Modify the code of the example Plane partition by Jang Soo Kim from “texample.net”
2
From the answer to this question, is it possible to make a lower cube position void, to create this drawing
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
tikz-pgf
asked 1 hour ago
HanyHany
1,141416
add a comment |
2
From the answer to this question, is it possible to make a lower cube position void, to create this drawing
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
tikz-pgf
asked 1 hour ago
HanyHany
1,141416
add a comment |
2
2
2
From the answer to this question, is it possible to make a lower cube position void, to create this drawing
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
tikz-pgf
asked 1 hour ago
HanyHany
1,141416
From the answer to this question, is it possible to make a lower cube position void, to create this drawing
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
tikz-pgf
tikz-pgf
asked 1 hour ago
HanyHany
1,141416
asked 1 hour ago
HanyHany
1,141416
asked 1 hour ago
HanyHany
1,141416
asked 1 hour ago
HanyHany
1,141416
asked 1 hour ago
HanyHany
1,141416
1,141416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
1
In this case, it is sufficient to build layer by layer from the lowest to the highest.
To do this, I modified the loop by adding an optional argument that is set to 0 by default. This allows you to keep the old syntax for previous graphics.
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
% Definition of planepartition
% To draw the following plane partition, just write planepartition{ {a, b, c}, {d,e} }.
% a b c
% d e
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
begin{document}
begin{tikzpicture}% Old syntax is functional
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
end{tikzpicture}
begin{tikzpicture}% The optional argument allow to build layer by layer
planepartition{{1,1,0,1,0,1,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,1,0,1,0,1,1}}
planepartition[1]{{1,0,1,0,1,0,1},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{1,0,1,0,1,0,1}}
end{tikzpicture}
end{document}
answered 1 hour ago
AndréCAndréC
8,55111446
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
add a comment |
1
Yes. I added an optional parameter which indicates how many boxes are to be skipped. So 2/1 means two boxes but skip the first one.
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b/d in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnumb>0
ifnumb=d
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
else
pgfmathtruncatemacro{cmin}{1+d}
addtocounter{z}{d}
foreach c in {cmin,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
fi
fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2/1,1,2/1,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
answered 1 hour ago
marmotmarmot
92.8k4109203
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the/1you will get the original result. But with this method you do not have to draw two partitions.
– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany Inplanepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},...there some2s are replaced by2/1. These are the positions at which the lower cube becomes "void".
– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
In this case, it is sufficient to build layer by layer from the lowest to the highest.
To do this, I modified the loop by adding an optional argument that is set to 0 by default. This allows you to keep the old syntax for previous graphics.
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
% Definition of planepartition
% To draw the following plane partition, just write planepartition{ {a, b, c}, {d,e} }.
% a b c
% d e
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
begin{document}
begin{tikzpicture}% Old syntax is functional
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
end{tikzpicture}
begin{tikzpicture}% The optional argument allow to build layer by layer
planepartition{{1,1,0,1,0,1,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,1,0,1,0,1,1}}
planepartition[1]{{1,0,1,0,1,0,1},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{1,0,1,0,1,0,1}}
end{tikzpicture}
end{document}
answered 1 hour ago
AndréCAndréC
8,55111446
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
add a comment |
1
In this case, it is sufficient to build layer by layer from the lowest to the highest.
To do this, I modified the loop by adding an optional argument that is set to 0 by default. This allows you to keep the old syntax for previous graphics.
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
% Definition of planepartition
% To draw the following plane partition, just write planepartition{ {a, b, c}, {d,e} }.
% a b c
% d e
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
begin{document}
begin{tikzpicture}% Old syntax is functional
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
end{tikzpicture}
begin{tikzpicture}% The optional argument allow to build layer by layer
planepartition{{1,1,0,1,0,1,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,1,0,1,0,1,1}}
planepartition[1]{{1,0,1,0,1,0,1},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{1,0,1,0,1,0,1}}
end{tikzpicture}
end{document}
answered 1 hour ago
AndréCAndréC
8,55111446
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
add a comment |
1
1
1
In this case, it is sufficient to build layer by layer from the lowest to the highest.
