Ternary parallelogram scales are MOS substitution: Difference between revisions

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This article proves the following theorem:

''Ternary parallelogram scale ~~words~~ are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''

''[[Ternary]] parallelogram scale [[word]]s are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''

== Definitions ==

=== Pitch-class group ===

The ''pitch-class group'' of a scale word ''w'' in letters {{nowrap|'''x'''1, ..., '''x'''''r''}} with [[step signature]] {{nowrap|'''e''' ∈ ℤ''r''{{angbr|'''x'''1, ..., '''x'''''r''}}}} is the abelian group {{nowrap|C(''w'') :{{=}} ℤ''r''{{angbr|'''x'''1, ..., '''x'''''r''}}/{{angbr|'''e'''}}.}} The pitch-class group is associated with a canonical map π that sends every step vector to its pitch class.

Below we take it as known that the gcd of the coordinates ''v''''i'' of '''v''' ∈ ℤ''r'' is 1 iff the quotient group ℤ''r''/{{angbr|'''v'''}} is torsion-free; this can be proven using Bézout's identity.

=== Parallelogram scale ===

A scale word ''w'' is a ''parallelogram scale word'' if C(''w'') is torsion-free (equiv. a free abelian group) and there ~~exists~~ integers {{nowrap|''m'', ''n'' > 1}} and linearly independent elements '''v''' and '''w''' in C(''w'') such that the π-image of

A scale word ''w'' is a ''parallelogram scale word'' if C(''w'') is torsion-free (equiv. a free abelian group) and there exist integers {{nowrap|''m'', ''n'' > 1}} and linearly independent elements '''v''' and '''w''' in C(''w'') such that the π-image of

<math>\mathcal{I}_w :=

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=== MOS substitution scale ===

~~See~~ [[MOS ~~substitution~~]].

A ternary scale word ''w''('''x'''1, '''x'''2, '''x'''3) is a ''MOS substitution'' scale word if there exists a permutation <math>\pi \in S_3</math> such that the following holds:

* identifying '''x'''π(1) and '''x'''π(2) results in a [[MOS]] scale word (called the ''template MOS'') and

* deleting all instances of '''x'''π(3) (called the ''slot letter'') results in a MOS scale word (called the ''filling MOS'').

== Proof ==

=== Step 1: Get a surjective homomorphism <math>\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> ===

=== Step 1: Get a surjective homomorphism <math>\varphi:\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> ===

The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} ~~Traversing~~ ''w'' step by step gives a traversal of {{nowrap|[0 : ''m''] × [0 : ''n'']}} where we label each grid point with the index of the current note in ''w''. We also recall that the pitch-class vector '''v''' has a representative that is a ''k'''''v'''-step interval in ''w'', {{nowrap|0 < ''k'''''v''' < ''mn'',}} and similarly for '''w'''.

The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} Because ''w'' is a parallelogram scale, traversing ''w'' step by step gives a traversal of {{nowrap|[0 : ''m''] × [0 : ''n'']}} where we label each grid point with the index of the current note in ''w''. We also recall that the pitch-class vector '''v''' has a representative that is a ''k'''''v'''-step interval in ''w'', {{nowrap|0 < ''k'''''v''' < ''mn'',}} and similarly for '''w'''.

We thus wish to constrain ways of labeling {{nowrap|[0 : ''m''] × [0 : ''n'']}} with {{nowrap|ℤ/''mn''ℤ}} elements such that

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as φ is a homomorphism. Hence corresponding elements of any two ''m'' × ''n'' windows get the same {{nowrap|ℤ/''mn''ℤ}} labeling under φ modulo a shift.

=== Step 3: By ternarity, ~~exactly~~ one of the 1-step vectors is parallel to a coordinate axis ===

=== Step 3: By ternarity, one of the 1-step vectors is parallel to a coordinate axis ===

Consider the following four "quadrants":

# ''Q''1 = [0 : ''m''] × [0 : ''n'']

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# ''Q''3 = [-''m'' + 1 : 1] × [-''n'' + 1 : 1]

# ''Q''4 = [0 : ''m''] × [0 : ''n'']

By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ2 is surjective, hence all of the four windows ''Q''1, ..., ''Q''4 have 1 somewhere in them. Call these positions '''u'''1, ..., '''u'''4 (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0}} by another application of the previous step we have '''u'''1, ..., '''u'''4 as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''''k'' = '''u'''''l''. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''''k'' = '''u'''''l'' is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both).

