【读书笔记】排列研究-模式避免-基础Pattern Avoidance

2021/6/6 10:31:27

编程Tag： Array left pattern end right begin 读书笔记 Avoidance conjecture

本文主要是介绍【读书笔记】排列研究-模式避免-基础Pattern Avoidance，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

模式避免的定义
避免Pattern q 的n-排列计数$S_n(q)$
- q长度是2
- q长度是3
- 对一些模式q,做$S_n(q)$的阶估计
- Backelin, West, and Xin给出的较为一般的Theorem
- q长度是4
证明Stanley-Wilf conjecture
- The Stanley-Wilf conjecture
- The Füredi-Hajnal conjecture

模式避免的定义

避免Pattern q 的n-排列计数$S_n(q)$

先扔结论，有时间把证明粘过来

q长度是2

\[S_n(12)=S_n(21)=1 \]

q长度是3

All patterns of length three are avoided by the same number of n-permutations.

\[S_n(123)=S_n(132)=S_n(213)=S_n(231)=S_n(312)=S_n(321)\\ S_{n}(132)=C_{n}=\frac{\left(\begin{array}{c} 2 n \\ n \end{array}\right)}{n+1} \]

对一些模式q,做$S_n(q)$的阶估计

Backelin, West, and Xin给出的较为一般的Theorem

举例，
$r=2,k=2,q=34$ it says, $S_n(1234)=S_n(2134)$
$r=2,k=2,q=43$ it says, $S_n(1243)=S_n(2143)$
$r=3,k=1,q=4$ it says, $S_n(1234)=S_n(3214)$

q长度是4

本来$q$有24种，可以证明最后归结为这三种代表,1234,1342,1324

\[\begin{aligned} &\text { for } S_{n}(1342) \text { we have } 1,2,6,23,103,512,2740,15485\\ &\text { for } S_{n}(1234) \text { we have } 1,2,6,23,103,513,2761,15767\\ &\text { for } S_{n}(1324) \text { we have } 1,2,6,23,103,513,2762,15793 \end{aligned} \]

aovid 1342 A022558

avoid 1234 A005802

avoid 1324 A061552

\[\begin{aligned} S_{n}(1342) &=(-1)^{n-1} \cdot \frac{\left(7 n^{2}-3 n-2\right)}{2} \\ &+3 \sum_{i=2}^{n}(-1)^{n-i} \cdot 2^{i+1} \cdot \frac{(2 i-4) !}{i !(i-2) !} \cdot\left(\begin{array}{c} n-i+2 \\ 2 \end{array}\right) \end{aligned} \]

\[S_{n}(1234)=2 \cdot \sum_{k=0}^{n}\left(\begin{array}{c} 2 k \\ k \end{array}\right)\left(\begin{array}{l} n \\ k \end{array}\right)^{2} \frac{3 k^{2}+2 k+1-n-2 n k}{(k+1)^{2}(k+2)(n-k+1)} \]

\[S_{n}(1234)=\frac{1}{(n+1)^{2}(n+2)} \sum_{k=0}^{n}\left(\begin{array}{c} 2 k \\ k \end{array}\right)\left(\begin{array}{c} n+1 \\ k+1 \end{array}\right)\left(\begin{array}{c} n+2 \\ k+1 \end{array}\right) \]