Genetic Algorithm for OneMax

In this section, we will apply the genetic algorithm to a binary string-based optimization problem.

The problem is called OneMax and evaluates a binary string based on the number of 1s in the string. For example, a bitstring with a length of 20 bits will have a score of 20 for a string of all 1s.

Given we have implemented the genetic algorithm to minimize the objective function, we can add a negative sign to this evaluation so that large positive values become large negative values.

The onemax() function below implements this and takes a bitstring of integer values as input and returns the negative sum of the values.

objective function

def onemax(x):
return -sum(x)

objective function

def onemax(x):
return -sum(x)
Next, we can configure the search.

The search will run for 100 iterations and we will use 20 bits in our candidate solutions, meaning the optimal fitness will be -20.0.

The population size will be 100, and we will use a crossover rate of 90 percent and a mutation rate of 5 percent. This configuration was chosen after a little trial and error.

define the total iterations

n_iter = 100

bits

n_bits = 20

define the population size

n_pop = 100

crossover rate

r_cross = 0.9

mutation rate

r_mut = 1.0 / float(n_bits)

define the total iterations

n_iter = 100

bits

n_bits = 20

define the population size

n_pop = 100

crossover rate

r_cross = 0.9

mutation rate

r_mut = 1.0 / float(n_bits)
The search can then be called and the best result reported.

perform the genetic algorithm search

best, score = genetic_algorithm(onemax, n_bits, n_iter, n_pop, r_cross, r_mut)
print(‘Done!’)
print(‘f(%s) = %f’ % (best, score))

perform the genetic algorithm search

best, score = genetic_algorithm(onemax, n_bits, n_iter, n_pop, r_cross, r_mut)
print(‘Done!’)
print(‘f(%s) = %f’ % (best, score))
Tying this together, the complete example of applying the genetic algorithm to the OneMax objective function is listed below.

genetic algorithm search of the one max optimization problem

from numpy.random import randint
from numpy.random import rand

objective function

def onemax(x):
return -sum(x)

tournament selection

def selection(pop, scores, k=3):
# first random selection
selection_ix = randint(len(pop))
for ix in randint(0, len(pop), k-1):
# check if better (e.g. perform a tournament)
if scores[ix] < scores[selection_ix]:
selection_ix = ix
return pop[selection_ix]

crossover two parents to create two children

def crossover(p1, p2, r_cross):
# children are copies of parents by default
c1, c2 = p1.copy(), p2.copy()
# check for recombination
if rand() < r_cross:
# select crossover point that is not on the end of the string
pt = randint(1, len(p1)-2)
# perform crossover
c1 = p1[:pt] + p2[pt:]
c2 = p2[:pt] + p1[pt:]
return [c1, c2]

mutation operator

def mutation(bitstring, r_mut):
for i in range(len(bitstring)):
# check for a mutation
if rand() < r_mut:
# flip the bit
bitstring[i] = 1 - bitstring[i]

genetic algorithm

def genetic_algorithm(objective, n_bits, n_iter, n_pop, r_cross, r_mut):
# initial population of random bitstring
pop = [randint(0, 2, n_bits).tolist() for _ in range(n_pop)]
# keep track of best solution
best, best_eval = 0, objective(pop[0])
# enumerate generations
for gen in range(n_iter):
# evaluate all candidates in the population
scores = [objective(c) for c in pop]
# check for new best solution
for i in range(n_pop):
if scores[i] < best_eval:
best, best_eval = pop[i], scores[i]
print(">%d, new best f(%s) = %.3f" % (gen, pop[i], scores[i]))
# select parents
selected = [selection(pop, scores) for _ in range(n_pop)]
# create the next generation
children = list()
for i in range(0, n_pop, 2):
# get selected parents in pairs
p1, p2 = selected[i], selected[i+1]
# crossover and mutation
for c in crossover(p1, p2, r_cross):
# mutation
mutation(c, r_mut)
# store for next generation
children.append(c)
# replace population
pop = children
return [best, best_eval]

