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Winter 2009 Assignment4 Soln

Winter 2009 Assignment4 Soln - MATH 239 ASSIGNMENT 4 Sample...

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MATH 239 ASSIGNMENT 4 Sample Solutions 1. 12 marks For each of the following sets, write down a decomposition that uniquely creates the elements of that set. (a) The { 0 , 1 } -strings that have no substring of 1s with length 3, and no substrings of 0s of length 2. Solution: Since we want to modify both the allowed lengths of strings of 1 and strings of 0s, let us start with the block decomposition { 0 } * ( { 1 }{ 1 } * { 0 }{ 0 } * ) * { 1 } * = { ², 0 , 00 , . . . } ( { 1 , 11 , 111 , . . . }{ 0 , 00 , 000 , . . . , } ) * { ², 1 , 11 , 111 , . . . } and remove all strings of 1s of length 3 or more, and remove all strings of 0s of length 2 or more. This leaves us with { ², 0 } ( { 1 , 11 }{ 0 } ) * { ², 1 , 11 } . One could also have started with the other block decomposition and arrived at { ², 1 , 11 } ( { 0 }{ 1 , 11 } ) * { ², 0 } . (b) The { 0 , 1 } -strings that have no blocks of 0s of size 2, and no blocks of 1s of size 3. Solution: Since we want to modify both the allowed lengths of blocks of 1 and blocks of 0s, let us start with the block decomposition { 0 } * ( { 1 }{ 1 } * { 0 }{ 0 } * ) * { 1 } * = { ², 0 , 00 , . . . } ( { 1 , 11 , 111 , . . . }{ 0 , 00 , 000 , . . . , } ) * { ², 1 , 11 , 111 , . . . } and remove all the blocks of 1s of length 3, and remove all blocks of 0s of length 2. This leaves us with { ², 0 , 000 , . . . } ( { 1 , 11 , 1111 , . . . }{ 0 , 000 , . . . , } ) * { ², 1 , 11 , 1111 , . . . } which can also be rewritten more compactly as ( { 0 } * \{ 00 } ) (( { 1 }{ 1 } * \{ 111 } )( { 0 }{ 0 } * \{ 00 } )) * ( { 1 } * \{ 111 } ) . A similar solution is possible starting with the other block decomposition. (c) The set of { 0 , 1 } -strings in which the substring 0111 does not occur. Solution: We start with the block decomposition where in the middle section the 0-blocks precede the 1-blocks (i.e. { 1 } * ( { 0 }{ 0 } * { 1 }{ 1 } * ) * { 0 } * ) and omit the 1-blocks of length 3 or more. This gives us { 1 } * ( { 0 }{ 0 } * { 1 , 11 } ) * { 0 } * . Alternatively, we could start with the 0-decomposition and remove all strings of 1s of length 3 or more that follow a 0. This gives us { 1 } * ( { 0 }{ ², 1 , 11 } ) * . 1
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(d) The set of { 0 , 1 } -strings in which the substring 0110 does not occur. Solution: We start with the block decomposition { 1 } * ( { 0 }{ 0 } * { 1 }{ 1 } * ) * { 0 } * . In the ( { 0 }{ 0 } * { 1 }{ 1 } * ) * part, we can only allow a 11 block if there are no 0’s following it. So we can write { 1 } * ( { 0 }{ 0 } * { 1 , 111 , 1111 , 11111 , . . . } ) * ( { 0 }{ 0 } * { 11 } ∪ { 0 } * ) . Alternatively, we could start with the 0-decomposition and in the ( { 0 }{ 1 } * ) part omit the string 011 unless it occurs at the very end of the string. This gives us { 1 } * ( { 0 }{ ², 1 , 111 , 1111 , . . . } ) * ( { ² } ∪ { 0 }{ 11 } ) .
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