Suppose you are given a weighted graph $\mathbf{G}\mathbf{=}\mathbf{(}\mathbf{V}\mathbf{,}\mathbf{E}\mathbf{)}$ with a distinguished vertex s and where all edge weights are positive and distinct. Is it possible for a tree of shortest paths from s and a minimum spanning tree in G to not share any edges? If so, give an example. If not, give a reason.

If somehow the weighted graph G is directed, a tree with shortest routes from "s " can exist, and the lowest spanning tree in graph "G " will just not share any edge weights regard as the following scenario. Moreover, judge the following counter-directed graph:

• Using Kruskal's approach, find the least spanning tree. The following is the minimal spanning tree for something like a directed graph:

• Therefore least expense of road $3\to 1is"1"$cost for path $3\to 2is"2,"$and cost of path $3\to 4is"3"$of the presented graph $1+2+3=6$.Consider the counter directed graph is given below

• Utilising Dijkstra's method, find the optimum from the vertex " 1." The shortest path tree for a directed graph is then shown as follows:

• Our smallest path's edges being, as well as the least cost of path $1\to 2is"6"$path $1\to 2is"6"$and path $4\to 3is"5"$of the given graph $6+4+5=15$ .Therefore, the directed graph is possible to find the minimum spanning tree without sharing any edge weights.

Assuming G seems to be an undirected weighted graph. Then, without expressing any edge weights, finding its minimum spanning tree is impossible. To determine the shortest path tree, MST must share at least one edge weight.

• If the weighted graph G is undirected, then use cut property to find the minimum cut edge for shortest paths from “s” and minimum spanning tree in graph “G”. Because, MST is unique and edge weights are distinct.

• According the Dijkstra’s algorithm, the minimum cut edge is needed to find the shortest path tree.