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

[Pleft(z, u_{1}, u_{2} ight)=frac{exp left(left(u_{1}-1 ight) z+left(u_{2}-1 ight) frac{z^{2}}{2} ight)}{1-z} ]

可以想象其实还乘了其他的(u_j),这些(u_j)都取1了

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} inom{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 ( binom{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} ight) ]

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

[P^{langle b angle}(z)=exp left(frac{z}{1}+frac{z^{2}}{2}+cdots+frac{z^{b}}{b} ight) ]

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

[egin{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 !} ight)^{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 angle}(z)=exp left(frac{z}{1 !}+frac{z^{2}}{2 !}+cdots+frac{z^{b}}{b !} ight) ]

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} imes operatorname{SEQ}(mathcal{G}) quad ext { 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} imes operatorname{SEQ}left(mathcal{G}^{[h]} ight) ]

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} imes mathcal{B} quad ext { 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) ight)^{2} ]

得到

[B^{[h]}(z)=z+left(z+left(z+(cdots)^{2} ight)^{2} ight)^{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 ext { 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

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