So far in this series, I have talked about “vanilla” siteswaps where one is allowed to throw only at most one object at every beat. Let’s now add some flavour and allow for more than one object to be thrown at each beat. As we did with vanilla siteswaps, we shall first review some mathematical results for these “flavoured siteswaps”, before worrying about how to actually juggle them.
If multiple objects are being thrown at a given beat, we need a siteswap number corresponding to each object being thrown at that beat. In other words, more than one siteswap number is required for one beat. We will indicate siteswap numbers corresponding to the same beat by enclosing them in curly brackets { }. For example, consider the two sequences “3 3 3 3” and “{3 3} 3 3”. The first one, “3 3 3 3”, is a vanilla siteswap sequence of period P = 4 (minimal period = 1). At each beat, exactly one object is thrown with a siteswap number “3”.
| Current Time (in beats) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Siteswap Number | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 3 |
| # of Objects thrown | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
In the second sequence “{3 3} 3 3”, the curly brackets around the first couple of 3’s indicate that two objects are to be thrown at that beat. Subsequently, one object is to be thrown at each of the next two beats. Only three beats of time are represented by this sequence, i.e., period P = 3.
| Current Time (in beats) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Siteswap Numbers | {3 3} | 3 | 3 | {3 3} | 3 | 3 | {3 3} | 3 | 3 | {3 3} |
| # of Objects thrown | 2 | 1 | 1 | 2 | 1 | 1 | 2 | 1 | 1 | 2 |
A comparison of Table 1 and Table 2 shows these differences between the sequences “3 3 3 3” and “{3 3} 3 3”.
Let us consider the sequence “{3 2} 3 1”. Again, two objects are to be thrown at beat #0 and one object each at beat #1 and beat #2. As worked out in Table 3, the sequence “{3 2} 3 1” is such that the same two objects (A and B) keep getting scheduled at the beat where a {3 2} throw is to be made. The choice of which object to throw as a “3” and which as a “2” is up to the juggler. For this reason, the order of the siteswap numbers within the brackets is immaterial[1].
| Current Time (in beats) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Siteswap Number | {3 2} | 3 | 1 | {3 2} | 3 | 1 | {3 2} | 3 | 1 | {3 2} |
| Objects thrown (option 1) | A:3, B:2 | C | B | A:3 B:2 | C | B | A:3, B:2 | C | B | A:3 B:2 |
| Objects thrown (option 2) | A:3, B:2 | C | B | B:3, A:2 | C | A | A:3, B:2 | C | B | B:3, A:2 |
| Objects thrown (option 3) | B:3, A:2 | C | A | B:3, A:2 | C | A | B:3, A:2 | C | A | B:3, A:2 |
Option 1 in Table 3 has object A being always thrown with siteswap number 3 and object B being always thrown with siteswap number 2. Option 3 does the opposite while option 2 alternates between {A:3, B:2} and {B:3, A:2}. Further, as before[2], the choice of “hand siteswap” is also up to the juggler. Some of these possibilities are demonstrated in the video below.
[Aside: See how in the video, the “3” throw for the yellow ball could be thrown to the other hand (Options 1a, 1b and 2a) or to the same hand (Options 1c and 2b). This is not surprising because the “hand siteswap” for options 1(b), 1(c) and 2(b) is not “2”[3].]
We now extend to flavoured siteswaps, the average theorem, the permutation test and the siteswap generating algorithm that we studied for vanilla siteswaps[2].
The Average Theorem
The number of numbers in a flavoured siteswap sequence is not equal to the period of the sequence because multiple numbers may be present for a single beat. However, the sum of all the numbers divided by the period of the siteswap will still yield the number of objects required to juggle a pattern corresponding to a valid flavoured siteswap[4].
| Siteswap Sequence | Sum of all the Numbers | Period | # of Objects |
|---|---|---|---|
| {3 3} 3 3 | 3+3+3+3 = 12 | 3 | 12/3 = 4 |
| {3 2} 3 1 | 3+2+3+1 = 9 | 3 | 9/3 = 3 |
| {3 1} | 3+1 = 4 | 1 | 4/1 = 4 |
| {4 2 2 3 1 4} 1 {2 1 1 1} {3 1 1 1} | 4+2+2+3+1+4+1+2+1+1+1+3+1+1+1 = 28 | 4 | 28/4 = 7 |
Table 4 shows the number of objects that would be required to juggle some flavoured siteswaps if they are valid siteswaps. Validity can be determined using the permutation test.
