# sln-HW01 - CSE450/598 Design And Analysis of Algorithms...

This preview shows pages 1–2. Sign up to view the full content.

CSE450/598 Design And Analysis of Algorithms HW01 Grading Keys 1. (10 pts) Let f ( n ) = 0 . 01 n 2 and g ( n ) = 1000 × n × log 2 ( n + 3). Find the smallest integer N 0 such that f ( N ) g ( N ) but f ( N + 1) > g ( N + 1). Show the values of N , f ( N ), g ( N ), f ( N + 1) and g ( N + 1). Grading Keys: 2 pts for each of the following five values. N = 2100210 , f ( N ) = 44108820441 , g ( N ) = 44108829304 . f ( N + 1) = 44108862445 , g ( N + 1) = 44108851749 . 2. (10 pts) Prove or disprove Θ( n 3 ) + O ( n 4 ) = Θ( n 4 ). Θ( n 3 ) + O ( n 4 ) = Θ( n 4 ) is not true. Proof: Let f ( n ) = n 3 , g ( n ) = n 2 . Then we have f ( n ) Θ( n 3 ) and g ( n ) O ( n 4 ). However, f ( n ) + g ( n ) 6∈ Θ( n 4 ). 2 Grading Keys: 4 pts for the correct conclusion. 6 pts each for the counter example. 3. (10 pts) Exercise 3.5 (p.108) Solutions: Let n ( T ) denote the number of nodes in T . Let n 0 ( T ) denote the number of leaf nodes in T . Let n 1 ( T ) denote the number of nodes in T that has exactly one child. Let n 2 ( T ) denote the number of nodes in T that has exactly two children. Then n ( T ) = n 0 ( T ) + n 1 ( T ) + n 2 ( T ). We need to show that for any binary tree T , we have n 2 ( T ) = n 0 ( T ) - 1 . We prove this by induction on the number of nodes in the binary tree T . When n ( T ) = 1, T has only one node, which is a leaf node. Therefore we have n 0 ( T ) = 1 and n 2 ( T ) = 0. Therefore the claim is true for the base case. Now we assume that the claim is true for any binary tree with k or fewer nodes, for some positive integer k . Let T be a binary tree such that n ( T ) = k + 1. Let v be a leaf node in T . Since n (

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.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern