Hi ashish and Hamster, First of all, thank you so much for your posts. They are very much appreciated.
Hamster, you mentioned that to show that an algorithm is O(2^n), I need to show that there is some other function (say f or g) that looks exactly the same. Thats where I lost you. How is this helpful? Oh, I see the point, hopefully I'm right on this one as well. That an algorithm must O(n^2) if there are some other (n^2) functions that are greater than that of the algorithm. This is logical in that if the algorithm is not larger than an (n^2) function, then it must be O(n^2) in nature. Hopefully i got you correctly. Nevertheless, thank you so much for the replies. --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
