Let G=(V,E) (where V(G) is a finite, non-empty set: the set of vertices (nodes)
ID: 3184064 • Letter: L
Question
Let G=(V,E) (where V(G) is a finite, non-empty set: the set of vertices (nodes) of G AND E(G) is the set of edges (links) of G) be a connected graph, and s,te V are two of its vertices where d-dc(s,t)21. We consider the graph H having as vertices all paths from s to t in G of length d. Two paths P and Q from s to t in G of length d are adjacent in H only if their sets of vertices V(P) and V(Q) satisfy the following condition IV(P)AV(Q)I I(V(P)-V(Q)U(V(Q)- V(P)) 2(they are different in only one vertex) Prove that if graph G is a chordal graph then graph H is connected and its diameter is at most d-1Explanation / 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:
vVdeg(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:
vVdeg(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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