20180815

Pitching Democracy

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 ≝ {pipj}. 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):

  • b66 subsets …
  • k2 elements …
  • v12 elements …
  • r11 times …
  • t2-form …
  • λ1 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 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 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 attached to this disc (for every disc supports three edges), 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 either side and just happens to land inconveniently close by.

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's no obvious pattern and one should not expect one. It would be a little startling 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 will ever stand out against each other in the design. But particular musical applications of the design may do so - for example by throwing some nominally equal citizens on the dole as we did above with 22. Other schemes may be more successful in employing everyone. Also, in an application of a particular design, any one of the (say) (12,44,11,3,2) block designs will clearly favour 44 of the 220 possible triads mentioned above. And other such designs will favour (draw attention to) a different 44. It's fair(-ish) to say, then, that musical democracy extends upwards in the aggregations (in this case of all possible 2-designs) but not in the particulars of any one of them. Naturally, you may actually democratise triples with 3-designs, quadruples with 4-designs, etc.

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[3]. 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.

[3] Other Harmony - beyond tonal and atonal, Tom Johnson, Editions 75 2014, block designs page 191.

20180705

All The Intervals

The so-called 'all interval tetrachord' is not exactly news to musicians, although that this news is only dozens, and not hundreds, of years old is possibly surprising. Its (or more accurately their, since there's more than one) 'interestingness' having been exposed only (relatively) recently is due to pitch class set theorists who invented the 'interval-class-vector' (scare-quotes because of course it's in only the loosest, most informal, mathematically-anathematical, way any kind of a vector). This intervalency is the thing which usually turns up in PC set theory as a comma separated list of numbers between a pair of angled brackets, such as <2,5,4,3,6,1>. It lists - in order - the frequency of occurence of differences between each pair of pitch classes in the set it's being used to characterise, where the differences range from 1 to 6 in the 12 tone system (because the shortest distance between any two 'hour points' on a clock is always going to be between 1 and 6).

History

Although it would have easily been possible for any (say) 15th century musician to notice that the tetrad comprising (say) the pitches B, C, E♭ and F carried (between B and C, B and E♭, B and F, C and E♭, C and F, E♭ and F) respective separations of 1, 4, 6, 3, 5, 2 semitones - which is to say exactly one of each possible separation between any pair of notes, it's not clear that this would have been considered in any way remarkable.

Indeed there seems little evidence that musicians - or even mathematicians - were considerate of the number of possible tetrachords (or chords of any size) possible within a universe of twelve pitches before the middle of the 19th century. Luigi Verdi's survey article of proto, pre-USAnian if you will, pitch class set theories "The History of Set Theory from a European point of view" is well worth a read in this regard. There are, by the way, 43 such tetrachords and only 4 of these have this property.

Note that when we say 43, it depends on what you are counting - in this case it's all of the distinct shapes, their reflections, and homometries. If you ignore reflections (i.e. if you accept the essential identity of congruent shapes regardless of mirroring) it drops to 29 since 15 of the quadrilaterals have (at least) bilateral symmetry, and the remaining 28 turn up as 14 asymmetric mirror-image pairs. If, further, you ignore homometries (differently-shaped tetrads with the same intervalency) it drops to 28 because 2 of those 14 pairs of mutually inversional tetrads - which happen to be the subject of this very post - carry the same interval-class distribution, which is to say one instance of each interval class 1 to 6.

The 'classic' case

The Four All-Interval Tetratonic sets out of 12
1 3 2 6 2 3 1 6 1 2 4 5 4 2 1 5
The above four PC sets applied as chords (with F ≡ Pitch Class 0)
PC Sets 4-Z15 and 4-Z29

1326 chord (rooted on 440Hz)

1245 chord (rooted on 440Hz)

Wherefore art thou audio?

So what exactly is so remarkable about these chords anyway? They don't sound all that great, especially when instantiated in their most compacted forms as a Fortean Prime Form Cluster (4-Z15 ≡ {0, 1, 4, 6} and 4-Z29 ≡ {0, 1, 3, 7}). One may, reasonably easily, see in the second of these forms certain jazziness since the root, minor 3rd, 5th and (the '1' being bumped up an octave) the minor 9th are applicable (sans the 7th). And an inversion of the first set (subtracting 1 from each element makes its second member a new root) {11, 0, 3, 5} carries a minor 3rd, a 4th and a major 7th in plain sight, as it were.

