[Solved] DataStructure – Assignment #2

Chapter 2

6 pts

1. In the definition of Big-O, why is the “for N >= n0” needed?

6 pts

2. If f1(N) = 2N and f2(N) = 3N, why are they both O(N), since 3N is larger than 2N for N>=1?

6 pts

3. a) For f1(N) = 2N and f2(N) = 3N:

calculate f1(5) and f2(5), then f1(10) and f2(10). When N was doubled in each case, what happened to the result? Explain why this happens.

1. b) For f1(N) = 2N*N and f2(N) = 3N*N: calculate f1(5) and f2(5), then f1(10) and f2(10). When N was doubled in each case, what happened to the result? Explain why this happens.

6 pts

4. Since Big-O notation is a mathematical tool for functions like f(N) or g(N), how is it applicable to algorithm analysis?

6 pts

5. Which grows faster, 2^n or n! ? Explain why.

10 pts (2 each)

6. Give the Big-O notation for the following expressions:
   1. 4n^5 + 3n^2 – 2
   2. 5^n – n^2 + 19
   3. (3/5)*n
   4. 3n * log(n) + 11
   5. [n(n+1)/2 + n] / 2

Questions 7-12 are 10 points each.

Assume numItems has the role of N, which may vary from one run to the next.

7. What is the Big-O running time for this code? Explain your answer.

for (int i=0; i<numItems; i++)

System.out.println(i+1);

8. What is the Big-O running time for this code? Explain your answer.

for (int i=0; i<numItems; i++) for (int j=0; j<numItems; j++)

System.out.println( (i+1) * (j+1) );

9. What is the Big-O running time for this code? Explain your answer.

for (int i=0; i<numItems+1; i++) for (int j=0; j<2*numItems; j++) System.out.println( (i+1) * (j+1) );

10. What is the Big-O running time for this code? Explain your answer.

if ( num < numItems ) for (int i=0; i<numItems; i++)

{

System.out.println(i);

} else

System.out.println(“too many”);

11. What is the Big-O running time for this code? Explain your answer.

int i = numItems; while (i > 0)

i = i / 2; // integer division will eventually reach zero

12. What is the Big-O running time for this code? Explain your answer. (You do not need to work out a recurrence formula).

public static int div(int numItems)

{

if (numItems == 0) return 0; else

return numItems%2 + div(numItems/2); }

Submit these files: hw2.doc (.doc can be .txt, .jpg, etc.)