ON THE EXISTENCE OF (v,k,t) TRADES E.S. MAHMOODIAN and NASRIN SOLTANKHAH Department of Mathematical Sciences Sharif University of Technology p. O. Box 11365-9415 Tehran, I.R. Iran ABSTRACT A (v, k, t) trade can be used to construct new designs with various support sizes from a given t-design. H.L. Hwang (1986) showed the existence of (v, k, t) trades of volume 2t and the non-existence of trades of volumes less than 2t or of volume 2t + 1. In thIs paper, first we show that there exist (v, k, t) trades of volumes 2t + 2t - 1 (t ~ 1), 2t + 2 t - 1 + 2t - 2 (t ~ 2), 2t + 2 t - 1 + 2t - 2 + 2t - 3 (t ~ 3), and 2t+l. Then we prove that, given integers v > k > t ~ 1, there does not exist a (v, k, t) trade of volume s, where 2t < s < 2t + 2t-l. 1. Introduction Let V be a set of elements with cardinality v, and let k, t and A be positive integers such that v > k > t. A k-subset of V will be called a block. A t-(v, k, A) design (or simply a t-design) is a collection of blocks such that every t-subset of V is contained in A blocks. The number of blocks in a t-design is denoted by b. A (v, k, t) trade of volume s consists of two disjoint collections Tl and T2 , each of s blocks, such that for every t-subset the number of blocks containing this subset is the same in both Tl and T2 . When s = 0 the trade is said to be void. The definitions of t-designs and (v, k, t) trades allow repeated blocks. At-design (or a trade) having no repeated block is called a uniform design (or trade). The number of distinct blocks in a t-design is referred to as the support size of the design, and is denoted by b*. It is an easy exercise to prove that at-design (or a (v, k, t) trade) is also a t'design (or a (v,k,t') trade), for all t' with 0 < t' < t. In a (v,k,t) trade both collections of blocks cover the same set of elements. This set of elements is called the foundation of the trade. Note that in a (v, k, t) trade, some of the elements of V may not appear at all. Let D be a t-( v, k, A) design which contains the collection of blocks in Tl of a (v, k, t) trade. Then by substituting the blocks of T2 for the blocks of Tl in the design, we obtain a new t-(v, k, A) design, with possibly a new support size. This method, called the method of trade off, was introduced in 1979 Australasian Journal of Combinatorics §.( 1992), pp.279-291 by A. Hedayat and S.-Y.R. Li [2J. For a reference on the application of this method see, for example, [1] or [5J. The structure of (v, k, t) trades has been studied by H.L. Hwang [3]. She has shown that for each t and any k 2 t + 1, v 2 k + t + 1, there exists a (v, k, t) trade of volume 2t , and there exists no (v, k, t) trade of volume less than 2t or of volume 2t + 1. In constructing 3-(9,4,A) and 3-(10,4,A) designs with various support sizes, one of the present authors noted that there might not exist a (v, 4, 3) trade of volume 10, 11 or 13 [5]. We shall prove a general theorem, which as a special case implies that there is no (v, k, 3) trade of volume 10 or 11. 