Combinatorics (Fall 2010)/Flow and matching

Flow and Cut

Flows

An instance of the maximum flow problem consists of:

• a directed graph [math]\displaystyle{ G(V,E) }[/math];
• two distinguished vertices [math]\displaystyle{ s }[/math] (the source) and [math]\displaystyle{ t }[/math] (the sink), where the in-degree of [math]\displaystyle{ s }[/math] and the out-degree of [math]\displaystyle{ t }[/math] are both 0;
• the capacity function [math]\displaystyle{ c:E\rightarrow\mathbb{R}^+ }[/math] which associates each directed edge [math]\displaystyle{ (u,v)\in E }[/math] a nonnegative real number [math]\displaystyle{ c_{uv} }[/math] called the capacity of the edge.

The quadruple [math]\displaystyle{ (G,c,s,t) }[/math] is called a flow network.

A function [math]\displaystyle{ f:E\rightarrow\mathbb{R}^+ }[/math] is called a flow (to be specific an [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] flow) in the network [math]\displaystyle{ G(V,E) }[/math] if it satisfies:

• Capacity constraint: [math]\displaystyle{ f_{uv}\le c_{uv} }[/math] for all [math]\displaystyle{ (u,v)\in E }[/math].
• Conservation constraint: [math]\displaystyle{ \sum_{u:(u,v)\in E}f_{uv}=\sum_{w:(v,w)\in E}f_{vw} }[/math] for all [math]\displaystyle{ v\in V\setminus\{s,t\} }[/math].

The value of the flow [math]\displaystyle{ f }[/math] is [math]\displaystyle{ \sum_{v:(s,v)\in E}f_{sv} }[/math].

Given a flow network, the maximum flow problem asks to find the flow of the maximum value.

The maximum flow problem can be described as the following linear program.

[math]\displaystyle{ \begin{align} \text{maximize} \quad& \sum_{v:(s,v)\in E}f_{sv}\\ \begin{align} \text{subject to} \\ \\ \\ \\ \end{align} \quad & \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\ \sum_{u:(u,v)\in E}f_{uv}-\sum_{w:(v,w)\in E}f_{vw} &=0 &\quad& \forall v\in V\setminus\{s,t\}\\ f_{uv}&\ge 0 &\quad& \forall (u,v)\in E \end{align} \end{align} }[/math]

Cuts

Definition

Let [math]\displaystyle{ (G(V,E),c,s,t) }[/math] be a flow network. Let [math]\displaystyle{ S\subset V }[/math]. We call [math]\displaystyle{ (S,\bar{S}) }[/math] an [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut if [math]\displaystyle{ s\in S }[/math] and [math]\displaystyle{ t\not\in S }[/math]. The value of the cut (also called the capacity of the cut) is defined as [math]\displaystyle{ \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv} }[/math].

A fundamental fact in flow theory is that cuts always upper bound flows.

Lemma

Let [math]\displaystyle{ (G(V,E),c,s,t) }[/math] be a flow network. Let [math]\displaystyle{ f }[/math] be an arbitrary flow in [math]\displaystyle{ G }[/math], and let [math]\displaystyle{ (S,\bar{S}) }[/math] be an arbitrary [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut. Then [math]\displaystyle{ \sum_{v:(s,v)}f_{sv}\le \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv} }[/math], that is, the value of any flow is no greater than the value of any cut.

Proof. By the definition of [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut, [math]\displaystyle{ s\in S }[/math] and [math]\displaystyle{ t\not\in S }[/math]. Due to the conservation of flow, [math]\displaystyle{ \sum_{u\in S}\left(\sum_{v:(u,v)\in E}f_{uv}-\sum_{v:(v,u)\in E}f_{vu}\right)=\sum_{v:(s,v)\in E}f_{sv}+\sum_{u\in S\setminus\{s\}}\left(\sum_{v:(u,v)\in E}f_{uv}-\sum_{v:(v,u)\in E}f_{vu}\right)=\sum_{v:(s,v)\in E}f_{sv}\,. }[/math] On the other hand, summing flow over edges, [math]\displaystyle{ \sum_{v\in S}\left(\sum_{u:(u,v)\in E}f_{uv}-\sum_{u:(v,u)\in E}f_{vu}\right)=\sum_{u\in S,v\in S\atop (u,v)\in E}\left(f_{uv}-f_{uv}\right)+\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}=\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}\,. }[/math] Therefore, [math]\displaystyle{ \sum_{v:(s,v)\in E}f_{sv}=\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu}\le\sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}\le \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv}\,, }[/math] [math]\displaystyle{ \square }[/math]

Augmenting paths

Definition (Augmenting path)

