Show how to find the maximum spanning tree of a graph, that is , the spanning tree of largest total weight.

1. Arrange the G 's edge with strength in decreasing order. Let T denote the collection of edges that make up the greatest weight spanning tree. T should be set to.
2. Insert the first edge to T.
3. Whenever or unless the following edge doesn't really create a cycle in T, add it to T. If no further edges are available, quit and report G as disconnected.
4. Stop and produce T if T contains $n1$ edges (where n is the number of vertices in G ). If not, go to step 3. Kruskal’s Minimum spanning tree algorithm is the polar opposite of this approach.

Multiply all edge values by -1 and use any of the traditional techniques to get the least spanning tree: Prim's, Kruskal's, etc.

a.)fake When the other edge is of smaller weight, the maximum weight edge can be a component of the Minimum spanning tree, but a cycle forms in the MST as a result. It will not be taken into account in that instance, and the maximum weight edge will indeed be inserted towards the Minimum spanning tree. A graph with only one edge is an example of a counter example.

b.) TRUE. An MST has no cycles, so at least one edge of the cycle e 0 is not in an MST T. If e 0 6=e then we could swap e 0 for e in T and get a lighter spanning tree.

c.) TRUE. The minimal weight edge of a cut equals trivially this same minimum weight edge of the that cut.

d.) TRUE. If indeed the lighter edge seems to be unique, therefore the lightest edge of the any cutting whose separate both endpoints is really the lightest edge of any cut.

e.) TRUE. They could substitute e for e' and then get a shorter MST if there had been a lighter edge e' across some cut of G.

f.) FALSE. Whenever a graph contains two cycles, as well as the shortest edge is now in the left cycle, it is excluded from the MST.

g.) FALSE. If somehow the hardest edge of either a cycle has been on the shortest path as from start s to a node t , Dijkstra's algorithm will employ it.

h.) FALSE. Since the quickest route between two edges may form a cycle in Graph, it is not always a member of any MST.

i.) TRUE. This same smallest edge between both the visiting vertices and indeed the unvisited vertices, which is really the lightest edge of this cut, is always added using Prim's method. Negative weights have no effect.

j.) TRUE. Assume that somehow a graph G has an r-path between node s through node t, but that there is no r-path through s to t in an MST T of G. T thus contains a route connecting s to t with a weighted edge e. Consider the vertices partitioned $\left(\overline{)S}Vs\right)$by subtracting e fom t. Because the r-path connects $sandt$ , one of its vertices e 0 must be along this cut. We can replace $\mathrm{e}0$fore to create a spanning tree that is lighter than T.

Applying Prime's technique, the shortest edge connecting both visiting and unvisited vertices, which really is the smallest edge of the this cut, has always been inserted. Because an MST does not have any cycles, at least one edge of the cycle $\mathrm{e}0$is not present in an MST T.