From: Charles Krug Newsgroups: comp.os.msdos.djgpp Subject: Re: The meaning of O(n)... Date: Mon, 26 Jan 1998 09:36:54 -0500 Lines: 26 Message-ID: <34CC9F86.166B3E3@pentek.com> References: <34C8B7D1 DOT A19B36EA AT pentek DOT com> <34C98CE6 DOT 861 AT prodigy DOT net> NNTP-Posting-Host: mail.pentek.com Mime-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit To: djgpp AT delorie DOT com DJ-Gateway: from newsgroup comp.os.msdos.djgpp Precedence: bulk Heutchy wrote: > > 2^n, n!, n^3, n^2, nlog(n), n, log(n), 1. > > At the risk of sounding like an idiot, isn't n! worse than 2^n? > > n! = 1 * 2 * 3 * 4 * ... * n-1 * n > 2^n = 2 * 2 * 2 * 2 * ... * 2 * 2 > > So the n! grows much faster. n^n would be worse however. I don't know > if there are any useful algorithms with O(n^n) though. I dunno, I was working from memory . . . . You're right, both by intuition, and by looking it up. You correctly guessed that I confused 2^n with n^n. So I stand (or sit) corrected. n^n is the complexity of multiplication of hypermatrices of order n in n dimensions. Sure, I do it all the time. Yeah, right! -- Charles Krug, Jr.