In computer science and optimization theory, the max-flow min-cut theorem states that in a flow network, the maximum amount of flow passing from the source to the sink is equal to the total weight of the edges in a minimum cut, i.e., the smallest total weight of the edges which if removed would disconnect the source from the sink.
This is a special case of the duality theorem for linear programs and can be used to derive Menger's theorem and the Kőnig–Egerváry theorem.[1]
The theorem equates two quantities: the maximum flow through a network, and the minimum capacity of a cut of the network. To state the theorem, each of these notions must first be defined.
A network consists of
c:E\to\R+
A flow through a network is a mapping
f:E\to\R+
fuv
f(u,v)
(u,v)\inE
fuv\lecuv.
v
s
t
\sum\nolimits\{
A flow can be visualized as a physical flow of a fluid through the network, following the direction of each edge. The capacity constraint then says that the volume flowing through each edge per unit time is less than or equal to the maximum capacity of the edge, and the conservation constraint says that the amount that flows into each vertex equals the amount flowing out of each vertex, apart from the source and sink vertices.
The value of a flow is defined by
|f|=\sum\nolimits\{v
where as above
s
t
The maximum flow problem asks for the largest flow on a given network.
Maximum Flow Problem. Maximize, that is, to route as much flow as possible from|f|
tos
.t
The other half of the max-flow min-cut theorem refers to a different aspect of a network: the collection of cuts. An s-t cut is a partition of such that and . That is, an s-t cut is a division of the vertices of the network into two parts, with the source in one part and the sink in the other. The cut-set
XC
XC:=\{(u,v)\inE : u\inS,v\inT\}=(S x T)\capE.
Thus, if all the edges in the cut-set of are removed, then no positive flow is possible, because there is no path in the resulting graph from the source to the sink.
The capacity of an s-t cut is the sum of the capacities of the edges in its cut-set,
c(S,T)=
| \sum\nolimits | |
| (u,v)\inXC |
cuv=\sum\nolimits(i,j)cijdij,
where
dij=1
i\inS
j\inT
0
There are typically many cuts in a graph, but cuts with smaller weights are often more difficult to find.
Minimum s-t Cut Problem. Minimize, that is, determine and such that the capacity of the s-t cut is minimal.
In the above situation, one can prove that the value of any flow through a network is less than or equal to the capacity of any s-t cut, and that furthermore a flow with maximal value and a cut with minimal capacity exist. The main theorem links the maximum flow value with the minimum cut capacity of the network.
Max-flow min-cut theorem. The maximum value of an s-t flow is equal to the minimum capacity over all s-t cuts.
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= and 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.
The max-flow problem and min-cut problem can be formulated as two primal-dual linear programs.[2]
| Max-flow (Primal) | Min-cut (Dual) | ||
|---|---|---|---|
| variables | fuv \forall(u,v)\inE | duv \forall(u,v)\inE zv \forallv\inV\setminus\{s,t\} | |
| objective | maximize \sum\nolimitsv:fsv [max total flow from source] | minimize \sum\nolimits(u,v)cuvduv [min total capacity of edges in cut] | |
| constraints | subject to \begin{align}fuv&\leqcuv&&\forall(u,v)\inE\\ \sumufuv-\sumwfvw&=0&&v\inV\setminus\{s,t\}\end{align} [a constraint per edge and a constraint per non-terminal node] | subject to \begin{align} duv-zu+zv&\geq0&&\forall(u,v)\inE,u ≠ s,v ≠ t\\ dsv+zv&\geq1&&\forall(s,v)\inE,v ≠ t\\ dut-zu&\geq0&&\forall(u,t)\inE,u ≠ s\\ dst&\geq1&&if(s,t)\inE \end{align} [a constraint per edge] | |
| sign constraints | \begin{align}fuv&\geq0&&\forall(u,v)\inE\\ \end{align} | \begin{align} duv&\geq0&&\forall(u,v)\inE\\ zv&\in\R&&\forallv\inV\setminus\{s,t\}\end{align} |
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:
duv=\begin{cases}1,&ifu\inSandv\inT(theedgeuvisinthecut) \ 0,&otherwise\end{cases}
zv=\begin{cases}1,&ifv\inS\ 0,&otherwise\end{cases}
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:[3]
duv-zu+zv\geq0
duv\geqzu-zv
duv\geq1
dsv+zv\geq1
dsv\geq1-zv
dut-zu\geq0
dut\geqzu
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.
See also: 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.[4] In this formulation, the limit of the current between the input terminals of the electrical network as the input voltage approaches
infty
| \lim | |
| Vin\toinfty |
(Iin)=
| min | |
| XC |
| \sum | |
| (u,v)\inXC |
cuv
In addition to edge capacity, consider there is capacity at each vertex, that is, a mapping
c:V\to\R+
\forallv\inV\setminus\{s,t\}: \sum\nolimits\{u\in
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.
See also: Menger's Theorem. In the undirected edge-disjoint paths problem, we are given an undirected graph and two vertices and, and we have to find the maximum number of edge-disjoint s-t paths in .
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.
In the project selection problem, there are projects and machines. Each project yields revenue and each machine costs to purchase. Each project requires a number of machines and each machine can be shared by several projects. The problem is to determine which projects and machines should be selected and purchased respectively, so that the profit is maximized.
Let be the set of projects not selected and be the set of machines purchased, then the problem can be formulated as,
max\{g\}=\sumir(pi)-
| \sum | |
| pi\inP |
r(pi)-
| \sum | |
| qj\inQ |
c(qj).
Since the first term does not depend on the choice of and, this maximization problem can be formulated as a minimization problem instead, that is,
min\{g'\}=
| \sum | |
| pi\inP |
r(pi)+
| \sum | |
| qj\inQ |
c(qj).
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, and the sink is connected by the machines with capacity . An edge with infinite capacity is added if project requires machine . The s-t cut-set represents the projects and machines in and 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:
| Project | Machine | |||
|---|---|---|---|---|
| 1 | 100 | 200 | Project 1 requires machines 1 and 2. | |
| 2 | 200 | 100 | Project 2 requires machine 2. | |
| 3 | 150 | 50 | Project 3 requires machine 3. |
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 and .
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.
See also: Maximum flow problem. In the image segmentation problem, there are pixels. Each pixel can be assigned a foreground value or a background value . There is a penalty of if pixels 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 be the set of pixels assigned to foreground and be the set of points assigned to background, then the problem can be formulated as,
max\{g\}=\sumifi+\sumibi-\sumipij.
This maximization problem can be formulated as a minimization problem instead, that is,
min\{g'\}=\sumipij.
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, and the sink (purple node) is connected by all the pixels with capacity . Two edges and with capacity are added between two adjacent pixels. The s-t cut-set then represents the pixels assigned to the foreground in and pixels assigned to background in .
An account of the discovery of the theorem was given by Ford and Fulkerson in 1962:[5]
"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.[6] A number of proofs have since appeared."[7] [8] [9]
Let be a network (directed graph) with and being the source and the sink of respectively.
Consider the flow computed for by Ford–Fulkerson algorithm. In the residual graph obtained for (after the final flow assignment by Ford–Fulkerson algorithm), define two subsets of vertices as follows:
Claim., where the capacity of an s-t cut is defined by
c(S,T)=\sum\nolimits(u,v)cuv
Now, we know,
value(f)=fout(A)-fin(A)
To prove the above claim we consider two cases:
(x,y),x\inA,y\inAc
(y,x),x\inA,y\inAc
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.