The term twelve tone music suggests we treat each of those tones equally to eliminate bias. But this does not limit us to any of the 836017[1] essential patterns of serial music's twelve tone rows, with their derivative inversions, retrogrades, retrograde-inversions and cyclic-shiftings, all regarded as equivalencies. There are numerous other ways of 'democratising' twelve elements, for example statistically rather than serially.
We could have equal numbers of instances of each of the twelve elements and throw them up into the air. The resultant scatter will be expected to be uniformly distributed. And very likely musically uninteresting. We need some way of introducing structure by building composite objects out of those elements, but without giving undue weight to any particular one of them.
… that all intervals are created equal
The next obvious candidate (many will argue it's actually the principal) is the interval, characterised as a set of pitch class pairs, a duple, pij ≝ {pi, pj}. We need hardly state explicitly that j ≠ i and pij ≡ pji (but there it is anyway). There are nC2 = n(n-1)/2 such pairs chooseable from n objects in general, so from 12 pitch classes (labelled 0 … 11) we have:
{0,1}, {1,2}, {2,3}, {3,4}, {4,5}, {5,6}, {6,7}, {7,8}, {8,9}, {9,10}, {10,11}, {0,2}, {1,3}, {2,4}, {3,5}, {4,6}, {5,7}, {6,8}, {7,9}, {8,10}, {9,11}, {0,3}, {1,4}, {2,5}, {3,6}, {4,7}, {5,8}, {6,9}, {7,10}, {8,11}, {0,4}, {1,5}, {2,6}, {3,7}, {4,8}, {5,9}, {6,10}, {7,11}, {0,5}, {1,6}, {2,7}, {3,8}, {4,9}, {5,10}, {6,11}, {0,6}, {1,7}, {2,8}, {3,9}, {4,10}, {5,11}, {0,7}, {1,8}, {2,9}, {3,10}, {4,11}, {0,8}, {1,9}, {2,10}, {3,11}, {0,9}, {1,10}, {2,11}, {0,10}, {1,11}, {0,11}
This is a well-defined set of sets in that it comprises a complete accounting of all possible 2-sets contained within a 12-set. As such, it is a simple enough animal. But simple things are ofttimes just manifestations of something more complicated.
A Confederacy of Duples
This animal is also an instance of a more complex artifice comprising - more generally - 66 subsets of 2 elements each, drawn from a set of 12 elements, each of which appears 11 times and where each and every 2-form appears exactly 1 time.
These specifications - which will first seem a rather over-engineered-for-purpose definition - make it more than just a set of sets. This structure also happens to be, and is known as, a balanced incomplete block design. This one is specifically a 2-design and the above specifying quantities, or enumerations, are usually notated with the following letters (ordered as they're found within the above description):
- b … 66 subsets …
- k … 2 elements …
- v … 12 elements …
- r … 11 times …
- t … 2-form …
- λ … 1 time …
Hitherto we've used n to denote the size of the pitch class universe, and also k for the smaller sizes of pitch class sets used for scales or chords. Block (or t-) design combinatorics (for that is where we are now) continues to use k for the subset size, but uses v instead of n. The structure itself is notated as the 5-tuple (v,b,r,k,λ) - where the t is understood to be 2. Often it's just the triple (v,k,λ) when b and r are inferred from the specific construction.
It's important to be aware that t is not the same as k, though they have - in this case coincidentally - the same value of 2. The t refers to the size of the duple (the size of the pitch class pair, the abstract 2-form - represented by another 2-set - it's what this structure is 'about'). The k, on the other hand, refers to the size of the subsets containing it (or, in general, them). The above set of 2-sets, each in one to one correspondence with its 2-form, is the simplest, most transparent design one could have.
The above structure carries each of the 66 possible duples exactly once. If we wished to instantiate a piece of music from that, then what could we do with such a collection? Again, like the elements themselves (the pitch classes available in the simpler structure, the universal 12-set that we started with), we could maybe play the 66 dichords, or (admittedly rather tiny) scale patterns, in some order with some rhythm and with some dynamics.
