Draw a checker pattern with a black X in the center





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{
margin-bottom:0;
}








12

















I am trying to recreate the following image in TikZ



enter image description here



Using some old code I was able to produce the following result



enter image description here



While I was able to produce the correct result, I feel that my solution was a bit strange as it required two passes. Any suggestions for alternative approaches, or improvements to the code are more than welcome.



documentclass[tikz]{standalone}

begin{document}

begin{tikzpicture}[x=1cm]
edefsize{4}
foreach x in {0,...,size} foreach y in {0,...,size}
{
pgfmathparse{mod(x+y,size) ? "none" : "black"}
edefcolour{pgfmathresult}
path[draw=black, fill=colour] (x,y) rectangle ++ (1,1);

pgfmathparse{x==y ? "black" : "none"}
edefcolour{pgfmathresult}
path[fill=colour] (x,y) rectangle ++ (1,1);
}
draw (0,0)--(0,size+1)--(size+1,size+1)--(size+1,0)--cycle;
end{tikzpicture}

end{document}









share|improve this question



































    12

















    I am trying to recreate the following image in TikZ



    enter image description here



    Using some old code I was able to produce the following result



    enter image description here



    While I was able to produce the correct result, I feel that my solution was a bit strange as it required two passes. Any suggestions for alternative approaches, or improvements to the code are more than welcome.



    documentclass[tikz]{standalone}

    begin{document}

    begin{tikzpicture}[x=1cm]
    edefsize{4}
    foreach x in {0,...,size} foreach y in {0,...,size}
    {
    pgfmathparse{mod(x+y,size) ? "none" : "black"}
    edefcolour{pgfmathresult}
    path[draw=black, fill=colour] (x,y) rectangle ++ (1,1);

    pgfmathparse{x==y ? "black" : "none"}
    edefcolour{pgfmathresult}
    path[fill=colour] (x,y) rectangle ++ (1,1);
    }
    draw (0,0)--(0,size+1)--(size+1,size+1)--(size+1,0)--cycle;
    end{tikzpicture}

    end{document}









    share|improve this question































      12












      12








      12


      1






      I am trying to recreate the following image in TikZ



      enter image description here



      Using some old code I was able to produce the following result



      enter image description here



      While I was able to produce the correct result, I feel that my solution was a bit strange as it required two passes. Any suggestions for alternative approaches, or improvements to the code are more than welcome.



      documentclass[tikz]{standalone}

      begin{document}

      begin{tikzpicture}[x=1cm]
      edefsize{4}
      foreach x in {0,...,size} foreach y in {0,...,size}
      {
      pgfmathparse{mod(x+y,size) ? "none" : "black"}
      edefcolour{pgfmathresult}
      path[draw=black, fill=colour] (x,y) rectangle ++ (1,1);

      pgfmathparse{x==y ? "black" : "none"}
      edefcolour{pgfmathresult}
      path[fill=colour] (x,y) rectangle ++ (1,1);
      }
      draw (0,0)--(0,size+1)--(size+1,size+1)--(size+1,0)--cycle;
      end{tikzpicture}

      end{document}









      share|improve this question
















      I am trying to recreate the following image in TikZ



      enter image description here



      Using some old code I was able to produce the following result



      enter image description here



      While I was able to produce the correct result, I feel that my solution was a bit strange as it required two passes. Any suggestions for alternative approaches, or improvements to the code are more than welcome.



      documentclass[tikz]{standalone}

      begin{document}

      begin{tikzpicture}[x=1cm]
      edefsize{4}
      foreach x in {0,...,size} foreach y in {0,...,size}
      {
      pgfmathparse{mod(x+y,size) ? "none" : "black"}
      edefcolour{pgfmathresult}
      path[draw=black, fill=colour] (x,y) rectangle ++ (1,1);

      pgfmathparse{x==y ? "black" : "none"}
      edefcolour{pgfmathresult}
      path[fill=colour] (x,y) rectangle ++ (1,1);
      }
      draw (0,0)--(0,size+1)--(size+1,size+1)--(size+1,0)--cycle;
      end{tikzpicture}

      end{document}






      tikz-pgf code-review






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question



      share|improve this question








      edited May 28 at 4:14









      Community

      1




      1










      asked May 27 at 14:00









      N3buchadnezzarN3buchadnezzar

      4,9576 gold badges44 silver badges99 bronze badges




      4,9576 gold badges44 silver badges99 bronze badges

























          4 Answers
          4






          active

          oldest

          votes


















          15


















          With tikz:



