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Overview of the Max-flow problem with sample code and example problem. Georgi Stoyanov Sofia University http://backtrack-it.blogspot.com Student at

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1. Definition of the problem 2. Where does it occur? 3. Max-flow min-cut theorem 4. Example 5. Max-flow algorithm 6. Run-time estimation 7. Questions 2

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Maximum flow problems Finding feasible flow Through a single -source, -sink flow network flow network Flow is maximum Many problems solved by Max-flow The problem is often present at algorithmic competitions

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Additional definitions Edge capacity – maximum flow that can go through the edge Residual edge capacity – maximum flow that can pass after a certain amount has passed residualCapacity = edgeCapacity – alreadyPassedFlow Augmented path – path starting from source to sink Only edges with residual capacity above zero 5

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In any kind of network with certain capacity Network of pipes – how much water can pass through the pipe network per unit of time? 7

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Electricity network – how much electricity can go through the grid? 8

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The internet network – how much traffic can go through a local network or the internet? 9

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In other problems Matching problem Group of N guys and M girls Every girl/guy likes a certain amount of people from the other group What is the maximum number of couples, with people who like each other? 10

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Converting the matching problem to a max-flow problem: We add an edge with capacity one for every couple that is acceptable We add two bonus nodes – source and sink We connect the source with the first group and the second group with the sink 11

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The max-flow min-cut theorem states that in a flow network, the maximum amount of flow passing from the source to the sink is equal to the minimum capacity that when removed in a specific way from the network causes the situation that no flow can pass from the source to the sink.flow networksourcesink 13

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Example 15 min( cf(A,D), cf(D,E), cf(E,G)) = min( 3 – 0, 2 – 0, 1 – 0) = min( 3, 2, 1) = 1 maxFlow = maxFlow + 1 = 1

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Example 16 min( cf(A,D), cf(D,F), cf(F,G)) = min( 3 – 1, 6 – 0, 9 – 0) = min( 2, 6, 9) = 2 maxFlow = maxFlow + 2 = 3

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Example 17 min( cf(A,B), cf(B,C), cf(C,D), cf(D,F), cf(F,G)) = min( 3 – 0, 4 – 0, 1 – 0, 6 – 2, 9 - 2) = min( 3, 4, 1, 4, 7) = 1 maxFlow = maxFlow + 1 = 4

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The flow in the previous slide is not optimal! Reverting some of the flow through a different path will achieve the optimal answer To do that for each directed edge (u, v) we will add an imaginary reverse edge (v, u) The new edge shall be used only if a certain amount of flow has already passed through the original edge! 18

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Example 19 min( cf(A,B), cf(B,C), cf(C,E), cf(E,D), cf(D,F), cf(F,g) ) = min( 3 – 1, 4 – 1, 2 – 0, 0 – -1, 6 – 3, 9 - 3) = min( 2, 3, 2, 1, 3, 6 ) = 1 maxFlow = maxFlow + 1 = 5 (which is the final answer)

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The Edmonds-Karp algorithm Uses a graph structure Uses matrix of the capacities Uses matrix for the passed flow 21

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The Edmonds-Karp algorithm Uses breadth-first search on each iteration to find a path from the source to the sink Uses parent table to store the path Uses path capacity table to store the value of the maximum flow to a node in the path 22

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23 #include

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The “heart” of the algorithm: 24 int edmondsKarp(int startNode, int endNode) { int maxFlow=0; int maxFlow=0; while(true) { while(true) { int flow=bfs(startNode, endNode); int flow=bfs(startNode, endNode); if(flow==0) break; if(flow==0) break; maxFlow +=flow; maxFlow +=flow; int currentNode=endNode; int currentNode=endNode; while(currentNode != startNode) { while(currentNode != startNode) { int previousNode = parentsList[currentNode]; int previousNode = parentsList[currentNode]; flowPassed[previousNode][currentNode] += flow; flowPassed[previousNode][currentNode] += flow; flowPassed[currentNode][previousNode] -= flow; flowPassed[currentNode][previousNode] -= flow; currentNode=previousNode; currentNode=previousNode; } } return maxFlow; return maxFlow;}

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Breadth-first search 25 int bfs(int startNode, int endNode) { memset(parentsList, UNINITIALIZED, sizeof(parentsList)); memset(parentsList, UNINITIALIZED, sizeof(parentsList)); memset(currentPathCapacity, 0, sizeof(currentPathCapacity)); memset(currentPathCapacity, 0, sizeof(currentPathCapacity)); queue q; queue q; q.push(startNode); q.push(startNode); parentsList[startNode]=-2; parentsList[startNode]=-2; currentPathCapacity[startNode]=INF; currentPathCapacity[startNode]=INF;...

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26... while(!q.empty()) { while(!q.empty()) { int currentNode = q.front(); q.pop(); int currentNode = q.front(); q.pop(); for(int i=0; i

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Breaking down the algorithm: The BFS will cost O(E) operations to find a path on each iteration We will have total O(VE) path augmentations (proved with Theorem and Lemmas) This gives us total run-time of O(VE*E) 27

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There are other algorithms that can run in O(V ³ ) time but are far more complicated to implement ! Note - this algorithm can also run in O(V ³ ) time for sparse graphs 28

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Perks of using the Edmonds-Karp algorithm Runs relatively fast in sparse graphs Represents a refined version of the Ford- Fulkerson algorithm Unlike the Ford-Fulkerson algorithm, this will always terminate It is relatively simple to implement 29

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Many problems can be transformed to a max-flow problem. So keep your eyes open! The Edmonds-Karp algorithm is: fairly fast for sparse graphs – O(V ³ ) easy to implement runs in O(VE ² ) time 30

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Don’t forget to add the reverse edges to your graph! The algorithm Looks for augmenting path from source to sink on each iteration Maximum flow == smallest residual capacity of an edge in that path 31

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Video lectures (in kind of English) http://www.youtube.com/watch?v=J0wzih3_5Wo http://www.youtube.com/watch?v=J0wzih3_5Wo http://en.wikipedia.org/wiki/Maximum_flow_problem http://en.wikipedia.org/wiki/Maximum_flow_problem http://en.wikipedia.org/wiki/Edmonds%E2%80%93Kar p_algorithm http://en.wikipedia.org/wiki/Edmonds%E2%80%93Kar p_algorithm http://en.wikipedia.org/wiki/Edmonds%E2%80%93Kar p_algorithm http://en.wikipedia.org/wiki/Matching_(graph_theory) http://en.wikipedia.org/wiki/Matching_(graph_theory) Nakov’s book: Programming = ++Algorithms; 32

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