MGF multivariate generating function 多变量生成函数

目录

• MGF多变量生成函数multivariate generating function
  • 定义
  • 例子
• Extremal parameters
  • III.8.1 largest components
    • 例子1 decomposition of permutations
    • 例子2 words over an m–ary alphabet
    • 例子3 set partitions
    • 例子4 the longest run in a sequence of binary draws
  • III.8.2 Height
    • 例子1 plane trees
    • 例子2 plane binary trees
    • 例子3 Cayley trees
  • III 8.3 Averages and moments

multivariate generating function MGF 多变量生成函数

MGF多变量生成函数multivariate generating function

定义

比如，以前可能让你计数有\(k-1\)个逆序对的n-排列个数\(A(n,k)\)

现在让你计数

指定了n-排列圆分解后各个长度\(j\)的cycles的个数\(\chi_j\)，让你计数

或者，指定了各字符出现频数的长n的字符串个数

例子

the exponential MGF of permutation with \(u_1,u_2\) marking the number of 1-cycles and 2-cycles to be

\[P\left(z, u_{1}, u_{2}\right)=\frac{\exp \left(\left(u_{1}-1\right) z+\left(u_{2}-1\right) \frac{z^{2}}{2}\right)}{1-z} \]

可以想象其实还有其他的\(u_j\),这些\(u_j\)都取1了

写完整了，就是

\[P\left(z, \mathbf{u} \right)=\exp \left(u_1 t+\frac{u_2 z^2}{2}+\frac{u_3 z^3}{3}+\ldots\right) \]

这正好是【The cycle index polynomials of the symmetric groups】

如果取$u_1=u_2=1 $ , 对应的是计数所有的n-排列个数的EGF

\[P(z,1,1)=(1-z)^{-1} \]

对应的计数是\(n!\)

 

如果\(u_1=0,u_2=1\)，给出derangement(错排；长度为1的cycles的个数为0)的EGF

\[P(z,0,1)=e^{-z}/(1-z) \]

对应的计数是\((n!)(\frac{1}{2!}-\frac{1}{3!}...+(-1)^{n}\frac{1}{(n)!})\)

 

如果\(u_1=u_2=0\) ,给出所有的cycles长度都大于2的n-排列个数的EGF

\[P(z,0,0)=e^{-z-z^2/2}/(1-z) \]

对应的计数是$
a(n) = \sum_{i=3..n} \binom{n-1}{i-1}\cdot (i-1)! \cdot a(n-i).a(0)=1,a(1)=0,a(2)=0$ https://oeis.org/A038205

 

如果置\(u_1=u,u_2=1\),给出有k个singleton cycles(k个位置正好不变)的n-排列的EBGF

\[P(z,u,1)=\frac{e^{(u-1)z}}{1-z} \]

对应的计数是 \(\tbinom{n}{k}(n-k)!(\frac{1}{2!}-\frac{1}{3!}...+(-1)^{n-k}\frac{1}{(n-k)!})\)

Extremal parameters

现在要研究这样的问题：

比如：

一个例子是the largest component in a combinatorial structure (for instance, the largest cycle of a permutation) ，指定了排列的长度\(n\)和最大的cycle长度\(k\),让你计数

另一个例子是 the maximum degree of nesting of constructions in a recursive structure (typically, the height of a tree). 制定了树的节点数\(n\)和树的的高度\(h\)，让你计数

面对这样的问题，(传统的)BGF无能为力。

III.8.1 largest components

例子1 decomposition of permutations

the cycle decomposition of permutations translated by

\[P(z)=\exp \left(\log \frac{1}{1-z}\right) \]

使用truncation后,得到the EGF of permutations with longest cycle≤ b

\[P^{\langle b\rangle}(z)=\exp \left(\frac{z}{1}+\frac{z^{2}}{2}+\cdots+\frac{z^{b}}{b}\right) \]

by the way，the EGF of permutations with smallest cycle of size > b

\[\begin{aligned} \exp(\log \frac{1}{1-z}-\frac{z}{1}-\frac{z^{2}}{2}-\cdots-\frac{z^{b}}{b})=\frac{1}{1-z}\exp(-\frac{z}{1}-\frac{z^{2}}{2}-\cdots-\frac{z^{b}}{b}) \end{aligned} \]

例子2 words over an m–ary alphabet

the EGF of words over an m–ary alphabet

\[W(z)=(e^z)^m \]

使用truncation后，得到the EGF of words such that each letter occurs at most b times

\[W^{(b)}(z)=\left(1+\frac{z}{1 !}+\frac{z^{2}}{2 !}+\cdots+\frac{z^{b}}{b !}\right)^{m} \]

例子3 set partitions

the EGF of set partitions

\[S(z)=exp(e^z-1) \]

the EGF of set partitions with largest block of size at most b

\[S^{\langle b\rangle}(z)=\exp \left(\frac{z}{1 !}+\frac{z^{2}}{2 !}+\cdots+\frac{z^{b}}{b !}\right) \]

例子4 the longest run in a sequence of binary draws

\(\mathcal{W}=\operatorname{SEQ}(a) \cdot \operatorname{SEQ}(b \operatorname{SEQ}(a))\)这么写有点正则的感觉，比如aabbbaabaa分解过来就是 aa|b|b|baa|baa

III.8.2 Height

例子1 plane trees

general plane trees定义是这样的

\[\mathcal{G}=\mathcal{N} \times \operatorname{SEQ}(\mathcal{G}) \quad \text { so that } \quad G(z)=\frac{z}{1-G(z)} \]

定义一个树的高度是最长的分支(branch)的边的个数，那么the set of trees of height\(\leq h\) 满足如下的递归关系

\[\mathcal{G}^{[0]}=\mathcal{N}, \mathcal{G}^{[h+1]}=\mathcal{N} \times \operatorname{SEQ}\left(\mathcal{G}^{[h]}\right) \]

对应的OGF满足

\[G^{[-1]}(z)=0, G^{[0]}(z)=z, G^{[h+1]}(z)=\frac{z}{1-G^{[h]}(z)} \]

展开得到（别问连分数怎么打的，问就是\frac套\frac）

\[G^{[h]}(z)=\frac{z}{1-\frac{z}{1-\frac{z}{\frac{\ddots}{1-z}}}} \]

例子2 plane binary trees

平面二叉树定义如下

\[\mathcal{B}=\mathcal{Z}+\mathcal{B} \times \mathcal{B} \quad \text { so that } \quad B(z)=z+(B(z))^{2} \]

递归关系是

\[B^{[0]}(z)=z, B^{[h+1]}(z)=z+\left(B^{[h]}(z)\right)^{2} \]

得到

\[B^{[h]}(z)=z+\left(z+\left(z+(\cdots)^{2}\right)^{2}\right)^{2} \]

例子3 Cayley trees

CayleyTree词条 所有非叶子节点的度都为\(n\)的树叫n-CayleyTree

Cayley树定义方程

\[\mathcal{T}=\mathcal{Z} \star \operatorname{SET}(\mathcal{T}) \quad \text { so that } \quad T(z)=z e^{T(z)} \]

有界高度的CayleyTrees的EGF的递归关系

\[T^{[0]}(z)=z, T^{[h+1]}(z)=z e^{T^{[h]}(z)} \]

得到\(T^{[h]}(z)\)的"连指数“形式

\[T^{[h]}(z)=z e^{z e^{z e^{\cdot^{\cdot^{\cdot^{ze^z}}}}} } \]

III 8.3 Averages and moments

没看懂

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资料来自网络

书用的是Analytic Combinatorics by Philippe Flajolet, Robert Sedgewick