          documentclass[tikz]{standalone}

          begin{document}
          begin{tikzpicture}[
          node distance = 0mm,
          box/.style = {draw, minimum size=10mm, fill=black,
          outer sep=0pt},
          ]
          edefsize{4}
          foreach y in {0,...,size}
          foreach x in {0,...,size}
          {ifnumx=y
          node[box] at (x,size-y) {};
          node[box] at (x,y) {};
          else
          node[box,fill=none] at (x,y) {};
          fi
          }
          end{tikzpicture}
          end{document}


          enter image description here



          Note: Value of size had to be zero or any even natural number (0, 2, 4, ...)



          addendum:




          • default baseline of above image is as (current bounding box.south)˙ For series of those images for different value ofsize` is:


          documentclass{article}
          usepackage{tikz}
          usepackage{tabularx}
          newcolumntype{C}{>{centeringarraybackslash}X}

          begin{document}
          begin{figure}
          begin{tabularx}{linewidth}{>{hsize=0.5hsize}C C >{hsize=1.5hsize}C}
          begin{tikzpicture}[baseline=(current bounding box.south),
          node distance = 0mm,
          box/.style = {draw, minimum size=10mm, fill=black,
          outer sep=0pt},
          ]
          edefsize{0} % in this MWE the meaning of `size` is changed
          foreach y in {0,...,size}
          foreach x in {0,...,size}
          {ifnumx=y
          node[box] at (x,size-y) {};
          node[box] at (x,y) {};
          else
          node[box,fill=none] at (x,y) {};
          fi
          }
          end{tikzpicture}
          caption{}
          &
          begin{tikzpicture}[%baseline=(current bounding box.south),
          node distance = 0mm,
          box/.style = {draw, minimum size=10mm, fill=black,
          outer sep=0pt},
          ]
          edefsize{1}
          foreach y in {0,...,2*size} % changed, now number of boxes is odd
          foreach x in {0,...,2*size} % changed,
          {ifnumx=y
          node[box] at (x,size-y) {};
          node[box] at (x,y) {};
          else
          node[box,fill=none] at (x,y) {};
          fi
          }
          end{tikzpicture}
          caption{}
          &
          begin{tikzpicture}[baseline=(current bounding box.south),
          node distance = 0mm,
          box/.style = {draw, minimum size=10mm, fill=black,
          outer sep=0pt},
          ]
          edefsize{2}
          foreach y in {0,...,size}
          foreach x in {0,...,size}
          {ifnumx=y
          node[box] at (x,size-y) {};
          node[box] at (x,y) {};
          else
          node[box,fill=none] at (x,y) {};
          fi
          }
          end{tikzpicture}
          caption{}
          end{tabularx}
          end{figure}
          end{document}


          enter image description here






          share|improve this answer





























          • How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33











          • With edefsize{1}? I'm not sure if I understood your comment correctly.

            – Zarko
            May 27 at 15:36











          • size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

            – N3buchadnezzar
            May 27 at 15:37













          • Indeed. It should be size{0}.

            – Zarko
            May 27 at 15:41






          • 1





            @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

            – Zarko
            May 27 at 16:06





















          10


















          A PSTricks solution only for fun purposes!



          enter image description here



          documentclass[border=1pt]{standalone}
          usepackage{pstricks}
          defobj#1{%
          pspicture[dimen=m](#1,#1)
          multips(0,0)(0,1){#1}{multips(0,0)(1,0){#1}{psframe(1,1)}}
          multips(0,0)(1,1){#1}{psframe*(1,1)}
          multips(0,#1)(1,-1){#1}{psframe*(1,-1)}
          endpspicture}

          begin{document}
          foreach i in {3,5,7}{obj{i}quad}
          end{document}


          Edit



          I invented the algorithm (that has not been patented yet) as follows. No nested loop is needed.



          enter image description here



          documentclass[border=12pt]{standalone}
          usepackage[nomessages]{fp}
          usepackage{xintexpr}
          usepackage{pstricks}
          psset{unit=5mm}
          defobj#1{%
          pspicture[dimen=m](#1,#1)
          FPevalN{#1*#1-1}
          foreach j in {0,...,N}
          {
          FPevaly{trunc(j/#1:0)}
          FPevalx{j-#1*y}
          xintifboolexpr{x=y||(x+y)=(#1-1)}
          {psframe[fillstyle=solid,fillcolor=black](x,y)(+x+1,y+1)}
          {psframe(x,y)(+x+1,y+1)}
          }
          endpspicture}
          begin{document}
          foreach i in {1,3,5,7,9}{obj{i}quad}
          end{document}





          share|improve this answer





























          • One downvote detected...