A pair of jazzy applications of 4-Z29A and 4Z-15A
A Displacement and an Inversion

But mainly the interest is in its very rarity as a musical (or geometric) object, in that out of all possible PC sets (351 distinct polygonal shapes, 223 distinct polygonal congruences, 200 distinct polygonal homometries) that one may pull out of a 12 pitch class universe, only these 4 (2 if you equivalence reflections, 1 if you equivalence intervalencies) have the property that the frequency distribution (or, more simply, 'counts') of differences between every single pair of pitch classes in the set occur exactly once.

Actually, the term 'all-interval set' is rather a weak description of the kind of object we're considering since, with enough pitch class pairs to play with (i.e. given sufficiently large k for a particular N), it's almost impossible to avoid all interval (classes) turning up between them. A better term would bring out not only the completeness of the coverage but also its parsimony, i.e. with exactly one instance of each. It's that property which makes these objects interesting, and this term doesn't really do it justice. The closest we seem to have is from Gamer and Wilson, in 2003, who recognise and define "a difference set (modulo n) to be a set of distinct integers c1, …, ck (modulo n) for which the differences ci − cj (for ij) include each non-zero integer (modulo n) exactly once" - but these are not the words you are looking for. Combinatorial Mathematics has the term "planar difference set", but this terminology would likely be completely opaque to a musician.

Such a property cannot simply happen for any set. Firstly, the things being counted are difference classes (i.e. the smaller of the two values |ci-cj| and N-|ci-cj| separating numbers ci and cj on an N-houred clock) between pairs of integers (ci, cj in the range 0 … N-1) drawn from the finite set N. In the above example, N is of course 12 and only (absolute) differences of 1 to 6 may turn up. In general, the number of possible values one may have for differences is N/2 for even N and (N-1)/2 for odd N. For instance, when N = 13 the differences from the 'top of the clock, at 0 [or any multiple of 13]' range from 1 - 0 to 6 - 0 clockwise, then 7 - 0 (having passed the halfway point of 6½) is the same distance or separation as 13 - 7 = 6; 8 - 0 is the same separation as 13 - 8 = 5, etc.

Secondly, the number of interval classes is not arbitrary. A set containing k pitch classes (i.e. a set Pk = { c1, c2, c3, … ck-1, ck }) can carry only k(k-1)/2 - a triangular number - pitch class differences |ci - cj| (i ≠ j). Thus if the distribution of these k(k-1)/2 differences is to be a k(k-1)/2 length list of 'all-exactly' 1s, it's clear that the only tonalities which can possibly carry these objects are modelled by 2, 3, 6, 7, 12, 13, 20, 21, 30, 31, etc where the corresponding all-interval k-sets are of dichords in 2 & 3, trichords in 6 & 7, tetrachords in 12 & 13, pentachords in 20 & 21, hexachords in 30 & 31 etc.

As it happens, 13 has a similar quartet:

The Four All-Interval Tetratonic sets out of 13
1 3 2 7 2 3 1 7 2 1 4 6 4 1 2 6

It's evident that the shapes are pretty similar to those found in 12. That they are in adjacent tonalities does not necessarily mean that 'all interval' shapes would be expected to be broadly similar.

Interval Strings

The white numerical annotations (along the polygonal edges) are simply the number of skips between each pitch class in the ordered set (the polygonal vertices). This interval skip, or interval string notation is a vastly superior method of denoting pitch class sets, far out-transparenting the structure-hiding and equivalence-hiding opacity of listing pitch class numbers between braces - which should only ever be needed when you're on the verge of committing music to paper (performers do - after all - usually need to know what actual notes to play).

