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# GENETIC ALGORITHMS by ewghwehws

VIEWS: 8 PAGES: 50

• pg 1
```									                   By
Prafulla S. Kota
Raghavan Vangipuram
 Genetic  Algorithms are a form of local search
that use methods based on evolution to make
small changes to a population of chromosomes
in an attempt to identify an optimal solution.

We discuss:
• Representations used for genetic algorithm.
• Idea of schemata.
• Genetic operators, crossover and mutations.
• Procedures used to run Genetic algorithms.

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 Genetic  programming can be used to
evolve S-expressions, which can be used
as LISP programs to solve problems.
 A string of bits is known as a
chromosome.
 Each bit is known as a gene.
 Chromosomes can be combined
together to form creatures.
 We will see how genetic algorithms can
be used to solve mathematical problems.
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1.    Generate a random population of
chromosomes.
2.   If the termination criteria are satisfied, stop.
Else, continue with step 3.
3.   Determine the fitness of each chromosome.
4.   Apply crossover and mutation to selected
chromosomes from the current generation, to
generate a new population of chromosomes –
the next generation.

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 The  size of the population should be
 The size of each chromosome must
remain the same for crossover to be
applied.
 Fittest chromosomes are selected in each
generation to produce offspring which
replace the previous generation.
 Each pair of parents produces two
offspring.

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 When    we use traditional genetic
algorithm, a metric is needed whereby
the fitness of a chromosome can be
objectively determined.
 When we consider real creatures, fitness
measure is based on the extent to which
the physical form (phenotype)
represented by the genetic information
(genotype) met certain criteria.
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 The crossover operator is applied to two
chromosomes of the same length as
follows:
1. Select a random crossover point.
2. Break each chromosome into two parts,
splitting at the crossover point.
3. Recombine the broken chromosomes by
combining the front of one with the back of the
other, and vice versa, to produce two new
chromosomes.

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example, consider the following two
 For
chromosomes:

110100110001001
010101000111101

A crossover point might be chosen
between the sixth and seventh genes.

110100 | 110001001
010101 | 000111101

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Now the chromosome parts are
recombined as follows:

110100 | 000111101 => 110100000111101

010101 | 110001001 => 010101110001001

point crossover is the most
 Single
commonly used form, but it is also
possible to apply crossover with two or
more crossover positions.

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two point crossover, two points are
 In
chosen that divide the chromosomes into
two sections, with the outer sections
considered to be joined together to turn
the chromosome into a ring. The two
sections are swapped with each other.
100110001        10 1100 001

011100110        01 0110 110

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Uniform Crossover:

• Here, a probability, p, is used to determine
whether a given bit from parent 1 will be used,
or from parent 2.
• In other words, a child can receive any random
bits from each of its parents.
Parent 1: 100011001
Parent 2: 001101110
100110001             1   1   0   10   0   1   01

011100110             0   0   1   11   0   0   10

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 Similarto Hill-Climbing, which involves
generating a possible solution to the
problem and moving toward a better
solution than the current one until a
solution is found from which no better
solution can be found.

 Mutation is a unary operator (it is applied
to just one argument – a single gene) that
is usually applied with a low probability.
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 Mutation  simply involves reversing the
value of a bit in a chromosome.
 For example, with a mutation rate of 0.01, it
might be expected that one gene in a
chromosome of 100 genes might be
reversed.

010101110001001

010101110101001
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 There  are typically two ways in which a run
of a genetic algorithm is terminated.
 Usually, a limit is put on the number of
generations, after which the run is
considered to have finished.
 The run can stop when a particular solution
has been reached, or when the highest
fitness level in the population has reached a
particular value.
 In some cases, genetic algorithms are used
to generate interesting pictures.
 In these cases, human judgment must be
used to determine when to terminate.
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 Use of Genetic Algorithm to maximize
value of a mathematic function.
 Maximize the function:
f(x)=sin(x)
over the range of ‘x’ from 1 to 15.
 Each chromosome represents a possible
value of ‘x’ using four bits.

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 Wewill use a population size of four
chromosomes.
• The first step is to generate a random
population, which is our first generation:
c1=1001
c2=0011
c3=1010
c4=0101
• To calculate the fitness of a chromosome, we
need to first convert it to a decimal integer and
then calculate f(x) for this integer.

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• We will assign a fitness numeric value from 0 to
100, where 0 is the least fit and 100 is the most fit.
• f(x) generates real numbers between -1 and 1.
• We will assign a fitness of 100 to f(x)=1 and
fitness of 0 to f(x)= -1.
