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Prove that in any graph with edge-weights that are distinct (no pair of edges exists with...

Prove that in any graph with edge-weights that are distinct (no pair of edges exists with the same weight), then both Prim and Kruskal must output the same MST, regardless of the starting node.

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Suppose there are two minimum trees, A and B. Let e be the edge in just one of A,B with the smallest cost. Suppose it is in A but not B. Suppose e is the edge PQ. Then B must contain a path from P to Q which is not simply the edge e. So if we add e to B, then we get a cycle. If all the other edges in the cycle were in A, then A would contain a cycle, which it cannot. So the cycle must contain an edge f not in A. Hence, by the definition of e (and the fact that all edge-costs are different) the cost of f must be greater than the cost of e. So if we replace f by e we get a spanning tree with smaller total cost. Contradiction.

Hence, it is proved that in any graph with edge-weights that are distinct (no pair of edges exists with the same weight), then both Prim and Kruskal must output the same MST, regardless of the starting node.

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