Suppose the balls all are traveling at 1 revolution/minute.

The first insight is that, ignoring colors, the problem is identical to a situation in which the balls are passing through each other instead of bouncing off. From this, we see that after 1 minute, there will be _some_ ball in each position that there was originally a ball, going in all the same directions as the original setup. That’s the coolest thing about this problem: you can have whatever crazy spacing you want, but it turns out to be irrelevant. From here on out, the problem has only to do with the _order_ of the balls around the circle.

Now, since balls are _not_ passing through each other, their order around the circle is preserved at all times. After a minute, they may have jointly rotated around the circle, but they are still in the same order as before. Call this 1-minute rotation R. Each time we apply R, it does the exact same rotation, shuffling the balls around from place to place until they come back to where they started. If you know a bit of (abstract) algebra, you are probably convinced of the answer by now. But for those who haven’t, some examples will make it clear.

First, suppose there are 3 balls. Number their starting positions 1,2,3. If R takes the ball in position 1 back to position 1, then it also must take 2 to 2 and 3 to 3, meaning that after a minute, everything is as it was.
Alternatively, R could take 1 to 2, in which case it must take 2 to 3 and 3 to 1. We can represent this by the cycle R=(1 2 3) (read “1 goes to 2 goes to 3 goes to 1″). Applying R twice gives R*R=(1 2 3)*(1 2 3) = (1 3 2), which is to say, letting the system run for 2 minutes will take ball 1 to position 3 and so forth. Applying R three times, we get back to where we started. To convince yourself of this, it helps to draw a picture and track what’s happening.

If you have more balls, say 6, you will get a cycle like (1 2 3 4 5 6) or (1 3 5)(2 4 6) (read “1 goes to 3 goes to 5 goes to 1, and 2 goes to 4 goes to 6 goes to 2″). Convince yourself that no matter what, you will always come back to where you started if you do it 6 times. (Of course, you may get back sooner, as in the second case).

Generalizing this, we see that for n balls, a snapshot taken after n minutes will always be identical to the starting position. What’s more, the balls will be going their original directions, ready to repeat the whole thing over again.

It helps to think about the problem in the reference frame that is corotating with, say, the counterclockwise beads. This makes it the same as a problem where some beads are stationary and some beads are moving clockwise.

Repeat but with 2 particles at the top and 2 at the bottom, they would collide every 15 minutes and reverse their track repeating the pattern again and again.

So obviously works with a perfectly symettrical start. I just don’t have the imagination to work out what happens in non symmetrical situations.

My gut feel is that given enough time you get repetition but then my understanding that pi never gets a repetitive pattern of decimals bites against that theory.

Consider the point particles to be identical (i.e. ignore their colors). Then two particles colliding and moving apart are equivalent to the two particles passing each other while each continuing at the same speed. So obviously, ignoring the colors, you just have particles looping around the circle, and you have infinitely many times at which the pattern repeats itself, where the interval between them is the time taken for each particle to make a trip around the circle.

Now, fill the colors back in. There are only finitely many ways that the colors can be assigned to the particles. That, by itself, doesn’t quite prove that your initial pattern I must repeat itself. But it proves that *some* pattern P must repeat itself.

Once you know that P repeats itself, look at the first occurrence of P and rewind time backwards to your initial pattern I. Since the perpetual motion machine is completely deterministic (forwards and backwards), you can start at the second occurrence of P and work backwards, and through exactly the same sequence of events, you must hit on a second occurrence of I.