yup, yours(binary gcd) is better. On Sat, Sep 10, 2011 at 9:34 PM, Neha Singh <neha.ndelhi.1...@gmail.com>wrote:
> How abt dis : > > The algorithm reduces the problem of finding the GCD by repeatedly applying > these identities: > 1. gcd(0, v) = v, because everything divides zero, and v is the largest > number that divides v. Similarly, gcd(u, 0) = u. gcd(0, 0) is not typically > defined, but it is convenient to set gcd(0, 0) = 0. > > 2. If u and v are both even, then gcd(u, v) = 2·gcd(u/2, v/2), because 2 is > a common divisor. > > 3. If u is even and v is odd, then gcd(u, v) = gcd(u/2, v), because 2 is > not a common divisor. Similarly, if u is odd and v is even, then gcd(u, v) = > gcd(u, v/2). > > 4. If u and v are both odd, and u > v, then gcd(u, v) = gcd((u - v)/2, v). > If both are odd and u < v, then gcd(u, v) = gcd((v - u)/2, u). > > > > > Which is better ? > > -- > 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 > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- 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 algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
