Recursive structure of MOS scales: Difference between revisions

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Note that the latter two words have at most k s's.

If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. It suffices to consider the case where w2 ∩ w3 contains at least k-1 complete chunks, since assuming otherwise would mean that W₂(Lr+1s, Lrs) has at least k+2 chunks, which would contradict the length of W₂(λ, σ).

If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. It suffices to consider the case where w2 ∩ w3 contains at least k-1 complete chunks, since assuming otherwise would mean that W₂(Lr+1s, Lrs) has at least k+1 chunks, which would contradict the length of W₂(λ, σ).

Hence it suffices to consider the following two cases, each split into subcases:

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w3: <L ... sXL ... sXL ... sXL ... s> (complete chunks only)

~~Truncate the strings as follows, to get three distinct K-steps in w:~~

So W₂(Lr+1s, Lrs) has k+1 chunks, a contradiction.

~~w1':~~ L ~~... [lop off the~~ s ~~at the beginning of w1~~, ~~so w1' has k s's]~~

~~w2':~~ <~~L ... sXL ... sXL ... sXL ... ] [lop off one s at the end of w2, so w2' has one fewer s than w3']~~

~~w3':~~ <~~L ... sXL ... sXL ... sXL ...~~ s~~> [lop off an L at the beginning of w3, so w3'~~ has ~~at most~~ k-1 ~~s's]~~

~~so this contradicts our original scale being~~ a ~~mos~~.

So we must have

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w2: <L ... sXL ... sXL ... sXL ... sXL ... s> (complete chunks only)

w3: [sXL ... sXL ... sXL ... sXL ... s> (s followed by complete chunks)

~~⇒ do~~ the ~~same trick~~ as in ~~Case~~ 1.2

Truncate the strings w1, w2, w3 as follows, to get three distinct K-steps in w:

w1': L ... [lop off the s at the beginning of w1, so w1' has k s's]

w2': <L ... sXL ... sXL ... sXL ... ] [lop off one s at the end of w2, so w2' has one fewer s than w3']

w3': <L ... sXL ... sXL ... sXL ... s> [lop off an L at the beginning of w3, so w3' has at most k-1 s's]

so this contradicts our original scale being a mos.

Case 2.2:

@@ Line 415: / Line 415: @@
 Note that the latter two words have at most k s's.
-If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. It suffices to consider the case where w2 ∩ w3 contains at least k-1 complete chunks, since assuming otherwise would mean that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) has at least k+2 chunks, which would contradict the length of W₂(λ, σ).
+If w₂ contains fewer complete chunks (<L...Ls preceded by where < is the left chunk boundary) or at least 2 more than w₃, we are done, since they must automatically have different numbers of s's. It suffices to consider the case where w2 ∩ w3 contains at least k-1 complete chunks, since assuming otherwise would mean that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) has at least k+1 chunks, which would contradict the length of W₂(λ, σ).
 Hence it suffices to consider the following two cases, each split into subcases:
@@ Line 434: / Line 434: @@
   w3:         <L ... sXL ... sXL ... sXL ... s> (complete chunks only)
-Truncate the strings as follows, to get three distinct K-steps in w:
+So W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) has k+1 chunks, a contradiction.
- w1':  L ... [lop off the s at the beginning of w1, so w1' has k s's]
- w2': <L ... sXL ... sXL ... sXL ... ] [lop off one s at the end of w2, so w2' has one fewer s than w3']
- w3':         <L ... sXL ... sXL ... sXL ... s> [lop off an L at the beginning of w3, so w3' has at most k-1 s's]
-so this contradicts our original scale being a mos.
 So we must have
@@ Line 461: / Line 457: @@
   w2: <L ... sXL ... sXL ... sXL ... sXL ... s> (complete chunks only)
   w3:               [sXL ... sXL ... sXL ... sXL ... s> (s followed by complete chunks)
-⇒ do the same trick as in Case 1.2
+Truncate the strings w1, w2, w3 as follows, to get three distinct K-steps in w:
+ w1':  L ... [lop off the s at the beginning of w1, so w1' has k s's]
+ w2': <L ... sXL ... sXL ... sXL ... ] [lop off one s at the end of w2, so w2' has one fewer s than w3']
+ w3':         <L ... sXL ... sXL ... sXL ... s> [lop off an L at the beginning of w3, so w3' has at most k-1 s's]
+so this contradicts our original scale being a mos.
 Case 2.2: