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The max flow problem

The max flow problem. 11. 7. 5. -2. 2. 1. 10. 5. 4. 9. Ford-Fulkerson method. Ford-Fulkerson( G ) f = 0 while ( 9 simple path p from s to t in G f ) f := f + f p output f. A cut. S. T. c(S , T )=26. Lemma 26.5 + Corollary 26.6.

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The max flow problem

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  1. The max flow problem 11 7 5 -2 2 1 10 5 4 9

  2. Ford-Fulkerson method Ford-Fulkerson(G) f = 0 while(9 simple path p from s to t in Gf) f := f + fp outputf

  3. A cut S T c(S,T)=26

  4. Lemma 26.5 + Corollary 26.6 • Let f be a flow in G and let (S,T) be a cut in G. Then |f| = f(S,T). • Let f be a flow in G and let (S,T) be a cut in G. Then |f| · c(S,T). This is a weak duality theorem.

  5. Max Flow – Min Cut Theorem Let f be a flow in G. The following three conditions are equivalent: 1. f is a maximum flow 2. Gf contains no augmenting path 3. There is a cut (S,T) so that |f|=c(S,T)

  6. Max Flow – Min Cut Theorem • The value of the maximum flow in G is equal to the capacity of the minimum cut in G. • This is a strong duality theorem.

  7. Remarks • The solution values agree, not the solutions themselves – flows and cuts are completely different objects. • Given a max flow we can easily find a min cut (follows from proof of max flow-min cut theorem). Going the other way is less obvious.

  8. Consequence • The Ford-Fulkerson method is partially correct, i.e., if it terminates it produces the flow with the maximum value.

  9. Local search checklist Design: • How do we find the first feasible solution? • Neighborhood design? • Which neighbor to choose? Analysis: • Partial correctness? (termination )correctness) • Termination? • Complexity? ٧ ٧ ٧

  10. Termination • Suppose all capacities are integers. • We start with a flow of value 0. • In each iteration, we get a new flow with higher integer value. • We always have a legal flow, i.e., one of value at most |f|. • Hence we can have at most |f| iterations.

  11. Correctness of Ford-Fulkerson • Since Ford-Fulkerson is partially correct and it terminates if capacities are integers it is a correct algorithm for finding the maximum flow if capacities are integers. • Exercise: It is also correct if capacities are rationals.

  12. Does Ford-Fulkerson always terminate? • In case of irrational capacities, not necessarily! (Exercise) • But we can’t give irrational capacities as inputs to digital computers anyway. • In case of floating point capacities, who knows?

  13. Integrality Theorem (26.11) If a flow network has integer valued capacities, there is a maximum flow with an integer value on every edge. The Ford-Fulkerson method will yield such a maximum flow. The integrality theorem is often extremely important when “programming” and modeling using the max flow formalism.

  14. Reduction: Maximum Matching ! Max Flow What is the maximum cardinality matching in G?

  15. G

  16. s t G’ All capacities are 1

  17. Finding a balanced set of Representatives (Ahuja, Application 6.2) • A city has clubs C1, C2,…,Cn and partiesP1, P2,…,Pm. A citizen may be a member of several clubs but may only be a member of one party. • A balanced city council must be formed by including exactly one member from each club and at most uk members from party Pk.

  18. Local search checklist Design: • How do we find the first feasible solution? • Neighborhood design? • Which neighbor to choose? Analysis: • Partial correctness? (termination )correctness) • Termination? • Complexity? ٧ ٧ ٧ ٧

  19. Complexity of Ford-Fulkerson • We have at most |f| improvement steps (iterations of the while-loop). • Is this the best possible bound?

  20. Complexity • We have at most |f| improvement steps (iterations of the while-loop) and this bound cannot be improved for the general Ford-Fulkerson method. • How fast can we implement a single improvement step?

  21. Complexity • Assume |V|-1 · |E|. Otherwise the graph is not connected. • Then, Ford-Fulkerson can be implemented to run in time at most O(|E| |f|). • Is this fast?

  22. Polynomial time algorithms • Defintion: A polynomial time algorithm is an algorithm than runs in time polynomial in n, where n is the number of bits of the input. • How we intend to encode the input influences if we have a polynomial algorithm or not. Usually, some “standard encoding” is implied. • In this course: Polynomial ¼ Fast Exponential ¼ Slow

  23. How to encode max flow instance? java MaxFlow ??????????

  24. How to encode max flow instance? java MaxFlow 6#0|16|13|0|0|0#0|0|10|12|0|0 #0|4|0|0|14|0#0|0|9|0|0|20 #0|0|0|7|0|4|#0|0|0|0|0|0

  25. Complexity of Ford-Fulkerson • With standard (decimal or binary) representation of integers, Ford-Fulkerson is an exponential time algorithm.

  26. java MaxFlow 111111 #|1111111111111111|1111111111111||| #||1111111111|111111111111|| #|1111|||11111111111111| #||111111111|||11111111111111111111 #|||1111111||1111 #|||||

  27. Complexity of Ford-Fulkerson • With unary (4 ~ 1111) representation of integers, Ford-Fulkerson is a polynomial time algorithm. • Intuition: When the input is longer it is easier to be polynomial time as a function of the input length. • An algorithm which is polynomial if integer inputs are represented in unary is called a pseudo-polynomial algorithm. • Intuitively, a pseudo-polynomial algorithm is an algorithm which is fast if all numbers in the input are small.

  28. Edmonds-Karp Edmonds-Karp algorithm for Max Flow: Implement Ford-Fulkerson by always choosing the shortest possible augmenting path, i.e., the one with fewest possible edges.

  29. Complexity of Edmonds-Karp • Each iteration of the while loop can still be done in time O(|E|). • The number of iterations are now at most O(|V||E|) regardless of capacities – to be seen next. • Thus, the total running time is O(|V| |E|2) and Edmonds-Karp is a polynomial time algorithm for Max Flow.

  30. Why at most O(|V| |E|) iterations? When executing Edmonds-Karp, the residual network Gfgradually changes (as f changes). This sequence of different residual networks Gf satisfies: Theorem (~ Lemma 26.8 and Theorem 26.9): • The distance between s and t in Gfnever decreases: After each iteration of the while-loop, it either increases or stays the same. 2) The distance between s and t in Gf can stay the same for at most |E| iterations of the while-loop before increasing. As the distance between s and t can never be more than |V|-1 and it starts out as at least 1, it follows from the theorem that we have at most (|V|-2)|E| iterations.

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