组合数学 (Spring 2016)/Sieve methods

Principle of Inclusion-Exclusion

Let [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] be two finite sets. The cardinality of their union is

[math]\displaystyle{ |A\cup B|=|A|+|B|-{\color{Blue}|A\cap B|} }[/math].

For three sets [math]\displaystyle{ A }[/math], [math]\displaystyle{ B }[/math], and [math]\displaystyle{ C }[/math], the cardinality of the union of these three sets is computed as

[math]\displaystyle{ |A\cup B\cup C|=|A|+|B|+|C|-{\color{Blue}|A\cap B|}-{\color{Blue}|A\cap C|}-{\color{Blue}|B\cap C|}+{\color{Red}|A\cap B\cap C|} }[/math].

This is illustrated by the following figure.

Generally, the Principle of Inclusion-Exclusion states the rule for computing the union of [math]\displaystyle{ n }[/math] finite sets [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math], such that

[math]\displaystyle{ \begin{align} \left|\bigcup_{i=1}^nA_i\right| &= \sum_{I\subseteq\{1,\ldots,n\}}(-1)^{|I|-1}\left|\bigcap_{i\in I}A_i\right|. \end{align} }[/math]

---

In combinatorial enumeration, the Principle of Inclusion-Exclusion is usually applied in its complement form.

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n\subseteq U }[/math] be subsets of some finite set [math]\displaystyle{ U }[/math]. Here [math]\displaystyle{ U }[/math] is some universe of combinatorial objects, whose cardinality is easy to calculate (e.g. all strings, tuples, permutations), and each [math]\displaystyle{ A_i }[/math] contains the objects with some specific property (e.g. a "pattern") which we want to avoid. The problem is to count the number of objects without any of the [math]\displaystyle{ n }[/math] properties. We write [math]\displaystyle{ \bar{A_i}=U-A_i }[/math]. The number of objects without any of the properties [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] is

[math]\displaystyle{ \begin{align} \left|\bar{A_1}\cap\bar{A_2}\cap\cdots\cap\bar{A_n}\right|=\left|U-\bigcup_{i=1}^nA_i\right| &= |U|+\sum_{I\subseteq\{1,\ldots,n\}}(-1)^{|I|}\left|\bigcap_{i\in I}A_i\right|. \end{align} }[/math]

For an [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math], we denote

[math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math]

with the convention that [math]\displaystyle{ A_\emptyset=U }[/math]. The above equation is stated as:

Principle of Inclusion-Exclusion

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be a family of subsets of [math]\displaystyle{ U }[/math]. Then the number of elements of [math]\displaystyle{ U }[/math] which lie in none of the subsets [math]\displaystyle{ A_i }[/math] is [math]\displaystyle{ \sum_{I\subseteq\{1,\ldots, n\}}(-1)^{|I|}|A_I| }[/math].

Let [math]\displaystyle{ S_k=\sum_{|I|=k}|A_I|\, }[/math]. Conventionally, [math]\displaystyle{ S_0=|A_\emptyset|=|U| }[/math]. The principle of inclusion-exclusion can be expressed as

[math]\displaystyle{ \left|\bar{A_1}\cap\bar{A_2}\cap\cdots\cap\bar{A_n}\right|= S_0-S_1+S_2+\cdots+(-1)^nS_n. }[/math]

Surjections

In the twelvefold way, we discuss the counting problems incurred by the mappings [math]\displaystyle{ f:N\rightarrow M }[/math]. The basic case is that elements from both [math]\displaystyle{ N }[/math] and [math]\displaystyle{ M }[/math] are distinguishable. In this case, it is easy to count the number of arbitrary mappings (which is [math]\displaystyle{ m^n }[/math]) and the number of injective (one-to-one) mappings (which is [math]\displaystyle{ (m)_n }[/math]), but the number of surjective is difficult. Here we apply the principle of inclusion-exclusion to count the number of surjective (onto) mappings.

Theorem

The number of surjective mappings from an [math]\displaystyle{ n }[/math]-set to an [math]\displaystyle{ m }[/math]-set is given by [math]\displaystyle{ \sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n }[/math].

Proof. Let [math]\displaystyle{ U=\{f:[n]\rightarrow[m]\} }[/math] be the set of mappings from [math]\displaystyle{ [n] }[/math] to [math]\displaystyle{ [m] }[/math]. Then [math]\displaystyle{ |U|=m^n }[/math]. For [math]\displaystyle{ i\in[m] }[/math], let [math]\displaystyle{ A_i }[/math] be the set of mappings [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] that none of [math]\displaystyle{ j\in[n] }[/math] is mapped to [math]\displaystyle{ i }[/math], i.e. [math]\displaystyle{ A_i=\{f:[n]\rightarrow[m]\setminus\{i\}\} }[/math], thus [math]\displaystyle{ |A_i|=(m-1)^n }[/math]. More generally, for [math]\displaystyle{ I\subseteq [m] }[/math], [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math] contains the mappings [math]\displaystyle{ f:[n]\rightarrow[m]\setminus I }[/math]. And [math]\displaystyle{ |A_I|=(m-|I|)^n\, }[/math]. A mapping [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is surjective if [math]\displaystyle{ f }[/math] lies in none of [math]\displaystyle{ A_i }[/math]. By the principle of inclusion-exclusion, the number of surjective [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is [math]\displaystyle{ \sum_{I\subseteq[m]}(-1)^{|I|}\left|A_I\right|=\sum_{I\subseteq[m]}(-1)^{|I|}(m-|I|)^n=\sum_{j=0}^m(-1)^j{m\choose j}(m-j)^n }[/math]. Let [math]\displaystyle{ k=m-j }[/math]. The theorem is proved. [math]\displaystyle{ \square }[/math]

Recall that, in the twelvefold way, we establish a relation between surjections and partitions.

