Neutral and interordinal intervals in MOS scales: Difference between revisions

Inthar (talk | contribs)
No edit summary
Tags: Mobile edit Mobile web edit
Inthar (talk | contribs)
No edit summary
 
(35 intermediate revisions by the same user not shown)
Line 1: Line 1:
: ''This page assumes that the reader is familiar with [[TAMNAMS]] mos interval and step ratio names.''
: ''This page assumes that the reader is familiar with [[TAMNAMS]] MOS interval and step ratio names.''


Given a tuning of a primitive (i.e. single-period) [[mos]] pattern aLbs{{angbr|E}} with arbitrary [[equave]] E in a specific tuning (i.e. with a specific [[hardness]] value for L/s), we may define two types of notes "in the cracks of" interval categories defined by aLbs{{angbr|E}}:
Given a tuning of a primitive (i.e. single-period) [[MOS]] pattern aLbs{{angbr|E}} with arbitrary [[equave]] E in a specific tuning (i.e. with a specific [[hardness]] value for L/s), we may define two types of notes "in the cracks of" interval categories defined by aLbs{{angbr|E}}:
# Given 1 ≤ ''k'' ≤ a + b − 1, the '''neutral''' ''k''-mosstep or ''k''-step (abbrev. n''k''ms, n''k''s) 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. sP''k''s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) ''k''-step. Depending on whether the imperfect generator is augmented or diminished, the corresponding semiperfect generator may be called '''semiaugmented''' or '''semidiminished'''.
# Given 1 ≤ ''k'' ≤ a + b − 1, the '''neutral''' ''k''-mosstep or ''k''-step (abbrev. n''k''ms, n''k''s) 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. sP''k''s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark) ''k''-step. Depending on whether the imperfect generator is augmented or diminished, the corresponding semiperfect generator may be called '''semiaugmented''' or '''semidiminished'''.  
#: The following ''always'' holds for a given interval class of ''k''-steps between 1-steps and (a + b − 1)-steps, inclusive:
#: We use '''semichroma''' for the quantity c/2 where c = L − s is the [[chroma]] of the MOS. The semichroma represents the difference between ordinary MOS intervals and their neutralized counterparts, in the sense that the following holds for a given interval class of ''k''-steps between 1-steps and (a + b − 1)-steps, inclusive:
#: neutral ''k''-step = smaller ''k''-step + c/2 = larger ''k''-step − c/2, where c = L − s is the [[chroma]] of the mos.
#: neutral ''k''-step = smaller ''k''-step + c/2 = larger ''k''-step − c/2
# Given 1 ≤ ''k'' ≤ a + b &minus; 2, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''×(''k'' + 1)(m)s (read "''k'' cross (''k'' + 1) (mos)step" or "''k'' inter (''k'' + 1) (mos)step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step.
# Given 0 ≤ ''k'' ≤ a + b &minus; 1, and assuming that the larger ''k''-step < the smaller (''k'' + 1)-step, the '''interordinal''' between ''k''-steps and (''k'' + 1)-steps, denoted ''k''×(''k'' + 1)(m)s (read "''k'' cross (''k'' + 1) (mos)step" or "''k'' inter (''k'' + 1) (mos)step"), is the interval exactly halfway between the larger ''k''-step and the smaller (''k'' + 1)-step.
#: If the smaller (''k'' + 1)-step is ''strictly larger'' than the larger ''k''-step in ''basic'' aLbs, ''k''×(''k'' + 1) is called a '''proper interordinal'''. If a > b, then aLbs{{angbr|E}} has b − 1 proper interordinals. The following holds for any proper interordinal interval ''k''-inter-(''k'' + 1)-step:
#: If the smaller (''k'' + 1)-step is ''strictly larger'' than the larger ''k''-step in ''basic'' aLbs, ''k''×(''k'' + 1) is called a '''proper interordinal'''; otherwise, it is called an '''improper interordinal'''. If a > b, then aLbs{{angbr|E}} has a + 1 proper interordinals, including 0×1ms and (a+b&minus;1)×(a+b)ms.
#: We call s/2 (or 0×1ms) the '''interizer'''{{idiosyncratic}}. The interizer is of note since the following holds for any proper interordinal interval ''k''-inter-(''k'' + 1)-step:
#: ''k''-inter-(''k'' + 1)-step = larger ''k''-step + s/2 = smaller (''k'' + 1)-step &minus; s/2.
#: ''k''-inter-(''k'' + 1)-step = larger ''k''-step + s/2 = smaller (''k'' + 1)-step &minus; s/2.
Neutral ''k''-steps generalize neutral interval categories based on the diatonic mos, which are:
Neutral ''k''-steps generalize neutral interval categories based on the diatonic MOS, which are:
* neutral 1-diastep = neutral 2nd (A-Bd)
* neutral 1-diastep = neutral 2nd (A-(Bd)
* neutral 2-diastep = neutral 3rd (A-Ct)
* neutral 2-diastep = neutral 3rd (A-Ct)
* semiperfect 3-diastep = semiperfect 4th (A-Dt)
* semiperfect 3-diastep = semiperfect 4th (A-Dt)
Line 16: Line 17:
* neutral 6-diastep = neutral 7th (A-Gt)
* neutral 6-diastep = neutral 7th (A-Gt)


