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The graph Ramsey number $R(s,t)$ is the least $N$ for which every graph on $N$ vertices has a clique of size $s$ or an independent set of size $t$; $R(s)=R(s,s)$ is the diagonal Ramsey number. The base cases are $R(2,t)=t$ and $R(s,2)=s$.

Assume the smaller parameters already have finite graph Ramsey numbers, and take $N=R(s-1,t)+R(s,t-1)$. For any vertex, either its neighbourhood has at least $R(s-1,t)$ vertices, or its nonneighbourhood has at least $R(s,t-1)$ vertices. In the first case it contains an independent $t$-set, or an $(s-1)$-clique which extends with the chosen vertex. In the second it contains an $s$-clique, or a $(t-1)$-independent set which extends with that vertex. Induction proves existence and

Use an ideal carrier-sense multiple access scheme on the interference graph. An active station finishes its transmission after an independent unit-rate exponential distribution waiting time. An inactive station has an independent attempt clock with an exponential distribution of rate $e^{{\theta }_r}$; an attempt succeeds only when all its neighbors are inactive. Attempts that are blocked do not change the state. Thus, on the independent sets represented by $S$, the continuous-time Markov chain has transition ratesThe empty independent set can be reached by successive deactivations, and any other independent set can be reached from it by successive activations. Hence this finite chain is an irreducible Markov chain. The weight $a_{\theta }(n)=e^{\theta \cdot n}$ satisfiesBy detailed balance for a continuous-time Markov chain, its unique stationary law isThis is a finite exponential family. Its natural parameter of an exponential family is $\theta$ and its mean parameter of an exponential family is the vector of active fractions.

The throughput region of an interference graph is exactly $\Lambda =\mathrm{conv}S$. To see the possibly less immediate inclusion, suppose a random feasible schedule has mean $v\ge \lambda$. Retain each of its active vertices $r$ independently with probability ${\lambda }_r/v_r$, interpreting zero coordinates as always deleted. A subset of an independent set remains an independent set; the thinned schedule has mean exactly $\lambda$. Thus $\lambda$ lies in the convex hull of $S$. The opposite inclusion follows from the definition using equality. Moreover $0,e_1,\dots ,e_R\in S$, so this convex hull has full dimension.

Now choose $\epsilon >0$ such that $\mu =\lambda +\epsilon 1$ remains in the interior of $\mathrm{conv}S$. Use interior moment matching in a finite exponential family. Consider the convex functionThere is a ball of radius $\delta >0$ around $\mu$ contained in $\mathrm{conv}S$. Maximizing a linear functional over the convex hull is the same as maximizing over its generating set, so, for every $\theta$,The last inequality uses the point $\mu +\delta \theta /\Vert \theta \Vert$ when $\theta \ne 0$. Hence $F$ is a coercive function and has a finite minimizer ${\theta }^{\ast }$. Differentiating the finite normalizing sum givesThis covariance matrix is a positive-definite matrix: a linear functional constant on all states of positive weight must be constant on $0,e_1,\dots ,e_R$, so its coefficients vanish. Thus $F$ is a strictly convex function; its minimizer is unique. At that minimizer the desired strict service margins areFor a backlogged station that transmits at unit speed whenever active, the long-run throughput is ${\mathbb{E}}_{\theta }[n_r]$, by the ergodic theorem for a finite continuous-time Markov chain. Therefore the ideal carrier-sense multiple access scheme can supply a strict service margin for every arrival vector in the interior of the throughput region of an interference graph. The local attempt rates can realize the whole interior capacity region in this sense. Boundary points may require parameters tending to infinity, and the argument gives no uniform delay or mixing-time bound near the boundary.

Symmetrization for an arbitrary real coefficient. Among graphs maximizing $f(G)=e(G)-c{k}_3(G)$, choose one maximizingFor a vertex $x$, put $L(x)=d(x)-ce(G[N(x)])$. This is the contribution of edges and triangles in a graph containing $x$. If $u,v$ are nonadjacent, Zykov symmetrization replacing $u$ by a clone of $v$ changes $f$ by $L(v)-L(u)$; the opposite replacement changes it by $L(u)-L(v)$. Maximality implies $L(u)=L(v)$, so both changes preserve $f$.