• Surjection to ordered partition:

For a surjective [math]\displaystyle{ f:[n]\rightarrow[m] }[/math], [math]\displaystyle{ (f^{-1}(0),f^{-1}(1),\ldots,f^{-1}(m-1)) }[/math] is an ordered partition of [math]\displaystyle{ [n] }[/math].

• Ordered partition to surjection:

For an ordered [math]\displaystyle{ m }[/math]-partition [math]\displaystyle{ (B_0,B_1,\ldots, B_{m-1}) }[/math] of [math]\displaystyle{ [n] }[/math], we can define a function [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] by letting [math]\displaystyle{ f(i)=j }[/math] if and only if [math]\displaystyle{ i\in B_j }[/math]. [math]\displaystyle{ f }[/math] is surjective since as a partition, none of [math]\displaystyle{ B_i }[/math] is empty.

Therefore, we have a one-to-one correspondence between surjective mappings from an [math]\displaystyle{ n }[/math]-set to an [math]\displaystyle{ m }[/math]-set and the ordered [math]\displaystyle{ m }[/math]-partitions of an [math]\displaystyle{ n }[/math]-set.

The Stirling number of the second kind [math]\displaystyle{ \left\{{n\atop m}\right\} }[/math] is the number of [math]\displaystyle{ m }[/math]-partitions of an [math]\displaystyle{ n }[/math]-set. There are [math]\displaystyle{ m! }[/math] ways to order an [math]\displaystyle{ m }[/math]-partition, thus the number of surjective mappings [math]\displaystyle{ f:[n]\rightarrow[m] }[/math] is [math]\displaystyle{ m! \left\{{n\atop m}\right\} }[/math]. Combining with what we have proved for surjections, we give the following result for the Stirling number of the second kind.

Proposition

[math]\displaystyle{ \left\{{n\atop m}\right\}=\frac{1}{m!}\sum_{k=1}^m(-1)^{m-k}{m\choose k}k^n }[/math].

Derangements

We now count the number of bijections from a set to itself with no fixed points. This is the derangement problem.

For a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math], a fixed point is such an [math]\displaystyle{ i\in\{1,2,\ldots,n\} }[/math] that [math]\displaystyle{ \pi(i)=i }[/math]. A derangement of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] is a permutation of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] that has no fixed points.

Theorem

The number of derangements of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] given by [math]\displaystyle{ n!\sum_{k=0}^n\frac{(-1)^k}{k!}\approx \frac{n!}{\mathrm{e}} }[/math].

Proof. Let [math]\displaystyle{ U }[/math] be the set of all permutations of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math]. So [math]\displaystyle{ |U|=n! }[/math]. Let [math]\displaystyle{ A_i }[/math] be the set of permutations with fixed point [math]\displaystyle{ i }[/math]; so [math]\displaystyle{ |A_i|=(n-1)! }[/math]. More generally, for any [math]\displaystyle{ I\subseteq \{1,2,\ldots,n\} }[/math], [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math], and [math]\displaystyle{ |A_I|=(n-|I|)! }[/math], since permutations in [math]\displaystyle{ A_I }[/math] fix every point in [math]\displaystyle{ I }[/math] and permute the remaining points arbitrarily. A permutation is a derangement if and only if it lies in none of the sets [math]\displaystyle{ A_i }[/math]. So the number of derangements is [math]\displaystyle{ \sum_{I\subseteq\{1,2,\ldots,n\}}(-1)^{|I|}(n-|I|)!=\sum_{k=0}^n(-1)^k{n\choose k}(n-k)!=n!\sum_{k=0}^n\frac{(-1)^k}{k!}. }[/math] By Taylor's series, [math]\displaystyle{ \frac{1}{\mathrm{e}}=\sum_{k=0}^\infty\frac{(-1)^k}{k!}=\sum_{k=0}^n\frac{(-1)^k}{k!}\pm o\left(\frac{1}{n!}\right) }[/math]. It is not hard to see that [math]\displaystyle{ n!\sum_{k=0}^n\frac{(-1)^k}{k!} }[/math] is the closest integer to [math]\displaystyle{ \frac{n!}{\mathrm{e}} }[/math]. [math]\displaystyle{ \square }[/math]

Therefore, there are about [math]\displaystyle{ \frac{1}{\mathrm{e}} }[/math] fraction of all permutations with no fixed points.

Permutations with restricted positions

We introduce a general theory of counting permutations with restricted positions. In the derangement problem, we count the number of permutations that [math]\displaystyle{ \pi(i)\neq i }[/math]. We now generalize to the problem of counting permutations which avoid a set of arbitrarily specified positions.

It is traditionally described using terminology from the game of chess. Let [math]\displaystyle{ B\subseteq \{1,\ldots,n\}\times \{1,\ldots,n\} }[/math], called a board. As illustrated below, we can think of [math]\displaystyle{ B }[/math] as a chess board, with the positions in [math]\displaystyle{ B }[/math] marked by "[math]\displaystyle{ \times }[/math]".

