First index is not integer using foreach loop from 0
I am trying to draw a tree using tikzpicture like this:
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = 1 - floor(nnodes / 2) - 1; }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[node] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
I get the tree like the following picture. The texts are nodes' isibling
within each layer. Most nodes are integers, but all leftmost nodes are not.
tikz-pgf foreach
New contributor
add a comment |
I am trying to draw a tree using tikzpicture like this:
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = 1 - floor(nnodes / 2) - 1; }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[node] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
I get the tree like the following picture. The texts are nodes' isibling
within each layer. Most nodes are integers, but all leftmost nodes are not.
tikz-pgf foreach
New contributor
1
Welcome to TeX.SX. When you post a question, please provide a "Minimal Working Example" (MWE) that starts withdocumentclass
, includes all relevantusepackage
commands, ends withend{document}
and compiles without errors, even if it does not produce your desired output.
– Sandy G
2 days ago
add a comment |
I am trying to draw a tree using tikzpicture like this:
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = 1 - floor(nnodes / 2) - 1; }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[node] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
I get the tree like the following picture. The texts are nodes' isibling
within each layer. Most nodes are integers, but all leftmost nodes are not.
tikz-pgf foreach
New contributor
I am trying to draw a tree using tikzpicture like this:
documentclass{article}
usepackage{tikz}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = 1 - floor(nnodes / 2) - 1; }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[node] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
I get the tree like the following picture. The texts are nodes' isibling
within each layer. Most nodes are integers, but all leftmost nodes are not.
tikz-pgf foreach
tikz-pgf foreach
New contributor
New contributor
edited yesterday
landings
New contributor
asked 2 days ago
landingslandings
414
414
New contributor
New contributor
1
Welcome to TeX.SX. When you post a question, please provide a "Minimal Working Example" (MWE) that starts withdocumentclass
, includes all relevantusepackage
commands, ends withend{document}
and compiles without errors, even if it does not produce your desired output.
– Sandy G
2 days ago
add a comment |
1
Welcome to TeX.SX. When you post a question, please provide a "Minimal Working Example" (MWE) that starts withdocumentclass
, includes all relevantusepackage
commands, ends withend{document}
and compiles without errors, even if it does not produce your desired output.
– Sandy G
2 days ago
1
1
Welcome to TeX.SX. When you post a question, please provide a "Minimal Working Example" (MWE) that starts with
documentclass
, includes all relevant usepackage
commands, ends with end{document}
and compiles without errors, even if it does not produce your desired output.– Sandy G
2 days ago
Welcome to TeX.SX. When you post a question, please provide a "Minimal Working Example" (MWE) that starts with
documentclass
, includes all relevant usepackage
commands, ends with end{document}
and compiles without errors, even if it does not produce your desired output.– Sandy G
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
You could just tell TikZ explicitly that you want an integer.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = int(1 - floor(nnodes / 2) - 1); }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
Or
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {int nnodes,leftnum,rightnum;
nnodes = 3 ^ ilayer;
leftnum = 1 - floor(nnodes / 2) - 1;
rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15;
x = isibling * d;
y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
In principle you do not need the math library here.
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer [evaluate=ilayer as nnodes using {int(3 ^ ilayer)},
evaluate=nnodes as leftnum using {int(1 - floor(nnodes / 2) - 1)},
evaluate=nnodes as rightnum using {int(nnodes - floor(nnodes / 2) - 1)}]
in {0,...,3} {
foreach isibling
[evaluate=ilayer as d using {3 ^ (- ilayer) * 15},
evaluate=isibling as x using {isibling * d},
evaluate=ilayer as y using {- ilayer * 2}]
in {leftnum,...,rightnum} {
node[mynode] (node_ilayer_isibling) at (x cm, y cm)
{isibling};
}
}
end{tikzpicture}
end{document}
Thanks a lot. I finally get where the problem starts. Why1 - floor(nnodes / 2) - 1
can be non-integer? Even1 - int(nnodes / 2) - 1
is problematic.
– landings
2 days ago
1
@landings It is due to the wayforeach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression intoint
.
– marmot
2 days ago
add a comment |
As @marmot said you do not need tikzmath
here, but if you use it you can do it in more efficient way :
- You can have a single
tikzmath
command with loops inside it. - You can declare your integer variables as
int
so you do not need to doint()
afterward. - As
nnodes
is odd you do not need separaterightnum
andleftnum
asrightnum = - leftnum
; - Why you use
1-floor(nnodes/2)-1
in place of-floor(nnodes/2)
? - The value
d
can be calculated in the outer loop. - Instead of using
x=isibling*d
you can say[x=d cm]
and then useisibling
asx
. And in the same wayy
can be replaced byilayer
using[y=-2cm]
.
