Symmetries
The symmetries of the dodecagon, the regular 12-sided polygon used in models of Pitch Class Set Theory, are well known. Viewed as a dihedral group (of order 24), it has a class index
$$P_{D_{12}}(\textbf{x}) = \frac{1}{24} \left( x_1^{12} + 7 x_2^6 + 2 x_3^4 + 2 x_4^3 + 2 x_6^2 + 4 x_{12} + 6 x_1^2 x_2^5 \right)$$which sheds some light upon the 2, 3, 4 and 6-fold symmetries of 12-sided figures. This can be used to enumerate all the essentially different (equivalent under rotations and reflections) shapes of all (mostly) irregular polygons of all orders - from 0 to 12 - inscribable within it. One simply replaces every $x_i$ with $(1 + t^i)$ in the above (Pólya Enumeration) to recover the following degree 12 polynomial in t:
$$E_{12}(t) = 1 + t + 6 t^2 + 12 t^3 + 29 t^4 + 38 t^5 + 50 t^6 + 38 t^7 + 29 t^8 + 12 t^9 + 6 t^{10} + t^{11} + t^{12}$$which counts the number of essentially different k-gonal shapes - as the coefficient of tk - corresponding to all 224 Pitch Class Sets of different sizes available to Twelve Tone Music.
If you count non-symmetric shapes twice (i.e. as distinct from their reflections) rather than just once, the Cyclic Group polynomial enumerator - $1 + t + 6 t^2 + 19 t^3 + 43 t^4 + 66 t^5 + 80 t^6 + 66 t^7 + 43 t^8 + 19 t^9 + 6 t^{10} + t^{11} + t^{12}$ - is the appropriate one to use. This aggregates 352 distinct shapes (evaluate the polynomial at t = 1). Because it counts mirror-pairs (musically inverted PC sets) as distinct, this enumeration reveals that there are (352 - 224) = 128 pairs of non-symmetric shapes and 352 - 2 × 128 = 96 symmetric ones.
So far, so standard (this blog also catalogues them all here).
Most of these tk coefficients are rather haphazard (like, say, 29 and 38) and have no particularly friendly relations with the polygon they inhabit - rather to be expected. However, we see that the number of 3-sided shapes is 12, a number very friendly indeed to the enclosing dodecagon. It means that there's a possibility that all the different 3-sided shapes - and here they all are - might be enticed to exactly fit, four at a time, into three separate dodecagons.
Symmetric 3-sets
(equilateral and isosceles triangles)
Asymmetric 3-sets
(scalene triangles)
Set Coverings
Note that the coefficient 6 (of t2) is also '12 friendly'. But tilt and move them as you will, it's impossible to fit the 6 different 'diangles' into one dodecagon, so that each occupies a different dodecagonal vertex pair. It's not possible to cover (or partition) the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} with 6 non-intersecting 2-sets where each such 2-set represents each of the six possible differences (on the dodecagon, i.e. modulo 12 differences) between the 12-set's points.
Octatonality
If we drop back to a tonality in which only 8 Pitch Classes are available, the Class Index of this system is
$$P_{D_{8}}(\textbf{x}) = \frac{1}{16} \left( x_1^8 + 4 x_1^2 x_2^3 + 5 x_2^4 + 2 x_4^2 + 4 x_8 \right)$$we find its catalogue of (30) essentially different sub-polygons is enumerated by
$$E_{8}(t) = 1 + t + 4 t^2 + 5 t^3 + 8 t^4 + 5 t^5 + 4 t^6 + t^7 + t^8$$One notices that the t2 term has coefficient 4, indicating that octatonicity's 4 interval classes 1, 2, 3 and 4, might be capable of covering itself in a single bound. And indeed it is. Interval class 1 can cover pitch classes 2, 3 and (switching to abbreviations) IC 2 can cover PCs 6, 8, IC 3 can cover PCs 4, 7 and finally IC 4 can cover PCs 1, 5. It's the only way of doing it with regard to equivalence of rotational and reflective symmetries.
