The Edge Set Grown in Kruskal's Algorithm [closed]

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Let G = (V, E) be a weighted, connected and undirected graph. Let T be the edge set that is grown in Kruskal's algorithm and stopped after k iterations (so T might contain less than |E|-1 edges). Let W(T) be the weighted sum of this set. Let T’ be an acylic edge set such that |T| = |T’|. Prove that W(T) <= W(T’)

I understand the original proof of the algorithm and I’ve tried several approaches to tackle this, neither worked.

For example: I thought an induction on |T| might work. For |T| = 1 it’s obvious.

We assume correctness for |T|=k and prove (or not…) for k+1. Assume by contradiction that there exists an edge set T’ such that |T’|=k+1 and W(T’) < W(T).

Let e be the last edge added by Kruskal algorithm. So for any edge f in T’, W(f) < W(e) (otherwise we remove the edges from the 2 sets and get a contradiction).

This can only happen if every edge in T’ is already in T or forms a cycle with T – {e}.

…

Note: It's not the same proof as in Kruskal's algorithm. We don't even know whether T' is connected.

I have no idea what to do next. I would really appreciate any help, Thanks in advance

Answer 1

Let T’ be an edge set such that |T| = |T’|. Prove that W(T) <= W(T’).

You'll have a hard time doing that, since it's false in general.

Consider

   1
 A---B
2 \ / 3
   C
   | 4
   D

Kruskal's algorithm produces the edge set T = { (A,B), (A,C), (C,D) }, which is the unique minimal spanning tree.

But the edge set T' = { (A,B), (A,C), (B,C) } has the same cardinality as T, and

W(T') = 6 < W(T) = 7

There's some condition missing in the problem statement (like that T' should connect the graph).

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You're right. I forgot to mention that there is no cycle in T'

In that case, T' spans a tree⁽¹⁾. And since |T'| = |T| is assumed, the tree that T' spans connects the graph, i.e. is a spanning tree.

⁽¹⁾ From the absence of cycles, it follows directly that each connected component of T' is a tree. A tree with n vertices has n-1 edges. Thus if T' has k connected components, the number of vertices in the graph is

V = |T'| + k

But T is a spanning tree, and |T| = |T'|, hence

V = |T| + 1 = |T'| + 1

which implies k = 1.

Thus you are asked to simply prove the correctness of Kruskal's algorithm. You can find proofs in the literature easily, for example on wikipedia.

A proof of correctness (by induction on the number of vertices):

Lemma: Let G be a connected graph with N > 1 vertices, and T a minimal spanning tree of G. Let e be an edge in T. Then T \ {e} projects to a minimal spanning tree of the graph G' obtained from G by identifying the two endpoints a and b of e. Conversely, if T' is a set of edges of G that projects to a minimal spanning tree of G', then T' ∪ {e} is a minimal spanning tree of G.

Proof: Let p : G -> G' be the projection identifying a and b.

Then p(T \ {e}) has no cycles.

Suppose p(T \ {e}) contained a cycle C. Then p^(-1)(C) must be a path connecting a and b. But then T would contain the cycle p^(-1)(C) ∪ {e}, contradicting the premise that T is a tree.

Thus p(T \ {e}) is a cycle-free set of edges of G' with cardinality N - 2, and that implies (see above) that it is a spanning tree.

Let T'' be a minimal spanning tree of G' and S = p^(-1)(T'').

Then S ∪ {e} has no cycles.

If there were a cycle in S, that would project to a cycle in T'', so every cycle in S ∪ {e} must contain e. Suppose C were a cycle in S ∪ {e}. Then C \ {e} is a path connecting a and b, thus C \ {e} projects to a cycle in G', since a and b project to the same vertex of G'. That contradicts the premise that T'' is a tree.

So S ∪ {e} is an edge set of cardinality N - 1 without cycles, and hence (see above) a spanning tree of G.

Then W(T) <= W(S ∪ {e}) since T is a minimal spanning tree, and thus

W(p(T \ {e})) = W(T \ {e}) <= W(S) = W(T'')

Since T'' is assumed to be a minimal spanning tree of G', it follows that equality holds, and that p(T \ {e}) is a minimal spanning tree of G', and that S ∪ {e} is a minimal spanning tree of G.

Now to the induction to prove the correctness of Kruskal's algorithm:

For a graph with at most two vertices, it is obvious that the algorithm produces a minimal spanning tree.

For n >= 2, assume the correctness of the algorithm for all connected graphs with at most n vertices. (Induction hypothesis)

Let G be a connected graph with n+1 vertices. Let e be the first edge chosen in the algorithm, and a and b its endpoints.

Let G' be the graph obtained from G by identifying a and b, and p :: G -> G' the projection.

Let T be the edge set selected by the algorithm.

Then p(T \ {e}) is the edge set selected by Kruskal's algorithm on G'. Thus, by the Lemma above, T is a minimal spanning tree of G.

(Okay, probably the proof in wikipedia is simpler, but I wanted to produce a different one.)