Or we could decide we need more instances of the pairs, to give us a little more scope to play with. Such as two of each duple - we can't single any one of them out, so each duple should appear in equal quantity. Naturally this could mean just using two of these structures and we'd have 132 little musical objects to deploy in some interesting way.
Repackaging without Bias
However, we might notice that 3-sets can carry 3 duples (i.e. p1~p2, p2~p3, p3~p1) in one bag, as it were. So instead of toting around 132 bags we can manage with only a third of them - i.e. 44 (admittedly 50% bigger) bags. We can't toss just any old trio of elements into each bag (which we'll call a block from now on) but must arrange their contents in such a way as to ensure that exactly two instances of each of our 66 duples (t-forms) turn up altogether. In other words we must construct a design out of b=44 blocks of k=3-sets (drawn from our universe of the v=12 set) so that exactly λ=2 of each duple/interval appear in it. Thus the t-set is no longer explicitly visible, whereas the k-set remains very much so.
It should be relatively easy to see that, when forming an exhaustive b sized list of k-sized subsets of v elements, you'll need r = bk/v each of the elements to build it. We have seen that vr = 12×11 = 132 = 2×66 = kb. With 3-sets we can instead have 132 = 3×44 = kb. Which is an illustration of the invariance of vr = kb within such systems.
We note that there are 12C3 = 220 possible triples, which is 5 times as many as we want/need. So how do we go about gathering a valid collection of exhaustively and non-preferentially distributed embedded duples? We could start with the whole 220, list all the duples implied in each, and (very carefully) remove 176 triples carrying exactly 8 each of the 66 duples. Or find some other algorithm to start from the bottom up. But sometimes the simple answer is that we just find them in the literature[2].
Redistribution of Assets
Consider this set of (so far, magically generated) 44 3-sets, which happens to be such a (12,44,11,3,2) block design:
{0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {4,5,7}, {5,6,8}, {6,7,9}, {7,8,10}, {0,8,9}, {1,9,10}, {0,2,10}, {4,5,9}, {5,6,10}, {0,6,7}, {1,7,8}, {2,8,9}, {3,9,10}, {0,4,10}, {0,1,5}, {1,2,6}, {2,3,7}, {3,4,8}, {2,6,8}, {3,7,9}, {4,8,10}, {0,5,9}, {1,6,10}, {0,2,7}, {1,3,8}, {2,4,9}, {3,5,10}, {0,4,6}, {1,5,7}, {7,10,11}, {0,8,11}, {1,9,11}, {2,10,11}, {0,3,11}, {1,4,11}, {2,5,11}, {3,6,11}, {4,7,11}, {5,8,11}, {6,9,11}
If you cared to, you could verify that the 11 instances of each of our 12 pitch classes are present. For example pitch class 3 turns up in {0,1,3}, {2,3,5}, {3,4,6}, {3,9,10}, {2,3,7}, {3,4,8}, {3,7,9}, {1,3,8}, {3,5,10}, {0,3,11} and {3,6,11}. You might also check that each of the 66 pitch class pairs turn up twice each, e.g. {4,8} ⊂ {3,4,8} and {4,8} ⊂ {4,8,10} but in no others.
Abstraction to Application
One moderately musically interesting result of this latter property (a consequence of λ=2) is that we may conceivably write a piece of music comprising a sequence of triads (the blocks) in which each triad leads to the next carrying a common interval with the third pitch guaranteed as changing. For example the two successive bars with the above {4,8} (conventionally enough taking PC4 as E and PC8 as G#) in common:
We may then choose the common pair {3,4} or {3,8} for the preceding bar's common interval connection, and either {8,10} or {4,10} for the following bar's. After electing {3,8} to the left (introducing PC3 = E♭) and {8,10} to the right (introducing PC10 = B♭) we'd get:
Now we're forced to complete the left hand bar with the only remaining triad containing {3,8} that we're assured we have by the design (i.e. {1,3,8}, introducing PC1 = C#) and also to complete the right hand bar as the only remaining triad containing {8,10}, which is {7,8,10} (bringing in a PC7 = G):
Here's what (a), (b) and (c) sound like:
At this point we can pick either of {1,3} or {1,8} to continue the leftward process (since by now we have consumed both {3,8}) and either of {7,8} or {7,10} for the rightward (likewise both {8,10}).
