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The Maximum flow and the Minimum cut

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• Classic Theorem: the Max-flow min-cut Theorem
  • In this webpage, we will study (prove) the classic Max-flow min-cut Theorem

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  • Here is the Wikipedia page: click here

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• Cut
  • Definition: a cut of a network

    Cut = a set of edges whose removal will divide/separate the network into 2 halves X and Y where:        (source) S ∈ X        and          (sink) T ∈ Y In other words: Removing the edges in a cut will severe the source S from the sink T

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  • Example: 2 cuts in the network

    Cut 1: Cut 2:

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• Capacity of a cut
  • Definition: the capacity of a cut

    Capacity of a cut = the sum of the capacity of the edges in the cut that are oriented from a vertex ∈ X to a vertex ∈ Y Note: These edges can be used to "carry" flow from S → T

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    Example: capacity of cuts

    Cut 1: Cut 2: Note: The capacity of edge B → A is not included in the sum because B → A is directed from a vertex in Y to a vertex in X Only edges oriented from a vertex in X to a vertex in Y are used to "carry" flow from S → T !!!

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• Minimum cut (and: minimum cut capacity)
  • Definition: minimum cut and min. cut capacity

    Minimum cut = the cut in the network that has the smallest possible capacity         Minimum cut capacity = the capacity of the minimum cut

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• An upper bound for the maximum flow
  • Recall: maximum flow:

    Maximum flow = the largest possible flow from source S to sink T

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  • Proposition:

    Maximum flow ≤ Capacity of the min. cut of the network

    Proof:

    Consider any cut of the basic network: Note that: Any flow must pass through the edge of a cut We have that: Total flow = ∑ forward edge e f(e) - ∑ backward edge b f(b) ≤ ∑ forward edge e f(e) (f(e) ≤ c(e)) ≤ ∑ forward edge e c(e) Therefore: Total flow ≤ capacity of any cut This is true for every flow So it is true for the maximum possible flow also... Therefore: Maximum flow ≤ capacity of any cut

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  • If the maximum flow ≤ capacity of any cut, then maximum flow ≤ capacity of the minimum cut:

    Maximum flow ≤ capacity of minimum cut .......... (1)

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• The   max-flow, min-cut   theorem
  • Theorem:

    In any basic network, the value of the maximum flow is equal to the capacity of the minimum cut       I.e.: maximum flow = capacity of min. cut

    Proof:

    From the previous discussion, we have that: maximum flow ≤ capacity of min. cut To complete the proof, we show that: There exists a cut C such that: capacity of cut C = maximum flow We will construct the cut C from a maximum flow.... Assume that we have found a maximum flow in the network Now construct the following cut: X = set of vertices that can be reached from source S through a flow-augmenting path        Y = all the other vertices Example: Here is a basic network with a maximum flow: Nodes that can reached from source S through a flow-augmenting path: only the node: C This is the cut that we will construct: We must prove the following: The above construction yields a cut I.e.: The source S and the sink T are in different sets ! The total flow through the cut is equal to the capacity of the cut We will prove the above by proving 3 special properties of this construction Property 1: the construct yields a cut, meaning: S   ∈   X T   ∉   X        (i.e.: T   ∈   Y)            Proof: By the way the set X is formed, the source S is always included in X. I.e.: S   ∈   X         We will show T ∉ X (i.e., T ∈ Y) by contradiction Suppose: T   ∈   X            From the construction of the cut C: X = all vertices that can be reached from source S by a flow-augmenting path Because   T   ∈   X, then: There exists a flow augmenting path from S → T which means: the current flow is not maximum !!!!         That contradicts the given fact that the current flow is maximum !!! Taking stock: We have shown that: The source S and the sink T are in different parts of the cut C: Now we will show that: Total flow over the cut is equal to the capacity of the cut Recall: Capacity of a cut = sum of the capacity of all forward edges Property 2: For each forward edge e in the cut C: f(e) = c(e) Graphically illustrated: Proof: Select an arbitrary forward edge A→B of the cut A ∈ X,       and        B ∈ Y       (i.e., there is no flow augmenting path from S to B) Because A ∈ X, there is a flow augmenting path from S → A: If: f(e) < c(e): then: there will be a flow augmenting path from S → B (by increasing the flow through A → B) !!! which mean: B ∈ X !!!            This contradicts the fact that: B ∈ Y Property 3: For each backward edge e in the cut C: f(e) = 0 Graphically illustrated: Proof: Select an arbitrary backward edge B→A of the cut A ∈ X,       and        B ∈ Y       (i.e., there is no flow augmenting path from S to B) Because A ∈ X, there is a flow augmenting path from S → A: If: f(e) > 0: then: there will be a flow augmenting path from S → B (by increasing the flow through A → B) !!! If we decrease some flow on A→B, the which mean: B ∈ X !!!            This contradicts the fact that: B ∈ Y Therefore, we have the following situation at this particular cut: And therefore: total flow = sum of f(e) for every forward edge e (f(e) = c(e) !!!) = sum of c(e) for every forward edge e = capacity of the cut In other words: We found a cut that has the capacity equal to the max. flow !!!        That completes the proof...

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• Practice
  • Find a minimum cut in the following basic network:

    Solution: the max flow is achieved with this flow assignment

    Here is the solution with saturated edges thickened:

    Find all node that can be reach by flow augmenting paths from S:

    S   → C   → E   → B (through backward edge BE by decreasing flow !)                           → A (through backward edge AE by decreasing flow !)

    The minimum cut: