Max-flow min-cut theorem

Example

The figure on the right shows a flow in a network. The numerical annotation on each arrow, in the form f/c, indicates the flow (f) and the capacity (c) of the arrow. The flows emanating from the source total five (2+3=5), as do the flows into the sink (2+3=5), establishing that the flow's value is 5.

One s-t cut with value 5 is given by S={s,p} and T={o, q, r, t}. The capacities of the edges that cross this cut are 3 and 2, giving a cut capacity of 3+2=5. (The arrow from o to p is not considered, as it points from T back to S.)

The value of the flow is equal to the capacity of the cut, showing that the flow is a maximal flow and the cut is a minimal cut.

Note that the flow through each of the two arrows that connect S to T is at full capacity; this is always the case: a minimal cut represents a 'bottleneck' of the system.

Linear program formulation

The max-flow problem and min-cut problem can be formulated as two primal-dual linear programs.

The max-flow LP is straightforward. The dual LP is obtained using the algorithm described in dual linear program: the variables and sign constraints of the dual correspond to the constraints of the primal, and the constraints of the dual correspond to the variables and sign constraints of the primal. The resulting LP requires some explanation. The interpretation of the variables in the min-cut LP is:

${\displaystyle d_{uv}=\{\begin{array}{ll}1,& \text{if }u\in S\text{ and }v\in T\text{ (the edge }uv\text{ is in the cut) }\\ 0,& \text{otherwise}\end{array}}$

${\displaystyle z_v=\{\begin{array}{ll}1,& \text{if }v\in S\\ 0,& \text{otherwise}\end{array}}$

The minimization objective sums the capacity over all the edges that are contained in the cut.

The constraints guarantee that the variables indeed represent a legal cut:

• The constraints ${\displaystyle d_{uv}-z_u+z_v\ge 0}$ (equivalent to ${\displaystyle d_{uv}\ge z_u-z_v}$) guarantee that, for non-terminal nodes u,v, if u is in S and v is in T, then the edge (u,v) is counted in the cut (${\displaystyle d_{uv}\ge 1}$).
• The constraints ${\displaystyle d_{sv}+z_v\ge 1}$ (equivalent to ${\displaystyle d_{sv}\ge 1-z_v}$) guarantee that, if v is in T, then the edge (s,v) is counted in the cut (since s is by definition in S).
• The constraints ${\displaystyle d_{ut}-z_u\ge 0}$ (equivalent to ${\displaystyle d_{ut}\ge z_u}$) guarantee that, if u is in S, then the edge (u,t) is counted in the cut (since t is by definition in T).

Note that, since this is a minimization problem, we do not have to guarantee that an edge is not in the cut - we only have to guarantee that each edge that should be in the cut, is summed in the objective function.

The equality in the max-flow min-cut theorem follows from the strong duality theorem in linear programming, which states that if the primal program has an optimal solution, x*, then the dual program also has an optimal solution, y*, such that the optimal values formed by the two solutions are equal.

Application

Cederbaum's maximum flow theorem

The maximum flow problem can be formulated as the maximization of the electrical current through a network composed of nonlinear resistive elements. In this formulation, the limit of the current I_in between the input terminals of the electrical network as the input voltage V_in approaches ${\displaystyle \mathrm{\infty }}$, is equal to the weight of the minimum-weight cut set.

${\displaystyle \underset{V_{\text{in}}\to \mathrm{\infty }}{lim}(I_{in})=\underset{X_C}{min}\sum _{(u,v)\in X_C}{c}_{uv}}$

Generalized max-flow min-cut theorem

In addition to edge capacity, consider there is capacity at each vertex, that is, a mapping ${\displaystyle c:V\to {\mathbb{R}}^{+}}$ denoted by c(v), such that the flow f has to satisfy not only the capacity constraint and the conservation of flows, but also the vertex capacity constraint

${\displaystyle \mathrm{\forall }v\in V\setminus \{s,t\}:{\sum }_{\{u\in V\mid (u,v)\in E\}}{f}_{uv}\le c(v).}$

In other words, the amount of flow passing through a vertex cannot exceed its capacity. Define an s-t cut to be the set of vertices and edges such that for any path from s to t, the path contains a member of the cut. In this case, the capacity of the cut is the sum of the capacity of each edge and vertex in it.

In this new definition, the generalized max-flow min-cut theorem states that the maximum value of an s-t flow is equal to the minimum capacity of an s-t cut in the new sense.

Menger's theorem

In the undirected edge-disjoint paths problem, we are given an undirected graph G = (V, E) and two vertices s and t, and we have to find the maximum number of edge-disjoint s-t paths in G.

Menger's theorem states that the maximum number of edge-disjoint s-t paths in an undirected graph is equal to the minimum number of edges in an s-t cut-set.

Project selection problem

In the project selection problem, there are n projects and m machines. Each project p_i yields revenue r(p_i) and each machine q_j costs c(q_j) to purchase. We want to select a subset of the project, and purchase a subset of the machines, to maximize the total profit (revenue of the selected projects minus cost of the purchased machines). We must obey the following constraint: each project specifies a set of machines which must be purchased if the project is selected. (Each machine, once purchased, can be used by any selected project.)

