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.''

''[[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 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> ===

=== Step 2: ~~Ternarity~~ implies that ~~exactly~~ one of the step vectors is parallel to ~~an axis~~ ===

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

* {{nowrap|'''v''' {{=}} (1, 0)}} is consistently the π-image of a ''k'''''v'''-step interval of ''w'', {{nowrap|0 < ''k'''''v''' < ''mn'',}} so traveling one step east (while staying within the grid) always shifts the label by ''k'''''v'''

* {{nowrap|'''w''' {{=}} (0, 1)}} is consistently the π-image of a ''k'''''w'''-step interval, {{nowrap|0 < ''k'''''w''' < ''mn'',}} so traveling one step north (while staying within the grid) always shifts the label by ''k'''''w'''

* every element of {{nowrap|ℤ/''mn''ℤ}} is used exactly once in the labeling.

After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k'''''v'''}} and {{nowrap|φ('''w''') {{=}} ''k'''''w'''.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain.

=== Step 2: Any ''m'' × ''n'' window in ℤ2 has the same cyclic ordering of elements under the φ-labeling ===

For any integers ''i''0 and ''j''0, if {{nowrap|φ((''i''0, ''j''0)) {{=}} ''a'',}} then for any (''i'', ''j'') in [0 : ''m''] × [0 : ''n''], we have

<math>\begin{align*}\varphi((i_0 + i, j_0 + j)) &= \varphi((i_0, j_0)) + \varphi((i, j))

\\ &= a + \varphi((i, j)),\end{align*}</math>

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, one of the 1-step vectors is parallel to a coordinate axis ===

Consider the following four "quadrants":

# ''Q''1 = [0 : ''m''] × [0 : ''n'']

# ''Q''2 = [-''m'' + 1 : 1] × [0 : ''n'']

# ''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).

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

Relabel the axial vector as '''u''''''x''' (= π('''x''')) and the two nonaxial vectors as '''u''''''y''' = π('''y''') and '''u''''''z''' = π('''z'''). We shall now show that '''x''' is the slot letter of a MOS substitution scale.

Assume without loss of generality that

'''u''''''x''' = (''t'', 0), ''t'' > 0 (parallel to '''v''').

==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ====

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: 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 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'' 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'', 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 ====

Since the two non-axial steps differ by ''n'''''w''', to identify the two we wrap ℤ2 with the vector ''n'''''w''', identifying lattice points separated by ''n'''''w'''. This makes the space ℤ × ℤ/''n''ℤ. Now the image of <math>\pi(\mathcal{I}_w)</math> under this wrap is of the form [0 : ''m''] × ℤ/''n''ℤ. Note that ''m'' (its order being ''n'') corresponds to a generator of the ℤ/''n''ℤ factor, establishing (by minimality) that the period of the template word is ''m''. Now do another wrap, identifying ''m'' with 0, and we're left with [0 : ''m''] in ℤ. We're now in period-equivalent pitch-class space with one generator chain. Since the template word is binary we have a MOS with period ''m'' and period count ''n'', which is "the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram."

==== Filling word is MOS ====

Delete all instances of the axial step '''x''' and consider what the two remaining step sizes do to the '''w'''-coordinate.

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''')."

~~=== Step 3: The axial step~~ is ~~a MOS substitution slot letter ===~~

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}}

~~==== When~~ the ~~two non-axial steps are identified,~~ the ~~result is a MOS~~ ====

~~==== When~~ the ~~axial step is deleted~~, the ~~result is a~~ MOS ~~====~~

== 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.''
+''[[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 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> ===
-=== Step 2: Ternarity implies that exactly one of the step vectors is parallel to an axis ===
+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
+* {{nowrap|'''v''' {{=}} (1, 0)}} is consistently the π-image of a ''k''<sub>'''v'''</sub>-step interval of ''w'', {{nowrap|0 < ''k''<sub>'''v'''</sub> < ''mn'',}} so traveling one step east (while staying within the grid) always shifts the label by ''k''<sub>'''v'''</sub>
+* {{nowrap|'''w''' {{=}} (0, 1)}} is consistently the π-image of a ''k''<sub>'''w'''</sub>-step interval, {{nowrap|0 < ''k''<sub>'''w'''</sub> < ''mn'',}} so traveling one step north (while staying within the grid) always shifts the label by ''k''<sub>'''w'''</sub>
+* every element of {{nowrap|ℤ/''mn''ℤ}} is used exactly once in the labeling.
+After rotating ''w'', we may assume that (0, 0) is labeled 0. The labeling now extends to a surjective homomorphism <math>\varphi: \mathbb{Z}^2\langle \mathbf{v},\mathbf{w}\rangle \to \mathbb{Z}/mn\mathbb{Z},</math> where {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} and {{nowrap|φ('''w''') {{=}} ''k''<sub>'''w'''</sub>.}} φ has {{nowrap|[0 : ''m''] × [0 : ''n'']}} as a fundamental domain.
+=== Step 2: Any ''m'' × ''n'' window in ℤ<sup>2</sup> has the same cyclic ordering of elements under the φ-labeling ===
+For any integers ''i''<sub>0</sub> and ''j''<sub>0</sub>, if {{nowrap|φ((''i''<sub>0</sub>, ''j''<sub>0</sub>)) {{=}} ''a'',}} then for any (''i'', ''j'') in [0 : ''m''] × [0 : ''n''], we have
+<math>\begin{align*}\varphi((i_0 + i, j_0 + j)) &= \varphi((i_0, j_0)) + \varphi((i, j))
+\\ &= a + \varphi((i, j)),\end{align*}</math>
+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, 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'']
+# ''Q''<sub>2</sub> = [-''m'' + 1 : 1] × [0 : ''n'']
+# ''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).
+=== Step 4: The axial step is a MOS substitution slot letter ===
+Relabel the axial vector as '''u'''<sub>'''x'''</sub> (= π('''x''')) and the two nonaxial vectors as '''u'''<sub>'''y'''</sub> = π('''y''') and '''u'''<sub>'''z'''</sub> = π('''z'''). We shall now show that '''x''' is the slot letter of a MOS substitution scale.
+Assume without loss of generality that
+'''u'''<sub>'''x'''</sub> = (''t'', 0), ''t'' > 0 (parallel to '''v''').
+==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ====
+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: 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 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'' 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'', 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 ====
+Since the two non-axial steps differ by ''n'''''w''', to identify the two we wrap ℤ<sup>2</sup> with the vector ''n'''''w''', identifying lattice points separated by ''n'''''w'''. This makes the space ℤ × ℤ/''n''ℤ. Now the image of <math>\pi(\mathcal{I}_w)</math> under this wrap is of the form [0 : ''m''] × ℤ/''n''ℤ. Note that ''m'' (its order being ''n'') corresponds to a generator of the ℤ/''n''ℤ factor, establishing (by minimality) that the period of the template word is ''m''. Now do another wrap, identifying ''m'' with 0, and we're left with [0 : ''m''] in ℤ. We're now in period-equivalent pitch-class space with one generator chain. Since the template word is binary we have a MOS with period ''m'' and period count ''n'', which is "the number of rows of the parallelogram parallel to the unique step size parallel to a side of the parallelogram."
+==== Filling word is MOS ====
+Delete all instances of the axial step '''x''' and consider what the two remaining step sizes do to the '''w'''-coordinate.
+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''')."
-=== Step 3: The axial step is a MOS substitution slot letter ===
+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}}
-==== When the two non-axial steps are identified, the result is a MOS ====
-==== When the axial step is deleted, the result is a MOS ====
+== 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]]