Ternary parallelogram scales are MOS substitution: Difference between revisions

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This article proves the following theorem:

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=== 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 ===

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== Proof ==

=== 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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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'')"}} ((''n'' - 1)-many 1's) and stacking (sums of) ''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]]

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[[Category:Combinatorics on words]]

[[Category:Pages with open problems]]

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+{{expert}}
 This article proves the following theorem:
@@ Line 5: / Line 6: @@
 === 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 ===
@@ Line 23: / Line 26: @@
 == Proof ==
 === 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 77: / Line 80: @@
 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'')"}} ((''n'' - 1)-many 1's) and stacking (sums of) ''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]]