2. Notation To avoid triviality, unless otherwise stated, all the trades in this paper are nonvoid and v, k, t satisfy k 2 t + 1 and v 2 k + t + 1. We shall follow the notation used by Hwang [3]: (i) A (v, k, t) trade of volume s will be represented by T = Tl - T2 = 8 8 i=l i=l L Bli - L B 2i , where the Bli's and B 2i's are the blocks contained in Tl and T 2 , respectively. (ii) The foundation of a trade T will be denoted by found (T). (iii) Let D be collection of blocks and {Xl, ... , Xi} be an i-subset of V, 0 < i < k. We define AD{:J:l •... ,:J:i} number of blocks in D which contain {Xl, ... ,Xi}. To avoid messy notation, we shall use A(:J:l •...• :J:i) for AD{:J:l •...• a:d and r:J: for AD{:J:}' (iv) For a subset B of A, A \ B denotes the complement of B with respect to A. For two given (v, k, t) trades T = Tl - T2 and T* = Tl* - Tt, T \ T* = (Tl \ Tn - (T2 \ Tn and T + T* = TITI* - T2T; are also (v, k, t) trades. ( v) AB C will denote the set union of A, Band C. Now we state some of the results of Hwang [3] which are needed in the sequel. Lemma 1 [3]. Let T be a (v, k, t) trade of volume s, and 280 X E found (T) with T:J: < s, I: Tla: = I: T2:c = Bli and T;:c I: = B2i i:B2 iO):C i:B 1 i3:C T;:c Bli i:Bl~C = I: B 2i . i:B2~C Then: (i) Ta; = Tl:c - (ii) T; = Tta; - T;:c is a is a (v, k, t - 1) trade of volume Ta;; T 2:c (v -1, k, t - 1) trade of volume s - Ta;. Theorem 1 [3]. If T is a (v, k, t) trade, then: (i) Ifound (T)l ~ k +t + 1; (ii) the volume of T is at least 2t. Lemma 2 [3]. If T is a (v, k, t) trade of volume 2t , then for any i-subset of V, say (Xl, ... , Xi), 0 < i ::; t, \ I\Tda;l, ... ,:Ci} t l t i 0 = I\\ T 2{:Cl, ... ,a;i} = 2t , 2. - , ... , 2 - or. Lemma 3 [3]. If T is a (v, k, t) trade of volume s, then for any x E found (T), either Ta; = S or 2t - l ::; T:c ::; S - 2t-l. Lemma 4 [3]. If T is a (v, k, t) trade of volume s with 2t < s < 2t + 2t-l, then > 2t - l for some x E found (T). Consider the permutations on {I, ... , 2n}. For each l, l = 1, ... , n, let T:c So PI. is a permutation which sends i j into i j + 1 and i j + 1 to i j for j = 1, ... , l, and leaves all other elements fixed. Also, let Po = {(I)} consist of the identity permutation only. Define ~2n = U I.: PI. and ~2n = U even I.: Pl.' odd Theorem 3 [3]. All (v, k, t) trades of volume 2t are of the form: T = Tl - T2 = :E Su(1)Su(3) ... SU(2t+l)S2t+3 - uEA 2t +2 :E uE2i:2 t+2 281 SU(1)SU(3)",SU(2t+ 1)S2t+ 3 with Si ~ = 1, ... , 2t+3, SinSj = IS2i-11 = k. V for i and L:~i 0 for i -# j, IS2i-ll = IS2il ~ 1 for i = 1, ... , t+l, Theorem 5 [3]. There is no (v, k, 2) trade of volume 5. 3. The existence of some trades In this section we show the existence of some trades of a given volume s. First we state and prove the following useful lemma. Lemma 3.1. Let T = Tl - T2 be a (v, t + 1, t) trade of volume s. Then, based on + 2, t + 2, t + 1) trade T* of volume 2s can be constructed. T, a (v Proof. Let x and y be two new elements. Then we can construct the blocks of T* as follows. x x x x Tl T2 x x x x y y y y T2 Tl Y y Y y Figure 1 + 2, t + 2, t + 1) trade of volume 2s. Let k ~ t + 1. Then: Clearly T* is a (v Theorem 3.2. I + t + 1; + 2 1 + 2t - 2 (i) there exists a (v, k, t) trade of volume 2t for any v 2:: k (ii) there exist trades of volumes 2t + 2t - 1 (t ~ 1), 2t 2t + 2t - 1 + 2t - 2 + 2t - 3 (t ~ 3), for any