To do this, I modified the loop by adding an optional argument that is set to 0 by default. This allows you to keep the old syntax for previous graphics.
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
% Definition of planepartition
% To draw the following plane partition, just write planepartition{ {a, b, c}, {d,e} }.
% a b c
% d e
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
begin{document}
begin{tikzpicture}% Old syntax is functional
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
end{tikzpicture}
begin{tikzpicture}% The optional argument allow to build layer by layer
planepartition{{1,1,0,1,0,1,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,1,0,1,0,1,1}}
planepartition[1]{{1,0,1,0,1,0,1},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{1,0,1,0,1,0,1}}
end{tikzpicture}
end{document}
answered 1 hour ago
AndréCAndréC
8,55111446
In this case, it is sufficient to build layer by layer from the lowest to the highest.
To do this, I modified the loop by adding an optional argument that is set to 0 by default. This allows you to keep the old syntax for previous graphics.
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue,fill opacity=.7, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
% Definition of planepartition
% To draw the following plane partition, just write planepartition{ {a, b, c}, {d,e} }.
% a b c
% d e
newcommandplanepartition[2][0]{
setcounter{x}{-1}
foreach a in {#2} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b in a {
addtocounter{y}{1}
setcounter{z}{-1}
addtocounter{z}{#1}
ifnum b>0
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}
}fi
}
}
}
begin{document}
begin{tikzpicture}% Old syntax is functional
planepartition{{2,1,2,1,2,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2,1,2,1,2}}
end{tikzpicture}
begin{tikzpicture}% The optional argument allow to build layer by layer
planepartition{{1,1,0,1,0,1,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,1,0,1,0,1,1}}
planepartition[1]{{1,0,1,0,1,0,1},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{0,0,0,0,0,0,0},{1,0,1,0,1,0,1}}
end{tikzpicture}
end{document}
answered 1 hour ago
AndréCAndréC
8,55111446
edited 1 hour ago
answered 1 hour ago
AndréCAndréC
8,55111446
answered 1 hour ago
AndréCAndréC
8,55111446
answered 1 hour ago
AndréCAndréC
8,55111446
8,55111446
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
add a comment |
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
Thank you very much for your answer. Both your code and marmot`s work fine.
– Hany
44 mins ago
1
1
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
Thank you, my way of doing things allows you to place several holes on the same column.
– AndréC
44 mins ago
add a comment |
1
Yes. I added an optional parameter which indicates how many boxes are to be skipped. So 2/1 means two boxes but skip the first one.
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b/d in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnumb>0
ifnumb=d
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
else
pgfmathtruncatemacro{cmin}{1+d}
addtocounter{z}{d}
foreach c in {cmin,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
fi
fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2/1,1,2/1,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
answered 1 hour ago
marmotmarmot
92.8k4109203
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the/1you will get the original result. But with this method you do not have to draw two partitions.
– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany Inplanepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},...there some2s are replaced by2/1. These are the positions at which the lower cube becomes "void".
– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
|
show 3 more comments
1
Yes. I added an optional parameter which indicates how many boxes are to be skipped. So 2/1 means two boxes but skip the first one.
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b/d in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnumb>0
ifnumb=d
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
else
pgfmathtruncatemacro{cmin}{1+d}
addtocounter{z}{d}
foreach c in {cmin,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
fi
fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2/1,1,2/1,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
answered 1 hour ago
marmotmarmot
92.8k4109203
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the/1you will get the original result. But with this method you do not have to draw two partitions.
– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany Inplanepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},...there some2s are replaced by2/1. These are the positions at which the lower cube becomes "void".
– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
|
show 3 more comments
1
1
1
Yes. I added an optional parameter which indicates how many boxes are to be skipped. So 2/1 means two boxes but skip the first one.