Its superiority as an 'interval showcase' is achieved by first tabulating all of its k(k-1) substrings (k of length 1, k of length 2, k of length 3 … k of length k-2 and k of length k-1), e.g. for one of the modes of 2317, such as 1723 (a mode being just a rotation of the set's polygonal representation within its 'N hour clock' space, keeping its top, 'noon', slot occupied):

substrings of '1723' of length
123
117172
772723
223317
331231

Then we just 'sum' each substring (by adding up its digits):

substring sums
1810
7912
2511
346

It is then evident by inspection that each interval in the set 1, 2, … 10, 11, 12 turns up exactly once. You may satisfy yourself that this works also for the pattern 2146 and any of its rotations. Pick any other interval string consistent with 13 and you will easily see either missing or duplicated intervals in the sum table.

As a matter of (possibly minor - since the tonal universes are so small) interest, the tonal space 6 - in which the all-interval sets are perforce triads - is the only space where an all-interval set could be the same size as its complementary set. As it happens there are a pair of such sets, 132 and 123 (interval string-wise) which are indeed not only self-inverse but self-complementary. If you insist on explicit Pitch Class Set representations (where their inversional relationships are much less immediately evident), they are {0,1,4} and {0,1,3}. The heptaphonic space 7 also admits of a pair of all-interval sets 142 and 124 ({0,1,5} and {0,1,3}) - also clearly (from their interval string representations) mutual inverses.

More solutions

The next possible tonality where we could find an all-interval set would be where the triangular number is 10 (where k = 5), which would have to be half the number of possible interval classes carried by it (i.e. where N = 20 or 21). But it turns out that there are no such sets to be found in 20. There is, however a single pair of mutually inverse 5-sets in 21 and it is 2513A (and its inverse 3152A), in interval-string denotation (where 'A' stands for an interval of 10). These are PC sets {0, 2, 7, 8, 11} and {0, 3, 4, 9, 11} in what would be their prime form, had Forte considered tonalities other than dodecaphonic, with a corresponding intervalency (interval-vector »choke«) of <1,1,1,1,1,1,1,1,1,1>. We can draw their polygons, but cannot reasonably represent them on a musical staff without use of microtonal notations.

The only pair of all-interval pentachords in 21ville
2 5 1 3 A 3 1 5 2 A

By now, we can see a definite 'meat cleaver' shape common to many of these polygons (the triangles from 6 and 7, being so geometrically limited, could be said to resemble either 4-Z15 or 4-Z29).

Next up would be 30, but again there are no sets to be found. This is somewhat over-compensated for in 31 where we jump to 5 homometric pairs - 13278A and 87231A, 47215C and 51274C, 12546D and 64521D, 17324E and 42371E, and finally 13625E and 52631E. Here they are (though this time we'll place the inversions underneath rather than to the right), each carrying exactly one instance of interval classes 1 to 15:

Five pairs of all-interval hexachords in 31ville
1 3 2 7 8 A 8 7 2 3 1 A 4 7 2 1 5 C 5 1 2 7 4 C 1 2 5 4 6 D 6 4 5 2 1 D 1 3 6 2 5 E 5 2 6 3 1 E 1 7 3 2 4 E 4 2 3 7 1 E

13625E hexachord (rooted on 440Hz)

17324E hexachord (rooted on 440Hz)

12546D hexachord (rooted on 440Hz)

47215C hexachord (rooted on 440Hz)

13278A hexachord (rooted on 440Hz)

The polygons are ordered in, again, what would be their Fortean prime forms; i.e. interval strings are descending reverse alphabetically ordered horizontally (which is, essentially, how prime form is calculated) and vertically (the upper reverse-sorting before its corresponding lower inversion). Note that by 'reverse alphabetically' we mean the reversed strings are ordered descendingly. Blame Forte.

It is the author's fancy that the meat cleaver remains visible in one of these pairs.

Systematic Solutions

Jedrzejewski and Johnson's useful 2013 paper, The Structure of Z-Related Sets, presents polynomials capable of representing both pitch class sets and their consequent intervalic distributions. These derive from the realm of crystallography and Patterson Functions. Briefly, it means that a pitch class set {p1, p2, … pk-1, pk} - as usual of size k and drawn from a tonality of order n - and its inversion may be represented by a polynomial in x:

P(x; k, n) = xp1 + xp2 + … + xpk-1 + xpk

P-1(x; k, n) = P(x-1; k, n) = x-p1 + x-p2 + … + x-pk-1 + x-pk

where, without loss of generality, 0 ≤ p1 < p2 < … pk-1 < pk < n are the k pitch classes in the set and where the consequent interval distribution between those pairs of pitch classes is measured as the coefficients of the k(k-1) powers of x in the expression P(x; k, n)P-1(x; k, n). Note that all exponents of x are taken modulo n. Thus x-p2, for example, may be rewritten as xn-p2 to regain positive exponents.

In our particular case we seek pitch class sets which carry exactly one instance of each interval from 1 to n-1. For this purpose we don't particularly care whether or not the set is in prime form (since we can always turn it into its prime form after we have found it) and so we might as well fix the first two pitch classes as 0 and 1 - or in other words have our sets be at least in normal form (with pitch class 0 at the beginning) and with the shortest interval of 1 - between those two pitch classes - right at the beginning of the set. Thus we seek those particular P(x; k, n) looking like:

1 + x + xp3 + … + xpk-1 + xpk

with 3 ≤ p3 < … pk-1 < pk < n. We can assume p3 > 2 since we've already accounted for the interval of 1 which would otherwise appear twice due to x2 and x. And of course k fixes n (as either the even k(k-1) or the odd k(k-1) + 1) because we're looking specifically for the resultant interval polynomial P(x; k, n)P(x-1; k, n)

= (1 + x + xp3 + … + xpk-1 + xpk)(1 + x-1 + x-p3 + … + x-pk-1 + x-pk)
= (1 + x-1 + x-p3 + … + x-pk-1 + x-pk) + (x + 1 + x1-p3 + … + x1-pk-1 + x1-pk)
+ (xp3 + xp3-1 + 1 + … + xp3-pk-1 + xp3-pk) + … + (xpk + xpk-1 + xpk-p3 + … + xpk-pk-1 + 1)

which we would wish to have equal k + x + x2 + x3 + x4 + … + x-3 + x-2 + x-1 by finding the right k-2 values for the remaining pi

We note that when k = 7, the consequent embedding tonality will be either 42 or 43. All-interval sets from the latter would, one supposes, be especially interesting to fans of Harry Partch.

Here are the solutions we get for k = 3 to 9. The pi are the pitch classes in the set and the iString column presents the interval string representation of the set (letters A-Z representing intervals of 10 to 35) from which one may readily recover prime forms by rotating the largest letters to the end of the string, and thence unpacking into the explicit set, if desired.

k = 3, n = 6 k = 3, n = 7
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
0 1 3   123 0 1 3   124
0 1 4   132 0 1 5   142
k = 4, n = 12 k = 4, n = 13
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
0 1 3 7   1245 0 1 3 9   1264
0 1 4 6   1326 0 1 4 6   1327
0 1 6 10   1542 0 1 5 11   1462
0 1 7 9   1623 0 1 8 10   1723
k = 5, n = 20 k = 5, n = 21
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
no solutions 0 1 4 14 16   13A25
  0 1 6 8 18   152A3
k = 6, n = 30 k = 6, n = 31
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
no solutions 0 1 3 8 12 18   12546D
  0 1 3 10 14 26   1274C5
  0 1 4 6 13 21   13278A
  0 1 4 10 12 17   13625E
  0 1 6 18 22 29   15C472
  0 1 8 11 13 17   17324E
  0 1 11 19 26 28   1A8723
  0 1 14 20 24 29   1D6452
  0 1 15 19 21 24   1E4237
  0 1 15 20 22 28   1E5263
k = 7, n = 42 k = 7, n = 43
no solutions no solutions
k = 8, n = 56 k = 8, n = 57
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
no solutions 0 1 3 13 32 36 43 52   12AJ4795
  0 1 4 9 20 22 34 51   135B2CH6
  0 1 4 12 14 30 37 52   1382G7F5
  0 1 5 7 17 35 38 49   142AI3B8
  0 1 5 27 34 37 43 45   14M7362C
  0 1 6 15 22 26 45 55   15974JA2
  0 1 6 21 28 44 46 54   15F7G283
  0 1 7 19 23 44 47 49   16C4L328
  0 1 7 24 36 38 49 54   16HC2B53
  0 1 9 11 14 35 39 51   1823L4C6
  0 1 9 20 23 41 51 53   18B3IA24
  0 1 13 15 21 24 31 53   1C2637M4
k = 9, n = 72 k = 9, n = 73
p1 p2 p3 p4 p5 p6 p7 p8 p9 iString
no solutions 0 1 3 7 15 31 36 54 63 1248G5I9A
  0 1 5 12 18 21 49 51 59 14763S28E
  0 1 7 11 35 48 51 53 65 164OD32C8
  0 1 9 21 23 26 39 63 67 18C23DO46
  0 1 11 20 38 43 59 67 71 1A9I5G842
  0 1 12 20 26 30 33 35 57 1B86432MG
  0 1 15 23 25 53 56 62 69 1E82S3674
  0 1 17 39 41 44 48 54 62 1GM23468B

