Tema 3 - Algoritmos geneticos.pdf

Algoritmos genéticos

Algoritmos genéticos

Introduction

What are genetic algorithms?

What are genetic algorithms?

`

“…Search algorithms based on the mechanics of natural

l

i

d

l

i ”

selection and natural genetics.”

C

bi i l

f

h

fi

`

Combine survival of the fittest.

I

i

`

In every generation,

` A new set of artificial creatures is created

` Pieces of the fittest old ` Pieces of the fittest old

` Ocassional new parts tried for good measure.

`

Randomized but no random walk.

What are genetic algorithms?

What are genetic algorithms?...

`

Developed by John Holland and colleagues at the

University of Michigan.

`

Central research theme:

robustness

.

` Balance between efficiency and efficacy. ` Nature is better for robust performance. ` Computationally simple and powerful.

` Not fundamentally limited by restrictive assumptions about the search space

C f d d l

Traditional optimization methods

Traditional optimization methods

`

Search methods

` Random search

` Do not use information about the search space.

` Simulated Annealing is a randomized technique (not a random ` Simulated Annealing is a randomized technique (not a random

search)

`

Calculus-based

` Local in scopep

` Depend upon the existence of derivatives

`

Enumerative

Optimization

Optimization

`

Improve performance toward some optimal point or

points

1. Seek improvement to approach some …

` Process of improvement

2. … optimal point (destination)

` U l t i t d id if d i d h ` Usual metric to decide if a procedure is good enough.

` Is like asking a person to be perfect when solving problems.

`

Nature: the most important goal of optimization is

improvement

improvement

.

GA vs Traditional methods

GA vs. Traditional methods

`

Four differences:

1. GAs work with a coding of the parameter set, not the

parameters themselves.

2. GAs search from a population of points, not a single point

3. GAs use payoff (objective function) information, not

derivatives or other auxiliary knowledge derivatives or other auxiliary knowledge.

GA vs Traditional methods

GA vs. Traditional methods …

`

Natural parameter set of the optimization problem must

be coded as a

finite-lenght string over some finite

alphabet

.

`

GAs work in parallel.

`

GAs are blind.

Simple Genetic Algorithm

p

g

Simple Genetic Algorithm

Simple Genetic Algorithm

`

Simplicity of operation and power of effect two main

attractions of GA approach.

`

Composed of three operators:

Reproduction

Reproduction

`

Process in which individual strings are

copied

according to

their objective function values

f

their objective function values,

f

.

` Biologist call it fitness function.

` Measure of profit, utility of goodness.

` We want to maximize it.

`

Strings with higher value have a higher

probability

of

Strings with higher value have a higher

probability

of

contributing one or more offspring in the next generation.

`

Artificial version of natural selection (Darwinian survival of the

Reproduction

Reproduction …

`

Biased roulette

` Each current string has a slot sized in proportion to its fitness.

f

P l t

∑

=

f

f

Pselect

f

unt

ExpectedCo

=

A i l

d f

d

i i

li

d i h

_

f

p

`

A string selected for reproduction is replicated in the

Crossover

Crossover

`

Two steps:

1 Members of the new reproduced strings are mated at random 1. Members of the new reproduced strings are mated at random.

2. Each pair of strings undergoes crossing:

1. An integer position k along the string is selected between 1 and the

string length less one [1, l – 1]

T i d b i ll h b

2. Two new strings are created by swapping all characters between

positions k + 1 and l inclusively.

`

The process of reproduction and crossover is a kind of

exchange – much of the power of GA.

g

p

` High performance notions are repeatedly tested and exchanged in

Mutation

Mutation

` Secondary role in GAs.

` Mutation is the occasional random alteration of the value of a string position.g p

` Needed because some information, relative to the position of

h b l d

characters, may be lost during crossover.

` Is a random walk through the string space ` Is a random walk through the string space.

Example

Example

` Maximize f(x) = x2

` x varies between 0 and 31 ` x varies between 0 and 31.

1. Coding

1. Binary unsigned integer length 5.

2 Select initial population at random. 2. Select initial population at random.

3. Repeat

1. Reproduction

2. Crossover

Example

Example …

Initial Expected Actual

String No. Population x f(x) Pselect count Count

1 01101 13 169 0.14 0.58 1

2 11000 24 576 0.49 1.97 2

3 01000 8 64 0.05 0.22 0

4 10011 19 361 0.31 1.23 1

Sum 1170 1 4 4

Generation 1 Average 292.5 0.25 1 1

Max 576 0 49 1 97 2

Example

Example …

Mating Crossover New 

P l M t Sit P l ti

Pool Mate Site Population

0110|1| 2 4 01100

1100|0 1 4 11001

11|000 4 2 11011

Example

Example …

Initial Expected Actual

String No. Population x f(x) Pselect count Count

1 01100 12 144 0.08 0.33 0

2 11001 25 625 0.36 1.43 1

3 11011 27 729 0.42 1.66 2

4 10000 16 256 0.15 0.58 1

Sum 1754 1 4 4

Exercises

Exercises

`

The following exercises must be handed in at the end of

the class.

Exercise 1

Exercise 1

`

Create a list of

N

numbers that equal

X

when summed

together

`

For instance:

` N = 5 ` X = 100 ` Solutions:

` [20,20,20,20,20]

` [50,20,10,10,10]

[10 20 30 40 0]

Exercise 2

Exercise 2

Exercise 3

Exercise 3

B

25 52

25

63

52

A D

33

48 19

Lingo

Lingo

Nat ral

Genetic Algorithm

Natural

Genetic

Algorithm

chromosome

string

gene

feature,

character,

or

detector

allele

feature

value

locus

string

position

genotype

structure

Mathematical Foundations

Similarity templates (Schemata)

Similarity templates (Schemata)

`

What information is contained in a population of strings

d h i bj

i f

i h l id i

?

and their objective function to help guide improvement?

String Fitness

01101 169 11000 576 01000 64

A h

i il i i

i

?

01000 64 10011 361

`

Are there any similarities among strings?

A h

l l

h b

h

l

`

Are there causal relationships between these similarities

Schemata

Schemata …

`

Schema

: similarity template describing a subset of

strings with similarities at certain string positions.

` Uses the * or don’t care symbol.

` For instance:

{ }

` Binary alphabet {0,1}

` Schemata created from the ternary alphabet {0, 1, *}

` Some schemata of length 6: ` Some schemata of length 6:

… *01010

… 1*0*0*

Schemata

Schemata …

`

Schema

` Powerful and compact way to analyze well-defined similarities.

(

)

` Number of possible schemata in a string

` l = length of string

k di li f l h b

(

)

k

+

1

` k = cardinality of alphabet

` Example: l = 6, k = 2; 3*3*3*3*3*3 = 36 = (2+1)6 = 729

Schemata

Schemata …

`

All schemata are not analyzed, only those present in a

population

population.

` Number of schemata in a binary string are: 2l (26 = 64)

f 2l ₍₂l

` Binary population of size n contains between 2l and n(2l) schemata.

` Example. n = 5, between 64 to 320 schemata.

` Several schemata are processed at each iteration.

` Some schemata are better: ` Some schemata are better:

` 0****1 ` 11****

` Are these schemata preserved by reproduction, crossover and

Fundamental Theorem Notation

Fundamental Theorem – Notation

`

Basic notation

{ }

` Binary alphabet

` Strings

{ }

0

,

1

=

V

` a_i is a single feature (gene) that can take a value (allele).

7 6 5 4 3 2

1a a a a a a

a A =

` Order of genes need not to be sequential A′ = a₂a₇a₃a₁a₅a₆a₄

` Population of individual strings A_j, j = 1, 2, …, n

` A(t) population at time t (generation) ` A(t) population at time t (generation).

` Schema alphabet V + =

{ }

Fundamental Theorem Notation

Fundamental Theorem – Notation …

`

Notation …

` All schemata are not created equal.

