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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.

Let $X$ count $k$-cliques in a binomial random graph on $m$ vertices and $Y$ count distinct ordered pairs sharing an edge. The clique conflict graph gives $Z_k\ge X^2/(X+Y)$: order its vertices randomly and keep each one preceding all its neighbours, then apply the Cauchy-Schwarz inequality. Another application gives $\mathbb{E}{Z}_k\ge (\mathbb{E}X{)}^2/(\mathbb{E}X+\mathbb{E}Y)$. If $\mathbb{E}Y/(\mathbb{E}X{)}^2=O(k^4/m^2)$ and $1/\mathbb{E}X$ is no larger, then $\mathbb{E}{Z}_k=\Omega (m^2/k^4)$. Changing one edge changes $Z_k$ by at most one, so the Azuma-Hoeffding inequality for an edge-exposure martingale yields the displayed exponential absence bound. This supports a union bound over exponentially many adaptive candidate subsets.

A disconnected binomial random graph has a connected component of a graph with at most half its vertices. The union bound over its possible vertex sets givesThe final geometric series bound tends to zero for fixed $p>0$.

For fixed real $c$, the number of isolated vertices in a binomial random graph with $p=(\log n+c)/n$ converges to a Poisson distribution of mean $e^{-c}$. Its fixed factorial moments tend to $(e^{-c}{)}^k$. A union bound for small connected components, together with exclusion of intermediate components, shows that isolated vertices are asymptotically the only obstruction to connectivity. Thus the displayed limit is the zero-mass of that Poisson distribution.

Expose the independent edge indicators of a binomial random graph in a fixed order, and set $M_i=\mathbb{E}[Z\mid \text{the first }i\text{ indicators}]$ for an integrable statistic $Z$. This is a martingale from $\mathbb{E}Z$ to $Z$. If changing one edge changes $Z$ by at most $c$, coupling the unexposed indicators gives $|M_i-M_{i-1}|\le c$, permitting the Azuma-Hoeffding inequality.

For fixed $k\ge 2$, a binomial random graph has no $k$-cliques with probability tending to one if ${\mu }_n\to 0$, and has at least one with probability tending to one if ${\mu }_n\to \infty$. The first assertion uses the first moment method. The second uses the overlap formula for the variance of a clique count: overlapping $k$-sets with $l\ge 2$ contribute $O(n^{-l}{p}^{-\left(\genfrac{}{}{0ex}{}{l}{2}\right)})$, and each contribution tends to zero above this threshold. The fixed-size qualification matters.

Fix $k\ge 1$, ${\gamma }_0>1/(k+1)$, and ${\gamma }_0\le \gamma \le 1-\omega (n)/\log n$ with $\omega (n)\to \infty$. A binomial random graph has one giant component of the displayed order and all other graph components are tree components of orders at most $k$, with high probability. The expected value and variance of the isolated vertex count concentrate it around $n^{1-\gamma }\to \infty$. A Cayley formula upper bound for connected sets excludes component sizes $k+1$ through a small fixed fraction of $n$; the empty-graph cut bound excludes the remaining sizes up to $n/2$. An extra-edge count excludes cyclic small graph components. The expected value of the number of vertices in small nonisolated tree components is $O(nd{e}^{-2d})=o(n{e}^{-d})$, where $d=np$. The Markov inequality completes the count of vertices outside the unique large graph component.

Let $X$ count unordered vertex pairs with no common neighbour in a binomial random graph. Each other vertex provides the required pair of edges independently, with success probability $p^2$. A fixed pair has no common neighbour with probability $(1-p^2{)}^{n-2}$. Summing its indicators gives the displayed expectation.

For the number $X$ of $k$-cliques in a binomial random graph $G(n,p)$, with $0<p<1$ and $\mu =\mathbb{E}X$,Two $k$-sets overlapping in $j$ vertices share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges. Overlaps of zero or one vertex give independent indicators; $j=k$ includes the diagonal terms. The identity is useful for the second moment method.

