The document describes using a genetic algorithm to solve the problem of coloring a checkboard with no adjacent squares having the same color. It initializes a population of random colorings, calculates their fitness based on constraints, performs selection, crossover and mutation to generate a new population, and iterates until finding a solution with optimal fitness. After 47 generations, it achieves the optimal solution of no adjacent squares having the same color.

1. CHECKBOARD USING GENETIC
ALGORITHM

2. • We are given an n by n checkboard in which
every field can have a different colour from a
set of four colours.
• Goal is to achieve a checkboard in a way that there
are no neighbours with the same colour (not diagonal)
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10

3. The size of a population populationSize = 4
The size of the matrix we'll be using as a solution matrixSize = 4
Build a data structure to contain the initial population population =
zeros( matrixSize, matrixSize, populationSize );
• The checkboard contain 4 colors which shows by number 0 to 3.
it is a node-edge problem so the maximum fitness can easily be computed
optimalFitness = 2 * ( ( matrixSize - 1 ) * matrixSize )=2*3*4=24;
Performance
Initialize:
for( i = 1:populationSize )
population( :, :, i ) = fix( 4 * rand( matrixSize, matrixSize ) );
end

4. Initial population
population(:,:,1)
1 3 0 0
3 3 2 0
2 3 0 1
2 3 0 2
population(:,:,2)
3 0 0 0
2 1 0 1
0 1 2 2
2 2 3 2
population(:,:,3)
2 0 1 0
3 0 0 3
2 1 1 1
0 2 3 2
population(:,:,4)
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0

5. Fitness function calculation
population(:,:,1)
1 3 0 0
3 3 2 0
2 3 0 1
2 3 0 2
1 1 0
0 1 1
1 1 1
1 1 1
Row Column
1 + 1 + 0
0 + 1 + 1
1 + 1 + 1
1 + 1 + 1
1 + 1 + 0
0 + 0 + 0
1 + 1 + 0
0 + 1 + 1
10 6
10 + 6 = 16Fitness value

6. Fitness of the initial population
fitness( 1 ) = checkers( population( :, :, 1 ) ) = 16
fitness( 2 ) = checkers( population( :, :, 2 ) ) = 17
fitness( 3 ) = checkers( population( :, :, 3 ) ) = 20
fitness( 4 ) = checkers( population( :, :, 4 ) ) = 21
optimalFitness = 2 × N × (N-1)
here N is the size of matrix (N=4) optimalFitness=2×4×3=24

7. what's the best fitness in this
population?
bestCurrentFitness = max( fitness ) = 21;
meanCurrentFitness = mean( fitness) = 18.5;
• optimalFitness = 24

8. bestCurrentFitness = 21 > bestFitness = 0
bestFitness = bestCurrentFitness
And displaye checkboard
0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5

9. Generate newPopulation
newPopulation(:,:,1)
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
newPopulation(:,:,2)
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
newPopulation(:,:,3)
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
newPopulation(:,:,4)
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

10. Number of chromosome which bestindividual is memorized to do
crossovers a3 = fix (populationSize*rand(1))+1;
Number of chromosome which bestindividual is memorized after crossovers
a4 = fix (populationSize*rand(1))+1;
a3 = 4
a4 = 4
Generation = 1

11. i = 1 and i +1 = 2
mom = select( population, fitness );
1 3 0 0
3 3 2 0
2 3 0 1
2 3 0 2
dad = select( population, fitness );
3 0 0 0
2 1 0 1
0 1 2 2
2 2 3 2

12. dad
3 0 0 0
2 1 0 1
0 1 2 2
2 2 3 2
mom
1 3 0 0
3 3 2 0
2 3 0 1
2 3 0 2
Crossover for i = 1 and i +1 = 2 and rand(1) = 0.68 > 0.5

