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Thus, the string represents a solution to a 3-rotor problem where hole of the first rotor is aligned with hole of the second and hole of the third. Of course, it would be possible to encode the alleles as binary strings, but there seems little point in so doing—particularly if q is not a power of 2, as some binary strings would not correspond to any actual orientation. This seems very straightforward, but there is a subtle point here that could be overlooked. The assignment of labels to the holes is arbitrary, and this creates a problem of competing conventions.

THE ELEMENTS OF A GENETIC ALGORITHM 37 the top P% of the population are eligible to reproduce; within this group parents are chosen uniformly at random. Selection intensity is clearly an a posteriori measure, and one that is also specific to the current population; if we make some assumption about the population fitness distribution (commonly, that the distribution is stationary and Normal), the value becomes a constant I(t) = I. Nevertheless, the value of I will still depend on the control parameters used for selection in a non-trivial way.

483) means that we select string 2. Now we perform crossover and mutation on these strings. 798, the chosen crosspoint is 4. If we cross the strings and at the 4th crosspoint, the resulting 50 CHAPTER 2. BASIC PRINCIPLES strings are and again! This illustrates one of the problems that we mentioned above—crossover can produce offspring that are merely clones of their parents. 5, we choose the first, otherwise the second. e. Now we need to apply the mutation operator. 10 for each locus of the string.

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