Enumerating (multiplex) juggling sequences

Starting about 20 years ago, there has been increasing activity by discrete mathematicians and (mathematically inclined) jugglers in developing and exploring ways of representing various possible juggling patterns numerically (e.g., see [1, 2, 3, 4, 5, 6, 10, 12, 14]). Perhaps the most prominent of these is the idea of a juggling sequence (or “siteswap”, as it is often referred to in the juggling literature). The idea behind this approach is the following. For a given sequence $T=(t_{1},t_{2},\ldots,t_{n})$ of nonnegative integers, we associate a (possible) periodic juggling pattern in which at time $i$, a ball is thrown so that it comes down at time $i+t_{i}$. This is to be true for each $i,1\leq i\leq n$. Because we assume this is to be repeated indefinitely with period $n$, then in general, for each $i$ and each $k\geq 0$, a ball thrown at time $i+kn$ will come down at time $i+t_{i}+kn$. The usual assumption made for a sequence $T$ to be a valid juggling sequence is that at no time do two balls come down at the same time. This assumption results in many consequences, e.g., all of the quantities $i+t_{i}\pmod{n}$ must be distinct, the number of balls in the pattern is the average $({1}/{n})\sum_{k=1}^{n}t_{k}$, and the number of juggling sequences with period $n$ having fewer than $b$ balls is $b^{n}$ (see [1]).

An important object for understanding the relationships and transitions between various juggling sequences is the concept of a state diagram, developed independently (and almost simultaneously) by Jack Boyce and Allen Knutson [9]. This is a directed graph where each vertex is called a state or landing schedule, a $0$-$1$ vector indicating when the balls that are currently in the air will land, and edges represent possible transitions between states. The vertex and edge sets for the state diagram can be defined as follows:

More specifically, each juggling sequence $T$ is associated with a state $\sigma=\sigma_{T}=\langle\sigma_{1},\sigma_{2},\ldots,\sigma_{h},\ldots\rangle$ which can be found by imagining that the sequence has been executed infinitely often in the past, with a final throw $t_{n}$ being made at time $0$. Then $\sigma_{i}$ is $1$ if and only if there is some ball still in the air at time $0$ that will land at time $i$. In this case we say that $T$ is a $\sigma$-juggling sequence.

If we are now going to throw one more ball at time $1$, transitioning to a (possibly) new juggling state $\sigma^{\prime}$, then we are restricted to throwing it so that it lands at some time $j$ which has $\sigma_{j}=0$. The new state $\sigma^{\prime}=\langle\sigma_{1}^{\prime},\sigma_{2}^{\prime},\ldots\rangle$ then has $\sigma_{k}^{\prime}=\sigma_{k+1}$ for $k\geq 1,k\neq j-1$ and $\sigma_{j-1}^{\prime}=1$. The preceding remarks assume that $\sigma_{1}=1$. If $\sigma_{1}=0$, so that there is no ball available to be thrown at time $1$, then a “no-throw” occurs, and the new state vector $\sigma^{\prime}$ satisfies $\sigma_{k}^{\prime}=\sigma_{k+1}$ for all $k\geq 1$. These give the two basic transitions that can occur in the state diagram. With this interpretation, it is easy to see that a juggling sequence of period $n$ exactly corresponds to a walk of length $n$ in the state diagram.

In [5], the problem of enumerating $\sigma$-juggling sequences of period $n$ was studied, which by the above comments is equivalent to counting the number of directed closed walks of length $n$ starting at $\sigma$ in the state diagram. In the same paper, the related problem of counting the number of “primitive” closed walks of length $n$ was also solved. These are walks in which the starting state $\sigma$ is visited only at the beginning and the end of the walk.

A particular unsolved problem mentioned in [5] was that of extending the analysis to the much more complex situation of multiplex juggling sequences. In a multiplex juggling sequence, for a given parameter $m$, at each time instance up to $m$ balls can be thrown and caught at the same time, where the balls thrown at each time can have different landing times. Thus, ordinary juggling sequences correspond to the case $m=1$.