Let $S=N(v)\setminus N(u)$ and $T=N(u)\setminus N(v)$. In the first replacement the vertex degrees in $S$ increase by one and those in $T$ decrease by one; in the opposite replacement these changes reverse. The changes to $d(u{)}^2+d(v{)}^2$ cancel when the two replacements are added. Since $(d+1{)}^2+(d-1{)}^2-2d^2=2$,If the two vertex neighbourhoods differ, one replacement increases $Q$, a contradiction. Thus every pair of nonadjacent vertices has the same vertex neighbourhood. Nonadjacency, with equality allowed, is an equivalence relation: if $u$ and $v$ are nonadjacent and $v$ and $w$ are nonadjacent, their identical vertex neighbourhoods forbid $uw$ as well. Its classes are independent sets and every cross-class edge is present. ThereforeThis edge-triangle symmetrization works for either sign of $c$.

The exact supporting line. For $n\ge 3$, writeWe prove the triangle support line between bipartite and tripartite Turan graphs by maximizing $e-c_{\ast }{k}_3$. First $c_{\ast }\ge 2/n$. For $n=3,4,5$, its values are respectively $1,1/2,1/2$. For $n\ge 6$, the exact balancing of the Turan graph gives $e_3\ge n^2/3-1/3$, while $e_2\le n^2/4$ and the arithmetic-geometric mean inequality gives $t_3\le n^3/27$. Hence

By edge-triangle symmetrization a maximizing graph can be taken complete multipartite graph; among such maximizers use one with the fewest nonempty parts. If there are at least four parts, let $a,b$ be the two smallest sizes. Then $a+b\le n/2$. Merging them loses $ab$ edges and $ab(n-a-b)$ triangles in a graph, so the objective changes byThis contradicts the choice of the number of parts. Thus at most three parts remain.

For three part sizes $a\ge b\ge d>0$, the objective isIf $1-c_{\ast }b\le 0$, merging the parts of sizes $a,d$ does not decrease it, again contradicting minimality. If $1-c_{\ast }b>0$ and $a-d\ge 2$, transferring one vertex from the largest to the smallest part raises $ad$ by $a-d-1>0$ and strictly increases the objective. Thus a three-part maximizer has all sizes differing by at most one, and is $T_3(n)$. A maximizer with at most two parts has objective at most $e_2$, since it has no triangles in a graph and the Mantel theorem bounds its edges. Finally,Therefore every graph satisfies $e(G)-c_{\ast }{k}_3(G)\le e_2$. At the prescribed edge count this givesFor $n\le 2$, the conclusion is simply the nonnegative lower bound zero. All rounding in the Turan graphs has been retained.

One edge above the bipartite threshold. Write $n=2m$. We prove the Triangle lower bound one edge above the Mantel threshold by induction on $m\ge 2$, allowing at least $m^2+1$ edges. For $m=2$, five edges on four vertices give two triangles in a graph, and adding edges preserves this bound.

We will also use the following consequence of an inductive bound: on $2a$ vertices, at least $a^2+q$ edges, for an integer $q\ge 1$, force at least $a+q-1$ triangles in a graph. Delete surplus edges first to leave exactly $a^2+q$. Repeatedly delete an edge in a triangle in a graph until $a^2+1$ edges remain. The Mantel theorem guarantees such a triangle in a graph at every step; each deletion destroys at least one distinct triangle in a graph. Apply the inductive bound to the remaining graph.

Now take exactly $m^2+1$ edges. If every edge lies in a triangle in a graph, the edge-triangle incidence bound gives $3k_3(G)\ge m^2+1$. This is at least $3m$ for $m\ge 3$; the base case was handled separately.

Otherwise choose an edge $uv$ in no triangle in a graph. Its endpoints have disjoint vertex neighbourhoods, so $d(u)+d(v)\le 2m$. Deleting $u,v$ removes exactly $d(u)+d(v)-1$ edges and leaves at least $(m-1{)}^2+1$ edges. If $d(u)+d(v)\le 2m-1$, it leaves at least $(m-1{)}^2+2$ edges, and the strengthened inductive bound gives at least $m$ triangles in a graph already.

