Graphic representation of a triangle using ArrayPlot












2












$begingroup$


So I need to write a function which takes natural integer $n$ and returns graphical representation of a matrix $n times n$ using ArrayPlot.
This matrix has to be pixel approximation of equilateral triangle which get better and better as $n$ increases.



I figured out a set of equilateral triangle points which is $$P={(x,y) in mathbb{R}^2:y<sqrt{3}x+frac{asqrt{3}}{2},y<-sqrt{3}x+frac{asqrt{3}}{2},y>0}$$
where $a$ is side length of this triangle.



f1[x_, a_] := -Sqrt[3]*x + (a*Sqrt[3])/2
f2[x_, a_] := Sqrt[3]*x + (a*Sqrt[3])/2
matrix[n_] := ConstantArray[0, {n, n}]
(...)
drawapprox[n_] := ArrayPlot[matrix[n], Mesh -> True]


So I make zero $n times n$ matrix and I want to put $1$ if a point belongs to $P$ but I don't know how to put together points from the plane to this 0-1 matrix to make it works. After that I just want to use ArrayPlot function to draw new 0-1 matrix which represents triangle.



How do I make up the missing (...) part?










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    2












    $begingroup$


    So I need to write a function which takes natural integer $n$ and returns graphical representation of a matrix $n times n$ using ArrayPlot.
    This matrix has to be pixel approximation of equilateral triangle which get better and better as $n$ increases.



    I figured out a set of equilateral triangle points which is $$P={(x,y) in mathbb{R}^2:y<sqrt{3}x+frac{asqrt{3}}{2},y<-sqrt{3}x+frac{asqrt{3}}{2},y>0}$$
    where $a$ is side length of this triangle.



    f1[x_, a_] := -Sqrt[3]*x + (a*Sqrt[3])/2
    f2[x_, a_] := Sqrt[3]*x + (a*Sqrt[3])/2
    matrix[n_] := ConstantArray[0, {n, n}]
    (...)
    drawapprox[n_] := ArrayPlot[matrix[n], Mesh -> True]


    So I make zero $n times n$ matrix and I want to put $1$ if a point belongs to $P$ but I don't know how to put together points from the plane to this 0-1 matrix to make it works. After that I just want to use ArrayPlot function to draw new 0-1 matrix which represents triangle.



    How do I make up the missing (...) part?










    share|improve this question







    New contributor




    apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      2












      2








      2





      $begingroup$


      So I need to write a function which takes natural integer $n$ and returns graphical representation of a matrix $n times n$ using ArrayPlot.
      This matrix has to be pixel approximation of equilateral triangle which get better and better as $n$ increases.



      I figured out a set of equilateral triangle points which is $$P={(x,y) in mathbb{R}^2:y<sqrt{3}x+frac{asqrt{3}}{2},y<-sqrt{3}x+frac{asqrt{3}}{2},y>0}$$
      where $a$ is side length of this triangle.



      f1[x_, a_] := -Sqrt[3]*x + (a*Sqrt[3])/2
      f2[x_, a_] := Sqrt[3]*x + (a*Sqrt[3])/2
      matrix[n_] := ConstantArray[0, {n, n}]
      (...)
      drawapprox[n_] := ArrayPlot[matrix[n], Mesh -> True]


      So I make zero $n times n$ matrix and I want to put $1$ if a point belongs to $P$ but I don't know how to put together points from the plane to this 0-1 matrix to make it works. After that I just want to use ArrayPlot function to draw new 0-1 matrix which represents triangle.



      How do I make up the missing (...) part?










      share|improve this question







      New contributor




      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      So I need to write a function which takes natural integer $n$ and returns graphical representation of a matrix $n times n$ using ArrayPlot.
      This matrix has to be pixel approximation of equilateral triangle which get better and better as $n$ increases.



      I figured out a set of equilateral triangle points which is $$P={(x,y) in mathbb{R}^2:y<sqrt{3}x+frac{asqrt{3}}{2},y<-sqrt{3}x+frac{asqrt{3}}{2},y>0}$$
      where $a$ is side length of this triangle.



      f1[x_, a_] := -Sqrt[3]*x + (a*Sqrt[3])/2
      f2[x_, a_] := Sqrt[3]*x + (a*Sqrt[3])/2
      matrix[n_] := ConstantArray[0, {n, n}]
      (...)
      drawapprox[n_] := ArrayPlot[matrix[n], Mesh -> True]


      So I make zero $n times n$ matrix and I want to put $1$ if a point belongs to $P$ but I don't know how to put together points from the plane to this 0-1 matrix to make it works. After that I just want to use ArrayPlot function to draw new 0-1 matrix which represents triangle.



      How do I make up the missing (...) part?







      plotting matrix approximation






      share|improve this question







      New contributor




      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






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      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 4 hours ago









      apoxeiroapoxeiro

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      New contributor





      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Check out our Code of Conduct.






















          1 Answer
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          3












          $begingroup$

          Update: An alternative method using SparseArray:



          ClearAll[sa, plot2]
          sa[a_] := SparseArray[{i_, j_} /;
          a - i < f1[j - (a + Boole[OddQ[a]])/2, a] &&
          a - i < f2[j - (a + Boole[OddQ[a]])/2, a] -> 1, {a, a}]
          plot2[a_] := ArrayPlot[sa[a], Mesh -> All]

          Row[plot2 /@ Range[3, 21, 2]]


          enter image description here



          Original answer:



          aplot[a_] := ArrayPlot[Boole @ MapIndexed[
          a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
          a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
          matrix[a], {2}], Mesh -> All];

