Graphic representation of a triangle using ArrayPlot
$begingroup$
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
asked 5 hours ago
apoxeiroapoxeiro
132
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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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add a comment |
$begingroup$
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
asked 5 hours ago
apoxeiroapoxeiro
132
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.
$endgroup$
add a comment |
$begingroup$
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
asked 5 hours ago
apoxeiroapoxeiro
132
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
asked 5 hours ago
apoxeiroapoxeiro
132
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.
asked 5 hours ago
apoxeiroapoxeiro
132
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.
asked 5 hours ago
apoxeiroapoxeiro
132
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.
asked 5 hours ago
apoxeiroapoxeiro
132
asked 5 hours ago
apoxeiroapoxeiro
132
132
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.
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.
apoxeiro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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]]
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]]
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]}]
answered 3 hours ago
kglrkglr
188k10203421
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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]]
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]]
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]}]
answered 3 hours ago
kglrkglr
188k10203421
$endgroup$
add a comment |
$begingroup$
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]]
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]]
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]}]
answered 3 hours ago
kglrkglr
188k10203421
$endgroup$
add a comment |
$begingroup$
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]]
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]]
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]}]
answered 3 hours ago
kglrkglr
188k10203421
$endgroup$
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]]
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]]
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]}]
answered 3 hours ago
kglrkglr
188k10203421
edited 3 hours ago
answered 3 hours ago
kglrkglr
188k10203421
answered 3 hours ago
kglrkglr
188k10203421
answered 3 hours ago
kglrkglr
188k10203421
188k10203421
add a comment |
add a comment |
apoxeiro is a new contributor. Be nice, and check out our Code of Conduct.
apoxeiro is a new contributor. Be nice, and check out our Code of Conduct.
apoxeiro is a new contributor. Be nice, and check out our Code of Conduct.
apoxeiro is a new contributor. Be nice, and check out our Code of Conduct.
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