1

Let $A_1, A_2, \dots , A_n$ be subsets of $\{1, 2,\dots , n\}$ of size $3$. Prove that $\lfloor \frac{n}{3}\rfloor$ members of $\{1, 2,\dots , n\}$ can be colored such that each $A_i$ has at least $1$ member that is not colored.

I think of doing induction. For $n=3$, it is true. Now say it is true for $n=1,2,\dots,k$, we will show that it's true for $n=k+1$. But this doesn't go that far.

Rewording the problem, Prove that $\lceil \frac{2n}{3}\rceil$ members of $\{1, 2,\dots , n\}$ can be colored such that each $A_i$ has at least $1$ member that is colored.

Note that $$|A_1|+\dots+|A_n|=3n.$$ Moreover consider $A=A_1\cup A_2\dots \cup A_n$. If $|A|\le \lceil \frac{2n}{3}\rceil$ then just colour every element and we are done. So, let $|A|>\lceil \frac{2n}{3}\rceil$.

But then I am not sure what to do for $|A|>\lceil \frac{2n}{3}\rceil$.

Any solutions? If possible, can greedy algorithm not be used as I am not comfortable with it.

RobPratt
  • 50,938
Raheel
  • 1,791

1 Answers1

1

By applying the probabilistic method as in this question, you can obtain a stronger lower bound of $\lceil 2\sqrt3 n/9\rceil$. Taking $p = 1/\sqrt{3}$ yields $p - p^3 = 2\sqrt{3}/9$, and $\alpha(G)$ is an integer, so $\alpha(G) \ge 2\sqrt3 n/9$ implies $\alpha(G) \ge \lceil 2\sqrt3 n/9 \rceil$.

RobPratt
  • 50,938