# Fitness proportionate selection

(Redirected from Roulette wheel selection)
Example of the selection of a single individual

Fitness proportionate selection, also known as roulette wheel selection, is a genetic operator used in genetic algorithms for selecting potentially useful solutions for recombination.

In fitness proportionate selection, as in all selection methods, the fitness function assigns a fitness to possible solutions or chromosomes. This fitness level is used to associate a probability of selection with each individual chromosome. If ${\displaystyle f_{i}}$ is the fitness of individual ${\displaystyle i}$ in the population, its probability of being selected is

${\displaystyle p_{i}={\frac {f_{i}}{\Sigma _{j=1}^{N}f_{j}}},}$

where ${\displaystyle N}$ is the number of individuals in the population.

This could be imagined similar to a Roulette wheel in a casino. Usually a proportion of the wheel is assigned to each of the possible selections based on their fitness value. This could be achieved by dividing the fitness of a selection by the total fitness of all the selections, thereby normalizing them to 1. Then a random selection is made similar to how the roulette wheel is rotated.

While candidate solutions with a higher fitness will be less likely to be eliminated, there is still a chance that they may be eliminated because their probability of selection is less than 1 (or 100%). Contrast this with a less sophisticated selection algorithm, such as truncation selection, which will eliminate a fixed percentage of the weakest candidates. With fitness proportionate selection there is a chance some weaker solutions may survive the selection process. This is because even though the probability that the weaker solutions will survive is low, it is not zero which means it is still possible they will survive; this is an advantage, because there is a chance that even weak solutions may have some features or characteristics which could prove useful following the recombination process.

The analogy to a roulette wheel can be envisaged by imagining a roulette wheel in which each candidate solution represents a pocket on the wheel; the size of the pockets are proportionate to the probability of selection of the solution.[citation needed] Selecting N chromosomes from the population is equivalent to playing N games on the roulette wheel, as each candidate is drawn independently.

Other selection techniques, such as stochastic universal sampling[1] or tournament selection, are often used in practice. This is because they have less stochastic noise, or are fast, easy to implement and have a constant selection pressure [Blickle, 1996].

The naive implementation is carried out by first generating the cumulative probability distribution (CDF) over the list of individuals using a probability proportional to the fitness of the individual. A uniform random number from the range [0,1) is chosen and the inverse of the CDF for that number gives an individual. This corresponds to the roulette ball falling in the bin of an individual with a probability proportional to its width. The "bin" corresponding to the inverse of the uniform random number can be found most quickly by using a binary search over the elements of the CDF. It takes in the O(log n) time to choose an individual. A faster alternative that generates individuals in O(1) time will be to use the alias method.

Recently, a very simple algorithm was introduced that is based on "stochastic acceptance".[2] The algorithm randomly selects an individual (say ${\displaystyle i}$) and accepts the selection with probability ${\displaystyle f_{i}/f_{M}}$, where ${\displaystyle f_{M}}$ is the maximum fitness in the population. Certain analysis indicates that the stochastic acceptance version has a considerably better performance than versions based on linear or binary search, especially in applications where fitness values might change during the run.[3] While the behavior of this algorithm is typically fast, some fitness distributions (such as exponential distributions) may require ${\displaystyle O(n)}$ iterations in the worst case. This algorithm also requires more random numbers than binary search.

## Pseudocode

For example, if you have a population with fitnesses [1, 2, 3, 4], then the sum is (1 + 2 + 3 + 4 = 10). Therefore, you would want the probabilities or chances to be [1/10, 2/10, 3/10, 4/10] or [0.1, 0.2, 0.3, 0.4]. If you were to visually normalize this between 0.0 and 1.0, it would be grouped like below with [red = 1/10, green = 2/10, blue = 3/10, black = 4/10]:

0.1 ]

0.2 \
0.3 /

0.4 \
0.5 |
0.6 /

0.7 \
0.8 |
0.9 |
1.0 /

Using the above example numbers, this is how to determine the probabilities:

sum_of_fitness = 10
previous_probability = 0.0

[1] = previous_probability + (fitness / sum_of_fitness) = 0.0 + (1 / 10) = 0.1
previous_probability = 0.1

[2] = previous_probability + (fitness / sum_of_fitness) = 0.1 + (2 / 10) = 0.3
previous_probability = 0.3

[3] = previous_probability + (fitness / sum_of_fitness) = 0.3 + (3 / 10) = 0.6
previous_probability = 0.6

[4] = previous_probability + (fitness / sum_of_fitness) = 0.6 + (4 / 10) = 1.0

The last index should always be 1.0 or close to it. Then this is how to randomly select an individual:

random_number # Between 0.0 and 1.0

if random_number < 0.1
select 1
else if random_number < 0.3 # 0.3 - 0.1 = 0.2 probability
select 2
else if random_number < 0.6 # 0.6 - 0.3 = 0.3 probability
select 3
else if random_number < 1.0 # 1.0 - 0.6 = 0.4 probability
select 4
en

## Coding examples

### Java – linear O(n) version

// Returns the selected index based on the weights(probabilities)
int rouletteSelect(double[] weight) {
// calculate the total weight
double weight_sum = 0;
for(int i=0; i<weight.length; i++) {
weight_sum += weight[i];
}
// get a random value
double value = randUniformPositive() * weight_sum;
// locate the random value based on the weights
for(int i=0; i<weight.length; i++) {
value -= weight[i];
if(value < 0) return i;
}
// when rounding errors occur, we return the last item's index
return weight.length - 1;
}

// Returns a uniformly distributed double value between 0.0 and 1.0
double randUniformPositive() {
// easiest implementation
return new Random().nextDouble();
}

