Interleaving: Difference between revisions

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== Ternary interleaved scales ==

~~Conjecture:~~ Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there exists a ~~unique~~ single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.

Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there always exists a single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.

~~=== Lemma ===~~

Conjecture: It is the ''unique'' one with this property.

If ''S'' consists of the subwords '''XZ''' and '''YZ''' arranged in the pattern of a single-period binary circular word ''w''(''x'', ''y'') where {{pipe}}''w''{{pipe}} > 2, and ''k'' is an odd number greater than 1 and less than {{pipe}}''S''{{pipe}} - 1, then the class of ''k''-steps has more than 3 abstract intervals.

~~{{proof|contents=Denote by {{pipe}}''w''{{pipe}} the length of subword ''w'' in letters and by ‖''w''‖ the interval subtended by subword ''w'' in its circular word.~~

Let ''A'' be the set of all (''k'' - 1)/2-step intervals of ''w''. {{pipe}}''A''{{pipe}} = 1 implies that ''k'' = 1 mod {{pipe}}''w''{{pipe}}, so {{pipe}}''A''{{pipe}} ≥ 2 and contains at least two intervals '''w'''1 and '''w'''2.

Say {{pipe}}''A''{{pipe}} = 2. (If {{pipe}}''A''{{pipe}} ≥ 3 then the proof is easy.) Say ''B'' is the set of all subwords of ''w'' (not intervals) subtending '''w'''1, and ''C'' is the set of all subwords of ''w'' subtending '''w'''2.

~~Choose ''b'' in ''B'' and ''c'' in ''C'' and consider the subwords ''b''('''XZ''', '''YZ''') and ''c''('''XZ''', '''YZ''') of ''S''.~~

~~If ''k'' = 3, then we can just say ''b'' = ''x'' and ''c'' = ''y''. Suppose '''YZX''' does not occur in ''S''; then ''yx'' does not occur in ''w'' which is a contradiction, so ''YZX'' must occur.~~

~~=> '''ZYZXZ''' occurs~~

~~=> either '''XZYZXZ''' or '''YZYZXZ''' occurs (with suffix '''ZYZXZ''')~~

~~==> If {{pipe}}''w''{{pipe}} = 3, then these are the whole word ''w''; either way we have four abstract intervals.~~

~~==> If {{pipe}}''w''{{pipe}} > 3, then either '''ZYZXZXZ''' or '''ZYZXZYZ''' occurs (with prefix '''ZYZXZ''').~~

~~Case '''XZYZYZXZ''' => '''XZY''', '''YZY''', '''ZYZ''', '''ZXZ'''~~

~~Case '''YZYZYZXZ''' => '''YZY''', '''YZX''', '''ZYZ''', '''ZXZ'''~~

~~So the case ''k'' = 3 is done.~~

~~For any other ''k'', we similarly go through the cases (a) ''bx'' and ''cx'', (b) ''bx'' and ''cy'', (c) ''by'' and ''cx'', (d) ''by'' and ''cy''.~~

~~(a) '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''X''', ''c''('''XZ''', '''YZ''')'''X''' => done~~

~~(d) is similar to (a).~~

(c) We have '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X'''. If the sizes of ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X''' are different we are done. If they are the same, we have that ''by'' and ''cx'' subtend the same interval in ''w'', hence ''b'' has one more ''x'' and one fewer ''y'' than ''c''.

By scooting ''bx'' one letter to the left in ''w'', we find either (i) ''xb'' or (ii) ''yb''. The {{pipe}}''b''{{pipe}}-letter prefix ''p'' of this subword subtends either the same interval as (1) ''b'' or (2) ''c''.