I.e. what remains is:
{0,1}, {1,2}, {2,3}, {3,4}, {4,5}, {5,6}, {6,7}, {7,8}, {8,9}, {9,10}, {10,11}, {0,2}, {1,3}, {2,4}, {3,5}, {4,6}, {5,7}, {6,8}, {7,9}, {8,10}, {9,11}, {0,3}, {1,4}, {2,5}, {3,6}, {4,7}, {5,8}, {6,9}, {7,10}, {8,11}, {0,4}, {1,5}, {2,6}, {3,7}, {4,8}, {5,9}, {6,10}, {7,11}, {0,5}, {1,6}, {2,7}, {3,8}, {4,9}, {5,10}, {6,11}, {0,6}, {1,7}, {2,8}, {3,9}, {4,10}, {5,11}, {0,7}, {1,8}, {2,9}, {3,10}, {4,11}, {0,8}, {1,9}, {2,10}, {3,11}, {0,9}, {1,10}, {2,11}, {0,10}, {1,11}, {0,11}
{0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {4,5,7}, {5,6,8}, {6,7,9}, {7,8,10}, {0,8,9}, {1,9,10}, {0,2,10}, {4,5,9}, {5,6,10}, {0,6,7}, {1,7,8}, {2,8,9}, {3,9,10}, {0,4,10}, {0,1,5}, {1,2,6}, {2,3,7}, {3,4,8}, {2,6,8}, {3,7,9}, {4,8,10}, {0,5,9}, {1,6,10}, {0,2,7}, {1,3,8}, {2,4,9}, {3,5,10}, {0,4,6}, {1,5,7}, {7,10,11}, {0,8,11}, {1,9,11}, {2,10,11}, {0,3,11}, {1,4,11}, {2,5,11}, {3,6,11}, {4,7,11}, {5,8,11}, {6,9,11}
The obvious question is whether or not we can continue in this fashion to consume all 44 triads exactly once without breaking the common duple rule tying successive bars together. I.e is there a Hamiltonian path, with the 3-sets regarded as the nodes of a graph? Or - even better - a Hamiltonian circuit in which the final, 44th, bar links neatly back to the 1st bar with a common duple.
Since at each stage of this construction we have two choices at both left and right edges, and bearing in mind that there are 44 3-sets to consume, along with 44 of the 66 duples (leaving 22 of them unemployed - one of the hazards of democracy?), it seems unlikely that the current example will succeed since it was begun somewhat arbitrarily, and grown via what amounts to a sequence of coin-tosses.
So, rather than (possibly) waste time working blindly, we can attempt to draw a graph of the design, with the 44 3-sets as nodes and the 66 duples as edges. Each edge will therefore connect two nodes (a direct consequence of λ=2) and each node will carry three edges (since each 3-set contains 3 2-sets and the 2-sets match the edges). With such a graph (although it's a bit of a mess) it may be easier to visualise a path or circuit which visits each (blue) node exactly once. We've emphasised the three edges we've used so far with three heavy reddish lines.
But visual path detection seems by no means such an easy task. It would be helpful if we could find some software to do this. It doesn't take long to dig up an algorithm from the early 2000s (thank you Ashay Dharwadker). All this particular piece of software requires is a text file containing an incidence matrix for the graph, which takes only minutes to prepare.