v 2:: k + t + 2; (iii) there exists a trade of volume 2t+l for any v ~ 282 k + t + 3. t - (t > 2), Proof. The existence of a (v, k, t) trade of volume 2t is shown in [3]. Observing the following special trades and utilizing Lemma 3.1 the theorem is established. A (5,2,1) trade of volume 3: Tl = {13, 14, 25}, T2 = {12, 15, 34}; a (7,3,2) trade of volume 7: Tl = {123, 147, 245, 267, 347, 357, 456}, T2 = {124, 137, 256, 237, 345, 457, 467}; a (9,4,3) trade of volume 15: Tl = {1238, 1245, 1357, 1478,2359,2367,2489,2569, 2578, 3469, 3478, 3789, 4567, 4579, 4589}, T2 = {1235, 1248, 1378, 1457, 2369, 2378, 2459, 2567, 2589, 3467, 3489, 3579, 4569, 4578, 4789}; a (6,2,1) trade of volume 4: Tl = {13, 14, 25, 26}, T2 = {15, 16, 23, 24 }.I Remark. Construction of the above l~trades is easy and the 2-trade and 3-trade can be deduced from the tables of [5]. All the above trades are uniform trades. Similar non-uniform trades exist as well; for example, the following (5,2,1) trade of volume 4 is non-uniform: Tl = {13, 14, 23, 25}, T2 = {12, 12, 34, 35 }. 4. Non-existence of some trades In this section we state and prove our main theorem, which shows that, for certain volumes, there does not exist any trade. First we prove the following lemmas. Lemma 4.1. Let k ~ t + 1, and let T be a (v, k, t) trade of volume s, such that s = 2t + 2t - 2 + i, with 0 ~ i ::; 2t - 2 - 1, and let x, y E found (T). Then: (ii) if Tx = Ty = 2t - 1 + 2t - 2 + i, then Axy = 283 2t - 1 + i or 2t - 1 + 2t - 2 + i; (iii) if Tx = Ty = 2t - 1 ) then Axy = 2t - 2 or 2t - 1 , Proof. (i) Let x, y E found (T) such that Tx = 2t - 1 + 2t - 2 + i and Ty = 2t-l. Then T~ is a (v - 1, k, i-I) trade of volume 2t - I , By Lemma 2 [3], T~, the number of blocks in T~ containing y, is given by 2t - 2 or 0, Therefore Axy = 0, 2t - 2 or 2t - I , t But ifAxy 2 - l , then Tx - Ty is a (v - 1,k,i - 1) trade of volume Tx - Ty 2t - 2 + i < 2t - 1 , This is impossible by Theorem l(ii) [3]. (ii) Let x, y E found (T), Tx = Ty = 2t - 1 + 2t trade of volume 2t-l. Again by Lemma 2 [3], = 1 T'y 2t -' 2t - 2 2 + i. or Then T~ is a (v - 1, k, i-I) ° , which implies that Axy 2t - 2 + i, 2t - 1 + i or + T~) 1 2t - +2 t - 2 + i. But ifA xy = 2t - 2 + i , then T T~( is a (v, k, i - I ) trade of volume 2t - 2 + i < 2t - 2 + 2t - 2 = 2t - 1 , which is also impossible by Theorem l(ii) [3], (iii) Let x, y E found (T), Tx = Ty = 2t - 1 , Then Tx is a (v, k, i-I) trade of volume 2t-l. By taking i = 1 in Lemma 2 [3], But the impossibility ofAxy T - (Tx + Ty). I = °can be seen by considering the (v - 2, k, i-I) trade Lemma 4.2. Suppose that k > i + 1 and that T is a (v, k, i) trade of volume s. If x, Y E found (T), such that Tx + Ty = sand Axy = 0, then T' = Tt - T~, where s T; = I)Bli - {x,y}) s and T~ = ~)B2i - {x,y}), i=l i=l is a (v - 2, k - 1, i) trade, Proof. If {aI, .. " at} is a i-subset of V such that ai t/:. {x,y} for i = 1, ... , i, then we have 284 AT~(al, ... ,at) Since T is a (v, k, t) trade, AT1(al, ... ,at) = Thus T' is a (v - 2, k - 1, t) trade. I = AT2 (al, ... ,at)· AT2 (al, ... ,at)' Therefore AT{(al, ... ,at) = AT~(al, ... ,at)' Note that in Lemma 4.2, the volume of T' is less than or equal