Graphic representation of a triangle using ArrayPlot












3












$begingroup$


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?






plotting matrix approximation







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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$endgroup$

















    3












    $begingroup$


    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?






    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.







    $endgroup$















      3












      3








      3





      $begingroup$


      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?






      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.







      $endgroup$




      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




      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






      New contributor




      apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 7 hours ago









      apoxeiroapoxeiro

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          1 Answer
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          4












          $begingroup$

          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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            active

            oldest

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            4












            $begingroup$

            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











            $endgroup$


















              4












              $begingroup$

              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











              $endgroup$
















                4












                4








                4





                $begingroup$

                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 5 hours ago

























                answered 6 hours ago









                kglrkglr

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                188k10203421






















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