Ternary scale theorems: Difference between revisions

* Indices for all words are 1-indexed.

** If ''s'' is a circular word and {{nowrap|''i'' < 1}} or {{nowrap|''i'' > len(''s'')}}, we first replace ''i'' with {{nowrap|''i'' % len(''s'') + 1}} before using it as an argument in ''s''[-].

Proof: Consider the subword for the closing generator of ''s'' on index ''p'', which is {{nowrap|''I'' {{=}} ''s''[''p'' : ''p'' + ''i'' + ''j'']}}, and suppose the result of deleting all '''X''''s from ''I'' has a ''j''-step subword ''w''. Shifting ''I'' one step to the left and one step to the right, {{nowrap|''s''[''p'' − 1: ''p'' − 1 + ''i'' + ''j'']}} and {{nowrap|''s''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j'']}} are both detemperings of perfect generators of ''T'', and have one fewer non-'''X''' step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-'''X''' letter. Removing all the '''X''''s from ''I'' results in a word that is {{nowrap|''j'' + 1}} letters long and is the ''j''-step word ''w'' with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords of ''s'' can all be obtained by deleting '''X'''s from detempered perfect generators; take {{nowrap|''q'' ≠ ''p''}} to be the index of the first letter of one such ''j''-step subword (as contained in ''s'') and use {{nowrap|''s''[''q'' : ''q'' + ''i'' + ''j'']}}.

# ''s'' is a union of two copies of a primitive MOS ''M'' of size {{sfrac|''n''|2}} generated by '''g'''; thus it is a [[flought scale]] obtained by taking two offset copies of said primitive MOS.

Λ₂ has a chunk of β (after the first β′) of size ''x'' where {{nowrap|''x'' {{=}} {{floor|''n''/μ}}}} {{nowrap|≥ {{floor|''n''/{{floor|''n''/2}}}}}} = 2 or {{nowrap|''x'' {{=}} {{ceil|''n''/μ}}}} {{nowrap|{{=}} {{floor|''n''/μ}} + 1}}. Hence Λ₃ has a chunk of γ of size ''x''. Λ₃ also has a chunk that contains {{nowrap|Λ₃[''n'' : 2]}} as a subword. This chunk must be of size ''y'', where  

Λ₂ has only two chunks of size 1, {{nowrap|Λ₂[''n'' − 1]}} and Λ₂[1], since otherwise Λ₃ would have a chunk of size 1 within {{nowrap|Λ₃[1 : ''n'']}}. Thus Λ₂ has exactly one chunk of size 2. Thus {{nowrap|Λ₂ {{=}} ββ′βββ′ββ′}} and {{nowrap|Λ₃ {{=}} γγ′γγγ′γγ}}. Thus we have:

For 7.1.3: The ternary Fraenkel word may be verified as SV3 by inspection, and we have already shown in Theorem 1 that odd-regular balanced scales are SV3. To show that even-regular balanced scales are ''not'' SV3, observe that {{nowrap|(''a'' + ''c'')}}-steps come in only 2 sizes in such a scale ''s'': {{nowrap|{{floor|''a''/2}}'''X''' + {{ceil|''a''/2}}'''Y''' + ''c'''''Z'''}} and {{nowrap|{{ceil|''a''/2}}'''X''' + {{floor|''a''/2}}'''Y''' + ''c'''''Z'''}}, since the underlying MOS 2''a'''''X'''2''c'''''Y''' only has the {{nowrap|(''a'' + ''c'')}}-step {{nowrap|''a'''''X''' + ''c'''''Z'''}}. The construction replaces the '''X'''s in these subwords with alternating '''X'''s and '''Y'''s; either of '''X''' or '''Y''' may occur first, corresponding to the two possible sizes, since ''a'' is odd and thus the {{nowrap|(''a'' + ''c'')}}-step subword {{nowrap|''s''[''k'' : ''k'' + ''a'' + ''c'']}} becomes the subword {{nowrap|''s''[''k'' + ''a'' + ''c'' : ''k'' + 2''a'' + 2''c'']}} via interchanging '''X''' and '''Y'''.

#* Interchange a '''Z''' and an '''X''' at some (possibly more than one) of the boundaries of these copies of the MOS word ''w''. Here, the boundary of two consecutive copies of ''w'' is the last letter of the first word and the first letter of the second word. (At the ends of the whole multiMOS word, the boundaries are just the first and last letters of the word.) For example, let ''w'' be the multiMOS word 8'''X'''6'''Z''', '''XXZXZXZXXZXZXZ'''. Then the border between the copies of the MOS subword '''XXZXZXZ''' are ''w''[7]''w''[8] and ''w''[14]''w''[1] (using one-based numbering).