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

For a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ \{1,\ldots,n\} }[/math], define the graph [math]\displaystyle{ G_\pi(V,E) }[/math] as

[math]\displaystyle{ \begin{align} G_\pi &= \{(i,\pi(i))\mid i\in \{1,2,\ldots,n\}\}. \end{align} }[/math]

This can also be viewed as a set of marked positions on a chess board. Each row and each column has only one marked position, because [math]\displaystyle{ \pi }[/math] is a permutation. Thus, we can identify each [math]\displaystyle{ G_\pi }[/math] as a placement of [math]\displaystyle{ n }[/math] rooks (“城堡”，规则同中国象棋里的“车”) without attacking each other.

For example, the following is the [math]\displaystyle{ G_\pi }[/math] of such [math]\displaystyle{ \pi }[/math] that [math]\displaystyle{ \pi(i)=i }[/math].

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Now define

[math]\displaystyle{ \begin{align} N_0 &= \left|\left\{\pi\mid B\cap G_\pi=\emptyset\right\}\right|\\ r_k &= \mbox{number of }k\mbox{-subsets of }B\mbox{ such that no two elements have a common coordinate}\\ &=\left|\left\{S\in{B\choose k} \,\bigg|\, \forall (i_1,j_1),(i_2,j_2)\in S, i_1\neq i_2, j_1\neq j_2 \right\}\right| \end{align} }[/math]

Interpreted in chess game,

• [math]\displaystyle{ B }[/math]: a set of marked positions in an [math]\displaystyle{ [n]\times [n] }[/math] chess board.
• [math]\displaystyle{ N_0 }[/math]: the number of ways of placing [math]\displaystyle{ n }[/math] non-attacking rooks on the chess board such that none of these rooks lie in [math]\displaystyle{ B }[/math].
• [math]\displaystyle{ r_k }[/math]: number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math].

Our goal is to count [math]\displaystyle{ N_0 }[/math] in terms of [math]\displaystyle{ r_k }[/math]. This gives the number of permutations avoid all positions in a [math]\displaystyle{ B }[/math].

Theorem

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)! }[/math].

Proof. For each [math]\displaystyle{ i\in[n] }[/math], let [math]\displaystyle{ A_i=\{\pi\mid (i,\pi(i))\in B\} }[/math] be the set of permutations [math]\displaystyle{ \pi }[/math] whose [math]\displaystyle{ i }[/math]-th position is in [math]\displaystyle{ B }[/math]. [math]\displaystyle{ N_0 }[/math] is the number of permutations avoid all positions in [math]\displaystyle{ B }[/math]. Thus, our goal is to count the number of permutations [math]\displaystyle{ \pi }[/math] in none of [math]\displaystyle{ A_i }[/math] for [math]\displaystyle{ i\in [n] }[/math]. For each [math]\displaystyle{ I\subseteq [n] }[/math], let [math]\displaystyle{ A_I=\bigcap_{i\in I}A_i }[/math], which is the set of permutations [math]\displaystyle{ \pi }[/math] such that [math]\displaystyle{ (i,\pi(i))\in B }[/math] for all [math]\displaystyle{ i\in I }[/math]. Due to the principle of inclusion-exclusion, [math]\displaystyle{ N_0=\sum_{I\subseteq [n]} (-1)^{|I|}|A_I|=\sum_{k=0}^n(-1)^k\sum_{I\in{[n]\choose k}}|A_I| }[/math]. The next observation is that [math]\displaystyle{ \sum_{I\in{[n]\choose k}}|A_I|=r_k(n-k)! }[/math], because we can count both sides by first placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math] and placing [math]\displaystyle{ n-k }[/math] additional non-attacking rooks on [math]\displaystyle{ [n]\times [n] }[/math] in [math]\displaystyle{ (n-k)! }[/math] ways. Therefore, [math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)! }[/math]. [math]\displaystyle{ \square }[/math]

Derangement problem

We use the above general method to solve the derange problem again.

Take [math]\displaystyle{ B=\{(1,1),(2,2),\ldots,(n,n)\} }[/math] as the chess board. A derangement [math]\displaystyle{ \pi }[/math] is a placement of [math]\displaystyle{ n }[/math] non-attacking rooks such that none of them is in [math]\displaystyle{ B }[/math].

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Clearly, the number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math] is [math]\displaystyle{ r_k={n\choose k} }[/math]. We want to count [math]\displaystyle{ N_0 }[/math], which gives the number of ways of placing [math]\displaystyle{ n }[/math] non-attacking rooks such that none of these rooks lie in [math]\displaystyle{ B }[/math].

By the above theorem

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)!=\sum_{k=0}^n(-1)^k{n\choose k}(n-k)!=\sum_{k=0}^n(-1)^k\frac{n!}{k!}=n!\sum_{k=0}^n(-1)^k\frac{1}{k!}\approx\frac{n!}{e}. }[/math]

Problème des ménages

Suppose that in a banquet, we want to seat [math]\displaystyle{ n }[/math] couples at a circular table, satisfying the following constraints:

• Men and women are in alternate places.
• No one sits next to his/her spouse.

In how many ways can this be done?

(For convenience, we assume that every seat at the table marked differently so that rotating the seats clockwise or anti-clockwise will end up with a different solution.)

First, let the [math]\displaystyle{ n }[/math] ladies find their seats. They may either sit at the odd numbered seats or even numbered seats, in either case, there are [math]\displaystyle{ n! }[/math] different orders. Thus, there are [math]\displaystyle{ 2(n!) }[/math] ways to seat the [math]\displaystyle{ n }[/math] ladies.

