Re: juggling's combinatorics problem with siteswap

From: Steve Bennett (gg_at_stevage.com)
Date: 02/10/05 

Date: 10 Feb 2005 01:55:16 -0800

>the period number. (the period is the number of throws all colored
balls return to their position.)

Algorithm: Use jugglefactory.stevage.com, type in the pattern, and
repeat it until you have at least as many characters as the highest
throw, and until the first ball in the output is an A. That probably
doesn't make much sense till you try it :)

>whether the pattern is ambidextrous. (a pattern is ambidextrous if
left and right hand play the same role. (not necessarily synchronized
or lockstep.))

A couple of people have answered this slightly incorrectly. If the
pattern length is odd, it's certainly ambidextrous. However, it may
also be even with an even length pattern. Trivially: 33. Less
trivially, 8448. For an algorithm if the length is not odd, split the
pattern into the two hands(take every second throw), then write those
siteswaps in a canonical form (eg, in whatever rotation has the highest
throws first). So in this case, you might end up with 8040 and 8040 -
same pattern, hence "ambidextrous"

>* the period of the orbit (in the framework of the whole pattern. i.e.
> the period of a orbit is necessarily less or equal to the period of
the
> pattern.).

Not quite sure what you mean. Take the pattern 531 - the period of the
"3" orbit is 3, the period of the "5_1" orbit is 6 - presumably you can
add together all the throw values that make up the orbit.

As for notation to describe an orbit, how about what I just used? You
don't even need the _'s...

Steve

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