Permutation Test
Before discussing the permutation test, I will digress a little to introduce a “double subscript” notation, Tmn, to represent the flavoured siteswap numbers. A general flavoured siteswap sequence of period P will now be written as:
“{T01 T02 … T0N0} {T11 T12 … T1N1} … {Tm1 Tm2 … TmNm} … {T(P-1)1 T(P-1)2 … T(P-1)N(P-1)}”.
For example, consider the sequence “{4 2 2 3 1 4} 1 {2 1 1 1} {3 1 1 1}” with period P = 4. The various values of Tmn for this sequence are shown in Table 5.
| Beat # (m) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Siteswap Numbers at m | {4 2 2 3 1 4} | 1 | {2 1 1 1} | {3 1 1 1} |
| Number of siteswap numbers at m (Nm) | N0 = 6 | N1 = 1 | N2 = 4 | N3 = 4 = N(P-1) |
| Throw Names and Values at m (Tmn) | {T01=4, T02=2, T03=2, T04=3, T05=1, T06=4=T0N0} | T11=1=T1N1 | {T21=2, T22=1, T23=1, T24=1=T2N2} | {T31=3, T32=1, T33=1, T34=1=T3N3} |
With the flavoured siteswap sequence represented thus, the algorithm for the permutation test is presented in Table 6.
| Given Sequence | {T01 T02 … T0N0} | {T11 T12 … T1N1} | {Tm1 Tm2 … TmNm} | {T(P-1)1 T(P-1)2 … T(P-1)N(P-1)} |
|---|---|---|---|---|
| Add | {0 0 … N0 times} | {1 1 … N1 times} | {m m … Nm times} | {(P-1) (P-1) … (P-1) times} |
| Result | {(T01+0) (T02+0) … (T0N0+0)} | {(T11+1) (T12+1) … (T1N1+1)} | {(Tm1+m) (Tm2+m) … (TmNm+m)} | {(T(P-1)1+P-1) (T(P-2)2+P-1) … (T(P-1)N(P-1)+P-1)} |
| Remainder of (Result/P) | {R01 R02 … R0N0} | {R11 R12 … R1N1} | {Rm1 Rm2 … RmNm} | {R(P-1)1 R(P-1)2 … R(P-1)N(P-1)} |
| Quotient of (Result/P) | {Q01 Q02 … Q0N0} | {Q11 Q12 … Q1N1} | {Qm1 Qm2 … QmNm} | {Q(P-1)1 Q(P-1)2 … Q(P-1)N(P-1)} |
The given siteswap sequence is valid if in the Remainder Row, we find N0 0’s, N1 1’s, …, Nm m’s, …,and N(P-1) (P-1)’s. For a valid sequence, the sum of all the terms in the quotient row gives the number of objects required to juggle the corresponding juggling pattern.
Examples
Let us see if the sequences of Table 4 are valid.
“{3 3} 3 3”:
This represents 3 beats, so P = 3. At beat #0, there are two siteswap numbers, so N0 = 2. At beat #1 and beat #2, there is one siteswap number each, so N1 = N2 = 1.
| Given Sequence | {3 3} | 3 | 3 |
|---|---|---|---|
| Add | {0 0} | 1 | 2 |
| Result | {3 3}+{0 0} = {3 3} | 3+1 = 4 | 3+2 = 5 |
| Remainder of (Result/P) | {3 3} mod 3 = {0 0} | 4 mod 3 = 1 | 5 mod 3 = 2 |
| Quotient of (Result/P) | int({3 3}/3) = {1 1} | int(4/3) = 1 | int(5/3) = 1 |
Now N0 = 2 and in the Remainder Row of Table 7, there are two 0s. N1 = 1 and there is one 1 in the Remainder Row of Table 7. Finally, N2 = 1 and there is one 2 in the Remainder Row of Table 7. Thus, the sequence “{3 3} 3 3” is a valid siteswap sequence. The sum of all numbers in the Quotient row of Table 7 is {1+1}+1+1 = 4, so this sequence corresponds to a 4 object juggling pattern.
