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> 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>.
3. 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, instead of one step away, from each of the adjacent 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} = # of "improprieties".

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) and (2) take 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 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. These j's correspond to values of k such that larger k-step < smaller k+1-step. 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 finite, namely a+b-2. This proves parts (1) and (2).

Part (3) is also immediate now: when larger k-step = smaller k+1-step, larger k+1-step - smaller k-step = 2(L-s) = 2s = L. The step L is 4 steps in 2n-edo.