Counting Juggling Sequences

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Transcript Counting Juggling Sequences

The
Mathematics
of
Juggling
Matthew Wright
slides also by John Chase
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Basic Juggling Patterns
Axioms:
1. The juggler must juggle at a constant rhythm.
2. Only one throw may occur on each beat of the pattern.
3. Throws on odd beats must be made from the right hand; throws on
even beats from the left hand.
4. The pattern juggled must be periodic. It must repeat. It must repeat.
5. All balls must be thrown to the same height.
Example: basic 3-ball pattern
arcs
represent
throws
dots
represent
beats
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3
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7
8
9 ∙∙∙
Basic 3-ball Pattern
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9 ∙∙∙
Notice: balls land in the opposite hand from which they were thrown
Basic 4-ball Pattern
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9 ∙∙∙
Notice: balls land in the same hand from which they were thrown
The Basic 𝒃-ball Patterns
If 𝑏 is odd:
• Each throw lands in the opposite
hand from which it was thrown.
• These are called cascade throws.
If 𝑏 is even:
• Each throw lands in the same hand
from which it was thrown.
• These are called fountain throws.
Let’s change things up a bit…
Axioms:
1. The juggler must juggle at a constant rhythm.
2. Only one throw may occur on each beat of the pattern.
3. Throws on even beats must be made from the right hand; throws on
odd beats from the left hand.
4. The pattern juggled must be periodic. It must repeat. It must repeat.
5. All balls must be thrown to the same height.
What if we allow throws of different heights?
Axioms 1-4 describe the simple juggling patterns.
Example
Start with the basic 4-ball pattern:
Concentrate on the landing sites of two throws.
Now swap them!
• The first 4-throw will land a count later, making it a 5-throw.
• The second 4-throw will land a count earlier, making it a 3-throw.
This is called a site swap.
Juggling Sequences
Site swaps allow us to obtain many simple juggling patterns, starting from
the basic juggling patterns.
We describe each simple juggling pattern by a juggling sequence: a
sequence of integers corresponding to the sequence of throws in the
juggling pattern.
The length of a juggling sequence is its period. A juggling sequence is
minimal if it has minimal period among all juggling sequences representing
the same pattern.
Example: the juggling sequence 441
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4
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4
4
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1
2
3
4
1
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5
6
1
7
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9 ∙∙∙
Juggling Sequences
2-balls: 31, 312, 411, 330
3-balls: 441, 531, 51, 4413, 45141
4-balls: 5551, 53, 534, 633, 71
5-balls: 66661, 744, 75751
4
5
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4
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4
5
1
4
1
4
5 ∙∙∙
Is every sequence a juggling
sequence?
No.
Consider the sequence 54.
collision!
5
4
Clearly, a 5-throw followed by a 4-throw results in a collision.
In general, an 𝑛-throw followed by an (𝑛 − 1)-throw results in a
collision.
How do we know if a
given sequence is
jugglable?
For instance, is 6831445 a jugglable sequence?
A juggling function is a function:
𝑗 ∶ ℤ → ℤ≥0
This function tells us what throw to make on each beat.
That is, on beat 𝑖, we juggle a 𝑗(𝑖)-throw, for each integer 𝑖.
The sequence described by this function is jugglable if and only if
the function
𝑗+ : ℤ → ℤ
𝑗+ (𝑖) = 𝑖 + 𝑗(𝑖)
is a permutation of the integers.
Two important properties of juggling functions:
1. Height of the highest throw:
height 𝑗 = max 𝑗(𝑖)
𝑖∈ℤ
2. Number of balls required to juggle the corresponding sequence:
balls( 𝑗) = number of balls required to juggle 𝑗
How many balls are required to
juggle a given sequence?
The Average Theorem: Let 𝑗 be a juggling function with finite height.