Odd tonalities appear to be favoured systems for all-interval chords. Except for the absence of solutions for 43 - almost as if it had been singled out. Also, it has not escaped our notice that the above solution for 21 appears to violate the Prime Power Conjecture. Answers on a postcard, please, as to why it does not.

We can just squeeze out one more table for the 6 (inversional) pairs of all-interval sets using 10 pitch classes, in a 91 tonality (there are none in 90). This now breaks at least one 'limit' (albeit an artificial one of our own making), which is to say the one involved in displaying the PC set's interval string. The alphabet is no longer big enough to carry one of the intervals and consequently an asterisk (*) is employed to represent an interval of 36 - just over the range provided by a letter Z.

k = 10, n = 91
p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 iString
0139274956617781126IM75G4A
01413243038404573139B6825SI
0157273544677780142K89NA3B
0158182029435965143A29EG6Q
01610232634415355154D387C2*
01716275660687073169BT4823I
0111153136436583891A4G57MI62
0112152548576585871B3AN98K24
0119222432366576851I3284TB96
0119475254626879881IS5286B93
0127334963727484871Q6GE92A34
0137395158666982861*2C783D45

20180528

Unlucky 4 sum

From pitches …

C13 chord

A thirteenth chord is principally known as an all bells and whistles dominant which is expected to resolve to its tonic chord a fifth below. It usually turns up as a dominant seventh (say C7) topped off with its relative supertonic minor (which would be Dm) for something in F major.

If we ignore the key it's in, we see that this chord comprises seven pitches separated by - in sequence - 4, 3, 3, 4, 3, 4 semitones. A final 3 semitones would 'round it off' to the top keynote 24 semitones above the root note.

C13 chord

The chord's two-octave path is thus 4334343 - of length 7 and sum 24 (for a two octave span). We've got that eighth top note in parentheses to indicate that it's not intended to be included in the chord's 'definition' (the pitch class - of 0 - is already included).

C13 chord

Now let's look instead at a Cm13th chord, perhaps resolving to F minor. Its key-independent two-octave path is 3434343 (see below - again with a parenthesised closing note).

Cm13 chord
C13 chord

Finally let's consider a more adventurous Cmajor13#11 chord, with a Lydian flavour imparted by that sharpened 11th. Despite the F#, it can still resolve quite nicely to F. Its key-independent two-octave path is 4343433 (see below - again with a parenthesised closing note). This time, however, we're colour-coding the notes because we're going to switch things around a bit.

… to Pitch Class Sets

CM13#11 chord

To construct the pitch class set modelling this particular chord, we drop the second part of the chord (green note heads) by an octave - which essentially turns them into pitch classes alongside the first (red note headed) part.

pitch classing

We'll now slide the green pitch heads to the left, where we can see that they sit between the red pitch heads, and that none of the pitch classes are duplicated.

pitch class set

We now have a formal PC set, with the intervals - in semitones - between the PCs forming the key-independent interval path 2221221 (7 PCs in the set, summing correctly to 12).

Again, that final green-headed parenthesised note is not in the set - it's shown only to elucidate the final 'wraparound' step (of 1 semitone) back to pitch class 0.

