Given a siteswap of length $L$ with $n$ objects, what is the maximum possible theoretical global period length?

Prerequisites:

A siteswap of length $L$ is defined here by $s = {[s_{i}]}_{i = 0 : L - 1}$ . A description of siteswap notation can be found here, chapter 2 (Introduction to Siteswap Notation).

The global period length is the number of siteswap beats required to return to the exact same start configuration (same state and the same labels on the same beats).

Beat permutation and orbits.

A vanilla siteswap (asynchronous, one throw per beat) is valid iff the map

π : {0, \dots, L - 1} \to {0, \dots, L - 1}, π (i) \equiv i + s_{i} (\mod L) .

is a permutation. Its disjoint cycles are the orbits. For convenience, we will list orbits using absolute beat indices (e.g. [0,5] for $L = 3$ ), not just residues modulo $L$ . As per siteswap convention, whenever an index $i$ lies outside [0,L−1], we interpret $s_{i}$ as $s_{i mod L}$ .

Let $C_{0}, C_{1}, \dots, C_{n - 1}$ denote the orbits followed by balls $0, \dots, n - 1$ respectively. Define

S_{C_{j}} = \sum_{i \in C_{j}} s_{i}, t_{j} := \frac{S_{C_{j}}}{L} .

Since $π (i) \equiv i + s_{i} (\mod L)$ , we have $s_{i} \equiv π (i) - i (\mod L)$ . Noting also that $\sum_{i \in C_{j}} (π (i) - i) \equiv 0 (\mod L)$ (an orbit concludes when the orbit returns to the same starting position relative to the siteswap length), summing over $i \in C_{j}$ we obtain:

S_{C_{j}} \equiv \sum_{i \in C_{j}} (π (i) - i) \equiv 0 (\mod L),

thus $t_{j} \in ℤ_{\geq 1}$ . Here $t_{j}$ is the number of siteswap lengths (periods) accumulated along the throws on the $j$ th orbit; equivalently, it is the number of periods needed for ball $j$ to return to its starting beat.

Thus, in terms of $t = [t_{j}]_{j = 0 : n - 1}$ , the global period is

S_{L} = L \cdot lcm (t_{0}, \dots, t_{n - 1}) .

Example ( $L = 3$ , $s = [5, 3, 1]$ ).

We have

π (0) \equiv 2, π (1) \equiv 1, π (2) \equiv 0 (\mod 3) .

The orbits are: $C_{0}$ = [0,5], $C_{1}$ = [1], $C_{2}$ = [2,3].

Therefore:

S_{C_{0}} = s_{0} + s_{5} = 5 + s_{2} = 5 + 1 = 6, S_{C_{1}} = s_{1} = 3, S_{C_{2}} = s_{2} + s_{3} = 1 + s_{0} = 1 + 5 = 6 .

t_{0} = \frac{6}{3} = 2, t_{1} = \frac{3}{3} = 1, t_{2} = \frac{6}{3} = 2 .

In the example, $S_{L} = 3 \cdot lcm (2, 1, 2) = 6$ beats.

Cyclic equivalence (notation).

Two orbits $C_{p}$ and $C_{q}$ are cyclically equivalent if $C_{p} \sim_{L} C_{q}$ .

i.e. their index lists agree modulo $L$ up to cyclic rotation. For vectors $u$ , $v$ in $ℤ^{k}$ we write $u \sim_{L} v$ if $u$ and $v$ are congruent mod $L$ after a cyclic rotation of entries. In the example, [0,5] $\sim_{3}$ [0,2] $\sim_{3}$ [2,3] all represent the same orbit as {0,2}.

Let $Φ$ select one representative from each cyclic-equivalence class:

Φ = {j such that C_{j} ≁_{L} C_{k} for all k > j} .

Define the non-equivalent orbits and sizes by

P = (P_{0}, \dots, P_{M - 1}) := (C_{j})_{j \in Φ}, T = (T_{0}, \dots, T_{M - 1}) := (t_{j})_{j \in Φ} .

Then, as before,

S_{P_{m}} = \sum_{i \in P_{m}} s_{i} = T_{m} \cdot L .

Summing over the $M$ representatives and using $\sum_{i = 0}^{L - 1} s_{i} = n L$ ,

\sum_{m = 0}^{M - 1} S_{P_{m}} = \sum_{m = 0}^{M - 1} \sum_{i \in P_{m}} s_{i} = L \cdot \sum_{m = 0}^{M - 1} T_{m} = n L,

so $\sum_{m = 0}^{M - 1} T_{m} = n$ . (Each $T_{m}$ is the number of objects carried on that orbit class; balls whose $C_{j}$ fall in the same equivalence class share the same $t_{j} = T_{m}$ .)

Maximization problem.

Thus, the problem reduces to:

S_{L_{\max}} = L \cdot \max_{T} {lcm (T)},

with the constraint $\sum_{m}^{} T_{m} = n$ . Equivalently, since duplicating equal parts does not change the least common multiple, one may write the objective as $lcm (t)$ ; the natural sum constraint, however, lives on $T$ via $\sum_{m}^{} T_{m} = n$ .

Relation to Landau's function and feasibility.

The maximum of $lcm (T_{0}, \dots, T_{M - 1})$ over all partitions of $n$ is Landau's function $g (n)$ , so

S_{L_{\max}} \leq L \cdot g (n) .

with equality whenever a Landau-optimal partition is realisable with $L$ beats. A convenient feasibility model is the standard beat budget: an orbit of size $1$ can occupy a single beat (throw $L$ ), while an orbit of size $> 1$ can be realized on a 2-beat swap; therefore a partition $(T_{0}, \dots, T_{M - 1})$ is feasible if

# {m : T_{m} = 1} + 2 \cdot # {m : T_{m} > 1} \leq L .

Under this budget, $S_{L_{\max}} = L \cdot g (n)$ whenever a Landau-optimal partition fits the inequality above.

This table presents the pairs $(n, L)$ for which the maximum $S_{L,max}$ can be obtained according to the budget inequality. Shaded cells indicate the feasible cases.

n/L	1	2	3	4	5	6	7	8	9	10
1
2
3
4
5
6
7
8
9
10

Here are some examples for Landau-optimal siteswaps for given $n$ and $L$ pairs:

$n$	$L$	Landau-optimal partition	$g (n)$	Siteswap (period $L$ )	Return beats = $L \cdot g (n)$
1	1	$1$	1	1	1
2	2	$2$	2	31	4
3	2	$3$	3	33	6
4	3	$4$	4	480	12
5	5	$3 + 2$	6	77830	30
6	3	$6$	6	7b0	18
7	5	$4 + 3$	12	c7880	60
8	4	$5 + 3$	15	a6a6	60
9	5	$5 + 4$	20	c7dd0	100
10	6	$5 + 3 + 2$	30	f99f93	180

Given a siteswap of length L with n objects, what is the maximum possible theoretical global period length?

Given a siteswap of length $L$ with $n$ objects, what is the maximum possible theoretical global period length?