Prove that if every vertex in a graph is within distance n of a given vertex v,
ID: 3769774 • Letter: P
Question
Prove that if every vertex in a graph is within distance n of a given vertex v, then the diameter of the graph is less than or equal to 2n. Are the following two graphs G_1 and G_2 isomorphic to each other? If so, please define an isomorphism between them; otherwise, please prove that there is none. G_1 with V_1 = {1, 2, 3, 4, 5, 6}, E_1 = {12, 23, 34, 14, 15, 35, 45} G_2 with V_2 = {1, 2, 3, 4, 5, 6}, E_2 = {12, 23, 34, 45, 51, 24, 25} Prove that in every graph, there are an even number of vertices of odd degree.Explanation / Answer
it is import to observe that it need not be the case that corsponding edges of isomerphic graphs have the same weights.isomophism did not require that edge weights be preversed.therefore even If G1 and G2 are isomerphic graphs , a minimum-weight circuit in G1 would not have to also be an optimal solution in G2(although it is true that a Hamiltion circuit in one must certainly correspond to a Hamilton circuit in the other).
so that G1 with V1={1,2,3,4,6} and G2 with V2={1,2,3,4,5,6}
have same no of
It is isomorphic as the Number of vertices on both graphs are 6 and the number of edges on both of the graphs are both 7.
6.Proving that the number of vertices of odd degree in any graph G is even
We represent GG by a symmetric relation on the set of points PP, which we also call GG, so
G={(a,b),(b,a):there is an edge between a and b}
G={(a,b),(b,a):there is an edge between a and b}
Clearly, #G|2#G|2 where #G#G is the number of elements in GG. Now
deg(a)=#{(a,x):(a,x)G}
deg(a)=#{(a,x):(a,x)G}
Since we have
aPdeg(a)=aP#{(a,x):(a,x)G}=#{(x,y):(x,y)G}=#G
aPdeg(a)=aP#{(a,x):(a,x)G}=#{(x,y):(x,y)G}=#G
We know
aPdeg(a)|2
aPdeg(a)|2
From number theory we have
j=1naj|2#{aj:aj/|2}|2
j=1naj|2#{aj:aj|2}|2
(the number of odd numbers in a sum is even, iff the sum is even) and setting aj=deg(bj)aj=deg(bj) with bjPbjP an enumeration of PP, the statement follows.
Thank you
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