Someone can give me some hints so as to achieve the two following improvements ?
EASY: I would like to display the number position of each column and row. This needs to display numbers just at the left and above the grid.
Answer of Sašo Živanović does the job for this easy part.
- HARD: Sometimes, I only need to display only one rectangular piece of the sudoku. What are the changes to do so as to achieve this ?
Code
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{backgrounds,shapes}
\usepackage{graphicx}
\usepackage{xstring}
% 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
},
hint special/.style = {
blue,
font = \sf,
minimum width = 3mm,
minimum height = 3mm,
fill=red!20,
inner sep=0pt,
},
hint border/.style = {
blue,
font = \sf,
minimum width = 2mm,
minimum height = 3mm,
inner sep=0pt, outer sep=0pt,
draw=red, thick,
rounded corners=1pt,
}
}
% 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 :-)
% Sorry, it needed to be read to obtain solution :-)
\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/\Style in {#1} {
\ifnum\n=\m
\IfStrEqCase{\Style}{%
{}{\node[hint] at (\x,\y) {\n};}
{\n}{\node[hint] at (\x,\y) {\n};}
{border}{\node[hint border] at (\x,\y) {\n};}
{special}{\node[hint special] 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}
