answerswinter2008midterm

answerswinter2008midterm - Sheet1 Page 1 Winter 2008 Answer...

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Unformatted text preview: Sheet1 Page 1 Winter 2008 Answer key 1)D 2) B 3)D 4) D 5) D hmmm...it's always D apparently 6)E (correct output is 1 1 2 5 6 7 7 ) 7)A 8)C 9)B 10) D 11)A is no good because the second do loop if (as in the example) n=6, d=2, then we have k=1 to 4.....then we say vec(6-2+k For b, again, suppose d=2, n=6. then k goes from 1 to 6. Suppose k is 5. Then mod(5,6) = 5 . 5 + d = 7. ....but vec1(7) doesn' C is correct One way to think about D is the following: K ranges from d+1 until n+d. If we assume n=6 and d=2 again, then we have K rang vec2(n+d-k+1) then ranges from ( n+d - 3 + 1) = (6) until (n+d-8+1) = 1 so the left side ranges from vec2(6) until vec2(1) The right side ranges from 3-d = 1 until 8-d= 6 So the right side ranges from vec1(1) until vec1(6). This essentially means we So C is the only answer that works 12) C will also work for any non-negative integer since the only time d is used is in taking the mod 13) A, B, C 14) note: fall 2008 midterm will only have multiple choice so a question like this would not be asked...
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This note was uploaded on 01/15/2010 for the course COMP COMP 206 taught by Professor Vybihal during the Spring '04 term at McGill.

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answerswinter2008midterm - Sheet1 Page 1 Winter 2008 Answer...

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