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Use equitable regularity energy with exceptional vertices retained as singleton cells. The refinement variance identity for regularity energy shows that splitting along an irregular pair's witnesses raises energy by more than ${\epsilon }^4|V_i||V_j|/n^2$. More than $\epsilon k^2$ irregular ordered pairs therefore raise energy by more than ${\epsilon }^5/4$ when the exceptional set occupies at most half the vertices.

For equalization with a controlled exceptional set, cut each of the at most $k{2}^k$ witness atoms into pieces of size $\lfloor L/4^k\rfloor$ and put leftovers into singleton exceptional cells. This is a refinement, so it cannot decrease energy. At most $n{2}^{-k}$ vertices are lost. Start with enough cells that this loss, over $\lceil 4{\epsilon }^{-5}\rceil$ rounds, is at most $\epsilon n/2$; the initial exceptional set uses the other half. The number of cells grows by a bounded recurrence depending only on $\epsilon$ and the prescribed minimum. Energy lies in $[0,1]$, so the process terminates.

We use the following precise form of the Szemerédi regularity lemma. For every $0<\eta <1/2$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph on $n\ge n_0$ vertices has a partitionwith at most $\eta k^2$ irregular unordered pairs among the nonexceptional classes. Here $d(A,B)=e(A,B)/(|A||B|)$ for disjoint nonempty sets, and $(A,B)$ is an $\eta$-regular pair ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \eta |A|$, $|Y|\ge \eta |B|$.

For the proof, let $f(x,y)$ be the adjacency indicator, including $f(x,x)=0$, and define $d(A,B)$ by averaging $f$ on $A\times B$ even for diagonal cells. The regularity energy of a partition $\mathcal{P}$ isEvery exceptional vertex is represented by its own singleton cell in this energy. If a rectangle $A\times B$ is refined into smaller rectangles with densities $d_{ij}$ and weights $w_{ij}=|A_i||B_j|$, expanding the square gives the refinement variance identity for regularity energyThus refinement never decreases energy.

Suppose an equitable partition with $k$ classes of size $L$ fails the desired conclusion. For every irregular pair choose witness subsets $X\subseteq V_i$, $Y\subseteq V_j$, each of size at least $\eta L$, whose density differs from $d(V_i,V_j)$ by more than $\eta$. Split each class by all witness subsets involving it. Each class has at most $2^{k-1}$ atoms. On a witness rectangle, the weighted mean of $d_{ab}-d(V_i,V_j)$ is $d(X,Y)-d(V_i,V_j)$. The Cauchy-Schwarz inequality therefore bounds the energy increment on this rectangle below byThe different original cell pairs have disjoint rectangles. More than $\eta k^2$ irregular unordered pairs, with both orientations counted in the energy, thus increase it by more than $2{\eta }^5{k}^2{L}^2/n^2$. As long as $|V_0|\le \eta n<n/2$, this is greater than ${\eta }^5/2$, and in particular greater than ${\eta }^5/4$.

Restore equitability by dividing each atom into pieces of common size $\ell =\lfloor L/4^k\rfloor$, moving each leftover vertex into the exceptional set as a singleton energy cell. This remains a refinement, so the energy gain persists. There are at most $k{2}^k$ atoms, so the additional exceptional vertices number at mostIf $L\ge 2\cdot 4^k$, then $\ell \ge L/(2\cdot 4^k)$, and the new number $k^{\prime }$ of equal cells satisfies $k\le k^{\prime }\le 4k{4}^k$. For the lower bound, within each old class the leftover count is at most $2^k\ell <L$, so at least one full new cell survives. For the upper bound, $n/\ell \le 2(n/L)4^k\le 4k{4}^k$, using $kL\ge n/2$. This proves equalization with a controlled exceptional set without any energy loss.

