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Unformatted text preview: n i =1 ( logi + log 2) = n i =1 log (2 i ) = log 2 + log 4 + log 6 + ... + log 2 n nlog (2 n ) = nlog ( n ) + nlog 2 Hence : n i =1 d log 2 i e = O ( nlogn ) d x e give smallest integer larger than or equal to x . So n i =1 d log 2 i e is not always same as n i =1 log 2 i 8. (5 points) Answer: Since: d ( n ) c * f ( n ) for n n , c > 0, n 1 e ( n ) c * g ( n ) for n n , c > 0, n 1 Hence: d ( n ) + e ( n ) c * f ( n ) + c * g ( n ) for n max ( n ,n ) Assume: c 00 = max ( c,c ) ,n 00 = max ( n ,n ) Hence: d ( n ) + e ( n ) c 00 * ( f ( n ) + g ( n )) for n n 00 , c 00 > 0, n 00 1 That is: d ( n ) + e ( n ) = O ( f ( n ) + g ( n )) 2...
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This note was uploaded on 03/24/2008 for the course CSC 316 taught by Professor Rhee during the Spring '08 term at N.C. State.
 Spring '08
 Rhee
 Data Structures

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