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The balanced complete multipartite blow-up $K_r(t)$ replaces each vertex of the complete graph $K_r$ by an independent set of $t$ vertices, placing every possible edge between different sets and none inside a set. Every fixed graph of chromatic number at most $r$ embeds into $K_r(t)$ for sufficiently large $t$.

For the binomial random graph $G(n,1/2)$, the chromatic number is $(1+o(1))n/(2{\log }_{2}n)$ with high probability. A first moment method bounds the independence number above. Edge-disjoint clique packing in the complement graph, followed by an edge-exposure martingale and a union bound over moderately large vertex sets, supplies independent sets for greedy colouring by removing independent sets.

A vertex $v$ is colour-critical if deleting it lowers the chromatic number by one. If $\chi (F)=r+1$ and $\chi (F-v)=r$, an embedding of $F-v$ into a complete $r$-partite graph inside one neighbourhood extends to an embedding of $F$. This supplies maximum-degree obstructions in extremal graph theory.

The Lovász number of a finite simple graph is the semidefinite program valuewhere $J$ is the all-ones matrix. Its value on the complement graph gives a lower bound for the chromatic number.

For integers $n\ge 2k\ge 2$, the chromatic number of the Kneser graph is $n-2k+2$. The upper bound colours by the least element among the first $n-2k+1$ elements, with one final colour for all other sets. The Gale hemisphere lemma and the Lusternik-Schnirelmann-Borsuk theorem give the matching lower bound.

If a fixed graph $F$ has chromatic number $r+1$, every extremal $F$-free graph $G$ on $n$ vertices has minimum degree of a graph $(1-1/r+o(1))n$. Use a high-minimum-degree multipartite stability subgraph $H$: replacing any vertex by one adjacent to all but one class of $H$ preserves $F$-freeness. Any supposed new copy of $F$ can replace the new vertex by an unused common neighbour in the omitted class. Extremality therefore gives the lower bound on every degree of a vertex; the Erdős-Stone theorem gives the upper bound on the average degree of a vertex.

In this enumeration problem, monotone means subgraph-closed graph property: both deleting vertices and deleting edges preserve membership. This is a decreasing convention, rather than the increasing-edge convention often used for random-graph thresholds. Assume first that the property is proper and has graphs of arbitrarily large orders. These are the usual nondegenerate hypotheses for the formula.

Put $r+1=\min \{\chi (F):F\notin \mathcal{Q}\}$. Every graph of chromatic number at most $r$ belongs to $\mathcal{Q}$. In particular, all subgraphs of a fixed balanced Turan graph $T_r(n)$ belong. Choosing each of its edges independently as present or absent givesFor the upper bound, fix $F\notin \mathcal{Q}$ with $\chi (F)=r+1$. Every member of $\mathcal{Q}$ is $F$-free, by closure under subgraphs.

We use the following standard regularity consequences. For every sufficiently small regularity error $\eta$, large minimum cluster count $q_0$, and fixed positive density cutoff $d$, the Szemerédi regularity lemma gives an equitable partition into $q_0\le q\le M$ clusters and an exceptional set of at most $\eta n$ vertices, with at most $\eta q^2$ irregular pairs. The graph embedding lemma for regular pairs says that a fixed graph with a prescribed proper colouring embeds into the corresponding clusters if all its required cross pairs are regular with density at least $d$, provided $\eta$ is sufficiently small relative to $d$ and the graph. Hence the reduced graph of a regularity partition, joining regular pairs of density at least $d$, is $K_{r+1}$-free: such a clique would embed $F$.

By the Turan theorem, this reduced graph has at most $(1-1/r)q^2/2$ edges. On its dense pairs allow arbitrary choices of original edges, giving at most $(1-1/r)n^2/2+O(n)$ free binary choices. All edges within clusters, touching the exceptional set, or lying in irregular pairs contribute $O((\eta +1/q_0)n^2)$ further choices. On each remaining pair there are at most $d$ times its number of possible edges. The elementary binary entropy estimatebounds the total sparse-pair contribution by $H_2(d)n^2/2$. The partition and pair classifications have at most $(M+1{)}^n{2}^{O(M^2)}$ descriptions, whose logarithm is $O(n)$ since $M$ is fixed. Choose $q_0$ large, then $d$ small, then $\eta$ sufficiently small; their total error can be made arbitrarily small. This provesIt is the enumeration of subgraph-closed graph properties.