But as a musical transformer - say, as the PC permuter (1,5)(2,3)(4,7)(6,8) - it's not very interesting since all it can do is exchange pitch classes. As such a transformation is an interval-preserving operation, it's pretty much a do-nothing, musically speaking. This is why, in general, we don't regard any tonality covers with PC Sets containing fewer than 3 pitch classes as interesting
But we're not quite done yet with the octagon since there's another friendly coefficient in the 8t4 which indicates that we may have a quadruple covering of the octagon with its four pairs of tetragons (possibly more commonly referred to as quadrilaterals). However such hopes are quickly dashed since one of the 8 quadrilaterals must be a square, and placing a square inside an octagon requires a second square to complete the cover, thus violating our règle du jeu that each shape be used exactly once. We don't even need to bother with the other shapes as we're already dead in the water.
This is a general upper limit to the size of an 'inner polygon'. Once we reach the halfway point of an evenly sized polygon, a subset of exactly half the size which covers every other vertex can complete a cover only with a duplicate of itself to soak up the remaining 'every-other' vertices. Beyond the halfway point of course, there's no possibility of coverage since too few vertices remain.
Our mission, then, is to find the 'special' values of n and k (2 < k < n/2) where rotationally and reflectionally equivalent binary colourings of n-gons (creatures known as necklaces) carry Cn,k (generally irregular) k-sized sub-polygons. We require that k divides both n (permitting n-set coverings with p = n/k of its k-sized subsets) and Cn,k itself (permitting use of all equivalent k-gons) and where p divides Cn,k (permitting complete coverage of Cn,k/p separate n-gons).
The Triumph of the Dodecagon
And so, by this definition, the first polygon where there's even a possibility that a complete set of dihedrally equivalent Pitch Set Classes will be capable of covering its own tonality is the dodecatonic. It's the first one large enough.
It's reasonably obvious that tonalities with a prime number of pitch classes don't even get a look-in with regard to subset covering. So goodbye to 17TET, 19TET, 31TET etc. The 15TET, non-prime, system is quite popular but none of its subset shape collections occur in quantities appropriate to such coverages. The only candidates for tonalities up to 64TET are presented here:
Tonicity
nSubset Size
kMultiplicity
pCount
CCoverages
C/p12 3 4 12 3 16 4 4 72 18 24 3 8 48 6 32 4 8 624 78 32 8 4 165288 41322 36 3 12 108 9 36 6 6 27474 4579 40 4 10 1240 124 44 11 4 87161756 21790439 48 3 16 192 12 48 8 6 3936144 656024 48 12 4 725782644 181445661 50 10 5 102749880 20549976 56 4 14 3472 248 60 3 20 300 15 63 21 3 219201890450655 73067296816885 64 4 16 5216 326 64 8 8 34597680 4324710
Coverings and Applications
The power of the enumerative polynomial lies in its proof of existence of such potential coverings. But it's non-constructive and cannot tell us how to build the things it counts. Still less does it tell us anything about their set-covering capabilities.
So, given the unlikelihood of coverage indicated by the previously shown impossibility of 'dianglising a dodecagon' (in musical terms, 'interval-covering a chromatic scale'), and despite the very real lack of interest we have in covering polygons with anything smaller than triangles in any case, the big question is can all 12 triangles (musical triads) - taken four at a time - fit into three dodecagons so that none of their vertices occupy the same dodecagonal vertex (musical pitch class) as another's?
Covering the first dodecagon is easy since there's so much choice. Even with only 8 triangles left, a second cover remains only mildly troublesome. But with only four remaining triangles, a final cover is hard to find since there are so many ways to rotate and reflect them. By the time you have four left, if your earlier choices are unsuitable then the final cover will likely be impossible.
But there are many solutions. Thousands, in fact. Here is one.
In fact this particular solution is not found by trial-and-error but is rather special even in terms of the game itself. We've used the computer algebra system, GAP, to find solutions. Of the thousands of solutions, this is one from a tiny subset of 20 of them. All of the others, except one, which is really special, are of the same class.
The vast majority of triple-set coverings by triad quadruples are found as an example of the Alternating Group on 12 symbols (usually notated as A12). This is a simple group of order 12!/2 = 239500800. Despite the term 'simple' (a term of Group Theory meaning that it has a simple structure and is akin to a prime number in Number Theory) the way this group can fling things about is prodigious, as the number of its operations suggests.
But the configuration above is an example of an arguably more interesting group discovered in the 19th century. It's one of the sporadic simple groups, specifically the Mathieu Group M12. But it is somewhat less flingy as its order is a paltry 95040. And from an applicative point of view it's often nicer to have fewer choices (the apparently paradoxical idea behind 'constraint sets you free') since it's easier to actually make a beginning.
