2004 exam paper - question 1

can anybody help with question 1a of the 2004 exam paper....

what I get is the following .... (but i dont think its right)

The following six expressions represent the running times of various algorithms in terms of the input size N:

N/1000 + 23
1000/N + 23
3 log 3 N
log N2
N2 - 23N
1023


Answer:

1023, 1000/N + 23, 3 log 3 N, N/1000 + 23, log N2, N2 - 23N


It this right ?

From another post:

Hi here are some solutions for the 2004/2005 big oh questions

2004 question 1b)

algorithm A - O(n) as n increases by 10 the running time also increases by 10.

algorithm B - O(n2) (that's n squared) as n increases by 10 the running time increases by 10 squared (ie 100).

algorithm C - O(log n)(thats log to the base 10) the running time increases slowly. Running time T = 10 log n - 10.

When you see that the running time is increasing slowly it is most likely a log function.

2005 question 1a)

Algorithm A O(n) see A above

Algorithm B O(n log n) (thats log to the base 10) Running time T = 2n log n.

If you look at this algorithm you will notice that it is not quite O(n) nor is O(n squared). It is someting in between hence it most likey O(n log n). From this you can determine that T=2n log n.

regards
Lazarus

Hi,

I saw that post, but that only answers Q1B of the 2004 paper, not Q1A of the 2004 paper.

Exam 2004 Question 1 A:
The following six expressions represent the running times of various algorithms in terms of the input size N:

N/1000 + 23
1000/N + 23
3 log 3 N
log N2
N2 - 23N
1023


My answer:

1023, 1000/N + 23, 3 log 3 N, N/1000 + 23, log N2, N2 - 23N

Is that right or wrong ?