Put $T=\lceil 4{\eta }^{-5}\rceil +1$. Choose $k_0\ge m_0$ with $T{2}^{-k_0}\le \eta /2$. Start with $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Iterate the bounded recurrence $k\mapsto 4k{4}^k$ $T$ times to obtain a bound $M$. Choose $n_0$ so large that the initial remainder is at most $\eta n/2$ and, for every $k\le M$, $(1-\eta )n/k\ge 2\cdot 4^k$. The total exceptional set remains at most $\eta n$, since at most $T$ rounds each add at most $n{2}^{-k_0}$. Each failing round raises an energy in $[0,1]$ by more than ${\eta }^5/4$, so fewer than $T$ rounds can fail. This proves the lemma with all constants independent of the input graph.

For the application, write $h=|V(F)|$ and $\chi (F)=r+1$. We may replace the given tolerance by a smaller one, so assume $0<\epsilon <1$. Set $d=\epsilon /4$, take $m_0\ge 8/\epsilon$, and chooseApply the lemma and delete all edges incident with $V_0$, all edges inside a class, all edges across irregular pairs, and all edges across regular pairs of density below $d$. The resulting spanning subgraph $H$ loses at mostLet $R$ be the reduced graph of a regularity partition, whose edges represent the surviving regular pairs of density at least $d$. We show that a copy of $K_{r+1}$ in $R$ forces a copy of $F$ in $G$.

Choose a proper $(r+1)$-coloring of $F$ and assign its color classes to the corresponding clusters. Embed its vertices one at a time. For each unembedded vertex maintain the candidate set in its cluster consisting of unused vertices adjacent to all already embedded neighbours. After at most $h$ restrictions its size is at least $(d-\eta {)}^{h}L-h\ge (d/2{)}^{h}L-h$. This last quantity is at least $2h\eta L$ when $L$ is large, so every such candidate set is large enough for regularity. In an $\eta$-regular pair of density $p\ge d$, fewer than $\eta L$ vertices on one side have fewer than $(p-\eta )|Y|$ neighbours in any fixed set $Y$ on the other side of size at least $\eta L$: otherwise these vertices and $Y$ violate regularity. For the vertex currently being embedded, there are at most $h$ relevant future candidate sets. At most $h\eta L$ vertices are bad for one of them. Its own candidate set has at least $2h\eta L$ vertices, so an unused good vertex can be chosen. All future candidate sets shrink by at most the factor $d-\eta$, apart from the removal of a used vertex; the stated size bound follows inductively, since the accumulated removals are at most $h$. This completes the graph embedding lemma for regular pairs in the form needed here.

Since $G$ is $F$-free, $R$ is $K_{r+1}$-free. Any $K_{r+1}$ in $H$ would use distinct clusters, because intraclass edges were deleted, and project to a $K_{r+1}$ in $R$. ThereforeThe required threshold on $n$ depends only on $F$ and $\epsilon$, since $k\le M$ makes every nonexceptional cluster sufficiently large.

For disjoint vertex sets $A,B$, define their edge density of a bipartite graph by $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair with parameter $\epsilon$ if every $A^{\prime }\subseteq A,B^{\prime }\subseteq B$ with $|A^{\prime }|\ge \epsilon |A|$, $|B^{\prime }|\ge \epsilon |B|$ satisfies $|d(A^{\prime },B^{\prime })-d(A,B)|\le \epsilon$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and minimum cell count $k_0$, some $K,n_0$ have the following property. Every graph on $n\ge n_0$ vertices admitswith $|V_0|\le \epsilon n$, equal-sized $V_1,\dots ,V_k$, and at most $\epsilon k^2$ unordered irregular pairs. It is enough to consider $0<\epsilon <1/2$.

Prove this using equitable regularity energy. Treat exceptional vertices as singleton cells and setwith ordered pairs and the adjacency indicator also defining diagonal densities. Then $0\le q\le 1$. The refinement variance identity for regularity energy proves refinement cannot decrease $q$:An irregular pair supplies witness subsets of relative sizes at least $\epsilon$ whose density differs by more than $\epsilon$. Refining along those witnesses raises $q$ by more than ${\epsilon }^4{L}^2/n^2$ per orientation, where $L$ is the common cell size. If there are more than $\epsilon k^2$ irregular pairs, the increase is greater than ${\epsilon }^5(kL/n{)}^2\ge {\epsilon }^5/4$, provided the exceptional set has size at most $n/2$.