To solve the problem, let P be the set of projects not selected and Q be the set of machines purchased, then the problem can be formulated as,

${\displaystyle max\{g\}=\sum _{i}r(p_i)-\sum _{p_i\in P}r(p_i)-\sum _{q_j\in Q}c(q_j).}$

Since the first term does not depend on the choice of P and Q, this maximization problem can be formulated as a minimization problem instead, that is,

${\displaystyle min\{g^{\prime }\}=\sum _{p_i\in P}r(p_i)+\sum _{q_j\in Q}c(q_j).}$

The above minimization problem can then be formulated as a minimum-cut problem by constructing a network, where the source is connected to the projects with capacity r(p_i), and the sink is connected by the machines with capacity c(q_j). An edge (p_i, q_j) with infinite capacity is added if project p_i requires machine q_j. The s-t cut-set represents the projects and machines in P and Q respectively. By the max-flow min-cut theorem, one can solve the problem as a maximum flow problem.

The figure on the right gives a network formulation of the following project selection problem:

The minimum capacity of an s-t cut is 250 and the sum of the revenue of each project is 450; therefore the maximum profit g is 450 − 250 = 200, by selecting projects p₂ and p₃.

The idea here is to 'flow' each project's profits through the 'pipes' of its machines. If we cannot fill the pipe from a machine, the machine's return is less than its cost, and the min cut algorithm will find it cheaper to cut the project's profit edge instead of the machine's cost edge.

Image segmentation problem

In the image segmentation problem, there are n pixels. Each pixel i can be assigned a foreground value f_i or a background value b_i. There is a penalty of p_ij if pixels i, j are adjacent and have different assignments. The problem is to assign pixels to foreground or background such that the sum of their values minus the penalties is maximum.

Let P be the set of pixels assigned to foreground and Q be the set of points assigned to background, then the problem can be formulated as,

${\displaystyle max\{g\}=\sum _{i\in P}{f}_i+\sum _{i\in Q}{b}_i-\sum _{i\in P,j\in Q\vee j\in P,i\in Q}{p}_{ij}.}$

This maximization problem can be formulated as a minimization problem instead, that is,

${\displaystyle min\{g^{\prime }\}=\sum _{i\in P,j\in Q\vee j\in P,i\in Q}{p}_{ij}.}$

The above minimization problem can be formulated as a minimum-cut problem by constructing a network where the source (orange node) is connected to all the pixels with capacity f_i, and the sink (purple node) is connected by all the pixels with capacity b_i. Two edges (i, j) and (j, i) with p_ij capacity are added between two adjacent pixels. The s-t cut-set then represents the pixels assigned to the foreground in P and pixels assigned to background in Q.

History

An account of the discovery of the theorem was given by Ford and Fulkerson in 1962:

"Determining a maximal steady state flow from one point to another in a network subject to capacity limitations on arcs ... was posed to the authors in the spring of 1955 by T.E. Harris, who, in conjunction with General F. S. Ross (Ret.) had formulated a simplified model of railway traffic flow, and pinpointed this particular problem as the central one suggested by the model. It was not long after this until the main result, Theorem 5.1, which we call the max-flow min-cut theorem, was conjectured and established. A number of proofs have since appeared."

Proof

Let G = (V, E) be a network (directed graph) with s and t being the source and the sink of G respectively.

Consider the flow f computed for G by Ford–Fulkerson algorithm. In the residual graph (G_f ) obtained for G (after the final flow assignment by Ford–Fulkerson algorithm), define two subsets of vertices as follows:

1. A: the set of vertices reachable from s in G_f
2. A^c: the set of remaining vertices i.e. V − A

Claim. value( f ) = c(A, A^c), where the capacity of an s-t cut is defined by

${\displaystyle c(S,T)={\sum }_{(u,v)\in S\times T}{c}_{uv}}$.

Now, we know, ${\displaystyle value(f)=f_{out}(A)-f_{in}(A)}$ for any subset of vertices, A. Therefore, for value( f ) = c(A, A^c) we need:

• All outgoing edges from the cut must be fully saturated.
• All incoming edges to the cut must have zero flow.

To prove the above claim we consider two cases:

• In G, there exists an outgoing edge ${\displaystyle (x,y),x\in A,y\in A^c}$ such that it is not saturated, i.e., f (x, y) < c_xy. This implies, that there exists a forward edge from x to y in G_f, therefore there exists a path from s to y in G_f, which is a contradiction. Hence, any outgoing edge (x, y) is fully saturated.

• In G, there exists an incoming edge ${\displaystyle (y,x),x\in A,y\in A^c}$ such that it carries some non-zero flow, i.e., f (y, x) > 0. This implies, that there exists a backward edge from x to y in G_f, therefore there exists a path from s to y in G_f, which is again a contradiction. Hence, any incoming edge (y, x) must have zero flow.

Both of the above statements prove that the capacity of cut obtained in the above described manner is equal to the flow obtained in the network. Also, the flow was obtained by Ford-Fulkerson algorithm, so it is the max-flow of the network as well.

Also, since any flow in the network is always less than or equal to capacity of every cut possible in a network, the above described cut is also the min-cut which obtains the max-flow.

A corollary from this proof is that the maximum flow through any set of edges in a cut of a graph is equal to the minimum capacity of all previous cuts.