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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.

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.