Let [math]\displaystyle{ f }[/math] be a flow in [math]\displaystyle{ G }[/math]. An augmenting path to [math]\displaystyle{ u_k }[/math] is a sequence of distinct vertices [math]\displaystyle{ P=(u_0,u_1,\cdots, u_k) }[/math], such that [math]\displaystyle{ u_0=s\, }[/math]; and each pair of consecutive vertices [math]\displaystyle{ u_{i}u_{i+1}\, }[/math] in [math]\displaystyle{ P }[/math] corresponds to either a forward edge [math]\displaystyle{ (u_{i},u_{i+1})\in E }[/math] or a reverse edge [math]\displaystyle{ (u_{i+1},u_{i})\in E }[/math], and [math]\displaystyle{ f(u_i,u_{i+1})\lt c(u_i,u_{i+1})\, }[/math] when [math]\displaystyle{ u_{i}u_{i+1}\, }[/math] corresponds to a forward edge [math]\displaystyle{ (u_{i},u_{i+1})\in E }[/math], and [math]\displaystyle{ f(u_{i+1},u_i)\gt 0\, }[/math] when [math]\displaystyle{ u_{i}u_{i+1}\, }[/math] corresponds to a reverse edge [math]\displaystyle{ (u_{i+1},u_{i})\in E }[/math]. If [math]\displaystyle{ u_k=t\, }[/math], we simply call [math]\displaystyle{ P }[/math] an augmenting path.

Let [math]\displaystyle{ f }[/math] be a flow in [math]\displaystyle{ G }[/math]. Suppose there is an augmenting path [math]\displaystyle{ P=u_0u_1\cdots u_k }[/math], where [math]\displaystyle{ u_0=s }[/math] and [math]\displaystyle{ u_k=t }[/math]. Let [math]\displaystyle{ \epsilon\gt 0 }[/math] be a positive constant satisfying

• [math]\displaystyle{ \epsilon \le c(u_{i},u_{i+1})-f(u_i,u_{i+1}) }[/math] for all forward edges [math]\displaystyle{ (u_{i},u_{i+1})\in E }[/math] in [math]\displaystyle{ P }[/math];
• [math]\displaystyle{ \epsilon \le f(u_{i+1},u_i) }[/math] for all reverse edges [math]\displaystyle{ (u_{i+1},u_i)\in E }[/math] in [math]\displaystyle{ P }[/math].

Due to the definition of augmenting path, we can always find such a positive [math]\displaystyle{ \epsilon }[/math].

Increase [math]\displaystyle{ f(u_i,u_{i+1}) }[/math] by [math]\displaystyle{ \epsilon }[/math] for all forward edges [math]\displaystyle{ (u_{i},u_{i+1})\in E }[/math] in [math]\displaystyle{ P }[/math] and decrease [math]\displaystyle{ f(u_{i+1},u_i) }[/math] by [math]\displaystyle{ \epsilon }[/math] for all reverse edges [math]\displaystyle{ (u_{i+1},u_i)\in E }[/math] in [math]\displaystyle{ P }[/math]. Denote the modified flow by [math]\displaystyle{ f' }[/math]. It can be verified that [math]\displaystyle{ f' }[/math] satisfies the capacity constraint and conservation constraint thus is still a valid flow. On the other hand, the value of the new flow [math]\displaystyle{ f' }[/math]

[math]\displaystyle{ \sum_{v:(s,v)\in E}f_{sv}'=\epsilon+\sum_{v:(s,v)\in E}f_{sv}\gt \sum_{v:(s,v)\in E}f_{sv} }[/math].

Therefore, the value of the flow can be "augmented" by adjusting the flow on the augmenting path. This immediately implies that if a flow is maximum, then there is no augmenting path. Surprisingly, the converse is also true, thus maximum flows are "characterized" by augmenting paths.

Lemma

A flow [math]\displaystyle{ f }[/math] is maximum if and only if there are no augmenting paths.