Use the random alteration method. Take arbitrarily large $n$ divisible by $2k$, and sample a binomial random graph with edge probability $p=n^{-1+1/(2g)}$. Let $Z$ count cycles of lengths $3,\dots ,g-1$. For each length $\ell$, its expected count is at most $(np{)}^{\ell }/(2\ell )$, soThe sum is empty if $g=3$. By the Markov inequality, $Z\le n/2$ with probability tending to one.

Set $s=n/(2k)$. The expected number of independent $s$-vertex sets isbecause the negative exponent has order $n^{1+1/(2g)}$, dominating the term of order $n$. Thus with probability tending to one there is no independent set of size $s$.

Choose a realization having both properties. Delete one vertex from each of its short cycles, using at most $Z$ deletions. The induced remaining graph $H$ has at least $n/2$ vertices and girth at least $g$. Deletion cannot create a new independent set, so $\alpha (H)\le s-1$. Every colour class is independent, givingIn particular a graph of the required girth and chromatic number exists. This proves graphs of arbitrarily high girth and chromatic number rather than invoking that existence theorem.

Write $K_s(t)$ for a balanced complete multipartite blow-up with $s$ classes of size $t$. In the nonvacuous range $0<\epsilon <1/r$, the answer isWe prove both bounds, including the quantitative lower bound.

First, suppose a bipartite graph with parts $A,B$, of sizes $m,N$, has at least $pmN$ edges. For any integer $s\le pm/2$, counting pairs consisting of an $s$-subset of $A$ and a common neighbour in $B$ givesThe inequality follows from discrete convexity: $\left(\genfrac{}{}{0ex}{}{x+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{x}{s}\right)=\left(\genfrac{}{}{0ex}{}{x}{s-1}\right)$ is nondecreasing, so balancing integer degrees of vertices minimizes their sum. Consequently some $S$ hasThe product estimate uses $\lfloor pm\rfloor -j\ge pm-s$ for $0\le j<s$. This is a common neighbourhood from bipartite density bound.

We next prove by induction on $q\ge 1$ that, for every fixed $\alpha >0$, a graph $J$ of order $N$ with minimum degree at least $(1-1/q+\alpha )N$ contains $K_{q+1}(\lfloor c\log N\rfloor )$, for some $c=c(q,\alpha )>0$ and all sufficiently large $N$. Impossible minimum-vertex degree parameters need no argument. For $q=1$, take two copies of $V(J)$ with adjacency given by $J$. The preceding count with $p=\alpha$ and $s=\lfloor c\log N\rfloor$, where $c\log (2/\alpha )<1/2$, gives at least $\sqrt{N}$ common neighbours. The selected vertices and their common neighbours are disjoint in $J$, since a vertex is not its own neighbour. Selecting $s$ common neighbours gives $K_2(s)$.

For $q\ge 2$, the minimum degree hypothesis implies the induction hypothesis for $q-1$, for example with excess $1/[q(q-1)]$. Thus $J$ contains $K_q(m)$ with $m=\lfloor a\log N\rfloor$ for a positive constant $a$. Pair the vertices in its classes into $m$ disjoint transversal cliques $C_1,\dots ,C_m$, each of order $q$. By the union bound, every $C_i$ has at leastcommon neighbours. Form an auxiliary bipartite graph between these $m$ cliques and the $N$ vertices, using common-neighbour incidence, and put $p=q\alpha$. Choose $c>0$ small enough that $c<pa/3$ and $c\log (2/p)<1/2$. The count above, with $s=\lfloor c\log N\rfloor$, gives $s$ cliques with at least $\sqrt{N}$ simultaneous common neighbours. Their vertices form $q$ classes of size $s$, and $s$ of those common neighbours form the last class. None of the common neighbours belongs to a selected clique, because that would require a loop. This proves the logarithmic clique blow-up from minimum degree assertion.

