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1 Maximum flow sender receiver Capacity constraint Lecture 6: Jan 25

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2 Network transmission Given a directed graph G A source node s A sink node t Goal: To send as much information from s to t

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3 Flows An s-t flow is a function f which satisfies: (capacity constraint) (conservation of flows) An s-t flow is a function f which satisfies: (capacity constraint) (conservation of flows ( at intermediate vertices )

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4 Value of the flow st G: 6 Value = 19 Maximum flow problem: maximize this value

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5 Flow decomposition Any flow can be decomposed into at most m flow paths. The same idea applies to the Chinese postman problem

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6 An upper bound sender receiver

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7 Cuts An s-t cut is a set of edges whose removal disconnect s and t The capacity of a cut is defined as the sum of the capacity of the edges in the cut Minimum s-t cut problem: minimize this capacity of a s-t cut

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8 Flows ≤ cuts Let C be a cut and S be the connected component of G-C containing s. Then:

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9 Main result Value of max s-t flow ≤ capacity of min s-t cut (Ford Fulkerson 1956) Max flow = Min cut A polynomial time algorithm

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10 Greedy method? Find an s-t path where every edge has f(e) < c(e) Add this path to the flow Repeat until no such path can be found. Does it work?

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11 A counterexample Hint: Find an augmenting path

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12 Residual graph Key idea: allow flows to push back c(e) = 10 f(e) = 2 c(e) = 8 c(e) = 2 Advantage of this representation is not to distinguish send forward or push back (which are irrelevant) Can send 8 units forward or push 2 units back.

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13 Finding an augmenting path Find an s-t path in the residual graph Add it to the current flow to obtain a larger flow. Why? 1.Flow conservations 2.More flow going out from s Key: don’t think about flow paths!

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14 Ford-Fulkerson Algorithm 1. Start from an empty flow f 2. While there is an s-t path P in G update f along P 3. Return f

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15 Max-flow min-cut theorem Consider the set S of all vertices reachable from s So, s is in S, but t is not in S No incoming flow coming in S (otherwise push back) Achieve full capacity from S to T Min cut!

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16 Integrality theorem If every edge has integer capacity, then there is a flow of integer value.

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17 Complexity Assume edge capacity between 1 to C At most nC iterations Finding an s-t path can be done in O(m) time Total running time O(nmC)

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18 Speeding up Capacity scaling (find paths with large capacity) Find a shortest s-t path time Preflow-push

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19 Faster Algorithms

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20 Even Faster Algorithms

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21 Applications of the algorithm of the min-max theorem of the integrality theorem

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22 Multi-source multi-sink A set of sources S = {s1,…,sk} A set of sinks T = {t1,…,tm} Maximum flow from S to T

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23 Bipartite matching Bipartite matching <= Maximum flow

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24 Disjoint paths Find the maximum number of disjoint s-t paths directed edge => directed vertex (vertex splitting) directed vertex => undirected vertex (bidirecting) undirected vertex => undirected edge (line graph)

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25 Minimum Path Cover Given a directed graph, find a minimum number of paths to cover all vertices Directed graph => Bipartite graph

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26 Matrix Rounding Round the entries to “keep” the row sums and column sums

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27 League winner See if your favorite team can still win the leaque

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28 Bonus Question 3 (25%) In a soccer tournament of n teams, every pair of teams plays one match. The winner gets 3 points, the loser gets 0, while both teams receive 1 point in a draw. Is there a polynomial algorithm to decide whether a given score sequence (a score for each team) can be the score sequence at the end of a valid championship? League winner version is NP-hard.

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