What's the expected length of LIS of a random n-permutation.
According to my test, it approximates $$$O(\sqrt n)$$$.
But how to prove it?
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Ah, I want to know the proof, too. I just met it today.
Let $$$E_n$$$ be the expected value.
Based on the position of $$$n$$$ in a $$$n$$$-permutation, we have the formula: $$$E_n = \dfrac{E_0+E_1+...+E_{n-1}}{n} + 1$$$, which leads to $$$E_n = E_{n-1} + \dfrac{1}{n}$$$.
So $$$E_n = 1 + \dfrac{1}{2} + ... \dfrac{1}{n} \approx \log{n}$$$.
Edit: Wrong solution
I don't think that's true, one doesn't have to include $$$n$$$ in LIS
Sorry but I have no idea that how can get the conclusion above？
You are right! The answer is indeed $$$O(\sqrt{n})$$$ or more precisely $$$\approx 2\sqrt{n}$$$ when $$$n$$$ is large, according to this paper.
In this problem that observation is needed: 1017C - The Phone Number.
The tutorial says the formal proof, by Dilworth's theorem.
I can't get the point that we can prove it by this theory.
Can you be more specific please? XD