Then
𝑖∈𝐼 𝑗(𝑖)
lim
𝐼 →∞
|𝐼|
exists, is finite, and is equal to balls( 𝑗), where the limit is over all
integer intervals 𝐼 = {𝑎, 𝑎 + 1, 𝑎 + 2, … , 𝑏}, and 𝐼 = 𝑏 − 𝑎 + 1 is
the number of integers in 𝐼.
Proof:
balls(𝑗) 𝐼 + 1 − height(𝑗)
≤
|𝐼|
𝑖∈𝐼 𝑗(𝑖)
|𝐼|
balls(𝑗) 𝐼 − 1 + height(𝑗)
≤
|𝐼|
height( 𝑗)
height( 𝑗)
interval 𝐼, with 𝐼 > height( 𝑗)
minimum contribution of any
particular ball to 𝑖∈𝐼 𝑗(𝑖)
maximum contribution of any
particular ball to 𝑖∈𝐼 𝑗(𝑖)
Proof:
balls(𝑗) 𝐼 + 1 − height(𝑗)
≤
|𝐼|
𝑖∈𝐼 𝑗(𝑖)
|𝐼|
balls(𝑗) 𝐼 − 1 + height(𝑗)
≤
|𝐼|
The left and right expressions tend to balls( 𝑗) as |𝐼| tends to infinity.
How many balls are required to
juggle a given sequence?
The Average Theorem:
balls 𝑗 = lim
𝑖∈𝐼 𝑗(𝑖)
|𝐼|
Corollary: The number of balls necessary to juggle a juggling
sequence equals its average.
𝐼 →∞
Application: A finite juggling sequence must have an integer average.
Examples:
534
441
7531
75751
352
4-ball
pattern
3-ball
pattern
4-ball
pattern
5-ball
pattern
not
jugglable!
How can we change one juggling
sequence into another?
We could perform a site swap.
Consider the sequence 𝑆 of 𝑝 nonnegative integers:
𝑆: 𝑎0 , 𝑎1 , … , 𝑎𝑖−1 , 𝑎𝑖 , 𝑎𝑖+1 , … , 𝑎𝑖+𝑑−1 , 𝑎𝑖+𝑑 , 𝑎𝑖+𝑑+1 , … , 𝑎𝑝−1
If 0 < 𝑑 ≤ 𝑎𝑖 , we can swap the landing positions of the balls
thrown on beats 𝑖 and 𝑖 + 𝑑 to obtain the sequence 𝑆′:
𝑆 ′ : 𝑎0 , 𝑎1 , … , 𝑎𝑖−1 , 𝑎𝑖+𝑑 + 𝑑, 𝑎𝑖+1 , … , 𝑎𝑖+𝑑−1 , 𝑎𝑖 − 𝑑, 𝑎𝑖+𝑑+1 , … , 𝑎𝑝−1
Notice:
• The sequence 𝑆 is a juggling sequence if and only if 𝑆′ is.
• The average of 𝑆 is the same as the average of 𝑆′.
• If 𝑆 is a juggling sequence, then the number of balls used to juggle
𝑆 equals the number of balls used to juggle 𝑆′.
How can we change one juggling
sequence into another?
We could perform a cyclic shift.
Again, let 𝑆 be a sequence of 𝑝 nonnegative integers:
𝑆: 𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎𝑝−1
Now move the last element, 𝑎𝑝−1 , to the beginning of the sequence
to obtain the sequence 𝑆→ :
𝑆→ : 𝑎𝑝−1 , 𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎𝑝−2
Notice:
• The sequence 𝑆 is a juggling sequence if and only if 𝑆→ is.
• The average of 𝑆 is the same as the average of 𝑆→ .
• If 𝑆 is a juggling sequence, then the number of balls used to juggle
𝑆 equals the number of balls used to juggle 𝑆→ .
