Exercises 26. 2-1 In Figure 26. 1(b), what is the flow across the cut ? What is the capacity of this cut? Solution



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Solution

The edge connectivity of some graph is the minimum number of edges over some minimum cut of , where is some flow network in which and and for each . for each . For all , and remains constant where for each .

The edge connectivity of a graph depends on the connectivity between individual nodes. If the number of edges over the minimum cut where and is k, then there must be k paths between and that must be broken. This is due to the fact that each edge has a capacity of one, and paths cannot “share edges”; otherwise, removing the shared edge would break multiple paths. More precisely, capacity constraint will delete some edge from path in . The next path to consider need not include as and . Once the two nodes with the fewest distinct paths are found, the edge connectivity is found.

Every possible must be considered unless some minimum cut yields one edge. Therefore, at most flow networks must be constructed, each with a unique .



26.2-10

Suppose that a flow network has symmetric edges, that is, if and only if . Show that the Edmonds-Karp algorithm terminates after at most iterations.

Solution

From the time becomes critical to the next time it becomes critical, and increases by at least 2. The maximum number augmentations is , so iterations are possible.
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