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Week 15-16: Combinatorial Design

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Week 15-16: Combinatorial Design Week 15-16: Combinatorial Design May 8, 2017 A combinatorial design, or simply a design, is an arrangement of the objects of a set into subsets satisfying certain prescribed properties. The area of combinatorial design is highly developed, yet many interesting problems and fundamental questions still remain open. Many of the methods for constructing designs rely on the algebraic structure called ¯nite ¯eld and more general system of arithmetics. 1 Modular Arithmetic Let Z denote the set of all integers, i.e., Z = f:::; ¡2; ¡1; 0; 1; 2;:::g: We denote by + and £ the ordinary addition and multiplication of integers respectively. Let n be a positive integer with n ¸ 2 and let Zn = f0; 1; 2; : : : ; n ¡ 1g be the set of nonnegative integers which are less than n. We can think of the integers of the set Zn as the possible remainders when any integer is divided by n. If m is an integer, then there is a unique integers q (the quotient) and r (the remainder) such that m = q £ n + r; 0 · r · n ¡ 1: 1 Keep this in mind we introduce an addition, denoted ©, and a multiplica- tion, denoted ­, on Zn as follows: For any two integers a; b 2 Zn, a © b = the remainder of a + b when divided by n, a ­ b = the remainder of a £ b when divided by n. Example 1.1. For n = 2, we have Z2 = f0; 1g, and © 0 1 ­ 0 1 0 0 1 0 0 0 1 1 0 1 0 1 For n = 3, we have Z3 = f0; 1; 2g, and © 0 1 2 ­ 0 1 2 0 0 1 2 0 0 0 0 1 1 2 0 1 0 1 2 2 2 0 1 2 0 2 1 For n = 6, 4 © 5 = 3; 3 © 4 = 1; 4 © 2 = 0; 4 ­ 5 = 2; 3 ­ 4 = 0; 4 ­ 2 = 2: 2 Block Design Example 2.1. Suppose there are 7 varieties of a product to be tested for acceptability among consumers. The manufacturer plans to ask some random (or typical) customers to compare the di®erent varieties. Due to time consuming, a customer is not asked to compare all pairs of the varieties. Instead, each customer is asked to compare a certain 3 of the varieties. In order to draw a meaningful conclusion based on statistical analysis of the results, the test is to have the property that each pair of the 7 varieties is compared by exactly one person. Can such a testing experiment be designed? ¡7¢ ¡3¢ There are 2 = 21 comparisons. Each person can do 2 = 3 comparisons. 21 Then 3 = 7 person are needed for the design. This means that such a design 2 is possible. Can we actually construct one of such designs? Let us denote the 7 varieties by the set f0; 1; 2;:::; 6g. Then the following is one of such designs. f0; 1; 2g; f0; 3; 4g; f0; 5; 6g; f1; 3; 5g; f1; 4; 6g; f2; 3; 6g; f2; 4; 5g: These subsets are called the blocks of the design. Let us denote these block by B1;B2;:::;B7 respectively. Then there is an incidence relation R between the set f0; 1; 2;:::; 7g and the set B = fB1;B2;:::;B7g, de¯ned by (ai;Bj) 2 R if ai 2 Bj: The incidence relation R can be exhibited by the table (or 0-1 matrix) B1 B2 B3 B4 B5 B6 B7 0 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 2 1 0 0 0 0 1 1 3 0 1 0 1 0 1 0 4 0 1 0 0 1 0 1 5 0 0 1 1 0 0 1 6 0 0 1 0 1 1 0 Let X = fx1; x2; : : : ; xvg be a ¯nite set, whose elements are called varieties. A design on X is a collection B = fB1;B2;:::;Bbg of subsets of X; the objects Bi are called blocks, and the design is denoted by (X; B). A design (X; B) is called complete if X 2 B; otherwise, it is called incomplete. A design (X; B) is called balanced if any two members xi; xj 2 X (i 6= j) meet in B the same number of times, i.e., © ª # B 2 B : fxi; xjg ⊆ B = ¸; the number ¸ is called the index of the design. A design (X; B) is called a block design if all blocks of B have the same number of elements, i.e., there is a number k such that jBj = k for all B 2 B: The block design with the parameters v; b; k is also called a (b; v; k)-design. 