First index is not integer using foreach loop from 0
I am trying to draw a tree using tikzpicture like this:
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}
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:
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}
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:
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}
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:
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}
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
New contributor
asked 9 hours ago
landingslandings
413
413
New contributor
New contributor
add a comment |
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
9 hours ago
@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
8 hours 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
5 hours ago
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
3 hours 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
9 hours ago
@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
8 hours 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
9 hours ago
@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
8 hours 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 9 hours ago
answered 9 hours ago
marmotmarmot
112k5141268
112k5141268
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
9 hours ago
@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
8 hours 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
9 hours ago
@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
8 hours 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
9 hours 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
9 hours 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
8 hours 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
8 hours 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
5 hours ago
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
3 hours 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
5 hours ago
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
3 hours 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 hours ago
answered 6 hours ago
KpymKpym
17.1k24090
17.1k24090
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
5 hours ago
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
3 hours 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
5 hours ago
Don't worry, even some experts don't know how to usetikzmath
;)
– Kpym
3 hours 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
5 hours 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
5 hours ago
Don't worry, even some experts don't know how to use
tikzmath
;)– Kpym
3 hours ago
Don't worry, even some experts don't know how to use
tikzmath
;)– Kpym
3 hours 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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