Neutral and interordinal intervals in MOS scales

This page assumes that the reader is familiar with TAMNAMS mos interval naming.

Given a tuning of a primitive (single-period) mos pattern aLbs<E> with a > b and arbitrary equave E, we may define two types of notes "in the cracks of" interval categories defined by aLbs<E>:

1. Given 1 ≤ k ≤ a + b - 1, the neutral k-step (abbrev. nks) is the interval exactly halfway between the smaller k-step and the larger k-step of the mos. When the mos is generated by a (perfect) k-step, this may instead be called the semiperfect k-step (abbrev. sPks), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) k-step.
2. Given 1 ≤ k ≤ a + b − 2, and assuming that the larger k-step < the smaller (k + 1)-step, the interordinal between k-steps and (k + 1)-steps, denoted kx(k + 1)s or kX(k + 1)s (read "k cross (k + 1) step"), is the interval exactly halfway between the larger k-step and the smaller (k + 1)-step. The name comes from the fact that k-steps in the diatonic mos are conventionally called "(k + 1)ths".

Given such a mos, it's easy to observe the following properties of the simplest equal tunings for the mos, due to the way they divide the small step (s) and the chroma (L − s):

• The basic equal tuning (2a + b)-ed contains neither neutrals nor interordinals.
• The monohard equal tuning (3a + b)-ed contains neutrals of that mos but not interordinals.
• The monosoft equal tuning (3a + 2b)-ed contains interordinals but not neutrals.
• 2(2a + b)-ed, twice the basic equal tuning, contains both types of intervals.

Examples

Diatonic (5L2s)

 Basic 5L2s (diatonic, dia-): 12edo
 Parent mos: soft 2L5s (pentic, pt-)
 1\24
 2\24 m1dias
 3\24 n1dias
 4\24 M1dias    == m1pts
 5\24 1X2dias   == n1pts
 6\24 m2dias    == M1pts
 7\24 n2dias    == 1x2pts
 8\24 M2dias    == d2pts
 9\24 2X3dias   == sP2pts
10\24 P3dias    == P2pts
11\24 sP3dias
12\24 A3/d4dias == 2x3pts
13\24 sP4dias
14\24 P4dias    == P3pts
15\24 4X5dias   == sP3pts
16\24 m5dias    == A3pts
17\24 n5dias    == 3x4pts
18\24 M5dias    == m4pts
19\24 5X6dias   == n4pts
20\24 m6dias    == M4pts
21\24 n6dias
22\24 M6dias
23\24

m-chromatic (7L5s)

Basic 7L5s (m-chromatic, mchr-): 19edo
Parent mos 5L2s (diatonic, dia-)
 1\38
 2\38 m1s
 3\38 n1s
 4\38 M1s      == m1dias
 5\38 1x2s     == n1dias
 6\38 m2s      == M1dias
 7\38 n2s
 8\38 M2s/m3s  == 1x2dias
 9\38 n3s
10\38 M3s      == m2dias
11\38 3x4s     == n2dias
12\38 m4s      == M2dias
13\38 n4s
14\38 M4s/d5s  == 2x3dias
15\38 sPs
16\38 P5s      == P3dias
17\38 5x6s     == sP3dias
18\38 m6s      == A3dias
19\38 n6s      == 3x4dias
20\38 M6s      == d4dias
21\38 6x7s     == sP4dias
22\38 P7s      == P4dias
23\38 sP7s
24\38 A7s/m8s  == 4x5dias
25\38 n8s
26\38 M8s      == m5dias
27\38 8x9s     == n5dias
28\38 m9s      == M5dias
29\38 n9s
30\38 M9s/m10s == 5x6dias
31\38 n10s 
32\38 M10s     == m6dias
33\38 10x11s   == n6dias
34\38 m11s     == M6dias
35\38 n11s
36\38 M11s
37\38

Manual (4L1s)

 Basic 2/1-equivalent 4L1s (manual, man-): 9edo
 Parent mos: soft 1L3s (antetric, att-)
 1\18
 2\18 d1mans
 3\18 sP1mans
 4\18 P1mans  == P1atts
 5\18 1x2mans == sP1atts
 6\18 m2mans  == A1atts
 7\18 n2mans  == 1x2atts
 8\18 M2mans  == m2atts
 9\18 2x3mans == n2atts
10\18 m3mans  == M2atts
11\18 n3mans  == 2x3atts
12\18 M3mans  == d3atts
13\18 3x4mans == sP3atts
14\18 P4mans  == P3atts
15\18 sP4mans
16\18 A4mans 
17\18

Interordinal Theorem

The Interordinal Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent mos, bL(a − b)s.

Statement

Suppose a > b and gcd(a, b) = 1.