As before, we can describe a (multiplex) juggling sequence as a walk in a state diagram. Here a state $\alpha=\langle a_{1},a_{2},a_{3},\ldots\rangle$ can again be described as a landing schedule where $a_{i}$ are the number of balls currently scheduled to land at time $i$. We also have a state diagram which has as its vertices all possible states and for edges all ways to go from one state to another state (see [10]). The state diagram is thus a directed graph with two important parameters: $b$, the number of balls that are being juggled, and $m$, the maximum number of balls that can be caught/thrown at any one time. The vertex set and edge set are defined as follows:

Since each state will only have finitely many nonzero terms, we will truncate the terminal zero portions of the state vectors when convenient. The height of a state $\alpha$ will be the largest index $i$ for which $a_{i}>0$, and will be denoted by $h(\alpha)$. A small portion of the state diagram when $b=3$ and $m=2$ is shown in Figure 1.

One way to use the buckets to find a sequence of length $n$ that starts in state $\alpha=\langle a_{1},a_{2},\ldots\rangle$ and ends in state $\beta=\langle b_{1},b_{2},\ldots\rangle$ (if one exists) is to start with the balls arranged in the buckets as dictated by $\alpha$. We then modify the capacities of the buckets so that they are (starting at the first bucket)

Finally, take $n$ steps (such as shown in Figure 2) being careful not to exceed the capacity of any bucket and at the end, we will be forced into state $\beta$. On the other hand, every possible way to start in $\alpha$ and end in $\beta$ in $n$ steps can be done in this modified buckets approach.

Finding all of the walks of length $n$ in the state diagram between $\alpha$ and $\beta$ is thus equivalent to finding all of the walks that can be run using this modified bucket procedure. This is what we will actually enumerate. We start with the following matrix, where $h=\max\{h(\alpha)-n,h(\beta)\}$ (this is a $0$-$1$ matrix, where any unspecified entries are $0$’s).

Each block of $m$ columns will correspond to one transition in the state diagram/buckets procedure. In a block, each column corresponds to a single element in the multi-set describing our transition. The first $1$ in a column corresponds to a no-throw or a throw of height $0$, the second $1$ corresponds to a throw of height $1$, the third $1$ to a throw of height $2$ and so on. So we choose exactly one $1$ in each column and by reading the blocks of columns we will get the transitions $T_{i}$ between states.

Each row will correspond to a bucket, and to incorporate the modified buckets approach we specify a row sum for each row. The row sum will be the “unused capacity” of the buckets. Beginning at the first row and going down the row sums will be

As an example, suppose that we wanted a walk of length $4$ that starts and ends in the state $\langle 1,2\rangle$ in the state diagram shown in Figure 1. Then this corresponds to choosing one $1$ out of each column in the matrix below on the left so that the row sums are as dictated on the side of the matrix.

One possible solution is shown in the matrix on the right which corresponds to the walk

Here we have indicated the sets $T_{i}$ under each transition.

To see why this works we need to show that each multi-set given by a block of columns is a valid transition in our modified buckets procedure. Starting with the first block, the first row sum should be $m-\alpha_{1}$, this indicates that we currently have an excess capacity of $m-\alpha_{1}$ in the bottom bucket and so we need to make $m-\alpha_{1}$ no-throws, i.e., throws of height $0$, and by row sum restrictions we are forced to select exactly $m-\alpha_{1}$ of the $1$’s on the first row. This accounts for $m-\alpha_{1}$ of the columns in the first block. The remaining $\alpha_{1}$ columns must have $1$’s (i.e., balls) distributed among the rows (i.e., buckets) which still have extra capacity. Thus the first block must give a valid transition in the procedure. After selecting the $1$’s for the first block we then update the row sums according to our choices, remove the first block and then repeat the same process $n-1$ times.

