Overview of the Max-flow problem with sample code and example problem. Georgi Stoyanov Sofia University Student at

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

  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

 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

 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

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

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

 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

 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

  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

 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

 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

 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

 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

 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)

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

 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

23 #include<cstdio>#include<queue>#include<cstring>#include<vector>#include<iostream> #define MAX_NODES 100// the maximum number of nodes in the graph #define INF // represents infity #define UNINITIALIZED -1// value for node with no parent using namespace std; // represents the capacities of the edges int capacities[MAX_NODES][MAX_NODES]; // shows how much flow has passed through an edge int flowPassed[MAX_NODES][MAX_NODES]; // represents the graph. The graph must contain the negative edges too! vector graph[MAX_NODES]; //shows the parents of the nodes of the path built by the BFS int parentsList[MAX_NODES]; //shows the maximum flow to a node in the path built by the BFS int currentPathCapacity[MAX_NODES];

 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;}

 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;...

26... while(!q.empty()) { while(!q.empty()) { int currentNode = q.front(); q.pop(); int currentNode = q.front(); q.pop(); for(int i=0; i<graph[currentNode].size(); i++) { for(int i=0; i<graph[currentNode].size(); i++) { int to = graph[currentNode][i]; int to = graph[currentNode][i]; if(parentsList[to] == UNINITIALIZED if(parentsList[to] == UNINITIALIZED && capacities[currentNode][to] - flowPassed[currentNode][to] > 0) { && capacities[currentNode][to] - flowPassed[currentNode][to] > 0) { parentsList[to] = currentNode; currentPathCapacity[to] = min(currentPathCapacity[currentNode], capacities[currentNode][to] - flowPassed[currentNode][to]); capacities[currentNode][to] - flowPassed[currentNode][to]); if(to == endNode) return currentPathCapacity[endNode]; q.push(to); } } } return 0; return 0;}

 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

 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

 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

 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

 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

 Video lectures (in kind of English)    p_algorithm p_algorithm p_algorithm   Nakov’s book: Programming = ++Algorithms; 32

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