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A prescribed finite vertex partition has a bounded ϵ-regular refinement, equitable when the initial partition is equitable

Statement

For every 0<ϵ<1 and k01 there is K=K(ϵ,k0) such that every partition P0 of a finite graph into at most k0 nonempty parts has an ϵ-regular refinement P into at most K parts. If P0 is equitable, P may also be chosen equitable.

Facts & Assumptions

Given: ϵ,k0, a finite graph, and a partition P0 as in the Statement.

[L1]

Every nonregular k-part partition has a refinement with at most k2k+1 parts and energy gain greater than ϵ5 (Every nonregular k-part partition has a refinement with energy gain greater than ϵ5 and at most k2k+1 parts).

[L2]

Energy is nondecreasing under refinement and at most 1 (Energy lies in [0,1] and cannot decrease under refinement).

[L3]

An equitable partition has part sizes differing by at most one; a partition is ϵ-regular when its irregular ordered pairs carry total weight at most ϵn2; and the empty family, the unique partition of the null graph, is equitable and ϵ-regular by convention (ϵ-regular vertex partitions, equitable partitions, and refinement).

[L4]

q(P)=n2A,BPABd(A,B)2, a sum over ordered pairs of parts with nonnegative weights AB/n2 and densities in [0,1] (The mean-square density, or energy, of a vertex partition).

[L5]

(X,Y) is ϵ-regular when every AX and BY with AϵX and BϵY satisfies d(A,B)d(X,Y)ϵ (ϵ-regular pairs and self-regular vertex sets).

Proof

technique · direct
1.1

Arbitrary refinement. Starting from P0, apply [L1] whenever the current partition is not ϵ-regular; each application raises the energy by more than ϵ5 and replaces the part-count bound k by k2k+1. By [L2] the energy lies in [0,1], so at most ϵ5 applications occur, and iterating that map ϵ5 times from k0 bounds the number of parts of the resulting ϵ-regular refinement of P0.

givenL1L2inductionalgebra
1.2

Constants for the equitable case. Put K0=k0, Kr+1=Krϵ52Kr+5, R=2ϵ5, and T=KR. Each factor is at least 1, so K0K1KR=T.

givenalgebrachoose
2.1

Small graphs. Assume P0 is equitable and n<T. The partition of V(G) into singletons refines P0, is equitable, and has n<T parts. It is ϵ-regular: the only subset of a singleton with at least ϵ times its size is the singleton itself, so every pair of singletons has density difference 0 in [L5]; for n=0 the convention in [L3] gives the same conclusion.

step 1.2L3L5
2.2

Equitisation. Let P be equitable with k parts, let R refine P with at most K=k2k+1 parts, put p=ϵ52k+5, and suppose kpn. Order V(G) so that each part of P is an interval and each cell of R is an interval inside its part, and cut each part X into p consecutive pieces of sizes X/p or X/p. Writing a=n/k and t=a/p1, the sizes X{a,a+1} force X/pt and X/pt+1, so every piece has size t or t+1 and the result P is an equitable refinement of P with exactly kp parts.

step 1.2L3algebraconstruct
3.1

Energy loss. Call a piece dirty when it is not contained in a single cell of R, and let D be the union of the dirty pieces. Each of the at most Kk boundaries between consecutive R-cells of one part lies in exactly one piece, so at most K pieces are dirty, each of size at most X/p2X/p4n/(kp); hence D4Kn/(kp)=2k+3n/pϵ5n/4. Let S be the common refinement of P and R. It refines R, so q(S)q(R) by [L2], while every clean piece is already a cell of S, so by [L4] the two energies agree except on ordered pairs with an entry inside D, which carry total weight at most 2D/n and squared densities in [0,1]. Therefore q(P)q(S)2D/nq(R)ϵ5/2.

step 2.2L2L4algebra
4.1

Iteration. Assume P0 is equitable with k0K0 parts and nT. Given an equitable Pr that refines P0, has krKr parts, and is not ϵ-regular, apply [L1] to get a refinement Rr with at most kr2kr+1 parts and energy gain more than ϵ5, then apply steps 2.2 and 3.1 with pr=ϵ52kr+5; the hypothesis krprn holds because krKr makes krprKr+1Tn. The result Pr+1 is equitable, refines Pr and hence P0, has kr+1=krprKr+1 parts, and satisfies q(Pr+1)>q(Pr)+ϵ5/2.

step 1.2step 2.2step 3.1L1induction
5.1

Since energy lies in [0,1] by [L2], it cannot gain ϵ5/2 at R=2ϵ5 successive rounds, so some Pr with r<R is ϵ-regular; it is equitable, refines P0, and has at most KrT parts.

step 4.1L2algebra
6.1

Let K be the larger of T and the bound supplied by step 1.1; both depend only on ϵ and k0. Step 1.1 proves the first assertion, and steps 2.1 and 5.1 prove the equitable assertion for n<T and for nT respectively.

step 1.1step 2.1step 5.1choose

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