After sitting the wives, we label the remaining [math]\displaystyle{ n }[/math] places clockwise as [math]\displaystyle{ 0,1,\ldots, n-1 }[/math]. And a seating of the [math]\displaystyle{ n }[/math] husbands is given by a permutation [math]\displaystyle{ \pi }[/math] of [math]\displaystyle{ [n] }[/math] defined as follows. Let [math]\displaystyle{ \pi(i) }[/math] be the seat of the husband of he lady sitting at the [math]\displaystyle{ i }[/math]-th place.

It is easy to see that [math]\displaystyle{ \pi }[/math] satisfies that [math]\displaystyle{ \pi(i)\neq i }[/math] and [math]\displaystyle{ \pi(i)\not\equiv i+1\pmod n }[/math], and every permutation [math]\displaystyle{ \pi }[/math] with these properties gives a feasible seating of the [math]\displaystyle{ n }[/math] husbands. Thus, we only need to count the number of permutations [math]\displaystyle{ \pi }[/math] such that [math]\displaystyle{ \pi(i)\not\equiv i, i+1\pmod n }[/math].

Take [math]\displaystyle{ B=\{(0,0),(1,1),\ldots,(n-1,n-1), (0,1),(1,2),\ldots,(n-2,n-1),(n-1,0)\} }[/math] as the chess board. A permutation [math]\displaystyle{ \pi }[/math] which defines a way of seating the husbands, is a placement of [math]\displaystyle{ n }[/math] non-attacking rooks such that none of them is in [math]\displaystyle{ B }[/math].

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

We need to compute [math]\displaystyle{ r_k }[/math], the number of ways of placing [math]\displaystyle{ k }[/math] non-attacking rooks on [math]\displaystyle{ B }[/math]. For our choice of [math]\displaystyle{ B }[/math], [math]\displaystyle{ r_k }[/math] is the number of ways of choosing [math]\displaystyle{ k }[/math] points, no two consecutive, from a collection of [math]\displaystyle{ 2n }[/math] points arranged in a circle.

We first see how to do this in a line.

Lemma

The number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive objects from a collection of [math]\displaystyle{ m }[/math] objects arranged in a line, is [math]\displaystyle{ {m-k+1\choose k} }[/math].

Proof. We draw a line of [math]\displaystyle{ m-k }[/math] black points, and then insert [math]\displaystyle{ k }[/math] red points into the [math]\displaystyle{ m-k+1 }[/math] spaces between the black points (including the beginning and end). [math]\displaystyle{ \begin{align} &\sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \, \sqcup \\ &\qquad\qquad\qquad\quad\Downarrow\\ &\sqcup \, \bullet \,\, {\color{Red}\bullet} \, \bullet \,\, {\color{Red}\bullet} \, \bullet \, \sqcup \, \bullet \,\, {\color{Red}\bullet}\, \, \bullet \, \sqcup \, \bullet \, \sqcup \, \bullet \,\, {\color{Red}\bullet} \end{align} }[/math] This gives us a line of [math]\displaystyle{ m }[/math] points, and the red points specifies the chosen objects, which are non-consecutive. The mapping is 1-1 correspondence. There are [math]\displaystyle{ {m-k+1\choose k} }[/math] ways of placing [math]\displaystyle{ k }[/math] red points into [math]\displaystyle{ m-k+1 }[/math] spaces. [math]\displaystyle{ \square }[/math]

The problem of choosing non-consecutive objects in a circle can be reduced to the case that the objects are in a line.

Lemma

The number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive objects from a collection of [math]\displaystyle{ m }[/math] objects arranged in a circle, is [math]\displaystyle{ \frac{m}{m-k}{m-k\choose k} }[/math].

Proof. Let [math]\displaystyle{ f(m,k) }[/math] be the desired number; and let [math]\displaystyle{ g(m,k) }[/math] be the number of ways of choosing [math]\displaystyle{ k }[/math] non-consecutive points from [math]\displaystyle{ m }[/math] points arranged in a circle, next coloring the [math]\displaystyle{ k }[/math] points red, and then coloring one of the uncolored point blue. Clearly, [math]\displaystyle{ g(m,k)=(m-k)f(m,k) }[/math]. But we can also compute [math]\displaystyle{ g(m,k) }[/math] as follows: Choose one of the [math]\displaystyle{ m }[/math] points and color it blue. This gives us [math]\displaystyle{ m }[/math] ways. Cut the circle to make a line of [math]\displaystyle{ m-1 }[/math] points by removing the blue point. Choose [math]\displaystyle{ k }[/math] non-consecutive points from the line of [math]\displaystyle{ m-1 }[/math] points and color them red. This gives [math]\displaystyle{ {m-k\choose k} }[/math] ways due to the previous lemma. Thus, [math]\displaystyle{ g(m,k)=m{m-k\choose k} }[/math]. Therefore we have the desired number [math]\displaystyle{ f(m,k)=\frac{m}{m-k}{m-k\choose k} }[/math]. [math]\displaystyle{ \square }[/math]

By the above lemma, we have that [math]\displaystyle{ r_k=\frac{2n}{2n-k}{2n-k\choose k} }[/math]. Then apply the theorem of counting permutations with restricted positions,

[math]\displaystyle{ N_0=\sum_{k=0}^n(-1)^kr_k(n-k)!=\sum_{k=0}^n(-1)^k\frac{2n}{2n-k}{2n-k\choose k}(n-k)!. }[/math]

