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Basic frame of Genetic algorithm

Basic frame of Genetic algorithm. Initialise a population Evaluate a population While (termination condition not met) do Select sub-population based on fitness Produce offspring of the population using crossover Mutate offspring stochastically Select survivors based on fitness.

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Basic frame of Genetic algorithm

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  1. Basic frame of Genetic algorithm Initialise a population Evaluate a population While (termination condition not met) do Select sub-population based on fitness Produce offspring of the population using crossover Mutate offspring stochastically Select survivors based on fitness

  2. Evolution Runs Until: A perfect individual appears (if you know what the goal is), Or: improvement appears to be stalled, Or: you give up (your computing budget is exhausted).

  3. Simple GA Simulation Optimisation problem maximise the function f(x)=x2 x between 0 and 31 objective function code using binary representation 110112 = 1910 1*24+0*23+0*22+1*21+1*20 0 is then 00000 31 is then 11111

  4. Simple GA Simulation: Initial Population Lets assume that we want to create a initial population of 4 random flip coin 20 times (4 population size * string of 5) String Initial No. Population 1 01101 2 11000 3 01000 4 10011

  5. Simple GA Simulation: Fitness Function String Initial f(x) Prob. = fi/Sum No. Population 1 01101 169 0.14 2 11000 576 0.49 3 01000 64 0.06 4 10011 361 0.31 Sum 1170 Average 293 Max 576

  6. Simple GA Simulation: Roulette Wheel String Initial f(x) Prob Roulette. No. Population Wheel 1 01101 169 0.14 1 2 11000 576 0.49 2 3 01000 64 0.06 0 4 10011 361 0.31 1 Sum 1170 Average 293 Max 576

  7. Simple GA Simulation: Mating Pool String Initial f(x) Prob Roulette. Mating No. Population Wheel Pool 1 01101 169 0.14 1 01101 2 11000 576 0.49 2 11000 3 01000 64 0.06 0 11000 4 10011 361 0.31 1 10011 Sum 1170 Average 293 Max 576

  8. Simple GA Simulation: Mate String Initial f(x) Prob Roulette. Mating Mate No. Population Wheel Pool 1 01101 169 0.14 1 01101 2 2 11000 576 0.49 2 11000 1 3 01000 64 0.06 0 11000 4 4 10011 361 0.31 1 10011 3 Sum 1170 Average 293 Mate is Randomly selected Max 576

  9. Simple GA Simulation: Crossover String Initial f(x) Prob R. Mating Mate Crossover No. Population W Pool 1 01101 169 0.14 1 01101 2 4 2 11000 576 0.49 2 11000 1 4 3 01000 64 0.06 0 11000 4 2 4 10011 361 0.31 1 10011 3 2 Sum 1170 Average 293 Crossover is Randomly selected Max 576

  10. Simple GA Simulation: New Population String Initial f(x) Prob R. Mating Mate Cross New No. Population W Pool over Pop. 1 01101 169 0.14 1 01101 2 4 01100 2 11000 576 0.49 2 11000 1 4 11001 3 01000 64 0 .06 0 11000 4 2 11011 4 10011 361 0.31 1 10011 3 2 10000 Sum 1170 Average 293 Max 576

  11. Simple GA Simulation: Mutation String Initial f(x) Prob R. Mating Mate Cross New No. Population W Pool over Pop. 1 01101 169 0.14 1 01101 2 4 01100 2 11000 576 0.49 2 11000 1 4 11001 3 01000 64 0 .06 0 11000 4 2 11011 4 10011 361 0.31 1 10011 3 2 10000 Sum 1170 Average 293 Max 576

  12. Simple GA Simulation: Mutation String Initial f(x) Prob R. Mating Mate Cross New No. Population W Pool over Pop. 1 01101 169 0.14 1 01101 2 4 01100 2 11000 576 0.49 2 11000 1 4 10001 3 01000 64 0 .06 0 11000 4 2 11011 4 10011 361 0.31 1 10011 3 2 10000 Sum 1170 Average 293 Max 576

  13. Simple GA Simulation: Evaluation New Population String Initial f(x) Prob R. Mating Mate C New f(x) No. Population W Pool o Pop. 1 01101 169 0.14 1 01101 2 4 01100 144 2 11000 576 0.49 2 11000 1 4 10001 289 3 01000 64 0 .06 0 11000 4 2 11011 729 4 10011 361 0.31 1 10011 3 2 10000 256 Sum 1170 1418 Average 293 354 Max 576 729

  14. Problem & Representations Chromosomes represent problems' solutions as genotypes They should be amenable to: Creation (spontaneous generation) Evaluation (fitness) via development of phenotypes Modification (mutation) Crossover (recombination)

  15. How GAs Represent Problems' Solutions: Genotypes Bit strings -- this is the most common method Strings on small alphabets (e.g., C, G, A, T) Permutations (Queens, Salesmen) Trees (Lisp programs). Genotypes must allow for: creation, modification, and crossover.

