【读书笔记】组合计数中的行列式方法 专题4 Routings: the Lindstrm–Gessel–Viennot lemma

书用的是Handbook of Enumerative Combinatorics (Miklos Bona)

目录

• 专题4-Routings: the Lindstrm–Gessel–Viennot lemma
  • 一些定义
  • the Lindstrm–Gessel–Viennot lemma
  • Lindstrm–Gessel–Viennot lemma 应用
    • Example1 Binomial determinants
    • Example2 Counting permutations by descent set
    • Example3 Rhombus tilings and plane partitions
    • Example4 Catalan determinants, multitriangulations, and Pfaffian rings
      • the number of k-fans of Dyck paths of length \(2(n−2k)\)
      • the number of k triangulations of an n-gon
    • Example5 Schröder determinants and Aztec diamonds

专题4-Routings: the Lindstrm–Gessel–Viennot lemma

一些定义

• DAG 有向无环图
• \(\mathrm{wt}(e)\) 边的权重
• \(\mathrm{wt}(P)=\prod \mathrm {wt}(e)\) 路径的权重
• \(\mathrm{wt}(R)=\prod _{i=1}^n \mathrm {wt}(P)\) routing的权重
• 源，汇 Let S ={s1,...,sn} and T ={t1,...,tn} be two (not necessarily disjoint) sets of vertices,which we call sources and sinks,respectively
• Routing定义 A routing from S to T is a set of paths P1,..., Pn from the n sources s1,...,sn to the n sinks t1,...,tn such that no two paths share a vertex.
• Let π be the permutation of [n] such that \(P_i\) starts at source \(s_i\) and ends at sink \(t_{π(i)}\), and define sign(R) = sign(π).
• path matrix Q定义 \(q_{ij}=\sum\limits_{P\ path from\ s_i to\ t_j }\mathrm{wt}(P)\) 考虑Q的元素\(q_{ij}\)时只是看\(s_i\)到\(t_j\)的所有路径

原文如下

the Lindstrm–Gessel–Viennot lemma

维基百科词条-Lindstrm–Gessel–Viennot lemma

式子右边的求和是对\(S\)到\(T\)的所有routings

【In particular】那里是说，

如果正好图还有这样的性质：每条边权都是1，而且\(S\)到\(T\)的所有routing都是1->1,2->2...这样的形式，那么\(detQ=number \ of \ routings\ from\ S\ to \ T\)

Lindstrm–Gessel–Viennot lemma 应用

Example1 Binomial determinants

\(\begin{gathered} \begin{pmatrix} a_1,...,a_n \\ b_1,...,b_n \end{pmatrix} \end{gathered}\)记号表示的是那个行列式

...开始我还看不懂为什么there are \(\begin{gathered} \begin{pmatrix} a_i \\ b_j \end{pmatrix} \end{gathered}\) SE paths from \(A_i\) to \(B_j\)

SE path 是说south 和 east ，我以为是往东往北了

since every SE routing from A to B takes Ai to Bi for all i, 这是因为0≤a1 <···<an and 0≤b1 <···<bn ; points A ={A1,...,An} and B ={B1,...,Bn} where Ai = (0,ai) and Bi = (bi,bi) for1≤i≤n 然后你还要要走SE path

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Example2 Counting permutations by descent set

先定义一个n排列的decent set是说一个集合\(S\)，集合\(S\)的元素是index i使得\(\pi_i>\pi_{i+1}\)

然后问你 number of permutations of [n] with descent set\(\{c_1,...,c_k\}\)

可以构造一个A到B的SE routings构成bijeciton，答案是\(\begin{gathered} \begin{pmatrix} c_1,...c_k,n \\ 0,c_1,...,c_k \end{pmatrix} \end{gathered}\)(前面的行列式记号)

随手讲解图片里的例子，顺便讲一下符号
比如对于\(\pi=27351684\)(原书中给的是\(\pi=28351674\)，我认为应该是\(\pi=27351684\)，不然说不通)来说，
定义\(\pi_i>\pi_{i+1}\)的位置是\(c_i\),\(c_0\)往往取\(0\),\(c_{k+1}\)往往取\(n\)，这里\(c=\{0,2,4,7,8\}\)
定义了一个从permuation π到序列f的映射，\(f_i\)是满足$j\leq i \text { && } \pi_j\leq\pi_i $ 的 \(j\) 的数量，这里，\(f(\pi)=12231574\)
接着在那些\(c_i\)位置处做分割，得到
\(f(\pi)=12.23.157.4\)