This gives the number of ways of seating the [math]\displaystyle{ n }[/math] husbands after the ladies are seated. Recall that there are [math]\displaystyle{ 2n! }[/math] ways of seating the [math]\displaystyle{ n }[/math] ladies. Thus, the total number of ways of seating [math]\displaystyle{ n }[/math] couples as required by problème des ménages is

[math]\displaystyle{ 2n!\sum_{k=0}^n(-1)^k\frac{2n}{2n-k}{2n-k\choose k}(n-k)!. }[/math]

Inversion

Posets

A partially ordered set or poset for short is a set [math]\displaystyle{ P }[/math] together with a binary relation denoted [math]\displaystyle{ \le_P }[/math] (or just [math]\displaystyle{ \le }[/math] if no confusion is caused), satisfying

• (reflexivity) For all [math]\displaystyle{ x\in P, x\le x }[/math].
• (antisymmetry) If [math]\displaystyle{ x\le y }[/math] and [math]\displaystyle{ y\le x }[/math], then [math]\displaystyle{ x=y }[/math].
• (transitivity) If [math]\displaystyle{ x\le y }[/math] and [math]\displaystyle{ y\le z }[/math], then [math]\displaystyle{ x\le z }[/math].

We say two elements [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are comparable if [math]\displaystyle{ x\le y }[/math] or [math]\displaystyle{ y\le x }[/math]; otherwise [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are incomparable.

Notation

• [math]\displaystyle{ x\ge y }[/math] means [math]\displaystyle{ y\le x }[/math].
• [math]\displaystyle{ x\lt y }[/math] means [math]\displaystyle{ x\le y }[/math] and [math]\displaystyle{ x\neq y }[/math].
• [math]\displaystyle{ x\gt y }[/math] means [math]\displaystyle{ y\lt x }[/math].

The Möbius function

Let [math]\displaystyle{ P }[/math] be a finite poset. Consider functions in form of [math]\displaystyle{ \alpha:P\times P\rightarrow\mathbb{R} }[/math] defined over domain [math]\displaystyle{ P\times P }[/math]. It is convenient to treat such functions as matrices whose rows and columns are indexed by [math]\displaystyle{ P }[/math].

Incidence algebra of poset

Let

[math]\displaystyle{ I(P)=\{\alpha:P\times P\rightarrow\mathbb{R}\mid \alpha(x,y)=0\text{ for all }x\not\le_P y\} }[/math]

be the class of [math]\displaystyle{ \alpha }[/math] such that [math]\displaystyle{ \alpha(x,y) }[/math] is non-zero only for [math]\displaystyle{ x\le_P y }[/math].

Treating [math]\displaystyle{ \alpha }[/math] as matrix, it is trivial to see that [math]\displaystyle{ I(P) }[/math] is closed under addition and scalar multiplication, that is,

• if [math]\displaystyle{ \alpha,\beta\in I(P) }[/math] then [math]\displaystyle{ \alpha+\beta\in I(P) }[/math];
• if [math]\displaystyle{ \alpha\in I(P) }[/math] then [math]\displaystyle{ c\alpha\in I(P) }[/math] for any [math]\displaystyle{ c\in\mathbb{R} }[/math];

where [math]\displaystyle{ \alpha,\beta }[/math] are treated as matrices.

With this spirit, it is natural to define the matrix multiplication in [math]\displaystyle{ I(P) }[/math]. For [math]\displaystyle{ \alpha,\beta\in I(P) }[/math],

[math]\displaystyle{ (\alpha\beta)(x,y)=\sum_{z\in P}\alpha(x,z)\beta(z,y)=\sum_{x\le z\le y}\alpha(x,z)\beta(z,y) }[/math].

The second equation is due to that for [math]\displaystyle{ \alpha,\beta\in I(P) }[/math], for all [math]\displaystyle{ z }[/math] other than [math]\displaystyle{ x\le z\le y }[/math], [math]\displaystyle{ \alpha(x,z)\beta(z,y) }[/math] is zero.

By the transitivity of relation [math]\displaystyle{ \le_P }[/math], it is also easy to prove that [math]\displaystyle{ I(P) }[/math] is closed under matrix multiplication (the detailed proof is left as an exercise). Therefore, [math]\displaystyle{ I(P) }[/math] is closed under addition, scalar multiplication and matrix multiplication, so we have an algebra [math]\displaystyle{ I(P) }[/math], called incidence algebra, over functions on [math]\displaystyle{ P\times P }[/math].

Zeta function and Möbius function

A special function in [math]\displaystyle{ I(P) }[/math] is the so-called zeta function [math]\displaystyle{ \zeta }[/math], defined as

[math]\displaystyle{ \zeta(x,y)=\begin{cases}1&\text{if }x\le_P y,\\0 &\text{otherwise.}\end{cases} }[/math]

As a matrix (or more accurately, as an element of the incidence algebra), [math]\displaystyle{ \zeta }[/math] is invertible and its inversion, denoted by [math]\displaystyle{ \mu }[/math], is called the Möbius function. More precisely, [math]\displaystyle{ \mu }[/math] is also in the incidence algebra [math]\displaystyle{ I(P) }[/math], and [math]\displaystyle{ \mu\zeta=I }[/math] where [math]\displaystyle{ I }[/math] is the identity matrix (the identity of the incidence algebra [math]\displaystyle{ I(P) }[/math]).

There is an equivalent explicit definition of Möbius function.