  16. Binary-Encoding for Genetic Algorithms The original formulation of genetic algorithms relied on a binary encoding of solutions, i.e., the chromosomes consist of a string of 0’s and 1’s. No restriction on the phenotype so long as a good method for encoding/decoding exists. Other methods of encoding will come later.

  17. Genetic Binary Encoding/Decoding of Integers Initially, single parameters (one gene chromosomes) Integer parameters: p is an integer parameter to be encoded. 3 distinct cases to consider: Case 1 p takes values from {0, 1, 2, …..2^N-1} for some N. In this case, p can be encoded by its equivalent binary representation.

  18. Genetic Binary Encoding/Decoding of Integers Case 2 p takes values from {M, M+1, …., M+2^N-1} for some M, N. In this case (p-M) can be encoded directly by its equivalent binary representation. Case 3 p takes values from {0, 1, .., L-1} for some L such that there exist no N for which L=2^N. There are two possibilities. Clipping and Scaling

  19. Clipping Take N = log(L)+1 and encode all parameter values 0<= p <= L-2 by their equivalent binary representation, letting all other n-bit strings serve as encodings of p = L-1. Example: p from {0,1,2,3,4,5}, i.e., L=6. Then N=log(6)+1= 3 p 0 1 2 3 4 5 5 5 Code 000 001 010 011 100 101 110 111 Advantages: Easy to implement Disadvantages: Strong representational bias. All parameter values between 0 and L-2 have a single encoding, but the single value L-1 has 2^N-L+1 encodings

  20. Scaling Take N=log(L)+1 and encode p by the binary representation of the integer e such that p = e(L-1)/(2^N-1) Example: p from {0,1,2,3,4,5} i.e. L=6, N=log(6)+1=3. p 0 0 1 2 2 3 4 5 Code 000 001 010 011 100 101 110 111 Advantages: Easy to implement. Smaller representational bias than clipping (at most double representations) Disadvantages: Small representational bias. More computation than clipping.

  21. Gray Coding Desired: points close to each other in representation space also close to each other in problem space This is not the case when binary numbers represent floating point values m is number of bits in representation binary number b = (b1; b2;  ; bm) Gray code number g = (g1; g2; ; gm)

  22. Gray Coding Binary gray code 000 000 001 001 010 011 011 010 100 110 101 111 110 101 111 100

  23. Gray Coding PROCEDURE Binary-To-Gray g1=b1 for k=2 to m do gk=b k-1 XOR bk endfor

  24. Gray Coding PROCEDURE Gray-To-Binary value = g1 b1 = value for k = 2 to m do if gk = 1 then value = NOT value end if bk =value end for

  25. Binary Encoding and Decoding of Real-valued parameters Can be encoded as: fixed-point numbers integers using scaling and quantisation If p ranges over [min,max] encode p using N bits by using the binary representation of the integer part of (2^N-1)(p-min)/(max - min)

  26. Multiple parameters Vectors of parameters are encoded on multi-gene chromosomes by combining the encodings of each individual parameter. Let e_i =[b_i0,….b_iN] be the encoding of the ith of M parameters. There are two ways of combining the e_i’s into a chromosome. Concatenating: Individual encodings simply follow one another in some predefined order e.g. [b_10,…b_1N,..,b_M0,…b_MN] Interleaving: the bits of each individual encoding are interleaved e.g. [b_10,…,b_M0,b_11,…,b_1N] The order of parameters in the vector (i.e. genes in the chromosome) is important, especially for concatenated encodings.

  27. Initialization init( ) { for( i = 0; i < POP_SIZE; i++ ) for( j = 0; j < N; j++ ) p[i][j] = random_int( 2 ); }

  28. GA Main Program for( trial = 0; trial < LOOPS; trial++ ) { selection( ) crossover( ) mutation( ) for( who = 0; who < POP_SIZE; who++ ) fitness[who] = fv(who); }

  29. Selection Selects individuals for reproduction • randomly with a probability depending on the relative fitness of the individuals so that the best ones are more often chosen for reproduction rather than poor ones • Proportionate-based selection picks out individuals based upon their fitness values relative to the fitness of the other individuals in the population • Ordinal-based selection selects individuals based upon their rank within the population; independent of the fitness distribution

  30. Roulette Wheel Selection Here is a common technique: let F = j=1 to popsizefitnessj Select individual k to be a parent with probability fitnessk/F

  31. Roulette Wheel Selection assigns to each solution a sector of a roulette wheel whose size is proportional to the appropriate fitness measure chooses a random position on the wheel (spin the wheel) c Fitness a:1 b:3 c:5 d:3 e:2 f:2 g:8 b d a e g f