To pass from an edge count to minimum degree, put $b=1-1/r+\epsilon$ and $a=b-\epsilon /2$. Repeatedly remove a vertex whose current degree of a vertex is below $a$ times the current order. If the process reaches order $m$, its deleted edges and remaining edges giveThus $(1-a)m^2\ge (\epsilon /2)n^2-O(n)$. In particular, for a fixed sufficiently small $\gamma >0$, the process cannot reach $m=\lfloor \gamma n\rfloor$: the displayed inequality would fail. It stops earlier with a subgraph of order $N\ge \gamma n$ and minimum degree at least $(1-1/r+\epsilon /2)N$. The induction just proved supplies $K_{r+1}(\lfloor c\log N\rfloor )$, hence a lower bound $c^{\prime }\log n$ on $t(r,\epsilon ,n)$. In particular, the guaranteed part size tends to infinity.

For the upper bound, choose a fixed $p$ with $b<p<1$ and take a binomial random graph $G(n,p)$. Its edge count has expectation $p\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and variance $p(1-p)\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. The Chebyshev inequality shows that $e(G)\ge b\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ with probability tending to one. The expected number of labeled copies of $K_{r+1}(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/[r\log (1/p)]$, this tends to zero. The Markov inequality therefore shows that with positive probability both the required edge density holds and there is no such copy. This proves $t(r,\epsilon ,n)=O(\log n)$ and completes the random obstruction to larger clique blow-ups argument.

There is a necessary qualification to the density parameter. If $\epsilon =1/r$, the only eligible graph is the complete graph, and $t(r,1/r,n)=\lfloor n/(r+1)\rfloor$. If $\epsilon >1/r$ and $n\ge 2$, there are no eligible simple graphs; the universal assertion is vacuous for every $t$, so the printed maximum has no finite value in that range.

It suffices to find a tree component of order $k$. Let $X$ count such graph components in the binomial random graph. The Cayley formula and independent edges give the tree-component expectation in the Erdős-Rényi modelHere $k$ is fixed. The upper bound on $p$ implies $p=O(\log n/n)$, hence $n{p}^2=o(1)$ andWe first verify that this expected value diverges throughout the permitted interval, including when $\omega$ grows quickly. If $np\le 1$, the exponential factor is at least $e^{-k}$ and $n^k{p}^{k-1}\ge {\omega }^{k-1}$. If $1\le np\le (\log n)/(2k)$, then $n(np{)}^{k-1}{e}^{-knp}\ge \sqrt{n}$. Finally, if $np\ge (\log n)/(2k)$, the upper bound on $p$ givesThus $\mu \to \infty$ in all three cases.

For distinct $k$-element vertex sets $S,T$, write $I_S,I_T$ for their tree-component indicators. If they overlap, they cannot both be distinct connected components of a graph, so $\mathbb{E}(I_S{I}_T)=0$. If they are disjoint, the absent edges between the two sets are counted twice in the product of their marginal probabilities and once in the joint probability. ThereforeSumming and discarding the negative overlap contributions yieldsThe Chebyshev inequality now gives $\mathrm{Pr}(X=0)\le \mathrm{Var}X/{\mu }^2\to 0$. Consequently there is a component of order $k$ with high probability, in fact a tree component. The argument proves the entire fixed-order tree-component window, rather than only a particular choice of $p$.

Couple the uniform random graph process by giving every possible edge an independent uniform label in $[0,1]$ and revealing them in increasing order. The graph at label threshold $p$ is a binomial random graph. Define ${\tau }_0$ as the first step with no isolated vertices and ${\tau }_c$ as the first connected graph. Deterministically ${\tau }_c\ge {\tau }_0$. We prove the reverse inequality with high probability by tracking the same process through a short interval.

Put $w=\frac{1}{4}\log \log n$, $p_{\pm }=(\log n\pm w)/n$, and $L=n{p}_{-}$. At $p_{-}$ the isolated vertex count $Z$ hasThe variance estimate follows by comparing two isolated-vertex indicators, whose joint-to-product probability ratio is $(1-p_{-}{)}^{-1}$. Thus $Z>0$ with high probability. Also the Markov inequality gives $Z\le \log n$ with probability tending to one.

