Scalar complexity revised

One more thing on this: the formula I was using before is actually a special case for the rank 2 case. The correct one is

sqrt(det(MTM)/n**r)

where

sqrt is the square root

det is the determinant

MTM is the matrix you get by multiplying the weighted mapping by its transpose

n is the number of prime intervals

r is the rank of the temperament

** is exponentiation

This gives the obvious result for equal temperaments -- the complexity is roughly the number of notes to the octave. It's also dimensionally correct in that the number of sums that get multiplied when calculating the determinant is balanced by the correction factor. You can replace MTM with a matrix containing means instead of sums and the scalar complexity is simply the square root of the determinant of that matrix.

Starting with unison vectors, the formula is

sqrt(det(XTX)/n**r)/prime_prod

where XTX is the equivalent of MTM, prime_prod is the product of the sizes of the prime intervals, and r is the rank of the temperament (not the number of unison vectors).

It happens that this formula is also valid for a rank 0 temperament -- where the number of prime intervals equals the number of unison vectors. I'll leave you all to guess what value this gives. One hint is that the weights cancel out.

Graham