Genetic Algorithms CSCI-2300 Introduction to Algorithms David Goldschmidt, Ph.D. Rensselaer Polytechnic Institute April 29, 2013

Genetic AlgorithmsCSCI-2300 Introduction to Algorithms

David Goldschmidt, Ph.D.Rensselaer Polytechnic InstituteApril 29, 2013

Evolutionary Computing

Evolutionary computing produceshigh-quality partial solutions to problems throughnatural selection andsurvival of the fittest

– Compare to naturalbiological systems thatadapt and learn over time

Integer Binary code Integer Binary code Integer Binary code1 112 7 123 8 134 9 145 10 15

6 1 0 1 11 1 0 01 1 0 11 1 1 01 1 1 1

0 1 1 00 1 1 11 0 0 01 0 0 11 0 1 0

0 0 0 10 0 1 00 0 1 10 1 0 00 1 0 1

Genetic Algorithm Example

Find the maximum value of function f(x) = –x2 + 15x

– Represent problem using chromosomes built from four genes:

http://w

ww.webgr

aphing.c

om


Initial random population of size N = 6:

Chromosomelabel

Chromosomestring

Decodedinteger

Chromosomefitness

Fitnessratio, %

X1 1 1 0 0 12 36 16.5X2 0 1 0 0 4 44 20.2X3 0 0 0 1 1 14 6.4X4 1 1 1 0 14 14 6.4X5 0 1 1 1 7 56 25.7X6 1 0 0 1 9 54 24.8

x

50

40

30

20

60

10

00 5 10 15

f(x)

(a) Chromosome initial locations.x

50

40

30

20

60

10

00 5 10 15

(b) Chromosome final locations.


Determine chromosome fitness foreach chromosome:

218 100.0

Chromosomelabel

Chromosomestring

Decodedinteger

Chromosomefitness

Fitnessratio, %

X1 1 1 0 0 12 36 16.5X2 0 1 0 0 4 44 20.2X3 0 0 0 1 1 14 6.4X4 1 1 1 0 14 14 6.4X5 0 1 1 1 7 56 25.7X6 1 0 0 1 9 54 24.8

x

50

40

30

20

60

10

00 5 10 15

f(x)

(a) Chromosome initial locations.x

50

40

30

20

60

10

00 5 10 15


fitness function here issimply the original function

f(x) = –x2 + 15x

100 0

36.743.149.5

75.2

X1: 16.5%X2: 20.2%X3: 6.4%X4: 6.4%X5: 25.3%X6: 24.8%


Use fitness ratios to determine which chromosomes are selected for crossoverand mutation operations:


Converge on a near-optimal solution:

Chromosomelabel

Chromosomestring

Decodedinteger

Chromosomefitness

Fitnessratio, %

X1 1 1 0 0 12 36 16.5X2 0 1 0 0 4 44 20.2X3 0 0 0 1 1 14 6.4X4 1 1 1 0 14 14 6.4X5 0 1 1 1 7 56 25.7X6 1 0 0 1 9 54 24.8

x

50

40

30

20

60

10

00 5 10 15

f(x)

(a) Chromosome initial locations.

x

50

40

30

20

60

10

00 5 10 15


Convergence Example

local maximum

global maximum

Genetic Algorithms – Step 1

Represent the problem domain asa chromosome of fixed length– Use a fixed number of genes to represent a

solution– Use individual bits or characters for efficient

memory use and speed

– e.g. Traveling Salesman Problem (TSP) http://www.lalena.com/AI/Tsp/

1 10 1 0 1 0 0 0 0 0 1 0 1 10


Define a fitness function f(x) to measurethe quality of individual chromosomes

The fitness function determines– which chromosomes carry over to the next

generation – which chromosomes are crossed over with

one another– which chromosomes are individually mutated


Establish our genetic algorithm parameters:– Choose the size of the population, N – Set the crossover probability, pc

– Set the mutation probability, pm

Randomly generate an initial populationof chromosomes:– x1, x2, ..., xN

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

1 10 1 0 1 0 0 0 0 0 1 0 1 10

. . .


