Generator-offset property: Difference between revisions

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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 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, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [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".]

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 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, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [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 whenever the boundaries of the substring are moved 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".]

Any generator of aX 2bW must have an odd number of W steps; if a generator had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction since the generator of aX bW' can be generated by a generator of aX 2bW. This with (4) immediately gives (5).

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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 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, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [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".]
+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 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, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [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 whenever the boundaries of the substring are moved 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".]
 Any generator of aX 2bW must have an odd number of W steps; if a generator had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction since the generator of aX bW' can be generated by a generator of aX 2bW. This with (4) immediately gives (5).