Regular pair of vertex sets

A pair $(A,B)$ is $\epsilon$-regular whenwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$, and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma gives equal cells. If every regular cross-cell pair had density at most $1/4$, their total edge contribution would be at most $n^2/8$. Exceptional vertices, within-cell edges and irregular pairs contribute at most $(2\eta +1/(2m_0))n^2$. Choosing $\eta =1/100$ and $m_0=100$ contradicts $e(G)=n^2/6$. The dense pair is $1/8$-regular since it is already $\eta$-regular.

For disjoint nonempty vertex sets $U,V$ of a graph, an $\epsilon$-irregular pair is one that is not an $\epsilon$-regular pair of vertex sets. Equivalently there are witnesses $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$ with $|U^{\prime }|\ge \epsilon |U|$, $|V^{\prime }|\ge \epsilon |V|$ and $|d(U^{\prime },V^{\prime })-d(U,V)|>\epsilon$. Splitting both cells by these witnesses increases their contribution to the regularity energy by at least ${\epsilon }^4|U||V|/n^2$: the witness rectangle has relative weight at least ${\epsilon }^2$, its mean discrepancy has magnitude greater than $\epsilon$, and the Cauchy-Schwarz inequality bounds the weighted sum of squared discrepancies below by the rectangle weight times its squared mean. This is the local increment in the energy-increment proof of Szemerédi regularity.

For fixed $r$ and positive density threshold $\lambda$, sufficiently regular pairwise dense bipartite graphs between $r$ disjoint equal-sized vertex classes contain at least $\delta n^r$ transversal cliques, for some $\delta >0$. A greedy embedding maintains common neighbourhoods and discards the small set of vertices with atypical degrees into them. The constants depend only on $r,\lambda$.

If three disjoint vertex classes have size $L$, and their three pairs are $\epsilon$-regular pairs of vertex sets of edge density of a bipartite graph at least $d$, where $0<\epsilon \le d/2$ and $\epsilon <1/2$, they contain at least the displayed number of transversal triangles in a graph. All but $2\epsilon L$ vertices in the first class have at least $(d-\epsilon )L$ neighbours in both other classes. Regularity between those two neighbour sets supplies the third edge. This strengthens a mere triangle embedding lemma for regular pairs to a cubic lower count.

Let $X=X_1\bigsqcup \cdots \bigsqcup X_r$ and $Y=Y_1\bigsqcup \cdots \bigsqcup Y_s$ be set partitions of the two parts of a bipartite graph. The pair of partitions is $\epsilon$-regular whenThus a uniformly random pair in $X\times Y$ lies in an irregular cell pair with probability at most $\epsilon$.

For every $\epsilon >0$ there is a positive integer $K(\epsilon )$ such that every finite bipartite graph has an $\epsilon$-regular pair of partitions with at most $K(\epsilon )$ cells on each side.

To prove this, give $X\times Y$ the uniform probability measure, let $f$ be the indicator function of the edge set, and for partitions $\mathcal{P},\mathcal{Q}$ define their energy byThis conditional expectation is an orthogonal projection, so $0\le \mathcal{E}\le \Vert f{\Vert }_2^2\le 1$. If the partitions are not $\epsilon$-regular, choose witnesses $A_{ij}\subseteq X_i$ and $B_{ij}\subseteq Y_j$ in every irregular cell pair and refine each $X_i$ by all its $A_{ij}$ and each $Y_j$ by all its $B_{ij}$. The refined energy exceeds the old energy byOn $A_{ij}\times B_{ij}$ the absolute value function of the mean of the difference is greater than $\epsilon$, while this rectangle occupies at least an ${\epsilon }^2$ fraction of $X_i\times Y_j$. The Cauchy-Schwarz inequality therefore contributes more than ${\epsilon }^4|X_i||Y_j|/(|X||Y|)$ from that cell pair. Irregular pairs have total weight greater than $\epsilon$, so every refinement raises the energy by more than ${\epsilon }^5$. The process stops after at most $\lceil {\epsilon }^{-5}\rceil$ refinements. If the current cell counts are $r,s$, the construction gives at most $r{2}^s,s{2}^r$ new cells, so a finite iterated bound depending only on $\epsilon$ supplies $K(\epsilon )$.

For every $\epsilon >0$ and $m_0$ there are $M,n_0$ such that every graph on at least $n_0$ vertices has a partitionwhere $m_0\le m\le M$, $|V_0|\le \epsilon |V|$, the other parts have equal size, and all but at most $\epsilon m^2$ pairs $(V_i,V_j)$ are $\epsilon$-regular.

