2$-Gagola pair. Then every involution in $G\/N$ lies in ${\\bf O}_2 (G)\/N$.\n\\end{lemma}\n\n\\begin{proof}\nLet $t \\in G$ so that $tN$ is an involution. If $t \\in {\\bf O}_2 (G)$, then the result holds; so assume $t \\not\\in {\\bf O}_2 (G)$. In particular, we know that $t$ does not centralize $N$ by \\cite{Gagola}. By Theorem 2.13 of \\cite{isa}, $tN$ must invert some nontrivial element $cN \\in G\/N$ of odd prime order. Let $M = \\langle N, c \\rangle$ and let $K = \\langle M, t \\rangle$. Let $T = \\langle N, t \\rangle$, and observe that $T$ is a Sylow $2$-subgroup of $K$. Notice that $T$ is not abelian, so $T$ must contain an element of order $4$. Since $N$ is elementary abelian, this element must lie in $T \\setminus N$. Since $T:N| = 2$, we see that $T \\setminus N = Nt$. Since $(G,N)$ is a Gagola pair, we know that all the elements in $Nt$ are conjugate, and so, they all have order $4$. On the other hand, we may now apply Lemma \\ref{five} to see that $K \\setminus N$ must contain an involution $s$. This yields a contradiction since $s$ must be conjugate to some element of $T \\setminus N$, and we have seen that this set contains no involutions.\n\\end{proof}\n\nWe begin with the following observation which appears in my paper \\cite{pCam}.\n\n\\begin{lemma} \\label{seven}\nLet $(G,N)$ be a $2$-Gagola pair. If ${\\bf O}_2 (G)\/N$ does not have exponent $2$, then $|G:N| \\ge |N|_2$. In particular, if ${\\bf O}_2 (G)\/N$ is not elementary abelian, then $|G:N|_2 \\ge |N|^2$.\n\\end{lemma}\n\n\\begin{proof}\nConsider $x \\in {\\bf O}_2 (G)$ so that $x^2 \\not\\in N$. By \\cite{Gagola}, we have $N \\le C_G (x)$. Let $D (x)\/N = C_{G\/N} (x)$. We know that $C_G (x) \\le D (x)$ and $|D (x):N| = |C_G (x)|$, so $|D(x):C_G (x)| = |N|$. Suppose that there exists $x \\in {\\bf O}_2 (G)$ Consider $y \\in D (x)$. Then $[x,y] \\in N$. We observe that $[x^2,y] = [x,y]^x [x,y]$, and since $N$ is central in ${\\bf O}_2 (G)$, we have $[x,y]^x = [x,y]$. Thus, $[x^2,y] = [x,y]^2$, and since $N$ is elementary abelian, we conclude that $[x,y]^2 = 1$. It follows that $y \\in C_G (x^2)$. Thus, $D (x) \\le C_G (x^2)$. Since $x^2 \\in {\\bf O}_2 (G)$, we deduce that $|N|^2 = |D(x^2):C_G(x^2)||D(x):C_G (x)|$ divides $|G:N|$. This proves the result when ${\\bf O}_2 (G)\/N$ does not have exponent $2$.\n\\end{proof}\n\n\\section{$p=2$ and $G$ nonsolvable}\n\nWe continue to work on the case when $p = 2$. In this section, we consider the subcase where $G$ is nonsolvable. We begin with some number theoretic results. We would not be surprised if these results were known.\n\n\\begin{lemma} \\label{one}\nLet $a$ be a nonnegative integer. Then $2^{3^a} \\equiv -1 ~({\\rm mod}~3^{a+1})$ and $2^{3^a} \\not\\equiv -1 ~({\\rm mod}~3^{a+2})$.\n\\end{lemma}\n\n\\begin{proof}\nWe work by induction on $a$. Notice that $2^{3^0} = 2^1 = 2 \\equiv -1 ~({\\rm mod}~3^{0+1})$ and $2 \\not\\equiv -1 ~({\\rm mod}~3^{1+1})$. This proves the base case.\n\nWe now prove the inductive step. Suppose for some nonnegative integer $a$, we have $2^{3^a} \\equiv -1 ~({\\rm mod}~3^{a+1})$ and $2^{3^a} \\not\\equiv -1 ~({\\rm mod}~3^{a+2})$. This implies that $2^{3^a} = -1 + b 3^{a+1}$ where $b$ is an integer that is not divisible by $3$. Cubing, we obtain $2^{3^{a+1}} = (2^{3^a})^3 = (-1 + b 3^{a+1})^3$. Using the binomial theorem, we obtain $(-1 + b 3^{a+1})^3 = -1 + 3 b 3^{a+1} - 3 b^2 3^{2(a+1)} + b^3 3^{3(a+1)} = -1 + 3^{a + 2} c$ where $c = b (1 - b 3^{a+1} + b^2 3^{2a+1})$. It follows that $2^{3^{a+1}} \\equiv -1 ~({\\rm mod}~3^{a+2})$. Since $3$ does not divide $b$, it follows that $3$ does not divide $c$, and so, $2^{3^{a+1}} \\not\\equiv -1 ~({\\rm mod}~3^{a+3})$. This proves the inductive step, and hence, the lemma is proved.\n\\end{proof}\n\nWe make use of the following corollary.\n\n\\begin{corollary}\\label{onea}\nSuppose that $n_3 = 3^a$. Then $2^n \\equiv -1 ~({\\rm mod}~3^{a+1})$ if $n$ is odd and $2^n \\equiv 1 ~({\\rm mod}~3^{a+1})$ if $n$ is even. In particular, $\\left((2^n - 1)(2^n + 1)\\right)_3 = 3^{a+1}$.\n\\end{corollary}\n\n\\begin{proof}\nWrite $b$ so that $n = 3^a b$, and note that $3$ does not divide $b$. Using Lemma \\ref{one}, we have $2^n = (2^{3^a})^b \\equiv (-1)^b ~({\\rm mod}~3^{a+1})$. This gives the first conclusion. Observe that $2^n - 1$ and $2^n + 1$ are relatively prime. Thus, $3$ only divides one of these. Suppose now that $n$ is odd. Then $2^n + 1 = (2^{3^a} + 1) (\\sum_{i=0}^{b-1} (-1)^i (2^{3^a})^i)$. Let $c = \\sum_{i=0}^{b-1} (-1)^i (2^{3^a})^i$. It follows that $c \\equiv \\sum_{i=0}^{b-1} (-1)^i(-1)^i \\equiv \\sum_{i=0}^{b-1} 1 \\equiv b ~({\\rm mod}~3)$ (this uses the fact from Lemma \\ref{one} that $3^a \\equiv -1 ~({\\rm mod}~3)$. It follows that $3$ does not divide $c$, so $(2^n+1)_3 = (2^{3^a} + 1)_3$, and applying Lemma \\ref{one} again, we have that $3^{a+1}$ divides $2^n + 1$ and $3^{a+2}$ does not divide $2^n + 1$, so $(2^n + 1)_3 = 3^{a+1}$. This gives the result $n$ is odd. If $n$ is even, then $n = 2m$ where $m_3 = n_3$. Working by induction on $n$, we have $(2^n - 1)_3 = (2^m - 1)(2^m + 1)_3 = 3^{a+1}$, and this proves the result when $n$ is even.\n\\end{proof}\n\nWe now get an application to the order of $|{\\rm SL}_2 (2^n)|$.\n\n\\begin{corollary} \\label{two}\nSuppose that $n_3 = 3^a$. Then $|{\\rm SL}_2 (2^n)|_3 = 3^{a+1}$.\n\\end{corollary}\n\n\\begin{proof}\nWe know that $|{\\rm SL}_2 (2^n)| = (2^n - 1)2^n (2^n + 1)$. \nThis now follows immediately from Corollary \\ref{onea}.\n\\end{proof}\n\n\n\n\\begin{lemma}\\label{three}\nSuppose that $V$ is the natural module for $S = {\\rm SL}_2 (2^n)$ over $F = {\\rm GF} (2^n)$, and let $\\phi$ be the character afforded by $V$. If $P$ is a Sylow $3$-subgroup of $S$, then $\\phi_P = \\lambda + \\lambda^{-1}$ where $\\lambda$ is a faithful character of $P$ (over the algebraic closure of $F$).\n\\end{lemma}\n\n\\begin{proof}\nFirst, we note that $P$ is cyclic, so $P = \\langle x \\rangle$ for some element $x$. Let ${\\mathcal X}$ be the representation afforded by $V$, so its trace is $\\phi$. We know by Brauer-Nesbitt that $\\phi$ is absolutely irreducible. Let $b$ be the integer so $3^b = |S|_3$. Take $E$ to be an extension of $F$ that contains a $3^b$th root of unity $\\eta$. Since $\\phi$ is absolutely irreducible, we may work with $V^E$ and ${\\mathcal X}$. Conjugating if necessary, we may assume that ${\\mathcal X}^E (x) = \\left[ \\begin{array} {ll}\n \\eta & 0 \\\\\n 0 & \\eta^{-1} \\\\\n \\end{array}\\right] .$\nDefine $\\lambda$ to the character of $P$ over $E$ defined by $\\lambda (x) = \\eta$. Since $\\eta$ and $P$ have the same order, we see that $\\lambda$ is faithful. It is not difficult to see that $\\phi_P = \\lambda+\\lambda^{-1}$.\n\\end{proof}\n\nThe next result contains one of the main pieces of our argument. In particular, we get a restriction on the dimensions of irreducible modules of ${\\rm SL}_2 (2^n)$ over $Z_2$.\n\n\\begin{lemma} \\label{four}\nIf $V$ is an irreducible module for ${\\rm SL}_2 (2^n)$ over $Z_2$ and if ${\\rm dim}_{Z_2} (V) < 4n$, then ${\\rm dim}_{Z_2} (V)$ is either $1$, $2n$, or $8n\/3$, and $8n\/3$ occurs only when $3$ divides $n$. Furthermore, if ${\\rm dim}_{Z_2} (V) = 8n\/3$ and $P$ is a Sylow $3$-subgroup of ${\\rm SL}_2 (2^n)$, then $C_V (P) = 0$.\n\\end{lemma}\n\n\\begin{proof}\nLet $F$ be the field with $2^n$ elements. Let $C$ be the Galois group for $F\/Z_2$, so $C$ is cyclic of order $n$. Let $W$ be the natural module for $G = {\\rm SL}_2 (2^n)$. We know, viewed as a module for $G$ over $F$, that $W$ has dimension $2$. (Viewed as a module over $Z_2$, it follows that $W$ has dimension $2n$.) Let $\\phi$ be the $F$-character of $G$ afforded by $W$. By the Brauer-Nesbitt theorem \\cite{Nesb}, we know that every absolutely irreducible character for $G$ has the form $\\phi^S = \\prod_{\\sigma \\in S} \\phi^\\sigma$ where $S$ is a subset of $C$. Notice that this implies that $F$ will be the splitting field for $G$.\n\nWe now write $\\psi$ for the character afforded by $V^F$. By Theorem 9.21 of \\cite{text}, we know that $\\psi$ is a sum of Galois conjugates of an irreducible $F$-character whose multiplicity is $1$. Fix the subset $S$ of $C$ so that $\\phi^S$ is an irreducible $F$-constituent of $\\psi$. We see that $\\psi$ will be the sum of the Galois conjugates of $\\phi^S$.\n\nNotice that if $S$ is empty, then $\\psi = \\psi^S = 1$, and we conclude that the dimension of $V$ is $1$. Thus, we may assume that $S$ is not empty. Consider an element $c \\in C$. Then $(\\phi^S)^c = \\phi^{Sc}$. Hence, $(\\phi^S)^c = \\phi^{Sc}$ if and only if $S = Sc$, and so, the number of Galois conjugates of $\\phi^S$ is equal to the number of sets of the form $Sc$. Let $T = \\{ c \\in C \\mid S = Sc \\}$. By the Fundamental Counting Principle (see Theorem 1.4 of \\cite{isa}), the number of sets of the form $Sc$ equals $|C:T|$. Fix an element $s \\in S$. If $t \\in T$, then $st \\in S$. This implies that $t \\in s^{-1} S$, and so, $T \\subseteq s^{-1}S$. We conclude that $|T| \\le |S|$.\n\nWe see that ${\\rm dim}_{Z_2} (V) = {\\rm dim}_F (V^F) = {\\rm deg} (\\psi) = |C:T| ({\\rm deg} \\phi)^{|S|} = n2^{|S|}\/|T| \\ge n2^{|S|}\/|S|$. If $|S| = 1$, then $|T| = 1$, and we have ${\\rm dim}_{Z_2} (V) = n2^1\/1 = 2n$. If $|S| = 2$, then either $|T| = 1$ or $|T| = 2$. If $|T| = 1$, then ${\\rm dim}_{Z_2} (V) = n2^2\/1 = 4n$, and if $|T| = 2$, then ${\\rm dim}_{Z_2} (V) = n2^2\/2 = 2n$. If $|S| = 3$, then $|T|$ is either $1$, $2$, or $3$. If $|T| \\le 2$, then ${\\rm dim}_{Z_2} (V) = n2^3\/|T| \\ge n8\/2 = 4n$. If $|T| = 3$, then ${\\rm dim}_{Z_2} (V) = n2^3\/3 = 8n\/3$. Notice that $|T|$ divides $|C| = n$, so $3$ divides $n$. Finally, if $|S| \\ge 4$, then ${\\rm dim}_{Z_2} (V) \\ge n 2^|S|\/|S|$. Consider the function $f (x) = 2^x\/x$. Observe that $f'(x) = 2^x(x {\\rm ln} (2) - 1)\/x^2$, and so, $f' (x) > 0$ when $x > 1\/{\\rm ln} (2) \\cong 1.44$. This implies that $f (x)$ is increasing when $x \\ge 4$. In particular, $2^{|S|}\/|S| \\ge 2^4\/4 = 16\/4 = 4$, and conclude that ${\\rm dim}_{Z_2} (V) \\ge 4n$. This proves the first conclusion.\n\nWe now focus on the case where ${\\rm dim}_{Z_2} (V) = 8n\/3$. Notice that we must have $|T| = |S| = 3$. Let $P$ be a Sylow $3$-subgroup of $G$. We need to show that $C_V (P) = 0$. It suffices to show that $C_{V^F} (P) = 0$. Since $\\psi$ is the character afforded by $V^F$, this will follow if we can prove that $1_P$ is not a constituent of $\\psi_P$.\n\nNotice that $\\psi$ contains all the characters of the form $\\phi^{Sc}$, so we may assume that $1 \\in S$. It is difficult to see that this implies that $S = T$. Let $c$ be the Frobenius automorphism of $F$, so $c$ is a generator of $C$. Write $n = 3m$ where $m$ is an integer. It is not difficult to see that $S = \\{ 1, c^m, c^{2m} \\}$, and so, $\\psi = \\sum_{i=0}^{m-1} (\\phi \\phi^{c^m} \\phi^{c^{2m}})^{c^i}$.\n\nWe now consider $\\psi_P$. We wish to show that the principal character of $P$ is not a constituent of $\\psi_P$. We know that if $x \\in F$, then $x^c = x^2$. It follows that $\\phi^{c^i} = \\phi^{2^i}$, and so, $(\\phi \\phi^{c^m} \\phi^{c^{2m}})^{c^i} = (\\phi \\phi^{c^m} \\phi^{c^{2m}})^{2^i}$. Thus, if $1_P$ is not a constituent of $\\phi_P {\\phi_P}^{c^m} {\\phi_P}^{c^{2m}}$, then $1_P$ will not be a constituent of $(\\phi \\phi^{c^m} \\phi^{c^{2m}})^{c^i}$. Thus, to show that $1_P$ is not a constituent of $\\psi_P$, it suffices to show that $1_P$ is not a constituent of $\\phi_P {\\phi_P}^{c^m} {\\phi_P}^{c^{2m}}$.\n\nBy Lemma \\ref{three}, we know that $\\phi_P = \\lambda + \\lambda^{-1}$ where $\\lambda$ is a faithful character of $P$ over an extension of $F$. It follows that\n$$\n\\phi_P {\\phi_P}^{c^m} {\\phi_P}^{c^{2m}} = (\\lambda + \\lambda^{-1})(\\lambda^{2^m} + \\lambda^{-2^m})(\\lambda^{2^{2m}} + \\lambda^{-2^{2m}}).\n$$\nThus, all of the irreducible constituents of the character $\\phi_P {\\phi_P}^{c^m} {\\phi_P}^{c^{2m}}$ have the form $\\lambda^{a_1 + a_2 2^m + a_3 2^{2m}}$ where $a_1, a_2, a_3 \\in \\{ \\pm 1 \\}$.\n\nLet $n_3 = 3^a$, and note that this implies that $m_3 = 3^{a-1}$. We now apply Corollary \\ref{onea} to see that $a_1 +a_2 2^m + a_3 2^{2m} \\equiv a_1 + a_2 (-1) + a_3 (-1)^2 ~({\\rm mod}~3^a)$ when $m$ is odd, and $a_1 +a_2 2^m + a_3 2^{2m} \\equiv a_1 + a_2 + a_3 ~({\\rm mod}~3^a)$ when $m$ is even. This gives possible values of $\\pm 1, \\pm 3 ({\\rm mod} 3^a)$. Since $\\lambda$ has order $3^{a+1}$, it follows that $1_P$ will not be occur when $a > 1$. Suppose that $a = 1$. We see that either $m$ is congruent to either $1$ or $2$ modulo $3$. If $m$ is congruent to $1$ modulo $3$, then $2^m \\equiv 2 ~({\\rm mod}~9)$ and $2^{2m} \\equiv 4 ~({\\rm mod}~9)$. This yields $a_1 +a_2 2^m + a_3 2^{2m} \\equiv a_1 + 2a_2 + 4a_3 ~({\\rm mod} ~9)$, and we obtain the possible values $\\pm 3, \\pm 5, \\pm 7$ modulo $9$. If $m$ is congruent to $2$ modulo $3$, then $2^m \\equiv 4 ~({\\rm mod}~9)$ and $2^{2m} \\equiv 2~({\\rm mod}~9)$. Again we obtain the values $\\pm 3, \\pm 5, \\pm 7$ modulo $9$. As before, we conclude that $1_P$ does not occur, and this proves the result.\n\\end{proof}\n\nFinally, we come to the main result of this section. This proves Theorem \\ref{thm2} when $p = 2$ and $G$ is nonsolvable.\n\n\\begin{theorem}\nLet $(G,N)$ be a $2$-Gagola pair where $G$ is not solvable. Then $|G:N|_2 \\ge |N|^2$.\n\\end{theorem}\n\n\\begin{proof} by way of contradiction, we assume that a counterexample $G$ exists. We see the result holds if ${\\bf O}_2 (G)\/N$ is not elementary abelian by Lemma \\ref{seven}. Thus, ${\\bf O}_2 (G)\/N$ is elementary abelian. By Lemma \\ref{six}, we know that every involution in $G\/N$ lies in ${\\bf O}_2 (G)\/N$. By Theorem 5.5 \\cite{Gagola}, we know that there a $2$-power $q$ so that $|N| = q^2$ and a subgroup $S$ of $G$ so that ${\\bf O}_2 (G) \\le S \\le G$ where $S$ is normal in $G$, $|G:S|$ is a power of $2$, $S\/{\\bf O}_2 (G) \\cong {\\rm SL}_2 (q)$, $N$ is the natural module for $S\/{\\bf O}_2 (G) \\cong {\\rm SL}_2 (q)$ and $G\/{\\bf O}_2 (G) \\le {\\rm Aut} (S\/{\\bf O}_2 (G))$.\n\nWe now prove a series of claims.\n\nClaim 1: $Z ({\\bf O}_2 (G)) = N$. Let $Z = Z ({\\bf O})_2 (G)$. We know that $N \\le Z$ by \\cite{Gagola}. Suppose $N < Z$, and pick $z \\in Z \\setminus N$. It follows that ${\\bf O}_2 (G) \\le C_G (z)$. Let $D(z) = \\{ g \\in G | [g,z] \\in N \\}$. We know that $|D(z):C_G (z)| = |N| = q^2$ and that $|D (z):C_G (z)|$ divides $|G:{\\bf O}_2 (G)|$. We know that $|G:{\\bf O}_2 (G)|$ divides $|{\\rm Aut} ({\\rm SL}_2 (q)|$. Thus, $|G:{\\bf O}_2 (G)|_2$ divides $q {\\rm log}_2 (q)_2 < q^2$. We now have a contradiction. Thus, $Z = N$, and the claim is proved.\n\nSince we know that $N < {\\bf O}_2 (G)$, Claim 1 implies that ${\\bf O}_2 (G)$ is nonabelian. We now show that often we may assume that $q > 4$.\n\nClaim 2: If $|{\\bf O}_2 (G):N| > q^2$ and $q = 4$, then the theorem is proved. We know from \\cite{Gagola} that $|G:N|_2$ is a square. We see that $|S:{\\bf O}_2 (G)|_2 |{\\bf O}_2 (G):N|$ divides $|G:N|_2$. This implies that $2q^2 4 = 2^7$ divides $|G:N|_2$. Since $|G:N|_2$ is a square, $2^8$ divides $|G:N|_2$. We conclude that $|G:N|_2 \\ge 2^8 = 4^4 = q^4 = |N|^2$.\n\nWe now consider subgroups $A$ with $N \\le A \\le {\\bf O}_2 (G)$ and $[A,S] \\le N$.\n\nClaim 3: If there exists a subgroup $A$ with $N \\le A \\le {\\bf O}_2 (G)$ and $[A,S] \\le N$, then $A$ is abelian, and in particular, $A < {\\bf O}_2 (G)$. Observe that $N$ is irreducible under the action of $S$. Thus, either $A' = 1$ or $A' = N$. Suppose $A' = N$, and thus, $N < A$. Then it follows that $A$ is nonabelian. Let $p$ be an odd prime divisor of $|S|$, and write $P$ for a Sylow $p$-subgroup of $S$. We see that $[A,P] \\le [A,S] \\le N$, so $P$ centralizes $A\/N$. Hence, $A = C_A (P) N = C_A (P) A' \\le C_A (P) \\Phi (A) \\le A$. Thus, $A= C_A (P) \\Phi (A)$, and this implies $A = C_A (P)$. We conclude that $P$ centralizes $A$ which is a contradiction since $P$ act Frobeniusly on $1 < N \\le A$. Therefore, we must have $A' = 1$, and the claim is proved.\n\n\nClaim 4: There does not exist a subgroup $A$ with $N < A < {\\bf O}_2 (G)$ and $[A,S] \\le N$. By way of contradiction, assume such a subgroup $A$ exits. Suppose $a \\in A \\setminus N$. Observe that $[a,S] \\le N$, so $S\/N \\le C_{S\/N} (aN)$. Since $(G,N)$ is a Gagola pair, we know that $|C_G (a)| = |C_{G\/N} (aN)|$. Because $|G:S|$ is a power of $2$, this implies that $C_G (a)$ contains a full Sylow $p$-subgroup for every odd prime $p$ that divides $|G|$. Let $p$ be an odd prime divisor of $|G|$, and let $P$ be a Sylow $p$-subgroup of $C_G (a)$. Observe that $P \\le S$, so $P$ is a Sylow $p$-subgroup of $C_S (a)$. It follows that $C_S (a)$ contains a full Sylow subgroup for every odd prime. Since $S\/{\\bf O}_2 (G) \\cong {\\rm SL}_2 (q)$ has no proper subgroups with this property, we conclude that $S = {\\bf O}_2 (G) C_S (a)$.\n\nObserve that $[{\\bf O}_2 (G), a] \\le N$. Since $N$ is central in ${\\bf O}_2 (G)$, this implies that $[{\\bf O}_2 (G),a]$ is normal in ${\\bf O}_2 (G)$. Obviously, $C_S (a)$ normalizes $[{\\bf O}_2 (G), a]$, so $[{\\bf O}_2 (G), a]$ is normal in $S$. Since $a \\not\\in N = Z ({\\bf O}_2 (G))$, we see that $[{\\bf O}_2 (G), a] > 1$. Since $N$ is irreducible under the action of $S$, we conclude that $[{\\bf O}_2 (G), a] = N$. Hence, the map ${\\bf O}_2 (G) \\rightarrow N$ defined by $x \\mapsto [x,a]$ is onto. Since $N$ is central in ${\\bf O}_2 (G)$, this map is a homomorphism and its kernel is $C_{{\\bf O}_2 (G)} (a)$. By the isomorphism theorems, we conclude that $|{\\bf O}_2 (G): C_{{\\bf O}_2 (G)} (a)| = |N| = q^2$.\n\nWe note that $C_{{\\bf O}_2 (G)} (a)$ is normal in $C_S (a)$. Notice that $N \\le C_{{\\bf O}_2 (G)} (a)$, and we have ${\\bf O}_2 (G)\/N$ is elementary abelian, so $C_{{\\bf O}_2 (G)} (a)$ is normal in ${\\bf O}_2 (G)$. It follows that $C_{{\\bf O}_2 (G)} (a)$ is normal in $S = {\\bf O}_2 (G) C_S (a)$.\n\nWe can view $C_{{\\bf O}_2 (G)} (a)\/N$ as a module for $S\/{\\bf O}_2 (G)$. If this module has any nonprincipal constituents, then we may apply Lemma \\ref{four} to see that $|C_{{\\bf O}_2 (G)} (a)\/N| \\ge q^2$. This implies that $|{\\bf O}_2 (G):N| = |{\\bf O}_2 (G): C_{{\\bf O}_2 (G)} (a)| |C_{{\\bf O}_2 (G)} (a):N| \\ge q^2 q^2 = q^4 = |N|^2$, and the theorem is proved. Thus, we may assume that $[C_{{\\bf O}_2 (G)} (a), S] \\le N$.\n\nWe have $C_S (a)\/ C_{{\\bf O}_2 (G)} (a) \\cong S\/{\\bf O}_2 (G) \\cong {\\rm SL}_2 (q)$, and we have just shown that $C_{{\\bf O}_2 (G)} (a)\/N \\le Z (C_S (a)\/N)$. Let $C = C_S (a)$. It is not difficult to see that $C' \\cap {\\bf O}_2 (G) = C' \\cap C_{{\\bf O}_2 (G)} (a)$. If $N < C' \\cap C_{{\\bf O}_2 (G)} (a)$, then ${\\rm SL}_2 (q)$ must have a nontrivial Schur multiplier, and so, $q = 4$. Let $p$ be an odd prime divisor of $|S|$, and let $P$ be a Sylow $p$-subgroup of $S$. Since $P$ centralizes $A\/N$, and $P$ does not centralize ${\\bf O}_2 (G)\/N$, we conclude that $P$ does not centralize ${\\bf O}_2 (G)\/A$. Thus, $S$ does centralize ${\\bf O}_2 (G)\/A$, so we may apply Lemma \\ref{four} to see that $|{\\bf O}_2 (G):A| \\ge q^2$, and hence, $|{\\bf O}_2 (G):N| > q^2$. With $q = 4$, we have seen in Claim 2 that the Theorem is proved. Thus, we may assume that $C' \\cap C_{{\\bf O}_2 (G)} (a) = N$.\n\nNow, we have that $C'\/N \\cong S\/{\\bf O}_2 (G)$. Obviously, $C'\/N$ contains an involution. But any such involution will not lie in ${\\bf O}_2 (G)\/N$, and this contradicts our earlier observation. This contradiction proves the claim.\n\nClaim 5: $[{\\bf O}_2 (G),S]\/N$ is irreducible under the action of $S$. Suppose $N \\le M < [{\\bf O}_2 (G),S]$ so that $[{\\bf O}_2 (G),S]\/M$ is chief for $G$. By Lemma \\ref{four}, we have $|[{\\bf O}_2 (G),S]:M| \\ge q^2$. If $S$ acts nontrivially on $M\/N$, then $|M:N| \\ge q^2$ by Lemma \\ref{four}, and $|{\\bf O}_2 (G):N| \\ge q^4$ as desired. Thus, we may assume that $[M,S] \\le N$. By Claim 4, this implies that $M = N$. This proves the claim.\n\nWe now set $M = [{\\bf O}_2 (G),S]$.\n\nClaim 6: $S' \\cap {\\bf O}_2 (G) = M$. Observe that ${\\bf O}_2 (G)\/M \\le Z (S\/M)$. If $S' \\cap {\\bf O}_2 (G) > M$, then ${\\rm SL}_2 (q)$ has a nontrivial Schur multiplier, so $q = 4$. We have $|{\\bf O}_2 (G):N| > q^2$, so we are done in this case. This proves the claim.\n\nClaim 7: $|M:N| = q^2$. If $|M:N| \\ge q^4$, then we have the result. Thus, we may assume that $|M:N| < q^4$. By Lemma \\ref{four}, we know that if $|M:N| > q^2$, then $|M:N| = 2^{8n\/3}$ where $q = 2^n$ and $3$ divides $n$. We also know that if $P$ is a Sylow $3$-subgroup of $S'$, then $C_{M\/N} (P) = 1$. Let $D\/M = N_{S'\/M} (PM\/M)$. From Dickson's classification of the subgroups of ${\\rm SL}_2 (q)$, we see that $D\/M$ is a dihedral group. Thus, we may apply Lemma \\ref{five} to see that $D\/N$ contains an involution not in $M\/N$, but this yields an involution in $G\/N$ that is outside ${\\bf O}_2 (G)\/N$ which is a contradiction. This implies that $|M:N| = q^2$.\n\nClaim 8: Final contradiction. If $M$ is abelian, then the result holds by Lemma \\ref{abel}. Thus, $M$ is nonabelian. This implies that $M' = N$. Since $M\/M'$ is a chief factor, we obtain $N = Z (M)$. We know the exponent of $M$ must be $4$, so there exists $x \\in M \\setminus N$ so that $x^2 \\ne 1$. Notice that every element in $xN$ must have order $4$, since they are all conjugate to $x$. We know that $S$ acts transitively on $N \\setminus \\{ 1 \\}$. This implies that the $S$-orbit of $x^2$ has size $q^2 - 1$. It follows that the $S$-orbit of $xN$ must size at least $q^2 - 1$. Since $|M\/N \\setminus \\{ N \\}| = q^2 - 1$, we conclude that $S$ acts transitively on $M\/N$. In particular, $C{S'\/M}{xN} = C_{S'\/M} (x^2) = 1$. If $p$ is an odd prime divisor of $|S|$ and $P$ is a Sylow $p$-subgroup of $S'$, then $C_{M\/N}(P) = 1$. Let $D\/M = N_{S'\/M} (PM\/M)$. From Dickson's classification of the subgroups of ${\\rm SL}_2 (q)$, we see that $D\/M$ is a dihedral group. Thus, we may apply Lemma \\ref{five} to see that $D\/N$ contains an involution not in $M\/N$, but this yields an involution in $G\/N$ that is outside ${\\bf O}_2 (G)\/N$ which is a contradiction. This the final contradiction and the theorem is proved.\n\\end{proof}\n\n\\section{$p=2$ and $G$ solvable: part I}\n\nWe now start to handle the case where $p = 2$ and $G$ is solvable. Recall that a non-abelian $2$-group $H$ is a Suzuki $2$-group if $H$ has more than one involution and ${\\rm Aut} (H)$ has a solvable subgroup that acts transitively on the involutions of $H$. The Suzuki $2$-groups of type A are denoted by $A (n,\\Theta)$ where $n$ is a positive integer and $\\Theta$ is a nontrivial odd-order automorphism of the field of order $2^n$. The order of $A (n,\\Theta)$ is $2^{2n}$. We will give a more complete description of these groups in the next section. At this time, we have not been able to dispense with the solvable hypothesis.