Recursive structure of MOS scales: Difference between revisions
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# w₃(L, s) = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | # w₃(L, s) = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) | ||
If w₂ contains fewer complete chunks (L...Ls) than w₃, we are done. | If w₂ contains fewer complete chunks (L...Ls) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. | ||
If w₂ and w₃ (which have the same length) contain the same number of complete chunks, one case is (X denotes a chunk boundary, < > are chunk boundaries that are also the boundary of the word, [] are non-chunk-boundary word boundaries.) | If w₂ and w₃ (which have the same length) contain the same number of complete chunks, one case is (X denotes a chunk boundary, < > are chunk boundaries that are also the boundary of the word, [] are non-chunk-boundary word boundaries.) | ||
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(⇒ w₃ has one more s). | (⇒ w₃ has one more s). | ||
If w₂ has more complete | If w₂ has one more complete chunk than w₃: | ||
Case 2.1: | Case 2.1: | ||