With our graph, the program produces 45 results almost instantly. Searching those results for circuits with the particular choices made above - perhaps not surprisingly - reveals nothing. However, the single circuit presented employs our last three bars (blocks {3,8,4} via {4,8} to {4,8,10} via {8,10} to {7,8,10}). Had we chosen {3,4} rather than the {3,8} we did, the previous bar (the first of four, above) would have been {3,4,6} instead of {3,1,8}, and we would at that point have remained on course - doubtless to hit the rocks not much later. But now we know better and the complete circuit is
{0,1,3}, {1,3,8}, {1,7,8}, {1,5,7}, {4,5,7}, {4,5,9}, {2,4,9}, {1,2,4}, {1,2,6}, {1,6,10}, {5,6,10}, {3,5,10}, {2,3,5}, {2,5,11}, {2,10,11}, {0,2,10}, {0,4,10}, {0,4,6}, {3,4,6}, {3,4,8}, {4,8,10}, {7,8,10}, {7,10,11}, {4,7,11}, {1,4,11}, {1,9,11}, {1,9,10}, {3,9,10}, {3,7,9}, {2,3,7}, {0,2,7}, {0,6,7}, {6,7,9}, {6,9,11}, {3,6,11}, {0,3,11}, {0,8,11}, {5,8,11}, {5,6,8}, {2,6,8}, {2,8,9}, {0,8,9}, {0,5,9}, {0,1,5}
Which, as you can see, cycles from the final {0,1,5} via a {0,1} to the initial {0,1,3} (not that {0,1,3} must be initial - we could now start anywhere).
In the following figure, we present the circuit - labeling the edges only - with the triads occupying 44 nodes as blue discs. We do not need to clutter-label the nodes as the three pitch classes within each disc simply comprise the union of the pitch class pairs on the large circle's arcs on either side of it. For example, right at the top we have a disc/node with {0,1} to its left and {1,3} to its right, so we know this block is {0,1,3}. Note that the third edge's (every disc supports three edges) label, i.e. {0,3}, can be found halfway along the line tracing off down and to the left towards block/disc {0,3,11}.
Placing edge labels automatically (halfway) may occasionally lead to confusion. Beware, for example, the {1,5} in proximity to the {1,3,8} node at the top. This may appear to suggest, erroneously, that pitch class 5 should be in that block. In fact that {1,5} labels an edge from two completely different nodes on either side and just happens to land inconveniently close by. It's true third edge's label, {3,8} is halfway along a line tracing to the south east.
Here is a simple musical manifestation of the circuit, along with a pair of audio files
The second playable example is a faster manifestation with tuned percussion and a bit of rhythmic interest (just not very much). It lasts longer only because it's played twice around the Hamiltonian circuit.
Note that the above block design - essentially the set of 44 3-sets, from which we have serialised the blocks in a sequence of our own choosing to satisfy some musical whim is only one of the 242995846[3] purported (12,44,11,3,2) block designs. One may apprehend the size of the solution space by considering the 22 unused edges criss-crossing the large circle above. There is no obvious pattern and one should not expect one. Indeed it would be a little unnerving to find one.
The Higher Strata
Other factorisations of the vr=bk invariant (in this case equal to 132) allow us to propose designs of 33 blocks of 4-sets (tetrachords or tetratonic scales/arpeggios), for example the following (12,33,11,4,3) design, where our 66 duples each turn up 3 times (there are more than 17 million[3] of these, this is just one of them):
{0,1,3,7}, {1,2,4,8}, {2,3,5,9}, {3,4,6,10}, {0,4,5,7}, {1,5,6,8}, {2,6,7,9}, {3,7,8,10}, {0,4,8,9}, {1,5,9,10}, {0,2,6,10}, {2,4,9,10}, {0,3,5,10}, {0,1,4,6}, {1,2,5,7}, {2,3,6,8}, {3,4,7,9}, {4,5,8,10}, {0,5,6,9}, {1,6,7,10}, {0,2,7,8}, {1,3,8,9}, {5,6,8,11}, {6,7,9,11}, {7,8,10,11}, {0,8,9,11}, {1,9,10,11}, {0,2,10,11}, {0,1,3,11}, {1,2,4,11}, {2,3,5,11}, {3,4,6,11}, {4,5,7,11}
There's another invariant of such systems, viz. λ(v-1) = r(k-1). The first has 1×(12-1) = 11×(2-1) and the last 5×(12-1) = 11×(6-1). You may wish to confirm this with the other two designs.