Calculate the fitness of eachindividual chromosome using f(x):– f(x1), f(x2), ..., f(xN)

Order the population based on fitness values

100 0

36.743.149.5

75.2

X1: 16.5%X2: 20.2%X3: 6.4%X4: 6.4%X5: 25.3%X6: 24.8%


Using pc, select pairs of chromosomesfor crossover

Using pm, select chromosomes for mutation

Chromosomes are selectedbased on their fitnessvalues using aroulette wheel approach:

X6i 1 00 0 01 0 X2i

0 01 0X2i 0 11 1 X5i

0X1i 0 11 1 X5i1 01 0

0 10 0

11 101 0


Create a pair of offspring chromosomes by applying a crossover operation:

0 11 1X5'i 01 0

X6'i 1 00

0 01 0X2'i 0 1

0 0

0 1 111X5i

1 1 1 X1"i1 1

X2"i0 1 0

0X1'i 1 1 1

0 1 0X2i


Mutate an offspring chromosome by applyinga mutation operation:

Genetic Algorithms – Steps 7 & 8

Step 7:– Place all generated offspring

chromosomes in a new population

Step 8:– Go back to Step 5 until the size of the new

population is equal to the size of the initial population, N

1 01 0X1i

Generation i

0 01 0X2i

0 00 1X3i

1 11 0X4i

0 11 1X5i f = 56

1 00 1X6i f = 54

f = 36

f = 44

f = 14

f = 14

1 00 0X1i+1

Generation (i + 1)

0 01 1X2i+1

1 10 1X3i+1

0 01 0X4i+1

0 11 0X5i+1 f = 54

0 11 1X6i+1 f = 56

f = 56

f = 50

f = 44

f = 44

Crossover

X6i 1 00 0 01 0 X2i

0 01 0X2i 0 11 1 X5i

0X1i 0 11 1 X5i1 01 0

0 10 0

11 101 0

Mutation

0 11 1X5'i 01 0

X6'i 1 00

0 01 0X2'i 0 1

0 0

0 1 111X5i

1 1 1 X1"i1 1

X2"i0 1 0

0X1'i 1 1 1

0 1 0X2i

Genetic Algorithms – Steps 9 & 10

Step 9:– Replace the initial population with

the new population

Step 10:– Go back to Step 4 and repeat the process

until termination criteria are satisfied– Typically repeat this process for 50-5000+

generations

1 01 0X1i

Generation i

0 01 0X2i

0 00 1X3i

1 11 0X4i

0 11 1X5i f = 56

1 00 1X6i f = 54

f = 36

f = 44

f = 14

f = 14

1 00 0X1i+1

Generation (i + 1)

0 01 1X2i+1

1 10 1X3i+1

0 01 0X4i+1

0 11 0X5i+1 f = 54

0 11 1X6i+1 f = 56

f = 56

f = 50

f = 44

f = 44

Crossover

X6i 1 00 0 01 0 X2i

0 01 0X2i 0 11 1 X5i

0X1i 0 11 1 X5i1 01 0

0 10 0

11 101 0

Mutation

0 11 1X5'i 01 0

X6'i 1 00

0 01 0X2'i 0 1

0 0

0 1 111X5i

1 1 1 X1"i1 1

X2"i0 1 0

0X1'i 1 1 1

0 1 0X2i

Iteration

Crossword Puzzle Construction

Given:– Dictionary of valid words

and phrases– Empty crossword grid

Problem:– Fill the crossword grid such

that all words both acrossand down are valid

(assign clues later)

1 01 0X1i

Generation i

0 01 0X2i

0 00 1X3i

1 11 0X4i

0 11 1X5i f = 56

1 00 1X6i f = 54

f = 36

f = 44

f = 14

f = 14

1 00 0X1i+1

Generation (i + 1)