~~(i), (1) => ''b'' = ''b'x'', ''p'' = ''xb' '', have ''xby'' = ''xb'xy'', and we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.~~

~~(ii), (1) => ''b'' = ''b'y'', ''p'' = ''yb' '', continue scooting to the left until you find an ''x'' to the left.~~

~~(i), (2) => ''b'' = ''b'y'', ''p'' = ''xb' '', we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.~~

~~(ii), (2) => ''b'' =''b'x'', ''p'' = ''yb' '', we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.~~

~~(c) is similar to (b).~~

}}

~~=== Attempted proof of~~ Conjecture ~~===~~

~~{{proof|contents=Suppose the scale~~ is ~~made of two interleaved subsets offset by~~ the ~~abstract interval~~ '''δ'~~''.~~

~~Case 1: gcd(2(''a'' + ''b''), ''k'') = 1. If ''k'' = 1, then any '''Z'''''q'' a maximal subword of consecutive '''Z'''s has ''q'' odd, and they all must be the same length and separated by~~ one ~~non-'''Z''' letter '''W''':~~

~~'''ZZ ZZ ... ZW''' (in strand ''S''1)~~

~~'''Z ZZ ... ZZ WZ''' (in strand ''S''2)~~

~~Moreover, the offset '''δ''' = '''Z''', i.e. ''S''2 is separated by the interval '''Z''' to the right of ''S''1.~~

If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of length ''a'' + ''b'', ''w''1 with the maximal number of consecutive '''Z''''s and ''w''2 with the minimal number of '''Z'''s. Scoot ''w''1 to the right one step at a time until it loses one '''Z'''. This proves either that a non-'''Z''' letter is equal to '''Z''' or that the offset goes in the wrong direction. Hence ''q'' = 1, as desired.

If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T'', which traverses all notes of ''S'' since gcd(''k'', 2(''a'' + ''b'')) = 1. Since ''k'' is odd, the letters of this word alternate between beginning in ''S''1 and beginning in ''S''2. By a reasoning similar to the above, ''T'' has a letter '''δ''' between its two mutually interleaved strands. By the lemma this scale has step variety ''r'' > 3, say with letters '''W'''1, ..., '''W'''''r''. Let ''k' '' be the inverse of ''k'' mod 2(''a'' + ''b''). By stacking ''k' ''-step subwords of ''T'', we end up with at least 4 different linear equations with 3 unknowns '''X''', '''Y''', and '''Z''', implying a nontrivial linear relation between '''X''', '''Y''', and '''Z''' (?). This is a contradiction as '''X''', '''Y''', and '''Z''' are assumed to not have a linear relation.

~~Case 2: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving~~ property~~, hence ''k'' = ''a'' + ''b'' which must be odd.~~

Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''.

~~We have '''V'''w'''W''' where '''V''' != '''W''', ‖'''V'''''w''‖ = '''δ''', ‖''w'''''W'''‖ = ‖''s''‖ - '''δ'''~~

~~=> '''δ''' - '''V''' = ‖''w''‖ = ‖''s''‖ - '''δ''' - '''W'''~~

=> 2'''δ''' + '''W''' = ‖''s''‖ + '''V'''. If this linear relation is nontrivial, it's a contradiction. If this linear relation is trivial, the third letter '''U''' occurs an even number of times in ''s'', '''V''' occurs an odd number of times in ''s'', and '''W''' occurs an even number of times. Since gcd(''a'', ''b'') = 1, {'''U''', '''W'''} != {'''X''', '''Y'''}, and so either {'''U''', '''W'''} = {'''X''', '''Z'''} or {'''U''', '''W'''} = {'''Y''', '''Z'''}. But this is again a contradiction since gcd(''a'' + ''b'', ''b'') = gcd(''a'', ''a'' + ''b'') = gcd(''a'', ''b'') = 1.