Conversely, it is easy to check that given the transitions $T_{1},T_{2},\ldots,T_{n}$ joining $\alpha$ to $\beta$, we can find a selection of $1$’s in the matrix $\cal A$ which corresponds to this walk and satisfies the row/column sum restrictions.

Finally, since multi-sets are unordered, our choice of $1$’s in columns is unique up to permutation of the columns in a block, so we may assume that the heights of the $1$’s in a block are weakly decreasing (as shown in the example above). We now have the following general result.

We now count the number of ways to fill the matrix $\cal A$ according to the restrictions in Lemma 1. We will demonstrate the procedure by working through an example, namely counting the number of periodic multiplex juggling sequences of length $n$ that start and end in state $\langle 3\rangle$ when $m=3$.

The first thing to observe is that when the height of $\alpha$ is small compared to $n$ then the row sums have the following form:

We can form a recurrence based on this where we fill in the last block and then reduce it to the case when there is one fewer $m$ in the middle.

Without loss of generality we can assume that our noise at the end is a partition of the $b$ balls so that no part is larger than $m$, i.e., we can ignore any rows with row sum $0$ and by a simple correspondence we can assume that the row sums are weakly decreasing.

For each partition $\gamma$ of $b$ with each part at most $m$, let $x_{\gamma}(k)$ be the number of ways to fill the matrix ${\cal A}$ where the row sums are given by

In our example, there are $3$ such partitions, $3=2+1=1+1+1$, and so we will have the three functions $x_{3}(k)$, $x_{2,1}(k)$ and $x_{1,1,1}(k)$.

We now form a system of linear recurrences, $x_{\gamma}(k)=\sum_{\delta}a_{\gamma,\delta}x_{\delta}(k-1)$, by examining the different ways to fill in the last block of columns. Note that the row sums corresponding to the last block are $m,\gamma$. The $m$ comes from the no-throws and after we have filled in the last block it will become incorporated into the new terminal distribution. Thus at each stage we will decrease the number of middle $m$’s by $1$.

For example, if we are considering $x_{2,1}$ then we have the following $6$ different ways to fill in the last block satisfying the conditions of Lemma 1:

By looking at the new terminal distributions it follows that

Similar analysis shows that

We can rewrite this in matrix form as

In general, we will have that $\big{(}x_{\gamma}(k)\big{)}=A\big{(}x_{\gamma}(k-1)\big{)}$ where $A_{\gamma,\delta}=a_{\gamma,\delta}$ as given above. Note that the matrix $A$ will be independent of our choice of $\alpha$ and $\beta$, and depends only on $b$ and $m$. Varying $\alpha$ will change the initial conditions and varying $\beta$ will change which of the $x_{\gamma}$ we are interested in for our recurrence.

For our problem, we are interested in $x_{3}$ as our terminal state is $\langle 3\rangle$ (i.e., corresponding to the partition $3$). We want to transform our first-order system of linear recurrences into a single recurrence for $x_{3}$. Manipulating the recurrences in (1) it can be shown that

This gives a recurrence for the number of periodic juggling sequence of length $k$ for $k$ sufficiently large (i.e., we have shifted our sequences $x_{\gamma}(k)$ past the initial noise; when counting periodic juggling sequences we need to remember to account for this shift).

It remains to calculate enough initial terms to begin the recursion. Using Lemma 1 this can be handled by brute force or some careful case analysis. In general, we will need to calculate the first $h(\alpha)+r-1$ terms where $r$ is the number of partitions of $b$ with each part at most $m$. The first $h(\alpha)-1$ terms are to handle the initial noise caused by $\alpha$ and the next $r$ terms are to help start the recursion. A calculation shows that the sequence counting the number of periodic juggling sequences that start and end at $\langle 3\rangle$ starts $1,4,20$. Now applying the recursion we get the sequence

With the recursion and the initial values it is then a simple matter to derive a generating function for the number of juggling sequences of period $n$. The generating function for this series and several others are given in Table 1.