At this same threshold, exclude every graph component with order $2\le s\le n/2$. For $2\le s\le n/\log n$, any such component contains a spanning tree. The Cayley formula and a union bound therefore giveIn this range the bracket is at most $A=e^{2}L{e}^w/n=o(1)$. Summing over $s\ge 2$ gives $O(n{A}^2/L)=O(L{e}^{2w}/n)=o(1)$. For $n/\log n<s\le n/2$, even the cut-absence bound is enough:The bracket tends to zero, so this range also has total probability $o(1)$. Consequently the graph at $p_{-}$ consists of one component of order greater than $n/2$ and its isolated-vertex set $S$, where $1\le |S|\le \log n$ with high probability.

Conditional on the entire graph at $p_{-}$, every unrevealed edge label remains independently uniform on $(p_{-},1)$. In particular,There are no isolated vertices at $p_{+}$ with high probability, because their expected value is asymptotic to $e^{-w}\to 0$. On the intersection of these events, each vertex of $S$ loses isolation by joining the original large component, never by forming a separate small component with another vertex of $S$. At every step between these two thresholds there is therefore one nontrivial component and some isolated vertices. The final isolated vertex disappears precisely when the graph becomes connected. This proves the connectivity hitting time equals disappearance of isolated vertices result:

For the critical-window probability, let $p=(\log n+\alpha (n))/n$ with $|\alpha (n)|\le B$. For each fixed $j$, the falling factorial moment of the isolated-vertex count isuniformly for such shifts. Apply the Bonferroni inequalities to the zero-count event. Its odd and even truncations bracket $\mathrm{Pr}(Z=0)$ by partial sums of ${\sum }_{j\ge 0}(-1{)}^j{e}^{-j\alpha (n)}/j!$. Since $e^{-\alpha (n)}\le e^B$, their remainders tend uniformly to zero. Hence $\mathrm{Pr}(Z=0)=e^{-e^{-\alpha (n)}}+o(1)$. The process equality just proved changes this probability by at most its $o(1)$ failure probability. ThereforeThe ratio follows because the denominator is bounded below by $e^{-e^B}>0$; no convergence of $\alpha (n)$ is needed.

The printed second conclusion has the complementary denominator and is false as written. For $\alpha (n)=0$, the printed ratio tends to $e^{-1}/(1-e^{-1})=1/(e-1)\ne 1$. Its denominator $1-e^{-e^{-\alpha (n)}}$ is correct for the probability of being disconnected. Thus either remove the leading $1-$ in the denominator or replace “connected” by “disconnected”; the proof supplies both corrected versions.

Take $c>0$ and $e>0$, as required for the displayed sparse scaling and the later normalization. A strictly balanced graph satisfies $e(F)/v(F)<e/v$ for every proper nonempty subgraph $F\subset H$. This forces $H$ to be connected and to have no isolated vertices: otherwise a component of at least the average density, or deletion of the isolated vertices, violates strict balance. Copies here are subgraphs, not induced subgraphs.

For fixed $r$, the factorial moment ${\mathbb{E}}_r(X_H)$ counts ordered $r$-tuples of distinct copies all present in the binomial random graph. The contribution of vertex-disjoint copies is exactlyThe factor $a^{-r}$ removes the multiple embeddings of each copy coming from its graph automorphisms. For an overlapping tuple, let $U$ be its union. There are $O(n^{v(U)})$ placements of each fixed union type, and the probability its edges are present is $p^{e(U)}$. When adding a copy meeting the earlier union in $f$ vertices and $g$ edges, the exponent $v(U)-(v/e)e(U)$ changes byIt is strictly negative if this intersection is a proper nonempty subgraph, and zero only for an empty intersection or a whole copy. In particular the first vertex-overlapping copy has a proper nonempty intersection: the preceding copies are vertex-disjoint, and a connected copy contained in their union would have to equal one of them, contrary to distinctness. Its contribution is therefore strictly negative; later additions cannot undo that loss. Only finitely many union types occur for fixed $r$, so their total contribution is $o(1)$. We have proved all the factorial moment limits, and the allowed factorial-moment criterion for Poisson convergence givesThis is the Poisson limit for strictly balanced subgraph counts.

