Re: [math-fun] no-4-in-plane problem
Ed wrote: <<
From a 3x3x3 grid, select 8 points such that no 4 lie in a plane. The solution is unique.
I'm not sure what this is quoted from . . . but in any case this is reminiscent of the question (arising from the game "Set"): What's the size of the largest subset S of the group G = (Z_3)^4 such that S does not contain any "affine line". (Where an affine line means any subset g + H of G, where H is any subgroup of G isomorphic to Z_3, and g is any element of G.) By replacing 3 with any integer n >= 2, and replacing the direct sum of 4 copies of Z_n with any number k >= 2 of copies, and replacing "line" with a d-dimensional affine plane for some d, 1 <= d < k, we have a family of analogous questions Q(n; k; d), where the Set question is Q(3; 4; 1). (As Rich once proved, the answer to Q(3; 4; 1) is 20.) [Definition I]: (Where an affine d-plane is just g + H, where H is any subgroup of G = (Z_n)^k isomorphic to (Z_n}^d, and g is any element of G.) I understand these tend to be quite difficult, and only a few answers are currently known (computer searches quickly becoming out of range). * * * For n a prime power p^e, e >= 2, there's an alternative definition of affine plane: [Definition II]: Let V be the vector space of dimension k over the finite field of order p^e. Then an affine d-plane in V is just and subset of V of the form v+P where P is any d-dimensional subspace of V, and v is an arbitrary vector of V. For n = p^e, denote the version of Q(n; k; d) that uses this alternative definition of an affine d-plane by Q'(n; k; d). (possibly easy) QUESTION: For n = p^e, are Q(n; k; d) and Q'(n; k; d) combinatorily equivalent? More precisely, for n = p^e, does there always exist a bijection between (Z_n)^k and (Z_n)^k that induces a bijection between all affine d-place (Def. I) and all affine d-planes (Def. II) ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Dan Asimov