5.1. In the Path Edge Cover problem, we are given a directed acyclic graph A with two distinguished nodes s and t. We wish to find a minimum number of directed s t paths that cover all edges, that is, every edge must be in at least one of the selected paths. (If no solution exists at all, this shall be recognized, too.) Clearly, this is a special case of the Set Cover problem. Show that Path Edge Cover is also a special case of our minimum flow problem fom Exercise 3. More precisely: Turn A in polynomial time into a graph G with appropriate lower and upper edge capacities, and briefly show equivalence of the problems. Do not forget that an equivalence has two directions.
5.1. In the Path Edge Cover problem, we are given a directed acyclic graph A with two distinguished nodes s and t. We wish to find a minimum number of directed s t paths that cover all edges, that is, every edge must be in at least one of the selected paths. (If no solution exists at all, this shall be recognized, too.) Clearly, this is a special case of the Set Cover problem. Show that Path Edge Cover is also a special case of our minimum flow problem fom Exercise 3. More precisely: Turn A in polynomial time into a graph G with appropriate lower and upper edge capacities, and briefly show equivalence of the problems. Do not forget that an equivalence has two directions.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can u solve 5.1
Transcribed Image Text:Exercise 5.
5.1. In the Path Edge Cover problem, we are given a directed acyclic graph
A with two distinguished nodes s and t. We wish to find a minimum number
of directed s- - t paths that cover all edges, that is, every edge must be in
at least one of the selected paths. (If no solution exists at all, this shall be
recognized, too.) Clearly, this is a special case of the Set Cover problem.
Show that Path Edge Cover is also a special case of our minimum flow
problem fom Exercise 3. More precisely: Turn A in polynomial time into
a graph G with appropriate lower and upper edge capacities, and briefly
show equivalence of the problems. Do not forget that an equivalence has
two directions.
5.2.. Combine the previous results, to show that Path Edge Cover is solvable
in polynomial time. In particular, state and motivate some polynomial time
bound, as a function of the graph size. The time bound might depend on
your approach to Exercise 4, but in any case it must be polynomial.
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