\(B_i\)点的坐标形如\((c_{i-1},c_{i-1})\)，在y=x上
\(A_i\)点的坐标形如\((0,c_i)\)
路径如何确定？拿到序列\(f\)被分割后的片段\(f^i\)，比如第一段是{1,2}，那么在\(B_4\)到\(A_4\)的2步长path中，第1步和第2步就是N step,其他步是W step

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Example3 Rhombus tilings and plane partitions

让你用【1,1,1,1 60° 120°】的菱形密铺边长为\(n\)的正六边形(需要按照网格线摆放)。问你方案数\(R_n\)

有几种观点：

1. 看成正视图，如此，容易看到3种【\(60^°120^°\)的菱形】每种都是\(n^2\)个

2. 思路是把每个菱形都看成是两个正三角形拼接，每个正三角形的中心作为一个节点，一般而言(内部的)每个节点和最近的3个节点相连。构成hexagonal grid。
   相邻两个节点相连如果两个等边三角形上面正好是覆盖所用的菱形。问题转变为求完美匹配。

3. plane partition理解。从观点1的角度往前一步，给出【表明每个格子上垒有多少个cube】的俯视图。 an array of nonnegative integers(finitely many of which are non-zero)that is weakly decreasing in each row and column. We conclude that \(R_n\) is also the number of plane partitions whose non-zero entries are at most n, and fit inside an n×n square.

4. 从观点1的角度出发，看高度\(n-0.5\)，...，高度2.5，高度1.5，高度0.5，截cube stack的曲线。这对应于 n sources S1,...,Sn on the left to the sinks T1,...,Tn ，（S1到T1,S2到T2....）的routing。利用前面的Lindstrm–Gessel–Viennot lemma，矩阵元素\(\begin{gathered} \begin{pmatrix} 2n \\ n+i-j \end{pmatrix} \end{gathered}\)

   \[R_n=det\bigg[\begin{gathered} \begin{pmatrix} 2n \\ n+i-j \end{pmatrix}\bigg] \end{gathered}_{1\leq i,j\leq n}=\prod\limits_{i,j,k=1}^{n}\frac{i+j+k-1}{i+j+k-2} \]

Example4 Catalan determinants, multitriangulations, and Pfaffian rings

定义一个序列\(A=(a_0,a_1,a_2,...)\)的Hankel矩阵\(H_n(A)\)和\(H_n'(A)\)

\[\begin{equation} H_n(A) =\begin{pmatrix} a_0 & a_1 & \cdots\ &a_n\\ a_1 & a_2 & \cdots\ & a_{n+1}\\ \vdots & \vdots & \ddots & \vdots \\ a_n & a_{n+1} & \cdots\ & a_{2n}\\ \end{pmatrix} \end{equation} \]

和

\[\begin{equation} H_n'(A) =\begin{pmatrix} a_1 & a_2 & \cdots\ &a_{n+1}\\ a_2 & a_3 & \cdots\ & a_{n+2}\\ \vdots & \vdots & \ddots & \vdots \\ a_{n+1}& a_{n+2} & \cdots\ & a_{2n+1}\\ \end{pmatrix} \end{equation} \]

如果我们知道 the Hankel determinants \(det \ H_n(A)\) and \(det\ H_n'(A)\) ，而且对所有的n它们都是非零的，我们可以利用递推关系从 \(a_0,…, a_{k-1}\)来恢复每一个 \(a_k\) 。

如果序列是卡特兰序列\(C=(C_0,C_1,C_2,...)\)的话，恰好有一种很好的解释。

式(1.9)是许多事物的组合计数。

the number of k-fans of Dyck paths of length \(2(n−2k)\)

式（1.9）也是the number of k-fans of Dyck paths of length \(2(n-2k)\)

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the number of k triangulations of an n-gon

k-crossing 定义 a k-crossing in an n-gon to be a set of k diagonals that cross pairwise（两两相交）

k-triangulation定义 A k-triangulation is a maximal set of diagonals with no(k+1)-crossings.

式(1.9)也是 the number of k triangulations of an n-gon 。

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Example5 Schröder determinants and Aztec diamonds