COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA. COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
318 A. D. Keedwell 2f 3 = 0 (mod 5), we require that the integers 2, 3 be not adjacent in either array, and this is clearly impossible. Thus, the non-existence of orthogonal latin squares of order six appears to be due to a combinatorial accident. Our general method for the construction of a pair of mutually orthogonal latin squares of assigned order I may easily be extended to give a method for constructing triples. It is easy to see that the latin squares L1 = (So, Sl, * . > SI-r}, L,* = {MSo, MSr, . . . , MS,-,}, and G = {M2So, M2S1, . ..) M2S,-r), where A4 E (I- 1) (0 1 . . . r-2) and SO E 1, Sr, . . .) S,_, are permutations of the natural numbers 0, 1, . . . , Y - 1, will be mutually orthogonal provided that the two sets of permutations SJrlMSi andS;liWSi,i=O,l, . . . . r - 1, are both sharply transitive on the symbolsO,l, . . . . Y- 1. Since Sz:lMZS, = (S;rA4SJ2, it is clear from Diagram 3 that a sufficient condition for the existence of such a triple of mutually orthogonal latin squares of order 10 is that a 9X9 matrix A = (a,), i = 1 to 9, j = 0 to 8, exist with the properties: (i) each of the integers 0, 1, . . . , 9 occurs at most once in each row and column, and the integer i does not occur in the (i+ l)th column or the (9 - i)th row ; (ii) if ali, = a2 j2 = . . . = agj, = r then (a) the integers r+l, alj,+l, a2 j2+b . . . 2 a9 j,+l are all different (all addition being modulo 9), r = 0, 1,2, . ..) 8, and (b) the integers r+2, arj1+2, asj2+2 . . . , a9js+2 are all different ; (iii) if arj, = a2je= . . . = agis = 9 then (a) the integers 9, a, jl+I, a2jz+l, . . a9 a9jp+l are all different, and (b) the integers 9, a, jI+2, a2j2+2, . . . , a, je+2 are ail different. To the disappointment of the author, it turns out that the arrays AZ corresponding to the property D neofields of order 10 have properties (i), (ii) (a), (iii) (a), and (iii) (b), but fail to satisfy property (ii) (b). For the purpose of searching for 9X9 matrices of type A, a computer programme was written which would insert successively the integers alo, all, . . . , a98 and would backtrack to the preceding place in the event that a place could not be filled successfully. Details of the construction of this programme so as to require as few instructions as possible, of the computer time needed, and of the results appear in [l] and so need not be repeated here. S,-lMS,, = (9)(0 1 2 3 4 5 6 7 8) SilA!fS1 = @)(a10 all al2 al3 al4 al5 a16 al7 ad S,1MSg = (W90 a91 a92 a93 a94 a95 a96 a97 a981 Diagram 3 Property D neojields and latin squares REFERENCES 1. A. D. KEEDWELL: On orthogonal latin squares and a class of neofields. Rend. Mat. e A&. (5) 25 (1966), 519-561. 2. A. D. KEEDWELL: On property Dneofields. Rend. Mat. e AppZ.(5)26(1967), 384-402. 3. L. J. PAIGE: Neofields. Duke Math. J. 16 (1949), 39-60. 319
A projective coqfiguration J. W. P. HIRSCHFELD 1. In ageometry over a field, four skew lines not lying in a regulus have two transversals (which may coincide or lie only in a quadratic extension of the field). From this come the following theorems. The double-six theorem (Schlafli, 1858): Given five skew lines with a single transversal such that each set of four has exactly one further transversal, the five lines thus obtained also have a transversal-the completing line of the double-six. Grace’s extension theorem (Grace, 1898): Given six skew lines with a common transversal such that each set of five gives rise to a double-six, the six completing lines also have a transversal-the Grace line. Conjecture: Given seven skew lines with a common transversal such that each set of six gives rise to a Grace figure, the seven Grace lines also have a transversal. 2. The double-six is self-polar and lies on a unique cubic surface, which contains 27 lines in all. The configuration exists for all fields except GF(q) for q = 2, 3 and 5 [l]. The six initial lines in the Grace figure are chords of a unique twisted cubic and are polar to the completing lines, which are also chords of the cubic. However, the theorem as it stands is true only if the six completing lines of the double-sixes are skew to one another. This is not necessarily true, as the six lines may be concurrent. The configuration exists for GF(9) but not for GE;(q) with q < 9 [2]. The conjecture depends only on the incidences of the lines. So, if it is true over the complex field, it is true over any finite field large enough for the seven Grace lines to exist. N7ren [3] mentions that both he and Grace attempted the conjecture but were not able to achieve anything. Grace proved his theorem by, in fact, first establishing a slightly more general result. He proved a theorem for linear complexes, which was then applied to special linear complexes, which was in turn dualized to the theorem of the extension of the double-six. In all, no one was very hopeful of extending Grace’s theorem, which itself was regarded as something of a fluke. Further scepticism set in on finding that the conditions of linear independence on the initial set of lines were not even sufficient for Grace’s theorem. 321
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A projective coqfiguration<br />
J. W. P. HIRSCHFELD<br />
1. In ageometry over a field, four skew lines not lying in a regulus have<br />
two transversals (which may coincide or lie only in a quadratic extension<br />
of the field). From this come the following theorems.<br />
The double-six theorem (Schlafli, 1858): Given five skew lines with a<br />
single transversal such that each set of four has exactly one further transversal,<br />
the five lines thus obtained also have a transversal-the completing<br />
line of the double-six.<br />
Grace’s extension theorem (Grace, 1898): Given six skew lines with a<br />
common transversal such that each set of five gives rise to a double-six,<br />
the six completing lines also have a transversal-the Grace line.<br />
Conjecture: Given seven skew lines with a common transversal such<br />
that each set of six gives rise to a Grace figure, the seven Grace lines also<br />
have a transversal.<br />
2. The double-six is self-polar and lies on a unique cubic surface, which<br />
contains 27 lines in all. The configuration exists for all fields except GF(q)<br />
for q = 2, 3 and 5 [l].<br />
The six initial lines in the Grace figure are chords of a unique twisted<br />
cubic and are polar to the completing lines, which are also chords of the<br />
cubic. However, the theorem as it stands is true only if the six completing<br />
lines of the double-sixes are skew to one another. This is not necessarily<br />
true, as the six lines may be concurrent. The configuration exists for GF(9)<br />
but not for GE;(q) with q < 9 [2].<br />
The conjecture depends only on the incidences of the lines. So, if it is<br />
true over the complex field, it is true over any finite field large enough for<br />
the seven Grace lines to exist.<br />
N7ren [3] mentions that both he and Grace attempted the conjecture but<br />
were not able to achieve anything. Grace proved his theorem by, in fact,<br />
first establishing a slightly more general result. He proved a theorem for<br />
linear complexes, which was then applied to special linear complexes,<br />
which was in turn dualized to the theorem of the extension of the double-six.<br />
In all, no one was very hopeful of extending Grace’s theorem, which<br />
itself was regarded as something of a fluke. Further scepticism set in on<br />
finding that the conditions of linear independence on the initial set of<br />
lines were not even sufficient for Grace’s theorem.<br />
321