MOS substitution: Difference between revisions

Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{MOS}(a+c,b;0)(\mathbf{X},\mathbf{m}).</math> This is the mode of <math>T</math> that has the most <math>\mathbf{X}</math> steps near the end. If <math>T</math> is [[primitive]], let <math>r</math> be the count of <math>\mathbf{X}</math> steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>n,</math> <math>r</math>-steps in the filling MOS <math>F = \mathsf{MOS}(a,c;k)(\mathbf{L},\mathbf{s})</math> come in exactly 2 sizes, <math>i\mathbf{L}+j\mathbf{s}</math> and <math>(i-1)\mathbf{L}+(j+1)\mathbf{s}.</math> Since the detempering of the imperfect generator of <math>T</math> occurs only once in <math>S</math>, <math>S</math> admits a particularly elegant well-formed binary (using two distinct generators) [[generator sequence]] of length <math>q,</math> corresponding to the circle of <math>r</math>-steps in the filling MOS. Letting <math>\mathsf{GS}(g_1, ..., g_{q})</math> be this generator sequence, <math>g_j</math> is either <math>p\mathbf{m} + i\mathbf{L} + j\mathbf{s}</math> or <math>p\mathbf{m} + (i-1)\mathbf{L} + (j+1)\mathbf{s},</math> according as the <math>j</math>-th <math>r</math>-step in the sequence of stacked <math>r</math>-steps on the chosen mode of <math>F</math> is <math>i\mathbf{L} + j\mathbf{s}</math> or <math>(i-1)\mathbf{L} + (j+1)\mathbf{s}.</math> (We could have chosen to use the mode of <math>T</math> on the other extreme of its generator arc instead, which corresponds to taking the circle of <math>(n - r)</math>-steps in <math>F</math> and is thus also valid.) The generator of the template MOS serves as the "guide generator" for this generator sequence.

==Examples==