User blog comment:P進大好きbot/New Difference Sequence System/@comment-43345498-20190924135704/@comment-35470197-20190924215841

When Mountain(a,Trivial_L,i) is very simple, then it can be computed in an effective way. By the restriction of Royal(b^{(n)},s^{(n)}), we have only to compute b^{(n)} by repeting "the reconstruction from the difference sequences".

For example, when a = (1,3), we have (h,r,k) = (0,0,0), Royal(a,Trivial_L) = (1,2), and c = (1,1,…). The computation of b^{(3)} is computed in the following way: & & \begin{eqnarray*} \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & ? \\ 0 & 1 & ? & ? \\ 1 & ? & ? & ? \end{array} \\ & & \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & \color{red}{1+1} \\ 0 & 1 & \color{red}{1+1} & ? \\ 1 & \color{red}{1+1} & ? & ? \end{array} \\ & & \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 2 & \color{red}{2+2} \\ 1 & 2 & \color{red}{2+2} & ? \end{array} \\ & & \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 2 & 4 \\ 1 & 2 & 4 & \color{red}{4+4} \end{array} \\ & & \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 2 & 4 \\ 1 & 2 & 4 & 8 \end{array} \\ & & b^{(3)} = (1,2,4,8) \end{eqnarray*} Similarly, when a = (1,4), we have b^{(3)} = (1,9,27,81). However, it has a bug when Mountain is complicated. (The original Y-sequence system also has an unsolved problem in that case.) I need to fix the computation of b^{(n)} so that the computation of b^{(n)} actually terminates.