` Order of a schema H is the number of fixed positions: o(H)

` The defining length of a schema H is the distance between the first

and the last specific string position: p g p _δ

( )

` Examples:

( )

δ

… H₁ = *1**10*

Fundamental Theorem Reproduction

Fundamental Theorem – Reproduction

`

Effect of reproduction in schemata

` At time t there are m examples of a schema H in population

A(t)

( )

m = ,

` String A_i is selected with probability

( )

∑

p_i i

g _i p y

∑

` Number of elements of schema H at time t+1

(

H

t

+

)

=

m

( )

H

t

⋅

n

⋅

f

( )

H

m

1

` f(H) is the average fitness of strings representing schema H

(

)

( )

∑

⋅

⋅

=

+

f

n

t

H

m

t

H

m

,

1

,

Fundamental Theorem Reproduction

Fundamental Theorem – Reproduction …

f

∑

`

Average fitness of entire population:

n f

f =

∑

`

therefore:

(

)

( )

( )

( ) ( )

f H f t H m f H f n t H m t H

m , +1 = , ⋅ ⋅ = ,

∑

` Above population average: increasing number of samples in next population

f f

∑

next population

Fundamental Theorem Reproduction

Fundamental Theorem – Reproduction …

` Ab d li

` Above-average grow and live. ` Below-average die off.

` How is this growth? Exponential

` Assume schema H remains above average an amount

c

f

(

)

( )(

) ( ) ( )

f f c f t H m t H

m , +1 = , + = 1+ ⋅ ,

` Starting at t = 0 and assuming c stationary:

( )

(

) (

)

c

H

m

t

H

m

=

0

⋅

1

+

` Geometric progression with with common ratio greater than 1, i.e. exponential

growth

( )

H

t

m

(

H

) (

c

)

m

,

=

,

0

⋅

1

+

Fundamental Theorem Crossover

Fundamental Theorem – Crossover

`

Effect of crossover in schemata

` Crossover: structured yet randomized information exchange between strings.

` Affects schemata

` Examples:

… A = 1101010 … H₁ = *1****0 … H₂₂ = ***10**

… Crossover sites: 3 and 5

Fundamental Theorem Crossover

Fundamental Theorem – Crossover …

( )

H

δ

`

Schema H is destroyed with probability:

_{( )}

( )

1

−

=

l

H

p

δ

( )

H

δ

`

and survives with probability:

( )

( )

1

1

1

−

−

=

−

=

l

H

p

p

δ

`

If crossover is selected random with probability p

If crossover is selected random with probability p

( )

( )

1

−

⋅

≥

p

H

p

δ

( )

1

1

−

≥

l

p

Fundamental Theorem Crossover

Fundamental Theorem – Crossover …

`

The combined effect of reproduction and crossover is:

(

)

( ) ( )

_⎢⎣

⎡

( )

_⎥⎦

⎤

−

⋅

−

≥

+

1

1

,

1

,

l

H

p

f

H

f

t

H

m

t

H

m

δ

` Schema survives or dies based on a factor. This factor depends

⎦

⎣

l

−

1

f

p on:

` Reproduction: schema above or below the average.

` Crossover: schema has a short or long defining length.

` Ab d h t d fi i l ht l d

` Above average and short defining lenght are sampled

Fundamental Theorem Mutation

Fundamental Theorem – Mutation

` Effect of mutation in schemata

` Mutation is a random alteration of genes with probability p ` Mutation is a random alteration of genes with probability p_m

` A schema survives if all its fixed positions survive with the same value, i.e. all

alleles survive.

` A single allele survives with probability

(

)

m

p

−

1

` Therefore a schema with o(H) fixed positions has the following probability to

survive:

(

)

m p

−

1

` For small values of p_m (p_m << 1) this expression can be approximated by:

( )

o ⋅

−

1

` Thefore, the lower the order, the greater the probabilities of a schema to

survive

( )

o ⋅

1

Fundamental Theorem

Fundamental Theorem

`

Taking into consideration all three factors:

(

+

)

≥

( ) ( )

_⎢⎣

⎡

−

⋅

( ) ( )

−

o

H

p

⎤

_⎥⎦

l

H

p

f

H

f

t

H

m

t

H

m

1

1

,

1

,

δ

⎥⎦

⎢⎣

l

−

f

1

`

Schemata Theorem (Fundamental Theorem of GAs)

` Short (crossover) ` Short (crossover),

` Low order (mutation),

` Above average (reproduction) ` Above average (reproduction)

Implicit parallelism

Implicit parallelism

`

GA processes at each cycle

n

strings.

`

At the same time it processes n

p

schemata.

Building block hypothesis

Building block hypothesis

`

Building blocks

. Highly fit schemata of low defining

lenght and low order.