Refine each cell by all its witness subsets; this produces at most $2^k$ atoms per cell. Restore equal sizes by the equalization with a controlled exceptional set: cut each atom into blocks of size $\lfloor L/4^k\rfloor$ and declare its remainder exceptional, still retaining every such vertex as a singleton in the energy. This is another refinement, so no energy is lost. At most $k{2}^k\lfloor L/4^k\rfloor \le n{2}^{-k}$ vertices become exceptional, and, when $L\ge 2\cdot 4^k$, at most $2k{4}^k$ new nonexceptional cells occur.

Put $T=\lceil 4{\epsilon }^{-5}\rceil$ and start with $k\ge k_0$ so large that $T{2}^{-k}\le \epsilon /2$. An initial equitable partition has at most $\epsilon n/2$ exceptional vertices when $n$ is sufficiently large. Subsequent cell counts only increase, so the total added exceptional set has size at most $\epsilon n/2$. The recurrence $k\mapsto 2k{4}^k$, iterated at most $T$ times, supplies a finite bound $K$; choose $n_0$ large enough that every intermediate $L$ meets the size requirement. More than $T$ unsuccessful refinements would raise $q$ above one. The process must therefore stop with the required regular pairs. This proves the lemma.

Its usefulness comes from replacing an arbitrarily large graph by the bounded reduced graph of a regularity partition. For fixed positive cross-density, the graph embedding lemma for regular pairs lifts fixed configurations from the reduced graph to the original one. In particular, three sufficiently regular positive-density pairs contain many triangles in a graph. Deleting exceptional, intracell, irregular and sparse-pair edges then gives the triangle removal lemma: a graph with sufficiently few triangles can be made triangle-free by deleting $o(n^2)$ edges. A triangle surviving in the reduced graph would force a positive proportion of the original possible triangles, a contradiction. More generally this approach yields the clique removal lemma and the asymptotic Erdős-Stone theorem. Applied to the tripartite graph encoding $x+z=2y$, triangle removal also gives Roth theorem on three-term arithmetic progressions. The price is that the energy argument gives a very rapidly growing, iterated-exponential bound on $K(\epsilon )$; the lemma is a structural tool, not a promise of small practical constants.

Use the equitable form of the Szemerédi regularity lemma. For $0<\epsilon <1/2$ and a prescribed lower bound $m_0$, there are $M,n_0$ such that every sufficiently large graph admits a set partitionwith at most $\epsilon k^2$ unordered irregular pairs among the nonexceptional cells. A uniform pair here means a regular pair of vertex sets. Larger values of $\epsilon$ follow from a smaller parameter, so the stated restriction loses nothing.

For the proof, retain the exceptional vertices as singleton cells in an auxiliary set partition $\mathcal{P}$. Define its regularity energy by the ordered sumFor diagonal rectangles the adjacency matrix of a graph defines $d(A,A)$; diagonal terms are included. Thus $0\le q\le 1$. If $A\times B$ is split into rectangles of densities $d_{uv}$ and sizes $w_{uv}$, their weighted mean is $d=d(A,B)$. Expanding squares proves the refinement variance identity for regularity energy:Every refinement therefore increases or preserves energy.

Suppose equal cells $V_i,V_j$ of size $m$ are $\epsilon$-irregular. Choose witnesses $X\subseteq V_i$, $Y\subseteq V_j$ of sizes at least $\epsilon m$, whose density differs from $d(V_i,V_j)$ by more than $\epsilon$. Refine the cells so that these witnesses are unions of refined atoms. Applying the Cauchy-Schwarz inequality inside $X\times Y$ gives an increase, for this oriented rectangle, greater thanChoose witnesses simultaneously for every irregular pair. Each original cell is cut by at most $k-1$ witness sets and has at most $2^k$ atoms. If more than $\epsilon k^2$ unordered pairs are irregular, both orientations contribute, givingas long as the exceptional set has at most $n/2$ vertices.

