Printing inside Grid using Two Loops

Question

I am trying to rearrange the value inside grid. For example, the below code

 \documentclass[tikz,border=7mm]{standalone} 

\begin{document}

\tikz\draw grid(4,4)foreach[count=~]\l in{1,...,16}  
{({.5+mod(~-1,4},{3.5-div(~-1,4})node{\l}}; 

\end{document}

gives a 4*4 grid where each box has a number from 1 to 16. like-

1  2  3  4
5  6  7  8
9  10 11 12
13 14 15 16

How can I perform the same thing using two for loop such that the grid becomes-

1  2  2   1
3  4  4   3
3  4  4   3
1  2  2   1

For 8 rows and 8 column, each quadrant should start with 1 and end with 16.

    1  2  3  4     4  3  2  1  
    5  6  7  8     8  7  6  5  
    9  10 11 12    12 11 10 9
    13 14 15 16    16 15 14 13

    13 14 15 16    16 15 14 13  
    9  10 11 12    12 11 10 9
    5  6  7   8    8  7  6  5
    1  2  3   4    4  3  2  1

Also , How can I print inside grid such as-

a=1  a=2  a=2   a=1
a=3  a=4  a=4   a=3
a=3  a=4  a=4   a=3
a=1  a=2  a=2   a=1

I am trying to construct the general code that is valid for x rows and y columns instead of 4 rows and 4 columns.

I found the above code from here.

Answer 1

Here is how you get a double loop, you add a second foreach inside the expression {...} that the first foreach loops over.

\documentclass{article}

\usepackage{tikz}

\begin{document}

\tikz\draw grid(4,4) foreach \x  in {1,...,4}
  { foreach[evaluate={\z = int((\x*(5-\x)/2+\y*(5-\y))-5)}] \y in {1,...,4}
  {({.5+(\x-1)},{4.5-\y}) node{$a = \z$}}};

\end{document}

To change the values printed, just adjust the formula for \z in the evaluate statement. Note that int is used here, so that a decimal answer 1.0 etc. is not display.

Here is a version for your new pattern specification with variables \n and \m (even numbers) giving the total number of columns and rows, and a variable \s storing a dimension to scale the box size so the contents fit:

\documentclass{article}

\usepackage{tikz}

\begin{document}

\def\n{6}
\def\m{8}
\def\s{1.5cm}
\tikz\draw grid[step=\s](\n*\s,\m*\s) foreach[evaluate] \x  in {1,...,\n}
  { foreach[evaluate={\z = int(min(\x,\n+1-\x)+\n*min(\y-1,\m-\y)/2)}] \y in {1,...,\m}
  {({\s*(.5+(\x-1))},{\s*(\m+.5-\y)}) node{$a=\z$}}};

\end{document}

Answer 2

\documentclass[tikz,border=7mm]{standalone} 
\begin{document}
\tikz\draw grid(4,4)foreach
    [evaluate={\x=.5+mod(\l-1,4);\y=3.5-div(\l-1,4);\z=int(5-(\x-2)^2/2-(\y-2)^2)}]
    \l in{1,...,16}{(\x,\y)node{$a=\z$}};
\end{document}

Answer 3

Like this? (Now that I've corrected the symmetry...)

I produced this with the following Metapost code, wrapped up in luamplib. Compile with lualatex.

\RequirePackage{luatex85}
\documentclass[border=5mm]{standalone}
\usepackage{luamplib}
\begin{document}
\mplibtextextlabel{enable}
\begin{mplibcode}
beginfig(1);
    numeric u, v, N; N = 2; 
    u = 14mm; v = 9mm;
    for i=0 upto 3:
        for j=0 upto N-1:
            for k=0 upto N-1:
                label("$a=" & decimal (N**2-(k+N*j)) & "$", 
                      (k+2/3,j+2/3) xscaled u yscaled v
                      if (i mod 3) > 0: reflectedabout(up,down) fi
                      if i > 1: reflectedabout(left,right) fi
                );
            endfor
        endfor
    endfor
endfig;
\end{mplibcode}
\end{document}

If you set N=4 in the above you get:

and so on...

The values are generated by the label in the middle of the three loops. Specifically decimal (N**2-(k+N*j)) generates a string with the value of N^2-(k+N*j) -- if you want different numbers you can adjust this formula. So to get numbers starting at a=1 in the middle, you could change the formula to decimal (1+k+N*j).

If you wanted the numbers is some other order, then you could (for example) swap the values of j and k in the coordinates to get a column-first sequence, or subtract them from N to get the numbers in reverse order.