The pendant-edge graph has density $7/5$, whereas its proper $K_4$ core has density $6/4$. Thus $K_4^{+}$ is not strictly balanced. The complete graph $K_4$ is strictly balanced, with 24 graph automorphisms, so at $p=c{n}^{-2/3}$ its copy count $Z$ tends to $\mathrm{Po}(c^6/24)$.

Every $K_4^{+}$ copy has a unique four-vertex clique core and one selected edge joining it to a fifth vertex. For any four-set $C$, let $D_C$ count the ambient edges between $C$ and its complement. Then $D_C\sim \mathrm{Bin}(4(n-4),p)$, with mean ${\mu }_n\sim 4c{n}^{1/3}$. It is independent of the internal edges determining whether $C$ is a clique. Choose ${\delta }_n=n^{-1/12}$. A Chernoff bound and a union bound giveTherefore, simultaneously for every actual core, its number of pendant extensions is $(1+o(1))4c{n}^{1/3}$. Extra ambient edges cause no overcount: each selected seven-edge subgraph has exactly one core and one pendant edge. Summing over the cores yieldsbecause $Z$ is uniformly tight. This is the pendant extensions of sparse clique copies mechanism. For $0<\epsilon <1$, neither endpoint of $(k-\epsilon ,k+\epsilon )$ is an integer, and its only nonnegative integer is $k$. Convergence in distribution consequently givesThe positive-$c$ assumption matters: the quotient printed in the question is undefined when $c=0$.

If the random k-out graph with $k=2$ is disconnected, one of its connected components of a graph has size $3\le s\le n/2$. The lower bound follows because every vertex chooses two distinct neighbours inside its component. For a specified $s$-set $S$, absence of crossing edges requires every choice at a vertex in $S$ to stay in $S$, and every choice outside to stay outside. Independence of choices at different vertices gives exactlyFor $3\le s\le n/2$, the two ratios are at most $(s/n{)}^2$ and $(1-s/n{)}^2$. Indeed each of the factors $(s-j)/(n-j)$, $j=1,2$, is at most $s/n$, and the complementary statement is identical. Put $t=s/n$. The binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{s}\right)\le t^{-s}(1-t{)}^{-(n-s)}$ follows by bounding the probability of $s$ successes in $\mathrm{Bin}(n,t)$ by one. Thus the union bound givesFor $s\le \sqrt{n}$, the last summands are at most $n^{-s/2}$, whose sum from three tends to zero. For $s>\sqrt{n}$, they are at most $2^{-s}$, again summing to zero. ThereforeThe proof bounds all possible small components rather than assuming the dependent undirected edges behave like a binomial random graph.

Write $q=1-p$. A disconnected graph has a connected component of a graph of size $1\le s\le \lfloor n/2\rfloor$. For a fixed $s$-element vertex set $S$, all $s(n-s)$ edges between $S$ and its complement must be absent. Their independent absence has probability $q^{s(n-s)}$. The union bound givesSet $r_n=n{q}^{n/2}$. Since $p$ is fixed in $(0,1)$, $r_n\to 0$. For large $n$, the final sum is at most the geometric series $r_n/(1-r_n)$, which tends to zero. Hence the binomial random graph is connected with probability tending to one, as in connectivity of a fixed-density binomial random graph.

Count each unordered $k$-vertex set once. For each such set $A$, let $I_A$ be its clique indicator in the binomial random graph. All $\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ required edges are independent, soThe Markov inequality, or first moment method, givesConsequentlyThis argument works even if $k=k(n)$ varies. The binomial coefficient shows that cliques are counted as vertex sets, not as their different ordered listings.

