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How to obtain easily the following red line in the code given just after?

Is it possible to only give the start and end positions so as to obtain the two kinds of paths?

The question is more about paths than about Sudoku. I give the code for the Sudoku so as to test one solution working in a “real life” example.

First style

enter image description here

Second style

enter image description here

Code

\documentclass{article}
    \usepackage{tikz}
    \usetikzlibrary{backgrounds}
    \usepackage{graphicx}

% Some customizable styles
    \tikzset {
        highlight/.style = {
            yellow,
            opacity = 0.3
        },
        digit/.style = {
            minimum height = 5mm,
            minimum width = 5mm,
            anchor = center
        },
        circle/.style = {
            draw = green!80!black,
            dotted,
            very thick
        },
        circle number/.style = {
            draw = #1,
            very thick
        },
        cross/.style = {
            red,
            opacity = .5,
            shorten >= 1mm,
            shorten <= 1mm,
            very thick,
            line cap = round
        },
        hint/.style = {
            blue,
            font = \sf,
            minimum width = 3mm,
            minimum height = 3mm
        }
    }

% Modified the \node to give a unique name to each one, which is the
% row number, a dash and the column number. E.g: 1-1, 4-5, etc.
    \newcounter{row}
    \newcounter{col}

    \newcommand\setrow[9]{
        \setcounter{col}{1}
        \foreach \n in {#1, #2, #3, #4, #5, #6, #7, #8, #9} {
            \edef\x{\value{col} - 0.5}
            \edef\y{9.5 - \value{row}}
            \node[digit,name={\arabic{row}-\arabic{col}}] at (\x, \y) {\n};
            \stepcounter{col}
        }
        \stepcounter{row}
    }

% New code -------------------------------------------------------------
    \def\highlightcell#1#2{
        \begin{scope}[on background layer]
            \fill[highlight] (#1-#2.north west) rectangle (#1-#2.south east);
        \end{scope}
    }

    \def\circlecell#1#2{
        \draw[circle] (#1-#2) circle(4mm);
    }

    \def\crosscell#1#2{
        \draw[cross] (#1-#2.north west) -- (#1-#2.south east);
        \draw[cross] (#1-#2.north east) -- (#1-#2.south west);
    }

    \def\highlightrow#1{
        \begin{scope}[on background layer]
            \fill[highlight] (#1-1.north west) rectangle (#1-9.south east);
        \end{scope}
    }

    \def\highlighcolumn#1{
        \begin{scope}[on background layer]
            \fill[highlight] (1-#1.north west) rectangle (9-#1.south east);
        \end{scope}
    }

    \def\highlightrectangle#1#2#3#4{
        \begin{scope}[on background layer]
            \fill[highlight] (#1-#2.north west) rectangle (#3-#4.south east);
        \end{scope}
    }

    \def\hintcell#1#2#3{
        \node at (#1-#2) {\hintbox{#3}};
    }


% Command to circle numbers:
% #1: optional -> circle color
% #2: mandatory -> cell identifier
% #3: mandatory -> name of the cell
    \newcommand\circlenumber[3][red!80!black]{
        \draw[circle number=#1, radius=5mm] (#2) circle node[outer sep=1mm] (#3){};
    }

% UGLY code. Do not read :-)
    \def\hintbox#1{
        \resizebox{4.5mm}{4.5mm}{%
            \tikz[scale=0.3]{%
                \def\auxc{0}
                \foreach \m in {1,...,9} {
                    \pgfmathparse{mod(\auxc,3)}
                    \xdef\x{\pgfmathresult}
                    \pgfmathparse{-floor(\auxc/3)}
                    \xdef\y{\pgfmathresult}
                    \xdef\hintprinted{0}
                    \foreach \n in {#1} {
                        \ifnum\n=\m
                            \node[hint] at (\x,\y) {\n};
                            \xdef\hintprinted{1}
                        \fi
                    }
                    \ifnum\hintprinted=0
                        \node[hint, opacity=0.1] at (\x,\y) {\m};
                    \fi
                    \pgfmathparse{\auxc+1}
                    \xdef\auxc{\pgfmathresult}
                }
            }%
        }
    }


\begin{document}

\begin{tikzpicture}[scale=.5]
    \begin{scope}
        \draw (0, 0) grid (9, 9);
        \draw[very thick, scale=3] (0, 0) grid (3, 3);

        \setcounter{row}{1}
        \setrow { }{2}{ }  {5}{ }{1}  { }{9}{ }
        \setrow {8}{ }{ }  {2}{ }{3}  { }{ }{6}
        \setrow { }{3}{ }  { }{6}{ }  { }{7}{ }

        \setrow { }{ }{1}  { }{ }{ }  {6}{ }{ }
        \setrow {5}{4}{ }  { }{ }{ }  { }{1}{9}
        \setrow { }{ }{2}  { }{ }{ }  {7}{ }{ }

        \setrow { }{9}{ }  { }{3}{ }  { }{8}{ }
        \setrow {2}{ }{ }  {8}{ }{4}  { }{ }{7}
        \setrow { }{1}{ }  {9}{ }{7}  { }{6}{ }
    \end{scope}
\end{tikzpicture}

\end{document}
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1 Answer 1

up vote 11 down vote accepted

I don't have time now for the general solution which automatically deduces the appropiate intermediate coordinates from the starting and ending ones, but the following code draws manually the arrows you want, and can be a starting point for solving the general case:

\draw[->,very thick, red, rounded corners] 
    (1-1.center) -- (1-9.center) 
 |- (2-1.center) 
 |- (3-9.center) 
 |- (4-1.center) 
 |- (5-3.center);

enter image description here

And:

\draw[->,very thick, red] 
     (1-1.center) -- (1-9.center) 
  -- (2-1.center) -- (2-9.center) 
  -- (3-1.center) -- (3-6.center);

enter image description here

share|improve this answer
    
The syntax is easy enough to use that there is no real need for one general solution which automatically deduces the appropiate intermediate coordinates. Thanks for this ! –  projetmbc Jan 20 '13 at 9:59

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