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tikz drawing rectangle discretized with triangle lattices and its centroids
How to draw triangular grid in TikZ?Confused about TikZ draw[rotate=angle]Drawing Subgroup Lattices in TikZDrawing an altitude in a triangle and labeling its footDrawing an obtuse triangle, and marking its heightInput/Output Nodes - Specification and Description LanguageDrawing rectangle triangles with labels only in its sizesHow to draw stacked rectangles (in a third dimension), but with the intersected lines clipped + grid pattern for each rectangle + shaded fill + textDrawing this matrix using Tikzfading angle does not work?tikz put text under rectangle
Is there a way to draw the following figure using tikz 
This post is similar but the for loops are complicated for me to understand.
The rectangle domain is 2.0 x 1.0 (width x height) and there are 20 subdivisions along horizontal direction and 10 subdivisions along vertical directions.
The blue dots are centroids of the triangle
Any help/clue will be really appreciated.
Thank you
tikz-pgf
asked May 14 at 13:36
ggulguliaggulgulia
281110
add a comment |
Is there a way to draw the following figure using tikz
This post is similar but the for loops are complicated for me to understand.
The rectangle domain is 2.0 x 1.0 (width x height) and there are 20 subdivisions along horizontal direction and 10 subdivisions along vertical directions.
The blue dots are centroids of the triangle
Any help/clue will be really appreciated.
Thank you
tikz-pgf
asked May 14 at 13:36
ggulguliaggulgulia
281110
add a comment |
Is there a way to draw the following figure using tikz
This post is similar but the for loops are complicated for me to understand.
The rectangle domain is 2.0 x 1.0 (width x height) and there are 20 subdivisions along horizontal direction and 10 subdivisions along vertical directions.
The blue dots are centroids of the triangle
Any help/clue will be really appreciated.
Thank you
tikz-pgf
asked May 14 at 13:36
ggulguliaggulgulia
281110
Is there a way to draw the following figure using tikz
This post is similar but the for loops are complicated for me to understand.
The rectangle domain is 2.0 x 1.0 (width x height) and there are 20 subdivisions along horizontal direction and 10 subdivisions along vertical directions.
The blue dots are centroids of the triangle
Any help/clue will be really appreciated.
Thank you
tikz-pgf
tikz-pgf
asked May 14 at 13:36
ggulguliaggulgulia
281110
asked May 14 at 13:36
ggulguliaggulgulia
281110
asked May 14 at 13:36
ggulguliaggulgulia
281110
asked May 14 at 13:36
ggulguliaggulgulia
281110
asked May 14 at 13:36
ggulguliaggulgulia
281110
281110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
One needs to find an algorithm to draw it. The following is one of such algorithms.
documentclass[tikz,margin=3]standalone
begindocument
begintikzpicture[scale=0.5]
draw[gray] (0,0) grid (20,10);
foreach i in 0,...,9
% Draw the grid
draw[gray] (0,i) -- (10-i,10);
draw[gray] (10+i,0) -- (20,10-i);
ifnumi=0relaxelse
draw[gray] (i,0) -- (10+i,10);fi
% Draw the dots
foreach j in 0,...,19
% The center of each square is j+.5 and i+.5
coordinate (center) at (j+.5,i+.5);
fill[blue] ([shift=(-1/6,1/6)]center) circle (3pt);
fill[blue] ([shift=(1/6,-1/6)]center) circle (3pt);
endtikzpicture
enddocument
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
add a comment |
An alternative construction with a matrix and nodes in empty cells.
