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链加法 mod 10

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链加法 mod 10

是否可以在 LaTeX 中创建一个类似表格的结构，表示对各个单元进行模 10 的连续模 10 加法和环绕式？这称为链式添加。

例如，从 123456 开始，链式添加，我们有

┌────────── 转移 ── ──────────┐
| |
| 五
1┌2┌3┌4┌5┌6┌1
│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │
3 5 7 9 1 7

（模 10 加法）得到 357917。再次进行链式加法，我们得到

┌────────── 转移 ── ──────────┐
| |
| 五
3 ┌ 5 ┌ 7 ┌ 9 ┌ 1 ┌ 7 ┌ 3
│ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │
8 2 6 0 8 0

（对 10 进行模加）得到 826080。将其制成表格，我们得到：

1 2 3 4 5 6
3 5 7 9 1 7
8 2 6 0 8 0
0 8 6 8 8 8
. . . . . .

等等等等。

您会注意到它的添加没有进位，因此对 10 进行取模。

我希望能够只输入起始数字并生成包含其他数字的表格

我知道你可以在 LaTeX 中进行数学计算，但我不知道从哪里开始才能选择初始输入。

答案1

电子表格的作业：

\documentclass{article}
\usepackage{spreadtab}
\begin{document}
\begin{spreadtab}{{tabular}{*6{c}}}
\SThiderow 1 & 2 & 3 & 4 & 5 & 6\\
\STcopy{>5,v}{a1+b1-10*trunc((a1+b1)/10,0)}&&&&&\STcopy{v}{a1+f1-10*trunc((a1+f1)/10,0)}\\
&&&&&\\
&&&&&\\
&&&&&\\
&&&&&\\
&&&&&\\
\end{spreadtab}
\end{document}

在此处输入图片描述

答案2

第二次更新增加了最终循环的自动确定功能。添加在答案底部。

第一次更新添加表格：我不确定 的具体含义table，因此第一个版本只是将数字链输出为连续的段落。带有表格的版本位于答案的底部。

\documentclass{article}

\newcommand*{\modten}[1]{\the\numexpr #1+10-((#1+5)/10)*10\relax}

\makeatletter
% expandable implementation
% #1 = initial string
% #2 number of reps
\newcommand*{\chainadd}[2]{#1\endgraf\chainadd@i {#2}#1\relax }

\def\chainadd@i #1#2#3\relax {\ifnum #1=0 \expandafter\@gobble\else
                              \expandafter\@firstofone \fi
      {\expandafter\chainadd@ii\expandafter{\the\numexpr #1-1}{}#2#3#2\relax }}

\def\chainadd@ii #1#2#3#4{\ifx\relax #4\expandafter\@firstoftwo\else
                       \expandafter\@secondoftwo\fi
   {#2\endgraf\chainadd@i {#1}#2\relax}%
   {\expandafter\expandafter\expandafter\chainadd@iii\modten{#3+#4}{#2}{#1}#4}}

\def\chainadd@iii #1#2#3{\chainadd@ii {#3}{#2#1}}
\makeatother

\begin{document}

\chainadd {123456}{10}

\bigskip

\chainadd {1234567}{10}

\end{document}

在此处输入图片描述

将输出显示为表格行的代码：我不确定输入数据是否是输出的第一行。此外，表格中的列数必须由用户提供，如果需要，可以进行修改，不仅可以\chainadd生成行，还可以\begin{tabular}使用\end{tabular}合适的表格模板生成行。

由于核心实现是可扩展的，因此可以\chainadd最容易地将其自身插入表格中。

与第一个实现一样，\chainadd它有两个参数，第一个是任意长度的十进制数字链，第二个是该过程必须重复的次数。初始数据构成输出的第一行。然后，根据第二个参数的要求，还有尽可能多的附加行\chainadd。

\documentclass{article}

\newcommand*{\modten}[1]{\the\numexpr #1+10-((#1+5)/10)*10\relax}

\makeatletter

% expandable implementation
% #1 = initial string
% #2 number of reps

% Update with added \ChainRow to prepare tabular rows

\newcommand*{\chainadd}[2]{\ChainRow{#1}\chainadd@i {#2}#1\relax }

\def\chainadd@stop #1\relax {}

\def\chainadd@i #1#2#3\relax {\ifnum #1=\z@\expandafter\chainadd@stop\fi
    \expandafter\chainadd@ii\expandafter{\the\numexpr #1-\@ne}{}#2#3#2\relax }

\def\chainadd@ii #1#2#3#4{\ifx\relax #4\expandafter\@firstoftwo\else
                       \expandafter\@secondoftwo\fi
   {\ChainRow{#2}\chainadd@i {#1}#2\relax}%
   {\expandafter\expandafter\expandafter\chainadd@iii\modten{#3+#4}{#2}{#1}#4}}

