String Init x x select expected Roulette |
No. Pop. val Sqd count Wheel |
1 0 1 1 0 1 13 169 0.14 0.58 1 |
2 1 1 0 0 0 24 576 0.49 1.97 2 |
3 0 1 0 0 0 8 64 0.06 0.22 0 |
4 1 0 0 1 1 19 361 0.31 1.23 1 |
Sum 1179 |
- In the the strings above each gene (0 or 1 bit) was
randomly created by a "coin toss".
- The x value was gotten by simply treating each string
as a binary number and finding its decimal equivalent.
- It was squared because. . . . we were solving for a max
of x squared in the interval from 0 to 31 . . .
- Each number in the select column is the number divided
by the total. An evaluation of the individual in com-
parison to the total population.
- The expected count is the select column multiplied by 4 -
there are to be strings in the new population.
- Although the roulette wheel is random, it is weighted
so that some strings are more likely than others to be
chosen to mate.
Strings Mate Crossover New x x |
to Mate with. Site Population val Sqd |
0 1 1 0| 1 2 4 0 1 1 0 0 12 144
|
1 1 0 0| 0 1 4 1 1 0 0 1 25 625
|
1 1| 0 0 0 4 2 1 1 0 1 1 27 729
|
1 0| 0 1 1 3 2 1 0 0 0 0 16 256
|
Sum 1754
|
- The new mating pool with the cross site marked.
The pairings for mating are also chosen randomly,
and, in this case each string mates once and only once.
- The crossover site is randomly selected.
- Each new string in the population has some part
of a string from each parent.
- The x value for each string is calculated and the
cycle begins again.
The population in the new generation is stronger than the
old - the sum is higher.
If the mutation probability is assumed to be 0.001 then
no mutations were expected in a single generation and
none are shown.
taken from a book "Genetic Algorithms in Search,
Optimization & Machine Learning" by David E. Goldberg
|