Generator-offset property: Difference between revisions

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# ''S'' is of the form ''ax by bz'' for some permutation (''x'', ''y'', ''z'') of (L, M, s).

# The cardinality (size) of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').

# If |S| is odd, then the SGA scale aX bY bZ is obtained from the mos aX 2bW by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality).

[Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.]

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(The above holds for any odd ''n'' ≥ 3.)

~~Now~~ we only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning. <math>\square</math>

For (1), we now only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.

For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)

In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) the (“detempered”) imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to MV3. [Let S be the subword consisting only of Y's and Z's If there is a substring of length j in the string that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".] <math>\square</math>

=== Proposition 2 ===

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 # ''S'' is of the form ''ax by bz'' for some permutation (''x'', ''y'', ''z'') of (L, M, s).
 # The cardinality (size) of ''S'' is either odd, or 4 (and ''S'' is of the form ''xyxz'').
+# If |S| is odd, then the SGA scale aX bY bZ is obtained from the mos aX 2bW by replacing all the W's successively with alternating Y's and Z's (or alternating Z's and Y's for the other chirality).
 [Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.]
@@ Line 68: / Line 69: @@
 (The above holds for any odd ''n'' ≥ 3.)
-Now we only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning. <math>\square</math>
+For (1), we now only need to see that SGA + odd cardinality => unconditionally MV3. But the argument in case 2 above works for any interval class (unconditional MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
+For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. If Y’s and Z’s don't alternate perfectly, then (ignoring X's) you have two consecutive Y's somewhere and two consecutive Z's somewhere else. Assume that g_pf = iX + jW with j >=2. (If this is not true, invert the generator, since b > 1.)
+In aX bY bZ, consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) the (“detempered”) imperfect generator. Since a + 2b >= 5, are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to MV3. [Let S be the subword consisting only of Y's and Z's If there is a substring of length j in the string that's not contained in a perfect generator, you can go somewhere else to find it, since the numbers of Y's and Z's change one at a time and reach a maximum and a minimum somewhere, guaranteeing that intermediate values are reached "on the other side".] <math>\square</math>
 === Proposition 2 ===