}}

== Generalizations ==

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 == Ternary interleaved scales ==
-Conjecture: Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there exists a unique single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.
+Given a [[ternary]] [[step signature]] of the form {{nowrap|''a'''''X'''''b'''''Y'''(''a'' + ''b'')'''Z'''}} where gcd(''a'', ''b'') = 1, there always exists a single-period 2-interleaved ternary [[word|scale word]] with that step signature imposing no nontrivial linear relations between step sizes. This ternary scale word consists of ''a''-many '''XZ''' subwords and ''b''-many '''YZ''' subwords arranged in a MOS pattern (like the steps of ''a'''''L'''''b'''''s''') and consists of an interleaved pair of two {{nowrap|''a''('''X''' + '''Z''')''b''('''Y''' + '''Z''')}} subsets, offset by '''Z'''. [[Blackdye]] ('''sLmLsLmLsL''', 5'''L'''2'''m'''3'''s''') and [[whitedye]] ('''LsLsLsmsLsLsms''', 5'''L'''2'''m'''7'''s''') are examples of this.
-=== Lemma ===
+Conjecture: It is the ''unique'' one with this property.
-If ''S'' consists of the subwords '''XZ''' and '''YZ''' arranged in the pattern of a single-period binary circular word ''w''(''x'', ''y'') where {{pipe}}''w''{{pipe}} > 2, and ''k'' is an odd number greater than 1 and less than {{pipe}}''S''{{pipe}} - 1, then the class of ''k''-steps has more than 3 abstract intervals.
-{{proof|contents=Denote by {{pipe}}''w''{{pipe}} the length of subword ''w'' in letters and by ‖''w''‖ the interval subtended by subword ''w'' in its circular word.
-Let ''A'' be the set of all (''k'' - 1)/2-step intervals of ''w''. {{pipe}}''A''{{pipe}} = 1 implies that ''k'' = 1 mod {{pipe}}''w''{{pipe}}, so {{pipe}}''A''{{pipe}} &ge; 2 and contains at least two intervals '''w'''<sub>1</sub> and '''w'''<sub>2</sub>.
-Say {{pipe}}''A''{{pipe}} = 2. (If {{pipe}}''A''{{pipe}} &ge; 3 then the proof is easy.) Say ''B'' is the set of all subwords of ''w'' (not intervals) subtending '''w'''<sub>1</sub>, and ''C'' is the set of all subwords of ''w'' subtending '''w'''<sub>2</sub>.
-Choose ''b'' in ''B'' and ''c'' in ''C'' and consider the subwords ''b''('''XZ''', '''YZ''') and ''c''('''XZ''', '''YZ''') of ''S''.
-If ''k'' = 3, then we can just say ''b'' = ''x'' and ''c'' = ''y''. Suppose '''YZX''' does not occur in ''S''; then ''yx'' does not occur in ''w'' which is a contradiction, so ''YZX'' must occur.
-=> '''ZYZXZ''' occurs
-=> either  '''XZYZXZ''' or '''YZYZXZ''' occurs (with suffix '''ZYZXZ''')
-==> If {{pipe}}''w''{{pipe}} = 3, then these are the whole word ''w''; either way we have four abstract intervals.
-==> If {{pipe}}''w''{{pipe}} > 3, then either '''ZYZXZXZ''' or '''ZYZXZYZ''' occurs (with prefix '''ZYZXZ''').
-Case '''XZYZYZXZ''' => '''XZY''', '''YZY''', '''ZYZ''', '''ZXZ'''
-Case '''YZYZYZXZ''' => '''YZY''', '''YZX''', '''ZYZ''', '''ZXZ'''
-So the case ''k'' = 3 is done.
-For any other ''k'', we similarly go through the cases (a) ''bx'' and ''cx'', (b) ''bx'' and ''cy'', (c) ''by'' and ''cx'', (d) ''by'' and ''cy''.
-(a) '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''X''', ''c''('''XZ''', '''YZ''')'''X''' => done
-(d) is similar to (a).
-(c) We have '''Z'''''b''('''XZ''', '''YZ'''), '''Z'''''c''('''XZ''', '''YZ'''), ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X'''. If the sizes of ''b''('''XZ''', '''YZ''')'''Y''' and ''c''('''XZ''', '''YZ''')'''X''' are different we are done. If they are the same, we have that ''by'' and ''cx'' subtend the same interval in ''w'', hence ''b'' has one more ''x'' and one fewer ''y'' than ''c''.