Designs comprising 22 blocks of 6-sets (hexachords and/or hexatonic arpeggios) exist. E.g. the (12,22,11,6,5) design where our 66 duples each turn up 5 times (1 of 11603[3] possible designs):
{0,2,3,4,8,10}, {0,1,3,4,5,9}, {1,2,4,5,6,10}, {0,2,3,5,6,7}, {1,3,4,6,7,8}, {2,4,5,7,8,9}, {3,5,6,8,9,10}, {0,4,6,7,9,10}, {0,1,5,7,8,10}, {0,1,2,6,8,9}, {1,2,3,7,9,10}, {1,5,6,7,9,11}, {2,6,7,8,10,11}, {0,3,7,8,9,11}, {1,4,8,9,10,11}, {0,2,5,9,10,11}, {0,1,3,6,10,11}, {0,1,2,4,7,11}, {1,2,3,5,8,11}, {2,3,4,6,9,11}, {3,4,5,7,10,11}, {0,4,5,6,8,11}
In a block design such as this, transitioning between blocks via duples would be modelled by a graph with the 22 blocks (nodes) and 165 (duples) edges, with each node carrying 6C2 = 15 edges.
At first glance a design using 5-sets, with pentatonics, seems impossible since k=5 doesn't divide evenly into v=12. But we may accomodate pentatonic designs by relaxing the vr=132 constraint (it must still equal bk though!). I.e. have r=55 of each pitch class turn up in a (rather larger) structure of b=132 k=5-sets with λ=20 instances each of our intervalic duples in a (12, 132, 55, 5, 20) design.
Realpolitik
We finally note here that - with these 2-designs - no particular pitch classes or intervals stand out against each other in the design. But particular musical applications of it may - for example by keeping 22 nominally equal citizens out of work as we did above. Other musical economies may be more successful in employing everyone. Additionally, in any application of a particular design, it will be the case that one (12,44,11,3,2) block design will draw attention to only 44 of the 220 possible triads mentioned above. And another will favour a different 44. It's fair(-ish) to say, then, that musical democracy extends upwards in the aggregations of all possible 2-designs, i.e. not simply any one of them. Naturally, you may actually democratise triples by building 3-designs, quadruples with 4-designs, etc. The problem with 'equal opportunities' for a design employing all 220 3-set citizens in equal numbers (via that design's λ value) is that no pre-singularity human being will be able to apprehend the structure. It's certainly not likely that even the 'mere' 44-ness presented here would be noticed as some kind of integrable musical idea - the 40 parts available to listeners of Spem In Alium notwithstanding (thanks due to Hans Jacobi for this thought).
I'm extremely grateful to Tom Johnson, with whom I spent a pleasant and mathematically engrossing couple of hours at the end of June, 2018. For it was he who introduced me to the world of block designs, both at his fortress of solfège-étude and via his book[4]. The field presents a vast source of ideas, with which I'm now mildly obsessed.
[1] Combinatorial problems in the theory of music, R C Read, Elsevier Discrete Mathematics 1997, table 2 page 547.
[2] A Survey of Resolvable Solutions of Balanced Incomplete Block Designs, Kageyama Sanpei, Longman Int Stat Rev Vol 40#3 1972, table on page 270.
[3] Handbook of Combinatorial Designs, IIed, Charles J Colbourn and Jeffrey H Dinitz, CRC Press 2010, page 37.
[4] Other Harmony - beyond tonal and atonal, Tom Johnson, Editions 75 2014, block designs page 191.
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