So here is my proposal :
documentclass[tikz,border=7pt]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
tikzmath{
int ilayer,nnodes,rightnum,isibling;
nnodes = 1;
for ilayer in {0,...,3}{
rightnum = (nnodes-1)/2;
d = 15/nnodes;
for isibling in {-rightnum,...,rightnum}{
{
path[x=d cm,y=-2cm]
node[node] (node_ilayer_isibling) at (isibling, ilayer) {isibling};
};
};
nnodes = 3*nnodes;
};
}
end{tikzpicture}
end{document}
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could just tell TikZ explicitly that you want an integer.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = int(1 - floor(nnodes / 2) - 1); }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
Or
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {int nnodes,leftnum,rightnum;
nnodes = 3 ^ ilayer;
leftnum = 1 - floor(nnodes / 2) - 1;
rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15;
x = isibling * d;
y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
In principle you do not need the math library here.
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer [evaluate=ilayer as nnodes using {int(3 ^ ilayer)},
evaluate=nnodes as leftnum using {int(1 - floor(nnodes / 2) - 1)},
evaluate=nnodes as rightnum using {int(nnodes - floor(nnodes / 2) - 1)}]
in {0,...,3} {
foreach isibling
[evaluate=ilayer as d using {3 ^ (- ilayer) * 15},
evaluate=isibling as x using {isibling * d},
evaluate=ilayer as y using {- ilayer * 2}]
in {leftnum,...,rightnum} {
node[mynode] (node_ilayer_isibling) at (x cm, y cm)
{isibling};
}
}
end{tikzpicture}
end{document}
Thanks a lot. I finally get where the problem starts. Why1 - floor(nnodes / 2) - 1
can be non-integer? Even1 - int(nnodes / 2) - 1
is problematic.
– landings
2 days ago
1
@landings It is due to the wayforeach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression intoint
.
– marmot
2 days ago
add a comment |
You could just tell TikZ explicitly that you want an integer.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = int(1 - floor(nnodes / 2) - 1); }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
Or
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {int nnodes,leftnum,rightnum;
nnodes = 3 ^ ilayer;
leftnum = 1 - floor(nnodes / 2) - 1;
rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15;
x = isibling * d;
y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
In principle you do not need the math library here.
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer [evaluate=ilayer as nnodes using {int(3 ^ ilayer)},
evaluate=nnodes as leftnum using {int(1 - floor(nnodes / 2) - 1)},
evaluate=nnodes as rightnum using {int(nnodes - floor(nnodes / 2) - 1)}]
in {0,...,3} {
foreach isibling
[evaluate=ilayer as d using {3 ^ (- ilayer) * 15},
evaluate=isibling as x using {isibling * d},
evaluate=ilayer as y using {- ilayer * 2}]
in {leftnum,...,rightnum} {
node[mynode] (node_ilayer_isibling) at (x cm, y cm)
{isibling};
}
}
end{tikzpicture}
end{document}
Thanks a lot. I finally get where the problem starts. Why1 - floor(nnodes / 2) - 1
can be non-integer? Even1 - int(nnodes / 2) - 1
is problematic.
– landings
2 days ago
1
@landings It is due to the wayforeach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression intoint
.
– marmot
2 days ago
add a comment |
You could just tell TikZ explicitly that you want an integer.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = int(1 - floor(nnodes / 2) - 1); }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
Or
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {int nnodes,leftnum,rightnum;
nnodes = 3 ^ ilayer;
leftnum = 1 - floor(nnodes / 2) - 1;
rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15;
x = isibling * d;
y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
In principle you do not need the math library here.
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer [evaluate=ilayer as nnodes using {int(3 ^ ilayer)},
evaluate=nnodes as leftnum using {int(1 - floor(nnodes / 2) - 1)},
evaluate=nnodes as rightnum using {int(nnodes - floor(nnodes / 2) - 1)}]
in {0,...,3} {
foreach isibling
[evaluate=ilayer as d using {3 ^ (- ilayer) * 15},
evaluate=isibling as x using {isibling * d},
evaluate=ilayer as y using {- ilayer * 2}]
in {leftnum,...,rightnum} {
node[mynode] (node_ilayer_isibling) at (x cm, y cm)
{isibling};
}
}
end{tikzpicture}
end{document}
You could just tell TikZ explicitly that you want an integer.
