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# Graph Theory 1016 467

RIT

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This 2 page Class Notes was uploaded by Mr. Eladio Murphy on Monday October 19, 2015. The Class Notes belongs to 1016 467 at Rochester Institute of Technology taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/225062/1016-467-rochester-institute-of-technology in Applied Math And Statistics at Rochester Institute of Technology.

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Date Created: 10/19/15

122 124 126 128 Graph Theory 1016 467 pring 005 Set 1 Remarks and Solutions The graph is K 5 This can be shown by the brute force method of simply checking all of the vertex connections directly However to reduce the work you might consider making use of the symmetries in this problem S itself is symmetric about 0 that is S S This together with the given rules for joining vertices allows us to prove that v w is an edge if and only if v w is an edge Some direct checking is still needed but it is much less than if no notice is taken of the symmetries Let G be a disconnected graph Then VG l2 2 If IV G 2 then G E2 so G K2 which has diameter 1 Therefore suppose IV G I2 3 Claim If v and w are vertices in then dvw s 2 Proof If v w E then dvw 1 and our claim is satisfied Suppose therefore that v w E Then v w E As G is disconnected there must be a vertex 2 not joined to v by any path in G In particular vz E and so vz E Also were zw E then zwv would be a path joining z and v in G which is forbidden Therefore zw E and so zw E With this vzw is now a path in Gjoining v and w That is dvw 2 and the claim is proved As dvw s 2 for all vertices v and win we have diamG s 2 i The graph G1 is bipartite with bipartition A r1u1w1t1y1q1 and B x1v1s1zl The graph G2 is not bipartite Suppose it were with bipartition A B and v2 E A Now vzwzrzxzu2 v2 is a cycle oflength 5 in G2 So we have v2 r2u2E A and wz x2v2 E B As v2 E Am B we havea H i contradiction and so no bipartition of G2 is possible For a given nonempty set of integers V we construct a graph G as follows Put VG V and join vertices v and w with an edge if and only if v w E 1mod 2 Claim G is bipartite Proof Put AvvE0mod2andBwwE1mod2 If v wEAthen vwE0 1mod2 If v wE Bthen vwE2 1mod2 If vEA and wEB then v w E 1 mod2 Therefore a pair of vertices are joined by an edge in G if and only if one vertex is in A and the other in B If either A Q or B Q then G is edgeless If both A Q and B Q then G is the complete bipartite graph KIAHBI In this problem recall the usual labeling of Q n where adjacent bit strings differ in exactly one place Parts a and b can then be regarded as asking for special subgraphs of Q n Note A union of graphs is taken as a disjoint union In particular the expression G U G denotes the graph made up of two disjoint copies of G a i R2 Q 2K2UK2 ii Adjacent vertices in Q 3 differ in 2 or 3 places Removing from each vertex the edge that joins it to its binary complement gives R3 When you ve done this you see that R3 K 4 U K 4 shown 22 24 O 9quot 0 P below 000 1 1 1 110 100 101 010 If two 3 bit binary strings differ in exactly 3 places then each is the binary complement of the other So in S3 each vertex is joined only to its binary complement Therefore S3 K2 U K2 U K2 U K2 In the graphs C for n 2 3 every vertex has degree 2 and so no vertex has odd degree The graph K2 X C3 has six vertices all of degree 3 It is not a complete graph since it has just 9 edges while K 6 has 15 Think of the given condition this way Every edge joins vertices of different degrees 1 1 The easiest way is to take the complement of your answer in part c Vertices that are not adjacent will then have different degrees This complement looks a bit busy if you simply draw it using the above configuration It looks nicer if you move things about so that there are no unsightly crossings 3 3 Let G be a graph with 6 vertices and 10 edges If 6G 3 and AG 4 then every vertex has degree 3 or 4 Let x denote the number of degree 3 vertices y the number of degree 4 vertices Then x y 6 and 3x 4y 2 10 20 Solving this system we get x 4 and y 2 To get the desired example start with K2 XC3 a graph with six vertices all of degree 3 and add a single edge joining some pair of vertices that are not already adjacent

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