`

These are sampled, recombined, and resampled to form

Computer Implementation of GA

p

p

Data structures

Data structures

`

Construct a population as an array of individuals with:

` Phenotype (decoded parameter of parameters) ` Genotype (artificial chromosome or string)

` Fitness (objective function)

Performance of GAs

Good performance of GAs

Good performance of GAs

`

Depend of a good choice of (De Jong, 1975):

` High crossover probability. ` Low mutation probability

` Inversely proportional to the population size

` Moderate population size

` For the problem of x10 the selection was:

` P t ti = 0 0333 ` Pmutation = 0.0333

` Pcrossover = 0.6

Mapping Objective Functions to Fitness

Form

Form

`

A fitness function must be a

nonnegative

figure of merit.

`

Necessary to map a natural objective function to a fitness

function.

`

Cost-fitness (minimization) transformation in GAs

( )

_⎨

⎧

C

−

g

( )

x

when

g

( )

x

<

C

f

( )

( )

( )

⎩

⎨

⎧

=

otherwise

g

g

x

f

0

max max

`

C

may be taken as the largest value of the cost function

g(x)

Mapping

Mapping …

`

Profit or utility (maximization) in GAs:

( )

x

=

_⎨

⎧

u

( )

x

+

C

when

u

( )

x

+

C

>

f

( )

0

⎩

⎨

otherwise

f

0

`

C

may be the absolute value of the worst

u

value in the

Fitness scaling

Fitness scaling

`

Regulation of the number of copies important in small

populations

populations

` Start of GA

` Few extraordinary stringsy g

` Normal selection rule

E di i di id l k i ifi i f l i

` Extraordinary individuals take over a significant proportion of population

very quickly

` Causes premature convergence.

` Later in run

` Not enough diversity ` Not enough diversity

` Population average fitness close to best fitness

Fitness scaling

Fitness scaling …

`

Fitness scaling helps.

b

f

f

′

+

` Linear scaling is very useful.

` Coefficients a and b are calculated based on:

b

af

f

=

+

` Coefficients a and b are calculated based on:

` Selection for reproduction insures each average member contributes

one offspring to next generation.

f

f

′

` Number of offsprings of a member can be controlled by

avg avg

f

f

′

=

` Number of offsprings of a member can be controlled by

avg mult

f

C

f

′

=

⋅

` C_mult is the number of expected copies of the best member

Codings

Codings

`

Codes used in GA can be any.

` Programmer limited by imagination.

`

What other code can be assigned to the problem

f(x) = x

in [0,31]?

`

GA exploit similarities in codes as long as building blocks

p

g

g

lead to near optima.

Codings

Codings …

`

Principle of meaningful building blocks

` The user should select a coding so that short, low-order

schemata are relevant to the underlying problem and relatively

l t d t h t th fi d iti

unrelated to schemata over other fixed positions.

` Can be formaly verified but usually is done empirically ` Can be formaly verified but usually is done empirically.

`

Principle of minimal alphabets

Principle of minimal alphabets

` The user should select the smallest alphabet that permits a natural expression of the problem

Codings

Codings …

Binary String Value X Nonbinary String Fitness 0  1  1  0  1 13 N 169 1  1  0  0  0 24 Y 576

0 1 0 0 0 8 I 64

0  1  0  0  0 8 I 64 1  0  0  1  1 19 T 361

Coding Correspondence Table Coding Correspondence Table

Binary Nonbinary

0  0  0  0  0 A 0  0  0  0  1 B

. .

. .

. .

1  1  0  0  1 Z

1 1 0 1 0 1

1  1  0  1  0 1 1  1  0  1  1 2

Codings

Codings …

`

Binary code

` Search for similarities possible by small cardinality ` l = 5

` Number of strings = 25 = 32

` Number of schemata = 35 = 243

`

Nonbinary code

` No similarities to exploit ` l = 1

Multiparameter mapped fixed point coding

Multiparameter, mapped, fixed-point coding

`

Coding of multiparameter optimization problems of real

parameters.

`

Map linearly from

[0,2

]

to an interval

[U

, U

]

.

`

Precision calculated by:

U U

1 2

min max

− − = U _l U

Constraints

Constraints

` Usually checked when evaluating the objective function.

` In some cases we want to degrade fitness value based on the degree of constraint violation.

` Penalty method

` A constrainded problem in optimization associates a cost or penalty with all constraint

violations.

