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Let $V$ be a set of $n$ permutations of $m$ numbers. Let $m<n<<m!$ I want to prove the following lemma:

Lemma: For any set of permutations $V$ there exists a pair of numbers $(a,b)$ such that $\frac{|(v\in V: a\succcurlyeq b)|}{|(v\in V: b\succcurlyeq a)|}\approx 1$.

In other words, prove that there always exists a pair of numbers such that the number of partitions where $a$ comes before $b$ is approximately equal to the number of partitions where $b$ comes before $a$.

For the sake of completeness, I want to prove this lemma in order to be able to prove the following theorem:

Theorem: Let $n$ be the number of permutations of $m$ numbers. Let $m<n<<m!$. We now construct a binomial tree where at the root of the tree there is the set of all $n$ permutations. Each node we select a couple of elements of the permutations $(a,b)$ and split the permutations into two nodes containing two subsets of permutations: one with each permutation where $a$ comes before $n$ and the other viceversa. A node is a leaf if it contains equal permutations. Prove that the optimal height of the tree is $O(\log{n})$.

xavrojo
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  • Do you mean that the height of the tree can be $O(\log n)$ when we make good choices? If we make bad choices, it can definitely be worse. – Misha Lavrov May 08 '23 at 12:56

1 Answers1

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Your lemma and theorem are false. If you let $V$ be the set of permutations that switch any two adjacent elements. Then there’s $n=m-1$ permutations but for any pair $a,b$, at least $m-2$ permutations preserve their order. The ratio $(m-2)/1$ is not around 1 and the difference is $n-2$ which is not $O(\log n)$.

Similarly, the binary tree you’re constructing needs to have height $n-1$ since each test can only eliminate at most one permutation in the worst case scenario. This is not $O(\log(n))$.

Eric
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