Question

Let $\mathbf{G}\mathbf{=}\mathbf{(}\mathbf{V}\mathbf{,}\mathbf{E}\mathbf{)}$ be an undirected graph. Prove that if all its edge weights are distinct, then it has a unique minimum spanning tree

Short answer

Here, we get undirected graph G. Given here to prove non - negative real edge weights, and presume that even if you calculated a minimal spanning tree of G and the short distance for that kind of node, it accepts particular nodesV.

Step-by-step solution

Two alternative shortest path trees

If most of the parameters associated in such an undirected graph disagree, the graph has a single minimal spanning tree.

• Assume there are two alternative shortest path trees inside of an undirected graph, including such ${\mathrm{T}}_1\mathrm{and}{\mathrm{T}}_2$.

• Let $e_1$seems to be the minimal side weight which relates to a few of the trees, for example. $e_1\in T_1$.

• Including the side role="math" localid="1658904910955" ${\mathrm{e}}_1$into the tree $T_2$making of cycle. The cycle includes the edge of role="math" localid="1658904901130" ${\mathrm{e}}_2$in tree $T_2$, any one is bigger than ${\mathrm{e}}_1$.

• This seems to be an inconsistency for which an indifference curve has two distinct minimum spanning trees.

Proof of theoem.

Using the cut property, find the minimal spanning tree based on the graph's topology and edge weight order.

The minimal side value is already on the lowest spanning tree for each and every cut.

• Since all of the respect to the weights inside a graph are different, then all of the cuts are unique.

• If all of the edge weights in a graph are the same, the outcome may be unclear.

If all the edge weights of a graph are distinct, the smallest spanning tree is unique.

For example:

Consider the undirected graph G = (V, E) is given below: