Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
Math Topics
»
discretemath
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Graph Theory / Trees / Minimal Spanning Trees / Kruskal's algorithm
Replies:
1
Last Post:
Sep 10, 1999 4:46 AM




Graph Theory / Trees / Minimal Spanning Trees / Kruskal's algorithm
Posted:
Sep 9, 1999 1:35 PM


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.



