User blog:Koteitan/Categorizing of the rule sets for all sub versions of bashicu matrix

This is the table of the rule sets of (almost) all sub versions invented in 2015-2018. The aim of this entry is the comparison of the rule sets of them. The all sub versions can be categorized according to the three micro rules; Bad root searching rule, Bad part ascending rule and ascension modification.

In this article, the all sub versions are described based on mathematical definition (than their original define in some sub version).

Rule set

 * bad root
 * The bad root is the leftmost column of in bad part.

settings

 * Bashicu matrix consist a matrix whose element is the natural number and a natural number named 'bracket'.
 * The matrix $$\begin{pmatrix}c_{11}&c_{12}\\c_{21}&c_{22}\end{pmatrix}$$ in the Bashicu matrix is often described as $$(c_{11},c_{21})(c_{12},c_{22})$$. In this page it is describe so.
 * $$f(n)$$ is the function from a natural number into a natural number. in BM1, rTSS, BM1.1, Bubby3's fix, BM2, $$f(n)=n^2$$and in BM2.1, BM2.2, BM2.3 $$f(n)=n+1$$.

Common Rules
A expansion procedure of a Bashicu matrix in a step is shown as below: $$(a_{11},\cdots, a_{N1})\cdots(a_{1L},\cdots, a_{NL}) \otimes (b_{11},\cdots, b_{N1})\cdots(b_{1L},\cdots, b_{NL}) $$ $$= (a_{11}b_{11},\cdots, a_{N1}b_{N1})\cdots(a_{1L}b_{1L},\cdots, a_{NL}b_{NL})$$ That's all.
 * 1) Let the rightmost column be $$C=(c_1, c_2, \cdots c_N)$$.
 * 2) If the all elements of $$C$$ is $$0$$, The bad root is not found and the Jump to 6.
 * 3) Let the index of the lowermost non-zero element of the $$C$$ be $$t$$. ($$c_t\gt 0\land \forall k(k\gt t\rightarrow c_k=0)$$)
 * 4) Decide the bad root bad root $$R=(r_1, r_2, \cdots, r_N)$$ with the bad rule searching rule.
 * 5) remove the rightmost column $$C$$.
 * 6) If the bat root is not found, replace $$[n]$$ into $$[f(n)]$$ and end the expansion procedure.
 * 7) let the bad root $$R$$ and the partial matrix in the right side of it be the bad part $$B=(b_{11},b_{21},\cdots,b_{N1})(b_{12},b_{22},\cdots,b_{N1})\cdots(b_{1L},b_{2L},\cdots,b_{NL})$$.
 * 8) Though bad root is copied after this, $$\Delta=(\delta_{11}, \cdots, \delta_{N1})\cdots(\delta_{1L},\cdots,\delta_{NL})$$is defined as the ascension ranges as: $$\forall j~\delta_{ij}=\begin{cases}c_i-r_i&(i\lt t)\\0&(i \geq t)\end{cases}$$.
 * 9) As the flag if the each elements of the bad part ascend, decide a pre-ascending matrix $$A'=(a'_{11}, a'_{21}, \cdots, a'_{N1})(a'_{12}, a'_{22}, \cdots, a'_{N2})\cdots(a'_{1L}, a'_{2L}, \cdots, a'_{NL})$$ as the 'bad part ascending rule'.
 * 10) As the ascending modification rule, decide an ascending matrix $$A$$ with the pre-ascending matrix $$A'$$.
 * 11) With the value of the bracket $$n$$, combine the $$B_1, B_2, \cdots, B_n$$ which is defined as below into the right of the matrix.
 * $$B_1 = B + A \otimes \Delta$$
 * $$B_{k+1} = B_k + A \otimes \Delta$$
 * $$\otimes$$ is the element-wise product as below:

After the repetitions of the expansion above, The value of the bracket when the matrix is empty is the large number.

Left method

 * 1) If there exists a column $$M=(m_1, \cdots, m_N)$$ which meets $$m_k > c_k$$ in the row $$t$$ and any upper row $$k \leq t$$, and which is in left side of $$c$$, let the right-side column for $$M=(m_1, \cdots, m_N)$$ be the bad root $$R$$. If it is not found, the bad root is not found.

Upper-branch-ignoring-model
(1) $$p$$ is in the same row of $$c$$. (2) $$p$$ is in the left side of $$c$$. (3) $$p$$ is smaller than $$c$$. (4) $$p$$ is in the top row, or y is in the same column as the ancestor of $$c'$$. ($$c'$$ is the element in the next row up of $$c$$.)
 * 'the ancestor of the $$c$$' is defined as '$$c$$ itself' or 'the ancestor of the parent of $$c$$' recursively.
 * 'the parent of the $$c$$' is the rightmost element $$p$$ which meets all of below recursively:

Then, the bad root $$R$$ is the column, in which there is the parent of the lowermost non-zero element of the rightmost column. If the parent is not found, the bad root is not found and end.

Concestor method
(1) $$p$$ is in the same row of $$c$$. (2) $$p$$ is in the left side of $$c$$. (3) $$p$$ is smaller than $$c$$.
 * 'the ancestor of the $$c$$' is defined as '$$c$$ itself' or 'the ancestor of the parent of $$c$$' recursively.
 * 'the parent of the $$c$$' is the rightmost element $$p$$ which meets all of below recursively:

Then, the bad root $$R$$ is the column, in which there is the parent of the lowermost non-zero element of the rightmost column. If the parent is not found, the bad root is not found and end.

All branches enabled

 * $$\forall i,j(a'_{ij}=1)$$

BM2-based
According to the index of column $$j$$, define the column which is in the uppermost low and which is in the left side of the column $$j$$ and in which there is the leftmost element which is smaller than the value of the uppermost element of column $$j$$ be the 'parent column $$p(j)$$ of the column $$j$$'.


 * For all the combination of $$i, j$$ which meets $$j=1 \lor (r_i \lt b_{ij} \land p(j)\mathrm{exists} \land a'_{ip(j)}=1) $$, $$a'_{ij}=1$$, otherwise $$a'_{ij}=0$$

Upper-branch-ignoring-model
(1) $$p$$ is in the same row of $$c$$. (2) $$p$$ is in the left side of $$c$$. (3) $$p$$ is smaller than $$c$$. (4) $$p$$ is in the top row, or y is in the same column as the ancestor of $$c'$$. ($$c'$$ is the element in the next row up of $$c$$.)
 * 'the ancestor of the $$c$$' is defined as '$$c$$ itself' or 'the ancestor of the parent of $$c$$' recursively.
 * 'the parent of the $$c$$' is the rightmost element $$p$$ which meets all of below recursively:


 * For all $$i, j$$ which meet $$r_i$$ is the ancestor of $$b_{ij}$$, $$a'_{ij}=1$$, otherwise $$a'_{ij}=0$$.

no modify
$$\forall ij (a_{ij}=a'_{ij})$$

All 1 or (1,0,…,0)
For all $$j$$, define $$a_{ij}$$ from $$a'_{ij}$$ as below:
 * If $$\forall i(a'_{ij}=1)$$ then $$\forall i(a_{ij}=1)$$ otherwise $$a_{ij}=\begin{cases}1&(i=1)\\0&(i \gt 1)\end{cases}$$.

All 1 or All 0
For all $$j$$, define $$a_{ij}$$ from $$a'_{ij}$$ as below:
 * if $$\forall i(a'_{ij}=1)$$ then $$\forall i(a_{ij}=1)$$, otherwise $$\forall i(a_{ij}=0)$$.