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 S = aX bY bZ is obtained from the (single-period) 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).

# If |S| is odd, then S = aX bY bZ is obtained from the (single-period) 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). Moreover, the two alternants differ by replacing one Y with a Z.

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

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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 preimage of) the 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 T be the subword consisting only of Y's and Z's. If T has a substring of length j 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>

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 preimage of) the 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 T be the subword consisting only of Y's and Z's. If T has a substring of length j 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".]

The generator of aX 2bW must have an odd number of W steps; if it had an even number of steps, it would be generated by the geneartor of the mos 2X bW' with W' = 2W, a contradiction. Since we just proved that Y and Z alternate in ''S'', the claim about the alternants immediately follows. <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 S = aX bY bZ is obtained from the (single-period) 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).
+# If |S| is odd, then S = aX bY bZ is obtained from the (single-period) 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). Moreover, the two alternants differ by replacing one Y with a Z.
 [Note: This is not true with SGA replaced with GO; [[blackdye]] is a counterexample that is MV4.]
@@ Line 73: / Line 73: @@
 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 preimage of) the 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 T be the subword consisting only of Y's and Z's. If T has a substring of length j 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>
+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 preimage of) the 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 T be the subword consisting only of Y's and Z's. If T has a substring of length j 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".]
+The generator of aX 2bW must have an odd number of W steps; if it had an even number of steps, it would be generated by the geneartor of the mos 2X bW' with W' = 2W, a contradiction. Since we just proved that Y and Z alternate in ''S'', the claim about the alternants immediately follows. <math>\square</math>
 === Proposition 2 ===