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Course: MATH 443, Fall 2009
School: Sveriges...
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443 MATH Assignment #7 Below are short answers (hints) to assigned questions. 19A (i) There are 6 = 15 points in the incidence structure. Consider two 2 distinct edges of K6 . If they have a common endpoint, they belong to a unique triangle, and to no perfect matching. If the two edges do not have a common endpoint, they belong to a unique perfect matching, and to no triangle. Thus the design dened in part (i)...

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443 MATH Assignment #7 Below are short answers (hints) to assigned questions. 19A (i) There are 6 = 15 points in the incidence structure. Consider two 2 distinct edges of K6 . If they have a common endpoint, they belong to a unique triangle, and to no perfect matching. If the two edges do not have a common endpoint, they belong to a unique perfect matching, and to no triangle. Thus the design dened in part (i) is an S(2, 3, 15). To see the isomorphism of this design with the S(2, 3, 15) of Example 19.1 we label vertices and edges of K6 as suggested in Appendix 1. We now have a bijection between the edges of K6 and non-zero vectors of F4 . To show that 2 this bijection is an isomorphism of the two S(2, 3, 15)s we need to show that the labels of three edges forming a triangle add up to the zero vector, and that the labels of three edges forming a perfect matching add up to the zero vector. Consider the triangle of vertices labeled by vectors a, b, c. The labels of the edges are a + b, a + c, b + c and their sum is a + b + a + c + b + c = 0. Now consider the edges of a perfect matching. The sum of their labels is precisely the sum of the six labels of vertices of K6 , which is the zero vector. This nishes the proof for part (i). (ii) There are 7 = 21 points in the incidence structure. Blocks are all sets 2 of 5 edges of the following three types: To show that the design is an S3 (3, 5, 21) we have to show that each set of three edges belongs to exactly three blocks of the design. There are ve possible congurations of three edges which are denoted d, e, f, g, h below: Each set of type d is contained in exactly three blocks of type a, and in no other blocks. Each set of type e is contained in exactly three blocks of type c, and in no other blocks. Each set of type f is contained in exactly three blocks of type b, and in no other blocks. Each set of type g is contained in exactly two blocks of type b, in exactly one block of type c, and in no other blocks. Each set of type h is contained in exactly three blocks of type and c, no other blocks. We have shown that each set of three edges belongs to exactly three blocks of the design, i.e. the design is an S3 (3, 5, 21). 1 19B For any S(3, 6, v), the numbers b2 and b1 introduced in Theorem 19.3 are b2 = 1 b1 = 1 v2 32 62 32 v1 31 61 31 = = v2 4 (v 1)(v 2) 20 and they both must be integers. If (v 2)/4 is an integer then v mod 20 must be one of 2, 6, 10, 14, 18. After substituting these ve values into (v 1)(v 2)/20 we see that only the cases v 2, 6 (mod 20) yield integral values for both b2 and b1 . This nishes the proof. 19C (i) We will prove that each k-set is a block, thus showing that the design is trivial. Let P be the point set of the design, and let S be a k-subset of P. Let J = P \ S. We have |J| = v k t by the assumption. Thus Theorem 19.4 vt can be applied to J, with j = v k. We get bvk = k / kt > 0. Thus k there exists at least one block disjoint from J. Since |P \ J| = k, that block must be precisely P \ J = S. We have proved that an arbitrary k-subset of P is block, thus the design is trivial. (ii) Again let P be the point set of the design. Let J be an arbitrary t-subset of P. By Theorem 19.4 there are exactly bt blocks of the original design that a...

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