Definition (Möbius function)

[math]\displaystyle{ \mu(x,y)=\begin{cases} -\sum_{x\le z\lt y}\mu(x,z)&\text{if }x\lt y,\\ 1&\text{if }x=y,\\ 0&\text{if }x\not\le y. \end{cases} }[/math]

To see the equivalence between this definition and the inversion of zeta function, we may have the following proposition, which is proved by directly evaluating [math]\displaystyle{ \mu\zeta }[/math].

Proposition

For any [math]\displaystyle{ x,y\in P }[/math], [math]\displaystyle{ \sum_{x\le z\le y}\mu(x,z)=\begin{cases}1 &\text{if }x=y,\\ 0 &\text{otherwise.}\end{cases} }[/math]

Proof. It holds that [math]\displaystyle{ (\mu\zeta)(x,y)=\sum_{x\le z\le y}\mu(x,z)\zeta(z,y)=\sum_{x\le z\le y}\mu(x,z) }[/math]. On the other hand, [math]\displaystyle{ \mu\zeta=I }[/math], i.e. [math]\displaystyle{ (\mu\zeta)(x,y)=\begin{cases}1 &\text{if }x=y,\\ 0 &\text{otherwise.}\end{cases} }[/math] The proposition follows. [math]\displaystyle{ \square }[/math]

Note that [math]\displaystyle{ \mu(x,y)=\sum_{x\le z\le y}\mu(x,z)-\sum_{x\le z\lt y}\mu(x,z) }[/math], which gives the above inductive definition of Möbius function.

Computing Möbius functions

We consider the simple poset [math]\displaystyle{ P=[n] }[/math], where [math]\displaystyle{ \le }[/math] is the total order. It follows directly from the recursive definition of Möbius function that

[math]\displaystyle{ \mu(i,j)=\begin{cases}1 & \text{if }i=j,\\ -1 & \text{if }i+1=j,\\ 0 & \text{otherwise.} \end{cases} }[/math]

Usually for general posets, it is difficult to directly compute the Möbius function from its definition. We introduce a rule helping us compute the Möbius function by decomposing the poset into posets with simple structures.

Theorem (the product rule)

Let [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math] be two finite posets, and [math]\displaystyle{ P\times Q }[/math] be the poset resulted from Cartesian product of [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math], where for all [math]\displaystyle{ (x,y), (x',y')\in P\times Q }[/math], [math]\displaystyle{ (x,y)\le (x',y') }[/math] if and only if [math]\displaystyle{ x\le x' }[/math] and [math]\displaystyle{ y\le y' }[/math]. Then [math]\displaystyle{ \mu_{P\times Q}((x,y),(x',y'))=\mu_P(x,x')\mu_Q(y,y') }[/math].

Proof. We use the recursive definition [math]\displaystyle{ \mu(x,y)=\begin{cases} -\sum_{x\le z\lt y}\mu(x,z)&\text{if }x\lt y,\\ 1&\text{if }x=y,\\ 0&\text{if }x\not\le y. \end{cases} }[/math] to prove the equation in the theorem. If [math]\displaystyle{ (x,y)=(x',y') }[/math], then [math]\displaystyle{ x=x' }[/math] and [math]\displaystyle{ y=y' }[/math]. It is easy to see that both sides of the equation are 1. If [math]\displaystyle{ (x,y)\not\le(x',y') }[/math], then either [math]\displaystyle{ x\not\le x' }[/math] or [math]\displaystyle{ y\not\le y' }[/math]. It is also easy to see that both sides are 0. The only remaining case is that [math]\displaystyle{ (x,y)\lt (x',y') }[/math], in which case either [math]\displaystyle{ x\lt x' }[/math] or [math]\displaystyle{ y\lt y' }[/math]. [math]\displaystyle{ \begin{align} \sum_{(x,y)\le (u,v)\le (x',y')}\mu_P(x,u)\mu_Q(y,v) &=\left(\sum_{x\le u\le x'}\mu_P(x,u)\right)\left(\sum_{y\le v\le y'}\mu_Q(y,v)\right)=I(x,x')I(y,y')=0, \end{align} }[/math] where the last two equations are due to the proposition for [math]\displaystyle{ \mu }[/math]. Thus [math]\displaystyle{ \mu_P(x,x')\mu_Q(y,y')=-\sum_{(x,y)\le (u,v)\lt (x',y')}\mu_P(x,u)\mu_Q(y,v) }[/math]. By induction, assume that the equation [math]\displaystyle{ \mu_{P\times Q}((x,y),(u,v))=\mu_P(x,u)\mu_Q(y,v) }[/math] is true for all [math]\displaystyle{ (u,v)\lt (x',y') }[/math]. Then [math]\displaystyle{ \begin{align} \mu_{P\times Q}((x,y),(x',y')) &=-\sum_{(x,y)\le (u,v)\lt (x',y')}\mu_{P\times Q}((x,y),(u,v))\\ &=-\sum_{(x,y)\le (u,v)\lt (x',y')}\mu_P(x,u)\mu_Q(y,v)\\ &=\mu_P(x,x')\mu_Q(y,y'), \end{align} }[/math] which complete the proof. [math]\displaystyle{ \square }[/math]

Poset of subsets

Consider the poset defined by all subsets of a finite universe [math]\displaystyle{ U }[/math], that is [math]\displaystyle{ P=2^U }[/math], and for [math]\displaystyle{ S,T\subseteq U }[/math], [math]\displaystyle{ S\le_P T }[/math] if and only if [math]\displaystyle{ S\subseteq T }[/math].

