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Kruskal's algorithm query
Posted:
Sep 9, 1999 6:58 PM


I must confess I have not looked at the proof material you included below. It is not really necessary to look at it. The algorithm you give below will produce a minimum spanning tree if the edges are sorted ahead of time according to nondecreasing weights. That is, if w(e_i) \leq w(e_j) whenever i \leq j, then you get an MST. If the edges are not sorted in this way, the spanning tree need not have minimum weight.
Brian Alspach
> Hi! > Could you take a look at this ... can you find this book? Is there a > mistake or am I just too confused ... this is not supposed to be so > hard. I have seen some different proofs of this algorithms. > > Lately I saw it in the book called "Discrete Mathematics" by Kenneth > A. Ross and Charles R.B.Wright (paper back version, page 412413). > > Kruskal's algorithm in this book is like this: > Set E to empty set > For j=1 to E(G) > If E union {e_j} is acyclic replace E by E union {e_j} > End for > End > > Graph is G, T is minimal spanning tree whose edge set is E. > > I understand the first part of the proof that T is spanning tree, but > when one must show that T is minimal ... I get into trouble ... I even > started to think there is mistake in my book ... last paragraph of the > proof ... should I change S and T around ... > > Here is the version of the book (and questions) ... > To show that T is actually minimal spanning tree, consider a minimal > spanning tree S of G that has as many edges as possible in common with > T. We will show that S=T. Suppose not, and let e_k be the first edge > on the list e_1,...,e_m (are these the edges of T?) that is in T but > not in S.Let S*=S union e_k. In view of Theorem 2(d), S* contains a > cycle, say C, which must contain e_k because S itself is acyclic. > Since T is also acyclic, there must be some other edge e in C that is > not in T. Note that e is an edge of S. Now delete e from S* to get > U=S*\e=(S union e_)\e. By Theorem 1, U is connected, and since it has > the same number of edges as S, U is spanning tree by Theorem 3. > Moreover, U has one more edge, namely e_k, in common with T than S > has. Because of the choice of S, U is not minimal spanning tree. By > comparing the weights of S and U we conclude that W(e)<W(e_k), and so > e=e_i for some i<k. > (That part is OK, but at the very end ... e=e_i ... and e<k ... and e > is not in T ... a little bit of confusing, I suppose that at the very > start the edge list consists of edges in T, and in S there are other > edges as well, and then those edges of T that are in T before edge > e_k) > (Now comes the problem) > Now e is not in T, so it must have been rejected at the j=i stage for > the reason that at that time E union e contained a cycle, say C'. All > edges in E at that stage were in S, by our choice of e_k as the first > edge in S (should it be T??) but not in T (should it be S??). Since e > is also in S, C' is a cycle in S, which is a contradiction. Thus S=T, > as we wanted to show. > > Thanks. > >