It suffices to prove the assertion after replacing $\omega$ by $\min \{\omega ,\sqrt{\log n}\}$: that replacement gives a tighter interval. Thus assume $\omega \to \infty$ and $\omega =o(\log n)$, and putAt $p_{-}$, let $J$ count isolated vertices in $L$. The same indicator random variable calculation as for all isolated vertices givesBy the Chebyshev inequality, $J>0$ with high probability. Since $\ell \ge 2$ eventually, the marked vertices are then not all connected.

At $p_{+}$, the logarithmic-regime giant component result applies with $k=2$ and any fixed ${\gamma }_0\in (1/3,1/2)$. Its unique giant component misses $(1+o(1))n^{1/2}{e}^{-\omega /4}$ vertices. On the event that it misses at most twice that number, exchangeability makes the missed vertex set uniform conditional on its size. A union bound therefore givesThus all of $L$ is connected at $p_{+}$ with high probability.

For the uniform random graph process, assign uniform labels to the edges of the complete graph as independent random variables and reveal them in label order. The binomial random graph $G(n,p)$ is the prefix containing the $M(p)$ labels at most $p$, where $M(p)$ has a binomial distribution with parameters $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $p$. Its variance at $p_{\pm }$ is $O(n\log n)$, whereas the gap between $\mathbb{E}M(p_{\pm })$ and the relevant endpoint $m_{\pm }$ is $(1+o(1))\omega n/8$. The Chebyshev inequality gives $M(p_{-})\ge m_{-}$ and $M(p_{+})\le m_{+}$ with high probability, with harmless integer rounding. The monotone graph property that all marked vertices are connected now impliesThe marked-set connectivity threshold is lower than the threshold for connecting every vertex, because only about $\sqrt{n}$ specified vertices need to avoid the small graph components.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

Write $K_s(t)$ for the balanced graph blow-up of a complete graph, with $s$ classes of size $t$; containment here is as a subgraph, not necessarily an induced subgraph. The logarithmic form of the Erdős-Stone theorem is: for fixed $r\ge 1$ and $0<ϵ<1/r$, there are $d=d(r,ϵ)>0$ and $n_0$ such thatThus the guaranteed balanced part size is at least $\lfloor d\log n\rfloor$, with $d$ independent of $n$. We prove the logarithmic Erdős-Stone theorem by a density-to-cliques step and a constructive dense clique family blow-up lemma.

First choose a fixed $m$ large enough that $t_r(m)/\left(\genfrac{}{}{0ex}{}{m}{2}\right)\le 1-1/r+ϵ/2$, where $t_r(m)$ is the edge count of the Turán graph. A uniformly chosen $m$-vertex subset has expected edge count at least $(1-1/r+ϵ)\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. If its induced graph contains no $K_{r+1}$, the Turan theorem bounds that count by $t_r(m)$. Since the count is always at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the probability that the subset contains an $(r+1)$-clique is at least $ϵ/2$. Double counting the incidences between such subsets and their cliques gives, for $s=r+1$,for a fixed $\eta >0$ and all sufficiently large $n$. Here $(x{)}_s=x(x-1)\cdots (x-s+1)$. This is clique supersaturation by sampling.

We now prove the needed dense clique family blow-up lemma, with the following stronger induction invariant. Given any family $\mathcal{M}$ of at least $\eta n^s$ distinct $s$-cliques, there is a complete $s$-partite subgraph with each class of size $\lfloor a_s(\eta )\log n\rfloor$, containing that many pairwise vertex-disjoint transversal members of $\mathcal{M}$, each with one vertex in every class. Extra edges within its classes can be ignored. We may suppose $0<\eta <1/2$. For $s=1$, any $\eta n$ singletons suffice, and we may take $a_1(\eta )=1$ once $n$ is large.

