A Combinatorics Problem of Music
Alyssa Siedeman,Christian Courtial & Lindsey Gentry
Boise State University
Background
Definitions to Know
The question in music: How many possible rhythms, excluding triplets and dotted notes, using
lengths ( 21 )0, ( 12 )1, ( 12 )2, ( 12 )3, ( 21 )4, ( 12 )5, can be put in one 4/4 measure?
Definition 0.1 Music Music will be defined as any sonic phenomena, including silence
Definition 0.2 Rhythm The sequential arrangement of sounds and silences
in time is rhythm.
Definition 0.3 Time Signature The amount and type of notes allowed in a
measure is dictated by the time signature.
These lengths correspond to whole,
half, quarter notes,
etc.
In plain English:
How many different rhythms can be created using combinations of notes
including Whole, Half, Quarter, Eighth, Sixteenth, and Thirty-second
notes can be created in one 4/4 measure?
Statement of Problem
Fix an integer n. How many sequences of the form (a1, a2, ..., ak ) with the properties that
1. Each ai ∈ {( 12 )0, ( 12 )1, ( 21 )2, ( 21 )3, ..., ( 12 )n}
2.
k
X
ai = 1
Motivation
Why fix N to 5?
This question came about by
thinking about how many different ways you could put
rhythms into a measure.
There are some rare pieces of music that use 128th notes but generally music does note use a note
smaller than a 32nd, which corresponds ( 12 )n}
i=1
are there? Observe that k ≤ 2n.
Statement of Main Results
Indication of Methodology
•Writing out a list of combinations to understand what was happening is where the research started.
•This lead us to know about the permutations which is detailed in the counting section of the poster.
•Burnside’s lemma is ultimately a method of counting. In the case of music, we can apply it to
count all the possible permutations of a combination of note durations in a unit and use shared
orbits to determine which of these permutations repeat each other and can be excluded from the
overall count.
The final number of possible combinations for one measure of music is:
46,764,804
Counting
•We started with knowing that at ( 21 )5 representing a 32nd note meant the maximum number of notes per measure we could have is 2n, in this case, 32
•The total is the sum of all possibilities of 1 note per measure, then 2 notes... up to the max of 32 notes
•We wrote out every possible combination (without the same number of repeated notes between combinations) into a table, giving us 209 combinations
•We then accounted for repeated notes within the combination by using the permutation formula n!
k! where n = total amount of notes and k = number of notes repeated
5!
•Example: At k = 5 notes per measure, 3 quarter notes and 2 eighth notes is one possible combination (QQQEE). To calculate the total we use 3!2!
= 10 permutations.
•We were also able to apply group theory in counting orbits by hand and using Burnside’s Lemma
•To count the orbits by hand, take a set X = every possible combination of a combination listed in the table and a group G to have every element of Sym(X).
•Then applying G to X (every g ∈ G to every x ∈ X) will allow you to see the orbit of every x
•Then you can find duplicate orbits and remove them from the count of total orbits (note that total orbits = amount of permutations)
•A simpler method is to write out the permutations in disjoint cycle form and then analyze which cycles would fix which elements given the known contents of the set X.
sumof allf ixedelementsof X
•Then the number of orbits =
where |G| is the number of different permutations, or k! where k is the number of note durations used to find the initial combination X
|G|
is based off of
References:
2015.
•ChordWizard Software. ”www.howmusicworks.org”. Retrieved 24 May
e:
ens
•McRay, Graeme. Math Help. N.p., 9 Apr. 2012. Web. 10 Apr. 2015.
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