This exercise asks you to prove that a certain reduction from VertexCover to Ste

Question

This exercise asks you to prove that a certain reduction from VertexCover to SteinerTree is correct. Suppose we want to find the smallest vertex cover in a given undirected graph G = (V, E). We construct a new graph H = (V , E) as follows:

• V=V E {z}

• E={ve|vV is an end point of e W}{vz|vV}.

Equivalently, we construct H by subdividing each edge in G with a new vertex, and then connecting all the original vertices of G to a new apex vertex z.

Prove that G has a vertex cover of size k if and only if there is a subtree of H with k + |E| + 1 vertices that contains every vertex in E {z}.

Explanation / Answer

Every edge has two ends. Therefore the total number of edge ends is even: It is two times the number of edges.

On the other hand, the degree of a vertex is the number of edges that end at that vertex. And since all edges have a vertex at both ends, the sum of all vertex degrees obviously also is the number of vertex edges, and thus even.

However the sum of all vertex degrees is the sum of all even vertex degrees plus the sum of all odd vertex degrees. The first one is obviously even, therefore the second one also has to be even. But a sum of odd numbers is only even if there is an even number of them.

The handshaking lemma states that for every graph G=(V,E):

vV deg(v)=2m,

so the sum vV deg(v) has to be even. This sum can be decomposed in two sums:

vV deg(v)=vV|deg(v)|=2kdeg(v)+vV|deg(v)=2k+1 deg(v),

The first is clearly even, so the second one also has to be even. But if deg(v)=2k+1, than the number of such vertices has to be even (as an odd number of odd terms cannot be even).

Every edge has two ends. Therefore the total number of edge ends is even: It is two times the number of edges.

On the other hand, the degree of a vertex is the number of edges that end at that vertex. And since all edges have a vertex at both ends, the sum of all vertex degrees obviously also is the number of vertex edges, and thus even.

However the sum of all vertex degrees is the sum of all even vertex degrees plus the sum of all odd vertex degrees. The first one is obviously even, therefore the second one also has to be even. But a sum of odd numbers is only even if there is an even number of them.

The handshaking lemma states that for every graph G=(V,E):

vV deg(v)=2m,

so the sum vV deg(v) has to be even. This sum can be decomposed in two sums:

vV deg(v)=vV|deg(v)|=2kdeg(v)+vV|deg(v)=2k+1 deg(v),

The first is clearly even, so the second one also has to be even. But if deg(v)=2k+1, than the number of such vertices has to be even (as an odd number of odd terms cannot be even).

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