Möbius function for subsets

The Möbius function for the above defined poset [math]\displaystyle{ P }[/math] is that for [math]\displaystyle{ S,T\subseteq U }[/math], [math]\displaystyle{ \mu(S,T)= \begin{cases} (-1)^{|T|-|S|} & \text{if }S\subseteq T,\\ 0 &\text{otherwise.} \end{cases} }[/math]

Proof. We can equivalently represent each [math]\displaystyle{ S\subseteq U }[/math] by a boolean string [math]\displaystyle{ S\in\{0,1\}^U }[/math], where [math]\displaystyle{ S(x)=1 }[/math] if and only if [math]\displaystyle{ x\in S }[/math]. For each element [math]\displaystyle{ x\in U }[/math], we can define a poset [math]\displaystyle{ P_x=\{0, 1\} }[/math] with [math]\displaystyle{ 0\le 1 }[/math]. By definition of Möbius function, the Möbius function of this elementary poset is given by [math]\displaystyle{ \mu_x(0,0)=\mu_x(1,1)=1 }[/math], [math]\displaystyle{ \mu_x(0,1)=-1 }[/math] and [math]\displaystyle{ \mu(1,0)=0 }[/math]. The poset [math]\displaystyle{ P }[/math] of all subsets of [math]\displaystyle{ U }[/math] is the Cartesian product of all [math]\displaystyle{ P_x }[/math], [math]\displaystyle{ x\in U }[/math]. By the product rule, [math]\displaystyle{ \mu(S,T)=\prod_{x\in U}\mu_x(S(x), T(x))=\prod_{x\in S\atop x\in T}1\prod_{x\not\in S\atop x\not\in T}1\prod_{x\in S\atop x\not\in T}0\prod_{x\not\in S\atop x\in T}(-1)=\begin{cases} (-1)^{|T|-|S|} & \text{if }S\subseteq T,\\ 0 &\text{otherwise.} \end{cases} }[/math] [math]\displaystyle{ \square }[/math]

Note that the poset [math]\displaystyle{ P }[/math] is actually the Boolean algebra of rank [math]\displaystyle{ |U| }[/math]. The proof relies only on that the fact that the poset is a Boolean algebra, thus the theorem holds for Boolean algebra posets.

Posets of divisors

Consider the poset defined by all devisors of a positive integer [math]\displaystyle{ n }[/math], that is [math]\displaystyle{ P=\{a\gt 0\mid a|n\} }[/math], and for [math]\displaystyle{ a,b\in P }[/math], [math]\displaystyle{ a\le_P b }[/math] if and only if [math]\displaystyle{ a|b\, }[/math].

Möbius function for divisors

The Möbius function for the above defined poset [math]\displaystyle{ P }[/math] is that for [math]\displaystyle{ a,b\gt 0 }[/math] that [math]\displaystyle{ a|n }[/math] and [math]\displaystyle{ b|n }[/math], [math]\displaystyle{ \mu(a,b)= \begin{cases} (-1)^{r} & \text{if }\frac{b}{a}\text{ is the product of }r\text{ distinct primes},\\ 0 &\text{otherwise, i.e. if }a\not|b\text{ or }\frac{b}{a}\text{ is not squarefree.} \end{cases} }[/math]

Proof. Denote [math]\displaystyle{ n=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k} }[/math]. Represent [math]\displaystyle{ n }[/math] by a tuple [math]\displaystyle{ (n_1,n_2,\ldots,n_k) }[/math]. Every [math]\displaystyle{ a\in P }[/math] corresponds in this way to a tuple [math]\displaystyle{ (a_1,a_2,\ldots,a_k) }[/math] with [math]\displaystyle{ a_i\le n_i }[/math] for all [math]\displaystyle{ 1\le i\le k }[/math]. Let [math]\displaystyle{ P_i=\{1,2,\ldots,n_i\} }[/math] be the poset with [math]\displaystyle{ \le }[/math] being the total order. The poset [math]\displaystyle{ P }[/math] of divisors of [math]\displaystyle{ n }[/math] is thus isomorphic to the poset constructed by the Cartesian product of all [math]\displaystyle{ P_i }[/math], [math]\displaystyle{ 1\le i\le k }[/math]. Then [math]\displaystyle{ \begin{align} \mu(a,b) &=\prod_{1\le i\le k}\mu(a_i,b_i)=\prod_{1\le i\le k\atop a_i=b_i}1\prod_{1\le i\le k\atop b_i-a_i=1}(-1)\prod_{1\le i\le k\atop b_i-a_i\not\in\{0,1\}}0 =\begin{cases} (-1)^{\sum_{i}(b_i-a_i)} & \text{if all }b_i-a_i\in\{0,1\},\\ 0 &\text{otherwise.} \end{cases}\\ &=\begin{cases} (-1)^{r} & \text{if }\frac{b}{a}\text{ is the product of }r\text{ distinct primes},\\ 0 &\text{otherwise.} \end{cases} \end{align} }[/math] [math]\displaystyle{ \square }[/math]

Principle of Möbius inversion

We now introduce the the famous Möbius inversion formula.

Möbius inversion formula

Let [math]\displaystyle{ P }[/math] be a finite poset and [math]\displaystyle{ \mu }[/math] its Möbius function. Let [math]\displaystyle{ f,g:P\rightarrow \mathbb{R} }[/math]. Then [math]\displaystyle{ \forall x\in P,\,\, g(x)=\sum_{y\le x} f(y) }[/math], if and only if [math]\displaystyle{ \forall x\in P,\,\, f(x)=\sum_{y\le x}g(y)\mu(y,x) }[/math].