The alert reader will note that the PC set resulting directly from CΔ13#11 is, in fact, the Lydian mode of the C Major (Ionian) scale. But that's not why we've seemingly drawn attention to its 'Lydianicity' by colouring the F# in blue. No - the real reason for drawing attention to the F# is because in order to get the prime form of this PC set, we have noticed that the set's largest interval skips (the three consecutive 2s) bring us to that F#. Since the prime form requires that the largest interval skips are to be placed at the end of the interval path, this means that the F# must become the prime form's pitch class zero (in other words, its first note). Accordingly (by transposing from C to F#) we obtain the prime form - with its interval path of 1221222 - shown below.

CΔ13#11 as an inversion of the F# Locrian mode (F#13♭5♭9♭13)
PC set in prime form

It's fairly easy to demonstrate that the PC sets which embody both the ordinary 13th and the minor 13th (with which we opened) are all exactly the same.

All of the C13ths discussed thus far 'prime form' (as Forte PC set 7-35 - the Locrian mode of the diatonic scale) - to some inversion of a Locrian, as the following two expositions show.

C13 as diatonic
C13 as an inversion of the E Locrian mode (Em13♭9♭13)
Cm13 as diatonic
Cm13 as an inversion of the A Locrian mode (Am13♭5♭9♭13)

Can this mean that all 13th chords are some inversion of the Locrian mode of the diatonic scale?

How many 13ths are there?

It seems reasonable to proceed with such an enumeration only if we have some formal definition of exactly what constitutes a 13th chord. One of the most obvious qualities of the chord would appear to be that it comprises 7 distinct pitch classes. Another would appear to be that they should be constructed by stacking up six successive major or minor thirds after the initial root note.

The interval paths of such chords will thus be formed from six 3s or 4s and terminated by whatever value would take the path sum up to 24. Six '3s or 4s', being the same as six '3 + (0s or 1s)', it's reasonably clear that 64 distinct interval paths - labelled from 333333, 333334, 333343, 333344, … to 444433, 444434, 444443, 444444 will capture all possibilities.

It's also easy to see that any path containing either three consecutive 4s or four consecutive 3s must be rejected since the pitch classes on either side of such jumps are bound to be exactly an octave apart, thus being the same pitch class and violating the principle of distinctness. There are many other routes, within this procedure, which result in the duplication of pitch classes.

It turns out, therefore, that - after filtering out all of the interval paths which would result in pitch class duplication - only 28 ways of stacking major and minor thirds to build up some kind of 7 note 13th chord remain. These are, in ascending order of minor-major-thirdiness:

33343441212213343334412131223443343212222143433342212131
33344341221213343343412221223443433213122143433432212221
33433441212222343344312222124333434131212243434332221221
33434341221222343433421221224333443131221243443333121221
33434431221312343434321222124334334221212244333432213121
33443342121222343443321312124334343221221244334332222121
33443432121312344333421221314334433222121244343333122121

Also shown, in red to the right of each 3|4 construction, is the interval path signature of the PC set which contains the 7 distinct PCs of the resulting chord. These are - of course - in normal form because interval path signatures give you that for nothing. A quick inspection should demonstrate that these are by no means all just variations of the seven modes of the diatonic scale. The presence of 3s in such signatures is enough to disabuse one of such notions. There are also several with runs of four 2s.

By rotating all 28 of the (red) PC set signatures into their prime forms (with their largest skips packing to the right, as per Forte), we quickly uncover the fact that there are four distinct PC sets in play here:

Four Prime Form PC Sets
1212213121222212212131221222
7-32A7-347-32B7-35
<3,3,5,4,4,2><2,5,4,4,4,2><3,3,5,4,4,2><2,5,4,3,6,1>
Harmonic MinorLocrian SuperIndian/EthiopianLocrian Diatonic
33343441212213334334412122223334434122121333434341221222
33434431221312334433421212223344343212131234334341222122
34344332131212343344312222123433344121312234343342122122
34433342122131344334321222213443433213122134343432122212
43334341312122433433422121224333443131221243343432212212
43443333121221433443322212124343334221213143433432212221
44333432213121443343322221214434333312212143434332221221

