Unformatted text preview: CS180 Spring 2009 Homework 2 Solutions 2.3 √ 2 n , n + 10, n 2 log n , n 2 . 5 , 10 n , 100 n 2.6 (a) This algorithm is O ( n 3 ) (b) The number of additions made by this algorithm: ∑ n i =1 ( ∑ n j = i +1 ( j i + 1)) = ∑ n i =1 ( ∑ n j = i +1 j ( n i ) i + n i ) = ∑ n i =1 ( n ( n +1) 2 i ( i +1) 2 ni + i 2 + n i ) Now that the sum over j has been removed, all remaining sums are from i = 1 to n . This will make the rest a bit easier to read. = n 3 +3 n 2 2 + 1 2 ∑ ( i 2 ) 3 2 ∑ ( i ) n ∑ ( i ) = n 3 +3 n 2 2 + n ( n +1)(2 n +1) 12 3( n )( n +1) 4 n 3 + n 2 2 = n 3 +3 n 2 2 + 2 n 3 +3 n 2 + n 12 3 n 2 +3 n 4 n 3 + n 2 2 Let’s collect the n 3 terms, as they are the highest order terms. We have: ( 1 2 + 1 6 1 2 ) n 3 = n 3 6 , so we have Ω( n 3 ). (c) The better algorithm is based on the observation that the sum of A [ i..j ] is calculated for B [ i,j ] and again for B [ i,j + 1]. This is wasteful. Instead, let B [ i,i ] = A [ i ] and B [ i,j ] = B [ i,j...
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 Spring '09
 Mayer
 Graph Theory, Summation, DI, Consistency, highest order terms

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