For every $\epsilon >0$, all sufficiently large finite graphs admit a set partition into at most $M(\epsilon )$ nonempty cells, each of size at most $\epsilon n$, with the displayed ordered irregular-pair weight bound. Start with small cells. Use regularity energy including diagonal rectangles, refine simultaneously by the witness subsets of all irregular pairs, and apply the refinement variance identity for regularity energy. Each failing round raises energy by more than ${\epsilon }^5$, while energy lies in $[0,1]$. The recurrence $k\mapsto k{2}^k$ bounds the final cell count. Equitability is unnecessary for this weighted version.

A partition with at most $M$ cells has at most $\epsilon n$ vertices in cells smaller than $\epsilon n/M$. Deleting all edges incident to those vertices costs at most $\epsilon n^2$ edges. Every remaining cell has a fixed positive fraction of all vertices, which permits a regular triangle counting lemma to give a cubic count without requiring equal cell sizes.

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.

For fixed forbidden induced patterns, a regularity and Ramsey refinement uses internal clique/independent types and sparse, almost-complete, or intermediate cross pairs. A clique of intermediate pairs realizes every bounded induced pattern consistent with those internal types, using both edge and nonedge regularity. This is the induced embedding consequence used in hereditary graph enumeration theorem; the exceptional-pair budget can be arbitrarily small.

For fixed $r\ge 2$ and every $\epsilon >0$, some $\delta >0$ has the following property for all sufficiently large $n$: a graph with fewer than $\delta n^r$ copies of $K_r$ can be made $K_r$-free by deleting at most $\epsilon n^2$ edges. A Szemerédi regularity lemma partition and the regular clique counting lemma prove this. It converts sparse counts of a forbidden clique into a small edit distance from clique-freeness.

For every $\eta >0$, some $\rho >0$ and $N_0$ have this property: a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ unordered triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. A Szemerédi regularity lemma partition deletes exceptional, intraclass, irregular and sparse-pair edges cheaply. A surviving triangle in a graph would force a positive cubic count by the regular triangle counting lemma.

For a set partition $\mathcal{P}$ of the vertices of a graph of order $n$, definewhere the sum is ordered, the adjacency indicator defines the density also on diagonal pairs, and exceptional vertices are treated as singleton parts. Refinement cannot decrease this energy, by the Cauchy-Schwarz inequality. An irregular pair of equal cells of size $m$, with witnesses of relative size at least $\epsilon$ and density discrepancy greater than $\epsilon$, increases the energy by more than ${\epsilon }^4{m}^2/n^2$ for each orientation. Splitting refined atoms into equal chunks and leftover singleton parts is a further refinement, so restoring equitability causes no energy loss.

In a proof of the Szemerédi regularity lemma, suppose $k$ equal classes have size $L$ and witness splitting creates at most $2^k$ atoms per class. Divide every atom into blocks of size $\ell =\lfloor L/4^k\rfloor$, putting leftovers into the exceptional set. At most $k{2}^k\ell \le kL{2}^{-k}$ vertices are lost. When $L\ge 2\cdot 4^k$, at most $2k{4}^k$ blocks remain. For equitable regularity energy defined by omitting exceptional vertices, the loss from discarding a fraction $r$ is at most $2r$: only ordered pairs incident with discarded vertices are removed, and every squared edge density of a bipartite graph is at most one. This provides a quantitative way to restore equal class sizes after each energy increment.

If a set partition refines a rectangle $U\times V$ of a graph into rectangles $U_a\times V_b$, write $d=d(U,V)$ for its edge density of a bipartite graph. The weighted variance identity isIt follows by expanding the square and using ${\sum }_{a,b}|U_a||V_b|d(U_a,V_b)=|U||V|d$. In particular, equitable regularity energy cannot decrease under refinement. If a witness rectangle has relative sizes at least $\epsilon$ and edge density of a bipartite graph discrepancy greater than $\epsilon$, its contribution, after division by the squared total vertex count, exceeds ${\epsilon }^4|U||V|/N^2$.

For every finite graph $H$ and density $\delta >0$, sufficiently large vertex classes joined according to $H$ by regular pairs of density at least $\delta$ contain a copy of $H$. The regularity parameter must be sufficiently small in terms of $H$ and $\delta$.

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.

If the three pairs among vertex sets $A,B,C$ are $\epsilon$-regular and each has density at least $2\epsilon$, where $0<\epsilon <1/2$, then the graph contains a triangle with one vertex in each set.

Given a regularity partition and a density cutoff $\delta$, its reduced graph has one vertex for each nonexceptional class and joins two classes when their pair is regular and has density at least $\delta$.