Q1. Thursday, February 01, 2018 The following strings (over the ternary alphabet

Question

Q1. Thursday, February 01, 2018 The following strings (over the ternary alphabet 10, 1, 2) and fitness are contained in a population at generation t: Pop. Number String 2002000 210 10 12 100 1 2 10 0201020 fitness 80 20 25 75 2 4 a. Calculate the number of points represented by this coding. b. Calculate the number of schemata represented by this coding. c. Determine all possible number of candidate solutions that the schemas in part d matches. Determine expected number of copies in generation t+1 of the following schema under roulette wheel selection, simple crossover, and mutation if pc 0.6, pm- 0.0: d. H3_: * * 0 1 H4-2 0 HI =2 H2_0 Assume non-overlapping populations of constant size. Of the schemata H1 - H4 in part d, determine which, if any, are building blocks. e.

Explanation / Answer

Lets represent the strings as:-

So to compute the points for this encoding we need to find the difference in each symbol for 4 Strings for each position of the characters

ie. (2, 2, 1, 0) = Points = 3 (as there are 3 distinct symbols)

(0, 1, 0, 2) = Points = 3 (as there are 3 distinct symbols)

(0,0,0,0) = Points = 0 (as there are 0 distinct symbols)

(2,1,1,1) = Points = 2 (as there are 2 distinct symbols)

(0,0,2,0) = Points = 2 (as there are 2 distinct symbols)

(0,1,1,2) = Points = 3 (as there are 3 distinct symbols)

(0,2,0,0) = Points = 2 (as there are 2 distinct symbols)

Number of Points = 15 points (Add the Points above computed)

b) No of Schema represented by this encoding:-

The Schema is :- * * 0 * * * * , since at the 3rd position 0 is fixed and all positions have varying symbols (0,1,2)

Hence for each * we can have (0,1,2) inputs which means we have total of 6 * 3 = 18 schemas possible.

c) For H1:- 2 * * * * * *, which gives us 3*6 = 18 candidate solutions for H1

For H2:- 0 * * * * * * , which gives us 3*6 = 18 candidate solutions for H2

For H3:- * * 0 1 * * *, which gives us 3*5 = 15 candidate solutions for H3

For H4:- 2 * * * * * 0, which gives us 3*5 = 15 candidate solutions for H4

Please let me know in case of any clarifications required. Thanks!

2 0 0 2 0 0 0 2 1 0 1 0 1 2 1 0 0 1 2 1 0 0 2 0 1 0 2 0

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