For the hereditary extension, a hereditary graph property is closed under induced subgraphs. Define the clique-independent partition class $\mathcal{C}(a,b)$, for $0\le b\le a$, to consist of graphs partitionable into $a$ possibly empty classes, exactly $b$ designated as cliques and the other $a-b$ designated as independent sets; cross edges are unrestricted. Thus $a$ is the total number of classes. The colouring number of a hereditary graph property isUse value zero if this set is empty, and infinity for the class of all graphs.

$r(\mathcal{C}(a,b))=a$ for $a\ge 1$. The lower bound follows from the definition. For the upper bound, fix any $s\in \{0,\dots ,a+1\}$ and a balanced partition of a large $N$-vertex set into $a+1$ classes with $s$ clique classes. Independent choices of cross edges give $2^{(1-1/(a+1))N^2/2-O(1)}$ distinct graphs in $\mathcal{C}(a+1,s)$. On the other hand, all graphs in $\mathcal{C}(a,b)$ can be described by at most $a^N$ partitions, each with at most $(1-1/a)N^2/2$ unrestricted cross pairs. The former count is strictly larger for large $N$, so $\mathcal{C}(a+1,s)$ is not contained in $\mathcal{C}(a,b)$ for any $s$. A contained class with even more parts would contain one of these $(a+1)$-part classes by making unused parts empty, so it is also impossible.

The main changes for the hereditary graph enumeration theorem are as follows. The lower bound now uses a contained class $\mathcal{C}(r,b)$: within-class edges are fixed as complete or empty, and the balanced cross pairs still give $(1-1/r)n^2/2-O(1)$ free edge choices. For the upper bound, since no $\mathcal{C}(r+1,b)$ is wholly contained in $\mathcal{P}$, choose one forbidden induced graph $F_b\in \mathcal{C}(r+1,b)\setminus \mathcal{P}$ for each $b=0,\dots ,r+1$.

Use the standard induced regularity template lemma: for a fixed finite forbidden induced family and any error tolerance, a regularity refinement gives bounded-size templates with internal clique/independent types and cross pairs classified as sparse, almost complete, or intermediate. A clique of intermediate pairs on $a$ classes realizes every fixed induced pattern consistent with the internal types of those classes. This induced embedding conclusion uses a Ramsey refinement within clusters and regularity for both edges and nonedges; the total exceptional-pair cost can be made arbitrarily small. Applied to the $F_b$, a clique on $r+1$ intermediate classes would realize the forbidden $F_b$ whose number of clique types is $b$. Thus the graph of intermediate pairs is $K_{r+1}$-free. Only intermediate pairs contribute unrestricted binary choices; sparse pairs and the missing edges in almost-complete pairs each have the same entropy bound as before. The Turan theorem and the earlier counting argument now giveThis formula assumes $1\le r<\infty$. For the universal property the count is exactly $2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$; a hereditary property of bounded order has $r=0$ and has no graphs on sufficiently large vertex sets, so the displayed expression with $1/r$ is not applicable.

For a proper subgraph-closed graph property, any contained $\mathcal{C}(a,b)$ must have $b=0$: if $b\ge 1$, it contains arbitrarily large complete graphs, and closure under subgraphs would force every graph into the property. Consequently its colouring number is exactly one less than the least chromatic number of an excluded graph. The excluded family of an intersection is the union of the two excluded families, whose minimum chromatic number is the smaller of their minima. Therefore $r(\mathcal{P}\cap \mathcal{Q})=\min (r(\mathcal{P}),r(\mathcal{Q}))$ for subgraph-closed properties, with the universal case treated by $r=\infty$.

For hereditary properties take $\mathcal{P}=\mathcal{C}(2,0)$ and $\mathcal{Q}=\mathcal{C}(2,2)$, both of colouring number two. A graph in their intersection is both bipartite and a union of two cliques. A clique in a bipartite graph has at most two vertices; hence every such graph has at most four vertices. The intersection contains no complete clique-independent class on arbitrarily large orders, so $r(\mathcal{P}\cap \mathcal{Q})=0<2$. This also demonstrates why the bounded-order exception to the enumeration formula is necessary.

Expose the independent coordinates one at a time and define the Doob exposure martingaleFor a fixed exposed prefix, couple the future coordinates identically under two choices of $Z_i$. The coordinate-change hypothesis bounds the difference of the resulting conditional expectations by $c_i$. Thus the increment $D_i=M_i-M_{i-1}$ has conditional mean zero and lies in a conditional interval of length at most $c_i$. This range bound, rather than merely $|D_i|\le c_i$, gives the sharp constant in the McDiarmid inequality.