E l

` Example

` Minimize ` Subject to

,... 2 , 1 0 ) ( ) ( n i x h x g

i ≥ =

` Minimize g(x) r

[

]

n i Φ +

∑

[

]

∑

( )

[

]

( )

Φ

t. coefficien penalty

Example

p

Example

Example

`

Evolutionary Approach for Strategy Learning in RoboCup

Soccer.

Nakashima et al. (2004). IEEE International Conference on Systems, Man and Cybernetics.

` Evolutionary method to obtain team strategies for soccer agents.

M f fi d

Example

Example …

`

Each soccer agent has a set of action rules.

`

Action rules used only when agent keeps ball, i.e. ball is

y

g

p

Example …

Aj is one of the 48 subareas of the field

of the field.

Bj examines whether the nearest opponent is near the agent or not, i.e.near

or not near.

Example …

p

Example

Example …

`

Encoding scheme

Th 48 2 96 i l f

` There are 48 x 2 = 96 action rules for an agent.

` 10 agents need to calculate rules (not the keeper), therefore, 960 ` 10 agents need to calculate rules (not the keeper), therefore, 960

rules in a team.

T k E l i i f l h 960 b i i

` Task: Evolve integer strings of length 960 to obtain team strategies

with high performance.

Example

Example …

`

Evaluation of integer strings

` Scores of soccer games.

1. Scored goals 2. Goals against.

`

Evolutionary operation

Evolutionary operation

1. Initialization

2 Generation of new integer string uses mutation only 2. Generation of new integer string – uses mutation only. 3. Performance evaluation.

4 Generation update best strings of current and new 4. Generation update – best strings of current and new

Example …

Experimental settings:

Number of integer strings in a population: one.

Number of new strings generated by mutation: one.

Probability of mutation for Probability of mutation for each gene: 1/96.

Example …

First Evaluation Scheme:

Scheme:

Current integer string directly competes with the new integer string

generated by the mutation operator.

Example …

Second Evaluation Scheme:

Each current integer string and the new generated integer string compete against UvA Trilearn 2003 base team.

I i i h h

Integer string with the

Improvements in basic technique

p

q

Alternate Selection Schemes

Alternate Selection Schemes

`

Brindle’s (1981) schemes:

` Deterministic sampling

` Remainder stochastic sampling without replacement ` Remainder stochastic sampling without replacement

` Stochastic sampling without replacement ` Stochastic sampling without replacement

` Remainder stochastic sampling with remainder

` Stochastic sampling with replacement

Alternate Selection Schemes

Alternate Selection Schemes …

`

Deterministic sampling

S l i l l d l pselect = fi

` Selection calculated as usual:

` Expected number of individuals e_i is calculated as:

` The integer part of e_i is the number of samples for individuals.

∑

g p _i p

` Population is sorted according to fractional parts of e_i.

` The remainder strings needed are drawn from the top of the sorted list.

`

Remainder stochastic sampling without replacement

` Starts as deterministic sampling. ` Starts as deterministic sampling.

` Integer part of e<sub>i</sub> is used in the same way. ` Fractional parts are treated as probabilities.

` Weighted coin tosses (Bernoulli trials) are performed.

Alternate Selection Schemes

Alternate Selection Schemes …

`

Stochastic sampling without replacement

f

` Each time a string is selected for

` Mating and crossover its offspring count is decreased by 0 5

f

f

` Mating and crossover its offspring count is decreased by 0.5.

` Without mating and crossover its offspring count is decreased by 1.0.

` If the count was below 0.0 the string was not available for g selection.

R

i d

h i

li i h

l

`

Remainder stochastic sampling with replacement.

` Starts as deterministic sampling.

` Integer part of e is used in the same way ` Integer part of e_i is used in the same way.

Alternate Selection Schemes

Alternate Selection Schemes …

`

Stochastic sampling with replacement

` Fancy name for normal selection.

`

Stochastic tournament (Wetzel ranking)

` Selection probabilities are calculated normally.

` Pairs of individuals are drawn using roulette wheel selection. ` After drawing a pair,

` String with highest fitness is the winner and inserted in the new

population.

Scaling mechanisms

Scaling mechanisms

`

Done to keep appropiate levels of competition in a

simulation.

`

Without scaling

` Few superindividuals dominate the selection process early on the simulation.

` Scaling helps to scale back superindividuals to prevent them to

takeover the population takeover the population.