\n\n\\begin{theorem} \\label{2pairclass}\nIf $(G,N)$ is a $2$-Gagola pair where $G$ is solvable,\nthen either:\n\\begin{enumerate}\n\\item $|N|^2 \\le |G:N|_2$ or\n\\item\n \\begin{enumerate}\n \\item $G$ is not $2$-closed.\n \\item Writing $|N| = 2^n$ for an integer $n$, then $n$ is divisible by both $2$ and some odd prime.\n \\item $G = M N_G (K)$ with $M \\cap N_G (K) = 1$.\n \\item ${\\bf O}_ 2 (G) = M C$ where $M \\cap C = 1$,\n $M = [{\\bf O}_2 (G),K]$ and $C = C_{{\\bf O}_2 (G)} (K)$ where $K$ is a subgroup of order $q$ where if possible $q$ is a Zsigmondy prime divisor of $|N| - 1$ and otherwise $q = 3$ when $|N| = 2^6$.\n \\item $M$ is a Suzuki $2$-group of type A with $M' = Z(M) = N$.\n \\item $C_G (M) = N$, so $N_G (K)$ is isomorphic to a subgroup of ${\\rm Aut} (M)$.\n \\item If $H$ is Hall $2$-complement of $G$, then $MH$ is a Frobenius group.\n \\item If $x \\in N_G (K)$, then $C_G (x) = C_M (x) C_{N_G (K)} (x)$ and $C_M (x) \\le N$. In particular, if $x \\in C$, then $C_M (x) = N$.\n \\end{enumerate}\n\\end{enumerate}\n\\end{theorem}\n\n\\begin{proof}\n\nBy Lemma \\ref{seven}, we have the first conclusion if ${\\bf O}_2 (G)\/N$ is not elementary abelian. Thus, we may assume ${\\bf O}_2 (G)\/N$ is elementary abelian.If $G$ is $2$-closed, then we may use \\cite{ChMc} to see that $({\\bf O}_2 (G),N)$ is a Camina pair. Since ${\\bf O}_2 (G)\/N$ is abelian, we may use Lemma \\ref{abel pair} to see that $|{\\bf O}_2 (G):N| \\ge |N|^2$, and this gives the desired conclusion. Thus, $G$ is not $2$-closed.\n\nLet $H$ be a Hall $2$-complement of $G$. If $|N| - 1$ has a Zsigmondy prime divisor $q$, then let $K = K^*$ be the subgroup of $H$ of order $q$. Otherwise, $|N| = 2^6$ and we take $K$ to be the subgroup of $H$ of order $3$ and we take $K^*$ to be the subgroup of $H$ order $9$. By Lemma \\ref{aff}, we know that $G\/{\\bf O}_2 (G) \\le \\Gamma (N)$, and so, ${\\bf O}_2 (G)K$ is normal in $G$. Applying Lemma \\ref{comm}, we can find $M$ so that $N < M \\le [{\\bf O}_2 (G),K]$ with $M$ normal in $G$ and $M\/N$ a chief factor of $G$. Notice that $K$ will act Frobeniusly on $M\/N$ since $M\/N$ is abelian, so $K^*$ acts Frobeniusly on $M\/N$. Also, $M\/N$ will be a direct sum of irreducible $K^*$ modules. It follows that $|N| \\le |M:N|$. If $M\/N$ is not irreducible under the action of $K^*$, then $|N|^2 \\le |M:N|$ and we are done. Thus, $M\/N$ is irreducible under the action of $K^*$ and so, $|M:N| = |N|$. Also, if $M < [{\\bf O}_2 (G),K]$, then $|N| \\le |[{\\bf O}_2 (G),K]:M|$ and $|N|^2 \\le |P:N|$ where $P$ is a Sylow $2$-subgroup of $G$. We have $M = [{\\bf O}_2 (G),K]$. We obtain ${\\bf O}_2 (G) = M C_{{\\bf O}_2 (G)} (K)$ and by the Frattini argument, $G = {\\bf O}_2 (G) N_G (K) = M N_G (K)$. Let $C = C_{{\\bf O}_2 (G)} (K)$, so ${\\bf O}_2 (G) = MC$.\n\nIf $M$ is abelian, then we are done by Lemma \\ref{abel}. We may assume that $M$ is not abelian. This implies that $M' = \\Phi (M) = Z(M) = N$.\n\nWe have $C_G (M) \\cap M = Z(M) = N$. Let $B = C_G (M)$. We will prove that $B = N$. Note that $B$ is a normal subgroup of $G$ and $B \\cap H = 1$, so $B$ is a $2$-group, and we obtain $B \\le {\\bf O}_2 (G)$. We have $B = [B,K] C_{B} (K)$. Now, $N = [N,K] \\le [B,K] \\le [{\\bf O}_2 (G),K] \\cap B \\le M \\cap B = N$. Also, $C_{B} (K) \\cap N \\le C \\cap N = 1$. Now, $M$ centralizes $B$, so $N$ will centralize $C_{B} (K)$, and $N_G (K)$ normalizes both $B$ and $K$, so $N_G (K)$ normalizes $C_B (K)$. It follows that $G = M N_G (K)$ normalizes $C_B (K)$. Since $N$ is the unique minimal normal subgroup of $G$, this implies that $C_B (K) = 1$ and $B = [B,K] = N$. In particular, $G\/N$ is isomorphic to a subgroup of the automorphism group of $M$.\n\n\nLet $X$ be any subgroup of $H$ of prime order. By Lemma \\ref{solv Frob}, we know that $X$ is normal in $H$, so $K$ centralizes $X$. In particular, $N_G (K)$ normalizes $C_M (X)$. Now, $M$ normalizes $C_M (X) N$, and thus, $C_M (X) N$ is normal in $G$. If $M = C_M (X) N = C_M (X) \\Phi (M)$, then $M = C_M (X)$ which is a contradiction since $X$ acts Frobeniusly on $N$. Hence, $C_M (X) \\le N$, and since $C_N (X) = 1$, we have $C_M (X) = 1$. In particular, $H$ acts Frobeniusly on $M$. This implies that $C \\cap M = 1$. Notice that $N_G (K) \\cap {\\bf O}_2 (G) = C$, so this implies $N_G (K) \\cap M = 1$.\n\nSince $|M:N| = |N| = |H| + 1$, we see that $H$ acts transitively on $M\/N$. Since $M$ is not abelian, there exists $m \\in M \\setminus N$ of order $4$. It is easy to see that every element in the coset $mN$ will have order $4$ since $N = Z (M)$ and $N$ is elementary abelian. Because $H$ acts transitively on $M\/N$, this implies that every element in $M \\setminus N$ has order $4$. It follows that $N$ contains all the involutions in $M$. Since $H$ acts transitively on $N \\setminus \\{ 1 \\}$, $H$ acts transitively on the involutions of $M$. We deduce that that $M$ is a Suzuki $2$-group. By Theorem VIII.7.9 of \\cite{HBII}, $M$ is isomorphic to $A (m,\\Theta)$ as defined in Section VIII.6 of \\cite{HBII} where $|N| = 2^n$ and $\\Theta$ is an automorphism of the field of order $2^n$. By Theorem VIII.6.9 of \\cite{HBII}, we see that $\\Theta$ must have odd order not equal to $1$. On the other hand, by Lemma \\ref{aff}, we know that $2$ divides $n$.\n\nBecause $N$ is central in $M$ and elementary abelian, the map $bN \\mapsto b^2$ is a well-defined function from $M\/N$ to $N$. Notice that if $h \\in H$, then $bN^h = b^hN$ and $(b^2)^h = (b^h)^2$, so the action of $H$ commutes with this function. Since $H$ is acting transitively, it follows that the function must be a bijection.\n\nSuppose $x \\in N_G (K)$. Obviously, we have $C_M (x) C_{N_G (K)} (x) \\le C_G (x)$. Consider an element $g \\in C_G (x)$. We can write $g = mc$ where $m \\in M$ and $c \\in N_G (K)$. We know that $1 = [g,x] = [mc,x] = [m,x]^c[c,x]$, and this implies that $[m,x]^c = [c,x]^{-1}$. We know that $[m,x]^c \\in M$ and $[c,x]^{-1} \\in N_G (K)$. This implies that $[m,x]^c = [c,x]^{-1}$ lie in $N_G (K) \\cap M = 1$, and so $[m,x] = [c,x] = 1$, and thus, $m \\in C_M (x)$ and $c \\in C_{N_G (K)} (x)$. We conclude that $C_G (x) = C_M (x) C_{N_G (K)} (x)$.\n\nA similar argument shows that\n$$\nC_{G\/N} (xN) = C_{M\/N} (xN) C_{N_G(K) N\/N} (xN).\n$$\nMaking the observation that $C_{N_G (K) N\/N} (xN) \\cong C_{N_G (K)} (x)$, we have\n$$\n|C_{G\/N} (xN)| = |C_{M\/N} (xN)| |C_{N_G (K)} (x)|.\n$$\nRecalling that $|C_G (x)| = |C_{G\/N} (xN)|$, we conclude that $|C_{M\/N} (xN)| = |C_M (x)|$. Notice that the action of $x$ must commute with the map $bN \\mapsto b^2$ from $M\/N$ to $N$, and so, we must have $|C_{M\/N} (xN)| = |C_N (x)|$. Since this implies that $|C_N (x)| = |C_M (x)|$, we must have $C_M (x) \\le N$. If $x \\in C$, then $C_M (x) = N$.\n\\end{proof}\n\n\\section{Automorphisms of Suzuki groups}\n\nIn order to finish the proof of Theorem \\ref{thm1} we need to understand conclusion 2 of Theorem \\ref{2pairclass}, and to do this we compute the automorphism group of a Suzuki $2$-group of type A and how it acts on the group.\n\nThere are two papers in Russian which we found in translation that address the automorphism groups of Suzuki $2$-groups of type A. The first, \\cite{autos}, computes the Sylow $2$-subgroup of the quotient of the\nautomorphism group by its centralizer of the center. In particular,\nit show that the Sylow $2$-subgroup is cyclic, and it uses that to\nshow that the automorphism group is solvable.\n\nThe second one,\n\\cite{p-alg}, finds three subgroups of the automorphism group. We\nwill discuss the three subgroups later. They then prove results for\nan analog of the Suzuki $2$-groups for odd primes. They mention\nthat they consider odd primes so that they do not need the ``subtle\nconsiderations needed in the case of $p = 2$.''\n\n\nWe now recall the structure of a Suzuki $2$-group of type A. Since\n$M$ is isomorphic to $A (n,\\Theta)$, we can write the elements of $M$\nas $(a,b)$ where $a,b \\in F$ and $F = {\\rm GF} (2^n)$. The\nmultiplication is defined by $(a,c) \\cdot (b,d) = (a + b, c + d + b\n\\Theta (a))$. (Recall that $\\Theta$ is a nontrivial automorphism of\n$F$ having odd order.) The elements of $N$ are $(0,b)$ as $b$ runs\nthrough $F$. Observe that $(a,b) = (a,0)(0,b) \\in (a,0) N$. Since\n$N$ is central ${\\bf O}_2 (G)$, we have $C_{{\\bf O}_2 (G)} ((a,0)) =\nC_{{\\bf O}_2 (G)} ((a,b))$. Also, we see that $(a,b)N = (a,0)N$, so\n$C_{G\/N} ((a,0)N) = C_{G\/N} ((a,b)N)$.\n\nLet $M$ be a Suzuki $2$-group of Type A and use the notation of the\nprevious paragraph. In \\cite{p-alg}, they define three subgroups of\n${\\rm Aut} (M)$. They are:\n\n\\begin{enumerate}\n\\item $A_1 = \\{ \\phi = \\phi_\\psi \\mid (a,b)^\\phi = (a, \\psi (a) + b\n\\}$ where $\\psi$ runs over the linear transformations of $F$\nregarded as a vector space over $Z_2$.\n\n\\item $A_2 = \\{ \\phi = \\phi_x \\mid (a,b)^\\phi = (xa,x \\Theta (x) b) \\}$\nwhere $x$ runs over the nonzero elements of $F$.\n\n\\item $A_3 = \\{ \\phi = \\phi_\\tau \\mid (a,b)^\\phi = (a^\\tau,b^\\tau)\n\\}$ where $\\tau$ runs over the Galois automorphisms of $F$ with\nrespect to $Z_p$.\n\\end{enumerate}\n\nIt is not difficult to show that $|A_1| = 2^{n^2}$. (Each of the\n$n$ basis elements of $F$ can be sent to any element of $F$.) Also,\nwe see that $|A_2| = 2^n - 1$ and $|A_3| = n$. Suppose $\\psi$ is a\nlinear transformation of $F$, $x \\in F \\setminus \\{ 0 \\}$, and\n$\\tau$ is a Galois automorphism of $F$. Consider $(a,b) \\in M$.\nThen\n$$\n(\\phi_\\psi)^{\\phi_x} (a,b) = (a,x^{-1} \\Theta (x^{-1}) \\psi (xa) +\nb),\n$$\nand observe that the map $a \\mapsto x^{-1} \\Theta (x^{-1}) \\psi\n(xa)$ is a linear transformation of $F$. Thus,\n$(\\phi_\\psi)^{\\phi_x} \\in A_1$. Also, we have\n$$\n(\\phi_\\psi)^{\\phi_\\tau} (a,b) = (a,(\\psi (a^\\tau))^{\\tau^{-1}} + b),\n$$\nand since the map $a \\mapsto (\\psi (a^\\tau))^{\\tau^{-1}}$ is a\nlinear transformation of $F$, we have $(\\phi_\\psi)^{\\phi_\\tau} \\in\nA_1$. Finally,\n$$\n(\\phi_x)^{\\phi_\\tau} (a,b) = (x^{\\tau^{-1}}a,x^{\\tau^{-1}} \\Theta\n(x^{\\tau^{-1}}) b),\n$$\nand so, $(\\phi_x)^{\\phi_\\tau} = \\phi_{x^{\\tau^{-1}}} \\in A_2$. We\nconclude that $A_2$ and $A_3$ both normalize $A_1$ and $A_3$\nnormalizes $A_2$. We note that $A_2 A_3$ is isomorphic to $\\Gamma (N)$ where $A_2$ corresponds to $\\Gamma_o (N)$.\n\n\nWe now work to prove that ${\\rm Aut} (M) = A_1 A_2 A_3$. We begin with a general lemma about the automorphisms of $M$.\n\n\\begin{lemma} \\label{auts}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. If $\\phi$ is an automorphism of $M$, then there exist bijective linear transformations $f$ and $h$ of $F$ that satisfies $h (a \\Theta (a)) = f(a) \\Theta(f(a))$ for all $a \\in F$ and a map $g$ from $F$ to $F$ that satisfies $g (0) = 0$ so that $(a,b)^\\phi = (f(a), g(a) + h (b))$. If $h (\\Theta (a_1) a_2) = \\Theta(f(a_1)) f(a_2)$ for all $a_1, a_2 \\in F$, then $g$ is a linear transformation of $F$.\n\\end{lemma}\n\n\\begin{proof}\nSince $\\phi$ is an automorphism, we have $(a,b)^\\phi = (a,0)^\\phi (0,b)^\\phi$ for all $(a,b) \\in M$. We define $f$ and $g$ by $(a,0)^\\phi = (f(a),g(a))$. Since $\\phi$ determines an automorphism of $M\/N$, it follows that $f$ is a bijective linear transformation of $F$. It is not difficult to see that $g$ is a map from $F$ to $F$. Since $\\phi$ maps $(0,0)$ to $(0,0)$, we conclude that $g (0) = 0$. Notice that $N = Z(M)$ is characteristic, so $\\phi$ must map $N$ to $N$. It follows that we define $h$ by $(0,b)^\\phi = (0,h(b))$. Since the restriction of $\\phi$ to $N$ will be an automorphism of $N$, we conclude that $h$ is a bijective linear transformation of $F$. We have $(a,0)^\\phi (0,b)^\\phi = (f(a),g(a))(0,h(b)) = (f(a), g(a)+h(b))$.\n\nObserve that $(a,0)^2 = (0,\\Theta (a) a)$. Applying $\\phi$ to this equation yields $((a,0)^\\phi)^2 = (f(a),g(a))^2 = (0,\\Theta (f(a)) f(a))$ and $(0,\\Theta (a) a)^\\phi = (0, h (\\Theta (a) a)$. We obtain $(0,\\Theta (f(a)) f(a)) = (0,h(\\Theta (a) a))$, and we have the desired equality $h (\\Theta (a) a) = \\Theta (f (a)) f(a)$.\n\nWe know that $(a_1,0)(a_2,0) = (a_1 + a_2, \\Theta (a_1) a_2)$. Applying $\\phi$, we obtain $(a_1,0)^\\phi (a_2,0)^\\phi = (f(a_1),g(a_1)) (f(a_2),g(a_2))) = (f(a_1)+f(a_2), g(a_1) + g(a_2) + \\Theta (f (a_1)) f(a_2))$ and $(a_1 + a_2, \\Theta (a_1) a_2)^\\phi = (f(a_1 + a_2), h(\\Theta (a_1) a_2) + g (a_1 + a_2))$. Since $\\phi$ is an automorphism, we know that $g(a_1) + g(a_2) + \\Theta (f (a_1)) f(a_2)) = h(\\Theta (a_1) a_2) + g (a_1 + a_2)$. If $\\Theta (f (a_1)) f(a_2) = h(\\Theta (a_1) a_2)$, then $g (a_1 + a_2) = g (a_1) + g (a_2)$, and we conclude that $g$ is a linear transformation.\n\\end{proof}\n\nThis next fact is Theorem VIII.6.9(b) of \\cite{HBII}.\n\n\\begin{lemma}\\label{bijection}\nLet $F$ be a field of order $2^n$ and let $\\Theta$ be an automorphism of $F$ of order $l$ where $l$ is an odd integer. Then the map $a \\mapsto a \\Theta (a)$ is a bijection of $F$.\n\\end{lemma}\n\n\nWe now consider the subgroup $A_1$ in ${\\rm Aut} (M)$.\n\n\\begin{lemma} \\label{cent N}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Then $C_{{\\rm Aut} (M)} (N) = A_1$.\n\\end{lemma}\n\n\\begin{proof}\nIt is obvious that $A_1 \\le C_{{\\rm Aut} (M)} (N)$. Suppose that $\\phi \\in C_{{\\rm Aut} (M)} (N)$. We can write $(a,b)^\\phi = (f(a),g(a)+ h(b))$ as in Lemma \\ref{auts}. Since $\\phi$ centralizes $N$, we see that $h (b) = b$ for all $b \\in F$. We know that $f(a) \\Theta (f(a)) = h (a \\Theta (a)) = a \\Theta (a)$ for all $a \\in F$. By Lemma \\ref{bijection}, we obtain $f(a) = a$ for all $a \\in F$. We conclude that $(a,b)^\\phi = (a,b + g(a))$. Since $h (\\Theta (a_1) a_2) = \\Theta (a_1) a_2 = \\Theta (f (a_1)) f(a_2)$, we may use Lemma \\ref{auts} to conclude that $g$ is a linear transformation of $F$, and so, $\\phi = \\phi_g \\in A_1$.\n\\end{proof}\n\nThis next corollary encodes the result of \\cite{autos}.\n\n\\begin{corollary} \\label{Sylow 2}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Let $\\sigma$ generate the Sylow $2$-subgroup of the Galois group. Then $\\langle A_1, \\phi_\\sigma \\rangle$ is a Sylow $2$-subgroup of ${\\rm Aut} (M)$.\n\\end{corollary}\n\nWe now prove the desired result. Our intuition is that this has been proved before, but we cannot find any reference where it is explicitly computed.\n\n\\begin{theorem}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Then ${\\rm Aut} (M) = A_1 A_2 A_3$.\n\\end{theorem}\n\n\\begin{proof}\nThe map $\\rho$ from ${\\rm Aut} (M)$ to ${\\rm Aut} (N)$ defined by restriction is a homomorphism whose kernel is $C_{{\\rm Aut} (M)} (N)$. We know ${\\rm Aut} (M)$ is solvable, so $\\rho ({\\rm Aut} (M))$ is solvable. Let $X$ be a Hall $2$-complement of $\\rho ({\\rm Aut} (M))$. We know that $X$ acts transitively on the nonidentity elements of $N$. By Theorem VIII.3.5 of \\cite{HBII}, we see that $X$ is isomorphic to a subgroup of the affine group of $N$. In particular, we have that $|X| \\le (2^n - 1)n_{2'}$. We saw in Corollary \\ref{Sylow 2} that $|\\rho ({\\rm Aut} (M))|_2 = n_2$, and so, $|\\rho ({\\rm Aut} (M))| \\le (2^n - 1)n$. We know that $A_2 A_3 \\cap A_1 = 1$ and $|A_2 A_3| = (2^n - 1)n$. It follows that $\\rho ({\\rm Aut} (M)) \\cong A_2 A_3$, and hence, ${\\rm Aut} (M) = A_1 A_2 A_3$.\n\\end{proof}\n\nBefore we apply this to obtain results in our situation, we make an observation. Suppose $\\sigma$ is a Galois automorphism of $F$. It is not difficult to see that $\\phi_\\sigma \\in A_3$ will fix $(1,0) \\in M$. In particular, $\\phi_\\sigma$ will fix an element of $M$ that is outside of $N$. If $(G,N)$ is a pair satisfying the conclusion of \\ref{2pairclass}, then we see that $N_G (K)$ cannot contain any element that induces an automorphism which is conjugate to an element of $A_3$. Hence, if we wish to show that no such group $G$ can exist, then it suffices to prove that $N_G (K)$ would have to contain an element that induces an automorphism that is conjugate to an element of $A_3$.\n\n\\section{$p=2$ and $G$ is solvable: part 2} \\label{spec}\n\nWe now use the observation from the end of the previous section to show that the groups in Conclusion 2 of Theorem \\ref{2pairclass} do not exist. We begin by determining more information about the elements of $A_1$ that commute with elements of $A_2$. Since the elements of $C$ commute with $K$ and the image of $K$ is conjugate to a subgroup of $A_2$, this can be viewed as describing the image of $C$ in ${\\rm Aut} (M)$.\n\n\\begin{lemma} \\label{cent A1}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Let $x \\in F^x$. If $1 \\ne \\phi_\\psi \\in C_{A_1} (\\phi_x)$, then there is a positive integer $j$ so that $x \\Theta (x) = x^{2^j}$.\n\\end{lemma}\n\n\\begin{proof}\nSince $\\phi_x$ and $\\phi_\\psi$ commute, we have that $\\psi (a) = x^{-1} \\Theta (x^{-1}) \\psi (xa)$ for every element $a \\in F$. This implies that $\\psi (xa) = x \\Theta (x) \\psi (a)$ for all $a \\in F$. Since $\\phi_\\psi \\ne 1$, we know that $\\psi \\ne 0$. Thus, there is an element $a \\in F$ so that $\\psi (a) \\ne 0$. Applying this with $x^2a$, we obtain $\\psi (x^2a) = x \\Theta (x) \\psi (xa) = x^2 \\Theta (x^2) \\psi (a)$. Inductively, we obtain $\\psi (x^ia) = x^i \\Theta (x^i) \\psi (a)$ for every positive integer $i$.\n\nLet $f (y)$ be the minimal polynomial in $Z_2 [y]$ for $x$. We write $f (y) = \\sum_{i=0}^r a_i y^i$ where $a_i \\in Z_2$. Notice that $f (y)$ is fixed by the Galois automorphisms of $F$. Thus, the roots of $f (y)$ have the form $x, x^2, x^4, \\dots, x^{2^n}$. We have\n$$\n0 = \\psi (0) = \\psi (f(x)a) = \\psi (\\sum_{i=1}^r a_i x^i a) = \\sum_{i=1}^r a_i \\psi (x^i a) $$\n$$= \\sum_{i=1}^r a_i (x\\Theta (x))^i \\psi (a) = f(x \\Theta (x)) \\psi (a).\n$$\nSince $\\psi (a) \\ne 0$, we conclude that $x \\Theta (x)$ is a root of $f (y)$, and thus, $x \\Theta (x) = x^{2^j}$ for some positive integer $j$.\n\\end{proof}\n\nWe refine the condition of the last lemma to a modular congruence.\n\n\\begin{corollary}\\label{numcond}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Let $x \\in F^x$. Suppose $\\Theta (a) = a^{2^h}$ for some positive integer $h$. If $2^h + 1 \\not\\equiv 2^j ~({\\rm mod}~o(x))$ for all positive integers $j$, then $C_{A_1} (\\phi_x) = 1$.\n\\end{corollary}\n\n\\begin{proof}\nSuppose $C_{A_1} (\\phi_x) > 1$. By Lemma \\ref{cent A1}, we know that $x \\Theta (x) = x x^{2^h} = x^{2^h + 1} = x^{2^j}$ for some positive integer $j$. It follows that $2^h + 1 \\equiv 2^j ~({\\rm mod}~o(x))$, a contradiction.\n\\end{proof}\n\nWe now come to our key observation.\n\n\\begin{lemma} \\label{nontrivcent}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Let $(G,N)$ be as in the conclusion of Theorem \\ref{2pairclass}. Let $\\rho$ be the homomorphism from $N_G (K)$ to ${\\rm Aut} (M)$. If the element $g \\in C_G (K)$ has odd order (if $n = 6$, then assume $g \\in C_G (K)$ has order $7$), then $\\rho (g)$ is conjugate to an element in $A_2$ and $C_{A_1} (\\rho (g)) > 1$.\n\\end{lemma}\n\n\\begin{proof}\nRecall that $G\/{\\bf O}_2 (G) \\cong N_G (K)\/C$ is isomorphic to a subgroup of the affine group of $N$. If $n > 6$, then the centralizer of $K$ in the affine group is normal and cyclic of order $2^n - 1$, so $C \\langle g \\rangle$ is a characteristic subgroup of $C_{N_G (K)\/C} (K)$. (In particular, the centralizer of the image of $K$ in $\\Gamma (N)$ is $\\Gamma_o (N)$.) If $n = 6$, it is not difficult to see that the subgroup of order $7$ is normal in the affine group. This implies that $C \\langle g \\rangle$ is normal in $N_G (K)$. We know that $\\langle g \\rangle$ is a Hall subgroup of $C \\langle g \\rangle$. By the Frattini argument, we know that $N_G (K) = C N_{N_G (K)} (\\langle g \\rangle )$. In particular, $N_{N_G (K)} (\\langle g \\rangle)$ has a nontrivial Sylow $2$-subgroup.\n\nObserve that $\\rho (\\langle K, g \\rangle)$ has odd order, so it is conjugate to a subgroup of $A_2 A_3$. Since $A_2A_3$ is isomorphic to the affine group of $N$, it follows that $C_{A_2A_3} (\\rho (K))$ is conjugate to $A_2$, and we conclude that $\\rho (g)$ is conjugate to some element of $A_2$.\n\nLet $x \\in F$ be the element so that $\\rho (g)$ is conjugate to $\\phi_x$. Let $\\sigma$ be the generator for the Sylow $2$-subgroup of the Galois group of $F$. Let $P = A_1 \\langle \\phi_\\sigma \\rangle$, and we know that $P$ is a Sylow $2$-subgroup of ${\\rm Aut} (M)$. Since $\\sigma$ will normalize $\\langle x \\rangle$, it follows that $\\phi_\\sigma$ will normalize $\\langle \\phi_x \\rangle$. Hence, $\\phi_\\sigma$ will normalize $A_1 \\langle \\phi_x \\rangle$. Let $A = P \\langle \\phi_x \\rangle$. We conclude that $N_A (\\langle \\phi_x \\rangle) = C_{A_1} (\\phi_x) \\langle \\phi_x, \\phi_\\sigma \\rangle$.\n\nSuppose $C_{A_1} (\\rho (g)) = 1$, then $C_{A_1} (\\phi_x) = 1$. This implies that $N_A (\\langle \\phi_x \\rangle) = \\langle \\phi_x, \\phi_\\sigma \\rangle$. It follows that a Sylow $2$-subgroup of $N_{{\\rm Aut} (M)} (\\langle \\rho (g) \\rangle)$ is conjugate to a subgroup of $\\langle \\phi_\\sigma \\rangle$. We next observe that $N_{N_G (K)} (\\langle g \\rangle)$ is isomorphic to a subgroup of $N_{{\\rm Aut} (M)} (\\langle \\rho (g) \\rangle)$. Since $N_{N_G (K)} (\\langle g \\rangle)$ contains a nontrivial Sylow $2$-subgroup, it contains an element that $\\rho$ maps to a nontrivial power of $\\phi_\\sigma$. However, $\\phi_\\sigma$ centralizes elements of $M$ outside of $N$, and thus, $N_G (K)$ contains an element that centralizes elements of $M$ outside of $N$. This is a contradiction of Theorem \\ref{2pairclass}.\n\\end{proof}\n\n\\begin{corollary} \\label{nonex}\nLet $M$ be the Suzuki $2$-group $A (n,\\Theta)$ with notation as above. Let $(G,N)$ be as in the conclusion of Theorem \\ref{2pairclass}. Suppose $\\Theta$ has order $l$, $n = kl$, and $\\Theta (a) = a^{2^h}$ for all $a \\in F$. \nIf $d$ divides $n$ and does not divide $k$, then $C_G (K)$ does not contain a subgroup of order $2^d - 1$.\n\\end{corollary}\n\n\\begin{proof}\n\nSuppose now that $d$ divides $n$ and does not divide $k$. (We note that when $n = 6$, it suffices to assume that $d = 3$ since we must have $l = 3$ and $k = 2$ implies that $d = 3$ or $d = 6$. However, if we can prove that $C_G (K)$ has no element of order $2^3 - 1 = 7$, then it will contain no element of order $2^ - 1 = 63$.) We have ${\\rm gcd} (h,n) = k$, so $d$ does not divide $h$. Observe that $d$ is the order of $2$ modulo $2^d - 1$. It follows that $2^h \\cong 2^{h'} ~({\\rm mod}~2^d - 1)$ where $h'$ is the remainder upon dividing $h$ by $d$, and so $1 \\le h' < d$. Now, $2^{h'} + 1$ will not be congruent to $2^j$ modulo $2^d - 1$ for all integers $j$ with $0 \\le j < d$ since both $2^{h'} + 1$ and $2^j$ are less than $2^d - 1$ and not equal. We conclude that $2^h + 1$ is not congruent modulo $2^d - 1$ to any power of $2$.