Proof. We have already proved the "only if" direction above. Now we prove the "if" direction. Let [math]\displaystyle{ S=\{u\in V\mid \exists\text{an augmenting path to }u\} }[/math]. Clearly [math]\displaystyle{ s\in S }[/math], and since there is no augmenting path [math]\displaystyle{ t\not\in S }[/math]. Therefore, [math]\displaystyle{ (S,\bar{S}) }[/math] defines an [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut. We claim that [math]\displaystyle{ \sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv} }[/math], that is, the value of flow [math]\displaystyle{ f }[/math] approach the value of the cut [math]\displaystyle{ (S,\bar{S}) }[/math] defined above. By the above lemma, this will imply that the current flow [math]\displaystyle{ f }[/math] is maximum. To prove this claim, we first observe that [math]\displaystyle{ \sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu} }[/math]. This identity is implied by the flow conservation constraint, and holds for any [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut [math]\displaystyle{ (S,\bar{S}) }[/math]. We then claim that [math]\displaystyle{ f_{uv}=c_{uv} }[/math] for all [math]\displaystyle{ u\in S,v\not\in S, (u,v)\in E }[/math]; and [math]\displaystyle{ f_{vu}=0 }[/math] for all [math]\displaystyle{ u\in S,v\not\in S, (v,u)\in E }[/math]. If otherwise, then the augmenting path to [math]\displaystyle{ u }[/math] apending [math]\displaystyle{ uv }[/math] becomes a new augmenting path to [math]\displaystyle{ v }[/math], which contradicts that [math]\displaystyle{ S }[/math] includes all vertices to which there exist augmenting paths. Therefore, [math]\displaystyle{ \sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}f_{uv}-\sum_{u\in S,v\not\in S\atop (v,u)\in E}f_{vu} = \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv} }[/math]. As discussed above, this proves the theorem. [math]\displaystyle{ \square }[/math]

The max-flow min-cut theorem

Max-Flow Min-Cut Theorem

In a flow network, the maximum value of any [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] flow equals the minimum value of any [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut.

Proof. Let [math]\displaystyle{ f }[/math] be a flow with maximum value, so there is no augmenting path. Again, let [math]\displaystyle{ S=\{u\in V\mid \exists\text{an augmenting path to }u\} }[/math]. As proved above, [math]\displaystyle{ (S,\bar{S}) }[/math] forms an [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut, and [math]\displaystyle{ \sum_{v:(s,v)}f_{sv}= \sum_{u\in S,v\not\in S\atop (u,v)\in E}c_{uv} }[/math], that is, the value of flow [math]\displaystyle{ f }[/math] equals the value of cut [math]\displaystyle{ (S,\bar{S}) }[/math]. Since we know that all [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] flows are not greater than any [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut, the value of flow [math]\displaystyle{ f }[/math] equals the minimum value of any [math]\displaystyle{ s }[/math]-[math]\displaystyle{ t }[/math] cut. [math]\displaystyle{ \square }[/math]

Flow Integrality Theorem

Let [math]\displaystyle{ (G,c,s,t) }[/math] be a flow network with integral capacity [math]\displaystyle{ c }[/math]. There exists an integral flow which is maximum.

Proof. Let [math]\displaystyle{ f }[/math] be an integral flow of maximum value. If there is an augmenting path, since both [math]\displaystyle{ c }[/math] and [math]\displaystyle{ f }[/math] are integral, a new flow can be constructed of value 1+the value of [math]\displaystyle{ f }[/math], contradicting that [math]\displaystyle{ f }[/math] is maximum over all integral flows. Therefore, there is no augmenting path, which means that [math]\displaystyle{ f }[/math] is maximum over all flows, integral or not. [math]\displaystyle{ \square }[/math]

Unimodularity

Integer Programming

Consider the maximum integral flow problem: given as input a flow network [math]\displaystyle{ (G(V,E),c,s,t) }[/math] where for every [math]\displaystyle{ uv\in E }[/math] the capacity [math]\displaystyle{ c_{uv} }[/math] is integer. We want to find the integral flow [math]\displaystyle{ f:E\rightarrow\mathbb{Z} }[/math] with maximum value.

The mathematical programming for the problem is:

[math]\displaystyle{ \begin{align} \text{maximize} \quad& \sum_{v:(s,v)\in E}f_{sv}\\ \begin{align} \text{subject to} \\ \\ \\ \\ \\ \end{align} \quad & \begin{align} f_{uv}&\le c_{uv} &\quad& \forall (u,v)\in E\\ \sum_{u:(u,v)\in E}f_{uv}-\sum_{w:(v,w)\in E}f_{vw} &=0 &\quad& \forall v\in V\setminus\{s,t\}\\ f_{uv}&\in\mathbb{N} &\quad& \forall (u,v)\in E \end{align} \end{align} }[/math]

where [math]\displaystyle{ \mathbb{N} }[/math] is the set of all nonnegative integers. Compared to the LP for the max-flow problem, we just replace the last line [math]\displaystyle{ f_{uv}\ge 0 }[/math] with [math]\displaystyle{ f_{uv}\in\mathbb{N} }[/math].

Integrality of polytopes

A point in an [math]\displaystyle{ n }[/math]-dimensional space is integral if all its coordinates are integers.

A polyhedron is said to be integral if all its vertices are integral.

There always exists an optimal solution which is a vertex in [math]\displaystyle{ P }[/math]. For integral [math]\displaystyle{ P }[/math], all vertices are integral.