By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ2 is surjective, hence all of the four windows ''Q''1, ..., ''Q''4 have 1 somewhere in them. Call these positions '''u'''1, ..., '''u'''4 (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0,}} by another application of the previous step we have '''u'''1, ..., '''u'''4 as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''''k'' = '''u'''''l''. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''''k'' = '''u'''''l'' is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both).

~~=== A technical lemma ===~~

Statement: If ''m'' > 1, ''n'' > 2, ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, and {{nowrap|ℤ/''mn''ℤ}} is partitioned into bins where one bin consists of ≤ 2''m'' - 1 adjacent elements and the rest of the bins consist of 2''m'' - 1, then {{nowrap|{0, ''a'', 2''a'', ..., (''n'' - 1)''a''}}} meets some bin at least twice.

~~Proof: Apply the pigeonhole principle: the number of bins is~~

~~<math>~~

~~\begin{align*}~~

~~\bigg\lceil \frac{mn}{2m-1} \bigg\rceil~~

~~&=\bigg\lceil \frac{n}{2-1/m} \bigg\rceil~~

~~\\ &\le \bigg\lceil \frac{2n}{3} \bigg\rceil < n.~~

~~\end{align*}</math>~~

~~The bin where two multiples of ''a'' coincide must have two distinct multiples of ''a'', because the order of ''a'' is > ''n''~~.

=== Step 4: The axial step is a MOS substitution slot letter ===

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Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k'''''v'''}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk'''''v''' {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k'''''v''') = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}}

Claim: ''k'''''w''' has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}

Claim: ψ(''k'''''w''') (thus ''k'''''w''' as well) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}

Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk'''''w''')) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. If ''n'' = 2, then ~~clearly~~ the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k'''''w''' = ''m'' which is of order 2. If ''m'' > 3, ~~then~~ by ~~the lemma,~~ {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k'''''w'''), are ~~not~~ disjoint, ~~contradicting injectivity~~.

Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk'''''w''')) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases.

* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ψ(''k'''''w''') = ''m'' which is of order 2.

* If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k'''''w'''), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}}

** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''.

** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows.

The claim establishes that

# the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order ''mn'', nor on ~~the axis opposite to the first~~

# the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order ''mn'' for ''k'''''w''', nor on {{nowrap|[-''m'' + 1 : 0] × {0}}} as that contradicts ''k'''''v''' having order ''mn''

# φ is minimally periodic under translation by ''n'''''w'''. Hence the two non-axial 1's on the grid from the 0 at the origin are translated by ''n'''''w'''.

# Since ''m'' is of order ''n'', ''m'' must be found on the '''w'''-axis as well.

# Since ''m'' is of order ''n'', and the cyclic group {{nowrap|ℤ/''mn''ℤ}} has at most one subgroup of any given order, ''m'' must be found on the '''w'''-axis as well.

==== Template word is MOS ====

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Without loss of generality assume that {{nowrap|'''u''''''y''' {{=}} (''b'', ''c''), ''c'' > 0,}} and {{nowrap|'''u''''''z''' {{=}} (''b'', ''c'' - ''n'').}} As the '''v'''-coordinates of both vectors are equal, we only need to look at the '''w'''-coordinate. Since the '''w'''-coordinate of a point must stay within {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c ≥ ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')."

This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} and stacking ~~(abelianized)~~ ''c''-step subwords. As ~~there is~~ only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ~~the latter word~~ for {{nowrap|2 ≤ ''k'' ≤ length - 1.}} {{Qed}}

This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|''v'' {{=}} "1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking sums <math>\sum_{i=0}^{c-1} v[i_0+i]</math> of ''c''-step subwords. As the resulting word has only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ''v'' for {{nowrap|2 ≤ ''k'' ≤ length - 1.}} {{Qed}}

== Open problems ==

# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 1}} are MOS substitution.

# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 2}} are MOS substitution.

