Prove that Prim's algorithm produces a minimum spanning tree of a connected weighted graph.
Ans: Suppose G be a connected, weighted graph. At each iteration of Prim's algorithm, an edge should be found that connects a vertex in a subgraph to a vertex outside the subgraph. As G is connected, there will all time be a path to each vertex. The output T of Prim's algorithm is a tree, as the edge and vertex added to T are connected. Suppose T1 be a minimum spanning tree of G. If T1=T then T is a minimum spanning tree. If not, let e be the first edge added throughout the construction of T that is not in T1, and V be the set of vertices connected by the edges added previous to e. After that one endpoint of e is in V and the other is not. As T1 is a spanning tree of G, there is a path in T1 joining the two endpoints. As one travels along with the path, one should encounter an edge f joining a vertex in V to one that is not in V. Now here, at the iteration while e was added to T, f could as well have been added and it would be added in place of e if its weight was less than e. As f was not added, we conclude that w(f) ≥ w(e).
Suppose T2 be the graph acquired by removing f and adding e from T1. It is simple to show that T2 is connected, has similar number of edges as T1, and the total weights of its edges is not larger as compared to that of T1, therefore it is as well a minimum spanning tree of G and it consists of e and all the edges added before it throughout the construction of V. Repeat the steps above and we will eventually acquired a minimum spanning tree of G that is similar to T. This depicts T is a minimum spanning tree.