可视化多个字符串段

Question 1

建议命令的语法是

\segments[<width>]{<string>}{<segments>;<segments>;...}

其中<width>是单元格的宽度（默认为 1em），<string>是要排版的字符串，<segments>是要加下划线的单元格列表。此类列表由单个数字组成，或<number>-<number>表示要跨越的单元格。

请注意，这将忽略两个参数中的空格（但输入范围时不应有空格），因此也可以像这样输入示例

\segments{1 0 1 0 0 1 0 0}{
  1, 2, 3, 4, 5, 6, 7, 8 ;
  1-3, 4-5, 7-8 ;
  4-8
}

如果愿意的话。

\documentclass{article}
\usepackage{xparse,array,booktabs}

\ExplSyntaxOn
\NewDocumentCommand{\segments}{O{1em}mm}
 {% #1 = width of cells, #2 = string, #3 = list of underlines
  \group_begin:
  \renewcommand{\arraystretch}{0}
  \fizruk_segments:n { #3 }
  \begin{tabular}[t]{ @{} *{ \tl_count:n { #2 } }{ w{c}{#1} @{} } }
  \seq_set_split:Nnn \l__fizruk_string_seq { } { #2 }
  \seq_use:Nn \l__fizruk_string_seq { & } \\
  \tl_use:N \l__fizruk_segments_tl
  \end{tabular}
  \group_end:
 }

\seq_new:N \l__fizruk_string_seq
\seq_new:N \l__fizruk_stages_seq
\tl_new:N \l__fizruk_segments_tl

\cs_new_protected:Nn \fizruk_segments:n
 {
  \tl_clear:N \l__fizruk_segments_tl
  \seq_set_split:Nnn \l__fizruk_stages_seq { ; } { #1 }
  \seq_map_function:NN \l__fizruk_stages_seq \__fizruk_segment_row:n
 }

\cs_new_protected:Nn \__fizruk_segment_row:n
 {
  \clist_map_function:nN { #1 } \__fizruk_add_segment:n
  \tl_put_right:Nn \l__fizruk_segments_tl { \morecmidrules }
 }

\cs_new_protected:Nn \__fizruk_add_segment:n
 {
  \str_if_in:nnTF { #1 } { - }
   {
    \tl_put_right:Nn \l__fizruk_segments_tl { \cmidrule(l{2pt}r{2pt}){#1} }
   }
   {
    \tl_put_right:Nn \l__fizruk_segments_tl { \cmidrule(l{2pt}r{2pt}){#1-#1} }
   }
 }

\ExplSyntaxOff

\begin{document}

\[
S=\segments{10100100}{1,2,3,4,5,6,7,8;1-3,4-5,7-8; 4-8}
\]

\end{document}

以下是输出

\[
S=\segments{10100100}{
    1-3,4-8;
    1,2-7,8;
    2-4,5-7;
    2,3-6,7-8;
    3,4-5,6,7;
    4,5
  }
\]

通过以下修改，您可以在两个符号之间插入一条线，.3如果您希望它跟在第三个符号后面，则用表示。

\documentclass{article}
\usepackage{xparse,array,booktabs}

\ExplSyntaxOn
\NewDocumentCommand{\segments}{O{1em}mm}
 {% #1 = width of cells, #2 = string, #3 = list of underlines
  \group_begin:
  \renewcommand{\arraystretch}{0}
  \dim_set:Nn \l__fizruk_cell_dim { #1 }
  \fizruk_segments:n { #3 }
  \tl_show:N \l__fizruk_segments_tl
  \begin{tabular}[t]{ @{} *{ \tl_count:n { #2 } }{ w{c}{#1} @{} } }
  \seq_set_split:Nnn \l__fizruk_string_seq { } { #2 }
  \seq_use:Nn \l__fizruk_string_seq { & } \\
  \tl_use:N \l__fizruk_segments_tl
  \end{tabular}
  \group_end:
 }

\dim_new:N \l__fizruk_cell_dim
\seq_new:N \l__fizruk_string_seq
\seq_new:N \l__fizruk_stages_seq
\tl_new:N \l__fizruk_segments_tl

\cs_new_protected:Nn \fizruk_segments:n
 {
  \tl_set:Nn \l__fizruk_segments_tl { \addlinespace[3pt] }
  \seq_set_split:Nnn \l__fizruk_stages_seq { ; } { #1 }
  \seq_map_function:NN \l__fizruk_stages_seq \__fizruk_segment_row:n
 }

\cs_new_protected:Nn \__fizruk_segment_row:n
 {
  \clist_map_function:nN { #1 } \__fizruk_add_segment:n
  \tl_put_right:Nn \l__fizruk_segments_tl { \morecmidrules }
 }

\cs_new_protected:Nn \__fizruk_add_segment:n
 {
  \str_if_in:nnTF { #1 } { - }
   {
    \tl_put_right:Nx \l__fizruk_segments_tl { \__fizruk_cmidrule:nn { /5 } { #1 } }
   }
   {
    \str_if_in:nnTF { #1 } { . }
     {
      \tl_put_right:Nx \l__fizruk_segments_tl
       {
        \__fizruk_cmidrule:ne { *4/5 } { \tl_tail:n { #1 }-\int_eval:n { \tl_tail:n { #1 }+1 } }
       }
     }
     {
      \tl_put_right:Nx \l__fizruk_segments_tl { \__fizruk_cmidrule:nn { /5 } { #1-#1 } }
     }
   }
 }
\cs_new:Nn \__fizruk_cmidrule:nn
 {
  \exp_not:N\cmidrule(l{\dim_eval:n{\l__fizruk_cell_dim #1}}r{\dim_eval:n{\l__fizruk_cell_dim #1}}){#2}
 }
\cs_generate_variant:Nn \__fizruk_cmidrule:nn { ne }

\ExplSyntaxOff

\begin{document}

\[
S=\segments{10100100}{1,2,3,4,5,6,7,8;1-3,4-5,7-8; 4-8}
\]

\[
S=\segments{10100100}{
    1-3,4-8;
    1,2-7,8;
    2-4,5-7;
    2,3-6,7-8;
    3,4-5,6,7;
    4,5
  }
\]

\[
S = \segments{1 0 1 0 0 1 0 0}{
      .3;
      3,4;
      2-4,5-7
    }
\]
\end{document}

Answer