It works as follows:
Prim's algorithm can be shown to run in time which is O (m + n log n) where m is the number of edges and n is the number of vertices.
Let P be a connected, weighted graph. At every iteration of Prim's algorithm, an edge must be found that connects a vertex in a subgraph to a vertex outside the subgraph. Since P is connected, there will always be a path to every vertex. The output Y of Prim's algorithm is a tree, because the edge and vertex added to Y are connected to other vertices and edges of Y and at no iteration is a circuit created since each edge added connects two vertices in two disconnected sets. Also, Y includes all vertices from P because Y is a tree with n vertices, same as P. Therefore, Y is a spanning tree for P.
Let Y_{1} be any minimal spanning tree for P. If Y = Y_{1}, QED. If not, there is an edge in Y that is not in Y_{1}. Let e be the first edge that was added when Y was constructed. Let V be the set of vertices of Y  e. Then one endpoint of e is in Y and another is not. Since Y_{1} is a spanning tree of P, there is a path in Y_{1} joining the two endpoints. As one travels along the path, one must encounter an edge f joining a vertex in V to one that is not in V. Now, at the iteration when e was added to Y, f could also have been added and it would be added instead of e if its weight was less than e. Since f was not added, we conclude that w(f) >= w(e).
Let Y_{2} be the tree obtained by removing f and adding e from Y_{1}. It shows that Y_{2} is a tree that is more common with Y than with Y_{1}. If Y_{2} equals Y, QED. If not, we can find a tree, Y_{3} with one more edge in common with Y than Y_{2} and so forth. Continuing this way produces a tree that is more in common with Y than with the preceding tree. Since there are finite number of edges in Y, the sequence is finite, so there will eventually be a tree, Y_{h}, which is identical to Y. This shows Y is a minimal spanning tree.
Other algorithms for this problem include Kruskal's algorithm, and Boruvka's algorithm.
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