Here is a proof of the needed Hoeffding lemma. For a mean-zero random variable $X$ in an interval of length $c$, let $h(u)=\log \mathbb{E}{e}^{uX}$. Its second derivative is the variance under the exponentially tilted distribution. A random variable in $[a,b]$ has variance at most $(b-a{)}^2/4$: the inequality $\mathbb{E}[(X-a)(b-X)]\ge 0$ gives $\mathrm{Var}X\le (\mathbb{E}X-a)(b-\mathbb{E}X)\le (b-a{)}^2/4$. Hence $h^{\prime \prime }(u)\le c^2/4$, and $h(0)=h^{\prime }(0)=0$ imply $h(u)\le u^2{c}^2/8$, for either sign of $u$.

Apply this conditionally to each increment and iterate the conditional expectation:For $u>0$, Markov's inequality bounds the upper tail by $\exp (-ut+u^2\sum c_i^2/8)$. Taking $u=4t/\sum c_i^2$ and then applying the same argument to $-f$ givesAt $t=0$ the bound is immediate. If all $c_i$ vanish, $f$ is constant on the product support and the positive tails are zero, so that degenerate case is handled directly.

For a binomial random graph, let coordinate $i$ be the entire vector of edges from vertex $i$ to vertices of smaller label. The coordinates are independent finite probability spaces. Changing coordinate $i$ changes only edges incident to that one vertex. Deleting the vertex gives the same graph under both outcomes, and each outcome's chromatic number is either that graph's chromatic number or one more. Thus the coordinate range is at most one. Taking all $c_i=1$ and $t=\lambda \sqrt{n}$ proves vertex exposure for chromatic number:Using individual edges as coordinates would give a weaker scale; grouping the incident edges is what yields the required bound.

For the expectation asymptotic, take fixed $0<p<1$, put $b=1/(1-p)$, and write $L={\log }_{b}n$. We outline both bounds and the step that turns probability estimates into an expectation estimate. For each fixed $\gamma >0$, the expected number of independent sets of size $k=\lceil (2+\gamma )L\rceil$ isIts logarithm is $k\log n-\frac{1}{2}k(k-1)\log b-O(k\log k)=-{\Omega }_{\gamma }((\log n{)}^2)$. Thus Markov's inequality gives $\alpha (G)\le (2+\gamma )L$ with probability tending to one, and $\chi (G)\ge n/\alpha (G)$ yields the corresponding lower bound on its expectation.

For the upper bound, set $m_0=\lceil n/(\log n{)}^2\rceil$ and $k=\lfloor (2-\gamma )L\rfloor$, with $0<\gamma <1$. The key uniform fact is that every vertex subset of size at least $m_0$ contains an independent $k$-set with probability tending to one. For a fixed subset of size $m$, let $X$ count its independent $k$-sets and let $\mu =\left(\genfrac{}{}{0ex}{}{m}{k}\right)b^{-\left(\genfrac{}{}{0ex}{}{k}{2}\right)}$. The normalized dependency sum in Janson inequality is bounded byThe $j=2$ term is $O(k^4/m^2)$. The remaining overlap terms give the same order or less: use $\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)/\left(\genfrac{}{}{0ex}{}{m}{k}\right)\le (2k^2/m{)}^j$ for large $n$, and split the sum at $k/2$. For the lower half the terms beyond $j=2$ decrease at the initial endpoint and are exponentially small at the other endpoint; for the upper half, $k\le (2-\gamma /2){\log }_{b}m$ makes every endpoint exponent negative of order $(\log n{)}^2$. Also $1/\mu$ is exponentially small on that scale. The exponential form of Janson inequality therefore gives, uniformly for $m\ge m_0$,for a positive constant depending only on $p,\gamma$. A union bound over at most $2^n$ subsets succeeds, because $m_0^2/k^4$ is of order $n^2/(\log n{)}^8\gg n$. This is independent sets in every large subset of a dense random graph.

On that event, repeatedly remove an independent $k$-set and assign it one new colour until fewer than $m_0$ vertices remain; colour the remainder individually. This gives $\chi (G)\le n/k+m_0$. The failure probability is exponentially small compared with $1/\log n$, and always $\chi (G)\le n$, so the failure event contributes negligibly to the expectation. Combining the upper and lower bounds and then letting $\gamma \downarrow 0$ provesThis is the chromatic number of a binomial random graph asymptotic. The slash in the printed expression must be read with $2{\log }_{1/(1-p)}n$ as the denominator; a multiplicative reading would eventually exceed $n$. At the endpoint probabilities $p=0$ and $p=1$, the chromatic numbers are respectively one and $n$ for $n\ge 1$, so this logarithmic formula is intended for $0<p<1$.

