Ternary scale theorems: Difference between revisions

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==== Proof ====

For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. ~~Thus it~~ remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced.

We will first prove that a balanced circular word is primitive iff rhe gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.

It remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.

(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.

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 ==== Proof ====
-For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced. Thus it remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.
+For 5.1.1: We showed previously that the Fraenkel, odd-regular, and even-regular circular words are balanced.
+We will first prove that a balanced circular word is primitive iff rhe gcd of the step signature is 1. Proof sketch: let ''d'' be the gcd of the step signature. (''n''/''d'')-step multisets come in 1 size, namely the equave divided by ''d'', because if some letter count differs, then we get 3 values for this letter count for (''n''/''d'')-step multisets by the discrete IVT.
+It remains to show that (a) ternary balanced words are pairwise-MOS (b) if ''a'' > ''b'' > ''c'', then ''s'' is equivalent to the Fraenkel word (c) assuming ''a'' != ''b'' = ''c'' any ''s'' that is not odd-regular or even-regular is not balanced.
 (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.