\n\nSuppose $C_G (K)$ contains an element $g$ of order $2^d - 1$. By Lemma \\ref{nontrivcent}, $\\rho (g)$ is conjugate to $\\phi_x$ for some element $x \\in F$ and $C_{A_1} (\\rho (g)) > 1$. (If $n = 6$, then assumption $d = 3$ implies that $g$ has order $7$, so the hypotheses of Lemma \\ref{nontrivcent} are met in this case.) Since $x$ must have order $2^d - 1$, we have by Corollary \\ref{numcond} that $C_{A_1} (\\phi_x) = 1$. Since $C_{A_1} (\\rho (g))$ and $C_{A_1} (\\phi_x)$ are congruent, this is a contradiction.\n\\end{proof}\n\nCombining Lemma \\ref{prime pow} with Theorem \\ref{2pairclass} and Corollary \\ref{nonex}, we obtain the conclusion. This is Theorem 2 when $p = 2$ and $G$ is solvable, and thus, it completes the proof of Theorem 2.\n\n\\begin{corollary}\nLet $(G,N)$ be a $2$-Gagola pair with $G$ solvable. Then $|G:N|_2 \\ge |N|^2$.\n\\end{corollary}\n\n\\begin{proof}\nIf we do not have the conclusion, then $(G,N)$ satisfies conclusion 2 of Theorem \\ref{2pairclass}. We use the notation from there. Let $l$ be the order of $\\Theta$. Let $p$ be a prime divisor of $l$, and take $d = n_p$. Thus, $d$ divides $n$, but $d$ will not divide $n\/l$. We apply Corollary \\ref{nonex} to see that $C_G (K)$ contains no element of order $2^d - 1$. On the other hand, we know by Lemma \\ref{aff} that $G\/{\\bf O}_2 (G)$ is isomorphic to a subgroup of $\\Gamma (N)$. Let $\\rho$ be the map from $G$ to $\\Gamma (N)$. It is not difficult to see $\\Gamma_o (N) \\le C_{\\Gamma (N)} (\\rho (K))$, so $\\rho (H) \\cap \\Gamma_o (N) \\le C_{\\rho (H)} (\\rho (K)) = \\rho (C_H (K))$. By Lemma \\ref{prime pow}, we know that $\\rho (H) \\cap \\Gamma_o (N)$ contains an element of order $2^d - 1$. This implies that $C_H (K) {\\bf O}_2 (G)\/{\\bf O}_2 (G) \\cong C_H (K)$ contains an element of order $2^d - 1$, and so, $C_G (K)$ contains an element of order $2^d - 1$. We now have a contradiction.\n\\end{proof}\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction} In this paper we consider graphs of girth at least $5$, i.e. graphs which have no 3-cycles or 4-cycles. In \\cite{Garnick93}, the maximum number of edges in graphs of girth at least $5$ with $v$ vertices, $f_4(v)$, were established for $v\\leq 30$, and $v=50$, and the number of unique (up to isomorphism) graphs with $f_4(v)$ edges, $F_4(v)$, was caculated for $v\\leq 21$ and $v=50$. Note that, once $f_4(v)$ has been established, it is straightforward to generate $F_4(v)$ for $n<20$ using nauty's geng tool \\cite{nauty}.In \\cite{comiprst1} and \\cite{comiprst2} we introduce two symmetry breaking constraints for combinatorial search and illustrated their effectiveness by applying it to the problem of finding graphs with girth at least $5$ (and other combinatorial problems). We were able to reproduce the results of \\cite{Garnick93} and find values $F_4(v)$ that were previously unpublished. Our symmetry breaking constraints alone were not sufficient to crack some of the harder cases. We were however able to reduce our search space by applying new theoretical results for cases $v>22$. In many cases we were able to identify non-trivial {\\it embedded stars} that must be present in a graph with the maximum number of edges and girth at least $5$, and fixing this embedded star made our search tractable. An outline of the proof of the existence of such stars is given in \\cite{comiprst2}, but for space reasons did not include full proofs. We give the full proofs in this paper. All of the graphs for $F_{4}(v)$, where $20\\leq v \\leq 32$ are presented as incidence arrays in our appendices and a full list in {\\it graph6} notation (as implemented in nauty \\cite{nauty}) can be obtained by contacting the authors directly. \n\n \n\n\\section{Preliminary Definitions and Results}\nIn this section we give some preliminary definitions and results that will be useful in this paper. \n\nThe girth of a graph $\\Gamma=(V,E)$ is the size of the smallest cycle contained in\nit. \nLet ${\\cal F}_k(v)$ denote the set of graphs with $v$ vertices and girth at\nleast $k+1$. Let $f_k(v)$ denote the maximum number of edges in a\ngraph in ${\\cal F}_k(v)$. A graph in ${\\cal F}_k(v)$ with $f_k(v)$ edges is\ncalled \\emph{extremal}. The number of non-isomorphic extremal graphs\nin ${\\cal F}_k(v)$ is denoted $F_k(v)$.\nExtremal graph problems involve discovering values of $f_k(v)$ and\n$F_k(v)$ and finding witnesses.\nIn~\\cite{AbajoD10} the authors attribute the discovery of values\n$f_4(v)$ for $v\\leq 24$ to \\cite{Garnick93} and for $25\\leq v\\leq 30$\nto \\cite{Garnick92}. In \\cite{Garnick93} the authors report values of\n$F_4(v)$ for $v\\leq 21$. In \\cite{Garnick93} and \\cite{WangDM01}\nalgorithms are applied to compute lower bounds on $f_4(v)$ for $31\\leq\nv\\leq 200$. Some of these lower bounds are improved in\n\\cite{AbajoBD10} and improved upper bounds for $33\\leq\nv\\leq 42$ are proved in \\cite{bong2017}.\nValues of $f_4(v)$ for $v\\leq 30$, and of $F_4(v)$ for $v\\leq 21$ are\navailable as sequences \\texttt{A006856} and \\texttt{A159847} of the\nOn-Line Encyclopedia of Integer Sequences~\\cite{oeis}. We have extended sequence \\texttt{A159847} to include values for $22\\leq v \\leq 32$, following the results presented in this paper.\n\nIn the remainder of this paper, as we are only interested in graphs with girth at least $5$, we abbreviate $f_4(v)$ and $F_4(v)$ to $f(v)$ and $F(v)$ respectively. \n\n\\begin{definition} A graph $\\Gamma=(V,E)$ is said to be {\\it extremal} if it has girth at least $5$ and any graph on $V$ vertices with more than $e=|E|$ edges has girth less than $5$. For any $v$, $f(v)$ and $F(v)$ denote the number of edges in an extremal graph on $v$ vertices and the number of unique extremal graphs on $v$ vertices respectively. For any extremal graph, $\\delta$ and $\\Delta$ denote the minimum and maximum degree respectively.\n\\end{definition}\n\nThe following lemma, and it's proof, is from \\cite{Garnick93}.\n\\begin{lemma}\\label{lem:garnickLemma}\nIf $\\Gamma$ is an extremal graph of order $v$, with $e=f(v)$ edges, whose minimum and maximum degrees are $\\delta$ and $\\Delta$ respectively, then \n$$e-f(v-1)\\leq \\delta \\leq \\sqrt(v-1)\\;{\\rm and}\\; \n\\lceil 2e\/v\\rceil\\leq \\Delta\\leq (v-1)\/\\delta$$\n\\end{lemma}\nValues of $f(v)$ for $22\\leq v \\leq 30$ can be found in \\cite{Garnick93}, and those for $v=31$ and $v=32$ are given in \\cite{comiprst1}. By applying Lemma \\ref{lem:garnickLemma}, and observing (by counting edges) that $\\delta=\\Delta$ is only possible when $v\\Delta = 2f(v)$ we obtain possible values of $(\\delta,\\Delta)$, for $22\\leq v \\leq 33$. These are given in Table \\ref{table:deltaTable}\n\n\\begin{table}\n\\begin{tabular}{l|l|l}\n$v$ & $f(v)$ & $(\\delta,\\Delta)$\\\\\n\\hline\n21&44&$(3,5)$, $(3,6)$, $(4,5)$\\\\\n22&47&$(3,5)$, $(3,6)$, $(3,7)$, $(4,5)$\\\\\n23&50&$(3,5)$, $(3,6)$, $(3,7)$, $(4,5)$\\\\\n24&54&$(4,5)$\\\\\n25&57&$(3,5)$, $(3,6)$, $(3,7)$, $(3,8)$, $(4,5)$, $(4,6)$\\\\\n26&61&$(4,5)$, $(4,6)$\\\\\n27&65&$(4,5)$, $(4,6)$\\\\\n28&68&$(3,5)$, $(3,6)$, $(3,7)$, $(3,8)$, $(3,9)$, $(4,5)$, $(4,6)$\\\\\n29&72&$(4,5)$, $(4,6)$, $(4,7)$\\\\\n30&76& $(4,6)$, $(4,7)$\\\\\n31&80& $(4,6)$, $(4,7)$, $(5,6)$\\\\\n32&85&$(5,6)$\\\\\n33&87&$(2,6)$, $(2,7)$, $(2,8)$, $(2,9)$, $(2,10)$, $(2,11)$, $(2,12)$,\\\\\n&& $(2,13)$, $(2,14)$, $(2,15)$, $(2,16)$, $(3,6)$, $(3,7)$, $(3,8)$,\\\\\n&& $(3,9)$, $(3,10)$, $(4,6)$, $(4,7)$, $(4,8)$, $(5,6)$\\\\ \n\\hline\n\\end{tabular}\n\\caption{Possible values of $(\\delta,\\Delta)$ for $22\\leq v \\leq 32$\\label{table:deltaTable}}\n\\end{table}\n\n\\begin{definition}\\label{def:embedded}\n If $\\Gamma$ is a graph of girth at least $5$ we say that $\\Gamma$\nhas a embedded $S_{D,[d_{0}-1,d_{1}-1, \\ldots, d_{D-1}-1]}$ star if $\\Gamma$ has a vertex of degree $D$ whose children have degrees at least $d_{0},d_{1},\\ldots,d_{D-1}$ respectively. \n\\end{definition}\nNote that the children and grandchildren of a vertex in a a graph of girth at least $5$ are distinct, and there are no edges between the children. If $\\Gamma$ is such a graph and has minimum and maximum degrees $\\delta$ and $\\Delta$, then any vertex $x$ of degree $\\Delta$ is the centre of a {\\it trivial} embedded \n$S_{\\Delta,[\\delta-1,\\delta-1, \\ldots, \\delta-1]}$ star. In order to reduce the number of isomorphic graphs produced, any combinatorial search for extremal graphs will assume a fixed position of such a star (for example, with central node as vertex $0$). The purpose of this paper is to show that in some cases an extremal graph must contain larger (non-trivial) embedded stars, in some cases containing all of the vertices in the graph. This allows us to fix such a star and reduce both our search space and the number of isomorphic graphs generated.\n\n\\begin{definition}\\label{def:sinkNode}\nA node $x$ of an extremal graph is said to be a {\\it sink node} if it has maximum degree and is the centre of an embedded star that contains all of the vertices (i.e. $x$ is at distance at most $2$ from all other vertices). \n\\end{definition}\n\nFrom a result in \\cite{Garnick93}, no two vertices in an extremal graph of girth at least $5$ are at distance greater than $3$ from each other. \n\n\\begin{definition}\\label{def:setsDifferent}\nFor a graph $\\Gamma=(V,E)$, for any $m>=2$, $S_{\\Gamma,m}$ is the set of sets of $m$ vertices that are at a distance of $3$ from each other. \n\\end{definition}\n\n\n\\begin{lemma}\\label{lem:sinknode} \nIf $\\Gamma$ is a graph of girth at least $5$ on $v$ vertices and contains a sink node $x$, then (i) $\\Gamma$ has an embedded $S_{\\Delta,[d_{0}-1,d_{1}-1,\\ldots, d_{\\Delta-1}-1]}$ star, where $\\Delta+1+\\Sigma_{i=0}^{\\Delta-1}(d_{i}-1) = v$. (ii) for $m\\geq 2$, no element of $S_{\\Gamma,m}$ contains $x$ or more than one child of $x$.\n\\end{lemma}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Follows immediately as $x$ is at distance at most $2$ from all other vertices and the children of $x$ are at distance $2$ from each other. \n\n\\begin{comment}\n\\begin{lemma}\\label{lem:deg4and5} If $\\Gamma$ is an extremal graph with at most $23$ vertices and $(\\delta,\\Delta)=(4,5)$ then any element of $S_{\\Gamma,4}$ contains only vertices of degree $4$.\n\\end{lemma}\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Suppose that $s\\in S_{\\Gamma,4}$ contains a vertex $x$ of degree $5$. Then $x$ is the centre of an embedded $S_{5,[3,3,3,3,3]}$ star, which contains $21$ vertices. Since the $3$ other elements of $s$ do not belong to this star, we have that $v\\geq 24$. \nThis is not even true!!\n\\end{comment}\n\n\n\n\n\\begin{lemma}\\label{lem:inductiveExtremal} If $\\Gamma=(V,E)$ is an extremal graph and $|V|=v$ with $\\delta(\\Gamma)=f(v)-f(v-1)$, then $\\Gamma=(V,E)$ is constructed from an extremal graph $\\Gamma^{\\prime}$ of order $v-1$ by adding a new vertex $x$ of degree $\\delta$ to $\\Gamma^{\\prime}$ and $\\delta$ edges from $x$ to a set $S_{\\Gamma^{\\prime},\\delta}$.\n\\end{lemma}\n\n\n\\begin{comment}\n$\\Delta(\\Gamma)\\leq \\Delta_{v-1}+1$. Suppose that all graphs $\\Gamma^{\\prime}=(V^{\\prime},E^{\\prime})\\in F(v-1)$ for which $S_{\\Gamma^{\\prime},\\delta}$ is non-empty have a sink vertex of degree $\\Delta_{v-1}$ with children of degrees $d_{0},d_{1},\\ldots, d_{\\Delta_{v-1}-1}$. Then $\\Gamma$ has an embedded $S_{\\Delta_{v-1},[d_{0}-1,d_{1}-1,\\ldots, d_{\\Delta_{v-1}-1}-1]}$ star. \n\\begin{itemize}\n\\item If, for every $\\Gamma^{\\prime}\\in F(v-1)$ for which $S_{\\Gamma^{\\prime},\\delta}$ is non-empty, every element of $S_{\\Gamma^{\\prime},\\delta}$ contains only vertices of degree less than $\\Delta_{v-1}$ then $\\Delta(\\Gamma)=\\Delta_{v-1}$. \n\\item If, for every $\\Gamma^{\\prime}$ for which $S_{\\Gamma^{\\prime},\\delta}$ is non-empty, every element of $S_{\\Gamma^{\\prime},\\delta}$ contains a child of a sink vertex, then $\\Gamma$ contains a sink vertex of degree $\\Delta_{v-1}$ with children of degrees $d_{0},\\ldots, d_{i-1}, d_{i}+1,d_{i+1}, \\ldots, d_{\\Delta_{v-1}-1}$, for some $1\\leq i \\leq \\Delta_{v-1}$.\n\\end{itemize}\n\\end{lemma}\n\\end{comment}\n\n\n\\begin{theorem} For each $20\\leq v \\leq 32$, an extremal graph $\\Gamma$ must have $(\\delta,\\Delta)$ taking one of the pairs of values indicated in Table \\ref{embeddedStarTable}. In each case, we show the number of edges ($f(v)$), the largest star known to be embedded in $\\Gamma$, the number of distinct graphs $(F(v))$, and the method of proof. In all cases Lemma \\ref{lem:inductiveExtremal} is employed. The ``method'' column indicates whether the graphs were obtained from \\cite{Garnick93} (G) or, if not, whether search is employed (based on the proven existence of embedded stars) (S), and whether some (or all) graphs are constructed by hand (H). Note that $S,H$ denotes that some of the graphs were found using search, and some by hand.\n\\end{theorem}\n\n\\begin{table}[t]\n\\label{embeddedStarTable}\n\\begin{center}\n\\setlength{\\tabcolsep}{3pt}\n\\begin{tabular}{|c|l|l|l|l|l|}\n\\hline\nv&$f(v)$&$(\\delta,\\Delta)$&Embedded star&$F(v)$&method\\\\\n\\hline\n20&$41$&$(3,5)$&$S_{5,[3,3,3,3,2]}$&1&G\\\\\n\\hline\n21&$44$&$(3,5)$&$S_{5,[3,3,3,3,3]}$&3&H\\\\\n & &$(4,5)$&$S_{5,[3,3,3,3,3]}$&&\\\\\n\\hline\n22&$47$&$(3,5)$&$S_{5,[4,3,3,3,3]}$&3&S\\\\\n & &$(4,5)$&$S_{5,[4,3,3,3,3]}$&&\\\\\n\\hline\n23&$50$&$(3,5)$&$S_{5,[4,4,3,3,3]}$&7&S\\\\\n& &$(4,5)$&$S_{5,[4,4,3,3,3]}$ or $S_{5,[4,3,3,3,3]}$&&\\\\\n\\hline\n24&$54$&$(4,5)$&$S_{5,[4,4,4,3,3]}$&1&S\\\\\n\\hline\n25&$57$&$(3,5)$&$S_{5,[4,4,4,4,3]}$ or $S_{5,[4,4,4,3,3]}$&6&S\\\\\n& & $(4,5)$& $S_{5,[4,4,4,4,3]}$ or $S_{5,[4,4,4,3,3]}$&&\\\\\n& & $(4,6)$&$S_{6,[3,3,3,3,3,3]}$&&\\\\\n\\hline\n26& $61$&$(4,5)$&$S_{5,[4,4,4,4,4]}$&2&H\\\\\n\\hline\n27&$65$& $(4,5)$&$S_{5,[4,4,4,4,4]}$&1&H\\\\\n\\hline\n28&$68$& $(3,6)$ & $S_{6,[4,4,4,4,3,2]}$&4&S,H\\\\ \n&& $(4,5)$ & $S_{5,[4,4,4,4,4]}$&&\\\\\n&& $(4,6)$ & $S_{6,[4,4,4,3,3,3]}$&&\\\\\n\\hline\n29&$72$&$(4,6)$ & $S_{6,[4,4,4,4,3,3]}$&1&H\\\\\n\\hline\n30&$76$&$(4,6)$ & $S_{6,[4,4,4,4,4,3]}$ and $S_{6,[5,4,4,4,3,3]}$&1&H\\\\\n\\hline\n31&$80$&$(4,6)$ & $S_{6,[4,4,4,4,4,4]}$ and $S_{6,[5,4,4,4,4,3]}$&2&H,S\\\\\n&&$(5,6)$ & $S_{6,[4,4,4,4,4,4]}$&&\\\\\n\\hline\n32&$85$&$(5,6)$&$S_{6,[5,4,4,4,4,4]}$&1&H\\\\\n\\hline\n\n\\end{tabular}\n\\end{center}\n\\caption{Embedded stars\\label{table:thetable} for \n $20 \\leq v \\leq 25$, and $v=32$}\n\\end{table}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Follows from Theorems \\ref{theorem:g20}, \\ref{theorem:g21}, \\ref{theorem:g22}, \\ref{theorem:g23}, \\ref{theorem:g24}, \\ref{theorem:g25}, \\ref{theorem:g26}, \\ref{theorem:g27}, \\ref{theorem:g28}, \\ref{theorem:g29}, \\ref{theorem:g30}, \\ref{theorem:g31} and \\ref{theorem:g32}. \n\n\n\n\\begin{definition}\\label{defn:linearspace}\nA linear space $\\Lambda$ on a set of $n$ points $V$ is a collection $B=\\{B_{1},\\ldots,B_{b}\\}$ of subsets of $V$ called blocks, such that every block has at least two points and each pair of points is in exactly one block. A prelinear space $\\Lambda$ is a set of blocks on $V$ in which every block has at least two points and each pair of points is in at most one block.\n\\end{definition}\n\n\\vspace{.25cm}\n\\noindent\nAny prelinear space can be extended to a linear space, by adding suitable blocks of size $2$, so we tend to use the term linear space and prelinear space interchangeably. A linear space with no blocks of size $2$ is called a {\\it proper} linear space. \n\n\\begin{comment}\n\\vspace{.25cm}\n\\noindent\n{\\bf Example 1:} The $5$ linear spaces on five points are shown graphically in Figure \\ref{linearspaces5points}. Blocks are represented as lines. They can also be represented as sets of tuples thus (note that the blocks of size $2$ are omitted):\n$\\Delta_{1}=\\{(0,1,2,3,4)\\}$, $\\Delta_{2}=\\{(0,1,2,3)\\}$, $\\Delta_{3}=\\{(0,1,2),(2,3,4)\\}$ $\\Delta_{4}=\\{(2,3,4)\\}$ and $\\Delta_{5}=\\{\\}$. We have arbitrarily labelled the points. A different labelling would lead to an isomorphic linear space.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{linearSpacesSmall}\n\\caption{Linear spaces on $5$ points\n\\label{linearspaces5points}}\n\\end{figure}\n\n\\vspace{.25cm}\n\\noindent\nOften (especially when there are several blocks of size greater than $2$) blocks are drawn vertically. For example, in consider a linear space $\\Delta$ on $16$ points whose blocks (of size $>2$) are $(0,1,2,3,4)$, $(0,5,6,7,8)$, $(0,9,10,11,12)$, $(1,5,9,13,14)$, $(2,6,10,13,15)$ and $(3,7,11,14,15)$. These blocks can be represented vertically as follows:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tabular}{cccccc}\n0&0&0&1&2&3\\\\\n1&5&9&5&6&7\\\\\n2&6&10&9&10&11\\\\\n3&7&11&13&13&14\\\\\n4&8&12&14&15&15\\\\\n\\end{tabular}\n\n\n\\vspace{.25cm}\n\\noindent\nA shorthand to represent a linear space $\\Delta$ on $v$ points containing $r_{i}$ blocks of size $i$, for $2\\leq i \\leq v$ is to say that $\\Delta$ is a $2^{r_{2}}3^{r_{3}}\\ldots v^{r_{v}}$ design (where the $i$th term is omitted if $r_{i}=0$, and we often omit the blocks of size $2$). The numbers of linear spaces on $n$ points, for $n\\leq 12$, for each possible profile, are given in \\cite{bettenbetten4}. \n\\end{comment}\n\n\\begin{lemma}\\label{lem:blocks} Let $\\Gamma=(V,E)$ be a graph of girth at least $5$ and $$V_{1} = \\{v_{0},v_{2},\\ldots, v_{m-1}\\}$$ a subset of $V$. If, for all $v_{i}\\in V_{1}$, $b_{i}$ is the set of elements of $V\\setminus V_{1}$ that are adjacent to $v_{i}$, then $B=\\{b_{i} : 0\\leq i \\leq m-1\\;{\\rm and}\\; |b_{i}|\\geq 2\\}$ is a (pre) linear space on $V\\setminus V_{1}$ and no set $b_{i}$ contains two vertices that are adjacent in $\\Gamma$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} No pair of elements in $V\\setminus V_{1}$ can appear in more than one set $b_{i}$ as $\\Gamma$ contains no $4$-cycles, so $B$ is a (pre) linear space. Similarly, no set $b_{i}$ contains two vertices that are adjacent in $\\Gamma$, as $\\Gamma$ contains no $3$-cycles. \n\n\\vspace{.25cm}\n\\noindent\nNote that in the context of linear spaces, we generally refer to the elements of the blocks as {\\it points}, whereas in the context of graphs the elements of edges are vertices. When applying Lemma \\ref{lem:blocks} this distinction is less clear. \n\n\n\\begin{comment}\n\\begin{lemma} Suppose that $\\Gamma$ is a triangle and square-free graph on $v$ vertices, and for any $x\\in V(\\Gamma)$ let $\\Pi(x)$ denote the set of vertices that are at a distance of at least two from $x$. If $x$ and $y$ are adjacent vertices in $\\Gamma$, then \n\\end{comment}\n\n\\begin{comment}\n\\begin{lemma}\\label{lemma:15.0} Let $\\Lambda$ be a $4^{8}$ on $15$ points in which the blocks can be arranged into $4$ parallel pairs. Then\n\\begin{enumerate}\n\\item At most $2$ points are in $4$ blocks.\n\\item If a point is in $0$ blocks, and at least one point is in $4$ blocks, then $2$ points are in $4$ blocks.\n\\item If a point is in $4$ blocks then at most $4$ other points are in $3$ blocks.\n\\item If $2$ points, $x$ and $y$ are in $4$ blocks, and two points $A$ and $B$ are in $1$, then all blocks contain either $x$ or $y$, and $A$ and $B$ are in different blocks, in different parallel pairs of blocks, one with $x$ and one with $y$.\n\\end{enumerate}\n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof}\n(1) If $x$ and $y$ are in $4$ blocks, then at most one block does not contain $x$ or $y$. So any other point can be in at most $3$ blocks. (2) Suppose that $x$ is in $4$ blocks, one from each parallel pair, and a point is in no blocks. There is one point, $y$ say, that is not in a block with $x$, and must be in all of the $4$ remaining blocks. (3) Let $x$ be as above and let $y$ and $z$ be the two points that are not in a block with $x$. Suppose there are $5$ points in $3$ blocks. At most one point from every block on $x$ can be in $3$ blocks and at most one from $\\{y,z\\}$. So w.l.o.g. one point from each block on $x$, and $y$ are in $3$ blocks. Each of the former is in $b$, the one block that does not contain $x$ or $y$. But one of the points in a block with $x$ that is in $3$ blocks is in the block on $x$ that is parallel to $b$, which is a contradiction. (4) Suppose that $x$ and $y$ are in a common block, and that the block parallel\nto the block containing $x$ and $y$ is $b$. Since the points in the block with $x$ and $y$ can be in no more blocks they must be $A$ and $B$. Let $b^{\\prime}$ with any block that is not in the two already mentioned, and suppose $b^{\\prime}$ contains $x$. Then the three other elements of $b^{\\prime}$ are each in a distinct other block, either a block on $y$ (not parallel to $b^{\\prime}$) or $b$. It follows that every other block has an element that is in $b$. Since there are $6$ such blocks, this is impossible. So $x$ and $y$ are not in a block together, and every parallel pair of blocks consists of a block on $x$ and a block on $y$. A similar argument shows that $A$ and $B$ are in different blocks, in different parallel pairs, and one on $x$ and one on $y$.\n\\end{comment}\n\n\\begin{comment}\n\\begin{lemma}\\label{blocks15} If $\\Lambda$ is a $4^{6}5^{2}$ on $15$ points in which the blocks of size $4$ can be arranged into $3$ parallel pairs, then either $0$ or $2$ points are in $4$ blocks. If $2$ points are in $4$ blocks then all other points are in $2$. \n\\end{lemma}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Any point $x$ that is in $4$ blocks must either be in two blocks of size $4$ and two of size $5$, or in $3$ of size $4$ and one of size $5$. In the first case, every block of size $4$ must intersect all of the blocks containing $x$, and so there is no block parallel to any of the blocks containing $x$. So $x$ is in $3$ blocks of size $4$ and one of size $5$. The $3$ remaining blocks of size $4$ must intersect all but one of the blocks on $x$, and contain the single point $y$ not contained in the blocks on $x$. The remaining block of size $5$ intersects all of the blocks on $x$ and also contains $y$. Since all blocks contain $x$ or $y$, no other point is in more than $2$ blocks. It follows that every other point is in exactly $2$ blocks. \n\\end{comment}\n\n\\begin{definition}\\label{definition:packing} A $(v,k,1)$-packing design is a linear space on a set $V$ where the blocks all have size $k$. For any $v,k$ function $P(k,v)$ is called the maximum packing number of $k$ and $v$, and is the maximum number of blocks in such a packing design.\n\\end{definition}\nThe following result follows from a simple counting argument (see \\cite{johnson1}):\n\\begin{lemma} For any $k\\leq v$, \n$$P(k,v)\\leq \\lfloor v\\lfloor(v-1)\/(k-1)\\rfloor\/k\\rfloor$$\n\\end{lemma}\nThe upper bound is not achievable in some cases. Absolute values for $P(k,v)$ for $3\\leq k\\leq 6$ and $1\\leq v\\leq 28$ are given in Table \\ref{table:packingTable}. These are obtained from \\cite{spence}, \\cite{brslsm}, and theoretical results in \\cite{brouwer1}.