SolutionsIV - ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT IV...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT IV Spring 2007 1. Start with the complicated one and work on it. First note that the k = 0-term is zero in the sum, so n X k =0 n k ! kp sel ( x ) k (1- p sel ( x )) n- k = n X k =1 n k ! kp sel ( x ) k (1- p sel ( x )) n- k = n X k =1 n- 1 k- 1 ! np sel ( x ) k (1- p sel ( x )) n- k , where the last equality follows from one of the helpful hints. Now change index of summation to r = k- 1, pull the n out of the sum, and you get thing above = n n- 1 X r =0 n- 1 r ! p sel ( x ) r +1 (1- p sel ( x )) n- 1- r = np sel ( x ) n- 1 X r =0 n- 1 r ! p sel ( x ) r (1- p sel ( x )) n- 1- r = np sel ( x )( p sel ( x ) + [1- p sel ( x )]) n- 1 = np sel ( x ) , where the second-to-last equality follows from the other helpful hint using m = n- 1. 2. Consider first the situation that the Suggestion alludes to. Let u be the output of your random-number generator. The probability that u lies in any sub-interval [ , ] or [ , ) of [0 , 1] is just ( - ). So, lets divide up the unit interval into three sub-intervals of the appropriate lengths, namely I 1 = [0 , 1 / 7), I 2 = [1 / 7 , 10 / 21), and I 3 = [10 / 21 , 1]. Select the string of fitness 3 when u I 1 , the string of fitness 7 when u I 2 , and the string of fitness 11 when u I 3 . Youre using your random-number generator as a linear roulette wheel; the spin of the wheel is the (random) value of u . More generally, if you have n strings x 1 ,x 2 ,...,x n in your population, divide up the unit interval into n sub-intervals I 1 ,I 2 ,...,I n , where I k has length f ( x k ) / ( P n i =1 f ( x i )), 1 k n . To pick a string for parenthood, first generate a value u with your random-number generator and select x k if u I k . Do this n times and youve implemented roulette-wheel selection of n parents for the next generation. 3. In parts (a) through (c), we have to check and see whether f 2 ( x ) / ( P y A f 2 ( y )) is the same for every x A as f 1 ( x ) / ( P y A f 1 ( y )). If so, the two fitness functions f 1 and f 2 yield the same p sel ( x ) for every string x A (as usual, Im referring to the population as A ). (a) If A has size n , then f 2 ( x ) P y A f 2 ( y ) = f 1 ( x ) + 37 37 n + P y A f 1 ( y ) . This is clearly not always equal to f 1 ( x ) / ( P y A f 1 ( y )) for every x A . If you do the algebra, you find that this requires f 1 ( x ) P y A f 1 ( y ) = 1 n for every x A , which means that every x A has the same fitness f 1 ( x ) (and also the same fitness f 2 ( x ) when f 2 ( x ) = f 1 ( x ) + 37). Accordingly, the answer 1 to (a) is: you get the same p sel ( x ) from f 2 as from f 1 if and only if f 1 ( x ) is the same for every x A ....
View Full Document

Page1 / 5

SolutionsIV - ECE 496 SOLUTIONS TO HOMEWORK ASSIGNMENT IV...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online