Problem 638. Uniform binary crossover

Created by Vitor Yano

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{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Given two binary vectors, return the two children by combining the genes of them using a binary crossover mask.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>More information:</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://en.wikipedia.org/wiki/Crossover_(genetic_algorithm)#Uniform_Crossover_and_Half_Uniform_Crossover\"><w:r><w:t>Uniform crossover</w:t></w:r></w:hyperlink></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>Parents and mask are row vectors of the same size</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>N</w:t></w:r><w:r><w:t>;</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>Output is a 2 x</w:t></w:r><w:r><w:t> </w:t></w:r><w:r><w:rPr><w:i/></w:rPr><w:t>N</w:t></w:r><w:r><w:t> matrix containing both children;</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>The first child must have genes from the first parent at the positions where the mask is true, and genes from the second parent where the mask is false;</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>The second child must have genes from the first parent at the positions where the mask is false, and genes from the second parent where the mask is true;</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>The mask is also supplied.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Example:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[mask = [1 0 0 0 1 1 1 0 0];\nparent1 = [0 0 0 1 1 1 0 0 0];\nparent2 = [1 1 0 0 1 0 1 1 0];\n\nchildren = [0 1 0 0 1 1 0 1 0;\n 1 0 0 1 1 0 1 0 0];]]></w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

Given two binary vectors, return the two children by combining the genes of them using a binary crossover mask.More information: <a href="http://en.wikipedia.org/wiki/Crossover_(genetic_algorithm)#Uniform_Crossover_and_Half_Uniform_Crossover">Uniform crossover</a><ul><li>Parents and mask are row vectors of the same size N;</li><li>Output is a 2 x N matrix containing both children;</li><li>The first child must have genes from the first parent at the positions where the mask is true, and genes from the second parent where the mask is false;</li><li>The second child must have genes from the first parent at the positions where the mask is false, and genes from the second parent where the mask is true;</li><li>The mask is also supplied.</li></ul>Example:<pre class="language-matlab">mask = [1 0 0 0 1 1 1 0 0];
parent1 = [0 0 0 1 1 1 0 0 0];
parent2 = [1 1 0 0 1 0 1 1 0];
</pre><pre class="language-matlab">children = [0 1 0 0 1 1 0 1 0;
 1 0 0 1 1 0 1 0 0];
</pre>

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Last Solution submitted on May 11, 2023

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