If $d(u)+d(v)=2m$, induction gives at least $m-1$ triangles in a graph after deletion. There must be an additional triangle in a graph containing $u$ or $v$. Otherwise both $N(u)$ and $N(v)$ are independent sets; since they are disjoint and cover all vertices, $G$ would be bipartite graph, contradicting $e(G)>m^2$ by the Mantel theorem. HenceThe bound is sharp: add one internal edge to one class of the complete bipartite graph $K_{m,m}$. Its triangles in a graph are precisely that edge together with each of the $m$ vertices in the other class.

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.

A $k$-colouring assigns one of $k$ colours to every vertex so that adjacent vertices have different colours. The chromatic number $\chi (G)$ is the least such $k$. An independent set contains no adjacent pair, and the independence number $\alpha (G)$ is the maximum size of such a set.

For a concrete first example, start with the five-cycle $C_5$ and apply the Mycielski construction $r-2$ times. Each application raises the chromatic number by one and preserves triangle-freeness. The resulting graph has chromatic number $r+1>r$ and contains no triangle, hence no complete graph $K_r$ for $r\ge 3$.

We next prove the stronger high-girth assertion by the probabilistic method. For a large integer $n$, takeIf $X$ counts cycles of lengths $3,\dots ,g$, thenBy the first moment method, with probability tending to one $X<n/2$.

Let $m=\lfloor n/(2k)\rfloor$. The expected number of independent sets of size $m$ is at mostbecause $p{m}^2=\Theta (n^{1+1/(2g)})$ dominates $m\log (en/m)=O(n)$. Hence with positive probability $G$ has fewer than $n/2$ short cycles and $\alpha (G)<m$.

Delete one vertex from each cycle of length at most $g$. The remaining graph $H$ has no such cycle, has at least $n/2$ vertices, and still has $\alpha (H)<n/(2k)$. Since each colour class is independent,This proves that there are graphs of arbitrarily high girth and chromatic number.

For the final claim, begin instead withThe expected number $T$ of triangles isso with probability tending to one $T<n$. Put $m=\lceil n^{0.7}\rceil$. The expected number of independent $m$-sets is bounded bybecause the positive term is $O(n^{0.7}\log n)$ whereas the negative term has order $n^{0.71}$.

Thus some such $G$ has fewer than $n$ triangles and no independent set of size $m$. Delete one vertex from every triangle; more than $n$ vertices remain and the graph is triangle-free. Take any induced subgraph on exactly $n$ vertices. Vertex deletion cannot increase the independence number, so the resulting graph satisfiesThis is the triangle-free graph with sub-power independence number construction.

If the edge $uv$ lies in no triangle, no vertex can be adjacent to both endpoints; hence the neighbourhoods satisfy $\Gamma (u)\cap \Gamma (v)=\varnothing$.

Suppose $|\Gamma (u)|+|\Gamma (v)|=2n$. The two disjoint neighbourhoods then partition $V(G)$. If both were independent, every edge would run between these two parts, so $G$ would be bipartite andcontrary to $e(G)=n^2+1$. Thus at least one neighbourhood contains an edge, and that edge forms a triangle with $u$ or $v$.

We now prove by induction on $n\ge 2$ the Triangle lower bound one edge above the Mantel threshold: every graph on $2n$ vertices with at least $n^2+1$ edges has at least $n$ triangles. Part (ii) is the base case. For $n\ge 3$, if every edge lies in a triangle, part (iii) givesOtherwise choose an edge $uv$ lying in no triangle and put $s=d(u)+d(v)\le 2n$. Deleting $u,v$ removes $s-1$ edges, so the graph $G^{\prime }=G-u-v$ hasIf $s=2n$, induction gives at least $n-1$ triangles in $G^{\prime }$, while the nonindependence proved above supplies a further triangle through $u$ or $v$.

If $s\le 2n-1$, then $e(G^{\prime })\ge (n-1{)}^2+2$. Induction first supplies a triangle $\Delta$ in $G^{\prime }$. Delete one edge of $\Delta$; the remaining graph still has at least $(n-1{)}^2+1$ edges, so induction supplies $n-1$ triangles avoiding that edge. Together with $\Delta$, these give at least $n$ distinct triangles in $G$.