            – Artificial Stupidity
            May 27 at 14:31











          • Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

            – Artificial Stupidity
            May 27 at 17:04





















          9


















          Edit:
          The following works for all values of size



          documentclass[tikz]{standalone}

          begin{document}

          begin{tikzpicture}
          edefsize{4}
          foreach x in {0,...,size} foreach y in {0,...,size} {
          pgfmathsetmacro{colour}{(x==y || x+y==size) ? "black" : "none"}
          draw[fill=colour] (x,y) rectangle ++ (1,1);
          }
          end{tikzpicture}

          end{document}


          enter image description here






          share|improve this answer





























          • How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33











          • The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

            – AboAmmar
            May 27 at 17:00





















          1


















          I do not see the reason for a double loop, nor complicated conditions.



          documentclass[tikz,border=3.14mm]{standalone}
          begin{document}
          begin{tikzpicture}[xboard/.style={insert path={
          (0,0) grid (#1,#1)
          foreach X in {1,...,#1}
          {(X-0.5,X-0.5) pic{bx} (X-0.5,#1-X+0.5) pic{bx}}}},
          pics/bx/.style={code={fill (-0.5,-0.5) rectangle (0.5,0.5);}}]
          draw[xboard=1] [xshift=2cm,xboard=3]
          [xshift=4cm,xboard=5] [xshift=6cm,xboard=7];
          end{tikzpicture}
          end{document}


          enter image description here






          share|improve this answer




























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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            15


















            With tikz:



            documentclass[tikz]{standalone}

            begin{document}
            begin{tikzpicture}[
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{4}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            end{document}


            enter image description here



            Note: Value of size had to be zero or any even natural number (0, 2, 4, ...)



            addendum:




            • default baseline of above image is as (current bounding box.south)˙ For series of those images for different value ofsize` is:


            documentclass{article}
            usepackage{tikz}
            usepackage{tabularx}
            newcolumntype{C}{>{centeringarraybackslash}X}

            begin{document}
            begin{figure}
            begin{tabularx}{linewidth}{>{hsize=0.5hsize}C C >{hsize=1.5hsize}C}
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{0} % in this MWE the meaning of `size` is changed
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[%baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{1}
            foreach y in {0,...,2*size} % changed, now number of boxes is odd
            foreach x in {0,...,2*size} % changed,
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{2}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            end{tabularx}
            end{figure}
            end{document}


            enter image description here






            share|improve this answer





























            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • With edefsize{1}? I'm not sure if I understood your comment correctly.

              – Zarko
              May 27 at 15:36











            • size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

              – N3buchadnezzar
              May 27 at 15:37













            • Indeed. It should be size{0}.

              – Zarko
              May 27 at 15:41






            • 1





              @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

              – Zarko
              May 27 at 16:06


















            15


















            With tikz:



            documentclass[tikz]{standalone}

            begin{document}
            begin{tikzpicture}[
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{4}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            end{document}


            enter image description here



            Note: Value of size had to be zero or any even natural number (0, 2, 4, ...)



            addendum:




            • default baseline of above image is as (current bounding box.south)˙ For series of those images for different value ofsize` is:


            documentclass{article}
            usepackage{tikz}
            usepackage{tabularx}
            newcolumntype{C}{>{centeringarraybackslash}X}

            begin{document}
            begin{figure}
            begin{tabularx}{linewidth}{>{hsize=0.5hsize}C C >{hsize=1.5hsize}C}
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{0} % in this MWE the meaning of `size` is changed
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[%baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{1}
            foreach y in {0,...,2*size} % changed, now number of boxes is odd
            foreach x in {0,...,2*size} % changed,
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{2}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            end{tabularx}
            end{figure}
            end{document}


            enter image description here






            share|improve this answer





























            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • With edefsize{1}? I'm not sure if I understood your comment correctly.

              – Zarko
              May 27 at 15:36











            • size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

              – N3buchadnezzar
              May 27 at 15:37













            • Indeed. It should be size{0}.

              – Zarko
              May 27 at 15:41






            • 1





              @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

              – Zarko
              May 27 at 16:06
















            15














            15










            15









            With tikz:



            documentclass[tikz]{standalone}

            begin{document}
            begin{tikzpicture}[
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{4}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            end{document}


            enter image description here



            Note: Value of size had to be zero or any even natural number (0, 2, 4, ...)



            addendum:




            • default baseline of above image is as (current bounding box.south)˙ For series of those images for different value ofsize` is:


            documentclass{article}
            usepackage{tikz}
            usepackage{tabularx}
            newcolumntype{C}{>{centeringarraybackslash}X}

            begin{document}
            begin{figure}
            begin{tabularx}{linewidth}{>{hsize=0.5hsize}C C >{hsize=1.5hsize}C}
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{0} % in this MWE the meaning of `size` is changed
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[%baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{1}
            foreach y in {0,...,2*size} % changed, now number of boxes is odd
            foreach x in {0,...,2*size} % changed,
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{2}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            end{tabularx}
            end{figure}
            end{document}


            enter image description here






            share|improve this answer
















            With tikz:



            documentclass[tikz]{standalone}

            begin{document}
            begin{tikzpicture}[
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{4}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            end{document}


            enter image description here



            Note: Value of size had to be zero or any even natural number (0, 2, 4, ...)



            addendum:




            • default baseline of above image is as (current bounding box.south)˙ For series of those images for different value ofsize` is:


            documentclass{article}
            usepackage{tikz}
            usepackage{tabularx}
            newcolumntype{C}{>{centeringarraybackslash}X}

            begin{document}
            begin{figure}
            begin{tabularx}{linewidth}{>{hsize=0.5hsize}C C >{hsize=1.5hsize}C}
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{0} % in this MWE the meaning of `size` is changed
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[%baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{1}
            foreach y in {0,...,2*size} % changed, now number of boxes is odd
            foreach x in {0,...,2*size} % changed,
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            &
            begin{tikzpicture}[baseline=(current bounding box.south),
            node distance = 0mm,
            box/.style = {draw, minimum size=10mm, fill=black,
            outer sep=0pt},
            ]
            edefsize{2}
            foreach y in {0,...,size}
            foreach x in {0,...,size}
            {ifnumx=y
            node[box] at (x,size-y) {};
            node[box] at (x,y) {};
            else
            node[box,fill=none] at (x,y) {};
            fi
            }
            end{tikzpicture}
            caption{}
            end{tabularx}
            end{figure}
            end{document}


            enter image description here







            share|improve this answer















            share|improve this answer




            share|improve this answer



            share|improve this answer








            edited May 27 at 18:04

























            answered May 27 at 15:01









            ZarkoZarko

            149k8 gold badges85 silver badges196 bronze badges




            149k8 gold badges85 silver badges196 bronze badges
















            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • With edefsize{1}? I'm not sure if I understood your comment correctly.

              – Zarko
              May 27 at 15:36











            • size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

              – N3buchadnezzar
              May 27 at 15:37













            • Indeed. It should be size{0}.

              – Zarko
              May 27 at 15:41






            • 1





              @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

              – Zarko
              May 27 at 16:06





















            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • With edefsize{1}? I'm not sure if I understood your comment correctly.

              – Zarko
              May 27 at 15:36











            • size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

              – N3buchadnezzar
              May 27 at 15:37













            • Indeed. It should be size{0}.

              – Zarko
              May 27 at 15:41






            • 1





              @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

              – Zarko
              May 27 at 16:06



















            How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33





            How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33













            With edefsize{1}? I'm not sure if I understood your comment correctly.

            – Zarko
            May 27 at 15:36





            With edefsize{1}? I'm not sure if I understood your comment correctly.

            – Zarko
            May 27 at 15:36













            size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

            – N3buchadnezzar
            May 27 at 15:37







            size{1} produces 4 black squares not 1. Here is how it looks for n=1 and n=2, not the same baseheight either. i.imgur.com/0OamEbM.png

            – N3buchadnezzar
            May 27 at 15:37















            Indeed. It should be size{0}.

            – Zarko
            May 27 at 15:41





            Indeed. It should be size{0}.

