University f Wllngng Research Online Faculty f Infrmatics - Papers (Archive) Faculty f Engineering and Infrmatin Sciences 992 On small defining sets fr sme SBIBD(4t -, 2t -, t - ) Jennifer Seberry University f Wllngng, jennie@uw.edu.au Publicatin Details Seberry, J, On small defining sets fr sme SBIBD(4t-, 2t-, t-), Bulletin f the ICA, 4, 992, 58-62; crrigendum, Bulletin f the ICA, 6, 992, 62-62. Research Online is the pen access institutinal repsitry fr the University f Wllngng. Fr further infrmatin cntact the UOW Library: research-pubs@uw.edu.au
On small defining sets fr sme SBIBD(4t -, 2t -, t - ) Abstract We cnjecture that 2t - specified sets f 2t - elements are enugh t define an SBIBD(4t -,2t -, t - ) when 4t - is a prime r prduct f twin primes. This means that in these cases 2t - rws are enugh t uniquely define the Hadamard matrix f rder 4t. We shw that the 2t - specified sets can be used t first find the residual BIBD(2t, 4t - 2, 2t -, t, t - ) fr 4t - prime. This can then be uniquely used t cmplete the SBIBD fr t =,2,3,5. This is remarkable as frmerly nly residual designs with λ = r 2 have been cmpletable t SBIBD. We nte that nt any set f elements will d as Marshall Hall Jr fund 3 sets frm 9 which culd nt be cmpleted t an (9,9,4). We will refer t a design and its incidence matrix, with treatments as rws and blcks as clumns, interchangeably. Disciplines Physical Sciences and Mathematics Publicatin Details Seberry, J, On small defining sets fr sme SBIBD(4t-, 2t-, t-), Bulletin f the ICA, 4, 992, 58-62; crrigendum, Bulletin f the ICA, 6, 992, 62-62. This jurnal article is available at Research Online: http://r.uw.edu.au/infpapers/03
On small defining sets fr sme SBIBD(4t -, 2t -, t - ) Jennifer Seberry Department f Cmputer Science The University f New Suth Wales Australian Defence Frce Academy Canberra ACT 2600 AUSTRALIA Abstract We cnjecture that 2t - specified sets f 2t - elements are enugh t define an SBIBD(4t -,2t -, t - ) when 4t - is a prime r prduct f twin primes. This means that in these cases 2t - rws are enugh t uniquely define the Hadamard matrix f rder 4t. We shw that the 2t - specified sets can be used t first find the residual BIBD(2t,4t - 2, 2t -, t, t - ) fr 4t - prime. This can then be uniquely used t cmplete the SBIBD fr t =,2,3,5. This is remarkable as frmerly nly residual designs with A = r 2 have been cmpletable t SBIBD. We nte that nt any set f elements will d as Marshall Hall Jr fund 3 sets frm 9 which culd nt be cmpleted t an (9,9,4). We will refer t a design and its incidence matrix, with treatments as rws and blcks as clumns, interchangeably. Cnject ure Let D be the quadratic residues mdul a prime pwer 4t -. Then the residual design with treatments given by the sets D, D + d,.., D + d 2 - di E D, i =,..., 2t - can be extended, uniquely, up t permutatin f treatments using the link prperty j blcks j an SBIBD (inner prduct j the clumns) t Jrm em SBIBD(4' -,2t - l,t - ). Therem 2 The cnjecture is true fr t = 2,3,5. Prperties f given sets Write Ei = D + di, i =,... 2t - where D is the set f quadratic residues mdul 4t -. Lemma 0 D; lr any i =,...,2t-. Prf. The elements f Ei are dj + di where dj,d; E D are bth quadratic residues. If d J + di = 0 fr sme j, i then d j = -di, that is, a quadratic residue equals a quadratic nn residue. This is nt pssible fr primes 4t -. 0 Bulletin r the lea, Vlume 4 (992) 58-62
We nte frm cycltmy, writing D fr the set f quadratic residues, N fr the set f quadratic nn residues and fld, fln fr the cllectin f distinct differences between elements f D and N respectively, that fld = fln = (t - )(V/{O}) where V is the elements f GF(4t - ). V = DuN u {O}. Als fl(d-n) = (t-l)d+tn fl(n-d) = td+(t-l)n where fl(a - B) is the cllectin f elements [a - b: a E A,b E B]. () Lemma 2 E, E 2,..., E 2t - can be cmpleted t be the residual design BIBD(2t, 4t- 2,2t - l,t,t -) f an SBIBD(4t -,2t - l,t - ). Prf. Since the sets E;, i =,..., 2t - d nt cntain 0 they are defined n a set f size 4t - 2. Clearly they each cntain 2t - elements. We write dwn the 0, incidence matrix f the sets E; and btain a (2t - ) x (4t - 2) matrix, A, with 2t - nes per rw. We will nw shw A has t r t - elements per clumn and inner prduct between its rws exactly t -. We first shw that in thse clumns f A which represent a quadratic residue in sme E; there are t - nes and in thse which represent a quadratic nn residue there are t nes. Let dh be a quadratic residue then we ask hw many times is d; + dj = dh, d;, dj E D. That is, hw many times