Got it.. basically i thought that
1. XOR at bit level is OK.. but for any number a, i was wondering what would ~a be.. (0?). But actually a XOR b would be nothing but their corresponding bitwise XORs. 2. Also, a XOR b = a. ~b + ~a.b. Thus if we expand the expression [a XOR b XOR c.... n numbers] it would give many terms, each having multiplications of n numbers, i.e. O(n) multiplication, which I thought is expensive. But then that would be a wrong way to evaluate the expression; we can easily iteratively find the XOR as mentioned above! --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups-beta.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
