On Tue, 1 May 2001, Al Borchers wrote: > Dan Drake <drake at lemongecko.org> wrote: > > 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. > > There is certainly a much smaller number than 26!. The order of an element > in the symmetric group is the LCM of its cycle lengths. For QWERTY->DVORAK > I get the cycle decomposition of > > (Q ' - [ / Z ; S O R P L N B X) (W ,) (E . V K T Y F U G I C J H D) (A) (M) That's what I was curious about. Did you sit down and map it? I guess there was nothing on T.V. > And you thought math would never help you out in real life. Oh no, math helps in real life quite often in a varitey of ways. What's unfortunate is that real life doesn't help more often in math! (But then I've a combinatorics test this afternoon...) -- "To misattribute a quote is unforgivable." --Anonymous