EE4047_Part4 - City University of Hong Kong Part 4: Theory...

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Unformatted text preview: City University of Hong Kong Part 4: Theory and Hypothesis Hyperplanes • Assume that we have a problem, where the solution can be encoded in 3 bits (X 1 ,X 2 ,X 3 ) X 2 X 1 X 3 000 100 010 110 101 001 011 111 **1 Terminology • Schema: binary string may contain “ * ” (don’t care or called wild symbol) • Each schema corresponds to a hyperplane in the search space • Each chromosome (binary encoded) corresponds to a corner in the hypercube and is a member of 2 L different hyperplanes (*** is included.) • Total number of hyperplanes is 3 L Hyperplanes in 3-dimensional cubic Corners Lines Planes Cubic 0 0 0 * 0 0 * * 0 * * * 0 0 1 * 0 1 * * 1 0 1 0 * 1 0 * 1 * 0 1 1 * 1 1 * 0 * 1 0 0 0 * 0 1 * * 1 0 1 0 * 1 0 * * 1 1 0 1 * 0 1 1 1 1 * 1 0 0 * 0 1 * 1 0 * 1 1 *----------------------------------------------------------------------- 8 12 6 1 total=27 Example Consider chromosome with 10 bits One wild symbol Two wild symbols [ * 1 1 1 0 0 1 0 0 1] [ * 1 * 1 1 0 0 1 0 0] It matches 2 strings It matches 4 strings [ 1 1 1 0 0 1 0 0 1]; [ 1 1 1 0 0 1 0 0]; [ 1 1 1 1 0 0 1 0 0 1] [ 1 1 1 1 0 0 1 0 0]; [ 1 1 1 1 0 0 1 0 0]; [ 1 1 1 1 1 0 0 1 0 0] • Every schema matches exactly 2 r strings where r = the number of wild symbols • For L= length of string, each string matches 2 L schemata . • The TOTAL number of schemata is 3 L Order of Schema The order of the schema, o(S) , is the number of “0”s or “1”s in a chromosome S1=[ * * * 0 0 1 * 1 1 0 ] o(S1)=6 S2=[ * * * * 0 0 * * *] o(S2)=3 S3=[ 1 1 1 0 1 * * 0 0 1 ] o(S3)=8 ) ( S o Defining Length The Defining Length of the schema S is the distance between the first and last fixed gene positions....
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This note was uploaded on 01/11/2011 for the course EE 4047 taught by Professor Kitsangtsang during the Fall '09 term at City University of Hong Kong.

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EE4047_Part4 - City University of Hong Kong Part 4: Theory...

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