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*In which we discuss the worst-case running of the Ford-Fulkerson algorithm, discuss plausible heuristics to choose an augmenting path in a good way, and begin analyzing the “fattest path” heuristic.*

In the last lecture we proved the Max-Flow Min-Cut theorem in a way that also established the optimality of the Ford-Fulkerson algorithm: if we iteratively find an augmenting path in the residual network and push more flow along that path, as allowed by the capacity constraints, we will eventually find a flow for which no augmenting path exists, and we proved that such a flow must be optimal.

Each iteration of the algorithm takes linear time in the size of the network: the augmenting path can be found via a DFS of the residual network, for example. The problem is that, in certain cases, the algorithm might take a very long time to finish. Consider, for example, the following network.

Suppose that, at the first step, we pick the augmenting path . We can only push one unit of flow along that path. After this first step, our residual network (not showing edges out of and into , which are never used in an augmenting path) is

Now it is possible that the algorithm picks the augmenting path along which, again, only one unit of flow can be routed. We see that, indeed, it is possible for the algorithm to keep picking augmenting paths that involve a link between and , so that only one extra unit of flow is routed at each step.

The problem of very slow convergence times as in the above example can be avoided if, at each iteration, we choose more carefully which augmenting path to use. One reasonable heuristic is that it makes sense to pick the augmenting path along which the most flow can be routed in one step. If we had used such an heuristic in the above example, we would have found the optimum in two steps. Another, alternative, heuristic is to pick the shortest augmenting path, that is, the augmenting path that uses the fewest edges; this is reasonable because in this way we are going to use the capacity of fewer edges and keep more residual capacity for later iterations. The use of this heuristic would have also resulted in a two-iterations running time in the above example.

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