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. Suppose for the sake of contradiction that Y’s and Z’s don't alternate perfectly. Assume that the perfect generator of aX 2bW is iX + jW with j ≥ 2. (If j = 1, we can invert the generator to make j ≥ 2, since b > 1.)

In ''S'', 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, having a different number of X’s than (1), (2), and (3). 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. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached.]

In ''S'', 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, having a different number of X’s than (1), (2), and (3). 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. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one root.]

Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). 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. Suppose for the sake of contradiction that Y’s and Z’s don't alternate perfectly. Assume that the perfect generator of aX 2bW is iX + jW with j ≥ 2. (If j = 1, we can invert the generator to make j ≥ 2, since b > 1.)
-In ''S'', 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, having a different number of X’s than (1), (2), and (3). 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. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached.]
+In ''S'', 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, having a different number of X’s than (1), (2), and (3). 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. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one root.]
 Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5).