documentclass[tikz,border=2mm]standalone
usetikzlibrarymatrix
begindocument
begintikzpicture[%
mymatrix/.style=
matrix of nodes,
nodes in empty cells,
row sep=-pgflinewidth,
column sep=-pgflinewidth,
nodes=draw=gray,
minimum size=1cm,
outer sep=0pt,
inner sep=0pt,
anchor=center,
path picture=%
draw (path picture bounding box.north east)--(path picture bounding box.south west);
fill[blue] ([shift=(-1/6,1/6)]path picture bounding box.center) circle(3pt);
fill[blue] ([shift=(1/6,-1/6)]path picture bounding box.center) circle(3pt);
]
matrix[mymatrix]%
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
;
endtikzpicture
enddocument
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
add a comment |
Yet another alternative: pics. And store the variables in pgf keys. Like Joule V's nice solution, this grid can be transformed (e.g. rotated) without resorting to transform canvas, which tends to mess up things.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[pics/cell/.style=code=draw (-#1,-#1) rectangle (#1,#1)
(-#1,-#1) -- (#1,#1);
fill[pgfkeysvalueof/tikz/cell/color]
(135:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius]
(-45:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius];,
cell/.cd,color/.initial=blue,radius/.initial=1/6,circle pos/.initial=sqrt(0.5)]
path foreach X in 1,...,20
foreach Y in 1,...,10 (X,Y) pic[cell/radius=1/5,cell/circle pos=1/2]cell=0.5;
endtikzpicture
enddocument
answered May 14 at 18:31
marmotmarmot
130k6165313
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
One needs to find an algorithm to draw it. The following is one of such algorithms.
documentclass[tikz,margin=3]standalone
begindocument
begintikzpicture[scale=0.5]
draw[gray] (0,0) grid (20,10);
foreach i in 0,...,9
% Draw the grid
draw[gray] (0,i) -- (10-i,10);
draw[gray] (10+i,0) -- (20,10-i);
ifnumi=0relaxelse
draw[gray] (i,0) -- (10+i,10);fi
% Draw the dots
foreach j in 0,...,19
% The center of each square is j+.5 and i+.5
coordinate (center) at (j+.5,i+.5);
fill[blue] ([shift=(-1/6,1/6)]center) circle (3pt);
fill[blue] ([shift=(1/6,-1/6)]center) circle (3pt);
endtikzpicture
enddocument
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
add a comment |
One needs to find an algorithm to draw it. The following is one of such algorithms.
documentclass[tikz,margin=3]standalone
begindocument
begintikzpicture[scale=0.5]
draw[gray] (0,0) grid (20,10);
foreach i in 0,...,9
% Draw the grid
draw[gray] (0,i) -- (10-i,10);
draw[gray] (10+i,0) -- (20,10-i);
ifnumi=0relaxelse
draw[gray] (i,0) -- (10+i,10);fi
% Draw the dots
foreach j in 0,...,19
% The center of each square is j+.5 and i+.5
coordinate (center) at (j+.5,i+.5);
fill[blue] ([shift=(-1/6,1/6)]center) circle (3pt);
fill[blue] ([shift=(1/6,-1/6)]center) circle (3pt);
endtikzpicture
enddocument
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
add a comment |
One needs to find an algorithm to draw it. The following is one of such algorithms.
documentclass[tikz,margin=3]standalone
begindocument
begintikzpicture[scale=0.5]
draw[gray] (0,0) grid (20,10);
foreach i in 0,...,9
% Draw the grid
draw[gray] (0,i) -- (10-i,10);
draw[gray] (10+i,0) -- (20,10-i);
ifnumi=0relaxelse
draw[gray] (i,0) -- (10+i,10);fi
% Draw the dots
foreach j in 0,...,19
% The center of each square is j+.5 and i+.5
coordinate (center) at (j+.5,i+.5);
fill[blue] ([shift=(-1/6,1/6)]center) circle (3pt);
fill[blue] ([shift=(1/6,-1/6)]center) circle (3pt);
endtikzpicture
enddocument
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
One needs to find an algorithm to draw it. The following is one of such algorithms.
