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-1-|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. 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

Jedrezejewski 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