to s. Indeed, T' may be a void trade; for example, consider the following (6,3,1) trade: T = (x12) + (y34) - (y12) (x34). Lemma 4.3. If a (v, i + 1, i) trade T contains x, y such that then Axy i=- o. Tx i=- Ty and Tx +Ty = s, Proof. Suppose Axy = O. Each i-subset {aI, ... , at} such that ai i=- x for i = 1, "" t, which appears with x in the blocks of TI must appear with y in the blocks of T 2 , This implies Tx = T y , a contradiction. I Theorem 4.4. Given integers v > k > i 2: 1, there exists no (v, k, t) trade of volume s, where 2t < s < 2t + 2t-l. Proof. We proceed by induction on i. For i = 1 there is nothing to prove. For = 2, by Theorem 5 [3], there exists no (v, k, 2) trade of volume 5. Assume that the theorem holds for all values less than i (i 2: 3). We show that it holds for i also. Suppose the theorem is not true for i, and let T be a (v, k, i) trade of volume s = 2t + i, where a < i < 2t-l. We show a contradiction. There are two possible cases: i (i) a < i < 2t - 2 (ii) 2t - 2 or ::; i < 2t-l. If 0 < i < 2t - 2 , then in any trade of volume s, there exists an element x in found (T) such that Tx < s. By Lemma 3 [3], we have Tx = 2t - 1 +j, where 0::; j ::; i. If j i=- 0, then Tx is a (v, k, i-I) trade of volume 2t - 1 + j, with a < j ::; i < 2t - 2 , which is contradictory to the induction hypothesis. If j 0, then T~ is a (v - 1, k, i-I) trade of volume 2t - 1 + i, where a < i < 2t - 2 , again contradicting the induction hypothesis. Now we show that (ii) is also impossible; that is, there exists no (v, k, i) trade of volume (1) s = 2t + 2t - 2 + i, where 0::; i ::; 2t - 2 1. There exists an element x in found (T) such that Tx < s. By Lemma 3 [3] we have Tx 2t - 2 +j, with 0::; j ::; 2t - 2 +i. If 0 < j < 2t - 2 +i, then either Tx or T~ is a (v, k, t -1) trade of volume 2t - 1 + l, where 1::; 1 < 2t - 2 , and this is impossible by the induction hypothesis. Therefore, if Tx i=- s, then either Tx = 2t - 1 + 2t - 2 + i or Tx 2t-l. For simplicity put a = 2t - 1 + 2t - 2 + i 285 and b 2t-l. We should note that the existence of a (v, k, t) trade of volume s, in which precisely c elements belong to all the blocks of T, is equivalent to the existence of a (v c, k c, t) trade of volume s, having no element x with rx s. Thus we may assume that, for each element, either rx or a = b. rx Therefore, there are three cases to be investigated: Case 1: rx = b for all x E found (T); Case 2: rx = a for all x E found (T); Case 3: there exist x, y E found (T), such that rx = a (2) and ry = b. Case 1 is impossible by Lemma 4 [3]. 1, k, t - 1) trade of volume 2t - \ For Case 2, let x E found (T). T~ is a (v which by Theorem 3 [3] is of the following form: T~ = Ttx - T~x where Sj ~ = 2:UEl:I.2t Su(I)Su(3)",Su(U-l)S2t+1 - 2:<fELl2t S<f(I)S<f(3)",S<f(2t-l)SU+1J V for j = 1, ... , 2t + 1, for j =I- I., Sj n St 0 t+l IS2j-ll IS2jl ~ 1 for j = 1,2,,,., t and L IS2j-11 k. j=l But S2t+1 = 0, for if y E S2t+l, then in trade T we have ).xy = 2t - 2 + i, and this contradicts Lemma 4.1(ii). For each pair of y, z E found (T~), by Lemma 2 [3] and from the fact that SUtl = 0, we have ).'yz = 2t - 2 " 2t - 3 or 0 , (3) and r'y (As earlier ).