0 01 1X2i+1

1 10 1X3i+1

0 01 0X4i+1

0 11 0X5i+1 f = 54

0 11 1X6i+1 f = 56

f = 56

f = 50

f = 44

f = 44

Crossover

X6i 1 00 0 01 0 X2i

0 01 0X2i 0 11 1 X5i

0X1i 0 11 1 X5i1 01 0

0 10 0

11 101 0

Mutation

0 11 1X5'i 01 0

X6'i 1 00

0 01 0X2'i 0 1

0 0

0 1 111X5i

1 1 1 X1"i1 1

X2"i0 1 0

0X1'i 1 1 1

0 1 0X2i

Crossword Puzzle Construction

Genetic Algorithm (GA)– Evolve a solution by crossovers and

mutations through many generations– Initial population of crossword grids:

Random letters? Random letters based on Scrabble® frequencies? Random words from dictionary?

– Fitness of each grid is number of valid words

Termination Criteria

When do we stop?– Pause a genetic algorithm after a

given number of generations, thencheck the fittest chromosomes

If the fittest chromosomes are fitbeyond a given threshold,terminate the genetic algorithm

– Also consider stopping when the highest fitness value does not change for a large number of generations

?

How long does it take for an algorithm toproduce a solution?– Depends on the size of the input and

the complexity of the algorithm

– The size of the input is n

– The complexity of the algorithm is classifiedbased on its expected run time

Computational Complexity

Big-O notation measures the expected run timeof an algorithm (i.e. its computational complexity)– Constant time: O(1) – Logarithmic time: O(log n) – Linear time: O(n) – Linearithmic time: O(n log n) – Quadratic time: O(n2) – Exponential time: O(c n) – Factorial time: O(n!)

Computational Complexity

P

NP

Genetic algorithms are often well-suited to producing reasonable solutions to intractable problems– Intractable problems are problems with

excessive computational complexity i.e. in the Nondeterministic Polynomial (NP) class

of problems

– A reasonable solution is a partial or inexact solution that adequately solves the problem in polynomial time

Genetic Algorithms

Consider the Traveling Salesman Problem (TSP) in which a salesman aims to visit n cities exactly once covering the least distance http://mathworld.wolfram.com/TravelingSalesmanProblem.html http://www.tsp.gatech.edu/games/index.html

– Starting at any given node, choose from n–1 remaining nodes, then choose from n–2 remaining nodes, etc.

– Testing every possible route takes (n–1)! stepssee http://bio.math.berkeley.edu/classes/195/2000/lec14/index.html

Genetic Algorithms Example

yikes!

Use a genetic algorithm to evolve a near-optimal solution to the TSP– Label cities A, B, C, D, E, F, etc.– Example circuits: ABCDEF, BDAFCE,

FBECAD

– How do we perform crossover operations? Basic crossovers might result in invalid members

of the population e.g. combining ABCDEF and BDAFCE may result

in ABCFCE


Key challenge of developing a genetic algorithm is often the representation of the problem– For TSP, consider a standard ordering

ABCDEF, assigning the code 123456– All other sequences encoded

based on the removal of letters

– Basic crossover works...


All other sequences encoded based on the removal of letters from standard ordering– Sequence BDAFCE has code 231311

B is 2 in ABCDEF D is 3 in ACDEF A is 1 in ACEF F is 3 in CEF C is 1 in CE E is 1 in E


Crossing ACEDB with ABCED...

Crossover Operation


Combining ACEDB with ABCED......yields ACBED

from A.K. Dewdney’s The (New) Turing Omnibus, Computer Science Press, New York, 1993


another approach: http://www.dna-evolutions.com/dnaappletsample.html

Genetic Algorithms

Advantages of genetic algorithms:– Often outperform “brute force” approaches by

randomly jumping around the search space– Ideal for problem domains in which near-

optimal (as opposed to exact) solutions are adequate

Disadvantages of genetic algorithms:– Might not find any satisfactory partial solutions– Tuning can be a challenge

Documents

Genetic Algorithms CSCI-2300 Introduction to Algorithms David Goldschmidt, Ph.D. Rensselaer Polytechnic Institute April 29, 2013