Theorem (Hoffman 1974)

If a polyhedron [math]\displaystyle{ P }[/math] is integral then for all integer vectors [math]\displaystyle{ \boldsymbol{c} }[/math] there is an optimal solution to [math]\displaystyle{ \max\{\boldsymbol{c}^T\boldsymbol{x}\mid \boldsymbol{x}\in P\} }[/math] which is integral.

Unimodularity and total unimodularity

Definition (Unimodularity)

An [math]\displaystyle{ n\times n }[/math] integer matrix [math]\displaystyle{ A }[/math] is called unimodular if [math]\displaystyle{ \det(A)=\pm1 }[/math]. An [math]\displaystyle{ m\times n }[/math] integer matrix [math]\displaystyle{ A }[/math] is called total unimodular if every square submatrix [math]\displaystyle{ B }[/math] of [math]\displaystyle{ A }[/math] has [math]\displaystyle{ \det(B)\in\{1,-1,0\} }[/math], that is, every square, nonsingular submatrix of [math]\displaystyle{ A }[/math] is unimodular.

Theorem

Let [math]\displaystyle{ A }[/math] be an [math]\displaystyle{ m\times n }[/math] integer matrix. If [math]\displaystyle{ A }[/math] is totally unimodualr, then for any integer vector [math]\displaystyle{ \boldsymbol{b}\in\mathbb{Z}^n }[/math] the polyhedron [math]\displaystyle{ \{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\} }[/math] is integral.

Proof. Let [math]\displaystyle{ B }[/math] be a basis of [math]\displaystyle{ A }[/math], and let [math]\displaystyle{ \boldsymbol{b}' }[/math] be the corresponding coordinates in [math]\displaystyle{ \boldsymbol{b} }[/math]. A basic solution is formed by [math]\displaystyle{ B^{-1}\boldsymbol{b}' }[/math] and zeros. Since [math]\displaystyle{ A }[/math] is totally unimodular and [math]\displaystyle{ B }[/math] is a basis thus nonsingular, [math]\displaystyle{ \det(B)\in\{1,-1,0\} }[/math]. By Cramer's rule, [math]\displaystyle{ B^{-1} }[/math] has integer entries, thus [math]\displaystyle{ B^{-1}\boldsymbol{b}' }[/math] is integral. Therefore, any basic solution of [math]\displaystyle{ A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0} }[/math] is integral, which means the polyhedron [math]\displaystyle{ \{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}=\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\} }[/math] is integral. [math]\displaystyle{ \square }[/math]

Theorem (Hoffman-Kruskal 1956)

Let [math]\displaystyle{ A }[/math] be an [math]\displaystyle{ m\times n }[/math] integer matrix. If [math]\displaystyle{ A }[/math] is totally unimodualr, then for any integer vector [math]\displaystyle{ \boldsymbol{b}\in\mathbb{Z}^n }[/math] the polyhedron [math]\displaystyle{ \{\boldsymbol{x}\in\mathbb{R}^n\mid A\boldsymbol{x}\ge\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\} }[/math] is integral.

Proof. Let [math]\displaystyle{ A'=\begin{bmatrix}A & -I\end{bmatrix} }[/math]. We claim that [math]\displaystyle{ A' }[/math] is also totally unimodular. Any square submatrix [math]\displaystyle{ B }[/math] of [math]\displaystyle{ A }[/math] can be written in the following form after permutation: [math]\displaystyle{ B=\begin{bmatrix} C & 0\\ D & I \end{bmatrix} }[/math] where [math]\displaystyle{ C }[/math] is a square submatrix of [math]\displaystyle{ A }[/math] and [math]\displaystyle{ I }[/math] is identity matrix. Therefore, [math]\displaystyle{ \det(B)=\det(C)\in\{1,-1,0\} }[/math], thus [math]\displaystyle{ A' }[/math] is totally unimodular. Add slack variables to transform the constraints to the standard form [math]\displaystyle{ A'\boldsymbol{z}=\boldsymbol{b},\boldsymbol{z}\ge\boldsymbol{0} }[/math]. The polyhedron [math]\displaystyle{ \{\boldsymbol{x}\mid A\boldsymbol{x}\ge\boldsymbol{b}, \boldsymbol{x}\ge \boldsymbol{0}\} }[/math] is integral if the polyhedron [math]\displaystyle{ \{\boldsymbol{z}\mid A'\boldsymbol{z}=\boldsymbol{b}, \boldsymbol{z}\ge \boldsymbol{0}\} }[/math] is integral, which is implied by the total unimodularity of [math]\displaystyle{ A'\, }[/math]. [math]\displaystyle{ \square }[/math]