I'm not sure if CS SE is the right place for this question, but since originally this question was in the CS area (and I translated it to a mathematical form), I will post it here.

I am given two parameters $m\in\mathbb{N}$ and $\frac{1}{2}< p\le 1$, and I want to find the maximal value $k\in \mathbb{N}$, such that there are $A_1,\dots,A_k\subseteq [m]$ for $[m]=\{1,\dots,m\}$ and, we have the two following properties:

  1. For every $1\le i\le k$, we have $|A_i| \ge p\cdot m$
  2. For every $i\neq j$, we have $|A_i\cap A_j|\le (1-p)\cdot m$

Essentially, I'm trying to find the maximal number of sets with size $p\cdot m$, who are $c$-far from being distinct in pairs, where $c:=(1-p)\cdot m$.

Equivalently, I'm interested in finding for every $k$ what is the highest $p$ with this property.

My attempt

I tried to think of an approach similar to one where $p=1$, and the sets would be completely disjoint, therefore every $A_i$ would add an additional $p\cdot m$ new elements, and thus $k=\frac{m}{p\cdot m}=\frac{1}{p}$. But this approach didn't work for me, since even if I say each set adds an additional $p\cdot m - (1-p) \cdot m = (2p-1)\cdot m$ new elements to every other set, I can't guarantee those elements are not in other sets.

Some (additional) thoughts

After thinking additionally a bit, I think that one thing that would suffice for me is to show that the solution doesn't depend on $m$.

This is because in my usage of this problem, I actually try to find a $k$ such that this property won't hold, and I can choose my $k$ to be arbitrarily large but it would also increase $m$ arbitrarily. So - if the solution here won't depend on $m$, I don't need to worry about increasing $m$ arbitrarily.

  • $\begingroup$ If $p > 2/3$ then any two sets have intersection at least $(2p-1)m > m/3 > (1-p)m$, and so $k \leq 1$. $\endgroup$ Commented May 23, 2021 at 14:04
  • $\begingroup$ Yes I already managed to prove this (and something a bit stronger than this, in fact, but I didn't include it here since it would change the problem's definition). The actual problem I'm trying to solve is for $p=\frac{2}{3}$, and I want to see what the maximal $k$ is in this case. If the maximal $k$ is too large (when compared to $m$) for my usage case, then I would be interested in even lower values of $p$, and hence I asked this general question. $\endgroup$
    – nir shahar
    Commented May 23, 2021 at 14:17
  • $\begingroup$ Just to note what I'm trying to achieve using this question: I'm trying to prove a lower bound on $dMA$ protocols for a certain problem, and solving this question would allow me to know if my thinking strategy could work in this case. $\endgroup$
    – nir shahar
    Commented May 23, 2021 at 14:20
  • $\begingroup$ If $|A_i| = pm$ then the condition on the intersection is equivalent to $|A_i \Delta A_j| \geq (4p-2)m$, and so you're interested in the size of constant-weight codes. $\endgroup$ Commented May 23, 2021 at 14:52
  • $\begingroup$ So, is this an open problem in the area of ECCs? $\endgroup$
    – nir shahar
    Commented May 23, 2021 at 15:21

1 Answer 1


This is a partial answer for $p > \frac{\sqrt{5}-1}{2} \approx 0.618.$ As Yuval Filmus points out in the comments, we are looking for the maximum number $k$ of binary codes with length $m$, weight $w = pm$ and pairwise distance at least $d = 2(2p-1)m$. If $p>2/3$ then $k\leq 1$.

Now let $\lambda = w-d/2$. Equation (3) in [1] tells us that if $w^2>\lambda m$ (which is the case when $p > \frac{\sqrt{5}-1}{2}$), then $$k \leq \frac{m\cdot d/2}{w^2-\lambda m}.$$ This works out to $$ k \leq \frac{2p-1}{p^2+p-1}.$$

In particular for the case which seems to interest you the most ($p=2/3$) this gives an upper bound of $k\leq 3$ (which is easily seen to be tight).

I wasn't able to apply the other formulas in [1] to get anything subexponential in $m$ when $\frac{1}{2} < p \leq \frac{\sqrt{5}-1}{2}$, but I applied the bounds very crudely, so it might be possible (might even be easy).

[1] Selmer Johnson, Upper Bounds for Constant Weight Error Correcting Codes, 1972.


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