          Row[Show[plot@#, Graphics[{FaceForm, EdgeForm[{Thick, Red}], SSSTriangle[#, #, #]}]] & /@
          Range[3, 21, 2]]


          enter image description here



          With a = 101; and Mesh -> None, we get



          a = 1001; 
          ap1001 = ArrayPlot[Boole@MapIndexed[
          a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
          a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
          matrix[a], {2}], Mesh -> None];

          Graphics[{ap1001[[1]], FaceForm, EdgeForm[{Thick, Red}],
          SSSTriangle[1001, 1001, 1001]}]


          enter image description here






          share|improve this answer











          $endgroup$













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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Update: An alternative method using SparseArray:



            ClearAll[sa, plot2]
            sa[a_] := SparseArray[{i_, j_} /;
            a - i < f1[j - (a + Boole[OddQ[a]])/2, a] &&
            a - i < f2[j - (a + Boole[OddQ[a]])/2, a] -> 1, {a, a}]
            plot2[a_] := ArrayPlot[sa[a], Mesh -> All]

            Row[plot2 /@ Range[3, 21, 2]]


            enter image description here



            Original answer:



            aplot[a_] := ArrayPlot[Boole @ MapIndexed[
            a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
            a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
            matrix[a], {2}], Mesh -> All];

            Row[Show[plot@#, Graphics[{FaceForm, EdgeForm[{Thick, Red}], SSSTriangle[#, #, #]}]] & /@
            Range[3, 21, 2]]


            enter image description here



            With a = 101; and Mesh -> None, we get



            a = 1001; 
            ap1001 = ArrayPlot[Boole@MapIndexed[
            a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
            a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
            matrix[a], {2}], Mesh -> None];

            Graphics[{ap1001[[1]], FaceForm, EdgeForm[{Thick, Red}],
            SSSTriangle[1001, 1001, 1001]}]


            enter image description here






            share|improve this answer











            $endgroup$


















              3












              $begingroup$

              Update: An alternative method using SparseArray:



              ClearAll[sa, plot2]
              sa[a_] := SparseArray[{i_, j_} /;
              a - i < f1[j - (a + Boole[OddQ[a]])/2, a] &&
              a - i < f2[j - (a + Boole[OddQ[a]])/2, a] -> 1, {a, a}]
              plot2[a_] := ArrayPlot[sa[a], Mesh -> All]

              Row[plot2 /@ Range[3, 21, 2]]


              enter image description here



              Original answer:



              aplot[a_] := ArrayPlot[Boole @ MapIndexed[
              a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
              a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
              matrix[a], {2}], Mesh -> All];

              Row[Show[plot@#, Graphics[{FaceForm, EdgeForm[{Thick, Red}], SSSTriangle[#, #, #]}]] & /@
              Range[3, 21, 2]]


              enter image description here



              With a = 101; and Mesh -> None, we get



              a = 1001; 
              ap1001 = ArrayPlot[Boole@MapIndexed[
              a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
              a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
              matrix[a], {2}], Mesh -> None];

              Graphics[{ap1001[[1]], FaceForm, EdgeForm[{Thick, Red}],
              SSSTriangle[1001, 1001, 1001]}]


              enter image description here






              share|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Update: An alternative method using SparseArray:



                ClearAll[sa, plot2]
                sa[a_] := SparseArray[{i_, j_} /;
                a - i < f1[j - (a + Boole[OddQ[a]])/2, a] &&
                a - i < f2[j - (a + Boole[OddQ[a]])/2, a] -> 1, {a, a}]
                plot2[a_] := ArrayPlot[sa[a], Mesh -> All]

                Row[plot2 /@ Range[3, 21, 2]]


                enter image description here



                Original answer:



                aplot[a_] := ArrayPlot[Boole @ MapIndexed[
                a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
                a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
                matrix[a], {2}], Mesh -> All];

                Row[Show[plot@#, Graphics[{FaceForm, EdgeForm[{Thick, Red}], SSSTriangle[#, #, #]}]] & /@
                Range[3, 21, 2]]


                enter image description here



                With a = 101; and Mesh -> None, we get



                a = 1001; 
                ap1001 = ArrayPlot[Boole@MapIndexed[
                a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
                a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
                matrix[a], {2}], Mesh -> None];

                Graphics[{ap1001[[1]], FaceForm, EdgeForm[{Thick, Red}],
                SSSTriangle[1001, 1001, 1001]}]


                enter image description here






                share|improve this answer











                $endgroup$



                Update: An alternative method using SparseArray:



                ClearAll[sa, plot2]
                sa[a_] := SparseArray[{i_, j_} /;
                a - i < f1[j - (a + Boole[OddQ[a]])/2, a] &&
                a - i < f2[j - (a + Boole[OddQ[a]])/2, a] -> 1, {a, a}]
                plot2[a_] := ArrayPlot[sa[a], Mesh -> All]

                Row[plot2 /@ Range[3, 21, 2]]


                enter image description here



                Original answer:



                aplot[a_] := ArrayPlot[Boole @ MapIndexed[
                a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
                a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
                matrix[a], {2}], Mesh -> All];

                Row[Show[plot@#, Graphics[{FaceForm, EdgeForm[{Thick, Red}], SSSTriangle[#, #, #]}]] & /@
                Range[3, 21, 2]]


                enter image description here



                With a = 101; and Mesh -> None, we get



                a = 1001; 
                ap1001 = ArrayPlot[Boole@MapIndexed[
                a - #2[[1]] < f1[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &&
                a - #2[[1]] < f2[#2[[2]] - (a + Boole[OddQ[a]])/2, a] &,
                matrix[a], {2}], Mesh -> None];

                Graphics[{ap1001[[1]], FaceForm, EdgeForm[{Thick, Red}],
                SSSTriangle[1001, 1001, 1001]}]


                enter image description here







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 hours ago

























                answered 3 hours ago









                kglrkglr

                188k10203421




                188k10203421






















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