            – Zarko
            May 27 at 15:41




            1




            1





            @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

            – Zarko
            May 27 at 16:06







            @N3buchadnezzar, images are aligned to their bottom side. also see "Note" in edited answer.

            – Zarko
            May 27 at 16:06















            10


















            A PSTricks solution only for fun purposes!



            enter image description here



            documentclass[border=1pt]{standalone}
            usepackage{pstricks}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            multips(0,0)(0,1){#1}{multips(0,0)(1,0){#1}{psframe(1,1)}}
            multips(0,0)(1,1){#1}{psframe*(1,1)}
            multips(0,#1)(1,-1){#1}{psframe*(1,-1)}
            endpspicture}

            begin{document}
            foreach i in {3,5,7}{obj{i}quad}
            end{document}


            Edit



            I invented the algorithm (that has not been patented yet) as follows. No nested loop is needed.



            enter image description here



            documentclass[border=12pt]{standalone}
            usepackage[nomessages]{fp}
            usepackage{xintexpr}
            usepackage{pstricks}
            psset{unit=5mm}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            FPevalN{#1*#1-1}
            foreach j in {0,...,N}
            {
            FPevaly{trunc(j/#1:0)}
            FPevalx{j-#1*y}
            xintifboolexpr{x=y||(x+y)=(#1-1)}
            {psframe[fillstyle=solid,fillcolor=black](x,y)(+x+1,y+1)}
            {psframe(x,y)(+x+1,y+1)}
            }
            endpspicture}
            begin{document}
            foreach i in {1,3,5,7,9}{obj{i}quad}
            end{document}





            share|improve this answer





























            • One downvote detected...

              – Artificial Stupidity
              May 27 at 14:31











            • Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

              – Artificial Stupidity
              May 27 at 17:04


















            10


















            A PSTricks solution only for fun purposes!



            enter image description here



            documentclass[border=1pt]{standalone}
            usepackage{pstricks}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            multips(0,0)(0,1){#1}{multips(0,0)(1,0){#1}{psframe(1,1)}}
            multips(0,0)(1,1){#1}{psframe*(1,1)}
            multips(0,#1)(1,-1){#1}{psframe*(1,-1)}
            endpspicture}

            begin{document}
            foreach i in {3,5,7}{obj{i}quad}
            end{document}


            Edit



            I invented the algorithm (that has not been patented yet) as follows. No nested loop is needed.



            enter image description here



            documentclass[border=12pt]{standalone}
            usepackage[nomessages]{fp}
            usepackage{xintexpr}
            usepackage{pstricks}
            psset{unit=5mm}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            FPevalN{#1*#1-1}
            foreach j in {0,...,N}
            {
            FPevaly{trunc(j/#1:0)}
            FPevalx{j-#1*y}
            xintifboolexpr{x=y||(x+y)=(#1-1)}
            {psframe[fillstyle=solid,fillcolor=black](x,y)(+x+1,y+1)}
            {psframe(x,y)(+x+1,y+1)}
            }
            endpspicture}
            begin{document}
            foreach i in {1,3,5,7,9}{obj{i}quad}
            end{document}





            share|improve this answer





























            • One downvote detected...

              – Artificial Stupidity
              May 27 at 14:31











            • Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

              – Artificial Stupidity
              May 27 at 17:04
















            10














            10










            10









            A PSTricks solution only for fun purposes!



            enter image description here



            documentclass[border=1pt]{standalone}
            usepackage{pstricks}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            multips(0,0)(0,1){#1}{multips(0,0)(1,0){#1}{psframe(1,1)}}
            multips(0,0)(1,1){#1}{psframe*(1,1)}
            multips(0,#1)(1,-1){#1}{psframe*(1,-1)}
            endpspicture}

            begin{document}
            foreach i in {3,5,7}{obj{i}quad}
            end{document}


            Edit



            I invented the algorithm (that has not been patented yet) as follows. No nested loop is needed.



            enter image description here



            documentclass[border=12pt]{standalone}
            usepackage[nomessages]{fp}
            usepackage{xintexpr}
            usepackage{pstricks}
            psset{unit=5mm}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            FPevalN{#1*#1-1}
            foreach j in {0,...,N}
            {
            FPevaly{trunc(j/#1:0)}
            FPevalx{j-#1*y}
            xintifboolexpr{x=y||(x+y)=(#1-1)}
            {psframe[fillstyle=solid,fillcolor=black](x,y)(+x+1,y+1)}
            {psframe(x,y)(+x+1,y+1)}
            }
            endpspicture}
            begin{document}
            foreach i in {1,3,5,7,9}{obj{i}quad}
            end{document}





            share|improve this answer
















            A PSTricks solution only for fun purposes!