\def\chainadd@iii #1#2#3{\chainadd@ii {#3}{#2#1}}

\newcommand*{\ChainRow}[1]{\chainrow@i #1\relax}

\def\chainrow@i  #1{#1\chainrow@ii }

\def\chainrow@ii #1{\ifx\relax #1\expandafter\@firstoftwo
                    \else 
                    \expandafter\@secondoftwo\fi
                    {\\}{&#1\chainrow@ii}}

\makeatother

\begin{document}

\begin{table}[ht]
  \centering
  \begin{tabular}{cccccc}
  \hline
  \chainadd {123456}{10}
  \hline
  \end{tabular}
\end{table}

\begin{table}[ht]
  \centering
  \begin{tabular}{ccccccc}
  \hline
  \chainadd {1234567}{10}
  \hline
  \end{tabular}
\end{table}

\end{document}

链式添加，表格形式

现在自动确定迭代过程的周期。

\documentclass{article}

\newcommand*{\modten}[1]{\the\numexpr #1+10-((#1+5)/10)*10\relax}

\makeatletter
% Looking for the cycle.
\def\gobtoexclam #1!{}

\newcommand*{\chainiterate}[1]
    {\romannumeral0\expandafter\chainiterate@i\romannumeral-`0#1!}
\def\chainiterate@i #1#2!{\chainiterate@ii {}#1#2#1!}
\def\chainiterate@ii #1#2#3{\gobtoexclam #3\chainiterate@end!%
     \expandafter\expandafter\expandafter\chainiterate@iii \modten{#2+#3}#3{#1}}
\def\chainiterate@iii #1#2#3{\chainiterate@ii {#3#1}#2}
\def\chainiterate@end!#1!#2{ #2}
% 
\newcommand*{\chainiteratetwice}[1]
    {\romannumeral-`0\chainiterate{\chainiterate{#1}}}

\newcommand*{\ChainCycle}{}
\newcommand*{\ChainCycleStart}{}

\newcommand*{\odef}[1]{\expandafter\def\expandafter#1\expandafter}

\newcommand*{\oodef}[1]{\expandafter\expandafter\expandafter
                        \def\expandafter\expandafter\expandafter 
                        #1\expandafter\expandafter\expandafter }

\def\FindCycle@DefTest #1{%
    \def\FindCycle@DoTest ##1,#1,##2##3?%
        {\if!##2\expandafter 0\else\expandafter 1\fi }%
}%

\newcommand*{\FindCycle}[1]{%
    \odef\FindCycle@X {#1}%
    \let\FindCycle@Y\FindCycle@X 
    \loop
        \oodef\FindCycle@X{\chainiterate{\FindCycle@X}}%
        \oodef\FindCycle@Y{\chainiteratetwice{\FindCycle@Y}}%
    \ifx\FindCycle@X\FindCycle@Y\else
    \repeat
    \odef\ChainCycle {\expandafter,\FindCycle@X}%
    \loop
        \oodef\FindCycle@Y{\chainiterate{\FindCycle@Y}}%
    \ifx\FindCycle@Y\FindCycle@X\else
        \oodef\ChainCycle {\expandafter\ChainCycle\expandafter,\FindCycle@Y}%
    \repeat
    \count@ \z@
    \odef\FindCycle@X {#1}%
    \loop
       \expandafter\FindCycle@DefTest\expandafter{\FindCycle@X}%
    % is this iterate in the cycle ?
    \if0\expandafter\expandafter\expandafter\FindCycle@DoTest
           \expandafter\ChainCycle\expandafter,\FindCycle@X,!?%
    % no, it isn't
      \oodef\FindCycle@X {\chainiterate{\FindCycle@X}}%
      \advance\count@\@ne
    \repeat
    The first iterate of #1 which starts the period is obtained after
    \the\count@{} iteration{\ifnum\count@>1 s\fi}. It is \FindCycle@X.\par
    \count@\z@
    \let\FindCycle@Y\FindCycle@X
    Here is the period: \FindCycle@X
    \loop
       \advance\count@\@ne
       \oodef\FindCycle@Y {\chainiterate{\FindCycle@Y}}%
    \ifx\FindCycle@Y\FindCycle@X
    \else
    , \FindCycle@Y
    \repeat
    .\par Its length is \the\count@.\par
}

