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Published byCeleste Barnett Modified over 2 years ago

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Max Flow Min Cut

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Theorem The maximum value of an st-flow in a digraph equals the minimum capacity of an st-cut. Theorem If every arc has integer capacity, then in a maximum flow every arc has integer flow.

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Find a maximum st-flow and a minimum st-cut s t

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3131 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7 STOP

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1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 Minimum Cut s t 1 3 5 2 4 8 6 9 7

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1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2121 3 2 1 2 3131 3 3232 3 1 1 1 1 2121 3 30 s t 1 3 5 2 4 8 6 9 7

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1 1 3 1 1 2 3 1 1 3 3 s t 1 3 5 2 4 8 6 9 7 So f is a max flow

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 Algorithm s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3 3 3 3 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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1 2 3 2 1 2 3232 3232 3 3131 1 1 1 1 2 3 30 s t 1 3 5 2 4 8 6 9 7

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Hall’s Theorem from Max Flow Min Cut

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11111 Direct all edges from s to t and assign all arcs unit capacity Adds and t, adjacent to all of A and B respectively.

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We have to show that Hall’s Condition gives a 1-factor.

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A flow of value is enough to guarantee a 1-factor.

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So all we have to do is show for each cut S.

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