Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Source. Published pp. 266–267, Theorem 2.1 (PDF).

Statement. If 0<β<10<\beta<1 and every F∈FF\in\mathcal F, G∈GG\in\mathcal G satisfies ∣F∩G∣>βn|F\cap G|>\beta n, then

∣F∣∣G∣≤22nH((1+β)/2).|\mathcal F||\mathcal G|\le2^{2nH((1+\beta)/2)}.

Proof. Empty families are immediate. Put r=min⁡F,G∣F∩G∣r=\min_{F,G}|F\cap G| and t=r−1t=r-1. Complementation shows that F\mathcal F and Gc={X−G:G∈G}\mathcal G^c=\{X-G:G\in\mathcal G\} are disjoint, so the smaller family has at most 2n−12^{n-1} members. Interchange the two families so this is F\mathcal F, and choose the integer a≤n/2a\le n/2 with

∑j<a(nj)<∣F∣≤∑j≤a(nj).\sum_{j<a}\binom nj<|\mathcal F|\le\sum_{j\le a}\binom nj.

The Hamming distance from FF to X−GX-G is ∣F∩G∣+∣X∖(F∪G)∣≥r|F\cap G|+|X\setminus(F\cup G)|\ge r. Thus the closed radius-tt neighborhood of F\mathcal F avoids Gc\mathcal G^c. The exact Harper input gives

∣F∣∣G∣≤(∑j≤a(nj))(∑j≥a+t(nj)).(1)|\mathcal F||\mathcal G| \le\left(\sum_{j\le a}\binom nj\right) \left(\sum_{j\ge a+t}\binom nj\right). \tag{1}

If a+t≥n/2a+t\ge n/2, the binomial bounds and concavity and symmetry of HH show that (1) is at most 22nH((1+t/n)/2)2^{2nH((1+t/n)/2)}: among pairs of arguments a distance t/nt/n apart, their entropy sum is maximized when they are symmetric about 1/21/2. If a+t<n/2a+t<n/2, bound the second factor by 2n2^n; then

H(a/n)+1≤H(1/2−t/n)+H(1/2)≤2H(1/2−t/(2n)).H(a/n)+1\le H(1/2-t/n)+H(1/2) \le2H(1/2-t/(2n)).

This proves the same bound in that case.

To remove the one-unit rounding loss in t=r−1t=r-1, take the Cartesian powers of the two families on NN disjoint copies of XX. Their minimum cross intersection is NrNr, and their size product is (∣F∣∣G∣)N(|\mathcal F||\mathcal G|)^N. Apply the proved bound with tN=Nr−1t_N=Nr-1, take NNth roots, and let N→∞N\to\infty. Continuity gives the bound with r/nr/n in place of β\beta. Since r/n>βr/n>\beta and HH decreases on [1/2,1][1/2,1], the stated inequality follows. □\square

Source precision. The source chooses the largest integer tt for which all intersections exceed tt, then compares it directly with the real parameter βn\beta n. The Cartesian-power step supplies the missing rounding justification. No integral-threshold assumption is imposed.

Dependencies. external_inputs, entropy_estimates.