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {nnodes = 3 ^ ilayer; }
tikzmath {leftnum = int(1 - floor(nnodes / 2) - 1); }
tikzmath {rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15; }
tikzmath {x = isibling * d; }
tikzmath {y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
Or
documentclass[tikz,border=3.14mm]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer in {0,...,3} {
tikzmath {int nnodes,leftnum,rightnum;
nnodes = 3 ^ ilayer;
leftnum = 1 - floor(nnodes / 2) - 1;
rightnum = nnodes - floor(nnodes / 2) - 1; }
foreach isibling in {leftnum,...,rightnum} {
tikzmath {d = 3 ^ (- ilayer) * 15;
x = isibling * d;
y = - ilayer * 2; }
node[mynode] (node_ilayer_isibling) at (x cm, y cm) {isibling};
}
}
end{tikzpicture}
end{document}
In principle you do not need the math library here.
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}
tikzset{mynode/.style={circle, fill=blue!25, minimum size=0.1 cm}}
foreach ilayer [evaluate=ilayer as nnodes using {int(3 ^ ilayer)},
evaluate=nnodes as leftnum using {int(1 - floor(nnodes / 2) - 1)},
evaluate=nnodes as rightnum using {int(nnodes - floor(nnodes / 2) - 1)}]
in {0,...,3} {
foreach isibling
[evaluate=ilayer as d using {3 ^ (- ilayer) * 15},
evaluate=isibling as x using {isibling * d},
evaluate=ilayer as y using {- ilayer * 2}]
in {leftnum,...,rightnum} {
node[mynode] (node_ilayer_isibling) at (x cm, y cm)
{isibling};
}
}
end{tikzpicture}
end{document}
edited 2 days ago
answered 2 days ago
marmotmarmot
113k5144273
113k5144273
Thanks a lot. I finally get where the problem starts. Why1 - floor(nnodes / 2) - 1
can be non-integer? Even1 - int(nnodes / 2) - 1
is problematic.
– landings
2 days ago
1
@landings It is due to the wayforeach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression intoint
.
– marmot
2 days ago
add a comment |
Thanks a lot. I finally get where the problem starts. Why1 - floor(nnodes / 2) - 1
can be non-integer? Even1 - int(nnodes / 2) - 1
is problematic.
– landings
2 days ago
1
@landings It is due to the wayforeach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression intoint
.
– marmot
2 days ago
Thanks a lot. I finally get where the problem starts. Why
1 - floor(nnodes / 2) - 1
can be non-integer? Even 1 - int(nnodes / 2) - 1
is problematic.– landings
2 days ago
Thanks a lot. I finally get where the problem starts. Why
1 - floor(nnodes / 2) - 1
can be non-integer? Even 1 - int(nnodes / 2) - 1
is problematic.– landings
2 days ago
1
1
@landings It is due to the way
foreach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression into int
.– marmot
2 days ago
@landings It is due to the way
foreach
is implemented, internally TikZ computes with dimensions and this can lead to slight inconsistencies. So it is better to wrap the full expression into int
.– marmot
2 days ago
add a comment |
As @marmot said you do not need tikzmath
here, but if you use it you can do it in more efficient way :
- You can have a single
tikzmath
command with loops inside it. - You can declare your integer variables as
int
so you do not need to doint()
afterward. - As
nnodes
is odd you do not need separaterightnum
andleftnum
asrightnum = - leftnum
; - Why you use
1-floor(nnodes/2)-1
in place of-floor(nnodes/2)
? - The value
d
can be calculated in the outer loop. - Instead of using
x=isibling*d
you can say[x=d cm]
and then useisibling
asx
. And in the same wayy
can be replaced byilayer
using[y=-2cm]
.
So here is my proposal :
documentclass[tikz,border=7pt]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
tikzmath{
int ilayer,nnodes,rightnum,isibling;
nnodes = 1;
for ilayer in {0,...,3}{
rightnum = (nnodes-1)/2;
d = 15/nnodes;
for isibling in {-rightnum,...,rightnum}{
{
path[x=d cm,y=-2cm]
node[node] (node_ilayer_isibling) at (isibling, ilayer) {isibling};
};
};
nnodes = 3*nnodes;
};
}
end{tikzpicture}
end{document}
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
2 days ago
add a comment |
As @marmot said you do not need tikzmath
here, but if you use it you can do it in more efficient way :
- You can have a single
tikzmath
command with loops inside it. - You can declare your integer variables as
int
so you do not need to doint()
afterward. - As
nnodes
is odd you do not need separaterightnum
andleftnum
asrightnum = - leftnum
; - Why you use
1-floor(nnodes/2)-1
in place of-floor(nnodes/2)
? - The value
d
can be calculated in the outer loop. - Instead of using
x=isibling*d
you can say[x=d cm]
and then useisibling
asx
. And in the same wayy
can be replaced byilayer
using[y=-2cm]
.
