I don't know if I'm on the right track, perhaps someone can give me some pointers. Thanks.
1b) Since algorithm A is of O(2n^3), and it can solve a problem of size n = 20 in one hr, then by substituting n = 20 into the equation, we have that A performs 2 * (20 ^ 3) = 16000 operations in 1 hr. thus, a machine 8 times faster would be able to perform: 16000 * 8 = 128000 operations in 1 hr. since 2n^3 = 128000, then n^3 = 128000 / 2 = 64000 and therefore n = 40 I hope I'm on the right track. As for part c, I only know that when n^2 gets sufficiently large, it becomes the domineering term and therefore b should be of O(2^n). Is that the way to show? Thanks. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
