User:Inthar/MV3: Difference between revisions

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Todo: Account for XYZYX

Assume the scale word S is not multiperiod. To eliminate words of the form X'Y'Z'Y'X' we manually check all words up to length 5... (todo)

Assume the scale word S is not multiperiod. To eliminate words of the form X'Y'Z'Y'X' we manually check all words up to length 5... (todo) Henceforth we assume len(S) >= 6. [This assumption must be used somewhere!]

~~Now assume len(S) >= 6.~~

Suppose that some class C in W', the word of Ys and Zs, has three sizes, T1, T2, T3. Assume WOLOG that len(C) ≤ 1/2*len(W').

Suppose that some class C in the word of Ys and Zs has three sizes, T1, T2, T3.

Assume T1, T2, T3 occur within a contiguous string of Y's and Z's. Then you get a 4th variant of this class (within the whole scale) by using a string of the same length including an X.

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This part needs to be more careful, taking into account where X can appear:

First assume for simplicity that len(C) = 2. Then there are only three possible lengths: 2Y, 2Z, Y+Z. Suppose all 3 occur. By (*), 2Y and 2Z cannot occur within the same contiguous string. We're assuming S has at least one X.

[Y X Y] [Y X Z] [Z X Z]

[_ _] [_ _] [_ _]

[_ X _] [_ X _] [_ X _]

[_ X X _] [_ X X _] [_ X X _]

[_ X X X _] [_ X X X _] [_ X X X _]

...

If the occurrence of any class has an X inserted in the middle, then we can scoot it left or right until we have one of T1(possibly with inserted X's)+X, T2(possibly with inserted X's)+X, or T3 (possibly with inserted X's)+X. Scoot the string with the least X's to the left and you lose the X on the right, and gain another non-X letter on the left so you get a fourth variant of this interval class that contains T1 + X, a contradiction. (Check this again...)

@@ Line 40: / Line 40: @@
 Todo: Account for XYZYX
-Assume the scale word S is not multiperiod. To eliminate words of the form X'Y'Z'Y'X' we manually check all words up to length 5... (todo)
+Assume the scale word S is not multiperiod. To eliminate words of the form X'Y'Z'Y'X' we manually check all words up to length 5... (todo) Henceforth we assume len(S) >= 6. [This assumption must be used somewhere!]
-Now assume len(S) >= 6.
+Suppose that some class C in W', the word of Ys and Zs, has three sizes, T1, T2, T3. Assume WOLOG that len(C) ≤ 1/2*len(W').
-Suppose that some class C in the word of Ys and Zs has three sizes, T1, T2, T3.
 Assume T1, T2, T3 occur within a contiguous string of Y's and Z's. Then you get a 4th variant of this class (within the whole scale) by using a string of the same length including an X.
@@ Line 51: / Line 49: @@
 This part needs to be more careful, taking into account where X can appear:
+First assume for simplicity that len(C) = 2. Then there are only three possible lengths: 2Y, 2Z, Y+Z. Suppose all 3 occur. By (*), 2Y and 2Z cannot occur within the same contiguous string. We're assuming S has at least one X.
+[Y X Y] [Y X Z] [Z X Z]
+[_ _] [_ _] [_ _]
+[_ X _] [_ X _] [_ X _]
+[_ X X _] [_ X X _] [_ X X _]
+[_ X X X _] [_ X X X _] [_ X X X _]
+...
 If the occurrence of any class has an X inserted in the middle, then we can scoot it left or right until we have one of T1(possibly with inserted X's)+X, T2(possibly with inserted X's)+X, or T3 (possibly with inserted X's)+X. Scoot the string with the least X's to the left and you lose the X on the right, and gain another non-X letter on the left so you get a fourth variant of this interval class that contains T1 + X, a contradiction. (Check this again...)