Though the term ''interordinal'' is intended to be JI-agnostic and generalizable to non-diatonic mosses, the term comes from the fact that ''k''-steps in the diatonic mos are conventionally called "(''k'' + 1)ths". Proper interordinals in other mosses generalize diatonic [[interordinal]]s (aka "interseptimals"), which are:
Though the term ''interordinal'' is intended to be JI-agnostic and generalizable to non-diatonic mosses, the term comes from the fact that ''k''-steps in the diatonic MOS are conventionally called "(''k'' + 1)ths". Proper interordinals in other mosses generalize [[interordinal]] categories that are novel with respect to diatonic (aka "interseptimals"), which are:
* 0-inter-1-diastep = "unison-inter-2nd" = s/2
* 1-inter-2-diastep = "2nd-inter-3rd" = semifourth = chthonic = ultramajor 2nd
* 1-inter-2-diastep = "2nd-inter-3rd" = semifourth = chthonic = ultramajor 2nd
* 2-inter-3-diastep = "3rd-inter-4th" = semisixth = naiadic = ultramajor 3rd
* 2-inter-3-diastep = "3rd-inter-4th" = semisixth = naiadic = ultramajor 3rd
* 4-inter-5-diastep = "5th-inter-6th" = semitenth = cocytic = inframinor 6th
* 4-inter-5-diastep = "5th-inter-6th" = semitenth = cocytic = inframinor 6th
* 5-inter-6-diastep = "6th-inter-7th" = semitwelfth = ouranic = inframinor 7th
* 5-inter-6-diastep = "6th-inter-7th" = semitwelfth = ouranic = inframinor 7th
* 6-inter-7-diastep = "7th-inter-octave" = octave &minus; s/2
Improper interordinals, in contrast, represent intervals that are technically between ordinal categories but occur within the MOS scale unlike proper interordinals which are wholly outside the interval categories defined by the MOS. The diatonic example of this is the tritone, which is interordinal but falls within diatonic interval categories as the [[12edo|basic tuning]] of diatonic tunes both the augmented 3-diastep and the diminished 4-diastep to 600 cents.


Given a primitive mos aLbs with a > b, 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 (c = L &minus; s). Note that s separates adjacent ordinal categories (i.e. [[interval class]]es) while c separates larger and smaller intervals in the same ordinal category.
Given a primitive MOS aLbs with a > b, one can 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 (c = L &minus; s). Note that s separates adjacent ordinal categories (i.e. [[interval class]]es) while c separates larger and smaller intervals in the same ordinal category.
* The basic equal tuning (2a + b)-edE contains neither neutrals nor interordinals, since both s and c are one edo step. (For the diatonic mos 5L2s, this tuning is [[12edo]].)
* The basic equal tuning (2a + b)-edE contains neither neutrals nor interordinals, since both s and c are one edo step. (For the diatonic MOS 5L2s, this tuning is [[12edo]].)
* The monohard equal tuning (3a + b)-edE contains neutrals of that mos but not interordinals, since c is two edo steps but s is one edo step. (For diatonic, this tuning is [[17edo]].)
* The monohard equal tuning (3a + b)-edE contains neutrals of that MOS but not interordinals, since c is two edo steps but s is one edo step. (For diatonic, this tuning is [[17edo]].)
* The monosoft equal tuning (3a + 2b)-edE contains interordinals but not neutrals, since c is one edo step and s is two edo steps. (For diatonic, this tuning is [[19edo]].)
* The monosoft equal tuning (3a + 2b)-edE contains interordinals but not neutrals, since c is one edo step and s is two edo steps. (For diatonic, this tuning is [[19edo]].)
* 2(2a + b)-edE, twice the basic equal tuning, contains both types of intervals, since both c and s are two edo steps. (For diatonic, this tuning is [[24edo]].)
* 2(2a + b)-edE, twice the basic equal tuning, contains both types of intervals, since both c and s are two edo steps. (For diatonic, this tuning is [[24edo]].)