  32. Roulette Wheel Example For each chromosome evaluate the fitness and the cumulative fitness For N times calculate a random number Select the chromosome where its cumulative fitness is the first value greater than the generated random number Individual Chromosome Fitness Cumulative x1 101100 20 20 x2 111000 7 27 x3 001110 6 33 x4 101010 10 43 x5 100011 12 55 x6 011011 9 64

  33. Roulette Wheel Example Individual Chromosome Fitness Cumulative Random Individual x1 101100 20 20 42.8 x4 x2 111000 7 27 19.78 x1 x3 001110 6 33 42.73 x4 x4 101010 10 43 58.44 x6 x5 100011 12 55 27.31 x3 x6 011011 9 64 28.31 x3

  34. Roulette Wheel Selection There are some problems here: fitnesses shouldn't be negative only useful for max problem probabilities should be “right” avoid skewing by super heros.

  35. Parent Selection: Rank Here is another technique. Order the individuals by fitness rank Worst individual has rank 1. Best individual has rank POPSIZE Let F = 1 + 2 + 3 +  + POP_SIZE Select individual k to be a parent with probability rankk/F Benefits of rank selection: the probabilities are all positive the probability distribution is “even”

  36. Parent Selection: Rank Power Yet another technique. Order the individuals by fitness rank Worst individual has rank 1. Best individual has rank POP_SIZE Let F = 1s + 2s + 3s +  + POP_SIZEs Select individual k to be a parent with probability rankks/F benefits: the probabilities are all positive the probabilities can be skewed to use more “elitist” selection

  37. Tournament Selection Pick k members of the population at random select one of them in some manner that depends on fitness

  38. Tournament Selection void tournament(int *winner, *loser) { int size = tournament_size, i, winfit, losefit; for( i = 0; i < size; i++ ) { int j = random_int( POP_SIZE );; if( j==0 || fitness[j] > winfit ) winfit = fitness[j],*winner = j; if( j==0 || fitness[j] < losefit ) losefit = fitness[j],*loser = j; } }

  39. Crossover Methods Crossover is a primary tool of a GA. (The other main tool is selection.) CROSS_RATE: determine if the chromosome attend the crossover Common techniques for bit string representations: One-point crossover: Parents exchange a random prefix Two-point crossover: Parents exchange a random substring Uniform crossover: Each child bit comes arbitrarily from either parent (We need more clever methods for permutations & trees.)

  40. 1-point Crossover Suppose we have 2 strings a and b, each consisting of 6 variables a1, a2, a3, a4, a5, a6 b1, b2, b3, b4, b5, b6 representing two solutions to a problem a crossover point is chosen at random and a new solution is produced by combining the pieces of the original solutions if crossover point was 2 a1, a2, b3, b4, b5, b6 b1, b2, a3, a4, a5, a6

  41. 1-point Crossover Parents Children

  42. 2-point Crossover With one-point crossover the head and the tail of one chromosome cannot be passed together to the offspring If both the head and the tail of a chromosome conatin good generic information, none of the offsprings obtained directly with one-point crossover will share the two good features A 2-point crossover avoids such a drawback Parents Children

  43. Uniform Crossover Each gene in the offspring is created by copying the corresponding gene from one or the other parent chosen according to a random generated binary crossover mask of the same length as the chromosomes where there is a 1 in the crossover mask the gene is copied from the first parent and where there is a 0 in the mask the gene is copied from the second parent a new crossover mask is randomly generated for each pair of parents

  44. Uniform Crossover Parents Child Crossover Mask 1 0 0 1 0 1 1 0

  45. Uniform Crossover make_children(int p1, p2, c1, c2) { int i, j; for( i = 0; i < N; i++ ) { if( random_int(2) ) p[c1][i] = p[p1][i],p[c2][i] = p[p2][i]; else p[c1][i] = p[p2][i],p[c2][i] = p[p1][i]; } }

  46. Another Clever Crossover Select three individuals, A, B, and C. Suppose A has the highest fitness and C the lowest. Create a child like this. for(i = 0; i < length; i++ ) { if( A[i] == B[i] ) child[i] = A[i]; else child[i] = 1 - C[i]; } We just suppose C is a “bad example.”

  47. Crossover Methods & Schemas Crossovers try to combine good schemas in the good parents. The schemas are the good genes, building blocks to gather. The simplest schemas are substrings. 1-point & 2-point crossovers preserve short substring schemas. Uniform crossover is uniformly hostile to all kinds of schemas.

  48. Limit Consistency of Crossover Operator

  49. Crossover for Permutations (A Tricky Issue) Small-alphabet techniques fail. Some common methods are: OX: ordered crossover PMX: partially matched crossover CX: cycle crossover We will address these and others later.

  50. Crossover for Trees These trees often represent computer programs. Think Lisp Interchange randomly chosen subtrees of parents.

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