They occur in four groups of seven arrangements - the seven modes of each of their common prime PC sets. The first (which includes the Harmonic Minor scale, 2122131, in its fourth row) and third (including Indian and Ethiopian scales in its third and sixth rows) column pairs are asymmetric PC sets (inverses of each other). The second and fourth are both symmetric PC sets, the first being the prime form of the half-diminished scale (arguably the next most popular heptatonic division of the octave, encompassing as it does the hindi, melodic minor, overtone, javanese, and both locrian natural and super scales). The fourth is the various rotations (i.e. modes) of PC Set 7-35, i.e. Locrian-Diatonic. The three types of 13th chords we dealt with above are in boldface.

Relationships between 13ths and the Diatonic Modes

The 13th chord patterns in the fourth, diatonic, group are - starting from the top, and not based on any particular scale

  • 3343434 ≡ m13♭5♭9♭13 ['Locrian 13th']
  • 3433434 ≡ m13♭9♭13 ['Phrygian 13th']
  • 3434334 ≡ m13♭13 ['Aeolian 13th']
  • 3434343 ≡ m13 ['Dorian 13th']
  • 4334343 ≡ 13 ['Mixolydian 13th' = the standard 'dominant 7th' mode]
  • 4343343 ≡ Δ13 ['Ionian 13th' = the major 7th dominant series]
  • 4343433 ≡ Δ13#11 ['Lydian 13th']

But naturally, upon actual transcription, one must commit to a key - say C:

the Modal bases for 7 diatonic 13ths

Stepping away from the diatonic 13ths, if we based a 13th chord on the Ethiopian scale (row 6 column 3) - with its PC Set interval path signature 2212131 - we would generate a major/minor 3rd stacking of 4343334 ≡ Δ13♭13. The Indian scale (1213122 → 3433344) would yield a distinctly weird m13♭9♭11♭13. We feel reasonably certain that there will be a circumstance where every one of these 28 possible 13ths will sound fantastic.

Squeezed 'Thirteenths'

The 28 13ths above are constrained to be contained within 3 or 4 semitones of a double octave span. As such, their top notes will always be a 'true' 13th, possibly flattened. The notes within the chord are not subject to undue 'stress' and the 7ths, 9ths and 11ths turn up in their expected places - perhaps occasionally bumped sideways as flattened or sharpened creatures as it were. However if this 'thirteenth pegging' is relaxed, there are - technically speaking - eight further 13ths. The first two are rather 'squashed' as the top notes are 5 semitones down from a double octave, giving us a double-flattened 13th - which a musician will consider illegitimate (a double-flattened 13th being - enharmonically - just a plain old 5th, the chord's 'internal' 5th already being flattened).

333433
m13♭5♭♭7♭9♭11♭♭13
e.g. Cm13♭5♭♭7♭9♭11♭♭13 = C–E♭–G♭–A–D♭–E–G
334333
m13♭5♭9♭11♭♭13
e.g. Cm13♭5♭9♭11♭♭13 = C–E♭–G♭–B♭–D♭–E–G

We note that the initial four notes of the first form a full-diminished chord. It thereby already contains the 13th as a pitch class, albeit an octave lower (so not actually a 13th but a 6th). It's as if the chord has been put under so much compressive force that the 9th, 11th, 13th (and even the 7th) begin to crash into each other (as pitch classes). In both cases, the top flat 11th and double-flat 13th are effectively the major 3rd and major 5th of the tonic. The flat nine is really the only 'novelty' in these chords and to call these chords 13ths is rather stretching a point (actually the opposite - it's compressing a point). As pitch class sets, these are mutual inverses, the first being characterised as an interval path signature 1212123 (Forte's 7-31A) and the second (its inverse, 7-31B) as 1212132 (which would of course 'prime form path' as 2121213), both sharing interval vector <3,3,6,3,3,3>, 'maxing out' with their 6 minor thirds.