\n\n\\begin{table}\n\\begin{tabular}{|l|l|l|l|l|}\n\\hline\n$v$ & $P(3,v)$ &$P(4,v)$ & $P(5,v)$ & $P(6,v)$\\\\\n\\hline\n4&1&1&0&0\\\\\n5&2&1&1&0\\\\\n6&4&1&1&1\\\\\n7&7&2&1&1\\\\\n8&8&2&1&1\\\\\n9&12&3&2&1\\\\\n10&13&5&2&1\\\\\n11&17&6&2&2\\\\\n12&20&9&3&2\\\\\n13&26&13&3&2\\\\\n14&28&14&4&2\\\\\n15&35&15&6&3\\\\\n16&37&20&6&3\\\\\n17&44&20&7&3\\\\\n18&48&22&9&4\\\\\n19&57&25&12&4\\\\\n20&60&30&16&5\\\\\n21&70&31&21&7\\\\\n22&73&37&21&7\\\\\n23&83&40&23&8\\\\\n24&88&42&24&9\\\\\n25&100&50&30&10\\\\\n26&104&52&30&13\\\\\n27&117&54&30&14\\\\\n28&121&63&33&16\\\\\n\\hline\n\\end{tabular}\n\\caption{Maximum packing numbers for $k=4$, $k=5$ and $k=6$\\label{table:packingTable}}\n\\end{table}\nThe following Lemma is a direct consequence of the definition of $P(k,v)$:\n\n\\begin{lemma}\\label{lemma:packingLemma} No linear space on $v$ points contains more than $P(k,v)$ blocks of size $k$. \n\\end{lemma}\n\nIn the lemma below, the {\\it regularity} of a point $p$ ($reg_{p}$) in a linear space is the number of blocks that contain it and, for $i>0$, $deg_{i}$ is the number of points that have regularity $i$. Note that, since all the other points in the blocks containing $p$ must be distinct, if all blocks have size $r$, then $reg_{p}\\leq (v-1)\/(r-1)$. \n\n\\begin{lemma}\\label{lemma:blockWeight} \nIf $\\Lambda=(V,B)$ is a linear space and, for any block $b\\in B$, the weight of $b$, $wt(b)$, is the sum of the regularities of the points in $b$, then (i) block $b$ intersects $wt(b)-|b|$ other blocks, and no block can have weight more than $|B|+|b|-1$ and, (ii) $\\Sigma_{b\\in B}wt(b) = \\Sigma_{i>0} i^{2}deg_{i}$.\n\\end{lemma}\n\n\\begin{comment}\n(ii) remember we are adding the weights, not adding the number of points, so this does work.\nIf we were adding the points in the blocks, then sum(b) = sum (i.deg(i))\n- a different result!\n\nOnce we get to $\\Sigma_{b\\in B}wt(b) = \\Sigma_{p} (reg_{p}^{2})$, expand RHS as\n(1^{2} + 1^{2} + ... + 1^{2}) (deg1 times) + (2^{2}+2^{2} + ... + 2^{2})(deg 2 times) ..... + (v^{2} + ...) (deg v times)\nwhich gives the result\n\\end{comment}\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} (i) Any point $p$ in $b$ with regularity $reg_{p}$ intersects $r_{p}-1$ other blocks, and no other block contains more than one element of $b$, so $b$ intersects $\\Sigma_{p\\in b}reg_{p}-1$ other blocks, and the result follows. (ii) If we sum the weights of the blocks, for every vertex $p$, $reg_{p}$ is counted $reg_{p}$ times. So $\\Sigma_{b\\in B}wt(b) = \\Sigma_{p} (reg_{p}^{2})$. For every $i>0$ there are $deg_{i}$ vertices of regularity $i$, so $\\Sigma_{b\\in B}wt(b) = \\Sigma_{i>0}deg_{i}.i^{2}$, as required. \n\n\\begin{lemma}\\label{lemma:12.1} If $\\Lambda$ is a linear space on $12$ points with $8$ blocks of size $4$, then all of the blocks intersect.\n\\end{lemma}\n\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} There are no points of regularity greater than $3$ (or the blocks containing such a point would contain $13$ distinct points), and at most $4$ of regularity less than $3$ (if there are $5$, then $22$ spaces must be filled by $7$ points, implying at least one point of regularity greater than $3$). Suppose that there are two non-intersecting blocks, then they must each have weight at most $10$ and each contain a pair of vertices of regularity at most $2$ (from Lemma \\ref{lemma:blockWeight}). It follows that $(deg_{2},deg_{3})=(4,8)$. But then two of the blocks contain only points of regularity $3$ and have weight $12$ which contradicts Lemma \\ref{lemma:blockWeight}.\n\n\\begin{comment}\n Let $V=V_{1}\\cup V_{2}$ where $V_{1}$ is the set of vertices that are in these blocks, and $V_{2}$ the remaining vertices. Then all other blocks contain a pair $V_{1}$ and a pair from $V_{2}$. Since no element of $V_{1}$ can be in more than $2$ of the remaining blocks, $4$ elements of $V_{1}$ are in $2$ of the remaining blocks. But there are only three pairs of remaining blocks that do not intersect. Hence two elements of $V_{1}$ are in two of the same remaining blocks, which is impossible. \n\\end{comment}\n\n\\begin{lemma}\\label{lemma:13.1} If $\\Lambda$ is a linear space on $13$ points with $9$ blocks of size $4$, then (1) there is no set of $3$ non-intersecting blocks. (2) There are at most $3$ pairs of non-intersecting blocks.\n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} \nFirst observe that if there is a set of $3$ parallel blocks or more than $1$ pair of parallel blocks then no point is in at most $1$ block, or removing the point leaves a linear space on at most $12$ points with $8$ blocks of size at least $4$ with a non-intersecting pair of blocks. There is no linear space on fewer than $12$ points with $8$ blocks of size at least $4$ \\cite{bettenbetten4}, and if $v=12$ we have a contradiction to Lemma \\ref{lemma:12.1}. So every point is in between $2$ and $4$ blocks and $(deg_{2},deg_{3},deg_{4})= (3,10,0)$, $(4,8,1)$, $(5,6,2)$ or $(6,4,3)$. By Lemma \\ref{lemma:blockWeight} no block can have weight more than $12$ and $\\Sigma_{b\\in B}wt(b) = 4deg_{2}+9deg_{3}+16deg_{4}$. \n\n\\vspace{.2cm}\n\\noindent\n(1) If there is a set of $3$ non-intersecting blocks, then there are $3$ blocks of weight at most $10$, and so $\\Sigma_{b\\in B}wt(b) \\leq 102$. This is only possible if $(deg_{2},deg_{3},deg_{4})= (3,10,0)$. But the set of $3$ parallel blocks must each contain two of the vertices of degree $2$, and so must intersect, which is a contradiction. \n(2) If there are $4$ pairs of non-intersecting blocks then there are $8$ blocks with weight at most $11$ and $\\Sigma_{b\\in B}wt(b) \\leq 100$, which is impossible. \n\n\\begin{lemma}\\label{lemma:16.1}\nIf $\\Lambda$ is a linear space on at most $16$ points with $6$ blocks of size $5$ then either $(deg_{1},deg_{2},deg_{3})=(3,12,1)$, $deg_{2}=15$, or $(deg_{1},deg_{2})=(2,14)$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} No point can be in $4$ blocks. If $p$ has regularity $3$ then for every block containing $p$ there is a point of regularity $1$ (as all blocks have weight at most $10$, by Lemma \\ref{lemma:blockWeight}). So we must solve the following set of equations\/inequalities:\n$deg_{1}+2deg_{2}+3deg_{3}=30$; $deg_{1}+deg_{2}+deg_{3}\\leq 16$, and $deg_{1}\\geq 3deg_{3}$. There are $3$ solutions, giving the result required.\n\n\\begin{comment}\n\\begin{lemma}\n\\label{lemma:10blocksSize6}\nThere is no linear space on $22$ points that has $10$ blocks of size $6$.\n\\end{lemma}\n\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} \nSuppose that $B_{6}$ is a set of $10$ blocks of size $6$ on $22$ points. \nNo point can be in $4$ blocks of $B_{6}$, and at least one point is in $3$ blocks of $B_{6}$ (as there are $60$ spaces in the blocks in $B_{6}$). Let $x$ be a point that is in $3$ blocks in $B_{6}$. These blocks miss exactly $6$ points. Let $V^{\\prime}$ denote this set of $6$ points. The remaining $7$ blocks of $B_{6}$ can contain at most one point from each of the blocks in $B_{6}$ containing $x$ and so must contain at least $3$ of $V^{\\prime}$. Hence, restricting the blocks in $B_{6}$ to $V^{\\prime}$ there is a linear space with $r_{i}$ blocks of size $i$, for $3\\leq i \\leq 6$ on $6$ points, where $r_{3}+r_{4}+r_{5}+r_{6}=7$. Counting the pairs from $V^{\\prime}$ in this linear space, we have that $3r_{3}+6r_{4}+10r_{5}+15r_{6}\\leq 15$. There is no solution.\n\\end{comment}\n\n\\begin{comment}\n\\begin{lemma}\nThe connected, acyclic graphs of order greater than $1$ and less than $7$ are given in Figure \\ref{fig:acyclic}.\n\n\\begin{figure}\n\\label{fig:acyclic}\n\\centering\n\\includegraphics[width=1\\textwidth]{smallAcyclic}\n\\caption{The connected acyclic graphs of order greater than $1$ and at most $7$}\n\\end{figure}\n\n\\end{lemma}\n\n\n\\begin{lemma}\\label{lemma:degs4&5} If $\\Gamma$ is a connected acyclic graph of order greater than $1$ and at most $7$, and where no vertex has degree more than $4$, then (i) \n$\\Gamma$ has a pair of adjacent vertices with combined degree less than $4$, or has a vertex $x$ for which \n$2(4-deg(x))+\\Sigma_{y\\in N(x)}(4-deg(y)) > 8$, where $N(x)$ is the neighbourhood of $x$.\n\\end{lemma}\n\n\n\\noindent\n{\\bf Proof} See Figure \\ref{fig:acyclic}.\n\\end{comment}\n\n\\begin{comment}\n\\begin{lemma}\\label{lem:connectedComponents} Let $\\Gamma$ be a graph with $v$ vertices and $E$ edges, where $E\\leq v$, and where $\\Gamma$ contains a cycle of length $\\mu > v\/2$ and no smaller cycle. Then (i) any vertex not in the cycle is in at most one edge with vertices that are. (ii) $\\Gamma$ has $r$ acyclic components, where $v-E\\leq r \\leq v-E+1$.\n\\end{lemma}\n\n \\noindent\n{\\bf Proof} (i) Let $S$ be the set of vertices not in the cycle. Any vertex in $S$ that is in two edges with vertices in the cycle are in a cycle of length at most $\\mu\/2+1$, which is a contradiction. (ii) Set $S$ is composed of $2$ sets: $S_{1}$, the set of vertices that are connected via a path to the cycle, and $S_{2}$, those that are not. Set $S_{2}$ contains $r$ acyclic components (there are no cycles smaller than $\\mu$), and $|S_{2}|-r$ edges. Remaining edges are between elements of $S_{1}$, or between elements of $S_{1}$ and the cycle. There are $E-\\mu-|S_{2}|+r$ of these. Either these edges contain a path of length at least $\\mu\/2$ connected to two vertices in the cycle, at distance $\\mu\/2$ apart (there is at most one such path), in which case there are $|S_{1}|+1$ such edges, or there is no such path and there are $|S_{1}|$ such edges. Since $|S_{1}|=v-\\mu-|S_{2}|$ the result follows.\n\\end{comment}\n\n\\begin{comment}\n\\begin{lemma}\\label{lemma:extremalAcyclic} If $\\Gamma$ is an extremal graph on $v$ vertices, where $v=22$ or $23$, and all vertices have degree at least $4$ there are no cycles amongst the vertices of degree $4$ of length less than $n$, where $n=7$ if $v=22$ and $n=8$ if $v=23$. In each case, if there is a cycle of length $n$ or $n+1$, then the graph $\\Gamma^{\\prime}$ formed by removing the vertices of the cycle and the edges on them is extremal.\n\\end{lemma}\n\n\n\\noindent\n{\\bf Proof} Let $\\Gamma_{4}$ denote the graph on the vertices of degree $4$. Clearly, as $\\Gamma$ is triangle and diamond free, it has no cycles of length less than $5$. Suppose that there is a cycle of length $n$, where $n\\geq 5$. Then the subgraph $\\Gamma^{\\prime}$ formed by removing the vertices of the cycle and the edges on them has $v-n$ vertices, and $f(v)-3n$ edges. If $v=22$, $f(v)-3n>f(v-n)$ if $n<7$, so $\\Gamma^{\\prime}$ is not triangle and square free unless $n<7$. If $n=7$ or $n=8$ then $f(v)-3n=f(v-n)$ and so $\\Gamma^{\\prime}$ is extremal. The proof for $v=23$ is similar. \n\\end{comment}\n\n\\vspace{.25cm}\n\\noindent\nIn the remainder of this paper we present all extremal graphs of orders $20\\leq v \\leq 32$. In all cases, for all graphs $\\Gamma$ we provide the adjacency matrix, degree sequence, edge set, relevant sets $S_{\\Gamma,n}$, and any other relevant material one of the appendices (A-M) to this paper.\n\n\n\n\\section{Case $v=20$}\n\n\\begin{theorem}\\label{theorem:g20} If $\\Gamma$ is a graph of girth at least $5$ on $20$ vertices with $41$ edges, then $(\\delta,\\Delta)=(3,5)$ and $\\Gamma$ has an embedded $S_{5,[3,3,3,3,2]}$ star.\n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} There is precisely one graph on $20$ vertices with $41$ edges. This is stated in \\cite{Garnick93} and the graph is given in \\cite{Garnick92}. This graph is given in Figure \\ref{fig:extremal20} (the numbering of the vertices is our own). There are $3$ embedded $S_{5,[3,3,3,3,2]}$ stars. Note that the sink nodes corresponding to these stars are precisely the vertices of degree 5, namely $0$, $1$ and $2$.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.5\\textwidth]{g20.jpg}\n\\caption{Extremal graph on $20$ vertices with $41$ edges\n\\label{fig:extremal20}}\n\\end{figure}\n\n\n\\begin{lemma}\\label{lemma:g20}\nIf $\\Gamma$ is a graph of girth at least $5$ on $20$ vertices with $41$ edges, then $(deg_{3},deg_{4},deg_{5}) =(1,16,3)$ and every element of $S_{\\Gamma,3}$ contains $2$ vertices of degree $4$ and a child of a sink node, namely the single vertex of degree $3$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\nThis graph is given in Figure \\ref{fig:extremal20}, and is represented as an embedded star with sink node $0$ in Figure \\ref{fig:extremal20A}. \n\\begin{comment}\n(i)\nThe set of edges is:\n\\begin{eqnarray*}\n&&(0,4),(0,7),(0,10),(0,13),(0,19),\n(1,5),(1,8),(1,11),(1,14),\n(1,19),\\\\\n&&(2,3),(2,6),(2,9),\n(2,12)),(2,19),(3,4),\n(3,14),(3,15),\n(4,5),(4,18),\\\\&&(5,6),\n(5,17),(6,7),(6,16),\n(7,8),(7,15),\n(8,9),(8,18),(9,10),\n(9,17),\\\\&&(10,11),(10,16),\n(11,12),(11,15),\n(12,13),(12,18),\n(13,14),(13,17),\\\\&& (14,16),(15,17),(16,18)\n\\end{eqnarray*}\n\\end{comment}\nNote that $$S_{\\Gamma, 3}=\\{\\{15,16,19\\}, \\{15,18,19\\}, \\{16,17,19\\}, \\{17,18,19\\}\\}$$\n Since vertices $15,16,17,18$ all have degree $4$, and $19$ is a child of all roots (i.e. $0$, $1$ and $2$), of degree $3$, we are done. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.8\\textwidth]{newstar20}\n\\caption{Extremal graph on $20$ vertices with $41$ edges, as a star with root $0$\\label{fig:extremal20A}}\n\\end{figure}\n\n\\section{Case $v=21$}\n\n\n\n\\begin{theorem}\\label{theorem:g21} If $\\Gamma$ is a graph of girth at least $5$ on $21$ vertices with $44$ edges, then either $(\\delta,\\Delta)=(3,5)$ or $(\\delta,\\Delta)=(4,5)$. In either case, $\\Gamma$ has an embedded $S_{5,[3,3,3,3,3]}$ star.\n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Let $\\Gamma$ be a graph of girth at least $5$ with $21$ vertices and $44$ edges. Then $\\Gamma$ is extremal and, by Table \\ref{table:deltaTable} it follows that the possible values for the minimum and maximum degrees, $\\delta$ and $\\Delta$ are $(\\delta,\\Delta)=(3,5)$, $(3,6)$, and $(4,5)$.\n\nSuppose that there is a vertex of degree $3$, $x$ say. Since $\\delta=f(21)-f(20)$, by Lemma \\ref{lem:inductiveExtremal}, $\\Gamma$ can be constructed from the extremal graph on $20$ vertices described in Lemma \\ref{lemma:g20}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $3$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},3}$. Since no element of $S_{\\Gamma^{\\prime},3}$ contains vertices of degree $5$, it follows that $\\Delta=5$. So we have $(\\delta,\\Delta)=(3,5)$. Since, by Lemma \\ref{lemma:g20} every element of $S_{\\Gamma^{\\prime},3}$ contains single vertex of $\\Gamma^{\\prime}$ of degree $3$ that is a child of every sink node, it follows that \n every sink node of $\\Gamma^{\\prime}$ is a sink node of $\\Gamma$\n of degree $5$ whose children all have degree $4$. Hence there is an embedded $S_{5,[3,3,3,3,3]}$ star. \n\nIf $(\\delta,\\Delta)=(4,5)$ then any node $x$ of degree $5$ must only have children of degree $4$ and the result follows. \n\n\n\\begin{lemma}\\label{lemma:g21}\nIf $\\Gamma$ is a graph on $21$ vertices of girth at least $5$ with $44$ edges, then one of the following holds: (i) $(deg_{3},deg_{4},deg_{5})=(1,15,5)$\n and $\\Gamma$ has $3$ embedded $S_{5,[3,3,3,3,3]}$ stars and is isomorphic to the graph shown in Figure \\ref{fig:extremal21A}. In this case every element of $S_{\\Gamma,3}$ contains $3$ vertices of degree less than $5$, and the child of a root of degree $4$. \n(ii) $(deg_{3},deg_{4},deg_{5})=(0,17,4)$, all vertices of degree $5$ are sink nodes and $\\Gamma$ is isomorphic to one of the graphs shown in Figure \\ref{fig:extremal21B}. In both cases $S_{\\Gamma,3}$ is empty. \n\\end{lemma}\n\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} From the proof of Theorem \\ref{theorem:g21}, $(\\delta,\\Delta)=(3,5)$ or $(4,5)$. If $(\\delta,\\Delta)=(3,5)$ then $(deg_{3},deg_{4},deg_{5})=(1,15,5)$ and $\\Gamma$ is obtained from the extremal graph $\\Gamma^{\\prime}$ on $20$ vertices shown in Figure \\ref{fig:extremal20}, by adding an additional vertex $x$ and edges from $x$ to one of the sets $S_{\\Gamma^{\\prime},3}$ identified in the proof of Lemma \\ref{lemma:g20}. \nUsing nauty \\cite{nauty}, it can be shown that all graphs constructed in this way are isomorphic to the graph shown in Figure \\ref{fig:extremal21A} (note that in the figure (and in Figure \\ref{fig:extremal21B}) neighbours of leaf nodes are indicated as sets). \n In this case the sink nodes are $0$, $1$ and $2$ and $S_{\\Gamma,3}=\\{ \\{4,9,20\\}, \\{5,12,20\\}, (8,13,20\\}, \\{17,18,19\\}\\}$. The result follows. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.6\\textwidth]{star21A}\n\\caption{Extremal graph on $21$ vertices with $44$ edges and $(\\delta,\\Delta)=(3,5)$, as a star with root $0$\\label{fig:extremal21A}.}\n\\end{figure}\n\n\nSuppose then that there is no vertex of degree $3$. Then $(\\delta,\\Delta)=(4,5)$ and $(deg_{4},deg_{5})=(17,4)$. All vertices of degree $5$ are sink nodes. By \\cite{Garnick93}, we know that there are $3$ extremal graphs $(21,44)$, so there are two graphs of this type. They must be isomorphic to the two graphs shown in Figure \\ref{fig:extremal21B}, which have been shown to be non-isomorphic. Note that graph (b) can be obtained from graph (a) by replacing edge $(8,20)$ with edge $(11,20)$ and in graph (a), the $3$ grandchildren of degree $5$ have the same parent, and in graph (b) two of the grandchildren of degree $5$ have the same parent, and the other does not. In both cases $S_{\\Gamma,3}$ is empty. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width=1.0\\textwidth]{star21Bcombined}\n\\caption{Extremal graphs on $21$ vertices with $44$ edges and $(\\delta,\\Delta)=(4,5)$, as stars with root $0$\\label{fig:extremal21B}}\n\\end{figure}\n\n\n\\section{Case $v=22$} \n\\begin{theorem}\\label{theorem:g22} If $\\Gamma$ is a graph of girth at least $5$ on $22$ vertices with $47$ edges, then either $(\\delta,\\Delta)=(3,5)$ or $(\\delta,\\Delta)=(4,5)$. In either case there is an embedded $S_{5,[4,3,3,3,3]}$ star. \n\\end{theorem}\n\n\n\n\n\\vspace{.25cm}\n\\noindent{\\bf Proof} \nSince $\\Gamma$ is extremal, it follows from Table \\ref{table:deltaTable} that the values of $(\\delta,\\Delta)$ are either $ (3,5)$, $(3,6)$, $(3,7)$ or $(4,5)$. \n\nIf there is a vertex, $x$, of degree $3$, then since $\\delta=f(22)-f(21)$, by Lemma \\ref{lem:inductiveExtremal}, \n$\\Gamma$ can be constructed from one of the extremal graphs on $21$ vertices described in Lemma \\ref{lemma:g21}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $3$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},3}$.\nSince $S_{\\Gamma^{\\prime},3}$ is non-empty, it follows that $\\Gamma^{\\prime}$ is the first graph defined in Lemma \\ref{lemma:g21}, which has a sink node of degree $5$ with $4$ children of degree $4$. Since every element of $S_{\\Gamma^{\\prime},3}$ only contains elements of degree less than $5$, it follows that, in $\\Gamma$, $\\Delta=5$. So $(\\delta,\\Delta)=(3,5)$. Since every element of $S_{\\Gamma^{\\prime},3}$ contains a child of a sink node of $\\Gamma^{\\prime}$ of degree $4$, $\\Gamma$ has an embedded $S_{5,[4,3,3,3,3]}$ star. \n\n\nSuppose then that there are no vertices of degree $3$. Thus $(\\delta,\\Delta)=(4,5)$ and there are $16$ vertices of degree $4$ and\n $6$ of degree $5$. \n\nIf there are no sink vertices, all vertices of degree $5$ have $5$ neighbours of degree $4$ (and\n $20$ grandchildren), and it follows from Lemma \\ref{lem:blocks} that there is a linear space on the vertices of degree $4$. This linear space consists of $6$ blocks of size $5$. Let $B$ denote this set of blocks and $\\Gamma_{4}$ the graph formed by the vertices of degree $4$ and the edges between them. It follows from Lemma \\ref{lemma:16.1} that, since any vertex in $V_{4}$ that is in $i$ blocks in $B$ has degree $4-i$ in $\\Gamma_{4}$, if $deg_{i}$ is the number of vertices of degree $i$ in $\\Gamma_{4}$, $(deg_{1},deg_{2},deg_{3},deg_{4})=(1,12,3,0)$, $(0,15,0,1)$ or $(0,14,2,0)$. If, in $\\Gamma_{4}$, there is a chain of $4$ vertices of degree $2$, $a-b-c-d$ say, then no two of $a$, $b$, $c$ can be in the same block of $B$, or there is a cycle of length at most $4$. Hence two of the blocks of $B$ contain $a$, two contain $b$ and two contain $c$. Now similarly $d$ is not in a block in $B$ with $b$ or $c$, so must be in both blocks of $B$ containing $a$, which is not possible. So we assume there is no such chain and it follows that $(deg_{1},deg_{2},deg_{3},deg_{4})\\neq (0,15,0,1)$. By a similar argument, if $a$ is a vertex of degree $1$ then there is no chain $a-b-c$ where $b$ and $c$ have degree $2$. Hence, if $(deg_{1},deg_{2},deg_{3},deg_{4})=(1,12,3,0)$ and $a$ the vertex of degree $1$, with neighbour $b$, either$b$ has degree $3$, or $b$ has degree $2$ and its other neighbour $c$ has degree $3$. In either case, there is a vertex of regularity $1$ in the blocks that is in no block with $a$, hence one of the blocks on $a$ contains no vertex of regularity $1$ which is not possible (see the proof of Lemma \\ref{lemma:16.1}). \n\n So $(deg_{1},deg_{2},deg_{3},deg_{4})=(0,14,2,0)$. \n\nLet $x$ and $y$ be the vertices of degree $3$. Since there can be no component consisting entirely of vertices of degree $2$ (or it would either contain a cycle of length $3$, or a chain of at least $4$ vertices of degree $2$) either $x$ and $y$ are in different components, or $\\Gamma_{4}$ consists of a single component. If $x$ and $y$ are in different components of $\\Gamma_{4}$ then their components contain a single vertex of odd degree and all other vertices of even degree, which is not possible (the sum of the degrees in a component must be even). So $\\Gamma_{4}$ consists of a single component and either there is a single path (of length at most $4$) from $x$ to $y$ and cycles containing $x$ and $y$ respectively, or $3$ paths from $x$ and $y$, together containing all of the vertices of degree $2$. In either case, there is a chain of at least $4$ vertices of degree $2$. \n\n\n\n\n\n\n\n\\begin{lemma}\\label{lemma:g22} There are $3$ nonisomorphic graphs of girth at least $5$ on $22$ vertices with $47$ edges. In one, $(\\delta,\\Delta)=(4,5)$ and $(deg_{4},deg_{5})=(16,6)$. In the other two, $(\\delta,\\Delta)=(3,5)$ and $(deg_{3},deg_{4},deg_{5})=(2,12,8)$ and $(1,14,7)$ respectively. In all cases, every element of $S_{\\Gamma,3}$ contains $3$ vertices of degree less than $5$, and contains a child of a sink node of $\\Gamma$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} By Theorem \\ref{theorem:g22} any such graph has $(\\delta,\\Delta)=(3,5)$ or $(\\delta,\\Delta)=(4,5)$ and there is an embedded $S_{5,[4,3,3,3,3]}$ star. Restricting search to those cases produced a set of graphs. Isomorph elimination using nauty \\cite{nauty} revealed there to be three such graphs, $\\Gamma_{0}$ with $(\\delta,\\Delta)=(4,5)$, and $(deg_{4},deg_{5})=(16,6)$, and $\\Gamma_{1}$ and $\\Gamma_{2}$ with $(\\delta,\\Delta)=(3,5)$ and $(deg_{3},deg_{4},deg_{5})=(2,12,8)$ and $(1,14,7)$ respectively. \n\n\nThese graphs are shown in Figures \\ref{fig:extremal22A0}-\\ref{fig:extremal22A2}. In each case the graph is drawn as a star about a sink node with $4$ children of degree $4$ and one of degree $5$. The vertices adjacent to each of the grandchildren is indicated as a set.