Also the offsets are the powers of 2 (mod 7). I like it! WFL On 6/24/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
I prefer shifting the indices of either the points or the lines so that {0,1,3} becomes {1,2,4}. Then we have:
P_i meets L_j if and only if (i-j) is a quadratic residue mod 7; L_i meets P_j if and only if (i-j) is a quadratic nonresidue mod 7.
Sincerely,
Adam P. Goucher
Sent: Wednesday, June 24, 2015 at 2:53 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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