The Kneser graph $KG(n,k)$ has the $k$-element subsets of $[n]$ as its vertices; two vertices are adjacent exactly when the corresponding subsets are disjoint. The Lovász theorem on Kneser graphs determines its chromatic number. A graph colouring is therefore a partition of the $k$-sets into intersecting families.

First construct a graph colouring with $d+2=n-2k+2$ colours. If a $k$-set meets $[d+1]$, give it the colour of its least element. Give every remaining $k$-set colour $d+2$. Two sets with one of the first $d+1$ colours share that colour's element. The last colour consists of $k$-sets in a $(2k-1)$-element ground set, so it too is an intersecting family. Thus $\chi (KG(n,k))\le d+2$.

For the converse we establish the Gale hemisphere lemma explicitly, using a signed moment curve. Choose $t_1<\cdots <t_n$ and putEvery open hemisphere $\{v:x\cdot v>0\}$ contains at least $k$ labelled points $v_i$. To see this, put $p(t)=x_0+x_{1}t+\cdots +x_d{t}^d$. It is a nonzero polynomial of polynomial degree at most $d$. First suppose none of the $p(t_i)$ vanishes. Write $b_i=\mathrm{sgn}((-1{)}^{i}p(t_i))$. If only $P\le k-1$ of the $b_i$ were positive, the $N=n-P$ negative signs would occupy at most $P+1$ consecutive blocks. There would be at leastadjacent pairs with both $b_i,b_{i+1}$ negative. Each such pair forces $p(t_i)$ and $p(t_{i+1})$ to have opposite signs. The intermediate value theorem would give $d+1$ distinct roots of a polynomial, a contradiction.

If $p$ vanishes at $z$ of the sample points, then $z\le d$. A Lagrange interpolation polynomial $q$ of polynomial degree at most $z-1$ can be chosen with $(-1{)}^{i}q(t_i)<0$ at all those points. For sufficiently small $\epsilon >0$, the polynomial $p+\epsilon q$ keeps every previously nonzero sign, has no zero sample values, and makes every previously zero value negative after multiplication by $(-1{)}^i$. Its positive count is exactly the positive count of $p$, so the preceding argument proves the open hemisphere assertion also in this case. For $d=0$, the labelled points alternate between the two points of $S^0$; distinct labels, rather than distinct positions, are what is needed.

Suppose now that $KG(n,k)$ had a graph colouring with at most $d+1$ colours, padding the palette with unused colours if necessary. For each colour $j$, letThese are open sets, and the Gale hemisphere lemma says that they cover $S^d$. If both $x$ and $-x$ belonged to $U_j$, two $k$-sets of colour $j$ would lie in opposite open hemispheres. They would be disjoint, hence adjacent in the Kneser graph, contradicting the graph colouring. The Lusternik-Schnirelmann-Borsuk theorem excludes this cover. Therefore the lower bound matches the explicit colouring:

The Erdős-Stone theorem says that, for a fixed graph $F$ with chromatic number $q\ge 2$,The lower bound comes from the balanced Turan graph with $q-1$ parts; the theorem supplies the matching asymptotic upper bound. Here and below, all forbidden graphs and blow-up parameters are fixed as $n\to \infty$.

First obtain an almost spanning, almost complete multipartite core. Put ${\alpha }_r=1-1/r$. We give the stability argument, rather than infer pointwise degree bounds by differentiating an asymptotic edge-count formula. Suppose first $r\ge 2$.

We need two elementary ingredients. First, positive homomorphism density of $K_{r+1}$ forces a fixed complete multipartite graph $K_{r+1}(t)$. To see this, perform vertex cloning in a graph, replacing one vertex by $t$ independent twins. Conditional on the images of the other vertices, its contribution to the homomorphism density changes from a number $p$ to $p^t$; Jensen inequality gives $\mathbb{E}{p}^t\ge (\mathbb{E}p{)}^t$. Duplicating each original vertex therefore givesA positive lower bound on the right supplies a positive proportion of all maps of the blow-up. Only $O(n^{(r+1)t-1})$ maps fail to be injective, so for large $n$ at least one map is an embedding. Consequently a $K_{r+1}(t)$-free sequence has $o(n^{r+1})$ copies of $K_{r+1}$. The clique removal lemma, proved in the next solution, then deletes $o(n^2)$ edges to give a $K_{r+1}$-free graph $J_n$ with $e(J_n)={\alpha }_r{n}^2/2-o(n^2)$.