To restore equal sizes without sacrificing this increase, use equalization with a controlled exceptional set. Cut every witness atom into blocks of size $\ell =\lfloor m/4^k\rfloor$, leaving fewer than $\ell$ unused vertices in each atom. Make each unused vertex a singleton exceptional cell. This is a further refinement, so there is no energy loss. The number newly exceptional is at mostWhen $m\ge 2\cdot 4^k$, the new nonexceptional cell count $k^{\prime }$ is at most $4k{4}^k$: indeed $\ell \ge m/(2\cdot 4^k)$ and $n/m\le 2k$. It is at least $k$, since each old cell supplies at least $4^k-2^k\ge 1$ blocks.

Let $L=\lceil 2{\epsilon }^{-5}\rceil +1$. Choose the initial count $k_0\ge m_0$ so large that $L{2}^{-k_0}\le \epsilon /2$, and begin with $k_0$ equal cells and fewer than $k_0$ leftover vertices. Take $n_0$ large enough that these initial leftovers are at most $\epsilon n/2$. Iterate the bound $k\mapsto 4k{4}^k$ for $L$ rounds to obtain a finite number $M$. Increase $n_0$ further to ensure $n\ge 4M{4}^M$. Then every cell size encountered satisfies $m\ge n/(2M)\ge 2\cdot 4^k$, so every equalization is legitimate. The total exception budget is at most $\epsilon n$, because the initial losses use half the budget and all later losses sum to at most $Ln{2}^{-k_0}$. Energy cannot increase by more than one, so $L$ failing rounds are impossible. The process must stop with at most $\epsilon k^2$ irregular pairs. This proves the Szemerédi regularity lemma, including the bounded cell count, equitability and exceptional-set control.

For the dense-pair application, apply this proof with $\eta =1/100$ and $m_0=100$. Suppose every $\eta$-regular pair of nonexceptional cells has density at most $1/4$. Edges touching $V_0$ number at most $\eta n^2$. Edges inside nonexceptional cells number at most $k{m}^2/2\le n^2/(2m_0)$. The at most $\eta k^2$ irregular cell pairs contribute at most $\eta n^2$. Finally the regular cross-cell pairs contribute at most $k(k-1)m^2/8\le n^2/8$. Thereforea contradiction. Some regular pair $(U,W)$ consequently has density greater than $1/4$. It is also a $1/8$-uniform pair, since subsets of relative size at least $1/8$ qualify in the stronger $\eta$-regularity condition. The common cell size satisfies $m\ge (1-\eta )n/M\ge n/(2M)$. Thus the dense regular pair from edge surplus givesThis first proves the assertion for $n\ge n_0$. For the finitely many smaller possible orders, choose the endpoints of any edge as singleton sets: the pair has density one and is regular for every positive parameter. Taking $c=\min \{1/(2M),1/n_0\}>0$ includes these cases as well.

Write $d(U,V)=e(U,V)/(|U||V|)$ for the edge density of a bipartite graph between nonempty disjoint vertex sets. A regular pair of vertex sets $(U,V)$ is $\epsilon$-regular if $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$ whenever $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$, $|U^{\prime }|\ge \epsilon |U|$, and $|V^{\prime }|\ge \epsilon |V|$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and positive integer $m_0$ there is $M=M(\epsilon ,m_0)$ such that every finite simple graph on $N\ge m_0$ vertices has a set partitionwhere $V_1,\dots ,V_k$ have equal positive size and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $i<j$, fail to be regular pairs of vertex sets. The version requiring only sufficiently large $N$ is equivalent, since the finitely many smaller admissible sizes can use singleton sets as classes. It suffices to prove this for $0<\epsilon \le 1/2$: reducing a larger parameter to $1/2$ gives stronger conclusions.