-By scooting ''bx'' one letter to the left in ''w'', we find either (i) ''xb'' or (ii) ''yb''. The {{pipe}}''b''{{pipe}}-letter prefix ''p'' of this subword subtends either the same interval as (1) ''b'' or (2) ''c''.
-(i), (1) => ''b'' = ''b'x'', ''p'' = ''xb' '', have ''xby'' = ''xb'xy'', and we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.
-(ii), (1) => ''b'' = ''b'y'', ''p'' = ''yb' '', continue scooting to the left until you find an ''x'' to the left.
-(i), (2) => ''b'' = ''b'y'', ''p'' = ''xb' '', we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.
-(ii), (2) => ''b'' =''b'x'', ''p'' = ''yb' '', we have ‖''p''('''XZ''', '''YZ''')'''X'''‖ as a fourth interval size.
-(c) is similar to (b).
-}}
-=== Attempted proof of Conjecture ===
-{{proof|contents=Suppose the scale is made of two interleaved subsets offset by the abstract interval '''δ'''.
-Case 1: gcd(2(''a'' + ''b''), ''k'') = 1. If ''k'' = 1, then any '''Z'''<sup>''q''</sup> a maximal subword of consecutive '''Z'''s has ''q'' odd, and they all must be the same length and separated by one non-'''Z''' letter '''W''':
-'''ZZ ZZ ... ZW''' (in strand ''S''<sub>1</sub>)
-'''Z ZZ ... ZZ WZ''' (in strand ''S''<sub>2</sub>)
-Moreover, the offset '''δ''' = '''Z''', i.e. ''S''<sub>2</sub> is separated by the interval '''Z''' to the right of ''S''<sub>1</sub>.
-If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of length ''a'' + ''b'', ''w''<sub>1</sub> with the maximal number of consecutive '''Z''''s and ''w''<sub>2</sub> with the minimal number of '''Z'''s. Scoot ''w''<sub>1</sub> to the right one step at a time until it loses one '''Z'''. This proves either that a non-'''Z''' letter is equal to '''Z''' or that the offset goes in the wrong direction. Hence ''q'' = 1, as desired.
-If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T'', which traverses all notes of ''S'' since gcd(''k'', 2(''a'' + ''b'')) = 1. Since ''k'' is odd, the letters of this word alternate between beginning in ''S''<sub>1</sub> and beginning in ''S''<sub>2</sub>. By a reasoning similar to the above, ''T'' has a letter '''δ''' between its two mutually interleaved strands. By the lemma this scale has step variety ''r'' > 3, say with letters '''W'''<sub>1</sub>, ..., '''W'''<sub>''r''</sub>. Let ''k' '' be the inverse of ''k'' mod 2(''a'' + ''b''). By stacking ''k' ''-step subwords of ''T'', we end up with at least 4 different linear equations with 3 unknowns '''X''', '''Y''', and '''Z''', implying a nontrivial linear relation between '''X''', '''Y''', and '''Z''' (?). This is a contradiction as '''X''', '''Y''', and '''Z''' are assumed to not have a linear relation.
-Case 2: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving property, hence ''k'' = ''a'' + ''b'' which must be odd.
-Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''.
-We have '''V'''w'''W''' where '''V''' != '''W''', ‖'''V'''''w''‖ = '''δ''', ‖''w'''''W'''‖ = ‖''s''‖ - '''δ'''
-=> '''δ''' - '''V''' = ‖''w''‖ = ‖''s''‖ - '''δ''' - '''W'''
-=> 2'''δ''' + '''W''' = ‖''s''‖ + '''V'''. If this linear relation is nontrivial, it's a contradiction. If this linear relation is trivial, the third letter '''U''' occurs an even number of times in ''s'', '''V''' occurs an odd number of times in ''s'', and '''W''' occurs an even number of times. Since gcd(''a'', ''b'') = 1, {'''U''', '''W'''} != {'''X''', '''Y'''}, and so either {'''U''', '''W'''} = {'''X''', '''Z'''} or {'''U''', '''W'''} = {'''Y''', '''Z'''}. But this is again a contradiction since gcd(''a'' + ''b'', ''b'') = gcd(''a'', ''a'' + ''b'') = gcd(''a'', ''b'') = 1.
-}}
 == Generalizations ==