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.