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.7\\textwidth]{newstar22_A0}\n\\caption{Extremal graph $\\Gamma_{0}$ on $22$ vertices with $44$ edges as a star, $(\\delta,\\Delta)=(4,5)$\\label{fig:extremal22A0}}\n\\end{figure}\n\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.7\\textwidth]{newstar22_A1}\n\\caption{Extremal graph $\\Gamma_{1}$ on $22$ vertices with $44$ edges as a star, $(\\delta,\\Delta)=(3,5)$\\label{fig:extremal22A1}}\n\\end{figure}\n\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.7\\textwidth]{newstar22_A2}\n\\caption{Extremal graph $\\Gamma_{2}$ on $22$ vertices with $44$ edges as a star, $(\\delta,\\Delta)=(3,5)$\\label{fig:extremal22A2}}\n\\end{figure}\n\n\n\n\\noindent\nExamination of these graphs using a computer reveals that in each case every element of $S_{\\Gamma,3}$ contains no vertex of degree greater than $4$, and contains at least one vertex that is the child of a sink node. See Appendix C for details. \n\n\\begin{lemma}\\label{lemma:g22A} If $\\Gamma$ is a graph of girth at least $5$ on $22$ vertices with $47$ edges then $S_{\\Gamma,3}$ contains at most $4$ non-intersecting sets. \nIf all vertices of degree $3$ (if any) are contained in a single element of $S_{\\Gamma,3}$ then $\\Gamma$ is $\\Gamma_{1}$ or $\\Gamma_{2}$ of Lemma \\ref{lemma:g22}. If in addition there are two vertices of degree $4$ which have fewer than $3-deg_{3}$ neighbours of degree $4$ between them, $\\Gamma$ is $\\Gamma_{2}$ . In this case there are three vertices of degree $4$ which have fewer than two neighbours of degree $4$ and they each have one such neighbour, $p$, $q$ and $r$ respectively. No two of $p$, $q$ and $r$ are in a common element of $S_{\\Gamma,3}$ with the vertex of degree $3$. \n\n\\begin{comment}\nIf $S_{\\Gamma,3}$ contains a set $X$ of $3$ non-intersecting sets and a further triple $t$ that either does not intersect any element of $X$ or does so only at vertices of degree $3$, then it is not possible to arrange the four vertices of degree less than $5$ that are not in $X$ or $t$ into two elements of $S_{\\Gamma,2}$ where neither pair has edges to both $t$ and to an element of $X$.\n\\end{comment}\n\\end{lemma}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} From examination of the graphs in Appendix C, $S_{\\Gamma,3}$ has at most $4$ non-intersecting sets and if $S_{\\Gamma,3}$ contains a set $X$ of $3$ non-intersecting sets then $\\Gamma$ is graph $\\Gamma_{0}$ or $\\Gamma_{2}$. Examination of the vertices of degree $4$ shows that all vertices of degree $4$ have at least two neighbours of degree $4$ unless $\\Gamma$ is $\\Gamma_{2}$ and the vertex is $1$, $16$, or $17$, which have neighbours of degree $4$: $12$, $13$ and $9$ respectively. No two of these vertices are in an element of $S_{\\Gamma,3}$ with the vertex of degree $3$ (i.e. vertex $0$).\n\n\\begin{comment}\nIf $\\Gamma=\\Gamma_{0}$ then, since $|S_{\\Gamma_{0},3}|=4$, $t\\cup X = S_{\\Gamma_{0},3}$. As shown in Appendix C the four vertices of degree less than $5$ that do not appear in $S_{\\Gamma_{0},3}$ can be uniquely arranged into two elements of $S_{\\Gamma_{0},2}$ and, whichever way we choose $t$, both pairs have edges to both $t$ and at least one element of $X=S_{\\Gamma_{0},3}\\setminus \\{t\\}$.\n\nIf $\\Gamma=\\Gamma_{2}$ then, since there are no $4$ parallel blocks, $t$ intersects $X$ at the vertex of degree $3$. As shown in Appendix C, there are $2$ sets of $4$ elements of $S_{\\Gamma_{2},3}$ such that the only vertex to appear twice is the vertex of degree $3$ and the remaining $4$ vertices of degree less than $5$ constitute two non-intersecting elements of $S_{\\Gamma_{2},2}$. In each case, there are edges from the elements of one of the pairs to both triples containing the vertex of degree $3$ and one of the other triples. The result follows. \n\\end{comment}\n\n\\section{Case $v=23$} \n\n\\begin{theorem}\\label{theorem:g23} If $\\Gamma$ is a graph of girth at least $5$ on $23$ vertices with $50$ edges then $(\\delta,\\Delta)=(3,5)$ and there is an embedded $S_{5,[4,4,3,3,3]}$ star or $(\\delta,\\Delta)=(4,5)$ and either there is an embedded $S_{5,[4,4,3,3,3]}$ star or there is no embedded $S_{5,[4,4,3,3,3]}$ star and there is an embedded $S_{5,[4,3,3,3,3]}$ star. \n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} \nSince $\\Gamma$ is extremal, it follows from Table \\ref{table:deltaTable} that the values of $(\\delta,\\Delta)$ are either $ (3,5)$, $(3,6)$, $(3,7)$ or $(4,5)$. \n\nIf there is a vertex, $x$, of degree $3$, then since $\\delta=f(23)-f(22)$, by Lemma \\ref{lem:inductiveExtremal}, \n$\\Gamma$ can be constructed from one of the extremal graph on $22$ vertices described in Lemma \\ref{lemma:g22}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $3$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},3}$. \n\n Since in each case $S_{\\Gamma^{\\prime},3}$ contains no vertex of degree greater than $4$, (by Lemma \\ref{lemma:g22}), in $\\Gamma$ $\\Delta=5$. Hence $(\\delta, \\Delta)=(3,5)$. In addition, since $S_{\\Gamma^{\\prime},3}$ contains a child of a sink node of $\\Gamma^{\\prime}$, it follows that $\\Gamma$ has a sink node of degree $5$ with $2$ children of degree $5$ and $3$ of degree $4$. \n\nIf there is no vertex of degree $3$ then $(\\delta,\\Delta)=(4,5)$ and $(deg_{4},deg_{5})=(15,8)$. If no vertex of degree $5$ has any neighbour of degree $5$ then there is a linear space on $15$ points with $8$ blocks of size $5$, contradicting Lemma \\ref{lemma:packingLemma}. Since every vertex of degree $5$ has at least one neighbour of degree $5$, the result follows.\n\n\n\\begin{lemma}\\label{lemma:g23} There are $7$ non-isomorphic graphs on $23$ vertices with $50$ edges. All of the graphs contain embedded $S_{5,[4,4,3,3,3]}$ stars. For $3$ of the graphs, $S_{\\Gamma,4}$ is empty. The remaining $4$ graphs consist of:\n\\begin{itemize}\n\\item $1$ graph with $(\\delta,\\Delta)=(4,5)$ and $(deg_{4},deg_{5})=(15,8)$. Every element of $S_{\\Gamma,4}$ contains $4$ vertices of degree $4$, including at least one child of a sink node \n\\item $2$ graphs with $(\\delta,\\Delta)=(3,5)$ and $(deg_{3},deg_{4},deg_{5})=(1,13,9)$. Every element of $S_{\\Gamma,4}$ contains the vertex of degree $3$ and $3$ vertices of degree $4$, including a child of a sink node.\n\\item $1$ graph with $(\\delta,\\Delta)=(3,5)$ and $(deg_{3},deg_{4},deg_{5})=(2,11,10)$. Every element of $S_{\\Gamma,4}$ contains the two vertices of degree $3$ and $2$ vertices of degree $4$, including a child of a sink node. \n\\end{itemize}\n\\end{lemma} \n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} (1) By Theorem \\ref{theorem:g23} any such graph has $(\\delta,\\Delta)=(3,5)$ and there is an embedded $S_{5,[4,4,3,3,3]}$ star or $(\\delta,\\Delta)=(4,5)$ and either there is an embedded $S_{5,[4,4,3,3,3]}$ star or there is no embedded $S_{5,[4,4,3,3,3]}$ star and there is an embedded $S_{5,[4,3,3,3,3]}$ star. Restricting search to those cases produced a set of\n graphs. Isomorph elimination using Nauty \\cite{nauty} revealed there to be $7$ such graphs, two with \n$(\\delta,\\Delta)=(4,5)$, and $(deg_{3},deg_{4},deg_{5})=(0,15,8)$, and $5$ with $(\\delta,\\Delta)=(3,5)$, two of which\n have $(deg_{3},deg_{4},deg_{5})=(2,11,10)$ and three of which have $(deg_{3},deg_{4},deg_{5})=(1,13,9)$. In all cases there is an embedded $S_{5,[4,4,3,3,3]}$ star, and using a computer we have verified the structure of $S_{\\Gamma,4}$ in each case. See Appendix D for details. \n\n\\begin{lemma}\\label{lemma:g23A} Suppose that $\\Gamma$ is a graph of girth at least $5$ on $23$ vertices with $(\\delta,\\Delta)=(4,5)$ and $\\Gamma_{4}=(V_{4},E_{4})$ the subgraph of $\\Gamma$ on the vertices of degree $4$, then there is at most one vertex in $V_{4}$ that is in no edges in $E_{4}$. If two vertices have degrees $0$ and $1$ in $V_{4}$, then they have a common neighbour of degree $5$ in $\\Gamma$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} From the proof of Lemma \\ref{lemma:g23} there are two such graphs with $(\\delta,\\Delta)=(4,5)$. Only the first of these contains a vertex of degree $4$ that has no neighbours of degree $4$, (vertex $a$ say). There are two vertices that have degree $1$, and each of these are in an edge in $\\Gamma$ with $a$. See Appendix D for details.\n\n\\begin{comment}\n (2) There are two graphs with $(deg_{3},deg_{4},deg_{5})=(0,5,18)$. They are shown as graphs $A$ and $B$ in Figure\n \\ref{fig:graphs23A}. In all cases there is a vertex of degree $5$ that has $2$ children of degree $5$ and $3$ of degree $4$. In each case they are drawn as a star with central node a sink node. The\n nodes of degree $5$ are circled, and (where necessary for our proof) for some of the grandchildren the set of\n neighbours of the grandchild are indicated as elements of a set, above the label for the grandchild. Consider graph\n $A$, and let $S$ be a set of $4$ vertices with the desired property. Then no element of $S$ is a child of vertex $22$\n (as vertex $22$ has two neighbours of degree $5$), and no two elements of $S$ can have the same parent (as all\n elements of $S$ are at a distance of at least $3$ from each other). It follows that the elements of $S$ consist of a child\n of each of the vertices $4$, $5$, $6$ and $7$. Call the corresponding elements of $S$, $x_{1}$, $x_{2}$, $x_{3}$ and\n $x_{4}$. As all elements of $S$ have degree $4$, it follows that $x_{1}$ is vertex $3$. Since vertex $3$ is adjacent to\n vertex $8$, we must have that $x_{2}$ is vertex $2$. But vertices $3$ and $2$ are both adjacent to vertex $21$, so\n are at a distance of $2$ apart, which is a contradiction. Consider graph $B$. Since no element of $S$ can be adjacent\n to either vertex $4$ or vertex $2$, two of the elements of $S$ must have the same parent, which is a contradiction. \n (3) There are three graphs with $(deg_{3},deg_{4},deg_{5})=(1,13,9)$. They are shown as graphs $C$, $D$ and $E$ in Figure\n \\ref{fig:graphs23B} in a similar way to the above, but with the vertex of degree $3$ indicated in bold face.\nConsider graph $C$. Set $S$ must contain the vertex of degree $3$, namely vertex $6$, and must include a child of at least one of vertices $19$ and $3$, so must contain vertex $9$ or vertex $4$. But vertices $4$ and $9$ both share a neighbour with vertex $3$, so this is impossible. \nFor graph $D$, by a similar argument $S$ must contain vertex $5$. Since $S$ can contain no child of vertex $22$ it must contain vertex $13$ and one of vertices $4$ and $14$. But vertex $4$ shares a neighbour with vertex $5$, and vertex $14$ shares a neighbour with vertex $13$. \nFor graph $D$, by a similar argument $S$ must contain vertex $3$. Since $S$ can not contain vertex $9$ (as it shares a neighbour with vertex $3$), $S$ can contain no child of vertex $19$. Hence $S$ must contain a child of vertex $18$, \nFor graph $E$, $S$ must contain vertex $3$. As $S$ can not contain vertex $9$ (which shares a neighbour with vertex $3$), it must contain a child of each of the vertices $17$, $18$ and $20$. The only child of $18$ of degree $4$ is vertex $12$, so $S$ contains $12$. Since $S$ contains a child of vertex $17$, it must contain either vertex $14$ (which is at a distance of $1$ from vertex $12$) or vertex $16$ (which shares a neighbour with vertex $12$). Either way we have a contradiction. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{stars23A}\n\\caption{Extremal graphs on $23$ vertices, with $(deg_{3},deg_{4},deg_{5})=(0,5,18)$\n\\label{fig:graphs23A}}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.8\\textwidth]{stars23B}\n\\caption{Extremal graphs on $23$ vertices, with $(deg_{3},deg_{4},deg_{5})=(1,13,9)$\n\\label{fig:graphs23B}}\n\\end{figure}\n\\end{comment}\n\n\n\n\\section{Case $v=24$}\n\n\\begin{theorem}\\label{theorem:g24} If $\\Gamma$ is a graph of girth at least $5$ on $24$ vertices with $54$ edges then $(\\delta,\\Delta)=(4,5)$ and there is an embedded $S_{5,[4,4,4,3,3]}$ star.\n\\end{theorem}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof}\nFrom Table \\ref{table:deltaTable} we know that in this case all vertices have degree $4$ or $5$. There are $12$ vertices of degree\n $4$ and $12$ vertices of degree $5$. Since $\\delta=f(24)-f(23)$, by Lemma \\ref{lem:inductiveExtremal}, \n$\\Gamma$ can be constructed from one of the extremal graph on $23$ vertices described in Lemma \\ref{lemma:g23}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $4$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},4}$.\nIn all cases, $S_{\\Gamma^{\\prime},4}$ includes precisely one child of degree $4$ of any central node of an embedded $S_{5,[4,4,3,3,3]}$ star. It follows that $\\Gamma$ contains an embedded $S_{5,[4,4,4,3,3]}$ star.\n\n\\begin{lemma}\\label{lemma:g24} (1) There is $1$ graph $\\Gamma$ of girth at least $5$ on $24$ vertices with $54$ edges. \n(2) Any element of $S_{\\Gamma,3}$ contains only vertices of degree $4$. (3) There are no $4$ non-intersecting sets in $S_{\\Gamma,3}$ for which there are are two pairs of sets with no edge between any two vertices in the same pair. (4) If $X$ is a set of $3$ non-intersecting elements of $S_{\\Gamma,3}$ then at least two of the three vertices of degree $4$ that are not in any set in $X$ are in an edge with an element in a set in $X$.\n\\end{lemma}\n\n\\begin{comment}\n(2) There is one set $X$ of $4$ elements of $S_{\\Gamma,3}$ that do not intersect. There is exactly one pair $X_{1},X_{2}\\in X$ such that for all $x_{1}\\in X_{1}$, $x_{2}\\in X_{2}$ there is no edge $(x_{1},x_{2}) \\in \\Gamma$. (3) If $e$ is an edge between two vertices of degree $4$ that are each in one edge only in $\\Gamma_{4}$, then $e$ intersects all but $2$ of the elements of $S_{\\Gamma,3}$. (4) There are no $4$ non-intersecting elements of $S_{\\Gamma,3}$ for which there are are two pairs with no edge between either of the pairs. \n\\end{comment}\n\n\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof}(1) By Theorem \\ref{theorem:g24} $(\\delta,\\Delta)=(4,5)$ and there is an embedded $S_{5,[4,4,4,3,3]}$ star. Restricting search to this case produced a set of graphs. Isomorph elimination using nauty \\cite{nauty} revealed there to be one such graph, with $2$ sink nodes. This is shown in Figure \\ref{fig:graph24} as a star about vertex $13$, together with set $S_{\\Gamma,3}$. Clearly all elements of $S_{\\Gamma,3}$ contain only vertices of degree $4$. (2) and (3) follow from observation, see Appendix E for details. \n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.8\\textwidth]{completeStar24b}\n\\caption{Extremal graph on $24$ vertices\n\\label{fig:graph24}}\n\\end{figure}\n\n\n\n\\begin{comment}\nThese sets are (0,7,18), (0,8,18), (1,4,23),(1,16,22),(4,14,17),(4,17,23),(7,14,21),(8,18,22)\n\\end{comment}\n\n\\section{Case $v=25$}\n\n\\vspace{.25cm}\n\\begin{theorem}\n\\label{theorem:g25}\nIf $\\Gamma$ is a graph of girth at least $5$ on $25$ vertices with $57$ edges then $(\\delta,\\Delta)=(3,5)$, $(4,5)$ or $(4,6)$. If $(\\delta,\\Delta)=(3,5)$ or $(4,5)$ then $\\Gamma$ has an embedded $S_{5,[4,4,4,4,3]}$ star or there is no embedded $S_{5,[4,4,4,4,3]}$ star and there is an embedded $S_{5,[4,4,4,3,3]}$ star. If $(\\delta,\\Delta)=(4,6)$ then $1\\leq deg_{6}\\leq 2$ and there is an embedded $S_{6,[3,3,3,3,3,3]}$ star. \n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} We know from Table \\ref{table:deltaTable} that in this case the possible values for $(\\delta,\\Delta)$ are $(3,5)$, $(3,6)$, $(3,7)$, $(3,8)$, $(4,5)$ and $(4,6)$. If there is a vertex, $x$, of degree $3$, then since $\\delta=f(25)-f(24)$, by Lemma \\ref{lem:inductiveExtremal}, \n$\\Gamma$ can be constructed from the extremal graph on $24$ vertices described in Lemma \\ref{lemma:g24}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $3$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},3}$.\nSince $S_{\\Gamma^{\\prime},3}$ only contains vertices of degree $4$, it follows that $(\\delta,\\Delta)=(3,5)$. Since $\\Gamma^{\\prime}$ contains a star $[5,[4,4,4,3,3])$, $\\Gamma$ does too, and the result for $\\delta=3$ holds. \n\n\nIf $(\\delta,\\Delta)=(4,5)$ then $(deg_{4},deg_{5})=(11,14)$ and the sets of vertices of degree $4$ adjacent to each vertex of degree $5$ form a linear space (see Lemma \\ref{lem:blocks}) $\\Lambda$. If no vertex of degree $5$ has more than $2$ neighbours of degree $5$ then $\\Lambda$ consists of $14$ blocks of size at least $3$. It follows from \\cite{bettenbetten4} that $\\Lambda$ has at least $11$ blocks of size $3$ (and, correspondingly, at least $11$ vertices of degree $5$ have two neighbours of degree $5$. \n\nLet $x$ be such a vertex of degree $5$. Let S be the embedded $S_{5,[4,4,3,3,3]}$ star with $x$ as root (see Figure \\ref{fig:star25false}). Note that $A$ and $B$ denote the sets of children of the two children of $x$ of degree $5$, and $R$ the set consisting of the remaining grandchildren of $x$, and the vertices $p_{1}$ and $p_{2}$ that are not contained in the star. Since we have assumed that all vertices of degree $5$ have at most one child of degree $5$, $A$ and $B$ contain at most $1$ vertex each of degree $5$ and $R$ contains at least $9$ vertices of degree $5$ and at most $2$ of degree $4$. Now if $A$ contains a vertex of degree 5 then it must have at least 2 neighbours in $R$ that do not have degree $5$. So $R$ contains exactly $9$ vertices of degree $5$ and $2$ of degree 4. Then $B$ must contain a vertex of degree $5$, which must be adjacent to the same two elements of $R$, and there is a square. It follows that $A$ and $B$ contain no vertices of degree $5$, and every element of $R$ has degree $5$. But then $p_{1}$ must be adjacent to $3$ elements of $A\\cup B$ (as it can have at most $2$ neighbours of degree $5$), which is impossible, as there will then be a square. So there is at least one vertex of degree $5$ that has more than $2$ neighbours of degree $5$ and so $\\Gamma$ either has an embedded \n$S_{5,[4,4,4,4,3]}$ star, or has no embedded $S_{5,[4,4,4,4,3]}$ star but has an embedded $S_{5,[4,4,4,3,3]}$ star.\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{star25_false}\n\\caption{$25$, embedded $S_{5,[4,4,3,3,3]}$ star\n\\label{fig:star25false}}\n\\end{figure}\n\n\nIf $(\\delta,\\Delta)=(4,6)$, any vertex of degree $6$ must have $6$ neighbours of degree $4$, so the result follows. \n\n\\begin{lemma}\\label{lemma:g25} There are $6$ graphs of girth at least $5$ on $25$ vertices with $57$ edges. Three of the graphs have $(\\delta,\\Delta)=(3,5)$ and $(deg_{3},deg_{4},deg_{5})=(1,19,5)$. One of these graphs has no sink node and set $S_{\\Gamma,4}$ is empty. The other two have an embedded $S_{5,[4,4,4,4,3]}$ star and non-empty sets $S_{\\Gamma,4}$ where every element of $S_{\\Gamma,4}$ contains a child of degree $4$ of a sink node, the vertex of degree $3$, and $2$ further vertices of degree $4$.\nIn the remaining three graphs, $(\\delta,\\Delta)=(4,5)$, $(deg_{4},deg_{5})=(11,14)$ and there is an embedded $S_{5,[4,4,4,4,3]}$ star. One of these graphs has empty set $S_{\\Gamma,4}$, and for the other two $S_{\\Gamma,4}$ contains a child of degree $4$ of a sink node and $3$ other vertices of degree $4$. In addition, for each of $\\Gamma_{3},\\ldots, \\Gamma_{5}$, $S_{\\Gamma,3}$ does not contain $3$ non-intersecting elements consisting entirely of vertices of degree $4$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Restricting our search to graphs with values of $(\\delta,\\Delta)$, with the identified embedded stars from Theorem \\ref{theorem:g25} produced a set of graphs. Isomorph elimination using nauty \\cite{nauty} revealed there to be six graphs, $A_{0},\\ldots, A_{5}$. Using a computer the graphs were shown to satisfy the stated properties. For details see Appendix F.\n\n\\begin{lemma}\\label{lemma:g25A} If $\\Gamma$ is a graph of girth at least $5$ on $25$ vertices with $57$ edges and \n$X_{1}$, $X_{2}$, $X_{3}$ parallel elements of of $S_{\\Gamma,3}$ then\n\\begin{enumerate}\n\\item if $(\\delta,\\Delta)=(4,5)$ one of the sets must contain a vertex of degree $5$\n\\item it is not possible that $X_{1}$ contains all vertices of degree $3$ and there are no edges between $X_{1}$ and $X_{2}$, or between $X_{1}$ and $X_{3}$.\n\\end{enumerate}\n\\end{lemma}\n\\begin{comment}\n, and any edge of $\\Gamma$ consisting of two vertices of degree $4$ intersects at least one of $X_{1}$, $X_{2}$, $X_{3}$.\n\\end{comment}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} Follows from examination of graphs $\\Gamma_{0}\\ldots \\Gamma_{5}$ (and corresponding sets $S_{\\Gamma,3}$) as shown in Appendix F. \n\n\\section{Case $v=26$}\n\n\\vspace{.25cm}\n\\begin{theorem}\n\\label{theorem:g26}\nIf $\\Gamma$ is a graph of girth at least $5$ on $26$ vertices with $61$ edges then $(\\delta,\\Delta)=(4,5)$ and $\\Gamma$ has an embedded $S_{5,[4,4,4,4,4]}$ star. \n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} We know from Table \\ref{table:deltaTable} that in this case the possible values for $(\\delta,\\Delta)$ are $(4,5)$ and $(4,6)$. Since in all cases $\\delta= f(26)-f(25)$, \nby Lemma \\ref{lem:inductiveExtremal}, $\\Gamma$ can be constructed from one of the extremal graphs on $25$ vertices described in Lemma \\ref{lemma:g25}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $4$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},4}$.\nBy Lemma \\ref{lemma:g25}, it follows that, in $\\Gamma$, $(\\delta,\\Delta)=(4,5)$ and $\\Gamma$ has an embedded $S_{5,[4,4,4,4,4]}$ star.\n\n\\begin{lemma}\\label{lemma:g26} There are $2$ graphs of girth at least $5$ on $26$ vertices with $61$ edges, both have $(\\delta,\\Delta)=(4,5)$. In one of these graphs, set $S_{\\Gamma,4}$ is empty. In the other, $S_{\\Gamma,4}$ contains $1$ set, which consists of $4$ vertices of degree $4$, none of which is a child of a sink node. In one graph there is a unique set $X=\\{X_{1},X_{2}\\}$ containing two non-intersecting elements of $S_{\\Gamma,3}$ that each consist of elements of degree $4$, and, for all $x_{1}\\in X_{1}$, $x_{2}\\in X_{2}$ there is no edge $(x_{1},x_{2}) \\in \\Gamma$. In the other graph there is no such set $X$. \n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} From Theorem \\ref{theorem:g26} all extremal graphs on $26$ vertices can be constructed by adding a new vertex $x$, and edges to a set of $4$ vertices that is a set in $S_{\\Gamma^{\\prime},4}$, where $\\Gamma^{\\prime}$ is one of the $6$ extremal graphs on $25$ vertices. This produced a set of $6$ graphs. Isomorph elimination using nauty \\cite{nauty} revealed there to be two distinct graphs, $A_{0}, A_{1}$. Using a computer the graphs were shown to satisfy the stated properties. See Appendix G for details. \n\n\\section{Case $v=27$}\n\\vspace{.25cm}\n\\begin{theorem}\n\\label{theorem:g27}\nIf $\\Gamma$ is a graph of girth at least $5$ on $27$ vertices with $65$ edges then $(\\delta,\\Delta)=(4,5)$ and $\\Gamma$ has an embedded $S_{5,[4,4,4,4,4]}$ star. \n\\end{theorem}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} We know from Table \\ref{table:deltaTable} that in this case the possible values for $(\\delta,\\Delta)$ are $(4,5)$ and $(4,6)$. Since in all cases $\\delta= f(27)-f(26)$, \nby Lemma \\ref{lem:inductiveExtremal}, $\\Gamma$ can be constructed from an extremal graph on $26$ vertices described in Lemma \\ref{lemma:g26}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $4$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},4}$.