The way the group features into these arrangements is by way of construction. If we take the first of the three coverings we can use it as a permutation - specifically (1,2,3)(4,5,7)(6,9,11)(8,10,12), the four clockwise tricycles in the covering - to generate a group. If we regard the numbers as pitch classes (it's usually 0 to 11 in musical PC Set theory but we can equivalence 12 and 0 with impunity) then we may operate upon any musical segment (the choice of what constitutes a segment is completely up to the applier) by permuting its pitch classes with that permutation. Each pitch class moves (conventionally clockwise) around the triad it finds itself on. For example we might have a segment with pitch classes { 1, 4, 8, 11 } - which could represent a $C\sharp m7$ chord. The aforementioned permutation moves 1 to 2, 4 to 5, 8 to 10 and 11 to 6 (the wraparound) and thereby produces the new set { 2, 5, 10, 6 } - re-presented as { 2, 5, 6, 10 }. This set may - if interpreted as rooted on pitch class 6 (PC Set theory conventionally $F\sharp$) - be perceived as $F\sharp^{+} Maj 7$.
Such a transformation may be applied to any musical segment, such as the one shown below.
A second use of the same permutation would then transform the PC Set from { 2, 5, 6, 10 } to { 3, 7, 9, 12 } (perhaps $Am7\flat5$?). By the way, that's 3 of the seven sevenths, so this - admittedly tiny - orbit fixes some kind of seventhiness, if you will.
And a third application will of course return us to the initial set, since all cycles are of the same size, i.e. the order of the permutation is 3. This single permutation has a class index of $x_3^4$ (being a product of 4 3-cycles) and the group it generates is the tiny simple cyclic group of order 3, C3.
The Group M12 is generated when all three coverings are used to construct permutations. The other two are - respectively from the above figure - (1,4,7)(2,5,9)(3,8,10)(6,11,12) and (1,5,9)(2,3,6)(4,8,10)(7,11,12) and although each is a simple $x_3^4$ cycle, when allowed to operate together by composition the group of permutations has the class index
\begin{split} P_{M_{12}}(\textbf{x}) = \frac{1}{95040} ( x_1^{12} + 495 x_1^4 x_2^4 &+ 2970 x_1^4 x_4^2 + 1760 x_1^3 x_3^3 + 396 x_2^6 + 11880 x_1^2 x_2 x_8\\ &+ 9504 x_1^2 x_5^2 + 15840 x_1 x_2 x_3 x_6 + 2970 x_2^2 x_4^2 + 2640 x_3^4 + 17280 x_1 x_{11}\\ &+ 9504 x_2 x_{10} + 11880 x_4 x_8 + 7920 x_6^2 ) \end{split}If we label these three permutations with the letters 'a', 'd', and 'm' - for no particular reason - then the successive application of, say m followed by a followed by d would be written as mad. It would be the permutation (2,4,3)(5,12,7,11,10,9)(6,8), which is one of the $15840 x_1 x_2 x_3 x_6$ permutations with that particular shape (the $x_1$ in this case representing the point 1, or pitch class $C\sharp$, which happens to be fixed by this particular permutation).
If we apply this mad operation to a fairly well-known melodic segment, we obtain the following:
To recover the original, one would apply the inverse operation $(mad)^{-1} = d^{-1}a^{-1}m^{-1}$ - which is of course the 'undoing' permutation (2,3,4)(5,9,10,11,7,12)(6,8). Because such transformations operate only upon pitch classes however, there can be no indication of which particular octave the original pitch class was in and such information is lost.
There are also relationships between the group's generators. One such relation is found when we look at dm-1 = (1,10,2)(3,4,12)(6,7,9), one of the $1760 x_1^3 x_3^3$ which - in this case - fixes the triad {5, 8, 11}. Since this is clearly an operation of order 3 (comprising exclusively 3-cycles) then any set acted upon by it thrice will reappear. This means that dm-1dm-1dm-1 = (dm-1)3 = (), the group's identity operation. It's also (dm2)3 because the square of any of the generators is also the generator's inverse (two clockwise turns gets you to the same place as one anticlockwise turn).
One might wish to verify that the rather unpleasant looking d2amad2m2a2md2ma2 is also, in fact, a 'do nothing'.
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