~z, Let y E for example, denotes ).T~ S2j-1 and zESt. = 2t - 2 (4) {yz}') Then by (3) we conclude that if I. = 2j then ).~z =0 (5) and if I. =I- 2j or 2j 1 then A~z = 2t - 3 . (6) Let y E S2j-l' By (5) there are as many as 2t-l +i blocks in T lx which contain y. By (6) and Lemma 4.1(ii) every element of S2j together with y, must be contained Since this is true for every y E S2j-l, all the elements in 2t - 1 + i blocks of 286 and S2j must be included in the same blocks of T1x ' For simplicity, let U S2j for j = 1, ... , t in T 1x . Since all the elements in Aj appear in the same blocks of T lix , we can consider every set Aj in T1x as an element. in Aj S2j-l = S2j-l By (7) and Lemma 4.1(ii), every pair of Aj and A l , for U -I- f), is included in exactly 3(2 t - 3 ) + i blocks in T1x ' Now TxAl is a (v, k, t - 2) trade of volume 2t - 1 + i (see Figure 2). 3( ~+i 1 2 1 t-l +i 1-----+-- a i x x -2 Figure 2 Moreover, (Tx)Al' that is, the blocks in Tx which do not contain AI) is also a 2) trade of volume 2t-2. Each Aj) for j -I- 1, is included in as many as 2t - 1 + i (3(2t-3) + i) = 2t - 3 blocks in (Tx)A 1 ' By Theorem 3 [3], the trade (TX)Al has the following unique structure: (v, k, t UE~2t- :E SU(1)SU(3) ... SU(2t-3)S2t-l :E uE"K2t - where Sj ~ V for j 1,2, ... , 2t 1, Sj n 287 SU(1)SU(3) ... SU(2t-3)S 2t-l 2 (7) for J J = 1,2, t 1,2, ... , t 1 and L j=l _ IS2j-11 = k. 2t-2 t j=l l=2 Since U 5 j C U A.l, ;.e. 15j l ~ 1 for 2t 2, and 5 j n 5 l = 0 for J Therefore, by the pigeonhole principle, some of the 5/s must be empty, which is a contradiction. "'j For Case 3, we will call an element of type a (or b) if r:r; = a (or r:r; = b) respectively. First we show that, for k = t + 1, this case is impossible. For, let T be a (v, t + 1, t) trade and r:r; = a, ry = b, for some x, Y E found (T). Then by Lemma 4.3, A:r;y i= 0 and, consequently by Lemma 4.1(i), A:r;y = 2t-2. Thus the number of appearances of y in the blocks of T{:r; is also 2t-2. We denote by ab the number of elements of type b in T. These elements occupy ab.2t-2 places in all the blocks of T{x' Since T{:r; is of volume b 2t - 1 , the remaining k.b - ab2t-2 places, are filled with elements of type a. By Lemma 4.1(ii), every element Z E found (T{:r;) of type a appears 2t - 2 times in the blocks of T{:r;. Therefore, the number of such elements is equal to This implies that (8) where aa denotes the number of elements of type a in found (T). We know that a.aa + b.ab = s.k. Hence (9) s.k - b.ab aa=---a Therefore, (10) Now by substituting the values of a, b, and s in (11) we obtain (ll) With a similar argument we can deduce that, at the same time aa> k. (12) Inequalities (12) and (13) contradict (10). Thus there does not exist a (v, t + 1, t) trade satisfying the condition s of case 3. Now by induction on k, we show that case 3 is always impossible. Suppose that for k - 1, there does not exist a (v, k - 1, t) trade (k > t + 1) satisfying the conditions of case 3, and of volume s, where 2t < s < 2t + 2t-1. Let T be a (v, k, t) 288 trade satisfying conditions of case 3. And let x, Y E found (T), such that b. We consider two cases. Tx = a and Ty (i) Suppose Axy 0. °: :; By Lemma 4.2, T' is a (v 