Creating 63 Idempotent Latin Squares of order 25 and then creating 3 Mutually Orthogonal Latin Squares of order 21

Question

For an assignment in my design theory class I am trying to construct idempotent latin squares and mutually orthogonal latin squares as stated in the question. I understand the process and can do this for small orders very easily, but for this particular problem I feel like it would be much quicker to write a program to generate these tables. I don't think any knowledge of design theory is necessary, but I certainly would appreciate someone helping me with the code to generate these tables.

Here is how the process works. I am given a pairwise balanced design of order 21 with blocks of size 5. For example, one block is {1,2,3,4,21}. I then use each block to create 3 idempotent latin squares of order 5. In each latin square

x<y<z<u<v

so for our particular block x=1,y=2,z=4,u=4,v=21. Here is the construction for the first idempotent latin square. (I'm confident that if I can get this coded I can modify it for the other 2). The outer row and column essentially create a multiplication table (think Cayley table from group theory) under the operation.

\documentclass[12pt]{article}
\usepackage{fullpage}

\begin{document}

\begin{tabular}{c|ccccc}
\(\circ_1\)&\(x\)&\(y\)&\(z\)&\(u\)&\(v\)\\
\hline
\(x\)&\(x\)&\(u\)&\(y\)&\(v\)&\(z\)\\
\(y\)&\(u\)&\(y\)&\(v\)&\(z\)&\(x\)\\
\(z\)&\(y\)&\(v\)&\(z\)&\(x\)&\(u\)\\
\(u\)&\(v\)&\(z\)&\(x\)&\(u\)&\(y\)\\
\(v\)&\(z\)&\(x\)&\(u\)&\(y\)&\(v\)
\end{tabular}

\end{document}

Using the idempotent latin squares for all 21 blocks, I can then create a latin square of order 21 where the (i,j) entry is found by locating the product in the previous latin squares. For example, in the construction given, y=2,z=3, and yz=21, so the (2,3) entry is 21.

Looking around, I found this topic Insert loop inside table? which mentions "This is a LaTeX solution, with a \forloop from forloop package. It builds a tabular array with a command \maketablerows{number_of_rows}{rowcontent}, where rowcontent is suspected to hold the values for the cells in a row, either direct or via a command." However, I have looked at the manual and I don't really see how to get {rowcontent} to generate what I want.

I know I could crank out a program to do this in C++ and make it export the tex table (aka the 21X21 latin square code, jeez!), but I was hoping someone could help guide me in how to do this in latex. I can figure out the programming logic, I just don't know the syntax to use.

TLDR: How do I use loops in latex to generate tables using specified formulas?

EDIT: Too long for a comment.

I believe I have found formulas for the other 2 and indeed one of them was 2(2x+y) mod 5, but the other I found was 2x-y mod 5. I am trying to convert these to formulas like you did for 3(x+y), which got me tinkering with your given formula of 3*(#2+#3)-5*((3*(#2+#3)+3)/5 -1)+1.

If I am understanding this correctly,

\expandafter\edef\csname Latin-\aBlock{#2+1}-\aBlock{#3+1}\endcsname {\aBlock{3*(#2+#3)-5*((3*(#2+#3)+3)/5 -1)+1}}% % this is crazy formula which simply does 3(a+b) modulo 5 % but recall it is a+1 and b+1 which serve as arguments to % the array \aBlock

is saying if I take the numbers 3,4 in the block 0,1,2,3,4, then the number 3*(3+4)-5((3*(3+4)+3)/5-1)+1=3 should appear in the ((3+1,4+1))=(4,5) entry? I must be misunderstanding the part where you are adding 1 to #2 and #3 as I've tried adding 1 then performing the operation, along with putting the calculated vaule into the (#2+1,#3+1) cell and still am not getting the values that are in the table.

Answer 1

For the example that you have given there is no need for a \forloop. Rather, you can just use a simple macro like:

\newcommand\LatinSquare[5]{
  \[
    \begin{array}{c|ccccc}
      \circ_1&#1&#2&#3&#4&#5\\
      \hline
      #1&#1&#4&#2&#5&#3\\
      #2&#4&#2&#5&#3&#1\\
      #3&#2&#5&#3&#1&#4\\
      #4&#5&#3&#1&#4&#2\\
      #5&#3&#1&#4&#2&#5
    \end{array}
  \]
}

that you would use as

 \LatinSquare{1}{2}{3}{4}{21}

As expected, this produces:

I suspect that this is not really what you want, however, so perhaps you need to add more detail to your question!