is dj = dh - d; fr fixed dj. Since D is the set f quadratic residues the equatin d j = dh - d; has t - slutins fr fixed dj. Hence the clumn f A which represents dj has t - nes. Next we cnsider the number f slutins f the equatin dj = -dh where (-dh ) is a quadratic nn residue. That is, hw many slutins are there t the equatin dj = (-dh) - d;, that is, the differences between a nn residue and a residue shuld be a residue. Equatin () frm the thery f cycltmy shws there are t slutins. We thus bserve that if we add anther rw t A which is in the clumns representing the quadratic residues we will have added the representatin f D and have a new matrix B which is 2t x (4t - 2), has 2t - nes per rw and t nes per clumn. T shw B is the required BIBD we need t shw that IEj n Ehl = IEj n DI = t - fr j, h =,..., 2t -. We first cnsider IEj n DI. Thus we want t knw hw many times an element dj + dh e Dj equals di e D, i :f: h. Hwever, di - dh = dj fr fixed d; and d; e {d,,d 2f - } has t - slutins. Hence IE; n DI = t -. Fr IEj n Ehl we need die + d; = d,l + dh, that is, the number f times w = dj - dh = dll - die has a slutin fr fixed dj, dh and this is again t -. Thus we have the required residual design BIBD(2t, 4t - 2, 2t -, t, t - ). 0 Example. ThesetsD={O,I,2,4,5,8,0}, D+l, D+2, D+4, D+5, D+ 8, D + 0 are defining sets fr an SBIBD(5, 7, 3). 59
Example. The 7 sets: 2 4 5 8 0 2 3 5 6 9 2 3 4 6 7 0 2 4 5 6 8 9 2 4 5 6 7 9 0 3 8 9 0 2 3 3 0 2 4 3 5 Example. Let D = {d,d 2,...,d 2t -d be the quadratic residues. Then D + d;, i =,2,..., 2t - are defining sets mdul 4t -, which can be uniquely cmpleted t SBIBD(4t -,2t - l,t - ) and an Hadamard matrix f rder 4t (up t permutatin f rws) fr t = 2,3,5. Case t = 2: The sets are {2,3,5}, {3,4,6}, {5,6, I}. Let us write them as a (0, ) incidence matrix giving: 23 4 5 6 U sing the fact that an SBIBD is a linked design, s each clumn has three nes and withut lss f generality the first clumn is (0,0,0,0,,, If, each ther clumn has ne in the last three rws, we see we can uniquely cmplete the furth rw s in each clumn we have a ttal f 2t - = 3 nes. S we have: 2 3 4 5 6 Nte the furth rw is the incidence matrix f D the set f quadratic residues and the first 2t = 4 rws and 4t - 2 = 6 clumns is the incidence matrix f the residual BIBD( 4,6,3,2,). Withut lss f generality we chse the (5,) = (6,2) elements ne. The inner prducts f the rws and clumns nw uniquely cmplete the design. Case t = 3: Let D = {,3,4,5,9} be the quadratic residues mdul. Then D + d;, d; ED, i =,...,5 are defining sets giving, as befre with D the residual 60
design BIBD(5,0,4,3,2) = H: ; H= 2 3 4 5 6 7 8 9 0 00 0 The first three clumns are easily cmpleted (up t permutatin f rws) and the inner prduct f the rws and clumns, as well as the rw and clumn sums, fix the remainder. Case t = 5: Let D be the quadratic residues and as befre use the sets D + d;, i =,...,9, d; E D. Again frm the incidence matrix and nte that withut lss f generality the first clumn can be written as 2t zers and 2t - nes in the 2t + t 4t - places. The fact that the SBIBD(9, 9, 4) is a linked design means each f the last 2t - rws has t - = 4 nes per clumn means the 2tth rw can be uniquely cmpleted t give a ttal f 2t - nes in each clumn f the design. Again the 2tth clmhn is the incidence matrix f D the set f quadratic residues. Nte the 2t x (4t - 2) matrix, which is the first 2t rws and the last 4t - 2 clumns, is the residual BIBD(2t,4t - 2, 2t -, t, t - ). It nw remains t shw the design can be uniquely cmpleted. \Ve have: H= 023456789ABCEFGHIJ 6
K= 0 2 3 4 5 6 7 8 9 A B C E F G H I J 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Again we have unique cmpletin up t permutatin f rws., References [] M. Hall Jr. Private cmmunicatin, 977. [2] Ken Gray. Further results n smallest defining sets f well knwn designs. Austral. J. Cmbinatrics, :9-00,990. [3] Ken Gray. On the minimum number f blcks defining a design. Bull. Austral. Math. Sc., 4:97-2, 990. [4] Ken Gray. Special Subsets f the Blck Sets f Designs. PhD thesis, University f Queensland, 990. [5] Thmas Strer. Cycltmy and Difference Sets. Lectures in Advanced Mathematics. Markham, Chicag, 967. [6] T. Tsuzuku. Finite Grups and Finite Gemetries. Cambridge University Press, Cambridge, 982. [7] J.S. Wallis. Hadamard matrices. In Cmbinatrics: Rm Squares, sum-free sets and Hadamard matrices, vlume 292 f Lecture Ntes in Mathematics. Springer-Verlag, Berlin-Heidelberg-New Yrk, 972. Part IV f W.D. Wallis, Anne Penfld Street, and Jennifer Seberry Wallis. 62