2, k - 1, t) trade of volume s', where s' :::; s. By Theorem l(ii) [3], it is impossible to have < Sf < 2t. If S' 2t , then T must be of the form given in Figure 3 below, with A and B identical. ° x x y y A r B i 2 b x y x y 1~ t -2 +i 1 y x x x y I Tl 1 a b j x x Y - j Figure 3 Next, we show that the rows of A (also B) are all identical. Since T' is a I, t) trade of volume 2t , then by Lemma 2 [3], for each Z E found (T') we have (13) (v - 2, k Since Z E found (T), if T z = 2t - 1 = b then it is obviously impossible to have T~ = 2t. Also, ~T i:- 0, since if T~ = 0, then z must belong only to the set of blocks in A or B (Figure 3), each of which have 2t - 2 + i blocks, where i < 2t-2. Therefore t 2 t = 2t - 1 = T Z ) implying T z :::; 2 - + i < 2 - \ a contradiction. Thus if T z = b, then T~ 289 that z does not appear either in the blocks of A or in the blocks of B. If T z = a, then a similar argument yields that z belongs to 2t -- 1 blocks of T' and therefore to all 2t - 2 + i blocks of A and also of B. Thus each set of blocks and containing 2t - 2 +i identical blocks. Therefore T~, which is a (v-l,k,t-l) trade ofvolume2 t -- 1 , must have 2t - 2 +i identical blocks in T~x, which is a contradiction of Theorem 3 [3]. If Sf 0, then any (k I)-subset S of elements (x 1- S) which appears with x in the blocks of it must appear with y in the blocks of Thus Tx = T y , a contradiction. Therefore, 2t < that T' is a (v, k - 1, t) trade of volume S' < Sf < 2t + ,"V~.L'U-JHM the induction hypothesis. Oi) Suppose Axy i- O. By Lemma 4.1(i), we have 2t-2. The non-existence of T can be proved in the same way as in the case k = t + 1. So we have established the impossibility of case 3. The given for the three cases establish the theorem. Remark. N ow that the existence and non-existence of some trades have been proven, one can ask a For what values of s does there exist a (v, k, t) trade of volume s? The answer for this question will help to attack the open question on the existence of with various support sizes questions 22 and 24 in [4]). iFrom the results of section 3 and 4, we conjecture that: 1. for every volume Si 2t +2 < S t - 1 + + 2t - i , i 0,1, ,." t, there exists a (v, k, t) trade of Si; 2. for every s, Si of volume s. < Si+l, i = 0,1, ... , t 1, there does not exist a (v, k, t) trade The authors thank Professors M. Behzad and M.A. Najafi for their reading part of the manuscript and for their helpful comments. The research is supported in part by the Sharif of Technology and the research of the second author is supported in part by the Iranian Ministry of Higher Education. References A. Hedayat and H.L. Hwang. Construction of BIB designs with various support sizes - with special emphasis for v = 8 and k = 4. J. Gombin. TheoTY SeT. A, 36:163-173, 1984. 290 [2] A. Hedayat and Shuo-Yen Robert Li. The trade-off method in the construction of BIB with variable sizes. Ann. 1979. [3] H.L. Hwang. On the structure of Inference, 13:179-191, 1986. k, t) trades. [4] S. and A. Flann. and Inference, [5] Ebadollah S. Mahmoodian. blocks. Ars Combinatoria, J. of Statist. Plann. and II. J. of On the 1990. Ql1rHVyrt 291 slze of ,)-ll.eSl PT1S statist.