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Genetic/ Evolutionary algorithms
Outline of the Basic Genetic Algorithm
1 [Start] Generate random population of n chromosomes (suitable solutions for the
problem)
2 [Fitness] Evaluate the fitness f(x) of each chromosome x in the population
3 [New population] Create a new population by repeating following steps until the new
population is complete
[Selection] Select two parent chromosomes from a population according to their
fitness (the better fitness, the bigger chance to be selected)
[Crossover] With a crossover probability cross over the parents to form a new
offspring (children). If no crossover was performed, offspring is an exact copy of
parents.
[Mutation] With a mutation probability mutate new offspring at each locus (position
in chromosome).
[Accepting] Place new offspring in a new population (keep population size constant)
4 [Replace] Use new generated population for a further run of algorithm
5 [Test] If the end condition is satisfied, stop, and return the best solution in current
population
[Loop] Go to step 3
CSE, IUT](https://image.slidesharecdn.com/genetic-algorithm-180209130349/85/Genetic-algorithm-7-320.jpg)
Genetic algorithm
- 2. 40 states (often encoded as a string of 0s and 1s) better states. crossover, and mutation Genetic/ Evolutionary algorithms Developed by John Holland in 1975 inspired by biological mutation & evolution stochastic (means non-deterministic) search techniques based on the mechanism of natural selection and genetics A successor state is generated by combining two parent Start with k randomly generated states (population) A state is represented as a string over a finite alphabet Evaluation function (fitness function). Higher values for Produce the next generation of states by selection, CSE, IUT
- 3. 43 “Select The Best, Discard The Rest” fitness function is upto user (problem specific)- means to evaluate the potential of a candidate in the population. e.g. for a binary representation fitness function could be: f(i) = ∑i f(i) ;where i is from 1 to l (l=total length of the sequence) And in the selection step, the probability of getting selection for a candidate is: Prob i = f(i) / ∑i f(i) CSE, IUT
- 4. 44 Encoding The process of representing the solution in the form of a string that conveys the necessary information. Binary Encoding – Most common method of encoding. Chromosomes are strings of 1s and 0s and each position in the chromosome represents a particular characteristic of the problem. Permutation Encoding – Useful in ordering problems such as the Traveling Salesman Problem (TSP). Example. In TSP, every chromosome is a string of numbers, each of which represents a city to be visited. Value Encoding – Used in problems where complicated values, such as real numbers, are used and where binary encoding would not suffice. CSE, IUT
- 5. 45 e.g. n-queen GA representation and encoding of the following arrangement can be written as: fitness of the chromosome v1 (24748552) is 28 – 4 = 24 (No. of non attacking pair of queens) • Encoding is given by the row no. of each column. That is because only 4 pairs of queens attack each other: The queens on 1st and 8th column The queens on 2nd and 4th column The queens on 6th and 7th column The queens on 3rd and 8th column CSE, IUT
- 6. 46 Genetic algorithms Fitness function: number of non-attacking pairs of queens (min = 0, max = 8c2 excluding same pair = 28) 24/(24+23+20+11) = 31% 23/(24+23+20+11) = 29% etc Here single point crossover & mutation occurred CSE, IUT
- 7. 41 Genetic/ Evolutionary algorithms Outline of the Basic Genetic Algorithm 1 [Start] Generate random population of n chromosomes (suitable solutions for the problem) 2 [Fitness] Evaluate the fitness f(x) of each chromosome x in the population 3 [New population] Create a new population by repeating following steps until the new population is complete [Selection] Select two parent chromosomes from a population according to their fitness (the better fitness, the bigger chance to be selected) [Crossover] With a crossover probability cross over the parents to form a new offspring (children). If no crossover was performed, offspring is an exact copy of parents. [Mutation] With a mutation probability mutate new offspring at each locus (position in chromosome). [Accepting] Place new offspring in a new population (keep population size constant) 4 [Replace] Use new generated population for a further run of algorithm 5 [Test] If the end condition is satisfied, stop, and return the best solution in current population [Loop] Go to step 3 CSE, IUT
- 9. 47 Genetic algorithms By doing this Crossover, it helps to accelerate the search at an early stage of evolution For detail about genetic algorithm refer to GA1.pdf in ftp://10.220.20.25/CSE 4701 CSE, IUT
- 10. 48 Another Example (Assignment 1: Deadline 15 Feb, 2018) Data population: RGB colours Aim: to obtain darkest colour represented by (0, 0, 0) This is a minimisation problem, i.e. a good colour is one that fits for (colour) --> 0. We now tabulate our data as shown CSE, IUT
- 11. 49 Start at a random pattern like this: Colour Red Green Blue C1 80 170 689 C2 130 690 15 C3 24 8 317 where Fitness for (C1) = 80 + 170 + 689 = 939 Fitness for (C2) = 130 + 690 + 15 = 835 Fitness for (C3) = 24 + 8 + 317 = 349 CSE, IUT
- 12. 50 Fitness (C2) = 130 + 690 + 15 = 835 C1 Start at a random pattern like this: Colour Red Green Blue C1 80 170 689 C2 130 690 15 C3 24 8 317 FITTEST PLACED TOP C3 Fitness (C1) = 80 + 170 + 689 = 939 C2 Fitness (C3) = 24 + 8 + 317 = 349 CSE, IUT
- 13. 51 After a Selection is done on the sample: ** Remember, this is a minimisation problem.. Colour Fitness C3 349 C2 835 C1 939 CSE, IUT
- 14. 52 C4 24 8 15C4 is crossover(C3,C2)= (24, 8, 15) C6 is crossover(C2,C1)= (130, 690, 689) GA: Reproduction & Crossover So far, we have this Colour Red Green Blue Colour Fitness C1 80 170 689 C3 349 C2 130 690 15 C2 835 C3 24 8 317 C1 939 Next step is to reproduce the pattern, like this, by crossing over: Colour Red Green Blue C5 is crossover(C3,C1)= (24, 8, 689) C5 24 8 689 C6 130 690 689 CSE, IUT
- 15. 53 perform mutation, and we have C7 is obtained by mutating(4) =(24, 8, 13) C8 is obtained by mutating(5) =(25, 9, 689 ) C9 is obtained by mutating(6) =(128, 688, 689) Colour Red Green Blue New population of C7 24 8 13 3 chromosomes C8 25 9 689 C9 128 688 689 CSE, IUT
- 16. 54 Conclusion (up to “mutation” to get a new data set) Some solutions have improved (after first iteration): Getting better rather fast Slightly improved of the answer Fitness for C7 = (24 + 8 + 13) = 45 Fitness for C8 = (25 + 9 + 689) = 743 Fitness for C9 = (128 + 688 + 689 ) = 1505 If the process is iterated, population will converge to have fitness near to zero (colour) --> 0. Continue.............. CSE, IUT