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

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

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So F x If yz is any edge of K and y # 0, z # 0, then there will be precisely one treatment x such that {x,y,z}e~, so yz belongs to precisely one factor, and clearly 0y belongs to precisely one factor. : {Fx: Therefore x~ V} is a one-factorization of K. 9. In ~he one-factorization derived from a Steiner triple system~, there is a 3-division corresponding to each 3-set. Each of these 3-d~v~sions is such that one of the components of the union of the three factors is a K~. PROOF. If {x,y,z}~ ~, then F x contains 0x and yz, F contains 0y and xz, Y F contains 0z and xy, z so {Fx,Fy,F z} is the required 3-division and the K 4 is spanned by {0,x,y,z}.

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.

Horton objects. [ii] found ways to prove similar theorems about other For Room squares he proved that the existence square of side v I and of a square of side v 2 with a subsquare original squares were skew and the subsquare The construction null subsquare was a skew subsquare, a n d to include Room squares of side v I exist. remains valid when v 3 = 0, even though we cannot have a of a Room square. So we obtain, as a corollary, Room squares of sides r I and r 2 can be used to construct For historical Subsequently, to show that a skew square results when the the case v2-v 3 : 6 whenever complementary first.

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