Even-length cycle laid out as a star
4
For demonstration purposes, I want to obtain a graph of C10 shaped as a star. I have been working on this for much longer than I should have. I currently have this:
begin{figure}
centering
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathmod{i}{10}
pgfmathadd{pgfmathresult}{1};
path (Ni) edge (Npgfmathresult);
}
end{tikzpicture}
end{figure}
Which outputs the following abomination:
I don't know if the same ugliness would happen had I typed out the nodes myself.
tikz-pgf tikz-node tikz-path
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
add a comment |
4
For demonstration purposes, I want to obtain a graph of C10 shaped as a star. I have been working on this for much longer than I should have. I currently have this:
begin{figure}
centering
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathmod{i}{10}
pgfmathadd{pgfmathresult}{1};
path (Ni) edge (Npgfmathresult);
}
end{tikzpicture}
end{figure}
Which outputs the following abomination:
I don't know if the same ugliness would happen had I typed out the nodes myself.
tikz-pgf tikz-node tikz-path
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00
add a comment |
4
4
4
For demonstration purposes, I want to obtain a graph of C10 shaped as a star. I have been working on this for much longer than I should have. I currently have this:
begin{figure}
centering
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathmod{i}{10}
pgfmathadd{pgfmathresult}{1};
path (Ni) edge (Npgfmathresult);
}
end{tikzpicture}
end{figure}
Which outputs the following abomination:
I don't know if the same ugliness would happen had I typed out the nodes myself.
tikz-pgf tikz-node tikz-path
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
For demonstration purposes, I want to obtain a graph of C10 shaped as a star. I have been working on this for much longer than I should have. I currently have this:
begin{figure}
centering
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathmod{i}{10}
pgfmathadd{pgfmathresult}{1};
path (Ni) edge (Npgfmathresult);
}
end{tikzpicture}
end{figure}
Which outputs the following abomination:
I don't know if the same ugliness would happen had I typed out the nodes myself.
tikz-pgf tikz-node tikz-path
tikz-pgf tikz-node tikz-path
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
asked Mar 7 at 14:34
ThoAppelsinThoAppelsin
330111
330111
The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00
add a comment |
The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00
The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00
add a comment |
2 Answers
2
active
oldest
votes
4
After the manipulations, the result won't be an integer any more. Rather, you'll get numbers like 1.0, where .0 is interpreted as an anchor. Therefore I suggest
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathtruncatemacro{j}{mod(i,10)+1}
path (Ni) edge (Nj);
}
end{tikzpicture}
end{document}
answered Mar 7 at 16:11
marmotmarmot
108k5133251
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
add a comment |
1
PGF's foreach has quite powerfull tools that you can use here:
begin{tikzpicture}[nodes={circle, draw}]
foreach[evaluate=i as j using isodd(i)] i in {1,...,10}
{
node (Ni) at (360/10*i:j?18mm:9mm) {i};
}
foreach[remember=i as j (initially 10)] i in {1,...,10}
{
draw (Ni) -- (Nj);
}
end{tikzpicture}
The optional [evaluate=i as j using isodd(i)] in the first loop computes isodd(i) and stores the result in macro j.
The optional [remember=i as j (initially 10)] in the second loop stores i's content in macro j at the end of the iteration, allowing for this content to be available for the next iteration.
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
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
4
After the manipulations, the result won't be an integer any more. Rather, you'll get numbers like 1.0, where .0 is interpreted as an anchor. Therefore I suggest
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathtruncatemacro{j}{mod(i,10)+1}
path (Ni) edge (Nj);
}
end{tikzpicture}
end{document}
answered Mar 7 at 16:11
marmotmarmot
108k5133251
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
add a comment |
4
After the manipulations, the result won't be an integer any more. Rather, you'll get numbers like 1.0, where .0 is interpreted as an anchor. Therefore I suggest
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathtruncatemacro{j}{mod(i,10)+1}
path (Ni) edge (Nj);
}
end{tikzpicture}
end{document}
answered Mar 7 at 16:11
marmotmarmot
108k5133251
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
add a comment |
4
4
4
After the manipulations, the result won't be an integer any more. Rather, you'll get numbers like 1.0, where .0 is interpreted as an anchor. Therefore I suggest
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathtruncatemacro{j}{mod(i,10)+1}
path (Ni) edge (Nj);
}
end{tikzpicture}
end{document}
answered Mar 7 at 16:11
marmotmarmot
108k5133251
After the manipulations, the result won't be an integer any more. Rather, you'll get numbers like 1.0, where .0 is interpreted as an anchor. Therefore I suggest
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[nodes={circle, draw}]
foreach i in {1,...,10}
{
pgfmathisodd{i};
node (Ni) at (360/10*i:pgfmathresult?18mm:9mm) {i};
}
foreach i in {1,...,10}
{
pgfmathtruncatemacro{j}{mod(i,10)+1}
path (Ni) edge (Nj);
}
end{tikzpicture}
end{document}
answered Mar 7 at 16:11
marmotmarmot
108k5133251
answered Mar 7 at 16:11
marmotmarmot
108k5133251
answered Mar 7 at 16:11
marmotmarmot
108k5133251
answered Mar 7 at 16:11
marmotmarmot
108k5133251
108k5133251
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
add a comment |
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
3
3
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
I am aware of the fact that one can shorten the code. This answer is to explain what happens and to provide a way that works which is very close to the code of the question.
