{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MidtermSolS04

# MidtermSolS04 - UML CS 91.404 Midterm Exam Spring 2004...

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

UML CS 91.404 Midterm Exam Spring, 2004 1 of 5 MIDTERM EXAM & SOLUTIONS 1. (20 points) What can you conclude? Given: ( ) ( ) n n n f lg 2 lg ) ( ) 2 Ω ( ) ) ( ) ( ) 3 lg lg 3 n n n f Θ 2 1 ) ( ) 1 n n f Ω a) (10 points) Can we conclude from statements (1)-(3) that ? )) ( ( ) ( 2 3 n f O n f Why or why not? Either prove or provide a counterexample. ) ( 1 n f 2 n ) ( 2 n f ) ( 3 n f () n n n n lg lg lg lg = SOLUTION : YES. First, observe that () . This can be verified by taking lg of both sides. Next, because lglgn > 2 for n > 16. From this we obtain the picture above. n n n n lg lg lg lg = () lg lg lg lg n n n n > = 2 n To show that , we first use the definition of Θ to conclude that . Since )) ( ( ) ( 2 3 n f O n f ) ( ) 3 lg lg n n f n ) ( ( ) ( lg lg 3 n n O n f Θ ( ) n n n n lg lg lg lg = , this implies, transitively, that . Now, () ) lg n n lg ( ) ( 3 O n f ( ) n n n lg 2 ) (lg ) ( Ω f implies, using

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

View Full Document
UML CS 91.404 Midterm Exam Spring, 2004 2 of 5 transpose symmetry, that ( ) ( ) ) ( lg 2 lg n f O n n . Thus, transitively, . )) ( ( ) ( 2 3 n f O n f b) (10 points) Can we conclude from statements (1)-(3) that ? )) ( ( ) ( 3 1 n f O n f Why or why not? Either prove or provide a counterexample. SOLUTION : NO. Counter-example: and . n n = n n n f lg lg 3 ) ( = n f ) ( 1 2. (20 points) Recurrence In this problem, you will find a tight upper and lower bound on the closed- form solution for the following recurrence: = > + + = 1 1 1 3 8 5 6 ) ( 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

MidtermSolS04 - UML CS 91.404 Midterm Exam Spring 2004...

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

View Full Document
Ask a homework question - tutors are online