• Fitness of 50 will be assigned to f(x)=0.
f’(x)= 50[f(x)+1]
= 50[Sin(x)+1]

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 The fitness ratio of a chromosome is that
chromosome’s fitness as a percentage of
the total fitness of the population.
Chromosome   Genes Integer   F(x)    Fitness   Fitness
value             F’(x)     ratio
C1           1001   9        0.41    70.61     46.3%
C2           0011   3        0.14    57.06     37.4%
C3           1010   10       -0.54   22.80     14.9%
C4           0101   5        -0.96   2.05      1.34%

Fitness values of First generation

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 Nowwe need to run a single step of our
genetic algorithm to produce the next
generation.
• First step is to select which chromosomes will
reproduce.

Roulette-wheel selection involves using the
fitness ratio to randomly select chromosomes to
reproduce.

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   The range of real numbers from 0 to 100 is
divided up between the chromosomes
proportionally to each chromosome’s fitness.
   A random number is now generated between 0
to 100. This number will fall in the range of one of
the chromosomes, and this chromosome has
been selected for reproduction.
   The next random number is used to select the
chromosome’s mate.
   Hence, fitter chromosomes will tend to produce
more offspring than less fit chromosomes.

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 Itis important that this method does not
stop less fit chromosomes from
reproducing at all.
 We will now need to generate four
random numbers to find the four parents
that will produce the next generation.
 We first choose 56.7, which falls in the
range of c2. So, c2 is parent 1.
 Next, 38.2 is chosen, so its mate is c1
(parent 2).
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 We  now combine c1 and c2 to produce
new offspring.
 Select a random crossover point.
10 | 01
00 | 11
 Crossover is applied to produce two
offspring, c5 and c6:
c5=1011
c6=0001
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 Similarly, we   calculate c7 and c8 from
parents c1 and c3:
c7=1000
c8=1011
 Observe that c4 did not get a chance to
reproduce. So its genes will be lost.
 c1 was the fittest of all chromosomes. So,
it could reproduce twice, thus passing on
its highly fit genes to all members of the
next generation.
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Chromosome   Genes   Integer   F(x)   Fitness   Fitness
value            F’(x)     ratio
C5           1011    11        -1     0         0%
C6           0001    1         0.84   92.07     48.1%
C7           1000    8         0.99   99.47     51.9%
C8           1011    11        -1     0         0%

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 It is possible to explain genetic
algorithms by comparison with natural
evolution: small changes that occur on a
selective basis combined with
reproduction will tend to improve the
fitness of the population over time.

 John Holland invented schemata to
provide an explanation for genetic
algorithms that is more rigorous.
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of numbers are used to represent
 Strings
input patterns in classifier systems. In these
patterns, * is used to represent “any value”
or “don’t care”, so that the following string:
1011*001*0
matches the following strings:
1011000100
1011000110
1011100100
1011100110

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 A schema is a string of bits that represents a
possible chromosome, using * to represent “any
value.”
 A schema is said to match a chromosome if the
bit string that represents the chromosome
matches the schema in the way shown above.

*11*
This matches the following four chromosomes

0110       0111       1110       1111

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A  schema with n *’s will match a total of
2^n chromosomes.
 Each chromosome of r bits will match 2^r
different schemata.

 TheDefining length of a schema is
defined as the distance between the first
and last defined bits in the schema.

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**10111*
1*0*1**
11111
1***1
*****10**1*****
The defining length for all the above
schemata is 4.

dL(S)<=L(S)
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of the schema is defined a the
 Order
number of defined bits in the schema.
**10*11*
1*0*1**1
1111
1***1***1***1
1*****10***1*****
The order of all the above schemata is 4.
 We denote the order of schema as O(S).
 Order of the schema tells us how specific it
is.
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   Consider the following population of 10
chromosomes, each of length 32.
C1=01000100101010010001010100101010
C2=10100010100100001001010111010101
C3=01010101011110101010100101010101
C4=11010101010101001101111010100101
C5=11010010101010010010100100001010
C6=00101001010100101010010101111010
C7=00101010100101010010101001010011
C8=11111010010101010100101001010101
C9=01010101010111101010001010101011
C10=11010100100101010011110010100001

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 Let us consider the following schema:
s0=11010*********************
 This schema is matched by three
chromosomes in our population:
c4,c5,c10.
 We say that schema s0 matches three
chromosomes in generation ‘i’ and write
this as follows:
m(s0,i) = 3
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   The fitness of a schema, S, in generation ‘i’ is
written as follows:
f(S,i)
   The fitness of a schema is defined as the average
fitness of the chromosomes in the population that
match the schema.
