08 - The Ford-Fulkerson algorithm for maximum flow

08 - The Ford-Fulkerson algorithm for maximum flow - And...

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2/25/08 - The Ford-Fulkerson algorith. .. reminders from Fri. Flow network is G = (V, E) directed with s, t V, c e 0 e Flow is f: E -> |R+ such that f(e) ce, ±n(v) = fout(v) (v s, t) Value of ²ow is v(f) = fout(s) Cut is a partition V = A A ̅ with s A, t A ̅ . Capacity Cap(A) = ce where E(A,B) = {(u, v) E|u A, v B} e E(A,A ̅ ) Def. An “augmenting path” is a path from s to t in G f bottleneck(P, F): minimum residual capacity of an edge in P. augment(f,p): Let f’ = f. Let b = bottleneck(P, f). For e P if e forward f’(e) <- f(e) + b if e backward f’(e) <- f(e) - b endfor return f’ 18-1 18
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Ford-Fulkerson Initialize f(e) = 0 e, G f = G 1 while an s-t path P in G f f’ = augment(f, P) update G f to G f’ , update f to f’. endwhile return f. Correctness 18-2
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(1) F.-F. terminates. Note that v(f) increases by 1 every iteration.
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Unformatted text preview: And its bounded above by Cap({s}). So # iterations is bounded. (2) F.-F. computes a feasible ow. Induction on # of iterations. Flow values remain between 0 and c e because bottleneck(P,f) was dened so that f(e) is between 0 and c e . Flow convervation 18-3 (3) f is a maximum Fow. If we can nd (A, A ) such that Cap(A) = v(f) we are done, because for every other Fow f v(f ) Cap(A) = v(f) 18-4 Let A = {vertices reachable from s in G f } Notice that s A, t A. Every edge in E(A, A ) is saturated, f(e) = c e . Recall: For every ow f, v(f) = f(e) - f(e) e Out(A) e In(A) = f(e) + (f(e) - f(e)) - f(e) e E(A,A ) e E(A,A) e E(A ,A) = c e- f(e) = Cap(A) e E(A,A ) e E(A ,A) 18-5...
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This note was uploaded on 10/02/2008 for the course CS 482 taught by Professor Kleinberg during the Spring '08 term at Cornell University (Engineering School).

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08 - The Ford-Fulkerson algorithm for maximum flow - And...

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