Let $\mathcal{A}$ be an independent set in the graph. Every member has size $p^2\equiv 0(\mathrm{mod}p)$, while for distinct $A,B\in \mathcal{A}$, independence means $|A\cap B|$ is nonzero modulo $p$. Apply part i withThis givesMoreover,

Put $k=p^{3p}$. Parts ii and iii give a graph with neither a clique nor an independent set of size $k$. Its number of vertices satisfies the standard binomial lower boundConsequently the modular-intersection graph Ramsey lower bound gives

For every fixed $C>0$,The first quantity exceeds the second when $p>3C$. Thus $p^{p^2}$ is eventually larger than $k^C$ for every fixed $C$, so this lower bound grows faster than every polynomial in $k$.

Colour every vertex of a graph independently and uniformly with three colours, and retain precisely the edges whose endpoints have different colours. Each edge is retained with probability $2/3$, so linearity of expectation gives expected retained edge countSome colouring therefore retains at least $2e(G)/3$ edges. Its retained subgraph is three-colourable, and deleting edges if necessary leaves exactlyedges without increasing its chromatic number. This is the three-colourable two-thirds subgraph lemma.

To prove sharpness, take $G=K_n$. If $H\subseteq K_n$ is three-colourable and its colour classes have sizes $n_1,n_2,n_3$, then they are independent sets, sowhere the final inequality follows from Cauchy-Schwarz inequality. Since $e(K_n)=n(n-1)/2$,Given $\epsilon >0$, choose $n$ with $2/[3(n-1)]\le \epsilon$. Then every three-colourable subgraph has at most $(2/3+\epsilon )e(G)$ edges.

Fix $k\ge 1$ and sample $G$ from the Erdős-Rényi model with $p=\frac{1}{2}{n}^{-2/3}$. Let $X$ count copies of the complete bipartite graph $K_{3,3}$. By the expected subgraph count in the Erdős-Rényi model,

Put $m=\lceil n/k\rceil$ and let $Z$ count independent set of size $m$. If the chromatic number satisfies $\chi (G)\le k$, some colour class has at least $m$ vertices, so $Z\ge 1$. Moreover,because the positive part of the exponent is $O(n)$ whereas $p\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ has order $n^{4/3}$. The first moment method therefore gives $\mathbb{P}(\chi (G)\le k)\le \mathbb{E}Z=o(1)$.

Using Markov inequality for $X$ and the union bound,for all sufficiently large $n$. Hence at least one such $G$ contains no $K_{3,3}$ and has $\chi (G)>k$.

Apply the Shearer independence bound for a triangle-free graph. It states that an $n$-vertex triangle-free graph of maximum degree at most $d$ has an independent set of sizefor an absolute $c>0$. The result is immediate for bounded $d$ after reducing $c$, while the theorem gives the asserted logarithmic gain for $d$ large. Thus

For completeness, the key input in Shearer's proof is to expose a random independent set one degree scale at a time. Triangle-freeness makes every neighbourhood independent, so conditioning on earlier choices creates no edges inside the available neighbours. The expected gain at degree scale $j$ is $\Omega (n_j/d)$; summing over the $\Theta (\log d)$ nonempty scales gives $\Omega (n\log d/d)$. The entropy, or hard-core-model, form of the argument makes this scale calculation rigorous without losing vertices counted at adjacent scales.

Letwhere the positive constant $c$ will be chosen small, and sample the binomial random graph $G(N,p)$. If $X$ counts its copies of $C_4$, thenIf $Y$ counts its independent $k$-sets, thenHere $pk\sim c^{-3/4}\log k$. Choosing $c$ sufficiently small makes the negative exponential term dominate $k\log (eN/k)$, so $\mathbb{E}Y=o(1)$. By Markov inequality, with positive probability $X<N/2$ and $Y=0$.

Starting from such a graph, delete one vertex from each remaining $C_4$. This random alteration method removes fewer than $N/2$ vertices, destroys every $C_4$, and cannot create an independent set of order $k$. The resulting graph has at least $N/2$ vertices, so its edges and nonedges give a red-blue colouring with neither a red $C_4$ nor a blue $K_k$. Consequently

For feasible independent set activation vectors $S$, the throughput region is their convex hull. Its downward closure is the same set because deleting active vertices preserves feasibility. Ideal carrier-sense multiple access can supply strict margins above every vector in its interior by adjusting attempt rates.