QB1 - Solution

# QB1 - Solution - COMP 271 Design and Analysis of Algorithms...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: COMP 271 Design and Analysis of Algorithms 2004 Fall Semester Solutions to Question Bank Number 1 (Selected Problems) Answer 1. The proof is by induction on n , the limit of the summation. For the basis case we consider the smallest legal value of n , namely 1. We have 1 summationdisplay i =1 i ( i − 1) = 0 = 1(1 − 1)(1 + 1) 3 , as desired. For the induction step, we will assume that the formula holds for all the values 1 , 2 , . . ., n − 1, then show that it holds for n . The standard method is to get rid of the last term of the sum, use the induction hypothesis to apply the formula to the the sum consisting of the first n − 1 terms, and then add the last term back in again and simplify. n summationdisplay i =1 i ( i − 1) = parenleftBigg n- 1 summationdisplay i =1 i ( i − 1) parenrightBigg + n ( n − 1) = ( n − 1)(( n − 1) − 1)(( n − 1) + 1) 3 + n ( n − 1) (by ind. hyp.) = ( n − 1)( n − 2) n 3 + n ( n − 1) = ( n − 2) n ( n − 1) + 3 n ( n − 1) 3 = ( n − 2 + 3) n ( n − 1) 3 = n ( n − 1)( n + 1) 3 , as desired. Answer 2. (a) True. Since T 1 ( n ) = O ( f ( n )) and T 2 ( n ) = O ( f ( n )), it follows from the def- inition that there exist constants c 1 , c 2 > 0 and positive integers n 1 , n 2 such that T 1 ( n ) ≤ c 1 f ( n ) for n ≥ n 1 and T 2 ( n ) ≤ c 2 f ( n ) for n ≥ n 2 . This implies that, T 1 ( n ) + T 2 ( n ) ≤ ( c 1 + c 2 ) f ( n ) for n ≥ max( n 1 , n 2 ). Thus, T 1 ( n ) + T 2 ( n ) = O ( f ( n )). (b) False. For a counterexample to the claim, let T 1 ( n ) = n 2 , T 2 ( n ) = n, f ( n ) = n 2 . Then T 1 ( n ) = O ( f ( n )) and T 2 ( n ) = O ( f ( n )) but T 1 ( n ) T 2 ( n ) = n negationslash = O (1). (c) False. We can use the same counterexample as in part (b). Note that T 1 ( n ) negationslash = O ( T 2 ( n )) Answer 3. A Relation: B (a) n 3 + n log n Ω , Θ , O n 3 + n 2 log n (b) log √ n Ω √ log n (c) n log 3 n Ω , Θ , O n log 4 n (d) 2 n Ω 2 n/ 2 (e) log(2 n ) Ω , Θ , O log(3 n ) Notes: (a) Both are Θ( n 3 ), the lower order terms can be ignored. Note that if A ( n ) = Θ( B ( n )), then automatically A ( n ) = O ( B ( n )) and A ( n ) = Ω( B ( n ))....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

QB1 - Solution - COMP 271 Design and Analysis of Algorithms...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online