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