So here is my proposal :
documentclass[tikz,border=7pt]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
tikzmath{
int ilayer,nnodes,rightnum,isibling;
nnodes = 1;
for ilayer in {0,...,3}{
rightnum = (nnodes-1)/2;
d = 15/nnodes;
for isibling in {-rightnum,...,rightnum}{
{
path[x=d cm,y=-2cm]
node[node] (node_ilayer_isibling) at (isibling, ilayer) {isibling};
};
};
nnodes = 3*nnodes;
};
}
end{tikzpicture}
end{document}
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
2 days ago
add a comment |
As @marmot said you do not need tikzmath
here, but if you use it you can do it in more efficient way :
- You can have a single
tikzmath
command with loops inside it. - You can declare your integer variables as
int
so you do not need to doint()
afterward. - As
nnodes
is odd you do not need separaterightnum
andleftnum
asrightnum = - leftnum
; - Why you use
1-floor(nnodes/2)-1
in place of-floor(nnodes/2)
? - The value
d
can be calculated in the outer loop. - Instead of using
x=isibling*d
you can say[x=d cm]
and then useisibling
asx
. And in the same wayy
can be replaced byilayer
using[y=-2cm]
.
So here is my proposal :
documentclass[tikz,border=7pt]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
tikzmath{
int ilayer,nnodes,rightnum,isibling;
nnodes = 1;
for ilayer in {0,...,3}{
rightnum = (nnodes-1)/2;
d = 15/nnodes;
for isibling in {-rightnum,...,rightnum}{
{
path[x=d cm,y=-2cm]
node[node] (node_ilayer_isibling) at (isibling, ilayer) {isibling};
};
};
nnodes = 3*nnodes;
};
}
end{tikzpicture}
end{document}
As @marmot said you do not need tikzmath
here, but if you use it you can do it in more efficient way :
- You can have a single
tikzmath
command with loops inside it. - You can declare your integer variables as
int
so you do not need to doint()
afterward. - As
nnodes
is odd you do not need separaterightnum
andleftnum
asrightnum = - leftnum
; - Why you use
1-floor(nnodes/2)-1
in place of-floor(nnodes/2)
? - The value
d
can be calculated in the outer loop. - Instead of using
x=isibling*d
you can say[x=d cm]
and then useisibling
asx
. And in the same wayy
can be replaced byilayer
using[y=-2cm]
.
So here is my proposal :
documentclass[tikz,border=7pt]{standalone}
usetikzlibrary{math}
begin{document}
begin{tikzpicture}
tikzstyle{node}=[circle, fill=blue!25, minimum size=0.1 cm];
tikzmath{
int ilayer,nnodes,rightnum,isibling;
nnodes = 1;
for ilayer in {0,...,3}{
rightnum = (nnodes-1)/2;
d = 15/nnodes;
for isibling in {-rightnum,...,rightnum}{
{
path[x=d cm,y=-2cm]
node[node] (node_ilayer_isibling) at (isibling, ilayer) {isibling};
};
};
nnodes = 3*nnodes;
};
}
end{tikzpicture}
end{document}
edited 2 days ago
answered 2 days ago
KpymKpym
17.2k24090
17.2k24090
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
2 days ago
add a comment |
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
2 days ago
1
1
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
Thanks for your help. I am new to LaTeX and need more practice. I didn't know I could loop and draw things just inside tikzmath code.
– landings
2 days ago
2
2
Don't worry, even some experts don't know how to use
tikzmath
;)– Kpym
2 days ago
Don't worry, even some experts don't know how to use
tikzmath
;)– Kpym
2 days ago
add a comment |
landings is a new contributor. Be nice, and check out our Code of Conduct.
landings is a new contributor. Be nice, and check out our Code of Conduct.
landings is a new contributor. Be nice, and check out our Code of Conduct.
landings is a new contributor. Be nice, and check out our Code of Conduct.
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