Xref: utzoo comp.misc:6470 sci.math:7169 Path: utzoo!utgpu!jarvis.csri.toronto.edu!rutgers!tut.cis.ohio-state.edu!pt.cs.cmu.edu!andrew.cmu.edu!jk3k+ From: jk3k+@andrew.cmu.edu (Joe Keane) Newsgroups: comp.misc,sci.math Subject: Re: Base 3 computers? (was: Divide by three?) Message-ID: Date: 2 Jul 89 00:21:00 GMT References: <6710021@hpcupt1.HP.COM> <6250@sunray.UUCP> <626@hrc63.co.uk> , <5787@rpi.edu> Organization: Mathematics, Carnegie Mellon, Pittsburgh, PA Lines: 15 In-Reply-To: <5787@rpi.edu> In article <5787@rpi.edu> puswad@pawl.rpi.edu (Math Student from Hell) writes: >In article >shafer@drynix.dfrf.nasa.gov writes: >>When I was at UCLA, majoring in CS, one of my professors told me that >>the Russians had tried to build a trinary computer. The reason being >>that Shannon proved that e (2.7...) was the most efficient base for >>information content and 3 was closer to e than 2 was. > >I've heard this before. Can someone in netland tell me >what it means? _If_ the cost of a base-b digit is proportional to b, then the cost of something with N possibilities in base b is b*log_b(N), which is minimized at b=e, or b=3 if you restrict b to integers. Of course storing bits is actually _much_ easier than trits, so you have to take into account the value of `much'.