Golden sequences and tuning: Difference between revisions

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+[[File:GoldenMOS.png|thumb|555x555px|A visualization of the golden generators. Each cell is a MOS, containing its step formula on the top and its generator's step formula on the bottom. For visual clarity, they have been organized into eight branches, the colored squares on the right of which contain the tunings of the generators in cents.]]
 Golden sequences (the generalization of the Fibonacci sequence to have any two starting values) have a number of interesting properties relating to the tuning of MOS scales, and potentially can be used to determine a way to "naturally" tune a MOS (and thus generate a line of daughter MOSes).
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 == Golden operations and MOS height ==
-Any MOS may be constructed from 1L&nbsp;1s using two "golden operations": chromaticizing (taking the soft child of the MOS, thus continuing the golden sequence) and inverting (inverting the number of large and small steps, switching to a new golden sequence). In effect, this makes taking the hard child "cost" 2 instead of 1. This table shows the number of such operations required to reach any MOS under 15 notes (that MOS' "height"): The total number of MOSes at each height corresponds to the Fibonacci sequence.
+Any MOS may be constructed from 1L&nbsp;1s using two "golden operations": chromaticizing, or "stepping" (taking the soft child of the MOS, thus continuing the golden sequence) and inverting, or "flipping" (inverting the number of large and small steps, switching to a new golden sequence). In effect, this makes taking the hard child "cost" 2 instead of 1. This table shows the number of such operations required to reach any MOS under 15 notes (that MOS' "height"): The total number of MOSes at each height corresponds to the Fibonacci sequence.
 {| class="wikitable sortable mw-collapsible mw-collapsed"
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 | 14
 |}
+== Finding the generator ==
+For any given MOS pattern, one can easily determine what generator creates it, by the power of golden sequences. This is done by working back from the MOS to 1L 1s (which has the generator of 1L + 0s) and then working forward with the generator size as if it is a MOS itself.
+Non-octave-periodic MOSes can be seen as their reduced patterns with a fractional-octave period (for example, 5L 5s can be seen as 240c-periodic 1L 1s); for this trick to work the numbers of large and small steps must be coprime.
+For example, let's take the MOS 11L 6s, and interpret it as the Fibonacci sequence fragment [6, 11]. Next, we will "step" backwards, moving our two-number window back so that 6 becomes the second element and the previous entry (which is trivial to calculate as 11-6 = 5) is the first element. So, we reach [5, 6]. Then, we proceed to [1, 5], and then [4, 1]. At this point, we've reached the "beginning" of a sequence, where the two elements are descending. So, we flip: [1, 4], then proceed back to [3, 1]. Continue on until you reach [1, 1], and log the steps you took:
+{| class="wikitable"
+|+
+!Step
+!Sequence fragment
+!Sequence
+|-
+|(Start)
+|6, 11
+|(4, 1)
+|-
+|Step
+|5, 6
+|(4, 1)
+|-
+|Step
+|1, 5
+|(4, 1)
+|-
+|Step
+|4, 1
+|(4, 1)
+|-
+|Flip
+|1, 4
+|(3, 1)
+|-
+|Step
+|3, 1
+|(3, 1)
+|-
+|Flip
+|1, 3
+|Lucas
+|-
+|Step
+|2, 1
+|Lucas
+|-
+|Flip
+|1, 2
+|Fibonacci
+|-
+|Step
+|1, 1
+|Fibonacci
+|}
+Then, we work back through our steps, starting with [0, 1] instead of [1, 1].
+{| class="wikitable"
+!Step
+!Sequence fragment
+!Sequence
+|-
+|(Start)
+|0, 1
+|Fibonacci
+|-
+|Step
+|1, 1
+|Fibonacci
+|-
+|Flip
+|1, 1
+|Fibonacci
+|-
+|Step
+|1, 2
+|Fibonacci
+|-
+|Flip
+|2, 1
+|Lucas
+|-
+|Step
+|1, 3
+|Lucas
+|-
+|Flip
+|3, 1
+|(3, 1)
+|-
+|Step
+|1, 4
+|(3, 1)
+|-
+|Step
+|4, 5
+|(3, 1)
+|-
+|Step
+|5, 9
+|(3, 1)
+|}
+Note that one flip operation leaves the ordered pair unchanged as it is [1, 1].
+If we take our result, [5, 9] as a number of small and large steps, we get 9L + 5s, which is the generator.
+For a simpler example, let's try diatonic, 5L 2s:
+{| class="wikitable"
+!Step
+!Sequence fragment
+!Sequence
+|-
+|(Start)
+|2, 5
+|(3, 2)
+|-
+|Step
+|3, 2
+|(3, 2)
+|-
+|Flip
+|2, 3
+|Fibonacci
+|-
+|Step
+|1, 2
+|Fibonacci
+|-
+|Step
+|1, 1
+|Fibonacci
+|}
+And then to find the generator:
+{| class="wikitable"
+!Step
+!Sequence fragment
+!Sequence
+|-
+|(Start)
+|0, 1
+|Fibonacci
+|-
+|Step
+|1, 1
+|Fibonacci
+|-
+|Step
+|1, 2
+|Fibonacci
+|-
+|Flip
+|2, 1
+|Lucas
+|-
+|Step
+|1, 3
+|Lucas
+|}
+And the generator is 3 large steps and 1 small step (which is correct).
+Note that the large steps are read from the ''second'' entry, which is opposite to the convention used on the wiki where the number of large steps comes first.
+Now we have the generator in steps, now how do we get to a range in cents? Well, for a generator (A)L + (B)s, and a scale (C)L + (D)s, the soft boundary (equalized tuning) is (A+B)\(C+D), and the hard boundary (collapsed tuning) is A\C. For example, the range for our first scale is between 9\11 (982 cents) and (9+5)\(11+6) = 14\17 (988 cents), and the range for diatonic is, as expected, between (3+1)\(5+2) = 4\7 (686 cents) and 3\5 (720 cents). For any hardness L/s = k, the tuning for the generator is (kA+B)/(kC+D). For the golden tuning, k is equal to the golden ratio.
 == Naming golden generators ==
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 | Magic
 | 5/4
-| 380.82{{c}}<ref group="note">Also has another tuning of 377.6 cents for 3L&nbsp;13s. This results in a fifth almost as flat as in 7edo, but is a simpler scale of 16 notes rather than 19.</ref>
+| 380.82{{c}}<ref group="note">Also has another tuning of 377.6 cents for 3L&nbsp;13s (interpretable as [[muggles]] rather than septimal magic). This results in a fifth almost as flat as in 7edo, but is a simpler scale of 16 notes rather than 19.</ref>
 | 3L&nbsp;16s
 |}