If a given bitvector has m
1s and n
0s, then a bit is flipped from
0 to 1 with probability 2p(m+1)/(m+n+2)
and from
1 to 0 with probability 2p(n+1)/(m+n+2)
. This is
equivalent with choosing bits uniformly at random with probability
2 * p
and drawing them from a bernoulli-distribution with parameter
(m+1)/(m+n+2)
.
mutBitflipCHW(ind, p = 0.1, ...)
[integer]
mutated binary individual.
[integer]
binary individual.
[numeric]
average flip probability, must be between 0
and 0.5.
further arguments passed on to the method.