# Prove a converse to this theorem.

#* <s>If the template MOS of a MOS substitution scale has m > 1 periods and the step signature's gcd = 1 => m × n parallelogram scale.</s> This is false as stated, since no 5L(3m7s) scale is a parallelogram scale.

#* Conjecture: If the template MOS of a MOS substitution scale has m > 1 periods, the step signature's gcd = 1, and the filling MOS is a multiMOS, then the scale is a parallelogram scale. (But the filling MOS need not be a multiMOS for a parallelogram scale: LLmLLsLLmLLsLLs is of type 10L(2m3s))

#* Conjecture: If a MOS substitution scale's template MOS is a multiMOS, and its step signature's gcd = 1, then its generator structure is a (possibly) *shifted* full parallelogram, i.e. either a full parallelogram or the first and last rows' lengths add up to the full row length.

[[Category:Math]]

[[Category:Pages with proofs]]

[[Category:Combinatorics on words]]

[[Category:Pages with open problems]]

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+{{expert}}
 This article proves the following theorem:
-''Ternary parallelogram scale words are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''
+''[[Ternary]] parallelogram scale [[word]]s are [[MOS substitution]] scale words, where the period count of the template MOS is the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram.''
 == Definitions ==
 === Pitch-class group ===
 The ''pitch-class group'' of a scale word ''w'' in letters {{nowrap|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}} with [[step signature]] {{nowrap|'''e''' ∈ ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}}} is the abelian group {{nowrap|C(''w'') :{{=}} ℤ<sup>''r''</sup>{{angbr|'''x'''<sub>1</sub>, ..., '''x'''<sub>''r''</sub>}}/{{angbr|'''e'''}}.}} The pitch-class group is associated with a canonical map π that sends every step vector to its pitch class.
+Below we take it as known that the gcd of the coordinates ''v''<sub>''i''</sub> of '''v''' ∈ ℤ<sup>''r''</sup> is 1 iff the quotient group ℤ<sup>''r''</sup>/{{angbr|'''v'''}} is torsion-free; this can be proven using Bézout's identity.
 === Parallelogram scale ===
-A scale word ''w'' is a ''parallelogram scale word'' if C(''w'') is torsion-free (equiv. a free abelian group) and there exists integers {{nowrap|''m'', ''n'' > 1}} and linearly independent elements '''v''' and '''w''' in C(''w'') such that the π-image of
+A scale word ''w'' is a ''parallelogram scale word'' if C(''w'') is torsion-free (equiv. a free abelian group) and there exist integers {{nowrap|''m'', ''n'' > 1}} and linearly independent elements '''v''' and '''w''' in C(''w'') such that the π-image of
 <math>\mathcal{I}_w :=
@@ Line 17: / Line 20: @@
 === MOS substitution scale ===
-See [[MOS substitution]].
+A ternary scale word ''w''('''x'''<sub>1</sub>, '''x'''<sub>2</sub>, '''x'''<sub>3</sub>) is a ''MOS substitution'' scale word if there exists a permutation <math>\pi \in S_3</math> such that the following holds:
+* identifying '''x'''<sub>π(1)</sub> and '''x'''<sub>π(2)</sub> results in a [[MOS]] scale word (called the ''template MOS'') and
+* deleting all instances of '''x'''<sub>π(3)</sub> (called the ''slot letter'') results in a MOS scale word (called the ''filling MOS'').
 == Proof ==
-=== Step 1: Get a surjective homomorphism <math>\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> ===
+=== Step 1: Get a surjective homomorphism <math>\varphi:\mathbb{Z}^2 \to \mathbb{Z}/mn\mathbb{Z}</math> ===
-The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} Traversing ''w'' step by step gives a traversal of {{nowrap|[0 : ''m''] × [0 : ''n'']}} where we label each grid point with the index of the current note in ''w''. We also recall that the pitch-class vector '''v''' has a representative that is a ''k''<sub>'''v'''</sub>-step interval in ''w'', {{nowrap|0 < ''k''<sub>'''v'''</sub> < ''mn'',}} and similarly for '''w'''.
