What is minimal edge cover?
What is minimal edge cover?
A minimal edge cover is an edge cover of a graph that is not a proper subset of any other edge cover. Every minimum edge cover is a minimal edge cover, but the converse does not necessarily hold.
What is an independent edge set?
An independent edge set (also called a matching) of a graph is a subset of the edges such that no two edges in the subset share a vertex of (Skiena 1990, p. 219). The counts of independent edge sets of size. in a graph are encoded through its matching-generating polynomial.
How do you find the largest set of independents?
Given a Binary Tree, find size of the Largest Independent Set(LIS) in it. A subset of all tree nodes is an independent set if there is no edge between any two nodes of the subset. For example, consider the following binary tree. The largest independent set(LIS) is {10, 40, 60, 70, 80} and size of the LIS is 5.
What is an independent set in a graph?
In graph theory, an independent set, stable set, coclique or anticlique is a set of vertices in a graph, no two of which are adjacent. That is, it is a set of vertices such that for every two vertices in , there is no edge connecting the two. Equivalently, each edge in the graph has at most one endpoint in. .
Is maximum independent set NP?
Maximum independent sets and maximum cliques The independent set decision problem is NP-complete, and hence it is not believed that there is an efficient algorithm for solving it. The maximum independent set problem is NP-hard and it is also hard to approximate.
How do you find the minimum vertex cover?
The size of the minimum vertex cover is 1 (by taking either of the endpoints). 3. Star: |V | − 1 vertices, each of degree 1, connected to a central node. The size of the minimum vertex cover is k − 1 (by taking any less vertices we would miss an edge between the remaining vertices).
Is edge cover NP-complete?
Computing total edge covers. We now consider the t-Total Edge Cover problem. For this problem becomes the Edge Cover problem, which has long been known to be solvable in polynomial time [8]. The problem is NP-complete for each fixed 2 ≤ t ≤ k [9, Theorem 3].