### Java – stochastic acceptance version

public class roulette {
/* program n_select=1000 times selects one of n=4 elements with weights weight[i].
* Selections are summed up in counter[i]. For the weights as given in the example
* below one expects that elements 0,1,2 and 3 will be selected (on average)
* 200, 150, 600 and 50 times, respectively.  In good agreement with exemplary run.
*/
public static void main(String[] args) {
int n = 4;
double[] weight = new double [n];
weight[0] = 0.4;
weight[1] = 0.3;
weight[2] = 1.2;
weight[3] = 0.1;
double max_weight = 1.2;
int[] counter = new int[n];
int n_select = 1000;
int index;
for (int i = 0; i < n_select; i++) {
while (true) {
index = (int) (Math.random() * n);
if (Math.random() < weight[index] / max_weight) break;
}
counter[index]++;
}
for (int i = 0; i < n; i++) {
System.out.printf("counter[%d]=%d\n", i, counter[i]);
}
}
/* The program uses stochastic acceptance instead of linear (or binary) search.
* More on https://arxiv.org/abs/1109.3627
*/
}
# Exemplary output:
# counter[0]=216
# counter[1]=135
# counter[2]=595
# counter[3]=54

### Ruby – linear O(n) search

# Normalizes an array that potentially contains negative numbers by shifting
#   all of them up to be positive (0 is left alone).
#
# +pop_fit+ array of each individual's fitness in a population to normalize
def norm_pop(pop_fit)
least_fit = pop_fit.min.abs + 1 # Absolute value so can shift up
# +1 so that it doesn't become 0

pop_fit.map! do |f|
(f != 0) ? (f + least_fit) : f
end

return pop_fit
end

# Returns an array of each individual's probability between 0.0 and 1.0 fitted
#   onto an imaginary roulette wheel (or pie).
#
# This will NOT work for negative fitness numbers, as a negative piece of a pie
#   (i.e., roulette wheel) does not make sense.  Therefore, if you have negative
#   numbers, you will have to normalize the population first before using this.
#
# +pop_fit+ array of each individual's fitness in the population
# +is_high_fit+ true if high fitness is best or false if low fitness is best
def get_probs(pop_fit,is_high_fit=true)
fit_sum  = 0.0 # Sum of each individual's fitness in the population
prob_sum = 0.0 # You can think of this in 2 ways; either...
# 1) Current sum of each individual's probability in the
#    population
# or...
# 2) Last (most recently processed) individual's probability
#    in the population
probs    = []
best_fit = nil # Only used if is_high_fit is false

# Get fitness sum and best fitness
pop_fit.each do |f|
fit_sum += f

if best_fit == nil or f > best_fit
best_fit = f
end
end

puts "Best fitness: #{best_fit}"
puts "Fitness sum:  #{fit_sum}"

best_fit += 1 # So that we don't get best_fit-best_fit=0

# Get probabilities
pop_fit.each_index do |i|
f = pop_fit[i]

if is_high_fit
probs[i] = prob_sum + (f / fit_sum)
else
probs[i] = (f != 0) ? (prob_sum + ((best_fit - f) / fit_sum)) : 0.0
end

prob_sum = probs[i]
end

probs[probs.size - 1] = 1.0 # Ensure that the last individual is 1.0 due to
#   decimal problems in computers (can be 0.99...)
return probs
end

# Selects and returns a random index using an array of probabilities that were
#   created to mirror a roulette wheel type of selection.
#
# +probs+ array of probabilities between 0.0 and 1.0 that total to 1.0
def roulette_select(probs)
r = rand # Random number between 0.0 and 1.0

probs.each_index do |i|
if r < probs[i]
return i
end
end

return probs.size - 1 # This shouldn't happen
end

pop_fit = [1,2,3,4]
pop_sum = Float(pop_fit.inject {|p,f| p + f})
probs = get_probs(pop_fit,true)

# These should all have the exact same output
puts probs.inspect
puts get_probs([4,3,2,1],false).inspect
puts get_probs(norm_pop([-4,-3,-2,-1]),true).inspect
puts get_probs(norm_pop([-1,-2,-3,-4]),false).inspect
puts

# Check the math
prev_prob = 0.0

puts "Math check:"
for i in 0..pop_fit.size-1
puts "%.4f|%.4f|%.4f" % [probs[i],probs[i] - prev_prob,pop_fit[i] / pop_sum]
prev_prob = probs[i]
end
puts

# Observe some random selections
observed_probs = Array.new(pop_fit.size,0)
observed_count = 1000

for i in 1..observed_count
observed_probs[roulette_select(probs)] += 1
end

puts "Observed:"
observed_probs.each_index do |i|
prob = observed_probs[i] / Float(observed_count)
puts "#{i}: #{prob}"
end

# Example output:
#
# Best fitness: 4
# Fitness sum:  10.0
# [0.1, 0.30000000000000004, 0.6000000000000001, 1.0]
# Best fitness: 4
# Fitness sum:  10.0
# [0.1, 0.30000000000000004, 0.6000000000000001, 1.0]
# Best fitness: 4
# Fitness sum:  10.0
# [0.1, 0.30000000000000004, 0.6000000000000001, 1.0]
# Best fitness: 4
# Fitness sum:  10.0
# [0.1, 0.30000000000000004, 0.6000000000000001, 1.0]
#
# Math check:
# 0.1000|0.1000|0.1000
# 0.3000|0.2000|0.2000
# 0.6000|0.3000|0.3000
# 1.0000|0.4000|0.4000
#
# Observed:
# 0: 0.108
# 1: 0.191
# 2: 0.296
# 3: 0.405