\makeatother

\begin{document}

\FindCycle {12}
\bigskip

\FindCycle {123}
\bigskip

\FindCycle {1234}
\bigskip

\FindCycle {12345}
\bigskip

\FindCycle {123456}
\bigskip

% SURPRISE !
\FindCycle {1234567}

\end{document}

第 1 页

第2页
第 3 页

答案3

和expl3：

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\chain}{mm}
 {% #1 is the starting number, #2 is the number of repetitions
  \egreg_chain_add:nn { #1 } { #2 }
 }

\tl_new:N \l_egreg_chain_tablebody_tl
\seq_new:N \l_egreg_chain_input_seq
\seq_new:N \l_egreg_chain_output_seq
\int_new:N \l_egreg_chain_length_int

\cs_new_protected:Npn \egreg_chain_add:nn #1 #2
 {
  % form the first input sequence
  \seq_set_split:Nnn \l_egreg_chain_input_seq { } { #1 }
  % the number of columns for the table
  \int_set:Nn \l_egreg_chain_length_int { \seq_count:N \l_egreg_chain_input_seq }
  % start building the table body
  \tl_clear:N \l_egreg_chain_tablebody_tl
  % first row with the input number
  \tl_put_right:Nx \l_egreg_chain_tablebody_tl
   {
    \seq_use:Nn \l_egreg_chain_input_seq { & } \exp_not:N \\
   }
  % repeat as many times as requested
  \prg_replicate:nn { #2 }
   {
    % clear the output sequence
    \seq_clear:N \l_egreg_chain_output_seq
    % build the output sequence
    \int_step_inline:nnnn { 1 } { 1 } { \seq_count:N \l_egreg_chain_input_seq - 1 }
     {
      \seq_put_right:Nx \l_egreg_chain_output_seq
       {
        \int_to_arabic:n
         {
          \int_mod:nn
           {
            \seq_item:Nn \l_egreg_chain_input_seq { ##1 } +
            \seq_item:Nn \l_egreg_chain_input_seq { ##1 + 1 }
           }
           { 10 }
         }
       }
     }
    \seq_put_right:Nx \l_egreg_chain_output_seq
     {
      \int_to_arabic:n
       {
        \int_mod:nn
         {
          \seq_item:Nn \l_egreg_chain_input_seq { -1 } +
          \seq_item:Nn \l_egreg_chain_input_seq { 1 }
         }
         { 10 }
       }
     }
    % fill the next row
    \tl_put_right:Nx \l_egreg_chain_tablebody_tl
     {
      \seq_use:Nn \l_egreg_chain_output_seq { & } \exp_not:N \\
     }
    % the output sequence becomes the input sequence for the next row
    \seq_set_eq:NN \l_egreg_chain_input_seq \l_egreg_chain_output_seq
   }
  \begin{tabular}{*{\l_egreg_chain_length_int}{c}}
  % deliver the table body
  \l_egreg_chain_tablebody_tl
  \end{tabular}
 }

\ExplSyntaxOff

\begin{document}

\chain{123456}{10}

\bigskip

\chain{42}{5}

\end{document}

该表使用两个序列逐步构建；在每个步骤中，输出序列成为下一步的输入序列。

在此处输入图片描述

答案4

我正在添加在 LuaTeX 中创建的代码片段和结果预览。

% lualatex mal-cells.tex
\documentclass[a4paper]{article}
\pagestyle{empty}
\usepackage{luacode}

\begin{document}

\begin{luacode*}
function malcompute() -- the core of computations
counter=0 -- number of columns
series={} -- actual values will be stored here
seriesdata=tex.toks[1] -- get data from TeX and parse them:
for entry in unicode.utf8.gmatch(seriesdata, "%d+") do
   counter=counter+1 -- move to the column+1
   series[counter]=entry -- save data to a Lua table
end -- for, entry
rows=tonumber(tex.toks[2]) -- how many lines is requested?

for row=1,rows do -- we need this number of rows and...
  for column=1,#series do -- ... and columns
    if column==1 then firstvalue=series[1] end -- repeat the computation from the beginning when you reach the last column
    newvalue=(series[column]+(series[column+1] or firstvalue)) % 10 -- get two numbers above the actual one, modulus is a percent sign in Lua
    tex.print(newvalue) -- print the result to TeX
    series[column]=newvalue -- store a new value instead of an old one
    end -- for, column
    tex.print("\\par") -- break a line when there is a new row coming
  end -- for, row
end -- function, malcompute
\end{luacode*}

\def\malcells#1#2{% pass arguments to Lua
   \toks1{#1}% initial values
   \toks2{#2}% number of rows
   \directlua{malcompute()}% get us to the Lua world
   } % End of the \malcells command.

% A small piece of code formatting the results at a TeX level
\ttfamily\Large

% An example as OP asked.
\malcells{1 2 3 4 5 6}{3}   

\end{document}

结果示例

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