1) Professor Sabatier conjectures the following converse of Theorem 23.1. Let G=(V,E) be a connected, undirected graph with a real-valued weight function w defined on E. Let A be a subset of E that is included in some minimum spanning tree for G, let (S,V−S) be any cut of G that respects A, and let (u,v) be a safe edge for A crossing (S,V−S). Then, (u,v) is a light edge for the cut. Show that the professor's conjecture is incorrect by giving a counterexample.
2) Let ee be a maximum-weight edge on some cycle of connected graph G=(V,E). Prove that there is a minimum spanning tree of G′=(V,E−{e}) that is also a minimum spanning tree of G. That is, there is a minimum spanning tree of G that does not include e.
2. Let T be a minimum spanning tree of G'. Then T is also a spanning tree of G since G' and G contain the same vertices.
Assume that T is not a minimum spanning tree.
The only difference between G and G' is the edge e. So if T is not a minimum spanning tree of G then there must be a tree T' in G that is a minimum spanning tree with weight less than T and containing the edge e.
But e is a maximum edge on a cycle. So remove the edge e from T' and add an edge (x,y) from the cycle that is not already in T' to make T''.
T'' must be a tree with weight less than T' since the edge (x,y) has weight less than e (since e is maximum).
But then T' is not a minimum spanning tree, which is a contradiction.
1) Professor Sabatier conjectures the following converse of Theorem 23.1. Let G=(V,E) be a connected, undirected...
2. Let G = (V, E) be an undirected connected graph with n vertices and with an edge-weight function w : E → Z. An edge (u, v) ∈ E is sparkling if it is contained in some minimum spanning tree (MST) of G. The computational problem is to return the set of all sparkling edges in E. Describe an efficient algorithm for this computational problem. You do not need to use pseudocode. What is the asymptotic time complexity of...
Let G = (V, E) be a weighted undirected connected graph that contains a cycle. Let k ∈ E be the edge with maximum weight among all edges in the cycle. Prove that G has a minimum spanning tree NOT including k.
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Problem 3's picture are given below. 5. (a) Let G = (V, E) be a weighted connected undirected simple graph. For n 1, let cycles in G. Modify {e1, e2,.. . ,en} be a subset of edges (from E) that includes no Kruskal's algorithm in order to obtain a spanning tree of G that is minimal among all the spanning trees of G that include the edges e1, e2, . . . , Cn. (b) Apply your algorithm in (a)...
Let (u, v) be a minimum-weight edge in a connected graph G. Show that (u, v) belongs to some minimum spanning tree of G.
Let G=(V, E) be a connected graph with a weight w(e) associated with each edge e. Suppose G has n vertices and m edges. Let E’ be a given subset of the edges of E such that the edges of E’ do not form a cycle. (E’ is given as part of input.) Design an O(mlogn) time algorithm for finding a minimum spanning tree of G induced by E’. Prove that your algorithm indeed runs in O(mlogn) time. A minimum...
IN JAVA Given is a weighted undirected graph G = (V, E) with positive weights and a subset of its edges F E. ⊆ E. An F-containing spanning tree of G is a spanning tree that contains all edges from F (there might be other edges as well). Give an algorithm that finds the cost of the minimum-cost F-containing spanning tree of G and runs in time O(m log n) or O(n2). Input: The first line of the text file...
Let G= (V, E) be a connected undirected graph and let v be a vertex in G. Let T be the depth-first search tree of G starting from v, and let U be the breadth-first search tree of G starting from v. Prove that the height of T is at least as great as the height of U
Let G = (V;E) be an undirected and unweighted graph. Let S be a subset of the vertices. The graph induced on S, denoted G[S] is a graph that has vertex set S and an edge between two vertices u, v that is an element of S provided that {u,v} is an edge of G. A subset K of V is called a killer set of G if the deletion of K kills all the edges of G, that is...
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