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I need to prove that there exists only one Minimum Spanning Tree that allows the graph to have equal edges. If there are any cycles in the Graph, the edges in each cycle has distinct weights. We have to prove it assuming the Minimum Spanning Tree is calculated using Kruskal's Algorithm.

Currently this is what I have, but I do not think it is correct:

Assume that we have two Minimum Spanning Tree (R1 & R2) for Graph G. Lets select t out of the minimum weighted edges throughout all edges. We will then say that t is an edge that is only in R1. Let’s say that t is the exact path from nodes P to Q, then we know that R2 must have a path from P to Q. We know that the path from P to Q in R2 is not the edge t or else we would have a cycle in R2. One of those edges m that would be in the cycle is not in R1. By the definition of t being the pool of the smallest edged and that all edge costs are unique in cycles, the cost of m must be greater than the cost of t. We then know that the total weight of R1 is less than the total weight of R2 because t < m. This then makes it a contradiction making our assumption incorrect because only R1 is the Minimum Spanning Tree for G, therefore there would only be one Minimum Spanning Tree.

Spencer Fronberg
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  • (1) "that allows the graph to have equal edges" - equal to what? (2) Is the second sentence an assumption about the input graph, or something to be proved about the MST? Is this also a correct formulation of what you're trying to prove? *Kruskal's Algorithm generates a unique MST for any graph whose cycles consist of edges with distinct weights.* – Patrick87 Oct 17 '17 at 12:49
  • (1) Meaning the edges in the graph can have the same weight values except if there is a cycle in the graph. (2) Yes we are assuming that if there is a cycle in the graph, it has distinct weights. Just a proof that will prove there exists a MST for this graph. Doesn't necessarily need to be the Kruskal's algorithm proof. – Spencer Fronberg Oct 17 '17 at 17:29

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