Question

Let G be a weighted undirected graph with n vertices. Let G' be a graph that is identical to G except that every edge weight is increased by a constant value c. Prove or disprove the following statement. Every MST in G is also an MST in G'.

Answer 1

Answer:

Given statement:

Every MST in G is also an MST in G'.

Given statement is true. We can prove this by using Kruskal's algorithm.

"Kruskal's algorithm processes the edges in order of their weight values (smallest to largest), taking for the MST (coloring black) each edge that does not form a cycle with edges previously added, stopping after adding V-1 edges. The black edges form a forest of trees that evolves gradually into a single tree, the MST."

Kruskal's algorithm accesses the edge weights only through the compareTo() method. Adding a positive constant to each weight (or multiplying by a positive constant) won't change the result of the compareTo() method.

And here G' graph is identical to G in which only every edge weight is increased by a constant value c. . So if we increase all weights, then order of edges won’t change.

So we can say that Every MST in G is also an MST in G'.

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