Ternary scale theorems: Difference between revisions

Line 278:

(a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.

(b) Assume that the projection ''p'''''YZ'''(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z''' is an ''r''-period MOS with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p'''''XY'''(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and ''p'''''XZ'''(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods. It suffices to show that ''r'' = 2. If ''r'' > 2, then...

(b) Assume that the projection ''p'''''YZ'''(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z''' is an ''r''-period MOS with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p'''''XY'''(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and ''p'''''XZ'''(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' > 2, then...

For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, ''a''''i'''''X''' + ''b''''i'''''Y''' + ''c''''i'''''Z''' = (''a''''i'', ''b''''i'', ''c''''i'') for ''i'' = 1, 2, 3. We may assume (''a''2, ''b''2, ''c''2) = (''a''1, ''b''1 + 1, ''c''1 - 1). Then either (''a''3, ''b''3, ''c''3) = (''a''1 + 1, ''b''1, ''c''1 - 1) or (''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.

@@ Line 278: / Line 278: @@
 (a) Let ''s'' be a ternary balanced word; then for any given letter '''y''' the number of '''y'''s in a subword of any given length ''L'' varies by at most 1. Thus the same is true when we count all non-'''y''' letters in any subword of length ''L''; thus when we equate '''x''' and '''z''', the count of the resulting letter in any subword of length ''L'' differs by 1. Being a binary balanced word is one characterization of the MOS property.
-(b) Assume that the projection ''p''<sub>'''YZ'''</sub>(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z'''  is an ''r''-period MOS with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p''<sub>'''XY'''</sub>(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and  ''p''<sub>'''XZ'''</sub>(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods. It suffices to show that ''r'' = 2. If ''r'' > 2, then...
+(b) Assume that the projection ''p''<sub>'''YZ'''</sub>(''s'') identifying '''Y''' and '''Z''' of a primitive PMOS scale ''s'' with signature ''ar'''''X''' ''b'''''Y''' ''c'''''Z'''  is an ''r''-period MOS with step signature ''ar'''''X''' ''dr'''''W'''. We claim that neither ''b'' nor ''c'' is divisible by ''r''. Since ''p''<sub>'''XY'''</sub>(''s'') is the MOS (''ar'' + ''b'')'''W''' ''c'''''Z''' and  ''p''<sub>'''XZ'''</sub>(''s'') is the MOS (''ar'' + ''c'')'''W''' ''b'''''Y''', if either ''b'' and ''c'' is divisible by ''r'', then the distributions of two of the letters have ''r'' periods. Then the distribution of the third letter also has ''r'' periods, meaning that ''s'' itself has ''r'' periods, a contradiction. It suffices to show that ''r'' = 2. If ''r'' > 2, then...
 For 7.1.2: Suppose ''s'' is balanced and has at least three sizes for ''k''-steps, ''a''<sub>''i''</sub>'''X''' + ''b''<sub>''i''</sub>'''Y''' + ''c''<sub>''i''</sub>'''Z''' = (''a''<sub>''i''</sub>, ''b''<sub>''i''</sub>, ''c''<sub>''i''</sub>) for ''i'' = 1, 2, 3. We may assume (''a''<sub>2</sub>, ''b''<sub>2</sub>, ''c''<sub>2</sub>) = (''a''<sub>1</sub>, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub> - 1). Then either (''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) = (''a''<sub>1</sub> + 1, ''b''<sub>1</sub>, ''c''<sub>1</sub> - 1) or (''a''<sub>3</sub>, ''b''<sub>3</sub>, ''c''<sub>3</sub>) =  (''a''<sub>1</sub> - 1, ''b''<sub>1</sub> + 1, ''c''<sub>1</sub>). In both cases, by balancedness applied to subwords of length ''k'', the three vectors represent the only possible interval sizes.