Proof of Kruskal’s Algorithm
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چکیده
(2) Minimality: The proof is by contradiction. Assume that Y is not a minimal spanning tree and among all minimum weight spanning trees, we pick Y1 which has the smallest number of edges which are not in Y . Consider the edge e which was first to be added by the algorithm to Y of those which are not in Y1. Y1 ∪ e forms a circuit. Being a tree, Y cannot contain all edges of this circuit. Therefore this circuit contains an edge f which is not in Y . The graph Y2 = Y1 ∪ e \ f is also a spanning tree and therefore its weight cannot be less than the weight of Y1(since Y1 is one of the minimal spanning tree with the smallest number of edges) and hence the weight of e cannot be less than the weight of f( i.e. e ≥ f). However, the edge e is selected at the first step by Kruskal’s algorithm which implies that it is of the minimal weight of all edges.(i.e. e ≤ f). Thus the weight of e and f are equal(e = f), and hence Y2 is also a minimal spanning tree. But Y2 has one more edge in common with Y and Y1, which contradicts to the choice of Y1.
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