Talk:Evolutionary algorithm: Difference between revisions

:::: Adding them up, we get the chance at step <math>x</math> as: <math>x(x-1)(2n-1)+2x(L-x)\over n^2L^2</math>. Summing the inverse of that should give you the answer of needed total mutations, unfortunately it's non-trivial to do. However, we can examine its tendency: when <math>x</math> is large (many unmatched chars), the expression is roughly <math>2 x^2\over nL^2</math>; recall that the single char mutation had <math>x\over nL</math>, which means double char mutation is initially faster than single char mutation, until <math>x\approx L/2</math>, i.e. half of the letters matched (this shouldn't be surprising).

:::: After that, when <math>x</math> becomes very small, the expression is <math>\approx {2x\over n^2L}</math>, about <math>2/n</math> of the single char mutation speed, meaning it requires about 13 (n=27)times as many mutations when close to perfect match. Since this is the slowest part of the evolution already, mutating two chars is overall probably about an order of magnitude slower. --[[User:Ledrug|Ledrug]] 18:34, 11 October 2011 (UTC)