Here's the full code:

\documentclass[12pt]{article}
\usepackage{fullpage}

\newcommand\LatinSquare[5]{
  \[
    \begin{array}{c|ccccc}
      \circ_1&#1&#2&#3&#4&#5\\
      \hline
      #1&#1&#4&#2&#5&#3\\
      #2&#4&#2&#5&#3&#1\\
      #3&#2&#5&#3&#1&#4\\
      #4&#5&#3&#1&#4&#2\\
      #5&#3&#1&#4&#2&#5
    \end{array}
  \]
}

\begin{document}

    \LatinSquare{1}{2}{3}{4}{21}

\end{document}

Answer 2

Table of contents

• general method

• proof of concept from Fano plane to obtain 7 by 7 case

• the projective plane over F4 has 16+4+1 = 21 points and lines. Thanks to an explicit enumeration found here I could apply the previously described method. I used the Latin square 5x5 of OP, which is simply 3(x+y) mod 5. Each pair of points lie on a unique one of the 21 projective lines. Each projective line has 5 points and we can thus use 5x5 Latin squares there. Using always the same we construct a 21x21 Latin square. Because the enumeration at https://www.uwyo.edu/moorhouse/pub/planes/pg24.txt of the points is from 0 to 20, I kept that in final result, hence rows and columns are indexed from 0 to 20, not 1 to 21, but this is detail which can be changed.

---

Here is how one could go about it: you have 21 blocks B each of size 5x5.

For each such block B, you use \@namedef{Latin-i-j}{the value at row i and column j for the 5x5 Latin square with letters from B}. Your small Latin squares are idempotent so one the diagonal at i, i we simply have i. Each small Latin block defines 20 values off-diagonal.

So you need a loop over the 21 blocks B for each to define 20+5 macros (diagonal ones will be defined again by other blocks, but that's ok). Once done your big Latin square which contains 21 times 20 = 420 off-diagional elements plus 21 diagonal elements for a total of 441 = 21x21 pairs is prepared. The diagonal is known because your small Latin squares are idempotent.

You can then typeset it using a construct like this:

\documentclass{article}
\usepackage{array}
\usepackage{xinttools}
% YOU NEED HERE TO DO A LOOP OVER THE BLOCKS TO DEFINE SUITABLY
% (POSSIBLY EVEN BY SOME \numexpr FORMULA) THE 21 x 21 MACROS
% \makeatletter
% % some loop here over the block to define for each suitable
% % \@namedef{Latin-i-j}
% \makeatother
\edef\TwentyOneNumbers{\xintSeq{1}{21}}
\begin{document}
\makeatletter
\begin{tabular}{*{22}{c|}}
  % first row
\xintFor* #1 in \TwentyOneNumbers\do{&#1}\\
\hline
\xintFor* #1 in \TwentyOneNumbers\do{% this will be row number #1
  #1
  \xintFor* #2 in \TwentyOneNumbers\do{% this indices the columns
     &\@nameuse{Latin-#1-#2}}
  \\\hline}
\end{tabular}
\makeatother
\end{document}

The mwe above is lacking all the needed \@namedef because you did not explain what the blocks were (the 5x5 Latin square you gave is at first sight simply addition in Z/5Z up to renaming of indices). So currently we only get this:

the above table spills over right margin, but I didn't fix the page layout (which is very narrow in LaTeX default, better use package geometry). Also the columns are of varying widths which should be fixed. Finally, I took over inclusion of first row and first column from your 5x5 example, perhaps as Latin square these decorations should be removed to really ahve a 21x21 not 22x22 grid.

---

As proof of concept here is a construction a 7x7 Latin square base on the geometry of the Fano plane. The lines give our 7 blocks of each 3 elements.

For 3x3 idempotent Latin squares I chose addition modulo 3 (more precisely -x-y mod 3 to get idempotence). The formula can be implemented by a \numexpr expression. The \numexpr is tacit, because I use \xintAssignArray from xinttools, which defines a macro whose argument is automatically evaluated via \numexpr. This argument starts at 1 (the value 0 gives the number of elements of the array, here 3 in this application as blocks are of size 3).

Each pair of distinct elements from 1 to 7 is contained in a unique Block. Applying the recipe we construct a 7 by 7 Latin square.

In the general description I mentioned \@namedef but here I needed to expand argument, sadly LaTeX is extremely limited in terms of "programming", so I had to go to deeper lying TeX primitives for this bit.

In your specific case you seem to be aiming at 3 distinct 21x21 squares, so you need to adapt or repeat the procedure here by looping again over blocks.

\documentclass{article}
\usepackage{array}
\usepackage{xinttools}

% Take a pairwise design from Fano Plane

% I picked up the blocks at https://fr.wikipedia.org/wiki/Plan_de_Fano
% but I needed to shift by 1 the enumeration to manipulate 1, 2, ..., 7

% For each block, having 3 elements, I will construct an idempotent Latin
% square from addition modulo 3.