For $s\ge 2$, put $\theta =\eta /2$. Repeatedly delete every member of the current family containing an $(s-1)$-clique whose number of extensions is at most $\theta n$. Each such face is processed at most once, and there are at most $n^{s-1}$ possible faces. At most $\theta n^s$ members are deleted. The remaining family $\mathcal{L}$ has size at least $(\eta /2)n^s$, and every face that remains has more than $\theta n$ extensions. Its family of $(s-1)$-faces has size at least $s|\mathcal{L}|/n\ge (\eta /2)n^{s-1}$, since each face is in at most $n$ members.

Apply the induction hypothesis to these faces. It gives a complete $(s-1)$-partite subgraph with $m=\lfloor a_{s-1}(\eta /2)\log n\rfloor$ vertices in each class and a matching $R_1,\dots ,R_m$ of transversal $(s-1)$-faces from the family. Form a bipartite graph whose left vertices are the $R_i$ and whose right vertices are the original $n$ vertices; join $R_i$ to $v$ when $R_i\cup \{v\}\in \mathcal{L}$. Every left degree exceeds $\theta n$.

The following common neighbourhood from bipartite density estimate is elementary. In a bipartite graph with class sizes $m,n$ and at least $\theta mn$ edges, averaging over left subsets $S$ of size $u$ and using convexity of the integer sequence $j\mapsto \left(\genfrac{}{}{0ex}{}{j}{u}\right)$ giveswhenever $1\le u\le \theta m/2$. The last inequality follows by comparing the $u$ factors in the two binomial coefficients; the integer convexity follows from the nondecreasing first differences $\left(\genfrac{}{}{0ex}{}{j}{u-1}\right)$.

ChooseFor large $n$, $m\ge \frac{1}{2}{a}_{s-1}(\eta /2)\log n$, so $u\le \theta m/4$. The estimate supplies $u$ selected faces with a common extension set of size at least $n(\theta /2{)}^u\ge \sqrt{n}$. This extension set is disjoint from the selected faces: a vertex in any one of them cannot extend that face. Their union has $u$ vertices in each of the old $s-1$ classes, with all required cross edges; choose $u$ distinct common extension vertices as a new class. Attaching one different extension vertex to each selected face also gives $u$ disjoint members of $\mathcal{L}$. This completes the induction, with positive constants independent of $n$. Apply it to the clique family above and take $d=a_{r+1}(\eta )$.

The logarithmic order cannot be increased in a uniform forcing result. Put $\rho =1-1/r+ϵ<1$, choose $p$ strictly between $\rho$ and one, and take a binomial random graph. Its edge density exceeds $\rho$ with probability tending to one, by the variance estimate for a sum of independent edge indicators. With $s=r+1$, the expected number of ordered embeddings of $K_s(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/((s-1)\log (1/p))$, the logarithm of this bound is negative of order $(\log n{)}^2$. Markov's inequality therefore makes the probability of any such copy tend to zero. Both events hold simultaneously for some graphs of every sufficiently large order. There are graphs meeting the density hypothesis whose largest balanced blow-up has part size $O(\log n)$. This upper bound concerns what density can force; it is not an upper bound for every dense graph, since a complete graph has much larger blow-ups.

Finally, $H_k$ is the square of a cycle, for $k\ge 3$. Any three consecutive vertices form a triangle in a graph. In a proper three-colouring, once the first three colours are fixed, each subsequent vertex must repeat the colour three positions earlier. Closing the cycle is possible precisely when $3\mid k$. Conversely, repeating the three colours gives such a colouring whenever $3\mid k$. In that case $H_k$ embeds in $K_3(k/3)$, and the density $0.51\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ exceeds the bipartite Turan theorem threshold by a fixed amount, so the theorem forces $H_k$ eventually. If $3\nmid k$, the balanced Turán graphs $T_3(n)$ have density at least $2/3$ and contain no graph of chromatic number greater than three. They give a counterexample sequence. Exactly the cycle lengths $k\ge 3$ divisible by three have the required property.