Computer ScienceGenetic Algorithms Slide 1 Random/Exhaustive Search l Generate and Test 1. Generate...

Computer Science Genetic Algorithms Slide 1

Random/Exhaustive Search Generate and Test

1. Generate a candidate solution and test to see if it solves the problem

2. Repeat

Information used by this algorithm• You know when you have found the solution


Hill Climbing

1. Generate a candidate solution by modifying the last solution, S

2. If the new solution, N, is “better” than S then S := N

3. Repeat

Local Search Information used by this algorithm

• Compare two candidate solutions and tell which is better


Population of Hill Climbers Randomly generate initial population of hill

climbers (Randomly generate initial candidate solutions)

Do hill climbing in parallel After time t, choose best solution in population

Information used by this algorithm• Same as hill climbing


Genetic Algorithms Population of information exchanging hill

climbers

Concentrates resources in promising areas of the search space

Information used:• Same as hillclimbing


Hard problems Computational complexity, problem size = n

• Binary Search O(log(n))

• Linear Search O(n)

• Bubble Sort O(n^2)

• Scheduling NP-complete (at least exponential == O(a^n)


Hard problems Poorly defined

Robotics How do we catch a ball, navigate, play basketball

User Interfaces

Predict next command, adapt to individual user

Medicine Protein structure prediction, Is this tumor benign, design drugs

Design Design bridge, jet engines, Circuits, wings

Control Nonlinear controllers


Search as a solution to hard problems

Strategy: generate a potential solution and see if it solves the problem

Make use of information available to guide the generation of potential solutions

How much information is available?• Very little: We know the solution when we find it

• Lots: linear, continuous, …

• Modicum: Compare two solutions and tell which is “better”


Search tradeoff Very little information for search implies we have no

algorithm other than RES. We have to explore the space thoroughly since there is no other information to exploit

Lots of information (linear, continuous, …) means that we can exploit this information to arrive directly at a solution, without any exploration

Modicum of information (compare two solutions) implies that we need to use this information to tradeoff exploration of the search space versus exploiting the information to concentrate search in promising areas


Exploration vs Exploitation

More exploration means• Better chance of finding solution (more robust)

• Takes longer

More exploitation means• Less chance of finding solution, better chance of getting stuck

in a local optimum

• Takes less time


Choosing a search algorithm The amount of information available about a

problem influences our choice of search algorithm and how we tune this algorithm

How does a search algorithm balance exploration of a search space against exploitation of (possibly misleading) information about the search space?

What assumptions is the algorithm making?


Genetic Algorithm Generate pop(0) Evaluate pop(0) T=0 While (not converged) do

• Select pop(T+1) from pop(T)

• Recombine pop(T+1)

• Evaluate pop(T+1)

• T = T + 1

Done






• T = T + 1

Done


Generate pop(0)

for(i = 0 ; i < popSize; i++){ for(j = 0; j < chromLen; j++){

Pop[i].chrom[j] = flip(0.5);}

}

Initialize population with randomly generated strings of 1’s and 0’s






• T = T + 1

Done


Evaluate pop(0)

EvaluateDecoded individual

Fitness

Application dependent fitness function


Genetic Algorithm Generate pop(0) Evaluate pop(0) T=0 While (T < maxGen) do




• T = T + 1

Done






• T = T + 1

Done


Selection

Each member of the population gets a share of the pie proportional to fitness relative to other members of the population

Spin the roulette wheel pie and pick the individual that the ball lands on

Focuses search in promising areas


Code

int roulette(IPTR pop, double sumFitness, int popsize){

/* select a single individual by roulette wheel selection */ double rand,partsum; int i;

partsum = 0.0; i = 0; rand = f_random() * sumFitness; i = -1; do{ i++; partsum += pop[i].fitness; } while (partsum < rand && i < popsize - 1) ; return i;}






• T = T + 1

Done


Crossover and mutation

Mutation Probability = 0.001

Insurance

Xover Probability = 0.7

Exploration operator


Crossover codevoid crossover(POPULATION *p, IPTR p1, IPTR p2, IPTR c1, IPTR c2){/* p1,p2,c1,c2,m1,m2,mc1,mc2 */ int *pi1,*pi2,*ci1,*ci2; int xp, i;

pi1 = p1->chrom; pi2 = p2->chrom; ci1 = c1->chrom; ci2 = c2->chrom; if(flip(p->pCross)){

xp = rnd(0, p->lchrom - 1); for(i = 0; i < xp; i++){ ci1[i] = muteX(p, pi1[i]); ci2[i] = muteX(p, pi2[i]); } for(i = xp; i < p->lchrom; i++){ ci1[i] = muteX(p, pi2[i]); ci2[i] = muteX(p, pi1[i]); } } else { for(i = 0; i < p->lchrom; i++){ ci1[i] = muteX(p, pi1[i]); ci2[i] = muteX(p, pi2[i]); } }}


Mutation code

int muteX(POPULATION *p, int pa){ return (flip(p->pMut) ? 1 - pa : pa);}

Computer ScienceGenetic Algorithms Slide 1 Random/Exhaustive Search l Generate and Test 1. Generate...

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