` Later on population is largely converged competition is less ` Later on, population is largely converged, competition is less

strong and simulation tends to wander.

Scaling mechanisms

Scaling mechanisms …

`

Linear scaling

b

af

f

′

=

+

` Works well.

` Can have problems with negative values.

`

Sigma truncation

` Constant c is chosen as a reasonable multiple of the population

)

(

´

=

f

−

f

−

c

⋅

σ

f

standard deviation (between 1 and 3)

` Negative results are set to 0.

` There is no danger of negative values ` There is no danger of negative values.

`

Power law scaling

g

f

´

=

f

` K is problem-dependent

May require change during a run to stretch or shrink the range.

Ranking procedures

Ranking procedures

`

Alternative to scaling.

`

Population sorted according to objective function value.

`

Individuals assigned offsprings as a function to their rank.

` Rank 1 is the maximum count ` Rank 1 is the maximum count.

`

Ranking has the same resistance to overselection and

g

underselection as normal selection in conjunction with scaling.

`

Problem

Generalized crossover model

Generalized crossover model

`

Proposed by De Jong (1975).

` Parameter: number of crossover points (CP).

` CP = 1 implies simple crossover.

` Even CP implies treating the string as a ring.

… Crossover points selected around the ring uniformily at random. … Crossover points selected around the ring uniformily at random.

` Odd CP implies

… Position 0 is always a default crossing point.

` Not very good results ` Not very good results.

Advanced Operators in GA

p

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance

`

Diploidy – Pair of chromosomes.

`

Haploidy – Single stranded chromosome.

`

Di l id f

`

Diploid form

` Genotype carries one or more pairs of chromosomes (homologous

chromosomes).)

` Contain information for the same functions (decode to the same

function) function).

` Example:

` AbCDe

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance …

`

AbCDe

` Each locus of a letter represents an allele.

` Capital letters and lowercase letters are different phenotypic characteristics.

` Only one allele can be used at each locus.

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance …

`

Dominance

` Primary mechanism for eliminating the conflict of redundancy.

` The dominant allele takes precedence over alternative alleles ` The dominant allele takes precedence over alternative alleles

(recesives).

` An allele is dominant if it is expressed when paired with some other

allele.

` Example: ` Example:

` If capital letters are dominant and lowercase are recessive then:

… AbCDe … aBCde

` will produce:

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance …

`

Heterozygous (mixed, Aa = A)

`

Homozygous (pure, aa = a)

yg

(p

)

`

Dominant are heterozygous or homozygous

Dominant are heterozygous or homozygous.

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance …

`

Diploidy (use of)

` Mechanism for remembering alleles and allele combinations that were previously useful

M l l l d l h l l

` Multiple solutions carried along with only one particular solution expressed.

`

Dominance (use of)

` O t t hi ld b d ll l f h f l l ti

` Operator to shield remembered alleles from harmful selection in a hostile environment.

` Dominance change allow old lessons to be remembered and ` Dominance change allow old lessons to be remembered and

Implementation of diploidy and dominance

p

p

y

H ll ti (1971) D i

Hollstien (1971) Dominance map

` Each binary gene described by

two genes two genes

` Modifier gene

` Functional gene

0M 0m 1M 1m

` Functional gene takes 0 or 1 as

normal.

0M

0

or m.

0m

1M

there is at least one M allele

Implementation …

p

Hollstein Holland (1975)

D

i

Hollstein-Holland (1975)

Dominance map

`

As in Hollstein (1971) but:

0

1

1

(

)

` 3-alphabet {0,1₀,1}

0

1

₀

1

` 1 is a dominant 1.

0

` 1₀ is a recesive 1.

1

₀

Dominance Diploidy and Abeyance

Dominance, Diploidy and Abeyance …

`

Dominance-diploidy

` For shielding or abeyance

Reordering and Inversion

Reordering and Inversion

`

In GAs, it is not clear if a building block will lead to a

solution of a problem.

`

Nature is also not sure about its codings

` Has operators to search for

` Better codings – good string arrengements

Reordering and Inversion

Reordering and Inversion …

`

Inversion operator

` Primary natural mechanism responsible for recoding a problem.

` Two points are chosen along the chromosome.

` The end points are swaped.

