3-Handed Siteswaps

August 2021

There are basically two options to design a 3-handed siteswap. The first, and more intuitive one, is to consider the three hands and enumerate them. In other words, we split a valid siteswap (for example 86277) between 3 hands. This will result in a globally equally long time between the throws.

Fig 1. If considering 3 hands to spilt the siteswap, it will result in a locally uneven time gab between throws for juggler \(A\), even if the global time gab is equally spaced.

However as shown in the time sequence in in Fig. 1. this will result in a locally uneven time gab between throws for juggler \(A\) using two hands, even if the time between throws is globally the same. Moreover, this will go against the familiar connotation of the more common 4-handed siteswaps. For example, this will result in that the \(7\) will not longer be a pass throw.

Another method to design a 3-handed siteswap is to regard a 4-handed siteswap, where one hand neither receiving nor throwing any objects. This is determent by a 0 on every second beat of juggler \(B\). Assuming we want to set every second beat of Juggler \(B\) to 0, which will result in a sequence like: \(? \: 0 \: ? \: 0 \: ? \: 0 \dots\), where the \(?\) is a placeholder for a throw. The throwing sequence of juggler \(A\) will have subsequential the same length \(n_A = n_B\). Combining that with the sequence of juggler \(B\), gives us a global patter with the following structure:

1 2 3 4 5 6 7 8 9
\(?_A\) \(?_B\) \(?_A\) \(0\) \(?_A\) \(?_B\) \(?_A\) \(0\) \(\dots\)

\(\Downarrow\)

global beats 1 2 3 4 5 6 7 8 9 10 11 12
\(A\) \(?\) \(?\) \(?\) \(?\) \(?\) \(?\)
\(B\) \(?\) 0 \(?\) 0 \(?\) 0

In general we can derive from the period length following statements:

  • \(p_g \mod 2 \equiv 1\)
    If the period length \(p_g\) of the global pattern is odd, every juggler is throwing the same pattern, but starts on different points on the sequence.
  • \(p_g \mod 2 \equiv 0\)
    If the period length \(p_g\) of the global pattern is even, both jugglers can have different patterns, their hands throwing the same pattern, but starting on different points on the sequence.
  • \(\frac{p_g}{2} \mod 2 \equiv 0\)
    If the period length \(p_g\) divided by two is even, it will give a unique sequence to each hand of each juggler.

To isolate one hand of Juggler \(B\) (meaning giving it a unique sequence for just that very hand), results in a total period length of four-time as long as the sequence of that very hand.

Since juggler \(B\) just has one hand to fill, just throws landing on the same hand or on one of juggler \(A's\) hands are possible. This excludes locally odd throws like \(1 \)(zip), \(3\) (self) or \(5 \), since those will land on the other hand. If we determinate for example, the missing beats of \(B\) with \(8\) (heff), \(7\) (pass), \(7\) (pass), we will get the full local sequence of juggler \(B\): \(807070\).

Adding the throws of juggler A will give us \(?8?0?7?0?7?0\). We can now calculate the interface. It can be determined by marking where the throws of B will land (counting the length of throws and mark where they land), which is: \(7--\:0\:7--\:0-8-0\).

throw \(?_A\) \(8_B\) \(?_A\) \(0\) \(?_A\) \(7_B\) \(?_A\) \(0\) \(?_A\) \(7_B\) \(?_A\) \(0\)
landing on \(7\) \(\) \(\) \(0\) \(7\) \(\) \(\) \(0\) \(\) \(8\) \(\) \(0\)
we need objects to arrive at \(\) x \(\) \(\) x \(\) \(\) \(\) \(\) \(\) \(\)
interface \(P\) \(P\) \(S\) \(S\) \(P\) \(P\) \(S\) \(S\) \(S\) \(S\) \(S\) \(S\)

We can see in line two at which point in time objects are arriving. To be able to trow the initially defined throws, further objects have to arrive at these beats. Meaning, we mark all beats where we need a club to throw, but none is arriving yet. We then translate the thrown into selfs (S) and passes (P) according to their value. The before marked beats are likewise arriving passes. This gives us the interface \(PPSSPPSSSSSSS\). Any valid siteswap with that same interface can be a valid 3-handed siteswap.

Here are 3-handed siteswaps, where juggler \(B\) has the pattern 807070:

  • 860677067706 \(\rightarrow B\):(H0P0P0); \(A\):(S S P S P S)
  • 860677067805 \(\rightarrow B\):(H0P0P0); \(A\):(S S P S Heff Zap)
  • 860678057706 \(\rightarrow B\):(H0P0P0); \(A\):(S S Heff Zap P S)
  • 860678057805 \(\rightarrow B\):(H0P0P0); \(A\):(S S Heff Zap Heff Zap)
  • 880477067706 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold P S P S)
  • 880477067805 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold P S Heff Zap)
  • 880478057706 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold Heff Zap P S)
  • 880478057805 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold Heff Zap Heff Zap)
  • 884067706770 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold S P S P)
  • 884067805770 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Hold S Heff Zap P)
  • 888027706770 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Heff Zip P S P)
  • 888027805770 \(\rightarrow B\):(H0P0P0); \(A\):(Heff Heff Zip Heff Zap P)