“{3 2} 3 1”:
This again represents 3 beats with two siteswap numbers at beat #0 and one siteswap number each at beats #1 and 2. So N0 = 2, N1 = 1 = N2.
| Given Sequence | {3 2} | 3 | 1 |
|---|---|---|---|
| Add | {0 0} | 1 | 2 |
| Result | {3 2}+{0 0} = {3 2} | 3+1 = 4 | 1+2 = 3 |
| Remainder of (Result/P) | {3 2} mod 3 = {0 2} | 4 mod 3 = 1 | 3 mod 3 = 0 |
| Quotient of (Result/P) | int({3 2}/3) = {1 0} | int(4/3) = 1 | int(3/3) = 1 |
Again, N0 = 2 and in the Remainder Row of Table 8, there are two 0s. N1 = 1 and there is one 1 in the Remainder Row of Table 8. Finally, N2 = 1 and there is one 2 in the Remainder Row of Table 8. Thus, the sequence “{3 2} 3 1” is a valid siteswap sequence. The sum of all numbers in the Quotient row of Table 8 is {1+0}+1+1 = 3, so this sequence corresponds to a 3 object juggling pattern.
“{3 1}”:
This represents only one beat, so P = 1 and N0 = 2 as there are two siteswap numbers at beat #0.
| Given Sequence | {3 1} |
|---|---|
| Add | {0 0} |
| Result | {3 1}+{0 0} = {3 1} |
| Remainder of (Result/P) | {3 2} mod 1 = {0 0} |
| Quotient of (Result/P) | int({3 1}/1) = {3 1} |
Again, N0 = 2 and in the Remainder Row of Table 9, there are two 0s. Thus, the sequence “{3 1}” is a valid siteswap sequence. The sum of all numbers in the Quotient row of Table 9 is {3+1} = 4, so this sequence corresponds to a 4 object juggling pattern.
In fact, just like the case of vanilla siteswaps, any flavoured siteswap with period P = 1 will be a valid siteswap: As any number when divided by 1 will give a remainder of 0, we will get as many 0’s as there are siteswap numbers at beat #0.
“{4 2 2 3 1 4} 1 {2 1 1 1} {3 1 1 1}”:
This sequence represents 4 beats. At beat #0, there are six siteswap numbers, so N0 = 6. At beat #1, there is one siteswap number so N1 = 1. At beat #2 and beat #3, there are four siteswap numbers each, so N2 = 4 = N3.
| Given Sequence | {4 2 2 3 1 4} | 1 | {2 1 1 1} | {3 1 1 1} |
|---|---|---|---|---|
| Add | {0 0 0 0 0 0} | 1 | {2 2 2 2} | {3 3 3 3} |
| Result | {4 2 2 3 1 4}+{0 0 0 0 0 0} = {4 2 2 3 1 4} | 1+1 = 2 | {2 1 1 1}+{2 2 2 2} = {4 3 3 3} | {3 1 1 1}+{3 3 3 3} = {6 4 4 4} |
| Remainder of (Result/P) | {4 2 2 3 1 4} mod 4 = {0 2 2 3 1 0} | 2 mod 4 = 2 | {4 3 3 3} mod 4 = {0 3 3 3} | {6 4 4 4} mod 4 = {2 0 0 0} |
| Quotient of (Result/P) | int({4 2 2 3 1 4}/4) = {1 0 0 0 0 1} | int(2/4) = 0 | int({4 3 3 3}/4) = {1 0 0 0} | int({6 4 4 4}/4) = {1 1 1 1} |
We see that in the Remainder Row of Table 10, there are six 0s and N0 = 6. Number of 1’s in the Remainder Row of Table 10 is 1 = N1. Number of 2’s in the Remainder Row of Table 10 is four, which is the same as the number of 3’s. And N2 = 4 = N3. Thus, the sequence “{4 2 2 3 1 4} 1 {2 1 1 1} {3 1 1 1}” is a valid siteswap sequence. The sum of all numbers in the Quotient row of Table 10 is {1+0+0+0+0+1}+0+{1+0+0+0}+{1+1+1+1} = 7, so this sequence corresponds to a 7 object juggling pattern.