+The π-image of any ''k''-step interval (abelianized slice) {{nowrap|ab(''w''[''i'' : ''i'' + ''k''])}} is identical to that of {{nowrap|ab(''w''[''i'' : ''i'' + ''k'' + ''mn'']).}} Hence there is a well-defined map from the pitch classes of intervals of ''w'' to {{nowrap|ℤ/''mn''ℤ.}} Because ''w'' is a parallelogram scale, traversing ''w'' step by step gives a traversal of {{nowrap|[0 : ''m''] × [0 : ''n'']}} where we label each grid point with the index of the current note in ''w''. We also recall that the pitch-class vector '''v''' has a representative that is a ''k''<sub>'''v'''</sub>-step interval in ''w'', {{nowrap|0 < ''k''<sub>'''v'''</sub> < ''mn'',}} and similarly for '''w'''.
 We thus wish to constrain ways of labeling {{nowrap|[0 : ''m''] × [0 : ''n'']}} with {{nowrap|ℤ/''mn''ℤ}} elements such that
@@ Line 36: / Line 42: @@
 as φ is a homomorphism. Hence corresponding elements of any two ''m'' × ''n'' windows get the same {{nowrap|ℤ/''mn''ℤ}} labeling under φ modulo a shift.
-=== Step 3: By ternarity, exactly one of the 1-step vectors is parallel to a coordinate axis ===
+=== Step 3: By ternarity, one of the 1-step vectors is parallel to a coordinate axis ===
 Consider the following four "quadrants":
 # ''Q''<sub>1</sub> = [0 : ''m''] × [0 : ''n'']
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 # ''Q''<sub>3</sub> = [-''m'' + 1 : 1] × [-''n'' + 1 : 1]
 # ''Q''<sub>4</sub> = [0 : ''m''] × [0 : ''n'']
-By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ<sup>2</sup> is surjective, hence all of the four windows ''Q''<sub>1</sub>, ..., ''Q''<sub>4</sub> have 1 somewhere in them. Call these positions '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0}} by another application of the previous step we have '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub>. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub> is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both).
+By the previous step, φ restricted to any ''m'' × ''n'' window in ℤ<sup>2</sup> is surjective, hence all of the four windows ''Q''<sub>1</sub>, ..., ''Q''<sub>4</sub> have 1 somewhere in them. Call these positions '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> (note that none of them are the zero vector). Since {{Nowrap|φ((0, 0)) {{=}} 0,}} by another application of the previous step we have '''u'''<sub>1</sub>, ..., '''u'''<sub>4</sub> as the images of 1-step vectors of ''w''. Since ''w'' is ternary, exactly two of these vectors will be pairwise equal, say '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub>. These four "quadrants" intersect in <math>[-m + 1 : m] \times \{0\} \cup \{0\} \times [-n + 1 : n],</math> entailing that '''u'''<sub>''k''</sub> = '''u'''<sub>''l''</sub> is on a coordinate axis (either the '''v'''-coordinate is 0 or the '''w'''-coordinate is 0 but not both).
-=== A technical lemma ===
-Statement: If ''m'' > 1, ''n'' > 2, ''a'' has order > ''n'' in {{nowrap|ℤ/''mn''ℤ}}, and {{nowrap|ℤ/''mn''ℤ}} is partitioned into bins where one bin consists of &le; 2''m'' - 1 adjacent elements and the rest of the bins consist of 2''m'' - 1, then {{nowrap|{0, ''a'', 2''a'', ..., (''n'' - 1)''a''}}} meets some bin at least twice.
-Proof: Apply the pigeonhole principle: the number of bins is
-<math>
-\begin{align*}
-\bigg\lceil \frac{mn}{2m-1} \bigg\rceil
-&=\bigg\lceil \frac{n}{2-1/m} \bigg\rceil
-\\ &\le \bigg\lceil \frac{2n}{3} \bigg\rceil < n.
-\end{align*}</math>
-The bin where two multiples of ''a'' coincide must have two distinct multiples of ''a'', because the order of ''a'' is > ''n''.
 === Step 4: The axial step is a MOS substitution slot letter ===
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 Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk''<sub>'''v'''</sub> {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k''<sub>'''v'''</sub>) = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}}
-Claim: ''k''<sub>'''w'''</sub> has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}
+Claim: ψ(''k''<sub>'''w'''</sub>) (thus ''k''<sub>'''w'''</sub> as well) has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}}
-Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. If ''n'' = 2, then clearly the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k''<sub>'''w'''</sub> = ''m'' which is of order 2. If ''m'' > 3, then by the lemma, {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are not disjoint, contradicting injectivity.
+Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases.
+* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ψ(''k''<sub>'''w'''</sub>) = ''m'' which is of order 2.
+* If ''n'' &ge; 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}}
+** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''.
+** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows.
 The claim establishes that
-# the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order ''mn'', nor on the axis opposite to the first
+# the two other 1-step vectors are not found on the axis orthogonal to the first vector, as that would imply order ''mn'' for ''k''<sub>'''w'''</sub>, nor on {{nowrap|[-''m'' + 1 : 0] × {0}}} as that contradicts ''k''<sub>'''v'''</sub> having order ''mn''
 # φ is minimally periodic under translation by ''n'''''w'''. Hence the two non-axial 1's on the grid from the 0 at the origin are translated by ''n'''''w'''.
-# Since ''m'' is of order ''n'', ''m'' must be found on the '''w'''-axis as well.
+# Since ''m'' is of order ''n'', and the cyclic group {{nowrap|ℤ/''mn''ℤ}} has at most one subgroup of any given order, ''m'' must be found on the '''w'''-axis as well.
 ==== Template word is MOS ====
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 Without loss of generality assume that {{nowrap|'''u'''<sub>'''y'''</sub> {{=}} (''b'', ''c''), ''c'' > 0,}} and  {{nowrap|'''u'''<sub>'''z'''</sub> {{=}} (''b'', ''c'' - ''n'').}} As the '''v'''-coordinates of both vectors are equal, we only need to look at the '''w'''-coordinate. Since the '''w'''-coordinate of a point must stay within {{nowrap|[0 : ''n''],}} at any point it must follow the rule: "If the current '''w'''-coordinate + c &ge; ''n'', then move by ''c'' - ''n'' units (using the letter '''z'''). Otherwise, move by ''c'' units (using the letter '''y''')."
-This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|"1 1 1 ... 1 (1 - ''n'')"}} and stacking (abelianized) ''c''-step subwords. As there is only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of the latter word for {{nowrap|2 &le; ''k'' &le; length - 1.}} {{Qed}}
+This pattern of movements is in fact the same as the one produced by taking the circular word {{nowrap|''v'' {{=}} "1 1 1 ... 1 (1 - ''n'')"}} ((''n'' - 1)-many 1's) and stacking sums <math>\sum_{i=0}^{c-1} v[i_0+i]</math> of ''c''-step subwords. As the resulting word has only one bad position per period, the filling word can easily be seen to be MOS by stacking ''kc''-step subwords of ''v'' for {{nowrap|2 &le; ''k'' &le; length - 1.}} {{Qed}}
+== Open problems ==
+# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 1}} are MOS substitution.
+# Conjecture: Ternary scales that are parallelogram substrings with full row length ''m'', full column length ''n'', and cardinality {{nowrap|''mn'' - 2}} are MOS substitution.
+# Prove a converse to this theorem.
+#* <s>If the template MOS of a MOS substitution scale has m > 1 periods and the step signature's gcd = 1 => m × n parallelogram scale.</s> This is false as stated, since no 5L(3m7s) scale is a parallelogram scale.
+#* Conjecture: If the template MOS of a MOS substitution scale has m > 1 periods, the step signature's gcd = 1, and the filling MOS is a multiMOS, then the scale is a parallelogram scale. (But the filling MOS need not be a multiMOS for a parallelogram scale: LLmLLsLLmLLsLLs is of type 10L(2m3s))
+#* Conjecture: If a MOS substitution scale's template MOS is a multiMOS, and its step signature's gcd = 1, then its generator structure is a (possibly) *shifted* full parallelogram, i.e. either a full parallelogram or the first and last rows' lengths add up to the full row length.
 [[Category:Math]]
 [[Category:Pages with proofs]]
 [[Category:Combinatorics on words]]
+[[Category:Pages with open problems]]