The functions [math]\displaystyle{ f,g:P\rightarrow\mathbb{R} }[/math] are vectors. Evaluate the matrix multiplications [math]\displaystyle{ f\zeta }[/math] and [math]\displaystyle{ g\mu }[/math] as follows:

[math]\displaystyle{ (f\zeta)(x)=\sum_{y\in P}f(y)\zeta(y,x)=\sum_{y\le x}f(y) }[/math],

and

[math]\displaystyle{ (g\mu)(x)=\sum_{y\in P}g(y)\mu(y,x)=\sum_{y\le x}g(y)\mu(y,x) }[/math].

The Möbius inversion formula is nothing but the following statement

[math]\displaystyle{ f\zeta=g\Leftrightarrow f=g\mu }[/math],

which is trivially true due to [math]\displaystyle{ \mu\zeta=I }[/math] by basic linear algebra.

The following dual form of the inversion formula is also useful.

Möbius inversion formula, dual form

Let [math]\displaystyle{ P }[/math] be a finite poset and [math]\displaystyle{ \mu }[/math] its Möbius function. Let [math]\displaystyle{ f,g:P\rightarrow \mathbb{R} }[/math]. Then [math]\displaystyle{ \forall x\in P, \,\, g(x)=\sum_{y{\color{red}\ge} x} f(y) }[/math], if and only if [math]\displaystyle{ \forall x\in P, \,\, f(x)=\sum_{y{\color{red}\ge} x}\mu(x,y)g(y) }[/math].

To prove the dual form, we only need to evaluate the matrix multiplications on left:

[math]\displaystyle{ \zeta f=g\Leftrightarrow f=\mu g }[/math].

Principle of Inclusion-Exclusion

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n\subseteq U }[/math]. For any [math]\displaystyle{ J\subseteq\{1,2,\ldots,n\} }[/math],

• let [math]\displaystyle{ f(J) }[/math] be the number of elements that belongs to exactly the sets [math]\displaystyle{ A_i, i\in J }[/math] and to no others, i.e.

[math]\displaystyle{ f(J)=\left|\left(\bigcap_{i\in J}A_i\right)\setminus\left(\bigcup_{i\not\in J}A_i\right)\right| }[/math];

• let [math]\displaystyle{ g(J)=\left|\bigcap_{i\in J}A_i\right| }[/math].

For any [math]\displaystyle{ J\subseteq\{1,2,\ldots,n\} }[/math], the following relation holds for the above defined [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math]:

[math]\displaystyle{ g(J)=\sum_{I\supseteq J}f(I) }[/math].

Applying the dual form of the Möbius inversion formula, we have that for any [math]\displaystyle{ J\subseteq\{1,2,\ldots,n\} }[/math],

[math]\displaystyle{ f(J)=\sum_{I\supseteq J}\mu(J,I)g(I)=\sum_{I\supseteq J}\mu(J,I)\left|\bigcap_{i\in I}A_i\right| }[/math],

where the Möbius function is for the poset of all subsets of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math], ordered by [math]\displaystyle{ \subseteq }[/math], thus it holds that [math]\displaystyle{ \mu(J,I)=(-1)^{|I|-|J|}\, }[/math] for [math]\displaystyle{ J\subseteq I }[/math]. Therefore,

[math]\displaystyle{ f(J)=\sum_{I\supseteq J}(-1)^{|I|-|J|}\left|\bigcap_{i\in I}A_i\right| }[/math].

We have a formula for the number of elements with exactly those properties [math]\displaystyle{ A_i, i\in J }[/math] for any [math]\displaystyle{ J\subseteq\{1,2,\ldots,n\} }[/math]. For the special case that [math]\displaystyle{ J=\emptyset }[/math], [math]\displaystyle{ f(\emptyset) }[/math] is the number of elements satisfying no property of [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math], and

[math]\displaystyle{ f(\emptyset)=\left|U\setminus\bigcup_iA_i\right|=\sum_{I\subseteq \{1,\ldots,n\}}(-1)^{|I|}\left|\bigcap_{i\in I}A_i\right| }[/math]

which gives precisely the Principle of Inclusion-Exclusion.

Möbius inversion formula for number theory

The number-theoretical Möbius inversion formula is stated as such: Let [math]\displaystyle{ N }[/math] be a positive integer,

[math]\displaystyle{ g(n)=\sum_{d|n}f(d)\, }[/math] for all [math]\displaystyle{ n|N }[/math]

if and only if

[math]\displaystyle{ f(n)=\sum_{d|n}g(d)\mu\left(\frac{n}{d}\right)\, }[/math] for all [math]\displaystyle{ n|N }[/math],

where [math]\displaystyle{ \mu }[/math] is the number-theoretical Möbius function, defined as

[math]\displaystyle{ \mu(n)=\begin{cases}1 & \text{if }n\text{ is product of an even number of distinct primes,}\\ -1 &\text{if }n\text{ is product of an odd number of distinct primes,}\\ 0 &\text{otherwise.}\end{cases} }[/math]

The number-theoretical Möbius inversion formula is just a special case of the Möbius inversion formula for posets, when the poset is the set of divisors of [math]\displaystyle{ N }[/math], and for any [math]\displaystyle{ a,b\in P }[/math], [math]\displaystyle{ a\le_P b }[/math] if [math]\displaystyle{ a|b }[/math].

Reference

• Stanley, Enumerative Combinatorics, Volume 1, Chapter 2.
• van Lin and Wilson, A course in combinatorics, Chapter 10, 25.