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b/d in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnumb>0
ifnumb=d
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
else
pgfmathtruncatemacro{cmin}{1+d}
addtocounter{z}{d}
foreach c in {cmin,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
fi
fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2/1,1,2/1,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
answered 1 hour ago
marmotmarmot
92.8k4109203
Yes. I added an optional parameter which indicates how many boxes are to be skipped. So 2/1 means two boxes but skip the first one.
documentclass{beamer}
setbeamertemplate{navigation symbols}{}% to suppresses (hide) navigation symbols bar
usepackage{tikz}
usepackage{verbatim}
% Three counters
newcounter{x}
newcounter{y}
newcounter{z}
% The angles of x,y,z-axes
newcommandxaxis{210}
newcommandyaxis{-30}
newcommandzaxis{90}
% The top side of a cube
newcommandtopside[3]{
fill[fill=yellow, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (0,1) --(150:1)--(0,0);
}
% The left side of a cube
newcommandleftside[3]{
fill[fill=green, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (0,-1) -- (210:1) --(150:1)--(0,0);
}
% The right side of a cube
newcommandrightside[3]{
fill[fill=blue, draw=black,shift={(xaxis:#1)},shift={(yaxis:#2)},
shift={(zaxis:#3)}, opacity=.7] (0,0) -- (30:1) -- (-30:1) --(0,-1)--(0,0);
}
% The cube
newcommandcube[3]{
topside{#1}{#2}{#3} leftside{#1}{#2}{#3} rightside{#1}{#2}{#3}
}
newcommandplanepartition[1]{
setcounter{x}{-1}
foreach a in {#1} {
addtocounter{x}{1}
setcounter{y}{-1}
foreach b/d in a {
addtocounter{y}{1}
setcounter{z}{-1}
ifnumb>0
ifnumb=d
foreach c in {1,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
else
pgfmathtruncatemacro{cmin}{1+d}
addtocounter{z}{d}
foreach c in {cmin,...,b} {
addtocounter{z}{1}
cube{value{x}}{value{y}}{value{z}}}
fi
fi}}}
begin{document}
begin{tikzpicture}
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},{2,1,2/1,1,2/1,1,2}}%1st column from back to front{row1,... from left to right}%0 is the same as 2% it does not allow void for 0
end{tikzpicture}
end{document}
answered 1 hour ago
marmotmarmot
92.8k4109203
answered 1 hour ago
marmotmarmot
92.8k4109203
answered 1 hour ago
marmotmarmot
92.8k4109203
answered 1 hour ago
marmotmarmot
92.8k4109203
92.8k4109203
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the/1you will get the original result. But with this method you do not have to draw two partitions.
– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany Inplanepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},...there some2s are replaced by2/1. These are the positions at which the lower cube becomes "void".
– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
|
show 3 more comments
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the/1you will get the original result. But with this method you do not have to draw two partitions.
– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany Inplanepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},...there some2s are replaced by2/1. These are the positions at which the lower cube becomes "void".
– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the
/1 you will get the original result. But with this method you do not have to draw two partitions.– marmot
1 hour ago
Let me mention that in the above the old syntax is, of course, still functional. So if you drop the
/1 you will get the original result. But with this method you do not have to draw two partitions.– marmot
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
Which /1 are you referring to?
– Hany
1 hour ago
@Hany In
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},... there some 2s are replaced by 2/1. These are the positions at which the lower cube becomes "void".– marmot
1 hour ago
@Hany In
planepartition{{2,1,2/1,1,2/1,1,2},{1,0,0,0,0,0,1},{1,0,0,0,0,0,1},... there some 2s are replaced by 2/1. These are the positions at which the lower cube becomes "void".– marmot
1 hour ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
I tried your suggestion, but unfortunately it did not work. Would you please give me the whole code.
– Hany
59 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
@Hany Did you run my above code? Which errors do you get? On my machine my code produces the output that is sown.
– marmot
58 mins ago
|
show 3 more comments
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