\nBy Lemma \\ref{lemma:g26} it follows that, in $\\Gamma$, $(\\delta,\\Delta)=(4,5)$ and (since any non-empty set $S_{\\Gamma^{\\prime},4}$ does {\\it not} contain the child of a sink node), $\\Gamma$ has an embedded $S_{5,[4,4,4,4,4]}$ star. \n\n\\begin{lemma}\\label{lemma:g27} There is $1$ graph of girth at least $5$ on $27$ vertices with $65$ edges. In this graph $(\\delta,\\Delta)=(4,5)$ and $S_{\\Gamma,3}$ contains $5$ sets, each of which contains $1$ vertex of degree $4$ and $2$ of degree $5$. All vertices of degree $5$ in elements of $S_{\\Gamma,3}$ are roots of an embedded $S_{5,[4,4,4,4,3]}$ star. Set $S_{\\Gamma,4}$ is empty.\n\\end{lemma}\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} From Theorem \\ref{theorem:g27} all extremal graphs on $27$ vertices can be constructed by adding a new vertex $x$, and edges to a set of $4$ vertices that is a set in $S_{\\Gamma^{\\prime},4}$, where $\\Gamma^{\\prime}$ is one of the $2$ extremal graphs on $26$ vertices. Since, by Lemma \\ref{lemma:g26} only one of the extremal graphs on $26$ vertices has a non-empty set $S_{\\Gamma^{\\prime},4}$, and this contains only one set, this produces a single graph. Using a computer the graph was shown to satisfy the stated properties. See Appendix H for details. \n\n\\section{Case $v=28$}\n\\vspace{.25cm}\n\\begin{theorem}\n\\label{theorem:g28}\nIf $\\Gamma$ is a graph of girth at least $5$ on $28$ vertices with $68$ edges then $(\\delta,\\Delta)=(3,6)$, $(deg_{3},deg_{4},deg_{5},deg_{6})=(1,4,21,2)$, and there is an embedded $S_{6,[4,4,4,4,3,2]}$ star, or $(\\delta,\\Delta)=(4,5)$ or $(4,6)$. \n\n If $(\\delta,\\Delta)=(4,5)$ then $(deg_{4},deg_{5})=(4,24)$ and there is an embedded $S_{5,[4,4,4,4,4]}$ star. If $(\\delta,\\Delta)=(4,6)$ then $(deg_{4},deg_{5},deg_{6})\\in \\{(i, 32-2i ,i-4):5\\leq i \\leq 13\\}$ and there is an embedded $S_{6,[4,4,4,3,3,3]}$ star.\n\\end{theorem}\n\n\n\\vspace{.25cm}\n\\noindent\n{\\bf Proof} We know from Table \\ref{table:deltaTable} that in this case the possible values for $(\\delta,\\Delta)$ are $(3,5)$, $(3,6)$, $(3,7)$, $(3,8)$, $(3,9)$, $(4,5)$ or $(4,6)$. If $\\delta=3$,\nby Lemma \\ref{lem:inductiveExtremal}, $\\Gamma$ can be constructed from the extremal graph on $27$ vertices described in Lemma \\ref{lemma:g27}, $\\Gamma^{\\prime}$, by adding a vertex $x$ of degree $3$ and edges from $x$ to all vertices in an element of $S_{\\Gamma^{\\prime},3}$.\nBy Lemma \\ref{lemma:g27} it follows that, in $\\Gamma$, $(\\delta,\\Delta)=(3,6)$, $(deg_{3},deg_{4},deg_{5},deg_{6})=(1,4,21,2)$, and there is an embedded $S_{6,[4,4,4,4,3,2]}$ star. \n\nIf $(\\delta,\\Delta)=(4,5)$ then $(deg_{4},deg_{5})=(4,24)$ and there are at most $16$ edges from the vertices of degree $4$ to the vertices of degree $5$. Hence at least one vertex of degree $5$ has no neighbours of degree $4$ and the result follows.\n\n\n If $(\\delta,\\Delta)=(4,6)$ then, since $deg_{4}>0$ and $deg_{6}>0$, $(deg_{4},deg_{5},deg_{6}) \\in \\{(i, 32-2i ,i-4):5\\leq i \\leq 16\\}$. A vertex of degree $6$ can have at most one neighbour of degree $6$, and so has at least $5$ edges to the set $S$ containing the vertices of degrees $4$ and $5$. It follows that there is a linear space on $deg_{4}+deg_{5}$ points containing at least $deg_{6}$ blocks of size $5$. It follows from Lemma \\ref{lemma:packingLemma} that this does not hold when $i\\geq 14$.\n\n If there is no embedded $S_{6,[4,4,4,3,3,3]}$ star then every vertex of size $6$ has at least $4$ neighbours of degree $4$. This is only possible if there is a linear space on $deg_{4}$ points that has at least $deg_{6}$ blocks of size at least $4$. By Lemma \\ref{lemma:packingLemma}, this does not hold if $5=8$ there are no $3-5$ or $4-4$ edges either (as then there is a graph on $31$ vertices with $80$ edges with a vertex of degree at least $7$).Actually, eliminated $3-5$ and $4-4$ except for the few cases identified below.\n\\item Every vertex of degree $5$ is adjacent to at most $1$ vertex of degree $4$ (as extremal on $30$ vertices has no $3$ parallel elements of $S_{\\Gamma,3}$). \n\\end{itemize}\n\n\\begin{lemma}\\label{lemma:g33} There are $?$ graphs of girth at least $5$ on $33$ vertices with $87$ edges. Of these \n\\begin{enumerate}\n\\item $3$ have $(deg_{2},deg_{5},deg_{6})=(1,20,12)$, (add new vertex $x$ and element of $S_{\\Gamma,2}$, $\\Gamma$ extremal on $32$ vertices. Potentially $41$, $3$ unique. \n\\item $5$ have a $3-5$ edge, and have $(\\delta,\\Delta)=(3,6)$. Of these $4$ have $(deg_{3},deg_{4},deg_{5},deg_{6})=(1,0,21,11)$ and \n\\item $1$ has $(deg_{3},deg_{4},deg_{5},deg_{6})=(1,1,19,12)$ (add two new vertices $x,y$ of degree $5$ and $3$ to extremal graph $\\Gamma^{\\prime}$ on $31$ vertices. Add edge $x-y$ and edges to sets $X$ and $Y$ from $S{\\Gamma^{\\prime},4}$, $S_{\\Gamma^{\\prime},2}$ that are distinct and have no edges between them. $\\Gamma^{\\prime}$ is graph $0$ (must have a vertex of degree less than $5$). There are $16$ possibilities, $5$ unique. \n\\item Rest?\n\\end{enumerate}\n\\end{lemma} **\n\n\\end{comment}\n\n\\section*{Appendix A}\\label{appendix20}\nThe extremal graph $(20,41)$. Note that $\\Gamma_{0}$ has $(deg_{3},deg_{4},deg_{5})=(1,16,3)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc}\n0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&1\\\\\n0&0&1&0&1&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0\\\\\n1&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&1&0&0\\\\\n0&1&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,5,5,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,3)$$\nEdge set is:\n\\begin{eqnarray*}\n(0,4), (0,7), (0,10), (0,13), (0,19), (1,5), (1,8),\n (1,11),\\\\\n (1,14), (1,19), (2,3), (2,6), (2,9), (2,12), (2,19),\\\\ (3,4),\n (3,14), (3,15), (4,5), (4,18), (5,6), (5,17),\\\\ (6,7),\n (6,16), (7,8), (7,15), (8,9), (8,18), (9,10), \\\\\n(9,17), (10,11), (10,16), (11,12), (11,15), (12,13), \\\\\n(12,18), (13,14), (13,17), (14,16), (15,17), (16,18)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{0},3}$ is given below. All elements contain vertex $19$ of degree $3$, and two other vertices of degree $4$. \n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(15,16,19) (15,18,19) (16,17,19) (17,18,19)\\}\\\\\n\\end{eqnarray*}\n\n\n\\section*{Appendix B}\\label{appendix21}\nThe $3$ extremal graphs $(21,44)$. Note that graph $\\Gamma_{0}$ has $(deg_{3},deg_{4},deg_{5})=(1,15,5)$, and graphs $\\Gamma_{1}$ and $\\Gamma_{2}$ have $(deg_{4},deg_{5})=(17,4)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc c}\n0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0\\\\\n1&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&1&0&0&0\\\\\n0&1&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&1\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,5,5,4,4,4,4,4,4,4,4,4,4,4,4,5,5,4,4,4,3)$$ Edge set is:\n\\begin{eqnarray*}\n(0,4), (0,7), (0,10), (0,13), (0,19), (1,5), (1,8), (1,11),\\\\ (1,14), \n(1,19), (2,3), (2,6), (2,9), (2,12), (2,19), (3,4),\\\\ (3,14), (3,15), (4,5), \n(4,18), (5,6), (5,17), (6,7),\\\\ (6,16), (7,8), (7,15), (8,9), (8,18), (9,10), \n(9,17),\\\\ (10,11), (10,16), (11,12), (11,15), (12,13), (12,18), (13,14),\\\\ (13,17), (14,16), \n(15,17), (15,20), (16,18), (16,20), (19,20)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{0},3}$ is given below. All elements contain vertices of degree $3$ or $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(4,9,20) (5,12,20) (8,13,20) (17,18,19) \\}\\\\\n\\end{eqnarray*}\n\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc c}\n0&1&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&1\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&0&0&1&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,4,4,4,4,4,5,5,5,4,4,4,4,4,4,4,4,4,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,1), (0,2), (0,3), (0,4), (0,5), (1,6), (1,7), (1,8),\\\\ (2,9), \n(2,10), (2,11), (3,12), (3,13), (3,14), (4,15), (4,16), \\\\(4,17), (5,18), (5,19), \n(5,20), (6,9), (6,12), (6,15),\\\\ (6,18), (7,10), (7,13), (7,16), (7,19), (8,11), \n(8,14),\\\\ (8,17), (8,20), (9,13), (9,17), (10,14), (10,15), (11,12),\\\\ (11,16), (12,19), \n(13,20), (14,18), (15,20), (16,18), (17,19)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{1},3}$ is empty.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{2}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc ccccc}\n0&1&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&1&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1\\\\\n0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&0&0&1&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&1&0&1&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,4,4,4,4,4,5,5,4,4,4,5,4,4,4,4,4,4,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,1), (0,2), (0,3), (0,4), (0,5), (1,6), (1,7), (1,8), \\\\(2,9), \n(2,10), (2,11), (3,12), (3,13), (3,14), (4,15), (4,16), \\\\(4,17), (5,18), (5,19), \n(5,20), (6,9), (6,12), (6,15),\\\\ (6,18), (7,10), (7,13), (7,16), (7,19), (8,11), \n(8,14),\\\\ (8,17), (9,13),(9,17), (10,14), (10,15), (11,12), (11,16),\\\\ (11,20), (12,19), \n(13,20), (14,18), (15,20), (16,18), (17,19)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma,3}$ is empty. \n\n\n\n\\section*{Appendix C}\\label{appendix22}\nThe $3$ extremal graphs $(22,47)$. Note that graph $\\Gamma_{0}$ has $(deg_{4},deg_{5})=(16,6)$, graph $\\Gamma_{1}$ has $(deg_{3},deg_{4},deg_{5})=(2,12,8)$ and graph $\\Gamma_{2}$ has $(deg_{3},deg_{4},deg_{5})=(1,14,7)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&1&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,4,5,4,4,4,5,4,4,4,5,4,5,4,4,4,4,4,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,9), (0,12), (0,15), (0,20), (1,6), (1,8), (1,11), (1,20),\\\\ (2,5), \n(2,7), (2,14), (2,20), (3,4), (3,10), (3,13), (3,20),\\\\ (4,7), (4,11), (4,15), \n(4,19), (5,8), (5,13), (5,19), (6,12),\\\\ (6,14), (6,19), (7,12), (7,18), (8,10), \n(8,15), (8,18), (9,11),\\\\ (9,14), (9,18), (10,14), (10,17), (11,17), (12,13), (12,17), \n(13,16),\\\\ (14,16), (15,16), (16,21), (17,21), (18,21), (19,21), (20,21)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $20$ and $21$, which have sets of children$\\{0,1,2,3,21\\}$ and $\\{16,17,18,19,20\\}$. \n\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{0},3}$ is given below. All elements of $S_{\\Gamma_{0},3}$ contain no vertex of degree $5$, and contain at least one child of a sink node. \n\\begin{comment}\nS_{\\Gamma_{0},2} &=& \\{(0,5), (0,10), (0,19), (1,7), (1,13), (1,16), (2,11),\\\\&&(2,15), (2,17), (3,6), \n(3,9), (3,18), (4,14), (5,9),\\\\&& (5,11), (5,17), (6,15), (6,18), (7,10), (7,16), \n(8,12),\\\\&& (9,13), (9,19), (10,19), (11,13), (11,16), (13,18), (15,17)\\} \\\\\n\\end{comment}\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(0,10,19), (1,7,16), (2,15,17), (3,6,18)\\}\\\\\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent \nNote the elements of $S_{\\Gamma_{0},3}$ do not intersect. \n\\begin{comment}\nThe vertices of degree less than $5$ that do not appear in any element of $S_{\\Gamma_{0},3}$ can be uniquely arranged into distinct pairs from $S_{\\Gamma_{0},2}$ thus: $(5,11)$, $(9,13)$. For any triple $t\\in S_{\\Gamma_{0},3}$ one of these pairs has edges to both $t$ and at least one of the other triples.\n\\end{comment}\n\nThe neighbourhoods of each vertex of degree $4$ are given below, with vertices of degree $5$ shown in bold. It can be seen that every vertex of degree $4$ has at least two neighbours of degree $4$.\n\\begin{eqnarray*}\nNb(0) &=& 9, {\\bf 12}, 15, {\\bf 20}\\\\\nNb(1) &=& 6, {\\bf 8}, 11, {\\bf 20}\\\\\nNb(2) &=& 5, 7, {\\bf 14}, {\\bf 20}\\\\\nNb(3) &=& {\\bf 4}, 10, 13, {\\bf 20}\\\\\nNb(5) &=& 2, {\\bf 8}, 13, 19\\\\ \nNb(6) &=& 1, {\\bf 12}, {\\bf 14}, 19\\\\\nNb(7) &=& 2, {\\bf 4}, {\\bf 12}, 18\\\\\nNb(9) &=& 0, 11, {\\bf 14}, 18\\\\\nNb(10) &=& 3, {\\bf 8}, {\\bf 14}, 17\\\\\nNb(11) &=& 1, {\\bf 4}, 9, 17\\\\\nNb(13) &=& 3,5,{\\bf 12}, 16\\\\\nNb(15) &=& 0, {\\bf 4}, {\\bf 8}, 16\\\\\nNb(16) &=&13, {\\bf 14}, 15, {\\bf 21}\\\\\nNb(17) &=& 10, 11, {\\bf 12}, {\\bf 21}\\\\\nNb(18) &=& 7,{\\bf 8},9,{\\bf 21}\\\\\nNb(19) &=&{\\bf 4},5,6,{\\bf 21}\n\\end{eqnarray*}\n\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&1&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,3,3,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,15), (0,20), (1,6), (1,10), (1,13), (1,20),\\\\ (2,5), \n(2,9), (2,12), (2,20), (3,4), (3,8), (3,14), (3,20),\\\\ (4,11), (4,13), (4,19), \n(5,10), (5,15), (5,19), (6,8), (6,12),\\\\ (6,19), (7,9), (7,14), (7,19), (8,15), \n(8,18), (9,13), (9,18),\\\\ (10,14), (10,18), (11,12), (11,18), (12,14), (12,17), (13,15), \n(13,17),\\\\ (14,16), (15,16), (16,21), (17,21), (18,21), (19,21), (20,21)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $18,19,20$ and $21$, which have sets of children$\\{8,9,10,11,21\\}$, $\\{4,5,6,7,21\\}$, $\\{0,1,2,3,21\\}$ and $\\{16,17,18,19,20\\}$. respectively. \n\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{1},3}$ is given below. All elements contain no vertex of degree $5$ and contains at least one child of a sink node. No element of $S_{\\Gamma_{1},3}$ contains both vertices of degree $3$ ($16$ and $17$). \n\\begin{eqnarray*}\nS_{\\Gamma_{1},3} &=&\\{(0,10,17), (1,11,16), (2,4,16), (3,5,17), (6,9,16), (7,8,17)\\}\n\\end{eqnarray*}\n\n\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{2}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&1&0&1&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(3,4,4,4,4,4,5,4,4,4,5,5,4,4,5,5,4,4,4,4,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,12), (0,15), (0,20), (1,6), (1,9), (1,11), (1,20), (2,5), \\\\(2,8), \n(2,14), (2,20), (3,4), (3,7), (3,10), (3,20), (4,11), \\\\(4,14), (4,19), (5,10), \n(5,15), (5,19), (6,7), (6,12), (6,13), \\\\(6,19), (7,15), (7,18), (8,11), (8,13), \n(8,18), (9,10), (9,14),\\\\ (9,18), (10,13), (10,17), (11,15), (11,17), (12,14), (12,17), \n(13,16),\\\\ (14,16), (15,16), (16,21), (17,21), (18,21), (19,21), (20,21)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $11$ and $21$, which have sets of children$\\{1,4,8,15,17\\}$ and $\\{16,17,18,19,20\\}$. \n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{2},3}$ is given below. All elements of degree $5$ are highlighted in bold. Every element of $S_{\\Gamma_{2},3}$ contains no vertex of degree $5$ and contains at least one child of a sink node.\n\n\\begin{eqnarray*}\nS_{\\Gamma_{2},3} &=&\\{(0,4,13), (0,4,18), (0,8,19), (0,9,19), \\\\&&(2,7,17), (3,8,12), (5,12,18)\\}\n\\end{eqnarray*}\n\\begin{comment}\nS_{\\Gamma_{2},2} &=& \\{(0,4), (0,8), (0,9), (0,10), (0,13), (0,18), (0,19),\\\\&& (1,5), (1,16), (2,6), \n(2,7), (2,17), (3,8), (3,12),\\\\&& (3,16), (4,13), (4,18), (5,12), (5,18), (7,14),\n(7,17), \\\\&&(8,12), (8,19), (9,15), (9,19), (12,18), (13,20)\\}\\\\ \n\\vspace{.25cm}\n\\noindent\nThere are $4$ sets of $4$ elements of $S_{\\Gamma_{2},3}$ such that the only vertex to appear twice is the vertex of degree $3$. Of these, there are only $2$ sets, \nsuch that the remaining $4$ vertices of degree less than $5$ constitute two non-intersecting elements of $S_{\\Gamma_{2},2}$. The sets are $\\{(0,4,13), (0,9,19), (2,7,17), (3,8,12)\\}$ and $\\{(0,4,13), (0,9,19), (2,7,17), (5,12,17)\\}$ and one of the pairs is $(1,16)$ in each case. Note that in each case, there are edges from the elements of pair $(1,16)$ to both triples containing $0$ and one of the other triples. \n\\end{comment}\n\nThe neighbourhoods of each vertex of degree $4$ are given below, with vertices of degree $5$ shown in bold. It can be seen that every vertex of degree $4$ has at least two neighbours of degree $4$ or $3$, except vertices $1$, $16$ and $17$.\n\\begin{eqnarray*}\nNb(1) &=& {\\bf 6}, 9, {\\bf 11}, {\\bf 20}\\\\\nNb(2) &=&5,8,{\\bf 14},{\\bf 20} \\\\\nNb(3) &=&4,7,{\\bf 10},{\\bf 20}\\\\\nNb(4) &=&3,{\\bf 11},{\\bf 14},19 \\\\\nNb(5) &=&2,{\\bf 10},{\\bf 15},19 \\\\\nNb(7) &=&3,{\\bf 6},{\\bf 15},18 \\\\\nNb(8) &=&2,{\\bf 11},13,18 \\\\\nNb(9) &=&1,{\\bf 10},{\\bf 14},18 \\\\\nNb(12) &=&0,{\\bf 6},{\\bf 14},17 \\\\\nNb(13) &=&{\\bf 6},8,{\\bf 10},16 \\\\\nNb(16) &=&13,{\\bf 14},{\\bf 15},{\\bf 21} \\\\\nNb(17) &=&{\\bf 10},{\\bf 11},12,{\\bf 21}\\\\\nNb(18) &=&7,8,9,{\\bf 21}\\\\\nNb(19) &=&4,5,{\\bf 6},{\\bf 21}\n\\end{eqnarray*}\n\nThe set of vertices of degree less than $5$ that are adjacent to $1$, $16$ and $17$ is $\\{9,13,12\\}$. No two of these vertices are in an element of $S_{\\Gamma_{2},3}$ with the vertex of degree $3$ ($0$). \n\n\\section*{Appendix D}\\label{appendix23}\nThe $7$ extremal graphs $(23,50)$. Note that graphs $\\Gamma_{0}-\\Gamma_{1}$ have $(deg_{3},deg_{4},deg_{5})=(0,15,8)$, graphs $\\Gamma_{2}-\\Gamma_{4}$ have $(deg_{3},deg_{4},deg_{5})=(1,13,9)$ and graphs $\\Gamma_{5}-\\Gamma_{6}$ have $(deg_{3},deg_{4},deg_{5})=(2,11,10)$.\n\n\\begin{comment}\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.7\\textwidth]{graph23_0}\n\\caption{Extremal graph $A_{0}$\n\\label{fig:graph23_0}}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.8\\textwidth]{graph23_1}\n\\caption{Extremal graph $A_{1}$\n\\label{fig:graph23_1}}\n\\end{figure}\n\n\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.75\\textwidth]{stars23A}\n\\caption{Extremal graphs on $23$ vertices, with $(deg_{3},deg_{4},deg_{5})=(0,5,18)$\n\\label{fig:graphs23A}}\n\\end{figure}\n\n\\begin{figure}\n\\centering\n\\includegraphics[width=0.8\\textwidth]{stars23B}\n\\caption{Extremal graphs on $23$ vertices, with $(deg_{3},deg_{4},deg_{5})=(1,13,9)$\n\\label{fig:graphs23B}}\n\\end{figure}\n\\end{comment}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&0&0&1&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,4,4,4,4,5,4,4,5,4,4,5,4,5,5,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,10), (0,16), (0,21), (1,6), (1,9), (1,15), (1,21), (2,5), \\\\\n(2,13), (2,14), (2,21), (3,4), (3,8), (3,12), (3,21), (4,10), (4,13),\\\\\n (4,20), (5,8), (5,15), (5,20), (6,11), (6,16), (6,20), (7,9), (7,12),\\\\ \n(7,14), (7,20), (8,16), (8,19), (9,13), (9,19), (10,11), (10,15), (10,19),\\\\\n (11,14), (11,18), (12,15), (12,18), (13,16), (13,18), (14,17),\\\\\n (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), (21,22)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $20$ and $22$, which have sets of children$\\{4,5,6,7,22\\}$ and $\\{17,18,19,20,21\\}$. \n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},3}$ and $S_{\\Gamma_{0},4}$ are given below. \n Neither set has elements containing vertices of degree $5$. Every element of $S_{\\Gamma_{0},4}$ contains at least one child of a sink node.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(0,5,18), (1,4,14), (1,8,14), (1,8,18), (2,6,12), (2,6,19),\\\\&& (2,12,19), (3,9,11), (3,9,17), (5,9,11), (6,12,19)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{(2,6,12,19)\\}\n\\end{eqnarray*}\n\n\n\\vspace{.25cm} The subgraph on vertices of degree $4$ has vertex set $\\{0,1,2,3,4,5,6,8,9,11,12,14,17,18,19\\}$ and degree sequence $$(0,2,2,3,1,2,2,3,2,3,2,3,1,2,2)$$. The only vertex of degree $0$ in this subgraph is vertex $0$. This vertex has a common neighbour of degree $5$ with the vertices of degree $1$ in the subgraph (vertices $4$ and $17$), namely $10$ and $17$ respectively. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&1&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,5,5,4,4,4,4,5,4,4,5,4,4,4,5,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,13), (0,16), (0,21), (1,6), (1,10), (1,12), (1,21), (2,5), \\\\\n(2,9), (2,15), (2,21), (3,4), (3,8), (3,11), (3,14), (3,21), (4,10),\\\\ (4,13), \n(4,15), (4,20), (5,8), (5,16), (5,20), (6,9), (6,11), (6,20),\\\\ (7,12), (7,14), \n(7,20), (8,12), (8,19), (9,13), (9,14), (9,19), (10,16),\\\\ (10,19), (11,16), (11,18), \n(12,15), (12,18), (13,18), (14,17),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{1},3}$ and $S_{\\Gamma_{1},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{1},3} &=&\\{ (0,6,8), (0,6,15), (0,15,19), (1,5,13), (1,5,14), (1,13,17), \\\\\n&&(2,7,10), (2,7,11), (2,10,18), (5,14,18), (6,8,17), (7,11,19),\\\\\n&& (8,13,17), ({\\bf 9},{\\bf 12},{\\bf 16}), (10,14,18), (11,15,19), \\}\\\\\nS_{\\Gamma_{1},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm} The subgraph on vertices of degree $4$ has vertex set $\\{0,1,2,5,6,7,8,10,11,13,14,15,17,18,19\\}$ and degree sequence $(2,2,2,2,2,2,2,2,2,2,2,2,2,2,2)$.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{2}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&1&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,4,4,4,3,4,5,4,5,4,5,5,5,4,5,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,10), (0,16), (0,21), (1,6), (1,9), (1,13), (1,21), (2,5), \\\\\n(2,12), (2,15), (2,21), (3,4), (3,8), (3,14), (3,21), (4,10), (4,12),\\\\ (4,20), \n(5,8), (5,11), (5,20), (6,16), (6,20), (7,13), (7,14), (7,20),\\\\ (8,13), (8,16), \n(8,19), (9,12), (9,14), (9,19), (10,11), (10,15), (10,19),\\\\ (11,14), (11,18), (12,16), \n(12,18), (13,15), (13,18), (14,17),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{2},3}$ and $S_{\\Gamma_{2},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{2},3} &=&\\{(0,5,9), (1,4,17), (1,5,17), (2,6,{\\bf 14}), (2,6,19), \\\\&&(2,7,19), (3,6,15), (3,6,18), (9,15,{\\bf 20})\\}\\\\\nS_{\\Gamma_{2},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{3}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&1&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,4,4,3,4,5,5,4,5,5,4,4,4,4,5,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,10), (0,13), (0,21), (1,6), (1,9), (1,16), (1,21), (2,5),\\\\ \n(2,8), (2,12), (2,15), (2,21), (3,4), (3,11), (3,14), (3,21), (4,13),\\\\ (4,16), \n(4,20), (5,10), (5,20), (6,8), (6,11), (6,20), (7,9), (7,12),\\\\ (7,14), (7,20), \n(8,13), (8,14), (8,19), (9,15), (9,19), (10,11), (10,16),\\\\ (10,19), (11,15), (11,18), \n(12,16), (12,18), (13,18), (14,17),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $21$ and $22$, which have sets of children$\\{0,1,2,3,22\\}$ and $\\{17,18,19,20,21\\}$\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{3},3}$ and $S_{\\Gamma_{3},4}$ are given below. Neither set has elements containing vertices of degree $5$. Every element of $S_{\\Gamma_{3},4}$ contains the vertex of degree $3$ at least one vertex of degree $4$ that is the child of a sink node.\n\\begin{eqnarray*}\nS_{\\Gamma_{3},3} &=&\\{(0,6,17), (1,5,13), (1,5,14), (1,5,18), (1,14,18), (3,5,9),\\\\&& (3,12,19), (5,9,13), (5,9,18), (5,13,17), (5,14,18)\\}\\\\\nS_{\\Gamma_{3},4}&= &\\{(1,5,14,18)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{4}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&1&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,5,4,3,4,4,4,4,5,4,5,5,4,5,4,5,4,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,10), (0,13), (0,21), (1,6), (1,9), (1,12), (1,16), (1,21), \\\\\n(2,5), (2,8), (2,11), (2,21), (3,4), (3,15), (3,21), (4,9), (4,13),\\\\ (4,20), \n(5,10), (5,16), (5,20), (6,8), (6,15), (6,20), (7,11), (7,14),\\\\ (7,20), (8,13), \n(8,14), (8,19), (9,11), (9,19), (10,12), (10,15), (10,19),\\\\ (11,15), (11,18), (12,14), \n(12,18), (13,16), (13,18), (14,17),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThere is one sink node, vertex $22$ which has set of children $\\{17,18,19,20,21\\}$.\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{4},3}$ and $S_{\\Gamma_{4},4}$ are given below. Neither set has elements containing vertices of degree $5$. All elements of $S_{\\Gamma_{4},4}$ contain the vertex of degree $3$ at least vertex of degree $4$ that is child of a sink node $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{4},3} &=&\\{(0,9,17), (2,4,12), (2,4,17), (3,5,14), (3,5,18),\\\\&& (3,7,16), (3,7,19), (3,16,19), (5,9,14), (7,16,19)\\}\\\\\nS_{\\Gamma_{4},4}&= &\\{(3,7,16,19)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{5}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(3,4,4,4,3,4,4,4,5,5,5,4,5,5,5,5,4,4,4,4,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,10), (0,13), (0,21), (1,7), (1,9), (1,16), (1,21), (2,6), (2,12), \\\\\n(2,15), (2,21), (3,5), (3,8), (3,14), (3,21), (4,10), (4,15), (4,20),\\\\ (5,9), \n(5,12), (5,20), (6,8), (6,11), (6,20), (7,13), (7,14), (7,20),\\\\ (8,13), (8,16), \n(8,19), (9,11), (9,15), (9,19), (10,12), (10,14), (10,19),\\\\ (11,14), (11,18), (12,16), \n(12,18), (13,15), (13,18), (14,17),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{5},3}$ and $S_{\\Gamma_{5},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{5},3} &=&\\{(0,5,17), (0,6,17), (0,11,16), (0,16,{\\bf 20}), (1,4,18),\\\\&& (1,6,{\\bf 10}), (2,7,19), (3,4,18), (4,11,16), (4,11,{\\bf 21})\\}\\\\\nS_{\\Gamma_{5},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{6}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,5,4,4,4,4,4,4,4,4,4,5,5,3,5,5,5,3,4,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,14), (0,21), (1,6), (1,10), (1,13), (1,16), (1,21),\\\\ \n(2,5), (2,9), (2,15), (2,21), (3,4), (3,8), (3,12), (3,21), (4,11),\\\\ (4,16), \n(4,20), (5,10), (5,14), (5,20), (6,8), (6,15), (6,20), (7,9),\\\\ (7,12), (7,20), \n(8,14), (8,19), (9,16), (9,19), (10,12), (10,19), (11,13),\\\\ (11,15), (11,19), (12,15), \n(12,18), (13,18), (14,16), (14,18),\\\\ (15,17), (16,17), (17,22), (18,22), (19,22), (20,22), \n(21,22)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $19$, $21$ and $22$, which have sets of children$\\{8,9,10,11,22\\}$, $\\{0,1,2,3,22\\}$ and $\\{17,18,19,20,21\\}$.\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{6},3}$ and $S_{\\Gamma_{6},4}$ are given below. Neither set has elements containing vertices of degree $5$. All elements of $S_{\\Gamma_{6},4}$ contain the two vertices of degree $3$ and at least one vertex of degree $4$ that is the child of a sink node.\n\\begin{eqnarray*}\nS_{\\Gamma_{6},3} &=&\\{(0,10,17), (2,4,18), (2,8,13), (3,5,13), (3,5,17),\\\\&& (3,9,13), (3,13,17), (5,13,17), (6,9,18), (7,8,13), \\\\&&(7,8,17), (7,13,17), (8,13,17), \\}\\\\\nS_{\\Gamma_{6},4}&= &\\{(3,5,13,17), (7,8,13,17)\\}\n\\end{eqnarray*}\n\n\n\n\n\n\n\\section*{Appendix E}\\label{appendix24}\nThe extremal graph $(24,54)$. Here $(deg_{4},deg_{5}) = (12,12)$\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&1&0&1\\\\\n0&0&1&0&0&0&0&0&0&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&1&0\\\\\n0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&1\\\\\n0&0&1&1&0&0&0&0&1&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&1&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1\\\\\n1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,5,4,5,5,4,4,5,5,5,5,5,4,5,4,4,4,5,5,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,11), (0,14), (0,19), (0,23), (1,5), (1,8), (1,13), (1,19), (2,4), \n(2,7), \\\\(2,10), (2,19), (2,22), (3,10), (3,13), (3,18), (3,21), (3,23), (4,11), \n(4,12), \\\\(4,18), (5,6), (5,9), (5,14), (5,18), (6,11), (6,17), (6,21), (6,22), \n(7,9),\\\\ (7,13), (7,17), (8,10), (8,12), (8,17), (9,12), (9,16), (9,23), (10,14),\\\\ \n(10,16),(11,13), (11,16), (12,15), (12,21), (13,15), (14,15), (15,20),\\\\ (15,22), (16,20), \n(17,20), (18,20), (19,20), (19,21), (22,23), \n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{0},3}$ is given below. Elements of $S_{\\Gamma_{0},3}$ contain only vertices of degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(0,7,18), (0,8,18), (1,4,23), (1,16,22), (4,14,17), \\\\\n&&(4,17,23), (7,14,21), (8,18,22)\\}\\\\\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe subgraph on the vertices of degree $4$, $\\Gamma_{4}$, has vertex set $$V_{4}=\\{0,1,4,7,8,14,16,17,18,21,22,23\\}$$ and edge set \n$E_{4}=\\{(0,14), (0,23), (1,8), (4,18), (7,17), (8,17), (22,23)\\}$.\n\n\\vspace{.25cm}\n\\noindent\n The only set of $4$ non-intersecting elements of $S_{\\Gamma_{0},3}$ is $$X=\\{(0,8,18),(1,16,22),(4,17,23),(7,14,21)\\}$$ The only pair $X_{1},X_{2}\\in X$ such that for all $x_{1}\\in X_{1}$, $x_{2}\\in X_{2}$ there is no edge $(x_{1},x_{2}) \\in \\Gamma$ is $X_{1}=(1,16,22)$ and $X_{2}=(7,14,21)$. \n\n\\vspace{.25cm}\n\\noindent\nThere are $10$ sets of $3$ non-intersecting elements of $S_{\\Gamma,3}$. These are shown in Table \\ref{missedTable}. In each case, at least $2$ of the vertices of degree $4$ not contained in any set are in an edge with a vertex that is in one of the sets. \n\n\\begin{table}\n\\begin{tabular}{|cc|}\n\\hline\nNon-intersecting sets & vertices missed\\\\\n\\hline\n$(0,7,18)$, $(1,16,22)$, $(4,14,17)$ & $8$, $21$, $23$\\\\ \n$(0,7,18)$, $(1,16,22)$, $(4,17,23)$ & $8$, $14$, $21$\\\\ \n$(0,8,18)$, $(1,4,23)$, $(7,14,21)$ & $16$, $17$, $22$\\\\ \n$(0,8,18)$, $(1,16,22)$, $(4,14,17)$ & $7$, $21$, $23 $\\\\\n$(0,8,18)$, $(1,16,22)$, $(4,17,23)$ & $7$, $14$, $21$\\\\ \n$(0,8,18)$, $(1,16,22)$, $(7,14,21)$ & $4$, $17$, $23$\\\\ \n$(0,8,18)$, $(4,17,23)$, $(7,14,21)$ & $1$, $16$, $22$\\\\ \n$(1,4,23)$, $(7,14,21)$, $(8,18,22)$ & $0$, $16$, $17$\\\\ \n$(1,16,22)$, $(4,17,23)$, $(7,14,21)$ & $0$, $8$, $18$\\\\ \n$(4,17,23)$, $(7,14,21)$, $(8,18,22)$ & $0$, $1$, $16$\\\\ \n\\hline\n\\end{tabular}\n\\caption{Sets of $3$ non-intersecting elements of $S_{\\Gamma,3}$ and the points missed by those elements \\label{missedTable}}\n\\end{table}\n\n\\section*{Appendix F}\\label{appendix25}\nThe $6$ extremal graphs on $25$ vertices. In each case there are $57$ edges. Graphs $\\Gamma_{0},\\ldots, \\Gamma_{2}$ have $(deg_{3},deg_{4},deg_{5})=(1,9,15)$ and graphs $\\Gamma_{3},\\ldots,\\Gamma_{5}$ have $(deg_{4},deg_{5})=(11,14)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&1\\\\\n0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1\\\\\n0&0&1&0&0&0&0&0&0&0&0&1&1&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1\\\\\n0&0&0&1&1&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&1&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1&0&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,4,5,3,4,5,5,4,5,4,4,5,5,5,5,5,5,4,4,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,11), (0,17), (0,22), (0,24), (1,7), (1,10), (1,14), (1,22), (2,6), \n(2,9), \\\\\n(2,13), (2,22), (3,5), (3,8), (3,12), (3,16), (3,22), (4,14), (4,16), \n(4,21),\\\\\n (5,9), (5,21), (5,24), (6,11), (6,12), (6,15), (6,21), (7,8), (7,13), \n(7,17),\\\\\n (7,21), (8,15), (8,20), (9,14), (9,17), (9,20), (10,12), (10,20),\n (10,24), \\\\\n(11,16), (11,20), (12,17), (12,19), (13,16), (13,19), (13,24), (14,15), (14,19),\\\\ (15,18), \n(15,24), (16,18), (17,18), (18,23), (19,23), (20,23), (21,23), (22,23) \n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},3}$ and $S_{\\Gamma_{0},4}$ are given below. Neither set has elements containing vertices of degree $5$.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(0,4,8), (0,8,19), (1,5,11), (1,5,18), (2,4,8), \\\\&&(2,4,10), (2,10,18), (5,11,19)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nNote that the only sets of $3$ parallel elements from $S_{\\Gamma_{0},3}$ which contain the vertex of degree $3$ (i.e. $4$) in the first set and for which neither of the other sets contain a neighbour of the vertex of degree $3$ (i.e. $14$, $16$ or $21$) are: are:\n$(0,4,8)$, $(1,5,11)$, $(2,10,18)$; \n$(0,4,8)$, $(2,10,18)$, $(5,11,19)$; \n$(2,4,10)$, $(0,8,19)$, $(1,5,11)$ and \n$(2,4,10)$, $(0,8,19)$, $(1,5,18)$. In all cases there is an edge from the first set to one of the other sets (via edge $(0,11)$ or $(1,10)$).\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,4,5,4,5,4,3,5,5,5,4,5,5,5,4,5,4,4,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (1,6), (1,10), (1,16), (1,22), (2,5), \n(2,9), \\\\\n(2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), (4,11), (4,14), \n(4,16),\\\\\n (4,21), (5,10), (5,12), (5,21), (6,8), (6,15), (6,21), (6,24), (7,9), \n(7,13), \\\\\n(7,21), (8,14), (8,20), (9,16), (9,20), (9,24), (10,13), (10,17), (10,20),\\\\ \n(11,12), (11,15), (11,20), (12,19), (12,24), (13,15), (13,19), (14,17), (14,19), \\\\(15,18), \n(16,18), (17,18), (17,24), (18,23), (19,23), (20,23), (21,23), (22,23) \n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $2$ and $15$ and $22$, which have sets of children $\\{5,9,14,15,22\\}$ and $\\{2,6,11,13,18\\}$.\n\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{1},3}$ and $S_{\\Gamma_{1},4}$ are given below. Vertices of degree $5$ are written in bold. Every element of $S_{\\Gamma_{1},4}$ contains the vertex of degre $3$ ($8$), a child of a sink node, and two other vertices of degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{1},3} &=&\\{(0,5,8), (0,5,16), (0,{\\bf 6},19), (0,8,16), (0,16,19), (1,7,12), (1,7,{\\bf 14}),\\\\ \n&&(3,5,8), (3,5,18), (3,8,18), (5,8,16), (5,8,18), (7,8,12), (7,8,18), \\\\&&(7,12,18), (8,12,16), (8,12,18), (8,12,{\\bf 22}), (8,{\\bf 13},16)\\}\\\\\nS_{\\Gamma_{1},4}&= &\\{(0,5,8,16)(3,5,8,18)(7,8,12,18)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nNote that the only sets of $3$ parallel elements from $S_{\\Gamma_{0},3}$ which contain the vertex of degree $3$ (i.e. $8$) in the first set and for which neither of the other sets contain a neighbour of the vertex of degree $3$ (i.e. $6$, $14$ or $20$) are:\n$(3,8,18)$, $(0,5,16)$, $(1,7,12)$; $(0,8,16)$, $(1,7,12)$, $(3,5,18)$; \n$(3,5,8)$, $(0,16,19)$, $(1,7,12)$; \n$(3,8,18)$, $(0,16,19)$, $(1,7,12)$;\n$(5,8,18)$, $(0,16,19)$, $(1,7,12)$; \n$(3,5,8)$, $(0,16,19)$, $(7,12,18)$; \n$(7,8,12)$, $(0,16,19)$, $(3,5,18)$; \n$(8,12,22)$, $(0,16,19)$, $(3,5,18)$; and \n$(8,13,16)$, $(1,7,12)$, $(3,5,18)$. \n\n In all cases there is an edge from the first set to one of the other sets (via edge $(16,18)$ or $(5,12)$).\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{2}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&1\\\\\n0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&1\\\\\n0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1\\\\\n1&0&0&0&0&0&1&0&0&0&1&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n0&1&0&0&1&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,4,4,4,5,5,3,4,5,5,5,4,5,5,5,5,4,4,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (1,6), (1,14), (1,22), (1,24), (2,5), \n(2,10),\\\\ (2,13), (2,16), (2,22), (3,4), (3,9), (3,15), (3,22), (4,10), (4,21), \n(4,24), \\\\(5,12), (5,15), (5,21), (6,9), (6,13), (6,17), (6,21), (7,8), (7,14), \n(7,16),\\\\ (7,21), (8,12), (8,20), (9,16), (9,20), (10,14), (10,17), (10,20), (11,13), \\\\\n(11,15), (11,20), (11,24), (12,17), (12,19), (12,24), (13,19), (14,15), (14,19), \\\\(15,18), \n(16,18), (16,24), (17,18), (18,23), (19,23), (20,23), (21,23), (22,23)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink node is vertex $10$ which has set of children $\\{2,4,14,17,20\\}$.\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{2},3}$ and $S_{\\Gamma_{2},4}$ are given below. Vertices of degree $5$ are written in bold. Every element of $S_{\\Gamma_{2},4}$ contains the vertex of degre $3$ ($8$), a child of a sink node, and two other vertices of degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{2},3} &=&\\{(0,4,19), (0,5,9), (0,9,19), (1,5,{\\bf 20}), (1,8,18), (3,{\\bf 7},13),\\\\&& (3,8,13), (4,8,13), (4,8,18), (4,13,18), (6,8,{\\bf 15}), (8,13,18), \\}\\\\\nS_{\\Gamma_{2},4}&= &\\{(4,8,13,18)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nNote that the only sets of $3$ parallel elements from $S_{\\Gamma_{0},3}$ which contain the vertex of degree $3$ (i.e. $8$) in the first set and for which neither of the other sets contain a neighbour of the vertex of degree $3$ (i.e. $7$, $12$ or $20$) are:\n$(6,8,15)$, $(0,5,9)$, $(4,13,18)$ and\n$(6,8,15)$, $(0,9,19)$, $(4,13,18)$. In both cases there is an edge from the first set to one of the other sets (via edge $(6,13)$.\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{3}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&1\\\\\n0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&1&0&0&1&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1\\\\\n0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1\\\\\n0&1&0&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n1&0&0&0&0&0&0&1&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,5,4,4,4,5,4,5,5,4,4,5,4,4,5,5,5,5,4,4,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,11), (0,17), (0,22), (0,24), (1,7), (1,10), (1,14), (1,16), (1,22), \n(2,6),\\\\\n (2,9), (2,15), (2,22), (3,5), (3,8), (3,13), (3,22), (4,11), (4,14), \n(4,15),\\\\\n (4,21), (5,10), (5,12), (5,17), (5,21), (6,8), (6,16), (6,21), (7,9), \n(7,13),\\\\\n (7,21), (7,24), (8,14), (8,20), (8,24), (9,17), (9,20), (10,15), (10,20), \\\\\n(11,13), (11,16), (11,20), (12,16), (12,19), (12,24), (13,19), (14,17), (14,19),\\\\\n (15,18), (15,24), (16,18), (17,18), (18,23), (19,23), (20,23), (21,23), (22,23)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nThe sink node is vertex $1$ which has set of children $\\{7,10,14,16,22\\}$\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{3},3}$ and $S_{\\Gamma_{3},4}$ are given below. Vertices of degree $5$ are written in bold. Every element of $S_{\\Gamma_{3},4}$ contains a child of the sink node, and three other vertices of degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{3},3} &=&\\{ (0,6,10), (0,6,19), (0,10,19), (2,5,{\\bf 11}), (3,4,9), (3,9,{\\bf 16}),\\\\&&\n (4,9,12), (4,12,{\\bf 22}), (6,10,13), (6,10,19), (6,13,{\\bf 17})\\}\\\\\n S_{\\Gamma_{3},4} &=&\\{(0,6,10,19)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nNote that \n\\begin{enumerate}\n\\item $S_{\\Gamma_{3},3}$ does not contain $3$ non-intersecting sets consisting entirely of vertices of degree $4$. \n\\item The only pairs of elements of $S_{\\Gamma_{3},3}$ that are parallel and have no edges between them are:\n\\begin{eqnarray*}\n&&\n((0,6,10),(3,4,9)),\\\\\n&& ((0,6,10),(4,9,12)), \\\\\n&&((0,6,19),(3,4,9)), \\\\\n&&((0,10,19),(3,4,9)), \\\\\n&&((0,10,19), (3,9,{\\bf 16})),\\\\\n&&((3,4,9),(6,10,19)), \\\\\n&&((4,9,12),(6,10,13)),\\\\\n&&((4,12,{\\bf 22}),(6,10,13))\\; {\\rm and} \\\\\n&&((4,12,{\\bf 22}),(6,13,17)). \n\\end{eqnarray*}\n\\end{enumerate}\nIf there is a set $X_{1}$, $X_{2}$, $X_{3}$ of mutually parallel elements of $S_{\\Gamma_{3},3}$ where there are no edges from $X_{1}$ to $X_{2}$ or $X_{3}$, then there would be two pairs of triples from the list above which intersect in one triple, and where the other two triples are distinct. This is not the case.\n\n\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{4}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&1\\\\\n0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&1\\\\\n0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1\\\\\n0&1&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n0&0&1&0&0&0&0&1&1&0&0&0&0&0&1&1&0&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,5,5,4,4,4,4,4,5,4,5,4,5,5,5,5,4,4,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (1,6), (1,14), (1,16), (1,22), (2,5), \n(2,10),\\\\\n (2,13), (2,22), (2,24), (3,4), (3,9), (3,12), (3,15), (3,22), (4,8), \n(4,13),\\\\\n (4,17), (4,21), (5,9), (5,16), (5,21), (6,10), (6,15), (6,21), (7,12), \n(7,21),\\\\\n (7,24), (8,16), (8,20), (8,24), (9,14), (9,20), (10,12), (10,17), (10,20), \\\\\n(11,13), (11,15), (11,20), (12,16), (12,19), (13,19), (14,17), (14,19), (14,24), \\\\(15,18), \n(15,24), (16,18), (17,18), (18,23), (19,23), (20,23), (21,23), (22,23)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nThe sink nodes are vertices $3$ and $10$ which have sets of children $\\{4,9,12,15,22\\}$ and $\\{2,6,12,17,17\\}$.\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{4},3}$ and $S_{\\Gamma_{4},4}$ are given below. Vertices of degree $5$ are written in bold. Every element of $S_{\\Gamma_{4},4}$ contains a child of the sink node, and three other vertices of degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{4},3} &=&\\{ (0,5,19), (0,6,8), (0,6,9), (0,6,19), (0,8,19), (1,7,13),\\\\&& (1,7,{\\bf 20}), (5,{\\bf 15},19), (6,8,19), (6,9,13), (7,9,13), (7,9,18), \\\\&&(7,13,18), (9,13,18), (11,{\\bf 14},{\\bf 21}), \\}\\\\\n S_{\\Gamma_{4},4} &=&\\{(0,6,8,19), (7,9,13,18)\\}\n\\end{eqnarray*}\nNote that \n\\begin{enumerate}\n\\item $S_{\\Gamma_{4},3}$ does not contain $3$ non-intersecting sets consisting entirely of vertices of degree $4$.\\item The only pairs of elements of $S_{\\Gamma_{4},3}$ that are parallel and have no edges between them are:\n\\newline \n$((0,6,8),(9,13,18))$, \n$((1,7,20),(5,15,19))$ and $((6,8,19),(7,9,18))$. \n\\end{enumerate}\n\nIf there is a set $X_{1}$, $X_{2}$, $X_{3}$ of mutually parallel elements of $S_{\\Gamma_{3},3}$ where there are no edges from $X_{1}$ to $X_{2}$ or $X_{3}$, then there would be two pairs of triples from the list above which intersect in one triple, and where the other two triples are distinct. This is not the case.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{5}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&1\\\\\n0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&1&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1\\\\\n1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0\\\\\n0&0&1&0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,5,4,4,5,5,4,5,4,5,4,5,5,5,5,4,5,4,4,4,5,5,5,5,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,14), (0,22), (1,6), (1,10), (1,13), (1,17), (1,22), \n(2,5),\\\\\n (2,12), (2,22), (2,24), (3,4), (3,9), (3,16), (3,22), (4,11), (4,13), \n(4,15), \\\\\n(4,21), (5,8), (5,14), (5,17), (5,21), (6,9), (6,12), (6,21), (7,10), \n(7,16),\\\\\n (7,21), (7,24), (8,13), (8,16), (8,20), (9,14), (9,20), (9,24), (10,15), \\\\\n(10,20), (11,12), (11,17), (11,20), (12,16), (12,19), (13,19), (13,24), (14,15),\\\\\n (14,19), \n(15,18), (16,18), (17,18), (18,23), (19,23), (20,23), (21,23), (22,23)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nThe sink node is vertex $21$ which has set of children $\\{4,5,6,7,23\\}$.\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{5},3}$ and $S_{\\Gamma_{5},4}$ are given below. Vertices of degree $5$ are written in bold. \n\\begin{eqnarray*}\nS_{\\Gamma_{5},3} &=&\\{(0,6,8), (0,6,18), (0,{\\bf 13},18), ({\\bf 1},{\\bf 14},{\\bf 16}), (3,{\\bf 5},10),\\\\&& (3,10,19), (3,17,19), (6,8,15), ({\\bf 7},17,19), (8,15,{\\bf 22})\\}\\\\\n S_{\\Gamma_{5},4} &=&\\{\\}\n\\end{eqnarray*}\nNote that \n\\begin{enumerate}\n\\item $S_{\\Gamma_{5},3}$ does not contain $3$ non-intersecting sets consisting entirely of vertices of degree $4$. \n\\item The only pairs of elements of $S_{\\Gamma_{3},3}$ that are parallel and have no edges between them are:\n\\begin{eqnarray*}\n&& ((0,6,8,),(3,10,19)),\\\\\n&&((0,6,8,),(3,17,19),\\\\ \n&&((0,6,18,),(3,5,10),\\\\\n&&((0,6,18,),(3,10,19),\\\\ \n&&((0,13,18,),(3,5,10),\\\\\n&&((3,17,19,),(6,8,15),\\\\\n&&((6,8,15,),(7,17,19),\\\\\n&&((7,17,19,),(8,15,22).\n\\end{eqnarray*}\n\\end{enumerate}\n\nIf there is a set $X_{1}$, $X_{2}$, $X_{3}$ of mutually parallel elements of $S_{\\Gamma_{3},3}$ where there are no edges from $X_{1}$ to $X_{2}$ or $X_{3}$, then there would be two pairs of triples from the list above which intersect in one triple, and where the other two triples are distinct. This is not the case.\n\n\\section*{Appendix G}\\label{appendix26}\nThe $2$ extremal graphs on $26$ vertices. In each case there are $61$ edges and $(deg_{4},deg_{5}) = (8,18)$.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,4,5,4,5,5,5,4,4,5,5,5,4,5,5,5,5,5,4,4,5,5,5,5,5,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (1,6), (1,10), (1,16), (1,22), (2,5), (2,9),\\\\ \n(2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), (4,11), (4,14), (4,16), (4,21), \\\\\n(5,10), (5,12), (5,21), (5,25), (6,8), (6,15), (6,21), (6,24), (7,9), (7,13), (7,21), \\\\\n(8,14), (8,20), (8,25), (9,16), (9,20), (9,24), (10,13), (10,17), (10,20), (11,12),\\\\ \n (11,15), (11,20), (12,19), (12,24), (13,15), (13,19), (14,17), (14,19),(15,18),\\\\ \n (16,18), (16,25), (17,18), (17,24), (18,23), (19,23), (20,23),\n(21,23), (22,23)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nThe sink node is vertex $2$ which has set of children $\\{5,9,14,15,22\\}$.\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},3}$ and $S_{\\Gamma_{0},4}$ are given below. Vertices of degree $5$ are written in bold. Note that no element of $S_{\\Gamma_{0},4}$ contains a child of the sink node.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{({\\bf 0}, {\\bf 6},19), (1,7,12), (1,7, {\\bf 14}), (3, {\\bf 5},18), (3,8,18), (7,8,12), (7,8,18),\\\\\n&&\\; (7,12,18), (8,12,18), (8,12, {\\bf 22}), \\}\\\\\nS_{\\Gamma_{0},4}&= &\\{(7,8,12,18)\\}\n\\end{eqnarray*}\nThe pairs of elements from $S_{\\Gamma_{0},3}$ that are parallel and have no edges between them are:\n$(({\\bf 0}, {\\bf 6},19), (3, {\\bf 5},18))$, $( (1,7,12), (3, 8,18))$, $((1,7, {\\bf 14}), (3, {\\bf 5},18))$ and $((1,7, {\\bf 14}), ( {\\bf 23}, {\\bf 24},25))$. Only one of these pairs consists entirely of vertices of degree $4$, namely $( (1,7,12), (3, 8,18))$.