Ttdfjt

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7

$begingroup$

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

share|improve this question

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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
  May 15 at 23:33
  

  
  
  
  

add a comment | 

7

$begingroup$

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

share|improve this question

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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
  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

share|improve this question

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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.

$endgroup$

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]]

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++]

share|improve this question

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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.

share|improve this question

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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 May 16 at 6:38

user64494

3,74711222

edited May 16 at 6:38

user64494

3,74711222

asked May 15 at 19:58

LilGregLilGreg

362

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 May 15 at 19:58

LilGregLilGreg

362

4 Answers
4

active

oldest

votes

7

$begingroup$

Try this:

R = DelaunayMesh[h[0, 0, 2]]

You may grab the edge indices with

MeshCells[R, 1]

share|improve this answer

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

$endgroup$

add a comment | 

6

$begingroup$

You can use NearestNeighborGraph as follows:

Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics

share|improve this answer

answered May 15 at 23:19

marchmarch

17.8k22870

$endgroup$

add a comment | 

6

$begingroup$

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]

share|improve this answer

answered May 16 at 0:47

MelaGoMelaGo

1,37017

$endgroup$

add a comment | 

5

$begingroup$

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]

share|improve this answer

edited May 16 at 7:14

answered May 16 at 7:08

kglrkglr

193k10214435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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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7

$begingroup$

Try this:

R = DelaunayMesh[h[0, 0, 2]]

You may grab the edge indices with

MeshCells[R, 1]

share|improve this answer

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

$endgroup$

add a comment | 

7

$begingroup$

Try this:

R = DelaunayMesh[h[0, 0, 2]]

You may grab the edge indices with

MeshCells[R, 1]

share|improve this answer

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

$endgroup$

add a comment | 

$begingroup$

Try this:

R = DelaunayMesh[h[0, 0, 2]]

You may grab the edge indices with

MeshCells[R, 1]

share|improve this answer

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

$endgroup$

R = DelaunayMesh[h[0, 0, 2]]

MeshCells[R, 1]

share|improve this answer

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

answered May 15 at 20:02

Henrik SchumacherHenrik Schumacher

62.9k587176

6

$begingroup$

You can use NearestNeighborGraph as follows:

Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics

share|improve this answer

answered May 15 at 23:19

marchmarch

17.8k22870

$endgroup$

add a comment | 

6

$begingroup$

You can use NearestNeighborGraph as follows:

Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics

share|improve this answer

answered May 15 at 23:19

marchmarch

17.8k22870

$endgroup$

add a comment | 

$begingroup$

You can use NearestNeighborGraph as follows:

Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics

share|improve this answer

answered May 15 at 23:19

marchmarch

17.8k22870

$endgroup$

Line[##] & @@@ EdgeList@NearestNeighborGraph[h[0, 0, 1]] // Graphics

share|improve this answer

answered May 15 at 23:19

marchmarch

17.8k22870

answered May 15 at 23:19

marchmarch

17.8k22870

answered May 15 at 23:19

marchmarch

17.8k22870

6

$begingroup$

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]

share|improve this answer

answered May 16 at 0:47

MelaGoMelaGo

1,37017

$endgroup$

add a comment | 

6

$begingroup$

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]

share|improve this answer

answered May 16 at 0:47

MelaGoMelaGo

1,37017

$endgroup$

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]

share|improve this answer

answered May 16 at 0:47

MelaGoMelaGo

1,37017

$endgroup$

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]

share|improve this answer

answered May 16 at 0:47

MelaGoMelaGo

1,37017

answered May 16 at 0:47

MelaGoMelaGo

1,37017

answered May 16 at 0:47

MelaGoMelaGo

1,37017

5

$begingroup$

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]

share|improve this answer

edited May 16 at 7:14

answered May 16 at 7:08

kglrkglr

193k10214435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

5

$begingroup$

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]

share|improve this answer

edited May 16 at 7:14

answered May 16 at 7:08

kglrkglr

193k10214435

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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]

share|improve this answer

edited May 16 at 7:14

answered May 16 at 7:08

kglrkglr

193k10214435

$endgroup$

NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]

RelationGraph[.1 < EuclideanDistance@## <= 1 &, h[0, 0, 1], VertexCoordinates -> h[0, 0, 1]]

Show @ NearestNeighborGraph[h[0, 0, 1], VertexCoordinates -> h[0, 0, 1], 
 VertexShapeFunction -> None]

share|improve this answer

edited May 16 at 7:14

answered May 16 at 7:08

kglrkglr

193k10214435

answered May 16 at 7:08

kglrkglr

193k10214435

answered May 16 at 7:08

kglrkglr

193k10214435