Question

(Station closing) Suppose you are in charge of a rail network connecting a large number of towns. Recently, several cases of a contagious disease have been reported in a town A. Your goal is to close down certain railway stations to prevent the disease from spreading to your hometown B. No trains can pass through a closed station. To minimize expense, you want to close down as few stations as possible. However, you cannot close the station A, as this would expose you to the disease, and you cannot close the station B, because then you couldn't visit his favorite pub. Describe and analyze an efficient algorithm, based either on max-flow or min-cut, to find the minimum number of stations that must be closed to block all rail travel from A to B. The rail network is represented by an undirected graph, with a vertex for each station and an edge for each rail connection between two stations. Two special vertices represent the stations in A and

Answer 1

The main idea is careful bookkeeping of the flows assigned to different routes from the origin node to the destination node. The steps in the method are:

1.Find any path from the origin node to the destination node that has a strictly positive flow capacity remaining. If there are no more such paths, exit.

2.Determine f, the maximum flow along this path, which will be equal to the smallest flow capacity on any arc in the path (the bottleneck arc).

3.Subtract f from the remaining flow capacity in the forward direction for each arc in the path. Add f to the remaining flow capacity in the backwards direction for each arc in the path.

4.Go to Step1.

Ford and Fulkerson method

The algorithm terminates after the last path is found in below Figure. No more strictly positive flow paths can be found between A and G. This is obvious since all paths must pass through the set of arcs B-E, D-E, F-E, and F-G, and these arcs have all had their flow capacities in the forward direction reduced to zero.

When the algorithm terminates, the maximum total simultaneous flow of vehicles from A to G is given by summing the flows on the 5 paths we selected: 4 + 3 + 4 + 2 + 1 = 14 vehicles per minute. But what is the actual

Pattern of flows that gives this optimum? How much flow should go on each arc, and in which direction? This is found by looking at the difference between the initial flow capacity and the final flow capacity: a positive difference indicates a flow in

When the algorithm terminates, the maximum total simultaneous flow of vehicles from A to G is given by summing the flows on the 5 paths we selected: 4 + 3 + 4 + 2 + 1 = 14 vehicles per minute. But what is the actual

Pattern of flows that gives this optimum? How much flow should go on each arc, and in which direction? This is found by looking at the difference between the initial flow capacity and the final flow capacity: a positive difference indicates a flow in direction (a negative difference isignored). The pattern of flows – and their directions – which gives the maximum imultaneous flow of vehicles per minute, is shown in Figure 9.7. The arc

labels in Figure 9.7 show the amount of flow in each arc. Note

that the principle of flow conservation at a node is respected. For example, the flows entering node F total 7 vehicles per minute, as do the flows leaving node F.

You probably noticed that it becomes harder and harder to find a strictly positive flow path as the algorithm progresses and all the easy-to-spot paths are used up. You might think this would be a problem in a computer implementation of the met hod, but it turns out that simple depth-first and

breadth-first searches are quite efficient for finding positive flow paths.

$kwOAAAAABJRU5ErkJggg==$

the flow capacities in the backwards directions on the arcs. This is because the backwards capacities that are added are a bookkeeping convention to indicate flow that can be undone if needed. This did not happen in our example, but 3^rd shows a simple example in which the backwards capacities are used in reaching a larger total flow. As you see, after the first path is chosen, the only way for the second path to route more flow from A to B is by undoing the flow placed on the vertical arc by the first path.