– marmot
Mar 7 at 16:12
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
– Sandy G
Mar 7 at 16:29
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
Great answer, thank you for being to the point. And to @SandyG, I already have fixed the size inconsistency via putting all nodes in boxes with zero width, and giving any size I like to the nodes.
– ThoAppelsin
Mar 7 at 19:10
add a comment |
1
PGF's foreach has quite powerfull tools that you can use here:
begin{tikzpicture}[nodes={circle, draw}]
foreach[evaluate=i as j using isodd(i)] i in {1,...,10}
{
node (Ni) at (360/10*i:j?18mm:9mm) {i};
}
foreach[remember=i as j (initially 10)] i in {1,...,10}
{
draw (Ni) -- (Nj);
}
end{tikzpicture}
The optional [evaluate=i as j using isodd(i)] in the first loop computes isodd(i) and stores the result in macro j.
The optional [remember=i as j (initially 10)] in the second loop stores i's content in macro j at the end of the iteration, allowing for this content to be available for the next iteration.
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
add a comment |
1
PGF's foreach has quite powerfull tools that you can use here:
begin{tikzpicture}[nodes={circle, draw}]
foreach[evaluate=i as j using isodd(i)] i in {1,...,10}
{
node (Ni) at (360/10*i:j?18mm:9mm) {i};
}
foreach[remember=i as j (initially 10)] i in {1,...,10}
{
draw (Ni) -- (Nj);
}
end{tikzpicture}
The optional [evaluate=i as j using isodd(i)] in the first loop computes isodd(i) and stores the result in macro j.
The optional [remember=i as j (initially 10)] in the second loop stores i's content in macro j at the end of the iteration, allowing for this content to be available for the next iteration.
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
add a comment |
1
1
1
PGF's foreach has quite powerfull tools that you can use here:
begin{tikzpicture}[nodes={circle, draw}]
foreach[evaluate=i as j using isodd(i)] i in {1,...,10}
{
node (Ni) at (360/10*i:j?18mm:9mm) {i};
}
foreach[remember=i as j (initially 10)] i in {1,...,10}
{
draw (Ni) -- (Nj);
}
end{tikzpicture}
The optional [evaluate=i as j using isodd(i)] in the first loop computes isodd(i) and stores the result in macro j.
The optional [remember=i as j (initially 10)] in the second loop stores i's content in macro j at the end of the iteration, allowing for this content to be available for the next iteration.
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
PGF's foreach has quite powerfull tools that you can use here:
begin{tikzpicture}[nodes={circle, draw}]
foreach[evaluate=i as j using isodd(i)] i in {1,...,10}
{
node (Ni) at (360/10*i:j?18mm:9mm) {i};
}
foreach[remember=i as j (initially 10)] i in {1,...,10}
{
draw (Ni) -- (Nj);
}
end{tikzpicture}
The optional [evaluate=i as j using isodd(i)] in the first loop computes isodd(i) and stores the result in macro j.
The optional [remember=i as j (initially 10)] in the second loop stores i's content in macro j at the end of the iteration, allowing for this content to be available for the next iteration.
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
answered Mar 7 at 16:37
Christoph FringsChristoph Frings
923211
923211
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
add a comment |
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
Thank you for your extended recommendations, but I think marmot's answer is more relevant to the issue/question, since it also explains the cause of the erroneous outcome in the question.
– ThoAppelsin
Mar 7 at 19:11
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
@ThoAppelsin I agree, this answer is also aimed at those who will find your question later and offer them a simple and robust solution.
– Christoph Frings
Mar 7 at 19:31
add a comment |
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The "abomination" is simply all numbers aligned the same way. Try instead of defining the points recursively like you did and then in a second step add the lines and then the numbers. This should make it better :)
– Superuser27
Mar 7 at 14:36
What I mean is: if you just look at how the numbers are positioned, they look fine. It's just the lines starting from weird places.
– Superuser27
Mar 7 at 14:37
I have given up and written every node one by one. It works! The question is even more curious to me now.
– ThoAppelsin
Mar 7 at 15:00