   Hence if we define the fitness of c4, c5, and c10
as follows:
f(c4,i)=10
f(c5,i)=22
f(c10,i)=40

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 Hence, thefitness of the schema s0 is
defined as the average of these three
values:
f(s0,i)=(10+22+40)/3
= 24

 We will now consider factors which affect
the likelihood of a particular schema
surviving from one generation to the
next.
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 Let us assume that there is a chromosome
that matches a schema, S, in the
population at time i.
 The number of occurrences of S in the
population at time i is:
m(S,i)
 Number of occurrences of S in the
population in the subsequent generation
is:
m(S,i+1)
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 The fitness of S in generation i is
f(S,i)
 We will calculate the probability that a given
chromosome, c, which matches the schema S at time
i, will reproduce and thus its genes will be present in
the population at time i+1.
   The probability that a chromosome will reproduce is
proportional to its fitness, so the expected number of
offspring of chromosome c is:
m(c,i+1)=f(c,i)/a(i)
Where a(i) is the average fitness of the
chromosomes in the population at time i.

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 Because   chromosome c is an instance of
schema S, we can thus deduce:
m(S, i+1)=f(c1,i)+…..f(Cn,i)/a(i)
where C1 to Cn are the chromosomes in
the population at time i that match
schema S.
 Let us compare this with the definition of
the fitness of schema S, f(S,i), which is
defined as follows:
f(S,i)=f(c1,i)+…..f(Cn,i)/m(S,i)
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   By combining the above stated formulas,
m(S,i+1)=f(S,i).m(S,i)/a(i)
   The more fit a schema is compared with the
average fitness of the current population, the
more likely it is that that schema will appear in a
subsequent population of chromosomes.
   There will be fewer occurrences of a given
schema whose fitness is lower than the average
fitness of the population and more occurrences
of a given schema whose fitness is higher than
average.

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 Both  mutation and crossover can destroy
the presence of a schema.
 A given schema can be said to have
survived crossover, if the crossover
operation produces a new chromosome
that matches the schema from a parent
that also matches the schema.
 For a schema to survive crossover, the
crossover point must be outside the
defining length.
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 Hence, the  probability that a schema S of
defining length dL(S) and of length L(S)
will survive crossover is
Ps(S)=1-[dL(S)]/L(S)-1
 This formula assumes that crossover is
applied to each pair of parents that
reproduce.
 Hence, after certain modifications of the
above formula:
Ps(S)>=1-(Pc)[dL(S)]/L(S)-1
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Effect of Mutation:
 Probability that mutation will be applied
is Pm.
 Hence, a schema will survive mutation if
mutation is not applied to any of the
defined bits. The probability of survival
can be defined as:
Ps(S)=(1-Pm)^O(s)
 Hence, a schema is more likely to survive
mutation if it has lower order
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   We can combine all the equations we have to
give one equation that defines the likelihood of a
schema surviving
reproduction using crossover and mutation.
   That equation represents the schema theorem,
developed by Holland, which can be stated as:
“Short, low order schemata which are fitter than
the average fitness of the population will appear
with exponentially increasing regularity in
subsequent generations.”

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   The short, low order, high-fitness schemata are
known as building blocks.
   Genetic algorithms work well when a small group of
genes that are close together represent a feature that
contributes to the fitness of a chromosome.
   Randomly selecting bit to represent particular
features of a solution is not good enough.
   Bits should be selected in such a way that they group
naturally together into building blocks, which
genetic algorithm are designed to manipulate.

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 Genetic  Algorithms can be mislead or
deceived by some building blocks into
 One way to avoid effects of deception is
to use inversion, which is a unary
operator that reverses the order of a
subset of the bits within a chromosome.
 Another way to avoid deception is to use
Messy Genetic Algorithms.
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 Developed    as an alternative to standard
genetic algorithms.
 Each bit is labeled with its position.
 A chromosome does not have to contain a
value for each position, and, a given
position in a chromosome can have more
than one value.
 Each bit in a chromosome is represented by
a pair of numbers: the first number
represents the position within the
chromosome, and the second number is the
bit value.
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 mGA’s  use the standard mutation
operation. But, instead of crossover, they
use splice and cut operations.
 Two chromosomes can be spliced
together by simply joining one to the end
of other.
 The cut operator splits one chromosome,
into two smaller chromosomes.

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 Theprocess for running this genetic
algorithm is as follows:
1.   Produce a random population of chromosomes. We
2.   Determine a score for each chromosome by
playing its strategy against a number of opponents.
3.   Select chromosomes for the population to
reproduce.
4.   Replace the previous generation with new
population produced by reproduction.

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 Most Combinatorial search problems can
be successfully solved using genetic
algorithms.
 Genetic Algorithms can be applied to:
•   Travelling Salesman Problem
•   The Knight’s tour.
•   The CNF- satisfiability problem.