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) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. | 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. Hence it suffices to consider the following two cases, each split into subcases: | ||
Case 1 | Case 1 | ||