Let’s now see a couple of examples where the permutation test determines that the flavoured siteswap is invalid. First, let’s see what happens if we move a few of the throws of the example of Table 10 from beat #0 to beat #1 to get the sequence
“{4 2 2} {3 1 4 1} {2 1 1 1} {3 1 1 1}”:
Again, Period P = 4. But now N0 = 3, and N1 = 4 = N2 = N3.
| Given Sequence | {4 2 2} | {3 1 4 1} | {2 1 1 1} | {3 1 1 1} |
|---|---|---|---|---|
| Add | {0 0 0} | {1 1 1 1} | {2 2 2 2} | {3 3 3 3} |
| Result | {4 2 2}+{0 0 0} = {4 2 2} | {3 1 4 1}+{1 1 1 1} = {4 2 5 2} | {2 1 1 1}+{2 2 2 2} = {4 3 3 3} | {3 1 1 1}+{3 3 3 3} = {6 4 4 4} |
| Remainder of (Result/P) | {4 2 2} mod 4 = {0 2 2} | {4 2 5 2} mod 4 = {0 2 1 2} | {4 3 3 3} mod 4 = {0 3 3 3} | {6 4 4 4} mod 4 = {2 0 0 0} |
| Quotient of (Result/P) | int({4 2 2}/4) = {1 0 0} | int({4 2 5 2}/4) = {1 0 1 0} | int({4 3 3 3}/4) = {1 0 0 0} | int({6 4 4 4}/4) = {1 1 1 1} |
For this pattern, N0 = 3 but the number of 0’s in the Remainder Row of Table 11 are 6 ≠ N0. Similarly, there is just one 1 in the Remainder Row but N1 = 4; there are five 2’s in the Remainder row, but N2 = 4 and there are three 3’s in the Remainder Row but N3 = 4. So the sequence “{4 2 2} {3 1 4 1} {2 1 1 1} {3 1 1 1}” is not a valid siteswap sequence. Therefore, we ignore the Quotient Row for Table 11.
Let us apply the permutation test to a rearrangement of the sequence of Table 8.
“{3 2} 1 3”:
Period P = 3. At beat #0, there are two siteswap numbers, so N0 = 2, while N1 = 1 = N2.
| Given Sequence | {3 2} | 1 | 3 |
|---|---|---|---|
| Add | {0 0} | 1 | 2 |
| Result | {3 2}+{0 0} = {3 2} | 1+1 = 2 | 3+2 = 5 |
| Remainder of (Result/P) | {3 2} mod 3 = {0 2} | 2 mod 3 = 2 | 5 mod 3 = 2 |
| Quotient of (Result/P) | int({3 2}/3) = {1 0} | int(2/3) = 0 | int(5/3) = 1 |
In the Remainder Row of Table 12, there is only one 0, so number of 0’s is not equal to N0 = 2. Similarly, the number 1 doesn’t appear in the Remainder Row at all while N1 = 1. And there are three 2’s in the Remainder Row while N2 = 1. Thus, this rearrangement of the sequence of Table 8 results in an invalid siteswap sequence and we again ignore the Quotient Row.
Generating Valid Siteswaps
As in the case of Vanilla Siteswaps, we can again follow the reverse process of the permutation test to generate Flavoured Siteswaps of Period P:
- Start with any permutation of numbers from 0 to P-1. Have as many of each number as you wish there to be objects to be thrown at that beat. For example, if you want 3 objects to be thrown at beat #1, then have three 1’s. Also group as many numbers in one beat as there should be objects thrown at that beat. For example, if three objects are to be thrown at beat #1, then group three numbers at beat #1. However, it is not necessary for any of the numbers at beat #1 to be 1.
