Diagonal table with numbers
I am trying to recreate the following table, which has a diagonal grid with regular numbers in it. Is there any efficient way of doing it without drawing / positioning everthing manually? I have looked through the TikZ/PGF manual, but couldn't find anything.
tikz-pgf tables diagrams
edited Jan 25 at 18:16
Skillmon
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
add a comment |
I am trying to recreate the following table, which has a diagonal grid with regular numbers in it. Is there any efficient way of doing it without drawing / positioning everthing manually? I have looked through the TikZ/PGF manual, but couldn't find anything.
tikz-pgf tables diagrams
edited Jan 25 at 18:16
Skillmon
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
add a comment |
I am trying to recreate the following table, which has a diagonal grid with regular numbers in it. Is there any efficient way of doing it without drawing / positioning everthing manually? I have looked through the TikZ/PGF manual, but couldn't find anything.
tikz-pgf tables diagrams
edited Jan 25 at 18:16
Skillmon
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
I am trying to recreate the following table, which has a diagonal grid with regular numbers in it. Is there any efficient way of doing it without drawing / positioning everthing manually? I have looked through the TikZ/PGF manual, but couldn't find anything.
tikz-pgf tables diagrams
tikz-pgf tables diagrams
edited Jan 25 at 18:16
Skillmon
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
edited Jan 25 at 18:16
Skillmon
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
edited Jan 25 at 18:16
Skillmon
22.1k11942
edited Jan 25 at 18:16
Skillmon
22.1k11942
edited Jan 25 at 18:16
Skillmon
22.1k11942
22.1k11942
asked Jan 25 at 18:02
MarkusMarkus
31
asked Jan 25 at 18:02
MarkusMarkus
31
asked Jan 25 at 18:02
MarkusMarkus
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Welcome to TeX.SE! The purpose of this site is to help you if you get stuck somewhere with your attempts. If you post an attempt, then the one answering does not have to punch in numbers from a screen shot. Anyway, here is a proposal that reproduces your screen shot. (If you play with transform canvas, please use it with care.) EDIT: Fixed the column and row seps, big thanks to @Skillmon!)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{matrix,shapes.geometric,positioning}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,transform canvas={rotate=45},%
nodes={regular polygon,regular polygon sides=4,draw,rotate=-45,shape border rotate=45},
row sep=-pgflinewidth,column sep=-pgflinewidth]
(mat)
{03 & 13 & 23 & 33 \
02 & 12 & 22 & 32 \
01 & 11 & 21 & 31 \
00 & 10 & 20 & 30 \
};
% transform canvas does not transform the bounding box, that's why we need
path ([xshift=-3mm,yshift=3mm]mat.south west |- mat.north west) rectangle
([xshift=3mm,yshift=-3mm]mat.south east -| mat.north east);
node at (mat.south west) {$Phi_1$};
node at (mat.south east) {$Phi_2$};
end{tikzpicture}
end{document}
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
2
I think (not sure about it, my viewer could trick me), that acolumn sep=-pgflinewidthandrow sep=-pgflinewidthis better and results in the same rule thickness as the surrounding one.
– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
Welcome to TeX.SE! The purpose of this site is to help you if you get stuck somewhere with your attempts. If you post an attempt, then the one answering does not have to punch in numbers from a screen shot. Anyway, here is a proposal that reproduces your screen shot. (If you play with transform canvas, please use it with care.) EDIT: Fixed the column and row seps, big thanks to @Skillmon!)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{matrix,shapes.geometric,positioning}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,transform canvas={rotate=45},%
nodes={regular polygon,regular polygon sides=4,draw,rotate=-45,shape border rotate=45},
row sep=-pgflinewidth,column sep=-pgflinewidth]
(mat)
{03 & 13 & 23 & 33 \
02 & 12 & 22 & 32 \
01 & 11 & 21 & 31 \
00 & 10 & 20 & 30 \
};
% transform canvas does not transform the bounding box, that's why we need
path ([xshift=-3mm,yshift=3mm]mat.south west |- mat.north west) rectangle
([xshift=3mm,yshift=-3mm]mat.south east -| mat.north east);
node at (mat.south west) {$Phi_1$};
node at (mat.south east) {$Phi_2$};
end{tikzpicture}
end{document}
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
2
I think (not sure about it, my viewer could trick me), that acolumn sep=-pgflinewidthandrow sep=-pgflinewidthis better and results in the same rule thickness as the surrounding one.
– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
add a comment |
Welcome to TeX.SE! The purpose of this site is to help you if you get stuck somewhere with your attempts. If you post an attempt, then the one answering does not have to punch in numbers from a screen shot. Anyway, here is a proposal that reproduces your screen shot. (If you play with transform canvas, please use it with care.) EDIT: Fixed the column and row seps, big thanks to @Skillmon!)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{matrix,shapes.geometric,positioning}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,transform canvas={rotate=45},%
nodes={regular polygon,regular polygon sides=4,draw,rotate=-45,shape border rotate=45},
row sep=-pgflinewidth,column sep=-pgflinewidth]
(mat)
{03 & 13 & 23 & 33 \
02 & 12 & 22 & 32 \
01 & 11 & 21 & 31 \
00 & 10 & 20 & 30 \
};
% transform canvas does not transform the bounding box, that's why we need
path ([xshift=-3mm,yshift=3mm]mat.south west |- mat.north west) rectangle
([xshift=3mm,yshift=-3mm]mat.south east -| mat.north east);
node at (mat.south west) {$Phi_1$};
node at (mat.south east) {$Phi_2$};
end{tikzpicture}
end{document}
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
2
I think (not sure about it, my viewer could trick me), that acolumn sep=-pgflinewidthandrow sep=-pgflinewidthis better and results in the same rule thickness as the surrounding one.
– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
add a comment |
Welcome to TeX.SE! The purpose of this site is to help you if you get stuck somewhere with your attempts. If you post an attempt, then the one answering does not have to punch in numbers from a screen shot. Anyway, here is a proposal that reproduces your screen shot. (If you play with transform canvas, please use it with care.) EDIT: Fixed the column and row seps, big thanks to @Skillmon!)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{matrix,shapes.geometric,positioning}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,transform canvas={rotate=45},%
nodes={regular polygon,regular polygon sides=4,draw,rotate=-45,shape border rotate=45},
row sep=-pgflinewidth,column sep=-pgflinewidth]
(mat)
{03 & 13 & 23 & 33 \
02 & 12 & 22 & 32 \
01 & 11 & 21 & 31 \
00 & 10 & 20 & 30 \
};
% transform canvas does not transform the bounding box, that's why we need
path ([xshift=-3mm,yshift=3mm]mat.south west |- mat.north west) rectangle
([xshift=3mm,yshift=-3mm]mat.south east -| mat.north east);
node at (mat.south west) {$Phi_1$};
node at (mat.south east) {$Phi_2$};
end{tikzpicture}
end{document}
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
Welcome to TeX.SE! The purpose of this site is to help you if you get stuck somewhere with your attempts. If you post an attempt, then the one answering does not have to punch in numbers from a screen shot. Anyway, here is a proposal that reproduces your screen shot. (If you play with transform canvas, please use it with care.) EDIT: Fixed the column and row seps, big thanks to @Skillmon!)
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{matrix,shapes.geometric,positioning}
begin{document}
begin{tikzpicture}
matrix[matrix of nodes,transform canvas={rotate=45},%
nodes={regular polygon,regular polygon sides=4,draw,rotate=-45,shape border rotate=45},
row sep=-pgflinewidth,column sep=-pgflinewidth]
(mat)
{03 & 13 & 23 & 33 \
02 & 12 & 22 & 32 \
01 & 11 & 21 & 31 \
00 & 10 & 20 & 30 \
};
% transform canvas does not transform the bounding box, that's why we need
path ([xshift=-3mm,yshift=3mm]mat.south west |- mat.north west) rectangle
([xshift=3mm,yshift=-3mm]mat.south east -| mat.north east);
node at (mat.south west) {$Phi_1$};
node at (mat.south east) {$Phi_2$};
end{tikzpicture}
end{document}
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
edited Jan 25 at 18:47
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
answered Jan 25 at 18:18
marmotmarmot
96.6k4111213
96.6k4111213
2
I think (not sure about it, my viewer could trick me), that acolumn sep=-pgflinewidthandrow sep=-pgflinewidthis better and results in the same rule thickness as the surrounding one.
– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
add a comment |
2
I think (not sure about it, my viewer could trick me), that acolumn sep=-pgflinewidthandrow sep=-pgflinewidthis better and results in the same rule thickness as the surrounding one.
– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
2
2
I think (not sure about it, my viewer could trick me), that a
column sep=-pgflinewidth and row sep=-pgflinewidth is better and results in the same rule thickness as the surrounding one.– Skillmon
Jan 25 at 18:43
I think (not sure about it, my viewer could trick me), that a
column sep=-pgflinewidth and row sep=-pgflinewidth is better and results in the same rule thickness as the surrounding one.– Skillmon
Jan 25 at 18:43
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
@Skillmon Yes, you are right. Thanks!
– marmot
Jan 25 at 18:51
add a comment |
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