define the total iterations

n_iter = 100

bits

n_bits = 20

define the population size

n_pop = 100

crossover rate

r_cross = 0.9

mutation rate

r_mut = 1.0 / float(n_bits)

perform the genetic algorithm search

best, score = genetic_algorithm(onemax, n_bits, n_iter, n_pop, r_cross, r_mut)
print(‘Done!’)
print(‘f(%s) = %f’ % (best, score))

genetic algorithm search of the one max optimization problem

from numpy.random import randint
from numpy.random import rand

objective function

def onemax(x):
return -sum(x)

tournament selection

def selection(pop, scores, k=3):
# first random selection
selection_ix = randint(len(pop))
for ix in randint(0, len(pop), k-1):
# check if better (e.g. perform a tournament)
if scores[ix] < scores[selection_ix]:
selection_ix = ix
return pop[selection_ix]

crossover two parents to create two children

def crossover(p1, p2, r_cross):
# children are copies of parents by default
c1, c2 = p1.copy(), p2.copy()
# check for recombination
if rand() < r_cross:
# select crossover point that is not on the end of the string
pt = randint(1, len(p1)-2)
# perform crossover
c1 = p1[:pt] + p2[pt:]
c2 = p2[:pt] + p1[pt:]
return [c1, c2]

mutation operator

def mutation(bitstring, r_mut):
for i in range(len(bitstring)):
# check for a mutation
if rand() < r_mut:
# flip the bit
bitstring[i] = 1 - bitstring[i]

genetic algorithm

def genetic_algorithm(objective, n_bits, n_iter, n_pop, r_cross, r_mut):
# initial population of random bitstring
pop = [randint(0, 2, n_bits).tolist() for _ in range(n_pop)]
# keep track of best solution
best, best_eval = 0, objective(pop[0])
# enumerate generations
for gen in range(n_iter):
# evaluate all candidates in the population
scores = [objective(c) for c in pop]
# check for new best solution
for i in range(n_pop):
if scores[i] < best_eval:
best, best_eval = pop[i], scores[i]
print(">%d, new best f(%s) = %.3f" % (gen, pop[i], scores[i]))
# select parents
selected = [selection(pop, scores) for _ in range(n_pop)]
# create the next generation
children = list()
for i in range(0, n_pop, 2):
# get selected parents in pairs
p1, p2 = selected[i], selected[i+1]
# crossover and mutation
for c in crossover(p1, p2, r_cross):
# mutation
mutation(c, r_mut)
# store for next generation
children.append(c)
# replace population
pop = children
return [best, best_eval]

define the total iterations

n_iter = 100

bits

n_bits = 20

define the population size

n_pop = 100

crossover rate

r_cross = 0.9

mutation rate

r_mut = 1.0 / float(n_bits)

perform the genetic algorithm search

best, score = genetic_algorithm(onemax, n_bits, n_iter, n_pop, r_cross, r_mut)
print(‘Done!’)
print(‘f(%s) = %f’ % (best, score))
Running the example will report the best result as it is found along the way, then the final best solution at the end of the search, which we would expect to be the optimal solution.

Note: Your results may vary given the stochastic nature of the algorithm or evaluation procedure, or differences in numerical precision. Consider running the example a few times and compare the average outcome.

In this case, we can see that the search found the optimal solution after about eight generations.

0, new best f([1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1]) = -14.000
0, new best f([1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0]) = -15.000
1, new best f([1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]) = -16.000
2, new best f([0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1]) = -17.000
2, new best f([1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -19.000
8, new best f([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -20.000
Done!
f([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -20.000000

0, new best f([1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1]) = -14.000
0, new best f([1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0]) = -15.000
1, new best f([1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1]) = -16.000
2, new best f([0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1]) = -17.000
2, new best f([1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -19.000
8, new best f([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -20.000
Done!
f([1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]) = -20.000000