\xintFor #1 in {{1, 2, 4}, {2, 3, 5}, {3, 4, 6}, {4, 5, 7}, {5, 6, 1},
   {6, 7, 2}, {7, 1, 3}}\do{%
  \xintAssignArray\xintCSVtoList{#1}\to\aBlock
  % arrays defined by \xintAssignArray are indexed starting at 1
  \xintFor #2 in {0, 1, 2}\do{%
   \xintFor #3 in {0, 1, 2}\do{%
   % unfortunately LaTeX has very few programming tools,
   % there is \@namedef but not even a \@nameedef
    \expandafter\edef\csname Latin-\aBlock{#2+1}-\aBlock{#3+1}\endcsname
    {\aBlock{2*(#2+#3)-3*((2*(#2+#3)+2)/3 -1)+1}}%
    % this is crazy formula which simply does -(a+b) modulo 3
    % which is same as 2(a+b) (we prefer non negative numbers with \numexpr)
    % but a = i-1, b = j-1, and modulo is complicated with \numexpr
    % (\numexpr is used by \aBlock to parse its argument)
    % (this is property of "arrays" defined by \xintAssignArray)
   }%
  }%
}

% done, all our macros defined

\edef\SevenNumbers{\xintSeq{1}{7}}
\begin{document}
\makeatletter
\begin{tabular}{*{8}{c|}}
  % first row
\xintFor* #1 in \SevenNumbers\do{&#1}\\
\hline
\xintFor* #1 in \SevenNumbers\do{% this will be row number #1
  #1
  \xintFor* #2 in \SevenNumbers\do{% this indices the columns
     &\@nameuse{Latin-#1-#2}}
  \\\hline}
\end{tabular}
\makeatother
\end{document}

---

Now the real thing, using the 21-plane.

\documentclass{article}
\usepackage{geometry}
\usepackage{array}
\usepackage{xinttools}

% The projective plane over the field of 4 elements has 
% 16+4+1 = 21 points.

% It also has 21 lines. Each pair of distinct points 
% is contained in a unique line.

% A projective line has 4+1 = 5 elements and we can
% construct an idempotent Latin square using the field
% with five elements and the rule 3(x+y) modulo 5
% as 6x is same as x modulo 5. This is indeed exactly the
% rule in the OP's example of Latin square.

% Thus we will get a 21x21 Latin square from that, simply
% by enumerating the 21 projective lines in the projective
% plane over F_4.

% We need a list of those lines, with the points of the plane
% are suitably enumerated. Thankfully, we found one at
% https://www.uwyo.edu/moorhouse/pub/planes/pg24.txt

% 0 1 2 3 4
% 0 5 6 7 8
% 0 9 14 15 16
% 0 10 12 17 19
% 0 11 13 18 20
% 1 5 9 10 11
% 1 6 14 17 18
% 1 8 13 16 19
% 1 7 12 15 20
% 2 5 14 19 20
% 4 5 13 15 17
% 3 5 12 16 18
% 2 6 9 12 13
% 2 7 11 16 17
% 2 8 10 15 18
% 3 6 11 15 19
% 4 6 10 16 20
% 4 7 9 18 19
% 3 8 9 17 20
% 4 8 11 12 14
% 3 7 10 13 14

% Our points are enumerated from 0 to 20.
% I manipulated it using search/replace in an Emacs buffer into
% nice format for input to \xintFor and \xintAssignArray

\xintFor #1 in {%
{0}{1}{2}{3}{4},
{0}{5}{6}{7}{8},
{0}{9}{14}{15}{16},
{0}{10}{12}{17}{19},
{0}{11}{13}{18}{20},
{1}{5}{9}{10}{11},
{1}{6}{14}{17}{18},
{1}{8}{13}{16}{19},
{1}{7}{12}{15}{20},
{2}{5}{14}{19}{20},
{4}{5}{13}{15}{17},
{3}{5}{12}{16}{18},
{2}{6}{9}{12}{13},
{2}{7}{11}{16}{17},
{2}{8}{10}{15}{18},
{3}{6}{11}{15}{19},
{4}{6}{10}{16}{20},
{4}{7}{9}{18}{19},
{3}{8}{9}{17}{20},
{4}{8}{11}{12}{14},
{3}{7}{10}{13}{14}}\do{%
  \xintAssignArray#1\to\aBlock
  % arrays defined by \xintAssignArray are indexed starting at 1
  \xintFor #2 in {0, 1, 2, 3, 4}\do{%
   \xintFor #3 in {0, 1, 2, 3, 4}\do{%
    \expandafter\edef\csname Latin-\aBlock{#2+1}-\aBlock{#3+1}\endcsname
    {\aBlock{3*(#2+#3)-5*((3*(#2+#3)+3)/5 -1)+1}}%
    % this is crazy formula which simply does 3(a+b) modulo 5
    % but recall it is a+1 and b+1 which serve as arguments to
    % the array \aBlock
   }%
  }%
}

% done, all our macros defined

\edef\TwentyOneNumbersStartingAtZero{\xintSeq{0}{20}}
\begin{document}
\makeatletter
\begin{tabular}{*{22}{c|}}
  % first row
\xintFor* #1 in \TwentyOneNumbersStartingAtZero\do{&#1}\\
\hline
\xintFor* #1 in \TwentyOneNumbersStartingAtZero\do{% this will be row number #1
  #1
  \xintFor* #2 in \TwentyOneNumbersStartingAtZero\do{% this indices the columns
     &\@nameuse{Latin-#1-#2}}
  \\\hline}
\end{tabular}
\makeatother
\end{document}