How to generate a triangular grid from a list of pointsPoint lattice leading to triangle latticeCombining 3 graphics of different coordinate systemsCreate triangular mesh from random list of pointsInterpolation on a regular square grid spanning a triangular domainHow to generate grid points on boundary of $[-1,1]^d$ for arbitrary dimension $d$ and specified resolution?How to generate the rows of data points from a model?How to apply $overliner(x,y)$ to shapes with straight lines or absolute values?How to make a cow smaller (in BubbleChart3D plot)How to plot a 2D triangular latticeHow to generate animation from manipulate?Point lattice leading to triangle lattice
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How to generate a triangular grid from a list of points
Point lattice leading to triangle latticeCombining 3 graphics of different coordinate systemsCreate triangular mesh from random list of pointsInterpolation on a regular square grid spanning a triangular domainHow to generate grid points on boundary of $[-1,1]^d$ for arbitrary dimension $d$ and specified resolution?How to generate the rows of data points from a model?How to apply $overliner(x,y)$ to shapes with straight lines or absolute values?How to make a cow smaller (in BubbleChart3D plot)How to plot a 2D triangular latticeHow to generate animation from manipulate?Point lattice leading to triangle lattice
$begingroup$
I am newbie with mathematica and the other day I saw a function that generates points from an original one defined as:
h[x_, y_, 0] := Prepend[Table[Cos[2 Pi k/6] + x, Sin[2 Pi k/6] + y, k,6], 0, 0]
h[x_, y_, n_] :=DeleteDuplicates[Flatten[Table[Cos[2 Pi k/6] + #1, Sin[2 Pi k/6] + #2, k, 6] & @@@h[x, y, n - 1], 1]]
So I started from this function and tried to create a triangle lattice with a new function definied as:
L[x_, y_, n_] :=Show@Graphics@While[j < Length[h[x, y, n] + 1],
For[i = 1, i < Length[h[x, y, n] + 1] , i++ ,
If[EuclideanDistance[h[x, y, n][[j]], h[x, y, n][[i]]] == 1,
Line[h[x, y, n][[j]], h[x, y, n][[i]]],
Point[h[x, y, n][[j]], h[x, y, n][[i]]]]]; j++]
But it doesn't work... I wanted to connect all the dots that were seperated by a distance of 1 and plot a graphic with them. It seems that i am not using for as it should properly be.
graphics lattices
edited 7 hours ago
march
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg 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 |
$begingroup$
I am newbie with mathematica and the other day I saw a function that generates points from an original one defined as:
h[x_, y_, 0] := Prepend[Table[Cos[2 Pi k/6] + x, Sin[2 Pi k/6] + y, k,6], 0, 0]
h[x_, y_, n_] :=DeleteDuplicates[Flatten[Table[Cos[2 Pi k/6] + #1, Sin[2 Pi k/6] + #2, k, 6] & @@@h[x, y, n - 1], 1]]
So I started from this function and tried to create a triangle lattice with a new function definied as:
L[x_, y_, n_] :=Show@Graphics@While[j < Length[h[x, y, n] + 1],
For[i = 1, i < Length[h[x, y, n] + 1] , i++ ,
If[EuclideanDistance[h[x, y, n][[j]], h[x, y, n][[i]]] == 1,
Line[h[x, y, n][[j]], h[x, y, n][[i]]],
Point[h[x, y, n][[j]], h[x, y, n][[i]]]]]; j++]
But it doesn't work... I wanted to connect all the dots that were seperated by a distance of 1 and plot a graphic with them. It seems that i am not using for as it should properly be.
graphics lattices
edited 7 hours ago
march
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago
add a comment |
$begingroup$
I am newbie with mathematica and the other day I saw a function that generates points from an original one defined as:
h[x_, y_, 0] := Prepend[Table[Cos[2 Pi k/6] + x, Sin[2 Pi k/6] + y, k,6], 0, 0]
h[x_, y_, n_] :=DeleteDuplicates[Flatten[Table[Cos[2 Pi k/6] + #1, Sin[2 Pi k/6] + #2, k, 6] & @@@h[x, y, n - 1], 1]]
So I started from this function and tried to create a triangle lattice with a new function definied as:
L[x_, y_, n_] :=Show@Graphics@While[j < Length[h[x, y, n] + 1],
For[i = 1, i < Length[h[x, y, n] + 1] , i++ ,
If[EuclideanDistance[h[x, y, n][[j]], h[x, y, n][[i]]] == 1,
Line[h[x, y, n][[j]], h[x, y, n][[i]]],
Point[h[x, y, n][[j]], h[x, y, n][[i]]]]]; j++]
But it doesn't work... I wanted to connect all the dots that were seperated by a distance of 1 and plot a graphic with them. It seems that i am not using for as it should properly be.