documentclass[tikz,margin=3]standalone
begindocument
begintikzpicture[scale=0.5]
draw[gray] (0,0) grid (20,10);
foreach i in 0,...,9
% Draw the grid
draw[gray] (0,i) -- (10-i,10);
draw[gray] (10+i,0) -- (20,10-i);
ifnumi=0relaxelse
draw[gray] (i,0) -- (10+i,10);fi
% Draw the dots
foreach j in 0,...,19
% The center of each square is j+.5 and i+.5
coordinate (center) at (j+.5,i+.5);
fill[blue] ([shift=(-1/6,1/6)]center) circle (3pt);
fill[blue] ([shift=(1/6,-1/6)]center) circle (3pt);
endtikzpicture
enddocument
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
edited May 14 at 14:16
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
answered May 14 at 13:55
The old JouleVThe old JouleV
19.5k33175
19.5k33175
add a comment |
add a comment |
An alternative construction with a matrix and nodes in empty cells.
documentclass[tikz,border=2mm]standalone
usetikzlibrarymatrix
begindocument
begintikzpicture[%
mymatrix/.style=
matrix of nodes,
nodes in empty cells,
row sep=-pgflinewidth,
column sep=-pgflinewidth,
nodes=draw=gray,
minimum size=1cm,
outer sep=0pt,
inner sep=0pt,
anchor=center,
path picture=%
draw (path picture bounding box.north east)--(path picture bounding box.south west);
fill[blue] ([shift=(-1/6,1/6)]path picture bounding box.center) circle(3pt);
fill[blue] ([shift=(1/6,-1/6)]path picture bounding box.center) circle(3pt);
]
matrix[mymatrix]%
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
;
endtikzpicture
enddocument
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
add a comment |
An alternative construction with a matrix and nodes in empty cells.
documentclass[tikz,border=2mm]standalone
usetikzlibrarymatrix
begindocument
begintikzpicture[%
mymatrix/.style=
matrix of nodes,
nodes in empty cells,
row sep=-pgflinewidth,
column sep=-pgflinewidth,
nodes=draw=gray,
minimum size=1cm,
outer sep=0pt,
inner sep=0pt,
anchor=center,
path picture=%
draw (path picture bounding box.north east)--(path picture bounding box.south west);
fill[blue] ([shift=(-1/6,1/6)]path picture bounding box.center) circle(3pt);
fill[blue] ([shift=(1/6,-1/6)]path picture bounding box.center) circle(3pt);
]
matrix[mymatrix]%
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
;
endtikzpicture
enddocument
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
add a comment |
An alternative construction with a matrix and nodes in empty cells.
documentclass[tikz,border=2mm]standalone
usetikzlibrarymatrix
begindocument
begintikzpicture[%
mymatrix/.style=
matrix of nodes,
nodes in empty cells,
row sep=-pgflinewidth,
column sep=-pgflinewidth,
nodes=draw=gray,
minimum size=1cm,
outer sep=0pt,
inner sep=0pt,
anchor=center,
path picture=%
draw (path picture bounding box.north east)--(path picture bounding box.south west);
fill[blue] ([shift=(-1/6,1/6)]path picture bounding box.center) circle(3pt);
fill[blue] ([shift=(1/6,-1/6)]path picture bounding box.center) circle(3pt);
]
matrix[mymatrix]%
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
;
endtikzpicture
enddocument
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
An alternative construction with a matrix and nodes in empty cells.
