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Kruskal's algorithm query
Posted:
Sep 9, 1999 6:58 PM
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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 non-decreasing 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 412-413).
>
> 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.
>
>
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