tikz grid without top edge
4
In this code from TikZ (finite) grid with character in each cell
documentclass{article}
usepackage{tikz}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
I would like to delete the top and left edges of the grid.
I have found solutions using nodes but much prefer this one with a matrix.
tikz-pgf
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
add a comment |
4
In this code from TikZ (finite) grid with character in each cell
documentclass{article}
usepackage{tikz}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
I would like to delete the top and left edges of the grid.
I have found solutions using nodes but much prefer this one with a matrix.
tikz-pgf
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
Off topic: You have a very nice reputation number :)
– JouleV
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday
add a comment |
4
4
4
In this code from TikZ (finite) grid with character in each cell
documentclass{article}
usepackage{tikz}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
I would like to delete the top and left edges of the grid.
I have found solutions using nodes but much prefer this one with a matrix.
tikz-pgf
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
In this code from TikZ (finite) grid with character in each cell
documentclass{article}
usepackage{tikz}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
I would like to delete the top and left edges of the grid.
I have found solutions using nodes but much prefer this one with a matrix.
tikz-pgf
tikz-pgf
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
asked yesterday
Ethan BolkerEthan Bolker
6,68832453
6,68832453
Off topic: You have a very nice reputation number :)
– JouleV
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday
add a comment |
Off topic: You have a very nice reputation number :)
– JouleV
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday
Off topic: You have a very nice reputation number :)
– JouleV
yesterday
Off topic: You have a very nice reputation number :)
– JouleV
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday
add a comment |
2 Answers
2
active
oldest
votes
4
We can use foreach:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
%draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
foreach i in {-0.5,0,0.5,1} {
draw[gray] (-1,-i)--(1,-i);
draw[gray] (i,-1)--(i,1);
}
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
Honestly I don't think nodes or matrices have any thing to do here.
This approach is based on marmot's creative way, which is nicer and more tricky:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] ({-1cm+0.2pt},-1) grid (1cm,{1cm-0.2pt});
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
answered yesterday
JouleVJouleV
8,15222153
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention alsodraw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99);which is shorter and gives (for all practical purposes) the same output.
– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
|
show 4 more comments
3
Another solution. It draws the grid with a matrix of drawn nodes. After that, left and top border are deleted with a white supperposed line.
documentclass[tikz,border=2mm]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,nodes={draw=gray, anchor=center, minimum size=.6cm}, column sep=-pgflinewidth, row sep=-pgflinewidth] (A) {
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
draw[white] ([xshift=.5pgflinewidth]A-4-1.south west)|-([yshift=-.5pgflinewidth]A-1-4.north east);
end{tikzpicture}
end{document}
answered yesterday
IgnasiIgnasi
95.3k4175319
add a comment |
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2 Answers
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active
oldest
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2 Answers
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4
We can use foreach:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
%draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
foreach i in {-0.5,0,0.5,1} {
draw[gray] (-1,-i)--(1,-i);
draw[gray] (i,-1)--(i,1);
}
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
Honestly I don't think nodes or matrices have any thing to do here.
This approach is based on marmot's creative way, which is nicer and more tricky:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] ({-1cm+0.2pt},-1) grid (1cm,{1cm-0.2pt});
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
answered yesterday
JouleVJouleV
8,15222153
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention alsodraw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99);which is shorter and gives (for all practical purposes) the same output.
– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
|
show 4 more comments
4
We can use foreach:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
%draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
foreach i in {-0.5,0,0.5,1} {
draw[gray] (-1,-i)--(1,-i);
draw[gray] (i,-1)--(i,1);
}
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
Honestly I don't think nodes or matrices have any thing to do here.
This approach is based on marmot's creative way, which is nicer and more tricky:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] ({-1cm+0.2pt},-1) grid (1cm,{1cm-0.2pt});
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
answered yesterday
JouleVJouleV
8,15222153
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention alsodraw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99);which is shorter and gives (for all practical purposes) the same output.
– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
|
show 4 more comments
4
4
4
We can use foreach:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
%draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
foreach i in {-0.5,0,0.5,1} {
draw[gray] (-1,-i)--(1,-i);
draw[gray] (i,-1)--(i,1);
}
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
Honestly I don't think nodes or matrices have any thing to do here.
This approach is based on marmot's creative way, which is nicer and more tricky:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] ({-1cm+0.2pt},-1) grid (1cm,{1cm-0.2pt});
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
answered yesterday
JouleVJouleV
8,15222153
We can use foreach:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
%draw[step=0.5cm,color=gray] (-1,-1) grid (1,1);
foreach i in {-0.5,0,0.5,1} {
draw[gray] (-1,-i)--(1,-i);
draw[gray] (i,-1)--(i,1);
}
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
Honestly I don't think nodes or matrices have any thing to do here.
