By W. D. Wallis, A. P. Street, J. S. Wallis

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Extra info for Combinatorics: Room Squares, Sum-Free Sets, Hadamard Matrices

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Identity. Clearly ai,b i and c i are - 53 - Suppose a. : a. for some i # j • Since x 2t -x 4t # O, we have x i = x j ; z ] is impossible. Similarly, b i : bj or c i = cj can never occur when i # j. Finally, say a i = bj for some i and j. x4t)2 : x J ( x 2 t _ l ) 2 that is xi+4t(l_x2t)2 = xJ(x2t-l) 2, whence x i+4t = x j. order 6t. This cannot happen, because 0 ~ i < t and 0 ~ j < t and x is of Similarly, a. : c. or b. = c. can never occur. z ] z ] This completes the proof that all 3t sums are different, and the starter (and consequently the set of Steiner difference blocks) is strong.

Suppose the set G forms one commutative idempotent quasigroup under the binary operation # and another under the operation A. groups we denote them (G,#) and (G,A); To distinguish these quasi- we write L(#) and L(A) for the Latin squares equivalent to (G,#) and (G,A) respectively. If ~ is the join of L(#) and L(A), then the ordered pairs (gi,gi) will occur once each on the diagonal o f ~ , groups are idempotent. (i) since the quasi- If, further, whenever x,y and p e G satisfy x#y = xay = p, x=y=p; (ii) whenever p,q E G, p # q, there exists at most one unordered pair of elements x and y of G such that x#y = p, x A y = q; then L(#) and L(A) will be orthogonal symmetric Latin squares.

PROOF. If we w~ite a i for -xi-Y i then xi+ai = -Yi Yi+ai = -x i so {xi+a i, Yi+ai } = {-xl x E G , x # O} = {xlx~, x ~ 0}. SO A X is an adder for X. We shall refer to an adder A X as skew if ae A X implies that - a ~ A X. 3. If there is an abelian group G of odd order r with starter and an adder, then there is a Room square of side r. If the adder is skew, then the square is skew Room. PROOF. ,g r. ,g r. Denote the ith pair in a starter X as {xi,Y i} and the ith men~ber of an adder A x for X as a i.

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