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A polynomial-time many-one reduction represents each variable by an adjacent literal pair and each clause by a triangle in a graph. Connect each occurrence vertex to its corresponding literal vertex. A vertex cover of size $n+2m$ must choose one literal vertex per variable and two vertices per clause, and its omitted clause vertex forces a true literal. Conversely a satisfying Boolean valuation supplies such a cover. This establishes NP-hardness of the vertex-cover decision problem from 3-SAT.

A pairwise intersecting family of two-element sets is contained in a star graph or in the three edges of a triangle in a graph. If two members are $\{a,b\},\{a,c\}$ and a third avoids $a$, the third must be $\{b,c\}$. Every member meeting all three belongs to that triangle.

For any fixed odd length $\ell \ge 3$, positive triangle in a graph density forces positive density of $\ell$-cycles. A Szemerédi regularity lemma partition retains a triangle in a graph of regular dense pairs after the sparse and exceptional pairs are removed. Count embeddings of a proper three-colouring of $C_{\ell }$ into those three clusters. The constants depend on $\beta$ and $\ell$, not on graph order.

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 Turan 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 Turan 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.

For a fixed uniform hypergraph $H$ with $h$ vertices, uniformity $\ell$, and at least one edge, define its Turán density bywhere $\mathrm{ex}(n,H)$ is the maximum edge count of an $H$-free hypergraph. The limit exists: averaging the edge count over all $m$-vertex subsets of an extremal $n$-vertex hypergraph, with $n\ge m\ge \ell$, shows that the displayed normalized extremal numbers are nonincreasing. Thus their limit is their infimum.

For hypergraph supersaturation by sampling, choose $m\ge h$ so that $\mathrm{ex}(m,H)/\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)\le \pi (H)+ϵ/2$. The expected edge count in a uniform $m$-vertex subset of $G$ is at least $(\pi (H)+ϵ)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. A subset containing no copy of $H$ has at most $(\pi (H)+ϵ/2)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$ edges, while any subset has at most $\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. Hence at least an $ϵ/2$ fraction of the subsets contain $H$. Every particular copy of $H$ lies in exactly $\left(\genfrac{}{}{0ex}{}{n-h}{m-h}\right)$ such subsets, so their number is at leastfor $n\ge N=\max (m,2h)$. TakeFor $n<N$, the requested integer lower bound is zero; for $n\ge N$, the calculation proves it. This positive $\delta$ depends only on $H$ and $ϵ$. If the density hypothesis is impossible there is nothing to prove. For an edgeless $H$, every $h$-set is already a copy and the conclusion follows directly.

For the edge-triangle objective, let $N(v)$ be the open neighbourhood of $v$ and define its local scoreIt counts the contribution to $k_2-c{k}_3$ of edges and triangles in a graph containing $v$. If $u,v$ are nonadjacent, making $u$ a false twin of $v$ changes the objective by $s(v)-s(u)$, since no edge or triangle in a graph uses both. More generally, partition vertices into false-twin classes, meaning equal open neighbourhoods. If distinct classes $U,V$ have no edges between them, making every vertex of $U$ a twin of a representative of $V$ changes the objective by $|U|(s(v)-s(u))$. Choose the direction with nonnegative change.

Among graphs maximizing the objective, choose one maximizing the sum of squared false-twin class sizes. The operation cannot split any old false-twin class and merges $U,V$, so it would strictly increase that secondary quantity. Therefore no two distinct classes can be nonadjacent. An objective maximizer is a complete multipartite graph. This edge-triangle symmetrization works for every real $c$, with no assumption on its sign.

For positive part sizes $a_1,\dots ,a_q$, the correct multipartite formula isThe first sum in the printed intermediate formula has inconsistent indices; the edge count is the pairwise-product sum above. Fix two part sizes with sum $A$ and let $B$ be the sum of all other sizes. All terms varying with the pair have the formChoose a maximizing partition with the fewest nonempty parts. If $1-cB\le 0$, merging those two parts cannot decrease $F$ and reduces the number of parts, a contradiction. Therefore this coefficient is positive for every pair. If two integer sizes differ by at least two, moving one vertex from the larger to the smaller increases their product and hence $F$, again a contradiction. All sizes differ by at most one, so a maximizer is a Turan graph. This is balancing a multipartite edge-triangle objective.

Next maximize the same polynomial on the compact simplex of nonnegative real sizes summing to $n$, starting with at least two possible parts. Choose a maximizer with the smallest positive support. The merging argument still applies, and now varying two positive unequal sizes towards equality improves the product. Thus every positive size is $n/q$ for some integer $q$. A two-part choice has value $n^2/4>0$ when $n>0$, so $q\ge 2$. The optimum is attained at rational sizes, and thereforeOne may take the initial simplex to have $\max (n,2)$ coordinates, so it includes every multipartite graph obtained above; zero coordinates are discarded.

Set $c=9/(4n)$. Dividing the right side by $n^2$ givesbecause $q$ is an integer at least two. Rearranging provesEquality in the continuous calculation occurs at two or three equal parts, explaining the triangle support line between bipartite and tripartite Turan graphs.

We prove edit-distance stability for clique-free graphs by induction on $r$, using the symmetric difference of edge sets as the distance. For $r=1$, a $K_2$-free graph is edgeless, and there is nothing to change. For $r\ge 2$, choose a vertex of maximum degree $d$, let $A$ be its neighbourhood and put $B=V(G)\setminus A$. Then $G[A]$ is $K_r$-free. WriteBoth deficits are nonnegative by the Turan theorem. Since every vertex of $B$ has degree at most $d=|A|$,The complete $r$-partite graph formed from an $(r-1)$-partite Turan graph on $A$ and the new class $B$ has at most $t_r(n)$ edges. Thus $L=t_r(n)-t_{r-1}(d)-|A||B|\ge 0$, and direct subtraction givesBy induction, edit $G[A]$ into a complete $(r-1)$-partite graph using at most $3D_A$ changes. Delete the $b$ edges inside $B$, and add the $m$ missing edges between $A$ and $B$. The resulting graph is complete $r$-partite on the same vertex set, andwhere the inequality uses $m\ge 2b$. The required edit distance is at most $3k$. Empty partition classes are allowed; a sufficiently dense nondegenerate case has the usual full complement of classes.

For the odd-cycle conclusion, use two standard consequences of the Szemerédi regularity lemma, stated explicitly. The triangle removal lemma says that for every $\alpha >0$, some $\beta >0$ ensures that a graph with at most $\beta n^3$ triangles in a graph can be made triangle-free by deleting at most $\alpha \left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges, for all sufficiently large $n$. Also, for each fixed odd length $\ell \ge 3$ and each $\beta >0$, there is $\zeta >0$ such that a graph with at least $\beta n^3$ triangles in a graph contains at least $\zeta n^{\ell }$ copies of $C_{\ell }$.

For clarity, the latter odd-cycle copies from positive triangle density consequence follows by applying regularity with error small relative to $\beta$, removing exceptional, irregular and very sparse pairs, and retaining a triangle in a graph among the remaining regular dense pairs. The graph embedding lemma for regular pairs counts a positive constant times $n^{\ell }$ embeddings of any fixed graph properly three-coloured into those three clusters, including $C_{\ell }$. Dividing by the fixed number of descriptions of a cycle gives the same conclusion for unlabelled copies.

Apply triangle in a graph removal with $\alpha =ϵ$, and take the resulting $\beta$. A $C_{2013}$-free graph cannot have $\beta n^3$ triangles in a graph for large $n$, by the preceding consequence. Delete at most $ϵ\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges to obtain a triangle-free $G^{\prime }$. With $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$,Apply the proved stability result with $r=2$. The resulting complete bipartite graph $H$ satisfiesChoose $n_0(ϵ)$ also large enough that $3n/4\le 4ϵN$. Then the distance is at most $11ϵ\left(\genfrac{}{}{0ex}{}{n}{2}\right)$.

For the unheaded continuation, use the same $\beta$ and its associated $\zeta$, and set $\delta =\zeta /2$. Having at most $\delta n^{2013}$ cycles rules out $\beta n^3$ triangles in a graph just as before. The identical deletion and stability calculation applies. A sufficiently small $\delta =\delta (ϵ)>0$ gives the same $11ϵ$ bound.

Use the specified construction as a polynomial-time many-one reduction from 3-SAT. There are $2n+3m$ vertices and $n+6m$ edges: one edge per variable pair, three edges per clause triangle in a graph, and three occurrence edges per clause. Any vertex cover must take at least one vertex from every variable pair and at least two from every clause triangle. Therefore every cover has at least $n+2m$ vertices.

Given a satisfying Boolean valuation, include the vertex corresponding to the true Boolean literal in each variable pair. In each clause choose a true occurrence and omit its clause vertex, including the other two. Every variable edge and triangle edge is covered. An occurrence edge with an included clause endpoint is covered automatically; the only omitted occurrence vertex is joined to the included true-literal vertex. This is a vertex cover with exactly $n+2m$ vertices.

Conversely, a cover of that size must use exactly one vertex in each variable pair and exactly two in each triangle. Declare a variable true precisely when its positive-literal vertex is included. Each clause has one omitted vertex. Its occurrence edge forces the corresponding literal vertex into the cover, so that literal is true. Every clause is therefore satisfied. ThusThe construction and target size are polynomial in the input length. Since 3-SAT is NP-complete, the decision problem is NP-hard. The usual “at most $k$” version has the same equivalence here because $n+2m$ is an unavoidable lower bound. More generally, a cover with fewer than $k\le |V|$ vertices can be padded to exactly $k$. Verifying a proposed cover is polynomial, so the usual vertex-cover decision problem is also in NP and hence NP-complete.

We derive the triangle removal lemma from the permitted Szemerédi regularity lemma, then turn a corner in an integer grid into a triangle in a graph.

The triangle removal lemma says that for every $\eta >0$ there are $\rho >0$ and $N_0$ such that a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. Here triangles in a graph are unordered triples of distinct vertices. We prove this consequence with the necessary quantitative dependence.

We may assume $0<\eta <1$. Put $d=\eta /4$, choose an integer $m_0\ge 4/\eta$, and chooseApply the Szemerédi regularity lemma to obtain an exceptional class $V_0$ of size at most $\epsilon N$ and equal-sized classes $V_1,\dots ,V_k$ of size $L$, where $m_0\le k\le M$, and at most $\epsilon k^2$ pairs are not regular pairs of vertex sets. The constant $M$ depends only on these chosen parameters. Delete all edges incident with $V_0$, all edges within a class, all edges between irregular pairs, and all edges between pairs whose edge density of a bipartite graph is less than $d$. The respective costs are at mostTheir sum is at most $\eta N^2/2$, hence certainly at most $\eta N^2$.

If a triangle in a graph survives, it lies in three distinct classes whose three pairs are $\epsilon$-regular pairs of vertex sets of edge density of a bipartite graph at least $d$ in the original graph. We now check the needed regular triangle counting lemma. In the first class all but at most $2\epsilon L$ vertices have at least $(d-\epsilon )L$ neighbours in each of the other two classes; otherwise the definition of a regular pair of vertex sets would be violated. For each such vertex $v$, its two neighbour sets each have size at least $\epsilon L$, so their mutual edge density of a bipartite graph is at least $d-\epsilon$. It follows that the original three classes contain at leasttriangles in a graph. The elementary lower bound holds because $\epsilon \le 1/4$ and $\epsilon \le d/4$. Also $L=(N-|V_0|)/k\ge N/(2M)$. Thus a surviving triangle in a graph implies at least $d^3{N}^3/(64M^3)$ original triangles in a graph. Set $\rho =d^3/(128M^3)$ and take $N_0$ large enough for the Szemerédi regularity lemma. Fewer than $\rho N^3$ original triangles in a graph force the cleaned graph to be triangle-free, proving the triangle removal lemma.

Now suppose $0<\delta \le 1$ and $A\subseteq [n{]}^2$. Use the tripartite graph encoding of a grid to construct a tripartite graph with disjoint labelled classesFor each $(a,b)\in A$, put in the three edges joining $a\in X$, $b\in Y$, and $a+b\in Z$. These yield $|A|$ canonical triangles in a graph. They are pairwise edge-disjoint triangles: an $XY$ edge determines $(a,b)$ directly, an $XZ$ edge determines $b=z-a$, and a $YZ$ edge determines $a=z-b$. Any deletion making the graph triangle-free must therefore remove at least $|A|$ edges.

On the other hand, an arbitrary triangle in a graph with labels $(x,y,z)$ gives three points of $A$:Writing $d=z-x-y$, these are $(x,y),(x,y+d),(x+d,y)$. If $d\ne 0$, they form the required corner in an integer grid. If there is no such corner in an integer grid, every triangle in a graph is canonical, and the total number is precisely $|A|\le n^2$.

Apply the triangle removal lemma with $\eta =\delta /32$. The constructed graph has $N=4n$ vertices. For sufficiently large $n$, $N\ge N_0$ andIn the absence of a corner in an integer grid, the triangle removal lemma would destroy all triangles in a graph by deleting at mostedges, contradicting the pairwise edge-disjoint triangles. Every sufficiently large grid therefore has the asserted corner in an integer grid in every subset of positive fixed density of a finite subset. For $\delta >1$ there is no eligible subset, so the assertion is vacuous. This proves the corners theorem, including the stipulated nonzero displacement; no positivity of $d$ was required.