` Example:

1

01

1

1

01

1

Reordering and Inversion

Reordering and Inversion …

`

Simple representation of GA

` Inversion has no meaning

` Value of an allele is dependent of its position

187

1

01

1

1

01

1

01

1

1

01

1

=

187

1

=

∧ ∧

159

10011111

I ll l ’ i i i i d d f i i

159

10011111

=

Reordering and Inversion

Reordering and Inversion …

`

New representation

1

01

11

01

1

12345678

`

changes to

1

01

11

01

1

∧ ∧

`

changes to

12654378

`

Allele value retains its meaning.

10011111

`

Allele value retains its meaning.

Reordering and Inversion

Reordering and Inversion …

`

Crossover with inversion

` Inversion has no problem with mutation, selection and decoding.

N bl

` Naive crossover constitutes a problem ` Example:

|

A A' 1234 | 43 8

1011 | 1011 5678 | 1234 = A 1111 | 1011 4378 | 1234 '= A 4378 | 1265 | =

B ' 1265 | 5678

| = B 1111 | 1001

Reordering and Inversion

Reordering and Inversion …

`

Reordering operators

` Partially Matched Crossover (PMX)

` Order Crossover (OX)

Reordering and Inversion

Reordering and Inversion …

` Partially Matched Crossover (PMX)

` Crossover of permutations ` Crossover of permutations

` Guarantees each position is found one in each chromosome.

1. Chromosomes are alligned. 1. Chromosomes are alligned.

2. Crossing sites are selected uniformily at random places (match section) 3. Matching section effect a crossover position by position.

4. Alleles are moved to their new positions in the chromosome. 4. Alleles are moved to their new positions in the chromosome.

` Example:

6

4

5

9

|

10

3

2

|

1

7

8

10

2

3

1

|

7

6

5

|

4

8

9

=

B

A

3

4

2

9

|

6

|

1

10

8

'

7

5

6

1

|

10

3

2

|

4

8

9

'

=

A

6

4

5

9

|

10

3

2

|

1

7

8

=

Reordering and Inversion

Reordering and Inversion …

`

Partially Matched Crossover (PMX)

` Useful for problems like Travel Salesman Problem (blind or normal versions)

` Chromosomes are a sequence of integers. ` Chromosomes are a sequence of integers.

` Each integer is a different city (permutation encoding)

87654321

` Permutation encoding

` Interest is on labels not on alleles

_12345678

00000000

12345678

` For problems where fitness function depend only on ordering of

Reordering and Inversion

Reordering and Inversion …

` Order Crossover (OX)

1 Stars similar to PMX – select a

A

=

9

8

4

|

5

6

7

|

1

3

2

10

matching section.

2. Map elements in the

6

4

5

9

|

10

3

2

|

1

7

8

=

B

3 Mapped elements leave a hole

H

9H4

|

10

23

|

1

H

8

=

B

3. Mapped elements leave a hole

(H) in the current chromosome.

4 Move holes with a sliding motion

_B

=

₂₃

₁₀

_H

_H

_H

₉₄₈₁

starting at the second crossing point.

9481

H

H

H

10

23

=

B

5. Fill in holes with the matching

Reordering and Inversion

Reordering and Inversion …

`

Order Crossover (OX)

` PMX and OX are very similar.

` PMX exploits absolute position similarities.

Reordering and Inversion

Reordering and Inversion …

` Cycle Crossover (CX)

1 Start with the first gene and fix it. C = 98217 4510 6 3 1. Start with the first gene and fix it.

2. The matching chromosome

defines the next position to be

10 9 8 7 6 5 4 3 2 1 D = p

considered using the value on the first gene (the one used first)

3 The corresponding gene in the

− − − − − − − − − = 9 ' C

3. The corresponding gene in the

matching chromosome defines

the next gene to consider. C'= 9− −1−− − − −−

4. Repeat until a cycle is found.

5. Crossover the remaining genes.

− − − − − −

Exercises Reordering

Exercises - Reordering

`

Results must be submitted at the end of the class.

`

Given the following pairs of chromosomes, what are the

g p

reorderings using?

` PMX

A

=

A

B

C

D

E

F

G

H

I

J

K

L

M

` OX ` CX

K

H

C

M

B

E

L

I

F

D

G

A

J

=

B

` Crossing points:

C

=

0

1

2

3

4

5

6

7

8

9

10

` A and B: 6, 11

` C and D: 2, 5