- Subtract “i” from all the numbers at the “i”th position in this permutation, where “i” ranges from 0 to P-1.
- Add various multiples of P to each number. Addition of the multiple “0” is also allowed as long as no negative numbers remain after this step.
The number of objects required to juggle the pattern corresponding to the siteswap generated as above, is equal to the number of multiples of P that got added in step 3.
Examples
Let us take the period P = 4 and have one object thrown at beats 0, 2 and 3 and three objects thrown at beat #1. So we need three 1’s to be present overall and three numbers to be present at beat #1. Let us start with the permutation as shown in Table 13.
| Initial arrangement of 0 to P-1 | 1 | {0 2 3} | 1 | 1 |
|---|---|---|---|---|
| Subtract | 0 | {1 1 1} | 2 | 3 |
| Result | 1 | {-1 1 2} | -1 | -2 |
| Adder (Multiple of P) | 0*4 = 0 | {1*4 0*4 0*4} = {4 0 0} | 1*4 = 4 | 1*4 = 4 |
| Siteswap (Result+Adder) | 1 | {3 1 2} | 3 | 2 |
Since we have added a total of 3 multiples of P (=4) in the Adder row, the generated sequence “1 {3 1 2} 3 2” corresponds to a 3-object juggling pattern.
Let’s take another example with period P = 3, two objects thrown at beat #0 and one object each to be thrown at beats #1 and 2. So we need two numbers at beat #0 and one number each at beat #1 and beat #2. Of these four numbers, two should be 0’s and we should have one 1 and one 2. Let us start with the permutation shown in Table 14 and generate a valid siteswap.
| Initial arrangement of 0 to P-1 | {0 2} | 1 | 0 |
|---|---|---|---|
| Subtract | { 0 0} | 1 | 2 |
| Result | {0 2} | 0 | -2 |
| Adder (Multiple of P) | {1*3 0*3} = {3 0} | 1*3 | 1*3 = 3 |
| Siteswap (Result+Adder) | {3 2} | 3 | 1 |
Thus, we have been able to generate the sequence “{3 2} 3 1” which we evaluated in Table 8. In the Adder Row of Table 14, we have added three multiples of the period (=3), so the sequence represents a 3 object juggling pattern which again agrees with what we had found in Table 8.
Siteswap Manipulations
As for vanilla siteswaps, once we have a valid flavoured siteswap, there are several ways in which it may be manipulated to generate other valid flavoured siteswaps. The technique of adding period P to any siteswap number works here too and generates a new siteswap corresponding to a juggling pattern for one more object than the original one. In fact, we can treat any number T in the siteswap sequence as two numbers {T 0}. Then we can add the period P to the “0” to create a multiple object throw at a beat where earlier only one object was being thrown. In fact, we can create an “N+1” object throw at a beat where earlier only N objects were being thrown. For example, the sequence “3 3 3” of period P = 3 can be written as “{3 0} 3 3”. Now if we add the period 3 to the “0”, we end up with the flavoured siteswap “{3 3} 3 3” corresponding to a 4 object juggling pattern after starting with the vanilla siteswap “3 3 3” corresponding to a 3 object juggling pattern.
The method of adding “1” to all the terms of the siteswap however, does not work in general in the case of flavoured siteswaps. However, several other manipulations are possible[5].
Footnotes
- We shall see, in a later blog on synchronous siteswaps, a case where the order of the siteswap numbers within the brackets does matter. But again, this is because of the imposition of a particular “hand siteswap” and not an inherent constraint of the siteswap notation itself.
- See the second part of this series.
- See the section on “Odd and Even Siteswap numbers” in the second part of this series.
- As with vanilla siteswaps, this “average” being a whole number is not a guarantee of the siteswap sequence being valid.
- See for example, Burkard Polster’s “Mathematics of Juggling”.