Ternary scale theorems: Difference between revisions

# If ''n'' is odd, {{nowrap|''s'' {{=}} ''a'''''X''' ''b'''''Y''' ''b''ℤ}} is obtained from some mode of the (primitive) MOS ''a'''''X''' 2''b'''''W''' by replacing all the '''W'''s successively with alternating '''Y'''s and ℤs (or alternating ℤs and '''Y'''s for the other chirality, fixing the mode of ''a'''''X''' 2''b'''''W'''). The two alternants differ by replacing one '''Y''' with a ℤ. In other words, ''s'' is ''odd-regular'' in our classification of MV3 scales.

By part (2), we have that ''s'' has step signature {{nowrap|''a'''''X''' ''b'''''Y''' ''b''ℤ}}, ''a'' odd. By part (4), we have that {{nowrap|''T''('''X''', '''W''') {{=}} ''s''('''X''', '''W''', '''W''')}} is a MOS scale ''a'''''X'''2''b'''''W'''. If {{nowrap|''b'' {{=}} 1}}, there's nothing to prove, so assume {{nowrap|''b'' > 1}}.

'''Claim 1''': Deleting '''X'''s from the generator subwords of ''s'' gives every ''j''-step subword in the scale ''E''_X(''s'')('''Y''', ℤ), the scale word obtained by deleting all '''X''''s from ''s''. These ''j''-step subwords are adjacent and alternating under the ordering induced by the AGS stack.

* B.2. By the hypothesis that ''s'' has an AGS, and since the AGS descends to stacking a single generator in the template MOS ''T'', the lifted generators ''g''₁ and ''g''₂ alternate in their counts of '''Y''' and also alternate in their counts of ℤ.

* C.1. Importantly, deleting '''X''''s gives windows of length ''j'', such that when you project adjacent lifted generators (by deleting '''X''''s) to the binary necklace {{nowrap|''U'' :{{=}} ''E''_'''X'''(''s'')('''Y''', ℤ)}}, the resulting ''j''-step windows in ''U'' are adjacent and do not overlap.

'''Claim 2''': If a binary necklace ''U'' has ''b'' '''Y'''s and ''b'' ℤs, {{nowrap|gcd(''j'', 2''b'') {{=}} 1}}, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then {{nowrap|''U'' {{=}} ('''YZ''')^''b''}}.

These two claims prove that {{nowrap|''E''_'''X'''(S) {{=}} ('''YZ''')^''b''}} and that the two GS generators' sizes differ by replacing one '''Y''' for a ℤ. {{Qed}}

A primitive ternary scale ''s'' is ''even-regular'' if len(''s'') is even and ''s'' is equivalent to a word constructed from taking the MOS 2''a'''''X''' 2''c''ℤ with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] equivalent to ''a'''''X''' ''a'''''Y''' ''b''ℤ with ''a'' odd and ''b'' even. For example, '''LsLsLmsLsLsm''' (achiral [[diachrome]], 5'''L''' 2'''m''' 5'''s''') is an even-regular scale.

If {{nowrap|''s'' {{=}} ''s''('''X''', '''Y''', ℤ)}} is even-regular, then:

# the generator has the same interval class as some generator of the MOS 2''a'''''W''' 2''c''ℤ;

The result of substituting '''Y''' with '''X''' (let us call this map ''p'') is the MOS {{nowrap|''M'' {{=}} 2''a'''''X''' 2''c''ℤ}}, which has exactly 2 periods since {{nowrap|gcd(''a'', ''c'') {{=}} 1}}. ''M'' thus consists of two generator chains separated by the period of ''M'', which has {{nowrap|''a'' + ''c'' {{=}} len(''s'')}} steps. It thus suffices for there to exist ''k'', {{nowrap|0 < ''k'' < ''a'' + ''c''}}, such that every perfect ''k''-step generator has the same preimage in ''s'', which will be our desired generator. Suppose that the perfect ''k''-step of ''M'' is {{nowrap|''i'''''W''' + ''j''ℤ}} where {{nowrap|0 < ''i'' < ''a''}}. Since ''a'' is odd, possibly after taking the period-complement we may assume that ''i'' is even. Hence each subword ''w'' of ''s'' such that its projection ''p''(''w'') subtends a perfect ''k''-step satisfies {{nowrap|{{abs|''w''}}_'''X''' {{=}} {{abs|''w''}}_'''Y''' {{=}} ''i''/2}}. It plainly follows that every such ''w'' satisfies {{nowrap|{{abs|''w''}}_'''X''' {{=}} {{abs|''w''}}_'''Y'''}} = {{sfrac|''i''|2}} and {{nowrap|{{abs|''w''}}_ℤ {{=}} ''j''}}.

It remains to show that ''s'' is balanced. Any ''k''-step subword has either ''j'' or ''j'' + 1 ℤs for some ''j'' since the result of conflating '''X''' and '''Y''' is a MOS, and ''k''-step subwords for both possibilities exist when 0 < ''k'' < len(''s'')/2. If the number of non-ℤ letters in a ''k''-step subword is even, then there is only one possibility for the number of '''X''' and the number of '''Y'''. If the number of non-ℤ letters in a ''k''-step subword is odd, then both the number of '''X'''s and the number of '''Y'''s differ by at most 1. {{qed}}

In the following, ''equivalent'' means "is the same circular word after permuting '''X''', '''Y''', and ℤ." This means that '''XYXZXYX''' is equivalent to '''YZYXYZY''', or '''XZXYXZX''', and so on.

## '''sporadic balanced''': ''s'' is equivalent to '''XYXZXYX''', the ternary [[Fraenkel word]], with step signature 4'''X'''2'''Y'''1ℤ.