This approach is based on marmot's creative way, which is nicer and more tricky:
documentclass[tikz]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
draw[step=0.5cm,color=gray] ({-1cm+0.2pt},-1) grid (1cm,{1cm-0.2pt});
matrix[matrix of nodes,nodes={inner sep=0pt,text width=.5cm,align=center,minimum height=.5cm}]{
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
end{tikzpicture}
end{document}
answered yesterday
JouleVJouleV
8,15222153
edited yesterday
answered yesterday
JouleVJouleV
8,15222153
answered yesterday
JouleVJouleV
8,15222153
answered yesterday
JouleVJouleV
8,15222153
8,15222153
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention alsodraw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99);which is shorter and gives (for all practical purposes) the same output.
– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
|
show 4 more comments
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention alsodraw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99);which is shorter and gives (for all practical purposes) the same output.
– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
1
1
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
Thanks. There is a "matrix" in the code. Other answers to my original question used nodes.
– Ethan Bolker
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
@EthanBolker No problem. You are welcome ;)
– JouleV
yesterday
Maybe mention also
draw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99); which is shorter and gives (for all practical purposes) the same output.– marmot
yesterday
Maybe mention also
draw[step=0.5cm,color=gray] (-0.99,-1) grid (1,0.99); which is shorter and gives (for all practical purposes) the same output.– marmot
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@marmot Ohh your way is wonderfully creative :)) I did not even think of that. However they are not mathematically correct.
– JouleV
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
@JouleV What do you mean by "not mathematically correct"? It is as correct as a truncated grid can be.
– marmot
yesterday
|
show 4 more comments
3
Another solution. It draws the grid with a matrix of drawn nodes. After that, left and top border are deleted with a white supperposed line.
documentclass[tikz,border=2mm]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,nodes={draw=gray, anchor=center, minimum size=.6cm}, column sep=-pgflinewidth, row sep=-pgflinewidth] (A) {
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
draw[white] ([xshift=.5pgflinewidth]A-4-1.south west)|-([yshift=-.5pgflinewidth]A-1-4.north east);
end{tikzpicture}
end{document}
answered yesterday
IgnasiIgnasi
95.3k4175319
add a comment |
3
Another solution. It draws the grid with a matrix of drawn nodes. After that, left and top border are deleted with a white supperposed line.
documentclass[tikz,border=2mm]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,nodes={draw=gray, anchor=center, minimum size=.6cm}, column sep=-pgflinewidth, row sep=-pgflinewidth] (A) {
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
draw[white] ([xshift=.5pgflinewidth]A-4-1.south west)|-([yshift=-.5pgflinewidth]A-1-4.north east);
end{tikzpicture}
end{document}
answered yesterday
IgnasiIgnasi
95.3k4175319
add a comment |
3
3
3
Another solution. It draws the grid with a matrix of drawn nodes. After that, left and top border are deleted with a white supperposed line.
documentclass[tikz,border=2mm]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,nodes={draw=gray, anchor=center, minimum size=.6cm}, column sep=-pgflinewidth, row sep=-pgflinewidth] (A) {
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
draw[white] ([xshift=.5pgflinewidth]A-4-1.south west)|-([yshift=-.5pgflinewidth]A-1-4.north east);
end{tikzpicture}
end{document}
answered yesterday
IgnasiIgnasi
95.3k4175319
Another solution. It draws the grid with a matrix of drawn nodes. After that, left and top border are deleted with a white supperposed line.
documentclass[tikz,border=2mm]{standalone}
usetikzlibrary{matrix}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,nodes={draw=gray, anchor=center, minimum size=.6cm}, column sep=-pgflinewidth, row sep=-pgflinewidth] (A) {
A & B & C & D \
E & F & & H \
I & J & K & L \
M & N & O & P\};
draw[white] ([xshift=.5pgflinewidth]A-4-1.south west)|-([yshift=-.5pgflinewidth]A-1-4.north east);
end{tikzpicture}
end{document}
answered yesterday
IgnasiIgnasi
95.3k4175319
answered yesterday
IgnasiIgnasi
95.3k4175319
answered yesterday
IgnasiIgnasi
95.3k4175319
answered yesterday
IgnasiIgnasi
95.3k4175319
95.3k4175319
add a comment |
add a comment |
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Off topic: You have a very nice reputation number :)
– JouleV
yesterday
@JouleV Indeed, It won't survive any votes on this question.
– Ethan Bolker
yesterday