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Suppose we have $x_3$ distinct $3$-element sets chosen arbitrarily among the $\binom{7}{3}=35 \ge x_3$ subsets of $U=\{1,2,3,4,5,6,7\}$. Consider the $x_4$ distinct $4$-element sets that can be obtained as a union of any two of the above $3$-element sets. I know the probability that a couple of $3$-element sets chosen at random in $U$ have $2$ elements in common and thus their union is a $4$-element set (as computed here). I would like to find a lower bound for $x_4$ as a function of $x_3$, in the worst case:

$$x_4 \ge f(x_3)$$

Ideally, I would like to later extend the result to any $j \lt k \le n$, i.e. find a lower bound:

$$x_k \ge g(x_j, j, k, n)$$

instead of the above case $j=3, k=4, n=7$.

Any hint?

RobPratt
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Fabius Wiesner
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1 Answers1

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You can solve the problem via integer linear programming as follows. For each $3$-subset $S \subset U$, let binary decision variable $y_S$ indicate whether $S$ is chosen. For a given $x_3$, the problem is to minimize $$\sum_{T \subset U: |T|=4} \max_{(R,S): R \cup S = T} y_R y_S \tag1$$ subject to $$\sum_S y_S = x_3. \tag2$$ The objective function $(1)$, which can be linearized, represents the number of $4$-subsets that are covered exactly by some pair of chosen $3$-subsets. The constraint $(2)$ enforces selection of exactly $x_3$ of the $3$-subsets.

The resulting optimal objective values are $$f(0)=\dots=f(7)=0, f(8)=2, f(9)=4, f(10)=f(11)=5, f(12)=6, f(13)=8, f(14)=11, f(15)=12, f(16)=f(17)=14, f(18)=f(19)=f(20)=15, f(21)=19, f(22)=22, f(23)=24, f(24)=f(25)=25, f(26)=28, f(27)=30, f(28)=f(29)=31, f(30)=33, f(31)=f(32)=34, f(33)=f(34)=f(35)=35.$$

To linearize $(1)$, introduce a binary (or nonnegative) variable $z_T$ for each $4$-subset $T \subset U$ and minimize $$\sum_{T \subset U: |T|=4} z_T \tag3$$ subject to $(2)$ and $$z_T \ge y_R + y_S - 1 \quad \text{for all $T$ and $(R,S): R \cup S = T$}. \tag4$$

RobPratt
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  • Thank you. Although I am looking for a lower bound, so I suppose I have to replace $max$ with $min$ in the objective function. For example $f(2)=0$ because in the worst case I can choose ${1,2,3}$ and ${4,5,6}$. I believe that $f(x_3)=0$ for $x_3<6$. Also, any suggestion of a reading for how to linearize the objective function, or in general, an introduction to linear programming? – Fabius Wiesner Apr 07 '22 at 05:13
  • And I want to minimize the objective function rather than maximise it, I think. – Fabius Wiesner Apr 07 '22 at 09:17
  • OK, I guess I misinterpreted what you meant by arbitrary and worst case. I thought that an adversary was picking the $3$-subsets to make the number of $4$-subsets large, but from your comments now it sounds like you get to pick the $3$-subsets to make the number of $4$-subsets small. – RobPratt Apr 07 '22 at 12:45
  • Yes, I want to make sure that in any case $x_4 \ge f(x_3)$ so that I can work out an inequality that I have for the background problem, like this $x3 + A \gt x_4 + B \ge f(x_3) +B$, for which I need that $f(x_3)$ be the lowest possible for any choice of the $3$-subsets. – Fabius Wiesner Apr 07 '22 at 12:55
  • I updated the answer just now to minimize instead of maximize. – RobPratt Apr 07 '22 at 14:51
  • Thank you very much. Do you think that it is computationally feasible finding a similar lower bound for $x_7$ and $x_6$ ($x_7 \ge f(x_6)$) with a $13$-element universe? – Fabius Wiesner Apr 07 '22 at 15:07
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    That would be $\binom{13}{6}=1716$ ILP problems (one per value of $x_6$), and most of them are not easy to solve. But the LP relaxation provides a lower bound, and for odd $n$, $j=\lfloor n/2\rfloor$, and $k=\lceil n/2\rceil$ the LP lower bound appears to be $\max(2x_j-\binom{n}{j}, 0)$. – RobPratt Apr 07 '22 at 21:30