@atul: on the first look, even I thought the same.. O(log N).. and this is may be true for the given precision.
*[-- the following may not be related to given qn.] -- but.. can u comment on this view point..* but.. I am thinking that, the complexity is dependent on the level of precision required. like, even for a value of sqrt(11), and the number of digits needed for precision is 100 digits, and the averaging method (bisecting method) just correctly find each digit precision every step, so... complexity in my case.. come to O(X), X - amount of precision needed. (i know this is of not much use for many cases.. a precision of 1000 digit's) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/ePb9eSEZlNAJ. 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.
