answers to Ed Carstens's Site Swap Problems
1) c,e,f To test whether or not a siteswap is valid, all one needs is the throw-heights mod L added to the identity. If any of the resulting numbers are equal, then a collision will occur. Examples 761 mod 3 = 101 101 + 123 = 224 Since there are two matching throw-heights, it is invalid. 6671 mod 4 = 2231 2231 + 1234 = 3465 Since there are no matches, it is valid. 2) a,f This is most easily seen by drawing diagrams. Let a,b,c,d, and e represent balls. Then look at how each siteswap affects the order of the balls. (a) abcde -- 64555 --> bacde (b) abcde -- 75355 --> cbade (c) abcde -- 66661 --> eabcd (d) abcde -- 66625 --> dabce Notice that in (a) the a & b balls are switched whereas in part (d) the d & e balls get switched. 3) b,c,d A ground state pattern is one that can be entered from the ground state. Thus, if a juggler is doing 44444.. and wants to get into 561561.. it is not possible to do this directly, for the ball thrown as a '1' would collide with the '4'. Thus, 561 is an excited-state pattern. Notice that the siteswap, 12345, is also excited-state even though it can be rotated to a ground state pattern, namely 34512 or 45123. It may be argued that the pattern is periodic so that 34512 is indeed the same as 12345. However, 333 12345 333 is not possible, whereas 333 34512 333 is! 4) Let t_1,t_2,...t_L denote throw-heights making up a siteswap. Obviously, if we let them all be equal, permutation has no effect, so this is a trivial solution. However, we know that because of periodicity (L) we can add or subtract multiples of L to any of the throw-heights (this is local translation). Thus, the general solution is t_i = k + m_i*L where k is constant for all i and m_i is arbitrary. Examples: 714, 744, 444, 333, 633, 663, 963, 26A6 Another way of looking at it is if all throw-heights mod L are equal, then any permutation of them is valid.