MHB How Does Unique Edge Weighting Guarantee a Singular Minimum Spanning Tree?

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In a weighted graph with unique edge weights, there is only one minimum spanning tree (MST). To prove this, one can use contradiction by assuming the existence of two MSTs with equal weight and examining an edge present in one but not the other. The application of Kruskal's Algorithm demonstrates that distinct edge costs lead to a unique sequence of edges, confirming the singularity of the MST. Additionally, properties of weighted graphs can support this conclusion. Therefore, unique edge weights guarantee a unique minimum spanning tree.
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Prove that if a weighted graph has unique weights on each edge that there is only 1 minimum spanning tree. Do I need to use an algorithm such as Prim's to show this or do I use properties of weighted graphs?
 
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Try contradiction. Assume you have two MST's of equal weight. Consider what happens with an edge that's in one MST but not the other (there must be at least one such edge if the MST's are not the same).
 
Kruskal's Algorithm can be used here as follows:

If edge costs are all distinct then running Kruskal's Algorithm on this graph G(say) creates only a unique sequence of (n-1) edges with edge costs in increasing order. Hence G has a unique minimum spanning tree.
 
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