\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&1&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(4,4,5,5,5,5,5,4,4,5,5,5,4,5,5,5,4,5,5,4,5,5,5,5,5,4)$$\nEdge set is: \n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (1,6), (1,10), (1,16), (1,22), (2,5), \\\\ \n(2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), (3,25),\\\\\n (4,11), (4,14), (4,16), (4,21), (5,10), (5,12), (5,21), (5,25), (6,8),\\\\\n (6,15), (6,21), (6,24), (7,9), (7,13), (7,21),(8,14), (8,20), (8,25), \\\\\n(9,16), (9,20), (9,24), (10,13), (10,17), (10,20), (11,12), (11,15), (11,20),\\\\\n (12,19), (12,24), (13,15), (13,19), (14,17), (14,19), (15,18), (16,18),\\\\\n (17,18),(17,24), (18,23), (18,25), (19,23), (20,23) \n(21,23), (22,23) \n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{1},3}$ and $S_{\\Gamma_{1},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{1},3} &=&\\{(0,{\\bf 5},16), (0,{\\bf 6},19), (0,8,16), (0,16,19), (0,19,25), (1,7,12), \\\\\n&&\\; (1,7,{\\bf 14}),(1,7,25), (1,{\\bf 11},25), (1,19,25), (7,8,12), (7,12,{\\bf 18}),\\\\&&\\; (8,12,16), (8,12,{\\bf 22}), (8,13,16), ({\\bf 9},19,25) \\}\\\\\nS_{\\Gamma_{1},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nNote that $S_{\\Gamma_{1},3}$ has no pair of distinct sets that consist of vertices of degree $4$ and have no edges between them. \n\n\\section*{Appendix H}\\label{appendix27}\nThe $1$ extremal graph on $27$ vertices. There are $65$ edges and $(deg_{4},deg_{5}) = (5,22)$.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,4,5,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,4,5,5,5,5,5,4,4)$$\nEdge set is: \n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (1,6), (1,10), (1,16), (1,22), (2,5),\\\\ \n(2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), (4,11), (4,14), \\\\\n(4,16), (4,21), (5,10), (5,12), (5,21), (5,25), (6,8), (6,15), (6,21), \n(6,24), \\\\\n(7,9), (7,13), (7,21), (7,26), (8,14),(8,20), (8,25),\n (8,26), (9,16), (9,20),\\\\ \n(9,24), (10,13), (10,17), (10,20), (11,12), (11,15), \n(11,20), (12,19), (12,24), \\\\(12,26), (13,15), (13,19), (14,17), (14,19), (15,18),\n (16,18), (16,25), \\\\(17,18), (17,24), \n(18,23), (18,26), (19,23), (20,23), (21,23), (22,23) \n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},3}$ and $S_{\\Gamma_{0},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{({\\bf 0},{\\bf 6},19), (1,{\\bf 7},{\\bf 14}), (3,{\\bf 5},{\\bf 18}), ({\\bf 4},{\\bf 10},26), ({\\bf 23},{\\bf 24},25)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{\\}\n\\end{eqnarray*}\n\nAll of the vertices of degree $5$ in $S_{\\Gamma_{0},3}$ have $4$ neighbours of degree $5$ and one of degree $4$.\n\n\\section*{Appendix I}\\label{appendix28}\nThe $4$ extremal graph on $28$ vertices. There are $68$ edges. Graph $\\Gamma_{0}$ has $(deg_{3},deg_{4},deg_{5}) =(1,4,21,2)$, \n$\\Gamma_{1}$ has $(deg_{4},deg_{5})=(4,24)$, $\\Gamma_{2}$ has $(deg_{4},deg_{5},deg_{6}) = (6,20,2)$ and $\\Gamma_{3}$ has $(deg_{4},deg_{5},deg_{6}) = (7,18,3)$. \n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&1\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,4,5,4,5,5,6,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,4,4,3)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,27), (1,6), (1,10), (1,16), \n(1,22),(2,5), \\\\\n(2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), (4,11),\n (4,14), (4,16),\\\\ \n(4,21), (5,10), (5,12), (5,21), (5,25), (6,8), (6,15), \n(6,21), (6,24),(6,27),\\\\\n(7,9), (7,13), (7,21), (7,26), (8,14), (8,20), (8,25), \n(8,26),\n (9,16),(9,20), \\\\\n(9,24), (10,13), (10,17), (10,20), (11,12), (11,15), (11,20), \n(12,19),\n (12,24),\\\\ (12,26), (13,15), (13,19), (14,17), (14,19), (15,18), (16,18), (16,25), \n(17,18),\\\\\n (17,24), (18,23),(18,26), (19,23), (19,27), (20,23), (21,23), (22,23), \n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n All vertices of degree $5$ are the root of an embedded $S_{5,[4,4,4,4,4]}$ star.\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},3}$ and $S_{\\Gamma_{0},4}$ are given below. Vertices of degree $5$ are written in bold. Note that both elements of $S_{\\Gamma_{0},4}$ contain the vertex of degree $3$ (i.e. $27$), which is adjacent to the vertex of degree $6$ (i.e. $0$).\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(1,{\\bf 7},{\\bf 14}),({\\bf 2},26,27),(3,{\\bf 5},{\\bf 18}),(3,{\\bf 5},27),(3,{\\bf 18},27),\n\\\\&&(3,{\\bf 20},27),(3,26,27),({\\bf 4},{\\bf 10},26),({\\bf 4},{\\bf 10},27),({\\bf 4},26,27),\\\\&&\n({\\bf 5},{\\bf 18},27),({\\bf 10},26,27),({\\bf 23},{\\bf 24},25)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{(3,{\\bf 5},{\\bf 18},27) ({\\bf 4},{\\bf 10},26,27)\\}\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&1\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,4,5,5,5,5,5,4,4,4)$$\nEdge set is: \n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (1,6), (1,10), (1,16), (1,22), \n(1,26),\n(2,5), \\\\ (2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), \n(3,27), \n(4,11), (4,14),\\\\ (4,16), (4,21), (5,10), (5,12), (5,21), (5,25), (6,8), \n(6,15),\n (6,21), \\\\(6,24), (7,9), (7,13), (7,21), (7,26), (8,14), (8,20), (8,25), \n(8,27),\n (9,16), \\\\(9,20), (9,24), (10,13), (10,17), (10,20), (11,12), (11,15), (11,20),\n(12,19),\\\\ (12,24), (12,26), (13,15), (13,19), (14,17), (14,19), (15,18), (16,18), (16,25), \\\\\n(17,18), (17,24), (18,23), (18,27), (19,23), (20,23), (21,23), (22,23), (26,27),\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{1},3}$ and $S_{\\Gamma_{1},4}$ are given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{1},3} &=&\\{({\\bf 0},{\\bf 6},19), ({\\bf 0},19,27), ({\\bf 8},{\\bf 12},{\\bf 22}), ({\\bf 9},19,27),\\\\&& ({\\bf 15},25,26), ({\\bf 23},{\\bf 24},25), ({\\bf 23},25,26)\\}\\\\\nS_{\\Gamma_{1},4}&= &\\{\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{2}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&1\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,5,5,5,5,6,5,5,4,5,5,5,4,5,6,5,5,5,5,4,5,5,5,5,5,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (1,6), (1,10), (1,16), (1,22), \n(1,26), (2,5),\\\\\n (2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), \n(3,27), (4,11), (4,14),\\\\\n (4,16), (4,21), (5,10), (5,12), (5,21), (5,25), (5,27), \n(6,8), (6,15), (6,21), \\\\\n(6,24), (7,9), (7,13), (7,21), (7,26), (8,14), (8,20), \n(8,25), (9,16), (9,20),\\\\\n (9,24), (10,13), (10,17), (10,20), (11,12), (11,15), (11,20), \n(12,19), (12,24), \\\\\n(13,15), (13,19), (14,17), (14,19), (14,26), (15,18), (16,18), (16,25), \n(17,18),\\\\\n (17,24), (18,23), (18,27), (19,23), (20,23), (21,23), (22,23), (26,27)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nVertices of degree $6$ are $5$ and $14$. Sets $S_{\\Gamma_{2},3}$ and $S_{\\Gamma_{2},4}$ are given below. Vertices of degree $5$ are written in bold.\nElements of both sets contain only vertices of degree $4$ and $5$. Note that for every vertex $p$ of degree $6$ ($5$ and $14$) all $3$ elements of $S_{\\Gamma_{2},4}$ contain precisely one vertex (of degree $4$) adjacent to $p$. (If $p=5$, vertices adjacent to $p$ in the elements of $S_{\\Gamma_{2},4}$ are $27$, $12$ and $25$ respectively, and if $p=14$, vertices adjacent to $p$ in the elements of $S_{\\Gamma_{2},4}$ are $19$, $8$ and $26$ respectively.\n\\begin{eqnarray*}\nS_{\\Gamma_{2},3} &=&\\{({\\bf 0},{\\bf 6},19),({\\bf 0},{\\bf 6},27),({\\bf 0},19,27),({\\bf 6},19,27),({\\bf 7},8,12),\\\\&&\n({\\bf 7},8,{\\bf 18}),({\\bf 7},12,{\\bf 18}),(8,12,{\\bf 18}),(8,12,{\\bf 22}),({\\bf 9},19,27),\\\\&&\n({\\bf 15},25,26),({\\bf 23},{\\bf 24},{\\bf 25}),({\\bf 23},{\\bf 24},26),({\\bf 23},25,26),({\\bf 24},25,26)\\}\\\\\nS_{\\Gamma_{2},4}&= &\\{({\\bf 0},{\\bf 6},19,27) ({\\bf 7},8,12,{\\bf 18}) ({\\bf 23},{\\bf 24},25,26)\\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{3}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{cccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&1\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&1&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,4,5,5,5,6,6,4,4,5,5,5,4,5,5,5,5,5,5,5,5,5,5,5,5,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,26), (1,6), (1,10), (1,16), \n(1,22), (2,5),\\\\\n (2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), \n(3,27), (4,11), (4,14),\\\\\n (4,16), (4,21), (5,10), (5,12), (5,21), (5,25), (5,27), \n(6,8), (6,15), (6,21),\\\\\n (6,24), (6,26), (7,9), (7,13), (7,21), (8,14), (8,20), \n(8,25), (9,16), (9,20),\\\\\n (9,24), (10,13), (10,17), (10,20), (11,12), (11,15), (11,20), \n(12,19), (12,24),\\\\\n (13,15), (13,19), (14,17), (14,19), (15,18), (16,18), (16,25), (17,18), \n(17,24),\\\\ (18,23), (18,27), (19,23), (19,26), (20,23), (21,23), (22,23), (26,27), \n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nVertices of degree $6$ are $0$, $5$ and $6$. Sets $S_{\\Gamma_{3},3}$ and $S_{\\Gamma_{3},4}$ are given below. Vertices of degree $5$ are written in bold.\nElements of both sets contain only vertices of degree $4$ and $5$. Note that for every vertex $p$ of degree $6$ ($0$, $5$ and $6$) both elements of $S_{\\Gamma_{3},4}$ contain precisely one vertex (of degree $4$) adjacent to $p$. (If $p=0$, vertices adjacent to $p$ in the elements of $S_{\\Gamma_{3},4}$ are $7$ in both cases, if $p=5$ or $p=6$ vertices adjacent to $p$ in the elements of $S_{\\Gamma_{3},4}$ are $1$, and $8$ respectively.\n\n\\begin{eqnarray*}\nS_{\\Gamma_{3},3} &=&\\{ (1,7,12),(1,7,{\\bf 14}),(1,7,27),(1,{\\bf 11},27),(1,{\\bf 14},27),\n({\\bf 4},{\\bf 10},26),(7,8,12),\\\\ &&(7,8,{\\bf 18}),(7,8,27),(7,12,{\\bf 18}),\n(7,{\\bf 14},27),(8,12,{\\bf 18}),(8,12,{\\bf 22}),({\\bf 23},{\\bf 24},25) \\}\\\\\nS_{\\Gamma_{3},4}&= &\\{(1,7,{\\bf 14},27) (7,8,12,{\\bf 18}) \\}\n\\end{eqnarray*}\n\n\n\n\n\\section*{Appendix J}\\label{appendix29}\nThe $1$ extremal graph on $29$ vertices. There are $72$ edges and $(deg_{4},deg_{5},deg_{6}) =(5,20,4)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccccccccccccccccccccccccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&1&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,4,5,5,5,6,6,5,5,5,5,5,5,5,5,5,5,5,6,5,5,5,5,5,5,4,4,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,27), (1,6), (1,10), (1,16), (1,22),\\\\\n (2,5), (2,9), (2,14), (2,15), (2,22), (3,4), (3,13), (3,22), (3,24), \n(3,28),\\\\(4,11), \n(4,14), (4,16), (4,21), (5,10), (5,12), (5,21), (5,25), \n(5,28),(6,8),\\\\(6,15), (6,21), (6,24), (6,27), (7,9), (7,13), (7,21), \n(7,26), (8,14), \\\\\n(8,20), (8,25), (8,26), (9,16), (9,20), (9,24), (10,13), \n(10,17), (10,20),\\\\ (11,12), \n(11,15), (11,20), (12,19), (12,24), (12,26), (13,15), (13,19), \\\\(14,17), (14,19),(15,18), \n(16,18), (16,25), (17,18), (17,24), (18,23), \\\\(18,26), (18,28), (19,23), (19,27), (20,23), \n(21,23), (22,23), (27,28)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma,3}$ and $S_{\\Gamma,4}$ are given below. Vertices of degree $5$ are written in bold, all other elements have degree $4$..\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{(1,{\\bf 7},{\\bf 14}),(1,{\\bf 7},28),(1,{\\bf 11},28),(1,{\\bf 14},28),\\\\&&({\\bf 2},26,27),\n({\\bf 4},{\\bf 10},26),({\\bf 4},{\\bf 10},27),({\\bf 4},26,27),\\\\&&({\\bf 7},{\\bf 14},28),(10,26,27)\n({\\bf 23},{\\bf 24},25)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{(1,{\\bf 7},{\\bf 14},28) ({\\bf 4},{\\bf 10},26,27)\\}\n\\end{eqnarray*}\n\n\\section*{Appendix K}\\label{appendix30}\nThe $1$ extremal graph on $30$ vertices. There are $76$ edges and $(deg_{4},deg_{5},deg_{6}) =(4,20,6)$. \n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc ccccc ccccc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&1&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,5,5,5,5,6,6,6,5,5,5,5,5,5,6,5,5,5,6,5,5,5,5,5,5,4,4,4,5,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,27), (1,6), (1,10),\\\\\n (1,16), (1,22), (1,29), (2,5), (2,9), (2,14), (2,15), (2,22),\\\\ \n(3,4), (3,13), (3,22), (3,24), (3,28), (4,11), (4,14), (4,16), \\\\\n(4,21), (5,10), (5,12), (5,21), (5,25), (5,28), (6,8), (6,15), \\\\\n(6,21), (6,24), (6,27), (7,9), (7,13), (7,21), (7,26), (7,29),\\\\\n(8,14), (8,20), (8,25), (8,26), (9,16), (9,20), (9,24), (10,13), \\\\\n(10,17), (10,20), (11,12), (11,15), (11,20), (12,19), (12,24), \\\\\n(12,26), (13,15), (13,19), (14,17), (14,19), (14,29), (15,18), \\\\\n(16,18), (16,25), (17,18), (17,24), (18,23), (18,26), (18,28), \\\\\n(19,23), (19,27), (20,23), (21,23), (22,23), (27,28), (28,29)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma,3}$ and $S_{\\Gamma,4}$ are given below. Vertices of degree $5$ are written in bold.\nElements of $S_{\\Gamma,4}$ contain two vertices of degree $4$ and two of degree $5$.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&\\{({\\bf 2},26,27),({\\bf 4},{\\bf 10},26),({\\bf 4},{\\bf 10},27),({\\bf 4},26,27),({\\bf 10},26,27),\\\\&&\n({\\bf 15},25,29),({\\bf 23},{\\bf 24},25),({\\bf 23},{\\bf 24},29),({\\bf 23},25,29),({\\bf 24},25,29)\\}\\\\\nS_{\\Gamma_{0},4}&= &\\{({\\bf 4},{\\bf 10},26,27) ({\\bf 23},{\\bf 24},25,29)\\}\n\\end{eqnarray*}\n\nLet $s_{1}=({\\bf 4},{\\bf 10},26,27)$ and $s_{2}=({\\bf 23},{\\bf 24},25,29)$ (the two elements of $S_{\\Gamma_{0},4}$ respectively). \nConsider the embedded $S_{6,[4,4,4,4,4,3]}$ stars with roots $6$ and $14$ respectively. The first star has children $27$ and $24$ which are from $s_{1}$ and $s_{2}$ respectively, with degrees $4$ and $5$, and the second star has children $4$ and $29$ which are from $s_{1}$ and $s_{2}$ respectively, with degrees $5$ and $4$ respectively. \n\n\\section*{Appendix L}\\label{appendix31}\nThe $2$ extremal graph on $31$ vertices. There are $80$ edges. Graph $\\Gamma_{0}$ has $(deg_{4},deg_{5},deg_{6}) =(3,20,8)$ and $\\Gamma_{1}$ has $(deg_{5},deg_{6})=(26 ,5)$.\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc ccccc ccccc c}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1&0&1&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,5,5,5,6,6,6,6,5,5,6,5,5,5,6,5,5,5,6,5,5,5,5,5,5,4,5,5,5,4,4)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,27), (1,6), (1,10), (1,16),\\\\ \n(1,22), (1,29), (2,5), (2,9), (2,14), (2,15), (2,22), (3,4), (3,13),\\\\ \n(3,22), (3,24), (3,28), (4,11), (4,14), (4,16), (4,21), (4,30), (5,10),\\\\ \n(5,12), (5,21), (5,25), (5,28), (6,8), (6,15), (6,21), (6,24),\\\\ (6,27),\n (7,9), (7,13), (7,21), (7,26), (7,29), (8,14), (8,20),\\\\ (8,25), (8,26),\n (9,16), (9,20), (9,24), (10,13), (10,17), (10,20),\\\\ (10,30), (11,12),\n (11,15), (11,20), (12,19), (12,24), (12,26), (13,15),\\\\ \n(13,19), (14,17),\n(14,19), (14,29), (15,18), (16,18), (16,25),\\\\ (17,18), (17,24), (18,23), \n(18,26), (18,28), (19,23), (19,27), \\\\(20,23), (21,23), (22,23), (26,30), (27,28), (27,30), \n(28,29)\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n\\begin{comment}\nDon't need this stuff now:\nSet $S_{\\Gamma_{0},3}$ is given below. Vertices of degree $4$ are written in bold. All sets contain at least one vertex of degree $4$. \n\n\\begin{eqnarray*}\nS_{\\Gamma_{0},3} &=&(15,{\\bf 25},{\\bf 29}) (15,{\\bf 25},{\\bf 30}) (15,{\\bf 29},{\\bf 30}) (23,24,{\\bf 25}) \\\\&&(23,24,{\\bf 29}) (23,24,{\\bf 30}) (23,{\\bf 25},{\\bf 29}) (23,{\\bf 25},{\\bf 30})\n\\\\&& (23,{\\bf 29},{\\bf 30}) (24,{\\bf 25},{\\bf 29}) \n(24,{\\bf 25},{\\bf 30}) (24,{\\bf 29},{\\bf 30}) ({\\bf 25},{\\bf 29},{\\bf 30})\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{0},4}$ is given below. Vertices of degree $4$ are written in bold. All sets contain at least one vertex of degree $4$. \n\\begin{eqnarray*}\nS_{\\Gamma_{0},4} &=&\\{(15,{\\bf 25},{\\bf 29},{\\bf 30}) (23,24,{\\bf 25},{\\bf 29}) (23,24,{\\bf 25},{\\bf 30})\\\\&& (23,24,{\\bf 29},{\\bf 30}) (23,{\\bf 25},{\\bf 29},{\\bf 30}) (24,{\\bf 25},{\\bf 29},{\\bf 30}) \\}\n\\end{eqnarray*}\n\n\\vspace{.25cm}\n\\noindent\n\\end{comment}\n\nSet $S_{\\Gamma_{0},5}$ is given below. Vertices of degree $5$ are written in bold, all other vertices have degree $4$.\n\\begin{eqnarray*}\nS_{\\Gamma_{0},5} &=&\\{({\\bf 23},{\\bf 24},25,29,30)\\}\n\\end{eqnarray*}\n\n\\begin{comment}\nDon't need this now\n\\vspace{.25cm}\nThere are $16$ pairs of sets, $(X,Y)$ where $X\\in S_{\\Gamma_{0},4}$, $Y\\in S_{\\Gamma_{0},2}$, where $X$ and $Y$ do not intersect and there is no edge from an element of $X$ to an element of $Y$. These $16$ sets are:\n\n\\vspace{.25cm}\n\\noindent\n$\n\\begin{array}{c|c}\n$X$ & $Y$ \\\\\n\\hline\n(15,25,29,30) & (9,19) \\\\\n(15,25,29,30) & (12,22) \\\\\n(15,25,29,30) & (17,21) \\\\\n(15,25,29,30) & (23,24) \\\\\n(23,24,25,29) & (2,26)\\\\\n(23,24,25,29) & (2,27)\\\\ \n(23,24,25,29) & (2,30) \\\\\n(23,24,25,29) & (15,30) \\\\\n(23,24,25,30) & (1,11)\\\\ \n(23,24,25,30) & (11,28) \\\\\n(23,24,25,30) & (11,29) \\\\\n(23,24,25,30) & (15,29) \\\\\n(23,24,29,30) & (8,13) \\\\\n(23,24,29,30) & (13,16) \\\\\n(23,24,29,30) & (13,25) \\\\\n(23,24,29,30) & (15,25) \\\\\n\\hline\n\\end{array}\n$\n\n\n\\vspace{.25cm}\nThere are $3$ distinct pairs of sets, $(X,Y)$ where $X,Y\\in S_{\\Gamma_{0},3}$, where $X$ and $Y$ do not intersect and there is no edge from an element of $X$ to an element of $Y$. These $5$ sets are:\n\n\\vspace{.25cm}\n\\noindent\n$\n\\begin{array}{c|c}\n$X$ & $Y$ \\\\\n\\hline\n(15,25,30) & (23,24,29)\\\\ \n(15,29,30) & (23,24,25)\\\\\n(23,24,30) & (15,25,29) \\\\\n\\hline\n\\end{array}\n$\n\\end{comment}\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{1}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc ccccc ccccc c}\n0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0\\\\\n0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0\\\\\n0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0\\\\\n1&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&1&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&1&0&0&1&0&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0\\\\\n0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&1&0&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&1&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&1&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(5,5,5,5,5,5,5,5,6,6,6,6,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6)$$ Edge set is:\n\\begin{eqnarray*}\n(0,11), (0,15), (0,19), (0,23), (0,29), (1,7), (1,10), (1,14), (1,18), \\\\\n(1,29), (2,6), (2,9), (2,17), (2,22), (2,29), (3,5), (3,8), (3,13),\\\\\n (3,21), (3,29), (4,10), (4,13), (4,16), (4,22), (4,28), (5,11), (5,17),\\\\\n (5,20), (5,28), (6,8), (6,15), (6,18), (6,28), (7,9), (7,12),\\\\ \n(7,23),(7,28), (8,14), (8,16), (8,23), (8,27), (9,13), (9,19),\\\\\n (9,20), (9,27), (10,15), (10,17), (10,21), (10,27), (11,12), (11,18),\\\\\n (11,22), (11,27), (12,16), (12,21), (12,26), (13,18), (13,26), (14,19),\\\\ \n(14,22), (14,26), (15,20), (15,26), (16,20), (16,25), (17,23), \\\\\n(17,25), (18,25), (19,21), (19,25), (20,24), (21,24), (22,24),\\\\\n (23,24), (24,30), (25,30), (26,30), (27,30), (28,30), (29,30)\n\\end{eqnarray*}\n\n\\begin{comment}\nDon't need this now\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{1},4}$ is given below. All pairs of sets intersect at least twice. \n\\begin{eqnarray*}\nS_{\\Gamma_{1},4} &=&\\{(3,7,15,22) (3,7,15,25) (3,7,22,25) \\\\&&(3,15,22,25) (7,15,22,25) \\}\n\\end{eqnarray*}\n\\end{comment}\n\n\\vspace{.25cm}\n\\noindent\nSet $S_{\\Gamma_{1},5}$ is given below. Vertices of degree $5$ are written in bold.\n\\begin{eqnarray*}\nS_{\\Gamma_{1},5} &=&\\{({\\bf 3},{\\bf 7},{\\bf 15},{\\bf 22},{\\bf 25})\\}\n\\end{eqnarray*}\n\n\\begin{comment}\nDon't need this now:\n\n\\vspace{.25cm}\n\\noindent\nThere are no pairs of sets, $(X,Y)$ where $X\\in S_{\\Gamma_{1},4}$, $Y\\in S_{\\Gamma_{1},2}$, where $X$ and $Y$ do not intersect and there is no edge from an element of $X$ to an element of $Y$. Similarly there are no such sets where $X,Y\\in S_{\\Gamma_{1},3}$. \n\\end{comment}\n\\section*{Appendix M}\\label{appendix32}\nThe $1$ extremal graph on $32$ vertices. There are $85$ edges and $(deg_{5},deg_{6})=(22 ,10)$.\n\n\n\\vspace{.25cm}\n\\noindent\n$\\Gamma_{0}$ has incidence array:\n\n\\vspace{.25cm}\n\\noindent\n\\begin{tiny}\n$\n\\begin{array}{ccccc ccccc ccccc ccccc ccccc ccccc cc}\n0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0\\\\\n0&0&0&0&0&1&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&1&0&0&0\\\\\n0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0\\\\\n0&0&1&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0\\\\\n0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&1&0&0&1&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&0&1&0&0\\\\\n0&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&1&1&0&0&0&0&0\\\\\n0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&0&0&0&1&0&0&0&0&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&0\\\\\n0&0&0&1&0&0&0&1&0&0&1&0&0&0&0&1&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&0&1&0&1&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&1&0&0\\\\\n0&0&1&0&0&0&1&0&0&0&0&1&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\\\\n0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0\\\\\n1&0&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&1&0&0&1&0&1&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&0&0&0&0&0&1&0&0&0&1&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n1&1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&1&1&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&1&0&0&1&0&0&1&0&0&1&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n1&0&0&0&0&1&0&0&1&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\\\\n0&0&0&0&0&0&0&1&1&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&1&0\\\\\n1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0\\\\\n0&0&0&1&0&1&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&1&0&1&0&0\\\\\n0&1&0&0&0&0&0&1&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&1\\\\\n0&0&0&0&1&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&0&0&0&1\\\\\n0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&1&1&0&0&0&1&1&0\\\\\n\\end{array}\n$\n\\end{tiny}\n\n\n\n\\vspace{.25cm}\n\\noindent\nand degree sequence: $$(6,5,5,5,6,6,6,6,5,5,6,5,5,5,6,5,5,5,6,5,5,5,5,6,6,5,5,5,5,5,5,5)$$ Edge set is:\n\\begin{eqnarray*}\n(0,7), (0,11), (0,17), (0,22), (0,25), (0,27), (1,6), (1,10), (1,16), \\\\\n(1,22), (1,29), (2,5), (2,9), (2,14), (2,15), (2,22), (3,4), (3,13), \\\\\n(3,22), (3,24), (3,28), (4,11), (4,14), (4,16), (4,21), (4,30), (5,10),\\\\\n (5,12), (5,21), (5,25), (5,28), (6,8), (6,15), (6,21), (6,24), (6,27),\\\\\n (7,9), (7,13), (7,21), (7,26), (7,29), (8,14), (8,20),\\\\ (8,25),\n(8,26), (9,16), (9,20), (9,24), (10,13), (10,17),\\\\ (10,20), (10,30),\n(11,12), (11,15), (11,20), (12,19), (12,24), \\\\(12,26), (13,15), (13,19),\n (14,17), (14,19), (14,29), (15,18),\\\\ (16,18), (16,25), (17,18), (17,24),\n (18,23), (18,26),(18,28),\\\\ (19,23), (19,27), (20,23), (21,23), (22,23),\n (23,31), (24,31),\\\\ (25,31), (26,30), (27,28), (27,30), (28,29), (29,31), (30,31)\n\\end{eqnarray*}\n\n\n\\vspace{.25cm}\n\\noindent\nSets $S_{\\Gamma_{0},j}$, for $j>2$ are empty. \n\n\n\\begin{eqnarray*}\nS_{\\Gamma_{0},2} &=&\\{\n(1,11), (1,12), (1,19), (1,26), (2,26), (2,27),\\\\ &&\n(2,30), (2,31), (3,8), (3,20), (3,25), (3,26),\\\\&& (8,13), (8,22), (8,28), (9,19), (9,27), (9,28), \\\\&&(9,30), (11,28), \n(11,29), (11,31), (12,16), (12,22),\\\\&& (12,29), (13,16), (13,25), (13,31), (15,25), (15,29), \\\\&&\n(15,30), (15,31), (16,19), (16,27), (17,21), (19,25), \\\\&&(20,27), (20,28), (20,29), (22,26),\n(22,30)\\}\n\\end{eqnarray*}\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}}