13. Mutation for i = 1 and i +1 = 2
child1
1 3 0 0
2 3 2 0
0 1 0 1
2 2 3 2
newPopulation( :, :, 1 ) =
mutate(child1 )
1 3 0 0
2 3 2 0
0 1 0 1
2 2 3 2
For each gene 1% probability exists to occur mutation.
child2
3 0 0 0
3 1 0 1
2 3 2 2
2 3 0 2
newPopulation( :, :, 2 ) =
mutate(child2 )
3 0 0 0
3 1 0 1
2 3 2 2
2 3 0 2

14. i = 3 and i +1 = 4
mom = select( population, fitness );
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0
dad = select( population, fitness );
3 0 0 0
2 1 0 1
0 1 2 2
2 2 3 2

15. dad
3 0 0 0
2 1 0 1
0 1 2 2
2 2 3 2
mom
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0
Crossover for i = 3 and i +1 = 4 and rand(1) = 0.38 < 0.5
Number of the Column for Crossover
a5 = fix (matrixSize*rand(1))+1 = 2;

16. Mutation for i = 3 and i +1 = 4
child1
3 0 3 1
1 1 2 3
3 1 3 0
0 2 3 0
newPopulation( :, :, 3 ) =
mutate(child1 )
3 0 3 1
1 1 2 3
3 1 3 0
0 2 3 0
For each gene 1% probability exists to occur mutation.
newPopulation( :, :, 4 ) =
bestindividual( population, fitness)
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0

17. Population = newPopulation
population(:,:,1)
1 3 0 0
2 3 2 0
0 1 0 1
2 2 3 2
population(:,:,2)
3 0 0 0
3 1 0 1
2 3 2 2
2 3 0 2
population(:,:,3)
3 0 3 1
1 1 2 3
3 1 3 0
0 2 3 0
population(:,:,4)
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0

18. Fitness changes
Old
fitness( 1 ) = checkers( population( :, :, 1 ) ) = 16
fitness( 2 ) = checkers( population( :, :, 2 ) ) = 17
fitness( 3 ) = checkers( population( :, :, 3 ) ) = 20
fitness( 4 ) = checkers( population( :, :, 4 ) ) = 21
New
fitness( 1 ) = checkers( population( :, :, 1 ) ) = 20
fitness( 2 ) = checkers( population( :, :, 2 ) ) = 16
fitness( 3 ) = checkers( population( :, :, 3 ) ) = 20
fitness( 4 ) = checkers( population( :, :, 4 ) ) = 21

19. After 47 generation the optimal fitness obtains
0.5 1 1.5 2 2.5 3 3.5 4 4.5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 5 10 15 20 25 30 35 40 45 50
16
17
18
19
20
21
22
23
24
Generations
Fitness
Best Fitness
Mean Fitness

20. Performance of Select Function
population(:,:,1)
1 3 0 0
2 3 2 0
0 1 0 1
2 2 3 2
population(:,:,2)
3 0 0 0
3 1 0 1
2 3 2 2
2 3 0 2
population(:,:,3)
3 0 3 1
1 1 2 3
3 1 3 0
0 2 3 0
population(:,:,4)
3 2 3 1
1 0 2 3
3 1 3 0
0 3 3 0
fitness( 1 ) = checkers( population( :, :, 1 ) ) = 20
fitness( 2 ) = checkers( population( :, :, 2 ) ) = 16
fitness( 3 ) = checkers( population( :, :, 3 ) ) = 20
fitness( 4 ) = checkers( population( :, :, 4 ) ) = 21

21. Performance of Select Function
function selected = select( population, fitness )
[m, n] = size( fitness ); % m = n = 4
s = sum( fitness ); % s = 77
roulette = s*rand( 1 ); % roulette = 48.7868
currentSum = 0;
picked = n; % n = 4
for i = 1:n
s1 = currentSum + fitness(i)
if s1 < roulette
picked = I
end
currentSum = currentSum + fitness(i);
end
selected = population( :, :, picked );
i currentSum fitness(i) s1 if s1 < roulette picked
1 0 20 20 Yes 1
2 20 16 36 Yes 2
3 36 20 56 No 2
4 56 21 77 No 2