3 De¯nition 2.1. Let b, v, k, and ¸ be positive integers such that v > k ¸ 2: Let (X; B) be a block design with X = fx1; x2; : : : ; xvg; B = fB1;B2;:::;Bbg; jBij = k (1 · i · b): If (X; B) is balanced, then the design is called balanced incomplete block design (or BIBD for short). For convenience we call such a BIBD a (b; v; k; ¸)- design or a BIBD(b; v; k; ¸). A design (X; B) can be completely described by the incidence matrix M, whose rows are indexed by the elements of X and whose columns are indexed by the blocks of B. The entry of the matrix M at (x; B) is de¯ned to be 1 if x 2 B; otherwise it is de¯ned to be 0. Theorem 2.2. Let (X; B) be a BIBD(b; v; k; ¸). Then each member of X meets exactly r blocks in B. More precisely, for each x 2 X, ¸(v ¡ 1) jfB 2 B : x 2 Bgj = r and r = : k ¡ 1 Proof. Let x be a member of X, and let Bl1;Bl2;:::;Blr be the blocks that contain the member x. We de¯ne a 0-1 matrix A whose rows are indexed by the members of X ¡ fxg and whose columns are indexed by the blocks Bl1;Bl2;:::;Blr; an entry (xi;Blj ) of A is 1 if and only if xi 2 Blj . We count the number of 1's in A in two ways (along the rows and along the columns): (1) Each member xi 2 X ¡fxg is contained in ¸ blocks since the pair fx; xig is contained in ¸ blocks. (2) Each block Blj contains k ¡ 1 elements of X ¡ fxg. Thus the number of 1's in the matrix A is (v ¡ 1)¸ = (k ¡ 1)r: ¸(v¡1) It follows that r = k¡1 , which is a constant for all x 2 X. 4 There are ¯ve parameters b, v, k, r, and ¸, not all independent, associated with any BIBD. Their meanings are as follows: b: the number of blocks; v: the number of varieties; k: the number of varieties contained in each block; r: the number of blocks containing any one particular variety; ¸: the number of blocks containing any one particular pair of varieties. Corollary 2.3. In any BIBD we have bk = vr; ¸ < r: Proof. Equality follows from counting the number of 1's in the incidence matrix of the design. As for the inequality, since k < v then k ¡ 1 < v ¡ 1, thus ¸(v ¡ 1) r = > ¸: k ¡ 1 Example 2.2. Is there any BIBD with the parameters b = 12, v = 16, k = 4, and r = 3? If there is such a BIBD then the index of the design is r(k ¡ 1) 3 £ 3 3 ¸ = = = v ¡ 1 15 5 which is not an integer. Thus there is no BIBD satisfying the given conditions. Example 2.3. Is there any BIBD with the parameters b = 12, v = 9, k = 3, and r = 4, and ¸ = 1? If yes, ¯nd such a BIBD? Note that ¸(v ¡ 1) 1 £ (9 ¡ 1) r = = = 4; k ¡ 1 3 ¡ 1 bk = 12 £ 3 = 9 £ 4 = vr: It seems that there is contradiction. Such a BIBD is displayed by the following 5 incidence matrix: 2 3 100100100100 6 7 6 100010001010 7 6 7 6 100001010001 7 6 7 6 010100010010 7 6 7 6 010010100001 7 6 7 6 010001001100 7 6 7 6 7 6 001100001001 7 4 001010010100 5 001001100010 Example 2.4. Let X be the set of the 16 squares of a 4-by-4 board; see below. ¤ ¤ ¤ ¤ x ¤ y x x ¤ ¤ ¤ x ¤ ¤ ¤ y ¤ y For each square x = (i; j), we take the 6 squares which are either in its row or in its column (but not the square itself) to be a block Bi;j, i.e., Bx = B(i; j) = f(i; q) 2 X : q 6= jg [ f(p; j) 2 X : p 6= ig: Let B = fB(i; j) : 1 · i; j · 4g. Then (X; B) is a BIBD with parameters b = 16, v = 16, k = 6, r = 6, and ¸ = 2. In fact, for any two squares x; y 2 X, there are two blocks shown above that contain both x and y. Thus (X; B) is a BIBD. Moreover, b = v. Theorem 2.4 (Fisher's Inequality). For any BIBD we have b ¸ v: Proof. Let A be the v-by-b incidence matrix of a BIBD. Since each variety belongs to r blocks and each pair of varieties is contained in ¸ blocks, then the (x; y)-entry of AAT is ½ X r if x = y a a = jfB 2 B : x; y 2 Bgj = xB yB ¸ if x 6= y B2B 6 where axB = 1 if x 2 B and axB = 0 if x 62 B.
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