== Examples ==
== Examples ==
These charts show where neutrals and interordinals corresponding to example 2/1-equivalent mosses are located in 2(2a+b)-edo (twice the basic edo tuning). Intervals larger than 600c have been omitted, as the structure is symmetrical about 600c.
These charts show where neutrals and interordinals corresponding to example 2/1-equivalent mosses are located in 2(2a + b)-edo (twice the basic edo tuning). Intervals larger than 600c have been omitted, as the structure is symmetrical about 600c.
=== Diatonic (5L2s) ===
=== Diatonic (5L2s) ===
<pre>
<pre>
  Basic 5L2s (diatonic, dia-): 12edo
  Basic 5L2s (diatonic, dia-): 12edo
  Parent mos: soft 2L3s (pentic, pt-)
  Parent MOS: soft 2L3s (pentic, pt-)
  1\24
  1\24 0×1dias  (1st×2nd)
  2\24 m1dias    (m2nd)
  2\24 m1dias    (m2nd)
  3\24 n1dias    (n2nd)
  3\24 n1dias    (n2nd)
Line 51: Line 55:
<pre>
<pre>
Basic 7L5s (m-chromatic, mchr-): 19edo
Basic 7L5s (m-chromatic, mchr-): 19edo
Parent mos: soft 5L2s (diatonic, dia-)
Parent MOS: soft 5L2s (diatonic, dia-)
  1\38
  1\38 0×1s
  2\38 m1s
  2\38 m1s
  3\38 n1s
  3\38 n1s
Line 76: Line 80:
<pre>
<pre>
  Basic 2/1-equivalent 4L1s (manual, man-): 9edo
  Basic 2/1-equivalent 4L1s (manual, man-): 9edo
  Parent mos: soft 1L3s (antetric, att-)
  Parent MOS: soft 1L3s (antetric, att-)
  1\18
  1\18 0×1mans
  2\18 d1mans
  2\18 d1mans
  3\18 sP1mans
  3\18 sP1mans
Line 90: Line 94:
<pre>
<pre>
  Basic 5L3s (oneirotonic, onei-): 13edo
  Basic 5L3s (oneirotonic, onei-): 13edo
  Parent mos: soft 3L2s (anpentic, apt-)
  Parent MOS: soft 3L2s (anpentic, apt-)
  1\26
  1\26 0×1oneis
  2\26 m1oneis
  2\26 m1oneis
  3\26 n1oneis
  3\26 n1oneis
Line 106: Line 110:
</pre>
</pre>


== Interordinal Theorem ==
== Interordinal-Neutral 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 &minus; b)s.
The Interordinal-Neutral Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent MOS, bL(a &minus; b)s, generalizing an observation by [[User:Godtone]] relating neutrals and interordinals of [[5L 2s]] to those of the parent MOS [[2L 3s]].
=== Statement ===
=== Statement ===
Suppose a > b and gcd(a, b) = 1.
Suppose a > b and gcd(a, b) = 1.
# Every interordinal of basic aLbs{{angbr|E}} is a neutral or semiperfect interval of the parent mos bL(a-b)s{{angbr|E}}.
# Every proper interordinal of basic aLbs{{angbr|E}} save for 0×1ms and (a+b&minus;1)×(a+b)ms is a neutral or semiperfect interval of the parent MOS bL(a &minus; b)s{{angbr|E}}. The interizer of aLbs{{angbr|E}}, 0x1ms = s/2, is the semichroma of the parent MOS.  
# Every interordinal interval of the parent mos bL(a-b)s{{angbr|E}} of basic aLbs{{angbr|E}} is a neutral or semiperfect interval of basic aLbs{{angbr|E}}.  
# Every interordinal interval of the parent MOS bL(a &minus; b)s{{angbr|E}} of basic aLbs{{angbr|E}} excluding 0×1ms and (a+b&minus;1)×(a+b)ms is a neutral or semiperfect interval of basic aLbs{{angbr|E}}.  
# Except the neutral/semiperfect 1-step and the neutral/semiperfect (a+b-1)-step, every neutral or semiperfect interval of basic aLbs{{angbr|E}} is a proper interordinal of bL(a-b)s{{angbr|E}}. The number (b &minus; 1) counts the places in 2(2a+b)edE (twice the basic mos tuning for aLbs{{angbr|E}}) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.
# Except the neutral/semiperfect 1-step and the neutral/semiperfect (a + b &minus; 1)-step, every neutral or semiperfect interval of basic aLbs{{angbr|E}} is a proper interordinal of bL(a &minus; b)s{{angbr|E}}. The number (b &minus; 1) counts the places in 2(2a + b)edE (twice the basic MOS tuning for aLbs{{angbr|E}}) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.
# aLbs{{angbr|E}} has a + 1 proper interordinals and b &minus; 1 improper interordinals.


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


Consider a primitive mos aLbs. Recall that (b &minus; 1) satisfies:
Consider a primitive MOS aLbs. Recall that (b &minus; 1) satisfies:


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


Also recall that the following are equivalent for a mos aLbs:
where ''potential improprieties'' are pairs of adjacent interval classes that witness the impropriety of a hard-of-basic tuning of the MOS. Part (4) immediately follows.
 
Also recall that the following are equivalent for a MOS aLbs:
* a > b.
* 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.
* The parent MOS, which is bL(a&minus;b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a&minus;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) &minus; floor(nx) ≥ floor(kx).
==== Proof ====
floor((n+k)x) &minus; floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) ≥ ceil((n+k)x &minus; nx) &minus; 1 = ceil(kx) &minus; 1 = floor(kx).
 
=== Discretizing Lemma ===
Consider an m-note [[maximal evenness|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 ===
Finally, recall that:
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 k×(k+1)ms. Recall that:
* In basic aLbs, s = 1\n = 2\2n.  
* 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.
* A concrete MOS tuning is improper if and only if its hardness is > 2/1 and the number of s steps it has is > 1.


Part (1) and (2) take some step size arithmetic:
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 k×(k + 1)ms. For parts (1) and (2):
* Smaller k+1-step of aLbs minus larger k-step of aLbs ≥ 0, with equality at improprieties (essentially by definition). At the values of k and k+1 that are proper, this equals s.  
* (Note that we are assuming the basic tuning of the MOS) smaller (k + 1)-step of aLbs minus larger k-step of aLbs ≥ 0, with equality at improprieties (essentially by definition). At the values of k and k+1 that are proper, this equals s. To see why, observe that the number of generators represented by the difference must be b mod (a+b), since L is obtained by stacking b bright generators. The difference is at most &minus;1 generators, since the larger interval of the difference between the larger k-step and the larger k+1 step, a chroma sharper, does occur (in the brightest mode of the MOS), and cannot be less than &minus;a generators, lest the gap be nonpositive in the basic MOS, a contradiction since kx(k+1) is a proper interordinal. Hence the difference must be &minus;a generators, corresponding to s.
** To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
* As s is the chroma of bL(a &minus; 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 k×(k + 1) would become neutral or semiperfect.
*** X = Larger (k+1)-step = (i+2)L + (k-i-1)s
* To show that these actually occur in bL(a &minus; b)s, consider smaller and larger j-steps (1 ≤ j ≤ a &minus; 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 &minus; 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 &minus; 1, so the total number of “junctures” to consider is finite, namely a + b &minus; 2. This proves parts (1) and (2).
*** 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:
*** <pre>Y=L^A sL...LsL...LsL...LsL...LsL^B</pre>
*** <pre>X=L^CsL...LsL...LsL^D</pre>
** 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) &minus; ceil(rμ)
*** = (A+B)-(C+D)-1 + floor((r+2)μ) &minus; 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) &minus; 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 k×(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 &minus; smaller k-step = 2(L-s) = 2s = L. The step L is 4 steps in 2n-edo.
Part (3) is also immediate now: when larger k-step = smaller (k + 1)-step, larger (k + 1)-step &minus; smaller k-step = 2(L &minus; s) = 2s = L. The step L is 4 steps in 2n-edo. {{qed}}
=== Corollary ===
If a > b, then aLbs{{angbr|E}} has b &minus; 1 proper interordinals.


[[Category:MOS scale]]
[[Category:MOS scale]]
[[Category:Pages with proofs]]