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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 May 16 at 6:38
user64494
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
$endgroup$
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 May 16 at 6:38
user64494
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
$endgroup$
$begingroup$
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
May 15 at 23:33
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 May 16 at 6:38
user64494
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
$endgroup$
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 May 16 at 6:38
user64494
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
edited May 16 at 6:38
user64494
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
edited May 16 at 6:38
user64494
3,90211323
edited May 16 at 6:38
user64494
3,90211323
edited May 16 at 6:38
user64494
3,90211323
3,90211323
asked May 15 at 19:58
LilGregLilGreg
362
asked May 15 at 19:58
LilGregLilGreg
362
asked May 15 at 19:58
LilGregLilGreg
362
362
$begingroup$
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
May 15 at 23:33
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
May 15 at 23:33
$begingroup$
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
May 15 at 23:33
$begingroup$
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
May 15 at 23:33
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered May 15 at 23:19
marchmarch
17.8k22970
$endgroup$
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 May 16 at 0:47
MelaGoMelaGo
1,53717
$endgroup$
add a comment |
$begingroup$
Another way to use NearestNeighborGraph:
NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
Alternatively, you can use RelationGraph:
RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
same picture
To remove the vertices and to get a Graphics object you can use:
Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1],
VertexShapeFunction -> None]
answered May 16 at 7:08
kglrkglr
195k10216438
$endgroup$
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
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 May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
add a comment |
$begingroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
$endgroup$
Try this:
R = DelaunayMesh[h[0, 0, 2]]
You may grab the edge indices with
MeshCells[R, 1]
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
answered May 15 at 20:02
Henrik SchumacherHenrik Schumacher
63.2k587176
63.2k587176
add a comment |
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered May 15 at 23:19
marchmarch
17.8k22970
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered May 15 at 23:19
marchmarch
17.8k22970
$endgroup$
add a comment |
$begingroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered May 15 at 23:19
marchmarch
17.8k22970
$endgroup$
You can use NearestNeighborGraph as follows:
Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics
answered May 15 at 23:19
marchmarch
17.8k22970
answered May 15 at 23:19
marchmarch
17.8k22970
answered May 15 at 23:19
marchmarch
17.8k22970
answered May 15 at 23:19
marchmarch
17.8k22970
17.8k22970
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 May 16 at 0:47
MelaGoMelaGo
1,53717
$endgroup$
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 May 16 at 0:47
MelaGoMelaGo
1,53717
$endgroup$
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 May 16 at 0:47
MelaGoMelaGo
1,53717
$endgroup$
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 May 16 at 0:47
MelaGoMelaGo
1,53717
answered May 16 at 0:47
MelaGoMelaGo
1,53717
answered May 16 at 0:47
MelaGoMelaGo
1,53717
answered May 16 at 0:47
MelaGoMelaGo
1,53717
1,53717
add a comment |
add a comment |
$begingroup$
Another way to use NearestNeighborGraph:
NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
Alternatively, you can use RelationGraph:
RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
same picture
To remove the vertices and to get a Graphics object you can use:
Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1],
VertexShapeFunction -> None]
answered May 16 at 7:08
kglrkglr
195k10216438
$endgroup$
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
add a comment |
$begingroup$
Another way to use NearestNeighborGraph:
NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
Alternatively, you can use RelationGraph:
RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
same picture
To remove the vertices and to get a Graphics object you can use:
Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1],
VertexShapeFunction -> None]
answered May 16 at 7:08
kglrkglr
195k10216438
$endgroup$
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
add a comment |
$begingroup$
Another way to use NearestNeighborGraph:
NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
Alternatively, you can use RelationGraph:
RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
same picture
To remove the vertices and to get a Graphics object you can use:
Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1],
VertexShapeFunction -> None]
answered May 16 at 7:08
kglrkglr
195k10216438
$endgroup$
Another way to use NearestNeighborGraph:
NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
Alternatively, you can use RelationGraph:
RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]
same picture
To remove the vertices and to get a Graphics object you can use:
Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1],
VertexShapeFunction -> None]
answered May 16 at 7:08
kglrkglr
195k10216438
edited May 16 at 7:14
answered May 16 at 7:08
kglrkglr
195k10216438
answered May 16 at 7:08
kglrkglr
195k10216438
answered May 16 at 7:08
kglrkglr
195k10216438
195k10216438
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
add a comment |
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
$begingroup$
Thank you for your help! I didn't know there was a function, I appreciate your help.
$endgroup$
– LilGreg
May 16 at 17:48
add a comment |
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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
May 15 at 23:33