assign1sol - Algorithms in the Real World (15-853), Fall...

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Unformatted text preview: Algorithms in the Real World (15-853), Fall 2009 Solutions for Assignment # 1 Problem 1 We know that: p ( w ) = p ( w | b ) p ( b ) + p ( w | w ) p ( w ) and that p ( b ) = 1- p ( w ) which lets us solve to p ( w ) = . 8 and p ( b ) = . 2 . Plugging this into the formula for unconditional entropy gives: . 8 lg(1 /. 8) + . 2 lg(1 /. 2) = . 7219 and for conditional entropy . 8( . 95 lg(1 /. 95) + . 05 lg(1 /. 05)) + . 2( . 2 lg(1 /. 2) + . 8 lg(1 /. 8)) = . 3735 The ratio is . 5174 . Problem 2 We have 00011011010 = 0 . 1064453125 . The initial intervals are partitioned by [0 , . 1 , . 3 , 1) . The message falls in the interval [ . 1 , . 3) so the first letter is b . On the second round the intervals are [.1,.12,.16,.2) so the second letter is a . On the third round the intervals are [.1,.102,.106,.12) so the third letter is c . On the final round the intervals are [.106,.1074,.1102,.12) so the fourth letter is a ....
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assign1sol - Algorithms in the Real World (15-853), Fall...

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