The Flattening Algorithm
Let 𝑆 be a sequence of 𝑝 ≥ 1 nonnegative integers:
𝑆: 𝑎0 , 𝑎1 , 𝑎2 , … , 𝑎𝑝−1
The flattening algorithm transforms 𝑆 into a new sequence as
follows:
1. If 𝑆 is a constant sequence, stop and output this sequence.
Otherwise,
2. use cyclic shifts to arrange the elements of 𝑆 such that a
maximum integer in 𝑆, say 𝑚, is at position 0 and a nonmaximum integer in 𝑆, say 𝑛, is at position 1. If 𝑚 = 𝑛 + 1, stop
and output this sequence. Otherwise,
3. perform a site swap of positions 0 and 1. Redefine 𝑆 to be the
resulting sequence, and return to step 1.
The Flattening Algorithm
Example: start with the sequence 642 also jugglable!
swap
642
shift
552
swap
525
shift
345
swap
534
444
jugglable!
Example: start with the sequence 514 also not jugglable
swap
514
shift
244
swap
424
shift
334
443
not jugglable
Observe:
• The Flattening Algorithm can be used to decide whether or not a
sequence is jugglable.
• If the input is a 𝑏-ball juggling sequence with period 𝑝, this
algorithm outputs the basic 𝑏-ball sequence of period 𝑝.
• If the input is not a juggling sequence, the program stops at step 2
and outputs a sequence of the form 𝑚, 𝑚 – 1, ….
How do we know if a given sequence is
jugglable?
Theorem: Let 𝑆 = {𝑎𝑘 }, for 𝑘 = 0, 1, … 𝑝 − 1, be a sequence of
nonnegative integers and let [𝑝] = {0, 1, 2, … , 𝑝 − 1}. Then, 𝑆 is a
juggling sequence if and only if the function 𝜙𝑆 : 𝑝 → 𝑝 defined
𝜙𝑆 (𝑖) = 𝑖 + 𝑎𝑖 (mod 𝑝)
is a permutation of the set 𝑝 .
Example: Show 534 is a valid juggling sequence.
Let 𝑆 = {5, 3, 4}. The period is 3, so 𝑝 = 3. Note 𝑝 = 0,1,2 .
Then 𝜙𝑆 0 , 𝜙𝑆 1 , 𝜙𝑆 2
= 0 + 5, 1 + 3, 2 + 4
mod 3
= 5, 4,6 (mod 3) = 2, 1,0
This is a permutation of 𝑝 , so 534 is a valid juggling sequence.
Theorem: Let 𝑆 = {𝑎𝑘 }, for 𝑘 = 0, 1, … 𝑝 − 1, be a sequence of
nonnegative integers and let [𝑝] = {0, 1, 2, … , 𝑝 − 1}. Then, 𝑆 is a
juggling sequence if and only if the function 𝜙𝑆 : 𝑝 → 𝑝 defined
(𝑖) = 𝑖 + 𝑎𝑖 (mod
Proof: The function 𝜙𝜙
if and𝑝)
only if the vector
𝑆 𝑆is a permutation
is a permutation
the
set 𝑝 .
𝑣 =of 𝜙
𝑆 0 , 𝜙𝑆 1 , 𝜙𝑆 2 , … , 𝜙𝑆 (𝑝 − 1)
contains all of the integers from 0 to 𝑝 – 1.
Suppose we apply site swaps and cyclic permutations to the
sequence 𝑆 to obtain sequence 𝑆′ with corresponding vector 𝑣′.
Then 𝑣′ contains all of the elements of 𝑝 if and only if 𝑣 does.
Therefore, given a sequence 𝑆, apply the flattening algorithm to
obtain 𝑆′. Then 𝑆 is a juggling sequence if and only if 𝑆′ is a constant
sequence, if and only if 𝑣′ contains all of the elements of 𝑝 .
How many ways are
there to juggle?
Infinitely many.
(Consider the basic 𝑏-ball sequences for each integer 𝑏 ∈ ℕ.)
How many 𝒃-ball juggling sequences
are there with period 𝒑?
