Genetic algorithm of AI

1. Genetic Algorithm

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states
(often encoded as a string of 0s and 1s)
better states.
crossover, and mutation
Genetic/ Evolutionary algorithms
 Developed by John Holland in 1975
 inspired by biological mutation & evolution
 stochastic (means non-deterministic) search techniques
based on the mechanism of natural selection and genetics
 A successor state is generated by combining two parent
 Start with k randomly generated states (population)
 A state is represented as a string over a finite alphabet
 Evaluation function (fitness function). Higher values for
 Produce the next generation of states by selection,
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“Select The Best, Discard The Rest”
 fitness function is upto user (problem specific)-
means to evaluate the potential of a candidate in
the population. e.g. for a binary representation
fitness function could be:
 f(i) = ∑i f(i) ;where i is from 1 to l (l=total length of the sequence)
 And in the selection step, the probability of getting
selection for a candidate is:
 Prob i = f(i) / ∑i f(i)
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Encoding
The process of representing the solution in the form of a string that
conveys the necessary information.
 Binary Encoding – Most common method of encoding.
Chromosomes are strings of 1s and 0s and each position in the
chromosome represents a particular characteristic of the problem.
Permutation Encoding – Useful in ordering problems such as the
Traveling Salesman Problem (TSP). Example. In TSP, every
chromosome is a string of numbers, each of which represents a city to
be visited.
 Value Encoding – Used in problems where complicated values, such
as real numbers, are used and where binary encoding would not suffice.
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e.g. n-queen
 GA representation and encoding of the following
arrangement can be written as: fitness of the chromosome
v1 (24748552) is 28 – 4 = 24 (No. of non attacking pair of queens)
• Encoding is given by the row no. of each column.
 That is because only 4 pairs of queens attack each other:
 The queens on 1st and 8th column
 The queens on 2nd and 4th column
 The queens on 6th and 7th column
 The queens on 3rd and 8th column
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Genetic algorithms
 Fitness function: number of non-attacking pairs of queens
(min = 0, max = 8c2 excluding same pair = 28)
 24/(24+23+20+11) = 31%

 23/(24+23+20+11) = 29% etc
 Here single point crossover & mutation occurred
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Genetic/ Evolutionary algorithms
 Outline of the Basic Genetic Algorithm
1 [Start] Generate random population of n chromosomes (suitable solutions for the
problem)
2 [Fitness] Evaluate the fitness f(x) of each chromosome x in the population
3 [New population] Create a new population by repeating following steps until the new
population is complete
 [Selection] Select two parent chromosomes from a population according to their
fitness (the better fitness, the bigger chance to be selected)
 [Crossover] With a crossover probability cross over the parents to form a new
offspring (children). If no crossover was performed, offspring is an exact copy of
parents.
 [Mutation] With a mutation probability mutate new offspring at each locus (position
in chromosome).
 [Accepting] Place new offspring in a new population (keep population size constant)
4 [Replace] Use new generated population for a further run of algorithm
5 [Test] If the end condition is satisfied, stop, and return the best solution in current
population
 [Loop] Go to step 3
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Genetic/ Evolutionary algorithms
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Genetic algorithms
By doing this Crossover, it helps to accelerate the search at an
early stage of evolution
For detail about genetic algorithm refer to GA1.pdf in
ftp://10.220.20.25/CSE 4701
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Another Example
(Assignment 1: Deadline 15 Feb, 2018)
 Data population: RGB colours
 Aim: to obtain darkest colour represented by
(0, 0, 0)
 This is a minimisation problem, i.e. a good
colour is one that fits for (colour) --> 0.
 We now tabulate our data as shown
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 Start at a random pattern like this:
Colour Red Green Blue
C1 80 170 689
C2 130 690 15
C3 24 8 317
where
Fitness for (C1) = 80 + 170 + 689 = 939
Fitness for (C2) = 130 + 690 + 15 = 835
Fitness for (C3) = 24 + 8 + 317 = 349
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Fitness (C2) = 130 + 690 + 15 = 835
C1
 Start at a random pattern like this:
Colour Red Green Blue
C1 80 170 689
C2 130 690 15
C3 24 8 317
FITTEST
PLACED TOP
C3
Fitness (C1) = 80 + 170 + 689 = 939
C2
Fitness (C3) = 24 + 8 + 317 = 349
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 After a Selection is done on the sample:
** Remember, this is a minimisation problem..
Colour Fitness
C3 349
C2 835
C1 939
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C4 24 8 15C4 is crossover(C3,C2)= (24, 8, 15)
C6 is crossover(C2,C1)= (130, 690, 689)
GA: Reproduction & Crossover
 So far, we have this
Colour Red Green Blue Colour Fitness
C1 80 170 689 C3 349
C2 130 690 15 C2 835
C3 24 8 317 C1 939
Next step is to reproduce the pattern, like this,
by crossing over:
Colour Red Green Blue
C5 is crossover(C3,C1)= (24, 8, 689)
C5 24 8 689
C6 130 690 689
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 perform mutation, and we have
C7 is obtained by mutating(4) =(24, 8, 13)
C8 is obtained by mutating(5) =(25, 9, 689 )
C9 is obtained by mutating(6) =(128, 688, 689)
Colour Red Green Blue
New population of C7 24 8 13
3 chromosomes C8 25 9 689
C9 128 688 689
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Conclusion (up to “mutation” to get a new data set)
 Some solutions have improved (after first iteration):
Getting better rather fast
Slightly improved of the answer
Fitness for C7 = (24 + 8 + 13) = 45
Fitness for C8 = (25 + 9 + 689) = 743
Fitness for C9 = (128 + 688 + 689 ) = 1505
If the process is iterated, population will converge to
have fitness near to zero (colour) --> 0.
Continue..............
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