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Re: Graph Theory / Trees / Minimal Spanning Trees / Kruskal's algorithm
Posted:
Sep 10, 1999 11:34 AM


> > 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 >
The edge set needs to be ordered according to edge weights here, so if (W(e) is the weight of the edge e, then W(e_1) <= W(e_2) <= .... <= W(e_{E(G)})
> 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
e_1, e_2, ... are the edges of the graph G, indexed as in the beginning so W(e_1)<= W(e_2)<=...<=W(e_m), where m=E(G).
So for every edge e_i where i< k one of the following holds: e_i is neither in S nor in T e_i is in both S and T e_i is in S but not in T and e_k is the first edge in the list such that e_k is in T but not in S.
> 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.
So you do not know at this point, what is the index of e, therefore you do not know how the weights if e and e_k compare. What you do know, however, is that e is in S but 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.
This is true, since if U would also be a minimal spanning tree, we would have chosen it instead of S (S was minimal spanning tree and had as many edges common with T as possible)
>........ 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.
This must be true, since S must have smaller total weight than U, and the only difference is that e is in S but not in U and e_k is in U but not in S. So W(e)<W(e_k), and because of the indexing, smaller weights come earlier in the list, 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'.
Yes, at the ith execution of the loop (which was before the kth, since i<k, so up till there we only have included edges in T which are edges in S) we must have rejected e=e_i, since it is not in T. The only reason we could have done it is that it induced a cycle with the already existing edges in T. But those edges are also in S, so e=e_i must induce a cycle in S as well. This gives you a contradiction with the fact that S is a tree. This imlies that e_k could not have existed, with other words there is no edge which is in T but not in S, S=T, the minimal spanning tree with as many common edges as possible with T is in fact T, so T itself is a minimal spanning tree. I hope it helped.
>.....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. >



