TIPS & TRICKS

 

Getting Back Around To Your Own Clubs

The Academic Juggler

 by Arthur Lewbel

 

Last Fall the Flying Karamazov Brothers appeared in Spingold Theatre, just 100 yards from my office at Brandeis University. Each F.K. Brother wears a different colored hat, and has clubs to match. They're often careful to repeat passing patterns just the right number of times to each end up with the same clubs they started with.

 

It's a nice visual effect, and I wondered if it's always possible. Does every possible passing pattern have the property that, if repeated enough times, everybody ends up with the same clubs they started with?

 

The answer is yes, and the proof is surprisingly simple, given that the number of possible passing patterns is infinite, and most have probably not been invented yet.

 

Consider any passing pattern, with any number of jugglers and any number of clubs. Let the integer k represent the position of all the clubs at the beginning of the 'k'th repetition of the passing pattern. Position zero is where everyone has their own clubs. For any k, if you know position k, you can figure out position k+1 by doing the pattern one more time, and you can figure out position k-1 by running the pattern in reverse.

 

Since the number of possible positions is finite (it equals N factorial, where N is the number of clubs), we know that sooner or later some position must get repeated. Let k1 be the smallest integer such that position k1 is the same as some earlier position, k0. (Notice that I'm only looking at the pattern at the beginning of each cycle, and am ignoring all the other positions that happen during the pattern).

 

If position k0 is the same as position zero, then everyone has their own clubs again, as desired. If position k0 isn't the same as position 0, then position k1-1 must be the same as position k0-1, but this would contradict the assumption that k1 was the first position that repeated an earlier one.

 

Therefore, position k0 must indeed be the same as position zero, and everyone does end up with the same clubs they started with.

 

This proof assumes there are no drops. Any break in the pattern, such as a drop or an unrepeated trick, may result in the clubs never ending up where they started.

 

This proof, which can be easily recast in general terms using the branch of mathematics called group theory, has many applications outside of juggling.

For example, the same argument can be used to prove that if you repeat a perfect riffle shuffle (or any other kind of shuffle) of a deck of cards enough times, you will end up with all the cards in the same position they started in.

 

Knowing that any juggling pattern will eventually get back to where it started isn't as interesting as knowing when it will get back. What does k0 equal for various common passing patterns? How many repetitions are required for everyone to get their own clubs back? I'll discuss this question more in the next installment of The Academic Juggler, but for now let's just consider an upper bound on the value.

 

In the proof, I used the fact that in a pattern with N clubs, the total number of possible positions is N factorial. N factorial is defined to equal 1 x2x3 x... xN, and is the general formula for the total number of ways to position N objects. It follows from the proof that number of repetitions before everyone gets their own clubs back, k0, must be less than N factorial. For any pattern, N factorial is an upper bound for k0.