Question: Problem 7.18 (a) Solve the following problems by reducing to the original max-flow problem Algorithms [Dasgupta, Papadimitriou, Vazirani][DPV] 1st edition There are many common variations

Problem 7.18 (a) Solve the following problems by reducing to the original max-flow problem

Algorithms [Dasgupta, Papadimitriou, Vazirani][DPV] 1st edition

There are many common variations of the maximum flow problem. Here are four of them.

Solve the following problems by reducing to the original max-flow problem. In other words, explain how to solve the new flow variant problem using an algorithm for solving max-flow as a black-box. Explain how to take an input for the new problem and define an input for the original max-flow problem. Then given a max-flow F* to this input you just defined, explain how to get the solution to the new problem.

describe your algorithm in words; no pseudocode.

(a) There are many sources and many sinks, and we wish to maximize the total flow from all sources to all sinks

my answer: (not sure if #5 is enough to satisfy the multi Source/Sink and is the steps for max-flow are 100% correct in 1-4)

1. Start with a directed graph Problem 7.18 (a) Solve the following problems by reducing to the original

2. Find fe for all max-flow problem Algorithms [Dasgupta, Papadimitriou, Vazirani][DPV] 1st edition There are many common a valid flow of maximum capacity

3. capacity constraint for all variations of the maximum flow problem. Here are four of them. Solve and flow-in to V = flow-out of V.

4. using BFS algorithm over a subset of the graph to find a valid flow of maximum capacity in the following problems by reducing to the original max-flow problem. In other path words, explain how to solve the new flow variant problem using an

5. to compensate for the multiple sources and sinks we add and artificial Source s and an artificial Sink algorithm for solving max-flow as a black-box. Explain how to take an which connects to all sources and sinks.

running time: this is reduced to the max-flow problem therefore the run time is O(|E|) for all edges explored and O(|V|) for all vertices explored so the total run time is O(|V|+|E|).

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