Lecture16

# Lecture16 - Lecture 16 Recall Standard(Simple Genetic...

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Unformatted text preview: Lecture 16: 03/26/2007 Recall: Standard (Simple) Genetic Algorithm (SGA) o Fixed-size populations (n) 0 Individuals in population are fixed-length (L) bit strings o Exogenous fitness function—each bit string x has a fitness f(x)—assume that f(x)>0, all x P(t)-->P(t+1) (populations) [that is P(t) evolves to P(t+1)] via: 1. Selection of parents for P(t+1) from P(t) 2. Generation of offspring from selected parents Talked briefly about various selection methods (fitness-prop, tournament, ranked, elitist.) Generation of offspring via crossover mutation. Crossover probability pc and mutation probability pm—pc“.75, pm“.01 and also different kinds of crossover (one-point, two-point, or uniform.) Do some back-of-envelope calculation regarding SGA with fitness-proportional selection with one-pt. crossover (w/ prob. PC), bitwise mutation (pm). What is fitness-prop. selection? A probabilistic selection method that gives every individual a non-zero probability of being selected as a parent (ie. don’t just pick the best!) and doesn’t guarantee ﬂ individual is a parent. To get the n-parents, draw from P9t); draw an individual xePsel(x) proportional to f(x) with replacement. @: there is only one way to do this— ( ) = 2(—)() all xeP(t) E () Easy to prove: want psel(x)=Cof(x) for some Co, for all x. For this to be a probability distribution, need: => 22500 Note: 1. Roulette-wheel selection is another name for this. Think of a roulette wheel with circumferencez E ( ) ( ). Assign each xeP(t) an arc length=f(x). 2. Say M>O; define a new fitness function g(x)=f(x) + M. g(x) defines same ordering offitness of individuals; however, for each xeP(t), U: () ()+ — ( ) = — Ze()() [ZE()()]+ Underf Under g As M gets larger, "under-g” distribution -->uniform—ie, psel(x)~>1/n, all x Moral: M->°o => "bleach out” selection pressure. Question: if using fitness proportional selection as above, given xeP(t), what is E(# of times x gets picked as a parent for P(t+1))? (Emphasize: two copies of same string are different x’s.) Say draw parents 1,..., n; let Note: =f(x)/avg fitness in P(t) In particular, if f(x)>avg fitness(P(t)), then E(# times x picked >1.) Alternative approach to calculating expected value above: you're picking n parents, for each k, the probability that x pts selected k times, is: ( )* (1 — ( )) =>E(# x-picks)=2 ( )(1 — ( D HW: reconcile these formulas; also look at a variation on fitness proportional selection—"expected III number contro Question: what if P(t) contains m copies of some string x? Then E(# times that string appears among parents)=m*E(#times individual x is a parent) 2% 12mm 3 m relatively |arge—> even if f(x) small-ish or mediocre, "expect" x to appear among parents. Question: Given any subset Q subset P(t), what is: o Prob{some member of Q appears among parents for P(t_1)}? o E(# times a member of Q appears among parents}? Prob{some member of Q picked MW}=—ZZ E (8) e () Hence, using "Zi-type argument,” find that E(# of Q-elements among parents for 2e () (+1): *Ze()() 126 () lZe()() Where m=size of Q=|Q| |f average fitness of strings in Q is bigger then the average fitness of strings in P(t), then “expect” at least |Q| members of Qto appear among parents. Won’t necessarily see "all members of Q” represented—Q might contain some low-f strings, e.g. Specialize this result-->Schema Theorem. Early motivation (Holland): biological evolutions "seems" to build viable genotypes from "lower-order blocks.” Maybe GAS do this, or can be made/designed to do this, or "generically" will do this, or something. Schema Theorem is a case of previous result where Q is set of representatives of some schema in P(t). Recall from the book that a schema H is a set of strings, where some bits are specified and some are free. Eg 1*0*={1000, 1001, 11000, 1101} [L=4] For any schema H, can compute its static average fitness: Where |H |=size of H Holland’s wishful thinking. Statis Building-Block Hypothesis: SGA will evolve populations where "highly fit schemas” have lots of representatives; start with short defining —length schemas, which, via crossover, assemble into longer ones, etc. Recall that defining length of a schema is distance between left most + right most defined bits. "Highly fit schemas.” Holland meant in terms of static average fitness. Holland used multi-armed bandit (slot machines with different probabilities of winning, want to maximize revenue, want to find best machine, but don’t want to move around too much) technique on this—analysis flawed because used static average H—fitness’s instead of actual average fitness’s of Hn ( ) ...
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• Spring '07
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