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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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For every n16 there is an n-vertex graph with hom(G)<3log2n

Statement

For every integer n16, there exists an n-vertex graph G such that hom(G)<3log2n.

Facts & Assumptions

Given: An integer n16 and k=3log2n.

[L1]

The homogeneous number of a graph is the larger of its clique and stable-set numbers (Homogeneous vertex sets and the homogeneous number hom(G)=max{ω(G),α(G)}).

[L2]

In G(n,1/2), all possible edges are independent Bernoulli variables on the labelled vertex set [n] (The Erdős-Rényi finite random graph G(n,p)).

[L3]

Prescribing r present edges and s absent edges in G(n,p) has probability pr(1p)s (A prescribed set of present and absent edges in G(n,p) has product probability).

[L4]

Expectations of a finite family of random variables add without an independence hypothesis (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L5]

If a nonnegative integer-valued random variable on a finite probability space has expectation below 1, some outcome makes it 0 (The first-moment method for avoiding or forcing a finite count of bad events).

[L7]

For b>0 with b1 and x>0, logbx:=logx/logb (The logarithm to a positive base other than one).

[L8]

log:(0,)R is strictly increasing, and for x,y>0 satisfies log(xy)=logx+logy, log(1/x)=logx, and log1=0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

technique · constructive
1.1

By [L8], log2>log1=0, so [L7] makes log2=log/log2 strictly increasing with log21=0. Applying log(xy)=logx+logy to the factors of an integer power, and log(1/x)=logx for a negative exponent, gives log2(2m)=m for every integer m and log2(xj)=jlog2x for every x>0 and every positive integer j.

L7L8algebra
2.1

Since n16=24, step 1.1 gives L:=log2n4, and k=3L satisfies k3L12.

givenstep 1.1algebra
3.1

In G(n,1/2) let X count the k-subsets that induce a clique or a stable set. For a fixed k-subset these two disjoint events each prescribe all (k2) pairs, so [L2] and [L3] give probability 22(k2).

step 2.1L2L3L6
4.1

By [L4] and [L6], EX=2(nk)2k(k1)/22nk2k(k1)/2.

step 3.1L4L6algebra
5.1

By step 1.1 the base-two logarithm of that last bound is 1+kLk(k1)/2=1+k(L(k1)/2)1+3L(1L)/2<0, where the first inequality uses k3L and L4, which make L(k1)/2(1L)/2<0. Since log2 is strictly increasing with log21=0, the bound itself is below 1, so EX<1.

step 1.1step 2.1step 4.1algebra
6.1

By [L5], choose an outcome graph G with X=0. It has no homogeneous k-subset, and any homogeneous set of order at least k would contain one, so hom(G)k1<3L=3log2n.

step 3.1step 5.1L1L5constructdischarge-construct

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