# key1sp01 - COP3530C.01, Spring 2001 S. Lang Solution Key to...

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COP3530C.01, Spring 2001 S. Lang Solution Key to Assignment #1 (40 pts.) 1/28/01 1. (12 pts.) Exercise 2.2 of the Text, do all parts and provide explanations as precisely as possible. That is, give a proof for each “yes” answer, give a counter example for each “no” answer. Suppose T 1 ( n ) = O( f ( n )) and T 2 ( n ) = O( f ( n )). (a) (3 pts.) Prove T 1 ( n ) + T 2 ( n ) = O( f ( n )). (Solution one) Since T 1 ( n ) = O( f ( n )) by assumption, there exist constants c , k such that T 1 ( n ) c f ( n ) when n k . Similarly, there exist constants c ’, k ’ such that T 1 ( n ) c f ( n ) when n k ’. When n max( k , k ’), adding the two inequalities yields T 1 ( n ) + T 2 ( n ) ( c + c ’) f ( n ), which implies T 1 ( n ) + T 2 ( n ) = O( f ( n )). (Solution two) T 1 ( n ) + T 2 ( n ) = max(O( f ( n )), O( f ( n ))), using Rule 1 of the Text = O( f ( n )). (b) (3 pts.) Disprove T 1 ( n ) – T 2 ( n ) = o( f ( n )) (the little-o). We define a counter example as follows: Let T 1 ( n ) = 2 n , T 2 ( n ) = n , and f ( n ) = n . Thus, T 1 ( n ) = O( f ( n )) because T 1 ( n ) = 2 f ( n ); T 2 ( n ) = O( f ( n )) because T 2 ( n ) = f ( n ). However, since , 1 lim ) ( ) ( ) ( lim 2 1 = = - n n n f n T n T n n so T 1 ( n ) – T 2 ( n ) o( f ( n )). (c) (3 pts.) Disprove O(1). ) ( ) ( 2 1 = n T n T We define a counter example as follows: Let T 1 ( n ) = n 2 , T 2 ( n ) = n , and f ( n ) = n 2 . Thus, T 1 ( n ) = O( f ( n )) because T 1 ( n ) = f ( n ); T 2 ( n ) = O( f ( n )) because T 2 ( n ) = n n 2 = f ( n ). However, , ) ( ) ( 2 2 1 n n n n T n T = = and n O(1) because n is not bounded by any constant. (d) (3 pts.) Disprove T 1 ( n ) = O( T 2 ( n )). We define a counter example as follows: Let T 1 ( n ) = n 2 , T 2 ( n ) = n , and f ( n ) = n 2 . Thus, T 1 ( n ) = O( f ( n )) because T 1 ( n ) = f ( n ); T 2 ( n ) = O( f ( n )) because T 2 ( n ) = n n 2 = f ( n ). However, T 1 ( n ) O( T 2 ( n )) because, if T 1 ( n ) = O( T 2 ( n )) were true, that is, if n 2 = O( n ), we show this leads to a contradiction. From n 2 = O( n ), there exist constants c , k such that n 2 cn when n k . That is, n c for n k . This is false when choosing n = max( c + 1, k ). 2.

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key1sp01 - COP3530C.01, Spring 2001 S. Lang Solution Key to...

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