1. Every interordinal in basic aLbs<E> (an interval that is exactly halfway between the larger k-step and the smaller (k+1)-step) is a neutral or semiperfect interval in the parent mos bL(a-b)s<E>.
2. Every interordinal interval in the parent mos bL(a-b)s<E> of basic aLbs<E> is a neutral or semiperfect interval in basic aLbs<E>. The number (b - 1) counts the places in 2(2a+b)edE (twice the basic mos tuning for aLbs<E>) where the parent's interordinal is two steps away from the ordinal categories.

Preliminaries for the proof

Below we assume that the equave is 2/1, but the proof generalizes to any equave.

Consider a primitive mos aLbs. Recall that (b - 1) satisfies:

(b - 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)| = #{k : 0 < k < a+b and larger k-step of basic aLbs = smaller (k+1)-step of basic aLbs}.

Also recall that the following are equivalent for a mos aLbs:

• a > b.
• The parent mos, which is bL(a-b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a-b)s has hardness 3/2 thus is strictly proper.

To prove the theorem, we need a couple lemmas.

Lemma 1

Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n+k)x) - floor(nx) ≥ floor(kx).

Proof

floor((n+k)x) - floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) ≥ ceil((n+k)x - nx) - 1 = ceil(kx) - 1 = floor(kx).

Discretizing Lemma

Consider an m-note maximally even mos of an n-equal division, and let 1 ≤ k ≤ m-1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n.

Proof

The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'<equave> or ceil(n'/m')\n'<equave>, implying the lemma.

Proof of Theorem

Let n = 2a + b (the basic edo tuning of aLbs) and suppose that m\(2n) is an interordinal (where m must be odd) between k-steps and (k+1)-steps, denoted kX(k+1)ms. Recall that:

• In basic aLbs, s = 1\n = 2\2n.
• A concrete mos tuning is improper iff its hardness is > 2/1 and the number of s steps it has is > 1.

Part (1) takes some step size arithmetic:

• Larger k+1-step of aLbs minus larger k-step of aLbs = Smaller k+1-step of aLbs minus smaller k-step of aLbs must be L, if 0 < k < k+1 < a+b. The reason is that larger 1-step = L, larger 2-step = LL, because there are more L’s than s’s.
• Smaller k+1-step of aLbs minus larger k-step of aLbs ≥ 0, with =0 at improprieties. At the values of k and k+1 that are proper, this equals s.
  • To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
    • X = Larger (k+1)-step = (i+2)L + (k-i-1)s
    • Smaller (k+1)-step = (i+1)L + (k-i)s
    • Larger k-step = iL + (k-i)s
    • Y = Smaller k-step = (i-1)L + (k-i+1)s
  • Since the smaller k-step has two more s steps than the larger (k+1)-step, Y has two more complete chunks of L's than X:

    • Y=L^A sL...LsL...LsL...LsL...LsL^B

    • X=L^CsL...LsL...LsL^D

  • Let r be the number of complete chunks (maximal contiguous strings of L's) in X, and let μ = n/(a+b). Note that the spacings between the s steps of a mos form a maximally even mos, and that since a > b, 3/2 < μ < 2. By the Discretizing Lemma, we have (|·| denotes length):
    • 1+A+B+floor((r+2)μ) ≤ |Y| ≤ 1+A+B+ceil((r+2)μ)
    • 1+C+D+floor(rμ) ≤ |X| ≤ 1+C+D+ceil(rμ)
    • -1 = |Y|-|X| ≥ (A+B)-(C+D) + floor((r+2)μ) - ceil(rμ)
    • = (A+B)-(C+D)-1 + floor((r+2)μ) - floor(rμ)
    • (Lemma 1) ≥ (A+B)-(C+D)-1 + floor(2μ)
    • Hence, (C+D)-(A+B) ≥ floor(2μ).
    • Also, (C+D)-(A+B) ≤ 2*ceil(μ)-2 = 2(ceil(μ)-1) = 2*floor(μ). This is a contradiction: as 3/2 < μ < 2, we have floor(2μ) - 2*floor(μ) = 1.
• As s is the chroma of bL(a-b)s, it *would* be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so kX(k+1) would become neutral or semiperfect.
• To show that these actually occur in bL(a-b)s, consider smaller and larger j-steps ( 1 ≤ j ≤ a-1) in the parent mos. These intervals also occur in the mos aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a-1. Note that we are considering “junctures” between k-steps and k+1-steps in aLbs, excluding k = 0 and k = a+b-1, so the total number of “junctures” to consider is a+b-2. This proves part (1).

Part (2) is easier to see: where basic aLbs is improper, larger k-step = smaller k+1-step, and larger k+1-step - smaller k-step = 2(L-s) = 2s = L. But the step L is 4 steps in 2n-edo.