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PS04-100323-solutions

# PS04-100323-solutions - MATH 682 Problem Set#4 This problem...

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MATH 682 Problem Set #4 This problem set is due at the beginning of class on March 23 . Below, “graph” means “simple finite graph” except where otherwise noted. 1. (10 points) The Petersen graph is shown below. a0 a1 a2 a3 a4 b0 b1 b2 b3 b4 (a) (5 points) Demonstrate that the Petersen graph is nonplanar by invoking Kura- towski’s Theorem. The above graph has been labeled with vertices a 0 , . . . , a 4 , b 0 , . . . , b 4 where each a i a i ± 1 and b i b i ± 2 . Let us rearrange this graph slightly, putting a 0 , b 2 , and a 3 on the top, and b 0 , a 2 , and a 4 on the bottom. This choice of 6 points was made to maximize the adjacencies between the top and bottom sets; specifically, we chose vertices lying on a cycle of size 6. Drawn this way, we see that this will specifically have the subgraph a0 b2 a3 a2 a4 b0 a1 b4 b3 which is a subdivision of K 3 , 3 . (b) (5 points) Demonstrate that the Petersen graph is nonplanar by invoking Euler’s theorem. First, let’s note that the Petersen graph contains no cycles of length 3 or 4. Let a 0 , . . . , a 4 , b 0 , . . . , b 4 be as above. Since every a i is adjacent to one and only one b i and vice versa, it is clear that the Petersen graph contains no C 3 , since if it were among the vertices { a i , a j , b k } , both a i and a j would need to be adjacent to b k ; likewise for { a i , b j , b k } . The triple { a i , a j , a k } also cannot be a C 3 since the only cycle among only the a i vertices is the outer C 5 , and likewise for { b i , b j , b k } . Showing no cycle of length 4 exists is only slightly more tedious. The possibility of a cycle with 4 a vertices of 4 b vertices is dispensed with above. Likewise, 3 a vertices and 1 b vertex is impossible, since b i would need to be adjacent to distinct a j and a k . A slight modification of this argument serves to eliminate the possibility of 3 b vertices and one a . Thus, there must be 2 a vertices and 2 b vertices. If they are in the order a i b j a k b a i , then we have the same problem as noted in the 3–1 split case; if, on the other hand, we have a i a j b k b a i , then it must be the case that j = k and i = , and likewise Page 1 of 4 Due March 23, 2010

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MATH 682 Problem Set #4 k ± 2 mod 5 while j i ± 1 mod 5, yielding ± 2 ≡ ± 1 mod 5, which is false. Thus, no cycle of length 4 lies in the Petersen graph. The Petersen graph has 15 edges and 10 vertices; if it were planar than by Euler’s Theorem it would have 15 - 10 + 2 = 7 faces. Since each face has a cycle on its boundary, so it is bounded by at least 5 edges, and since each edge can appear in at most 2 faces, in order to have 7 faces the Petersen graph would need at least 7 · 5 2 = 17 . 5 edges, which it does not have.
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