Stretched 'Thirteenths'

The remaining 6 chords are - in contrast - stretched, and their top notes are only two semitones below the double octave. Thus a dominant 7thness turns up at the top end, rather than in the middle, of the chord (where the internal 7thnesses are all Δ, i.e. major 7ths). The first couple of these are the antisymmetric pair (path signatures of 2131311 and 3131211, corresponding to prime form signatures of 1121313 (Forte 7-21A) and 1211313 (Forte 7-21B) respectively, sharing interval vector <4,2,4,6,4,1> with maximal major-thirdy content.

344344
mΔ13#11#13
e.g. CmΔ13#11#13 = C–E♭–G–B–D–F#–A#
443443
+Δ13#9##11#13
e.g. C+Δ13#9##11#13 = C–E–G#–B–D#–G–A#

In fact by now it's becoming rather difficult to 'spell' these as 13th chords, since the pitches tread on each others' toes so much. The above '+Δ13#9##11#13' is pure guesswork on this author's part.

The second pair of 'stretchy' 13ths have path signatures of 2221311 (prime path 1122213 ≡ Forte 7-30A) and 3122211 (prime path 1222113 ≡ Forte 7-30B) - again mutual inverses - with interval content <3,4,3,5,4,2>

434344
Δ13#11#13
e.g. CΔ13#11#13 = C–E–G–B–D–F#–A#
443434
+Δ13#9#11#13
e.g. C+Δ13#9#11#13 = C–E–G#–B–D#–F#–A#

There remain two further '13ths'. They each represent symmetric (self-inverting) PC sets, the first with a path signature of 3121311 (prime form path as 1131213 ≡ Forte 7-22, interval content <4,2,4,5,4,2>) and the second with path signature 2222211 (prime form path as 1122222 ≡ Forte 7-33, interval content <2,6,2,6,2,3> replete with major 2nds and major 3rds).

434434
Δ13#9#11#13
e.g. CΔ13#9#11#13 = C–E–G–B–D#–F#–A#
443344
+Δ13#11#13
e.g. C+Δ13#11#13 = C–E–G#–B–D–F#–A#

Terminological Conclusions

Are any of these eight stretched or squeezed chords really 13ths? Insofar as they comprise heptachords they are. Insofar as their construction involved upward skips of only major and minor thirds they are. But musically, they just aren't. They have too much internal compression or tension. This is due to not 'nailing down' - for the want of a better term - a (possibly flattened) 13th at the top of the runs of thirds and leaving the end of the chord flapping around in the breeze (as David Jones would say). Such freedom permits an absence of what one might expect of 'thirteenthness'.

So the 28 unstretched and unsqueezed 13ths above, distributed between only four PC Sets, are all (assuming fully populated ones) there really are. We note that this means there are only three distinct interval contents (two being identical since non-symmetric PC Sets' forms - A and B - always share a common interval vector) available to carry the various flavours of all 28. These are <3,3,5,4,4,2>, <2,5,4,4,4,2>, and <2,5,4,3,6,1>.

Interval classwise, then, we have that a 13th contains either 2 or 3 minor 2nds [or major 7ths], with - correspondingly - 5 or 3 major 2nds [minor 7ths] and (also correspondingly) 5 or 4 minor 3rds [major 6ths]. We also have that if the 13th contains only a single tritone then it must also contain the maximum number of perfect 4ths [or perfect 5ths] possible within a heptachord, i.e. 6, and must also contain 3 major 3rds [or minor 6ths]. Otherwise a 13th must contain 2 tritones and 4 each of major 3rds and perfect 4ths [or minor 6ths and perfect 5ths].

This means that if you elect a 13th chord constrained to containing only a single tritone then all else follows, i.e you cannot help but have 2 m2/M7, 5 M2/m7, 4m3/M6, 5 M3/m6 and 6 P4/P5, which further means that you have elected one of the 7 diatonic 13ths. If, however, you choose one with two tritones (the only remaining kind) then you have forced it to contain 4 P4/P5 and 4 M3/m6 and at least 4 m3/M6 and 2 m2/M7. The only choice you have left is a chord with 5 m3/M6, 3 M2/m7 and 3 m2/M7 or one with only 4 m3/M6 and 5 M2/m7 and 2 m2/M7.

It would appear, therefore, that there are - at most - two degrees of freedom available to anyone constructing a 13th chord from its interval content alone. Is that interesting or is that interesting?