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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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The Kővári–Sós–Turán common-neighbour count and the discrete convexity lower bound for degree sums

Statement

Let s,t1 be integers, let G be bipartite with parts A,B, where A=m, B=n, and let d(b)=N(b)A. If G contains no oriented Ks,t with s vertices in A, then

bB(d(b)s)(t1)(ms).

For n1 and any nonnegative integers d1,,dn with sum E, moving one unit from a value at least two larger than another cannot increase i(dis). Consequently the minimum occurs when the values differ by at most one. If E/ns1, this gives

i=1n(dis)ns!(Ens+1)s.

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

In z(m,n;s,t), the s-vertex part of the forbidden Ks,t lies on the left and the t-vertex part lies on the right (The Zarankiewicz number z(m,n;s,t) for a forbidden Ks,t in a bipartite graph).

[F2]

The open neighbourhood is NG(v)={u:{u,v}E} and degG(v)=NG(v) (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

[F3]

For a finite incidence relation, the sum of its row-fibre sizes equals the sum of its column-fibre sizes (Double counting: xXRx=R=yYRy for a relation between finite sets).

[F4]

(nk) is the number of k-element subsets of an n-element set (The set [A]k of k-element subsets and the binomial coefficient (nk):=[n]k).

Proof

technique · double-count common neighbours, then smooth integer degrees
1.1

Count pairs (S,b) with SA, S=s, and SN(b). Counting first by b gives the left side. For fixed S, at most t1 vertices of B contain S in their neighbourhood, or those vertices with S form the forbidden Ks,t. Counting first by S proves the upper bound.

givenF1F2F3
1.2

Partitioning the s-subsets of a d-element set according to whether they contain one distinguished element gives (ds)(d1s)=(d1s1), a nondecreasing function of d. Thus if didj+2, replacing (di,dj) by (di1,dj+1) does not increase the binomial sum. Repetition terminates at values q=E/n and q+1.

givenF4
2.1

For integers ds1, (ds)(ds+1)s/s!. Writing E/n=q+θ with 0θ<1, the balanced sum is the corresponding linear interpolation between (qs) and (q+1s); convexity of yys on nonnegative reals bounds it below by n(E/ns+1)s/s!.

step 1.2algebraF4
3.1

Steps 1.1-2.1 prove the common-neighbour upper count and the discrete smoothing lower count with the stated threshold.

step 1.1step 1.2step 2.1

Depends on

Used by

Dependency tree · next 3 levels

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Sources