Here is a complete energy proof, including restoration of equal class sizes. For a set partition with exceptional set $V_0$, use the following version of equitable regularity energy, omitting pairs incident with $V_0$:For $i=j$, define $d(V_i,V_i)$ using the ordered adjacency indicator function, with diagonal entries zero. Thus the same formula works on every rectangle. The refinement variance identity for regularity energy iswhere $A,B$ run over the refined cells. Expanding the square proves the identity because their weighted mean edge density of a bipartite graph is $d(U,V)$. In particular, refinement never decreases the energy. If a union of refined cells $U^{\prime }\times V^{\prime }$ has discrepancy greater than $\epsilon$, the Cauchy-Schwarz inequality bounds its contribution below by $|U^{\prime }||V^{\prime }|{\epsilon }^2/N^2$. For an irregular equal-class pair of size $L$, witnesses have sizes at least $\epsilon L$, so the increase exceeds ${\epsilon }^4{L}^2/N^2$ in each orientation.

Choose $k_0\ge \max \{m_0,2\}$ so large that $2^{-k_0}\le {\epsilon }^6/100$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. For sufficiently large $N$ this exceptional set has size at most $\epsilon N/2$. Suppose there are more than $\epsilon k^2$ irregular unordered pairs at a subsequent step, with equal class size $L$. Choose one witnessing pair of subsets for every such pair. Split each class according to membership in all its witness subsets; each class produces at most $2^k$ atoms. The preceding refinement variance identity for regularity energy shows that the energy increase exceedsas long as the exceptional set has size at most $\epsilon N\le N/2$.

Apply equalization with a controlled exceptional set. Set $\ell =\lfloor L/4^k\rfloor$, cut every atom into blocks of size $\ell$, and send its leftover points to $V_0$. Provided $L\ge 2\cdot 4^k$, the new number $k^{\prime }$ of classes satisfiesThe lower bound holds because every old class retains at least one full block: its leftovers total less than $2^k\ell <L$. The discarded fraction is at mostCutting blocks is a further refinement before discarding. Discarding changes only ordered pairs with at least one discarded vertex, so it loses at most $2r$ in the energy. The net increase is therefore at least ${\epsilon }^5/2-2{\epsilon }^6/100\ge {\epsilon }^5/4$.

There can be fewer than $T=\lceil 4{\epsilon }^{-5}\rceil$ such strict energy increases, because the energy stays in $[0,1]$. Through $T$ steps the exceptional fraction is at mostThis also verifies the exceptional-set bound used at every preceding step. Iterating the finite recursion $k\mapsto 2k{4}^k$ at most $T$ times bounds all class counts by some $K=K(\epsilon ,m_0)$. Choosing $N$ sufficiently large in terms of $K$, for instance with $N\ge 4K{4}^K$ and $N\ge 2k_0/\epsilon$, guarantees $L=(N-|V_0|)/k\ge N/(2K)\ge 2\cdot 4^k$ throughout. Thus the process terminates with the desired regular pairs of vertex sets. Choose a finite integer threshold $N_{\ast }$ for these requirements and set $M\ge \max \{K,N_{\ast }\}$. For $m_0\le N<N_{\ast }$, use $N$ singleton sets; every pair is regular because its only admissible nonempty subsets are the whole pair. This completes the Szemerédi regularity lemma for all admissible $N$.

For the half graph, a class of size $s$ has exactly $\lceil s/10\rceil$ elements with fewer than $s/10$ predecessors, and the same number with fewer than $s/10$ successors. Thus its boundary contains at most $2\lceil s/10\rceil$ elements. Let $k_X,k_Y$ be the numbers of classes. The boundary rank obstruction in a half graph gives the union boundbecause $k_X,k_Y\le n/100$. There is therefore an index $m$ away from both boundaries. In its classes $X_i,Y_j$, all four subsets $X^{-}=\{x<m\}$, $X^{+}=\{x>m\}$, $Y^{-}=\{y<m\}$, $Y^{+}=\{y>m\}$ have at least one tenth of their respective class sizes. These subsets are taken within $X_i,Y_j$, which need not be intervals.

In the half graph, $d(X^{-},Y^{+})=1$ and $d(X^{+},Y^{-})=0$. If $(X_i,Y_j)$ were a $1/10$-regular pair of vertex sets, both values would be within $1/10$ of $d(X_i,Y_j)$, implying $1\le 2/10$, a contradiction. At least one class pair is not $1/10$-regular.