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.
can u solve 5.2 in the exercise . here s the solution of 5.1:
In the Path Edge Cover problem, we are given a directed acyclic graph A with two distinguished nodes s (source) and t (sink).
The objective is to find the minimum number of directed s-t paths that cover all edges in A.
In other words, each edge in the graph must be included in at least one of the chosen paths.
To transform A into a graph G suitable for the minimum flow problem, we perform the following steps:
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Node Splitting: For each node v in A (except s and t), we split v into two nodes vin and vout. Then we add an edge from vin to vout with a lower capacity of 1 and an upper capacity of 1. This enforces that any flow passing through v must be part of exactly one path.
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Edge Transformation: For each edge (u,v) in A, we create an edge (uout,vin) in G with a lower capacity of 0 and an upper capacity of 1. This preserves the structure of the original graph while accommodating flow constraints.
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Source and Sink: Maintaining the original source s and sink t in G , the edges connected to these nodes are treated similarly to the other edges, with lower capacity 0 and upper capacity 1.
To demonstrate the equivalence of the Path Edge Cover problem in A and the minimum flow problem in G, we establish the following two directions:
From Path Edge Cover to Minimum Flow:
A solution to the Path Edge Cover problem in A can be directly mapped to a flow in G. Each s-t path in A corresponds to a unit of flow in G. The constraints of node splitting ensure that the flow through each node is part of exactly one path, thus satisfying the requirements of the Path Edge Cover problem.
From Minimum Flow to Path Edge Cover:
Conversely, a minimum flow in G corresponds to a set of s-t paths in A. The flow passing through an edge (uout,vin) in G indicates that the edge (u,v) in A is part of an s-t path. The constraints on the capacities ensure that each edge in A is covered at least once.
This transformation can be done in polynomial time since each node and edge in A is processed a constant number of times.
The resulting graph G has a size linear in the size of A, and the flow problem can be solved using standard minimum flow algorithms.
Through these steps, we have shown that the Path Edge Cover problem in a directed acyclic graph is a special case of the minimum flow problem.
The transformation from A to G is efficient and maintains the integrity of the original problem, ensuring that solutions in one domain can be translated into solutions in the other.
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