Using GA’s to Solve Problems
Amy Hoover
What is a Genetic Algorithm?
• Genetic Algorithm – “a search technique … to
find exact or approximate solutions to
optimization and search problems.”
-Wikipedia
Components of a GA
• Chromosome
– How we represent the individual
• Fitness function
– How we rate the individual
• Mutation
• Crossover
How do GAs work?
•
Randomize initial population (of 5)
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Note: These were arbitraily picked
A = 10111
B = 11100
C = 00110
D = 00001
E = 11011
• Evaluate the fitness of each individual (A, B, C,
D, E) in the population
– A = 10111
– B = 11100
– C = 00110
– D = 00001
– E = 11011
(fitness = 1+ 0 + 1 + 1+ 1 = 4)
(fitness = 1 + 1 + 1+ 0 + 0 = 3)
(fitness = 0 + 0 + 1+ 1 + 0 = 2)
(fitness = 0 + 0 + 0+ 0 + 1 = 1)
(fitness = 1 + 1 + 0+ 1 + 1 = 4)
• Repeat
– Select parents
• Who would we choose from the previous generation?
(A and E)
– Mate the parents, they plus their offspring form
next generation
• How do we mate them?
• What will the next generation look like?
Next Generation
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A = 10111 (parent 1 clone)
E = 11011 (parent 2 clone)
F=?
G=?
H=?
• Which generation are F, G, and H from?
Next Generation
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•
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A = 10111 (parent 1)
E = 11011 (parent 2)
F=?
G=?
H=?
• Which generation are F, G, and H from?
• How can we make F, G, H?
Crossover and Mutation
• Let’s make F, G, H (Gen 1 from initial generation 0)
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A = 10111 (parent 1)
E = 11011 (parent 2)
F = 11111 (Mutate A)
G = 11011(Crossover AE 2nd 3rd bits of E on A, also clone E )
H = 10011 (Crossover AE 3rd and 4th of E)
• Which generation are F, G, and H from?
• How can we make F, G, H?
Calculate the fitness of each individual
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•
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A = 10111 (parent 1)
E = 11011 (parent 2)
F = 11111 (Mutate A)
G = 11011 (Crossover AE)
H = 10011 (Crossover AE)
Calculate the fitness of each individual
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A = 10111 = 4
E = 11011 = 4
F = 11111 = 5
G = 11011 = 4
H = 10011 = 3
• Select parents (F,G)
• Repeat
– Select parents (A, E, F, G, H)
• How do we pick the parents?
– Pick the top 2 from A, E, F, G, H
– Mate parents (crossover, mutation)
– Calculate fitness
– (Repeat Select parents…)
Example
• Problem: I want to maximize the number of
0’s in a bit string
– (I want A = 00000)
• Let’s work through the psuedo code
Example
• Problem: I want to maximize the number of 0s
in a bit string
– (I want A = 00000)
• Step 1: Randomize the initial population
– A = 01101
– B = 11011
– C = 01010
– D = 01111
– E = 11111
Example
• Initial Population (gen 0)
– A = 01101
– B = 11011
– C = 01010
– D = 01111
– E = 11111
• Calculate fitness (recall 0’s are good, 1’s are
bad, we want to reward good genes!)
Example
• Initial Population
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– B = 11011 = 0 + 0 + 1 + 0 + 0 = 1
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– D = 01111 = 1 + 0 + 0 + 0 +0 = 1
– E = 11111 = 0 + 0 + 0 + 0 +0 = 0
• Reward 0’s, not 1’s
Example
• Initial Population
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– B = 11011 = 0 + 0 + 1 + 0 + 0 = 1
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– D = 01111 = 1 + 0 + 0+ 0+ 0 = 1
– E = 11111 = 0 + 0 + 0 + 0 + 0 = 0
Example
• Who should our parents be? (A and C)
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– B = 11011 = 0 + 0 + 1 + 0 + 0 = 1
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– D = 01111 = 1 + 0 + 0+ 0+ 0 = 1
– E = 11111 = 0 + 0 + 0 + 0 + 0 = 0
Example
• Who should our parents be?
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
• What next?
Mating!
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– F = 01100 = 3
– G = 01110 = 2
– H = 01011 = 2
Calculate Fitness!
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– F = 01100 = 3
– G = 01110 = 2
– H = 01011 = 2
Pick the top two!
– A = 01101 = 1 + 0 + 0 + 1 + 0 = 2
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– F = 01100 = 3
– G = 01110 = 2
– H = 01011 = 2
– Parents, C and F
Start Over!
– C = 01010 = 1 + 0 + 1 + 0 + 1 = 3
– F = 01100 = 3
–G=?
–H=?
–I=?