How to enter a solution for each square?

Question

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage[left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm]{geometry}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\draw (6,0) -- (6,-24);
\draw (6,-24) -- (12,-24);
\draw (12,-24) -- (12,0);
\draw (6,0) -- (12,0);
\draw (0,-6) -- (18,-6);
\draw (18,-6) -- (18,-12);
\draw (18,-12) -- (0,-12);
\draw (0,-6) -- (0,-12);
\draw (6,-18) -- (12,-18);
\draw (8,0) -- (8,-24);
\draw (10,0) -- (10,-24);
\draw (0,-8) -- (18,-8);
\draw (0,-10) -- (18,-10);
\draw (6,-2) -- (12,-2);
\draw (6,-4) -- (12,-4);
\draw (6,-14) -- (12,-14);
\draw (6,-16) -- (12,-16);
\draw (6,-18) -- (12,-18);
\draw (6,-20) -- (12,-20);
\draw (6,-22) -- (12,-22);
\draw (2,-6) -- (2,-12);
\draw (4,-6) -- (4,-12);
\draw (14,-6) -- (14,-12);
\draw (16,-6) -- (16,-12);
\end{tikzpicture}
\end{document}

I am trying to enter in each of the squares of this picture a number such that for each of the 6 "obvious" big squares, the entries are 0,1,2,3,4,5,6,7,8.

I thought maybe using a for loop, but I need to know how to enter the number in the center of the small squares.

Kind Regards.

On attached image I see 54 squares. Where exactly you like to have numbers? Pleas, add this information to your image. — Zarko, May 20 '21 at 08:16

score 4 · Accepted Answer · answered May 20 '21 at 08:58

I guessing that you after the following:

A bit rude but working MWE:

\documentclass[12pt,a4paper]{article}
\usepackage[margin=2.5cm]{geometry}
\usepackage{tikz}
\usetikzlibrary{chains, positioning}
\begin{document}
\begin{center}
    \begin{tikzpicture}[
node distance = 0pt,
  start chain = A going right,
  start chain = B going right,
  start chain = C going right,
  start chain = D going right,
S/.style = {draw, minimum size=15mm, outer sep=0pt}
                        ]
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n11\i) [S, on chain=A] {\i};
\node (n12\i) [S,below=of n11\i] {\j};
\node (n13\i) [S,below=of n12\i] {\k};
}
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n21\i) [S, on chain=A] {\i};
\node (n22\i) [S,below=of n21\i] {\j};
\node (n23\i) [S,below=of n22\i] {\k};
}
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n31\i) [S, on chain=A] {\i};
\node (n32\i) [S,below=of n31\i] {\j};
\node (n33\i) [S,below=of n32\i] {\k};
}
%vertical, upper part
    \begin{scope}[shift={(45mm,45mm)}]
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n21\i) [S, on chain=B] {\i};
\node (n22\i) [S,below=of n21\i] {\j};
\node (n23\i) [S,below=of n22\i] {\k};
}
    \end{scope}
%vertical, lover part
    \begin{scope}[shift={(45mm,-45mm)}]
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n21\i) [S, on chain=C] {\i};
\node (n22\i) [S,below=of n21\i] {\j};
\node (n23\i) [S,below=of n22\i] {\k};
}
    \end{scope}
    \begin{scope}[shift={(45mm,-90mm)}]
\foreach \i in {0,1,2}
{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n31\i) [S, on chain=D] {\i};
\node (n32\i) [S,below=of n31\i] {\j};
\node (n23\i) [S,below=of n32\i] {\k};
}
    \end{scope}
    \end{tikzpicture}
\end{center}
\end{document}

Quick question, could you make a modification such that I can enter a,b,c,d,e,f,g,h,i instead? Thanks! — Dedekid, May 24 '21 at 20:14

Vincent Kuhlmann · Answer 2 · 2021-05-20T12:03:25.833

Updated my answer to include the borders as well:

\documentclass[crop,border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[every node/.style={font={\Huge\bfseries\sffamily},scale=1.5}]
\foreach\zoneX/\zoneY in {0/0, 0/1, 0/2, 0/3, -1/1, 1/1}{
    \foreach\tilenumber in {0,1,...,8} {
        % localX = tilenumber % 3 (modulo)
        % localY = tilenumber / 3 (integer division)
        \pgfmathtruncatemacro{\localX}{\tilenumber/3}
        \pgfmathtruncatemacro{\localX}{\tilenumber-\localX3}
        \pgfmathtruncatemacro{\localY}{\tilenumber/3}
        \node at ($ (6\zoneX + 2\localX + 7, -6\zoneY - 2\localY - 1) $) {\tilenumber};
        \draw[line width=1.5pt] ($ (6\zoneX + 2\localX + 6, -6\zoneY - 2\localY) $) rectangle +(2,-2);
    }
    \draw[line width=6pt] ($ (6\zoneX + 6, -6*\zoneY) $) rectangle +(6,-6);
}
\end{tikzpicture}
\end{document}

The idea of using a tuple foreach is from Juan. The \usetikzlibrary{calc} is required for the calculations between dollar signs.

Looks nice, and the code is rather compact.

score 2 · Answer 3 · answered May 20 '21 at 09:20

If the numbers are correlative, as Zarko suggests, this can also be done like this.

\documentclass[border=2mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\foreach\x/\y in {0/0, 0/3, -3/6, 0/6, 3/6, 0/9} \foreach\i in {0,1,2} \foreach\j in {0,1,2}
{%
  \draw (\x+\i,\y+\j) rectangle (\x+\i+1,\y+\j+1);
  \pgfmathtruncatemacro\n{3*(2-\j)+\i};
  \node at (\x+\i+0.5,\y+\j+0.5) {$\n$};
}
\end{tikzpicture}
\end{document}

score 2 · Answer 4 · answered May 24 '21 at 22:01

Regarding your request, that instead of numbers you have letters in squares. This can be obtained by defining new command in preamble:

\newcommand\makealph[1]{% see https://tex.stackexchange.com/questions/595043/converting-numbers-to-letters-but-starting-with-0-a-instead-of-1-a
\ifcase\numexpr#1\relax a\or b\or c\or d\or e\or f\or g\or h\or i\fi}

and than each loop writes as (considering their chains names):

{
    \pgfmathsetmacro{\j}{int(\i + 3)}
    \pgfmathsetmacro{\k}{int(\i + 6)}
\node (n11\i) [S, on chain=A] {\makealph{\i}};   % <---
\node (n12\i) [S,below=of n11\i] {\makealph{\j}};% <---
\node (n13\i) [S,below=of n12\i] {\makealph{\k}};% <---
}

Result of this changes is:

How to enter a solution for each square?

4 Answers4