On Tue, 1 May 2001, Phil Mendelsohn wrote: > > If you accidentally apply it twice, just apply it 26! - 2 more times > > and that will fix the problem. :) > > I haven't looked at it; is it a permutation without any cycles? Are there > no keys that remain fixed? This could be a slightly more interesting > combinatorial problem than it appears... No, it's pretty trivial. The order of the group is 26!, and when you raise an element to the order of the group, you get the identity. It's quite possible that there's a smaller number that will get you back where you started, but 26! will do it. As a grad student in math, I really should know how to figure out the order of the permutation, since that's quite basic abstract algebra / combinatorics... Of course, if you start with QWERTY and then apply the permutation that gives you Dvorak 26! times, you'll get back to QWERTY. So perhaps, in the Dvorak-advocacy spirit of the thread, we should apply that permutation one more time to get back to Dvorak. Dan -- /*-------------------------------- Dan Drake <drake at lemongecko.org> http://lemongecko.org/drake/ --------pgp encouraged----------*/