Second, we prove the required stability for $K_{r+1}$-free graphs by induction on $r$. For $r=1$ such a graph is edgeless. In an $m$-vertex $K_{r+1}$-free graph $J$, take a vertex of maximum vertex degree $D$, put $A=N(v)$ and $B=V(J)\setminus A$, and note that $J[A]$ is $K_r$-free. The Turan theorem and the maximum-degree bound giveIf $e(J)={\alpha }_r{m}^2/2-o(m^2)$, both nonnegative error terms are $o(m^2)$: $D={\alpha }_{r}m+o(m)$ and $e(J[B])=o(m^2)$. The deficit in the bound for $e(J[A])$ is likewise $o(m^2)$. Induction partitions $A$ into $r-1$ classes with a total of $o(m^2)$ internal edges; adjoining $B$ gives an $r$-class partition with the same property. Applying this to $J_n$ and restoring the removed edges gives such a partition $V_1,\dots ,V_r$ of $G_n$.

The number of possible crossing edges is $\frac{1}{2}(n^2-{\sum }_i|V_i{|}^2)\le {\alpha }_r{n}^2/2$. Since the actual crossing-edge count is within $o(n^2)$ of this maximum, $|V_i|=n/r+o(n)$ for every $i$, and there are $o(n^2)$ missing crossing edges. Choose $a_n\to 0$ so slowly that the number of vertices missing more than $a_{n}n$ neighbours in other classes is $o(n)$. Delete those vertices and all internal edges. The remaining r-partite graph $Q_n$ has $n-o(n)$ vertices and every vertex has crossing vertex degree at least ${\alpha }_{r}n-o(n)$. The Turan theorem, or the size of a largest part, supplies the matching upper bound for its minimum vertex degree. ThusThis is a high-minimum-degree multipartite stability subgraph. The subgraph need not be spanning; deleting the exceptional $o(n)$ vertices is essential. For $r=1$, the spanning edgeless subgraph already proves the assertion.

Extremality forces the minimum degree. Choose $t\ge |V(F)|$. A proper $(r+1)$-graph colouring embeds $F$ in $K_{r+1}(t)$, so every $F$-free $H_n$ is also $K_{r+1}(t)$-free. By the Erdős-Stone theorem, its extremal edge count has the needed asymptotic size. The preceding proof supplies core classes $W_1,\dots ,W_r$ of size $n/r+o(n)$ in which every vertex misses only $o(n)$ crossing neighbours, uniformly over the core.

Delete an arbitrary vertex $x$ from $H_n$ and introduce a new vertex $z$ adjacent precisely to $W_2\cup \cdots \cup W_r$ with $x$ removed if necessary. The new graph is still $F$-free. Indeed any new copy would use $z$, all its neighbours in that copy would be in the specified core classes, and their set of common neighbours in $W_1$ has size $n/r-o(n)$. A vertex of this common neighbourhood outside the bounded vertex set of the copy replaces $z$, producing an existing copy of $F$ in $H_n$, a contradiction. For $r=1$ the new vertex is isolated and the same conclusion is immediate.

The replacement adds ${\alpha }_{r}n-o(n)$ edges. Since $H_n$ has the maximum possible number of edges, $d_{H_n}(x)\ge {\alpha }_{r}n-o(n)$. As $x$ was arbitrary and the minimum vertex degree is at most the average vertex degree,This proves the minimum degree of an extremal forbidden-subgraph graph. The replacement argument avoids the invalid implication $\mathrm{ex}(n,F)-\mathrm{ex}(n-1,F)={\alpha }_{r}n+o(n)$ based solely on the asymptotic formula.

A colour-critical vertex also bounds the maximum degree. Suppose a vertex $x$ of $H_n$ has $d(x)\ge {\alpha }_{r}n+\epsilon n$ for some fixed $\epsilon >0$. Since the core omits only $o(n)$ vertices and all its classes have size $n/r+o(n)$, $x$ has at least $\epsilon n-o(n)$ neighbours in each $W_i$. Otherwise, even counting every vertex in all other classes and outside the core would give too small a vertex degree.

Inside these $r$ neighbourhood sets, greedily choose a fixed number $t\ge |V(F)|$ of vertices in each class. Every previously chosen vertex excludes only $o(n)$ crossing nonneighbours, so each next candidate set still has positive linear size. This produces $K_r(t)$ inside $N(x)$. A proper $r$-graph colouring of $F-v$ embeds it in that complete multipartite graph; map $v$ to $x$. All required incident edges exist, so this embeds $F$, a contradiction. The case $r=1$ simply uses enough neighbours to embed the edgeless $F-v$. Hence $\Delta (H_n)\le {\alpha }_{r}n+o(n)$, and the average-degree lower bound completes the answer: