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Fix an odd cycle length. If the graph has edge density at least $1/2-ϵ$ and sufficiently few copies of that cycle, odd-cycle copies from positive triangle density forces few triangles in a graph. The triangle removal lemma deletes at most $ϵ\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges, after which the clique-free edit bound applies with two parts. Absorbing its linear rounding error into the density margin gives the displayed bound for large order.

For every $\delta >0$, all sufficiently large integers $n$ have the property that every $A\subseteq [n{]}^2$ with $|A|\ge \delta n^2$ contains a corner in an integer grid. The tripartite graph encoding of a grid turns the absence of a corner into a family of many edge-disjoint triangles but only quadratically many total triangles in a graph, contradicting the triangle removal lemma.

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.

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.