37-rule-is-optimal

Theorem: Any protocol of date problem has success probability less than \frac{u}{n}\sum\limits_{i=u}^{n-1}\frac1i which is about 37\% . Here u is the biggest number such that \sum_{i=u}^{n-1}\frac1i\leq 1 .

Proof: First, let's introduce some notation. \Pi_n is the set of permutations of \{1,2,\cdots, n\} .

For any two permutations \alpha\in \Pi_a and \beta\in \Pi_b , we say \alpha \leq \beta if a\leq b and \alpha fit with the first a-th entries with \beta (i.e. for any 1\leq i,j\leqa , \alpha_i < \alpha_j \Leftrightarrow \beta_i < \beta_j )

For \alpha\in\Pi_n , let F_k(\alpha) is the permutation\beta\in \Pi_k who \leq \alpha .

Because the expectation of any randomized protocol is no more than the maximum of some deterministic protocol (randomized protocol is actually a distribution on the set of deterministic protocols), it suffice to consider deterministic protocol.

Let's look out how a protocol works. The protocol check the permutation one by one and determine when to stop. When the protocol examine the permutation \alpha to the k -th entries, the protocol doesn't have all the information of \alpha but only knows F_k(\alpha) . Supposed S_k\in \Pi_k be the set of permutations who make the protocol stop. Let s_k=|S_k| .

Obviously, any \alpha\in S_k satisfies \alpha_k=k .

For any \alpha\in S_k , there are \frac{n!}{k!} permutations larger than \alpha . Among them, there are\frac{(n-1)!}{(k-1)!} success permutations (i.e. the k-th entry is n ). So the success probability is

P=\frac{1}{n!}\sum_k\frac{(n-1)!}{(k-1)!}\cdot s_k

Lemma 1: S_k is disorder with each other, i.e. for any i<j ,\alpha\in S_i and \beta\in S_j , \alpha couldn't be smaller than \beta .

Otherwise the protocol will stop at i -th entry when encounter permutation \beta , then \beta can't be in S_j .

From Lemma 1, we know that for any i ,

\sum_{j=1}^{j-1}s_j\frac{(i-1)!}{j!} + s_i\leq (i-1)!

by computing the number of permutation \gamma\in \Pi_i whose i -th entry is i .

Let t_i=s_i/i! and u be the largest number makes\sum_i=u^{n-1}\frac1i\leq 1 , then for any i ,

\sum_{j=1}^{i-1}t_j+it_i\leq 1

and

\begin{array}{rcl}P&=&\frac1n\sum\limits_{k=1}^n kt_k\\&\leq&\frac1n\sum\limits_{i=u}^n(\sum\limits_{j=1}^{i-1}t_j+it_i)(1-\sum\limits_{j=i}^{n-1}\frac1i)\\&=&\frac1n\sum\limits_{i=u}^n(1-\sum\limits_{j=i}^{n-1}\frac1i)\\&=&\frac{u}{n}\sum\limits_{i=u}^{n-1}\frac1i\end{array}

And the equality is made when the protocol ignores the first u entries then stop at the first entry who is larger than the firstu -th entries.

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