graphics lattices
edited 7 hours ago
march
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am newbie with mathematica and the other day I saw a function that generates points from an original one defined as:
h[x_, y_, 0] := Prepend[Table[Cos[2 Pi k/6] + x, Sin[2 Pi k/6] + y, k,6], 0, 0]
h[x_, y_, n_] :=DeleteDuplicates[Flatten[Table[Cos[2 Pi k/6] + #1, Sin[2 Pi k/6] + #2, k, 6] & @@@h[x, y, n - 1], 1]]
So I started from this function and tried to create a triangle lattice with a new function definied as:
L[x_, y_, n_] :=Show@Graphics@While[j < Length[h[x, y, n] + 1],
For[i = 1, i < Length[h[x, y, n] + 1] , i++ ,
If[EuclideanDistance[h[x, y, n][[j]], h[x, y, n][[i]]] == 1,
Line[h[x, y, n][[j]], h[x, y, n][[i]]],
Point[h[x, y, n][[j]], h[x, y, n][[i]]]]]; j++]
But it doesn't work... I wanted to connect all the dots that were seperated by a distance of 1 and plot a graphic with them. It seems that i am not using for as it should properly be.
graphics lattices
graphics lattices
edited 7 hours ago
march
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
march
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
march
17.7k22870
edited 7 hours ago
march
17.7k22870
edited 7 hours ago
march
17.7k22870
17.7k22870
asked 7 hours ago
LilGregLilGreg
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
LilGregLilGreg
212
asked 7 hours ago
LilGregLilGreg
212
212
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
LilGreg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago
add a comment |
$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago
$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago
$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered 4 hours ago
marchmarch
17.7k22870
$endgroup$
add a comment |
$begingroup$
Your code wasn't far off, though the other answers may be more elegant.
This works:
L2[x_, y_, n_] := Module[pts,
pts = h[x, y, n];
Show[Graphics[
Point[pts],
Table[
If[EuclideanDistance[pts[[i]], pts[[j]]] == 1,
Line[pts[[i]], pts[[j]]]], i, Length[pts], j, Length[pts]]
]]]
L2[0, 0, 2]
answered 3 hours ago
MelaGoMelaGo
1,21317
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
$endgroup$
add a comment |
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
$endgroup$
add a comment |
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
$endgroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
62.2k586172
62.2k586172
add a comment |
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered 4 hours ago
marchmarch
17.7k22870
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered 4 hours ago
marchmarch
17.7k22870
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered 4 hours ago
marchmarch
17.7k22870
$endgroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered 4 hours ago
marchmarch
17.7k22870
answered 4 hours ago
marchmarch
17.7k22870
answered 4 hours ago
marchmarch
17.7k22870
answered 4 hours ago
marchmarch
17.7k22870
17.7k22870
add a comment |
add a comment |
$begingroup$
Your code wasn't far off, though the other answers may be more elegant.
This works:
L2[x_, y_, n_] := Module[pts,
pts = h[x, y, n];
Show[Graphics[
Point[pts],
Table[
If[EuclideanDistance[pts[[i]], pts[[j]]] == 1,
Line[pts[[i]], pts[[j]]]], i, Length[pts], j, Length[pts]]
]]]
L2[0, 0, 2]
answered 3 hours ago
MelaGoMelaGo
1,21317
$endgroup$
add a comment |
$begingroup$
Your code wasn't far off, though the other answers may be more elegant.
This works:
L2[x_, y_, n_] := Module[pts,
pts = h[x, y, n];
Show[Graphics[
Point[pts],
Table[
If[EuclideanDistance[pts[[i]], pts[[j]]] == 1,
Line[pts[[i]], pts[[j]]]], i, Length[pts], j, Length[pts]]
]]]
L2[0, 0, 2]
answered 3 hours ago
MelaGoMelaGo
1,21317
$endgroup$
add a comment |
$begingroup$
Your code wasn't far off, though the other answers may be more elegant.
This works:
L2[x_, y_, n_] := Module[pts,
pts = h[x, y, n];
Show[Graphics[
Point[pts],
Table[
If[EuclideanDistance[pts[[i]], pts[[j]]] == 1,
Line[pts[[i]], pts[[j]]]], i, Length[pts], j, Length[pts]]
]]]
L2[0, 0, 2]
answered 3 hours ago
MelaGoMelaGo
1,21317
$endgroup$
Your code wasn't far off, though the other answers may be more elegant.
This works:
L2[x_, y_, n_] := Module[pts,
pts = h[x, y, n];
Show[Graphics[
Point[pts],
Table[
If[EuclideanDistance[pts[[i]], pts[[j]]] == 1,
Line[pts[[i]], pts[[j]]]], i, Length[pts], j, Length[pts]]
]]]
L2[0, 0, 2]
answered 3 hours ago
MelaGoMelaGo
1,21317
answered 3 hours ago
MelaGoMelaGo
1,21317
answered 3 hours ago
MelaGoMelaGo
1,21317
answered 3 hours ago
MelaGoMelaGo
1,21317
1,21317
add a comment |
add a comment |
LilGreg is a new contributor. Be nice, and check out our Code of Conduct.
LilGreg is a new contributor. Be nice, and check out our Code of Conduct.
LilGreg is a new contributor. Be nice, and check out our Code of Conduct.
LilGreg is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I've edited your question to include a link to what you saw the other day. In the future, make sure to do this so that you can give questions and answerers their proper credit! In addition, once you have enough rep (which I think you do), make sure to upvote questions and/or answers that you found useful (which includes the now-linked ones, I assume, since you asked a question about it!).
$endgroup$
– march
4 hours ago