How many 𝒏-ball juggling
sequences are there of period 𝒏?
𝑛 = 1: There is one unique sequence, namely, 1.
1
𝑛 = 2: Starting with the sequence 22, we can perform site swaps to
obtain two other sequences, 31 and 40 (unique up to cyclic shifts).
2 2
3 1
4 0
𝑛 = 3: Starting with 333 and performing site swaps, we (eventually)
obtain 13 sequences (unique up to cyclic shifts).
How many 𝒏-ball juggling
sequences are there of period 𝒏?
3
3
1
5
3 3
4 2
4 4
0 4
0
7 2
1
0
5 3
3 6
5
2 2
0
8
1
1
0 9
0 1
7 1
2 6
3
0 6
𝑛 = 3: Starting with 333 and performing site swaps, we (eventually)
obtain 13 sequences (unique up to cyclic shifts).
Is there a better way to count
juggling sequences?
Suppose we have a large number of each of the following juggling
cards:
These cards can be used to construct all juggling sequences that are
juggled with at most three balls.
Example: juggling sequence 441
juggling
diagram
∙∙∙ 4
4
1
4
4
1
4
4
1 ∙∙∙
4
1
4
4
1
4
4
1
constructed
with juggling
cards
4
Counting Juggling Sequences
With many copies of these four
cards, we can construct any (notnecessarily minimal) juggling
sequences that is juggled with at
most three balls.
Similarly, with many copies of 𝑏 + 1 distinct cards, we can construct
any (not-necessarily minimal) juggling sequence that is juggled with
at most 𝑏 balls.
Lemma: The number of all juggling sequences of period 𝑝, juggled
with at most 𝑏 balls, is:
𝑆 ≤ 𝑏, 𝑝 = 𝑏 + 1 𝑝
Counting Juggling Sequences
Lemma: The number of all juggling sequences of period 𝑝, juggled
with at most 𝑏 balls, is:
𝑆 ≤ 𝑏, 𝑝 = 𝑏 + 1 𝑝
It follows that:
Lemma: The number of all 𝑏-ball juggling sequences of period 𝑝 is:
𝑆 𝑏, 𝑝 = 𝑆 ≤ 𝑏, 𝑝 – 𝑆 ≤ 𝑏 − 1, 𝑝 = 𝑏 + 1
𝑝
− 𝑏𝑝
However, we have counted each cyclic permutation of every
sequence, as well as non-minimal sequences.
How can we count the minimal 𝒃-ball juggling sequences
of period 𝒑, not counting cyclic permutations of the same
sequence as distinct?
Counting Juggling Sequences
Theorem: The number of all minimal 𝑏-ball juggling sequences of
period 𝑝, with 𝑏 ≥ 1, is
1
𝑀 𝑏, 𝑝 =
𝑝
𝑑|𝑝
𝑝
𝜇
𝑑
𝑏+1
𝑑
− 𝑏𝑑
if cyclic permutations of juggling sequences are not counted as distinct.
Here, 𝜇 denotes the Möbius function:
1 if 𝑛 has an even number of distinct prime factors,
𝜇 𝑛 = −1 if 𝑛 has an odd number of distinct prime factors,
0
if 𝑛 has repeated prime factors.
Proof: If 𝑑 divides 𝑝, then each minimal juggling sequence of period 𝑑
gives rise to exactly 𝑑 sequences of period 𝑝. Thus,
𝑆 𝑏, 𝑝 = 𝑏 + 1
𝑝
− 𝑏𝑝 =
𝑑𝑀 𝑏, 𝑑
𝑑|𝑝
The expression for 𝑀 𝑏, 𝑝 follows by Möbius inversion.
Questions?
Reference:
Burkard Polster. The Mathematics
of Juggling. Springer, 2003.
Juggling Simulators:
• www.quantumjuggling.com
• jugglinglab.sourceforge.net