            enter image description here



            documentclass[border=1pt]{standalone}
            usepackage{pstricks}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            multips(0,0)(0,1){#1}{multips(0,0)(1,0){#1}{psframe(1,1)}}
            multips(0,0)(1,1){#1}{psframe*(1,1)}
            multips(0,#1)(1,-1){#1}{psframe*(1,-1)}
            endpspicture}

            begin{document}
            foreach i in {3,5,7}{obj{i}quad}
            end{document}


            Edit



            I invented the algorithm (that has not been patented yet) as follows. No nested loop is needed.



            enter image description here



            documentclass[border=12pt]{standalone}
            usepackage[nomessages]{fp}
            usepackage{xintexpr}
            usepackage{pstricks}
            psset{unit=5mm}
            defobj#1{%
            pspicture[dimen=m](#1,#1)
            FPevalN{#1*#1-1}
            foreach j in {0,...,N}
            {
            FPevaly{trunc(j/#1:0)}
            FPevalx{j-#1*y}
            xintifboolexpr{x=y||(x+y)=(#1-1)}
            {psframe[fillstyle=solid,fillcolor=black](x,y)(+x+1,y+1)}
            {psframe(x,y)(+x+1,y+1)}
            }
            endpspicture}
            begin{document}
            foreach i in {1,3,5,7,9}{obj{i}quad}
            end{document}






            share|improve this answer















            share|improve this answer




            share|improve this answer



            share|improve this answer








            edited May 27 at 16:55

























            answered May 27 at 14:23









            Artificial StupidityArtificial Stupidity

            6,2861 gold badge14 silver badges50 bronze badges




            6,2861 gold badge14 silver badges50 bronze badges
















            • One downvote detected...

              – Artificial Stupidity
              May 27 at 14:31











            • Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

              – Artificial Stupidity
              May 27 at 17:04





















            • One downvote detected...

              – Artificial Stupidity
              May 27 at 14:31











            • Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

              – Artificial Stupidity
              May 27 at 17:04



















            One downvote detected...

            – Artificial Stupidity
            May 27 at 14:31





            One downvote detected...

            – Artificial Stupidity
            May 27 at 14:31













            Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

            – Artificial Stupidity
            May 27 at 17:04







            Accessing 2 dimensional arrays with a single-indexed pointer is much faster.

            – Artificial Stupidity
            May 27 at 17:04













            9


















            Edit:
            The following works for all values of size



            documentclass[tikz]{standalone}

            begin{document}

            begin{tikzpicture}
            edefsize{4}
            foreach x in {0,...,size} foreach y in {0,...,size} {
            pgfmathsetmacro{colour}{(x==y || x+y==size) ? "black" : "none"}
            draw[fill=colour] (x,y) rectangle ++ (1,1);
            }
            end{tikzpicture}

            end{document}


            enter image description here






            share|improve this answer





























            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

              – AboAmmar
              May 27 at 17:00


















            9


















            Edit:
            The following works for all values of size



            documentclass[tikz]{standalone}

            begin{document}

            begin{tikzpicture}
            edefsize{4}
            foreach x in {0,...,size} foreach y in {0,...,size} {
            pgfmathsetmacro{colour}{(x==y || x+y==size) ? "black" : "none"}
            draw[fill=colour] (x,y) rectangle ++ (1,1);
            }
            end{tikzpicture}

            end{document}


            enter image description here






            share|improve this answer





























            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

              – AboAmmar
              May 27 at 17:00
















            9














            9










            9









            Edit:
            The following works for all values of size



            documentclass[tikz]{standalone}

            begin{document}

            begin{tikzpicture}
            edefsize{4}
            foreach x in {0,...,size} foreach y in {0,...,size} {
            pgfmathsetmacro{colour}{(x==y || x+y==size) ? "black" : "none"}
            draw[fill=colour] (x,y) rectangle ++ (1,1);
            }
            end{tikzpicture}