documentclass[tikz,border=2mm]standalone
usetikzlibrarymatrix
begindocument
begintikzpicture[%
mymatrix/.style=
matrix of nodes,
nodes in empty cells,
row sep=-pgflinewidth,
column sep=-pgflinewidth,
nodes=draw=gray,
minimum size=1cm,
outer sep=0pt,
inner sep=0pt,
anchor=center,
path picture=%
draw (path picture bounding box.north east)--(path picture bounding box.south west);
fill[blue] ([shift=(-1/6,1/6)]path picture bounding box.center) circle(3pt);
fill[blue] ([shift=(1/6,-1/6)]path picture bounding box.center) circle(3pt);
]
matrix[mymatrix]%
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
&&&&&&&&&&&&&&&&&&&\
;
endtikzpicture
enddocument
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
answered May 14 at 15:11
IgnasiIgnasi
97.1k6177327
97.1k6177327
add a comment |
add a comment |
Yet another alternative: pics. And store the variables in pgf keys. Like Joule V's nice solution, this grid can be transformed (e.g. rotated) without resorting to transform canvas, which tends to mess up things.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[pics/cell/.style=code=draw (-#1,-#1) rectangle (#1,#1)
(-#1,-#1) -- (#1,#1);
fill[pgfkeysvalueof/tikz/cell/color]
(135:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius]
(-45:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius];,
cell/.cd,color/.initial=blue,radius/.initial=1/6,circle pos/.initial=sqrt(0.5)]
path foreach X in 1,...,20
foreach Y in 1,...,10 (X,Y) pic[cell/radius=1/5,cell/circle pos=1/2]cell=0.5;
endtikzpicture
enddocument
answered May 14 at 18:31
marmotmarmot
130k6165313
add a comment |
Yet another alternative: pics. And store the variables in pgf keys. Like Joule V's nice solution, this grid can be transformed (e.g. rotated) without resorting to transform canvas, which tends to mess up things.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[pics/cell/.style=code=draw (-#1,-#1) rectangle (#1,#1)
(-#1,-#1) -- (#1,#1);
fill[pgfkeysvalueof/tikz/cell/color]
(135:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius]
(-45:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius];,
cell/.cd,color/.initial=blue,radius/.initial=1/6,circle pos/.initial=sqrt(0.5)]
path foreach X in 1,...,20
foreach Y in 1,...,10 (X,Y) pic[cell/radius=1/5,cell/circle pos=1/2]cell=0.5;
endtikzpicture
enddocument
answered May 14 at 18:31
marmotmarmot
130k6165313
add a comment |
Yet another alternative: pics. And store the variables in pgf keys. Like Joule V's nice solution, this grid can be transformed (e.g. rotated) without resorting to transform canvas, which tends to mess up things.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[pics/cell/.style=code=draw (-#1,-#1) rectangle (#1,#1)
(-#1,-#1) -- (#1,#1);
fill[pgfkeysvalueof/tikz/cell/color]
(135:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius]
(-45:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius];,
cell/.cd,color/.initial=blue,radius/.initial=1/6,circle pos/.initial=sqrt(0.5)]
path foreach X in 1,...,20
foreach Y in 1,...,10 (X,Y) pic[cell/radius=1/5,cell/circle pos=1/2]cell=0.5;
endtikzpicture
enddocument
answered May 14 at 18:31
marmotmarmot
130k6165313
Yet another alternative: pics. And store the variables in pgf keys. Like Joule V's nice solution, this grid can be transformed (e.g. rotated) without resorting to transform canvas, which tends to mess up things.
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[pics/cell/.style=code=draw (-#1,-#1) rectangle (#1,#1)
(-#1,-#1) -- (#1,#1);
fill[pgfkeysvalueof/tikz/cell/color]
(135:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius]
(-45:#1*pgfkeysvalueof/tikz/cell/circle pos)
circle[radius=#1*pgfkeysvalueof/tikz/cell/radius];,
cell/.cd,color/.initial=blue,radius/.initial=1/6,circle pos/.initial=sqrt(0.5)]
path foreach X in 1,...,20
foreach Y in 1,...,10 (X,Y) pic[cell/radius=1/5,cell/circle pos=1/2]cell=0.5;
endtikzpicture
enddocument
answered May 14 at 18:31
marmotmarmot
130k6165313
edited May 14 at 18:46
answered May 14 at 18:31
marmotmarmot
130k6165313
answered May 14 at 18:31
marmotmarmot
130k6165313
answered May 14 at 18:31
marmotmarmot
130k6165313
130k6165313
add a comment |
add a comment |
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