## '''odd-regular''': len(''s'') is odd, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS ''c'''''X'''''b''ℤ with ''c'' even and {{nowrap|gcd(''c'', ''b'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. We assume {{nowrap|'''X''' &gt; ℤ}} when constructing the MOS. In particular, ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b''ℤ where ''b'' is odd (with {{nowrap|''a'' {{=}} ''c''/2}}).

## '''even-regular''': len(''s'') is even, and ''s'' is equivalent to a word constructed from taking the brightest mode of the MOS 2''a'''''X'''2''c''ℤ with ''a'' odd and {{nowrap|gcd(''a'', ''c'') {{=}} 1}}, and replacing every other '''X''' with '''Y'''. In particular,  ''s'' has [[step signature]] ''a'''''X'''''a'''''Y'''''b''ℤ with ''a'' odd and ''b'' even.

(Condensed: All single-period balanced ternary scales that are not the Fraenkel word are a'''X''' a'''Y''' bℤ. In this case, if b is odd, then the scale is odd-regular. If b is even, then the scale is even-regular.)

There are two consecutive '''X'''s with no '''Y''' in between since ''a'' > ''b''. This means either '''XX''' or '''XZX''' appears. If '''XX''' appears, then a ℤ is necessarily surrounded by two '''X'''s.

There exists a pair of consecutive '''Y'''s with no ℤ in between. Thus we have a subword of the form '''YX'''^''n'''''Y'''. Now, ''n'' &le; 1 is not possible because of the presence of '''XZX''' and '''Y'''-balance. ''n'' &ge; 3 implies the existence of '''X'''^''n''-1'''ZX'''^''n''-1 by '''X'''-balance which is incompatible with '''YX'''^''n'''''Y''' because of '''Y'''-balance. Therefore, ''n'' = 2. Note that this also implies the presence of subwords '''XX''' and '''XYXXYX'''.

The sequence ''W'' must contain a ℤ. This ℤ is necessarily surrounded by two '''X'''s since '''XX''' exists by Step (ii). This group is necessarily surrounded by two '''Y'''s since '''YXXY''' exists, and consequently, necessarily surrounded by two '''X'''s because '''XYXXYX''' exists. We get the sequence '''XYXZXYX'''.

No letter around this word can be a ℤ because '''YXXY''' exists. None can be a '''Y''' since '''XZX''' exists. Therefore, they have to be two '''X'''s. Then note that the two surrounding letters cannot be ℤ (because of the existence of '''XYXXYX''') nor '''X''' (because of the existence of '''YXZ''') so they are '''Y''', then followed by '''X''' (because '''XX''' exists). At this point, we have the sequence

“_'''XYXXYXZXYXXYX'''_”. Both _s are necessarily ℤs. To end the proof, note that we have obtained the configuration around every ℤ and this determines the whole sequence. Thus ''W'' = ('''XYXZXYX''')^ω.

(c) The scale made by taking ''s'' and conflating '''Y''' and ℤ into the letter '''W''' must be a MOS. To this scale we may imagine substituting a scale made of an equal amount of '''Y''' and ℤ letters into the "slot letters" '''W''' letter by letter. Let ''t''₁ be a length-''k'' subword of the form '''YX'''^''k''-2'''Y''' under the projection. We may assume that the chunk sizes of the MOS are ''k'' - 2 and ''k'' - 1, or ''k'' - 2 and ''k'' - 3. Either way, there exists some subword with (''k'' - i)-many '''X'''s, i = 1 or 2, and two ℤs. This violates balance because ''t''₁ contains zero ℤs.

For 5.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, {{nowrap|''a''_''i'''''X''' + ''b''_''i'''''Y''' + ''c''_''i''ℤ {{=}} (''a''_''i'', ''b''_''i'', ''c''_''i'')}} for {{nowrap|''i'' ∈ {{(}}1, 2, 3{{)}}}}. We may assume {{nowrap|(''a''₂, ''b''₂, ''c''₂) {{=}} (''a''₁, ''b''₁ + 1, ''c''₁ − 1)}}. Then either {{nowrap|(''a''₃, ''b''₃, ''c''₃) {{=}} (''a''₁ + 1, ''b''₁, ''c''₁ − 1)}} or {{nowrap|(''a''₃, ''b''₃, ''c''₃) {{=}} (''a''₁ − 1, ''b''₁ + 1, ''c''₁)}}. In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.

For 5.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c''ℤ}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c''ℤ}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c''ℤ}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' &minus; 1 : ''k'' + ''a'' + ''c'' &minus; 1]}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' &minus; 1 : ''k'' + 2''a'' + 2''c'' &minus; 1]}} via interchanging '''X''' and '''Y'''.

#* Start with the brightest multiMOS word ''kc'''''X'''''kb''ℤ with ''c'' being an even number.

#* Interchange a ℤ and an '''X''' at some (possibly more than one) of the boundaries of these copies of the MOS word ''w''. Here, the boundary of two consecutive copies of ''w'' is the last letter of the first word and the first letter of the second word. (At the ends of the whole multiMOS word, the boundaries are just the first and last letters of the word.) For example, let ''w'' be the multiMOS word 8'''X'''6ℤ, '''XXZXZXZXXZXZXZ'''. Then the border between the copies of the MOS subword '''XXZXZXZ''' are ''w''[6]''w''[7] and ''w''[13]''w''[0] (using 0-based numbering).

#* Replace every other '''X''' with '''Y''' in ''w''. (Thus in particular, twisted MV3 scales have step signature ''ka'''''X'''''ka'''''Y'''''kb''ℤ)

# Conjecture: If a twisted MV3 is not SV3, then it is constructed from ''ka'''''X'''''kb''ℤ where ''k'' is composite.