            end{document}


            enter image description here






            share|improve this answer
















            Edit:
            The following works for all values of size



            documentclass[tikz]{standalone}

            begin{document}

            begin{tikzpicture}
            edefsize{4}
            foreach x in {0,...,size} foreach y in {0,...,size} {
            pgfmathsetmacro{colour}{(x==y || x+y==size) ? "black" : "none"}
            draw[fill=colour] (x,y) rectangle ++ (1,1);
            }
            end{tikzpicture}

            end{document}


            enter image description here







            share|improve this answer















            share|improve this answer




            share|improve this answer



            share|improve this answer








            edited May 27 at 17:45

























            answered May 27 at 14:50









            AboAmmarAboAmmar

            37.1k3 gold badges31 silver badges91 bronze badges




            37.1k3 gold badges31 silver badges91 bronze badges
















            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

              – AboAmmar
              May 27 at 17:00





















            • How can this produce the correct case with only one black box?

              – N3buchadnezzar
              May 27 at 15:33











            • The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

              – AboAmmar
              May 27 at 17:00



















            How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33





            How can this produce the correct case with only one black box?

            – N3buchadnezzar
            May 27 at 15:33













            The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

            – AboAmmar
            May 27 at 17:00







            The mod function was not a good choice, pgfmathparse{x+y==size ? "black" : "colour"} is better.

            – AboAmmar
            May 27 at 17:00













            1


















            I do not see the reason for a double loop, nor complicated conditions.



            documentclass[tikz,border=3.14mm]{standalone}
            begin{document}
            begin{tikzpicture}[xboard/.style={insert path={
            (0,0) grid (#1,#1)
            foreach X in {1,...,#1}
            {(X-0.5,X-0.5) pic{bx} (X-0.5,#1-X+0.5) pic{bx}}}},
            pics/bx/.style={code={fill (-0.5,-0.5) rectangle (0.5,0.5);}}]
            draw[xboard=1] [xshift=2cm,xboard=3]
            [xshift=4cm,xboard=5] [xshift=6cm,xboard=7];
            end{tikzpicture}
            end{document}


            enter image description here






            share|improve this answer































              1


















              I do not see the reason for a double loop, nor complicated conditions.



              documentclass[tikz,border=3.14mm]{standalone}
              begin{document}
              begin{tikzpicture}[xboard/.style={insert path={
              (0,0) grid (#1,#1)
              foreach X in {1,...,#1}
              {(X-0.5,X-0.5) pic{bx} (X-0.5,#1-X+0.5) pic{bx}}}},
              pics/bx/.style={code={fill (-0.5,-0.5) rectangle (0.5,0.5);}}]
              draw[xboard=1] [xshift=2cm,xboard=3]
              [xshift=4cm,xboard=5] [xshift=6cm,xboard=7];
              end{tikzpicture}
              end{document}


              enter image description here






              share|improve this answer





























                1














                1










                1









                I do not see the reason for a double loop, nor complicated conditions.



                documentclass[tikz,border=3.14mm]{standalone}
                begin{document}
                begin{tikzpicture}[xboard/.style={insert path={
                (0,0) grid (#1,#1)
                foreach X in {1,...,#1}
                {(X-0.5,X-0.5) pic{bx} (X-0.5,#1-X+0.5) pic{bx}}}},
                pics/bx/.style={code={fill (-0.5,-0.5) rectangle (0.5,0.5);}}]
                draw[xboard=1] [xshift=2cm,xboard=3]
                [xshift=4cm,xboard=5] [xshift=6cm,xboard=7];
                end{tikzpicture}
                end{document}


                enter image description here






                share|improve this answer














                I do not see the reason for a double loop, nor complicated conditions.



                documentclass[tikz,border=3.14mm]{standalone}
                begin{document}
                begin{tikzpicture}[xboard/.style={insert path={
                (0,0) grid (#1,#1)
                foreach X in {1,...,#1}
                {(X-0.5,X-0.5) pic{bx} (X-0.5,#1-X+0.5) pic{bx}}}},
                pics/bx/.style={code={fill (-0.5,-0.5) rectangle (0.5,0.5);}}]
                draw[xboard=1] [xshift=2cm,xboard=3]
                [xshift=4cm,xboard=5] [xshift=6cm,xboard=7];
                end{tikzpicture}
                end{document}


                enter image description here







                share|improve this answer













                share|improve this answer




                share|improve this answer



                share|improve this answer










                answered May 28 at 3:26







                user121799









































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Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029