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3

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

share|improve this question

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have given up and written every node one by one. It works! The question is even more curious to me now.
  
  – ThoAppelsin
  5 hours ago
  

  
  
  
  

add a comment | 

3

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

share|improve this question

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have given up and written every node one by one. It works! The question is even more curious to me now.
  
  – ThoAppelsin
  5 hours ago
  

  
  
  
  

add a comment | 

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

share|improve this question

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

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}

share|improve this question

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

share|improve this question

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

asked 5 hours ago

ThoAppelsinThoAppelsin

325111

2 Answers
2

active

oldest

votes

2

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}

share|improve this answer

answered 4 hours ago

marmotmarmot

106k5129242

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
  
  – Sandy G
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  

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.

share|improve this answer

answered 3 hours ago

Christoph FringsChristoph Frings

901211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  54 mins ago
  
  
  

  
  
  
  

add a comment | 

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2

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}

share|improve this answer

answered 4 hours ago

marmotmarmot

106k5129242

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
  
  – Sandy G
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  

add a comment | 

2

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}

share|improve this answer

answered 4 hours ago

marmotmarmot

106k5129242

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably a good idea to fix the circle size so the 10 node isn't bigger than the others.
  
  – Sandy G
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  

add a comment | 

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}

share|improve this answer

answered 4 hours ago

marmotmarmot

106k5129242

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}

share|improve this answer

answered 4 hours ago

marmotmarmot

106k5129242

answered 4 hours ago

marmotmarmot

106k5129242

answered 4 hours ago

marmotmarmot

106k5129242

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.

share|improve this answer

answered 3 hours ago

Christoph FringsChristoph Frings

901211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  54 mins ago
  
  
  

  
  
  
  

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.

share|improve this answer

answered 3 hours ago

Christoph FringsChristoph Frings

901211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @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
  54 mins ago
  
  
  

  
  
  
  

add a comment | 

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.

share|improve this answer

answered 3 hours ago

Christoph FringsChristoph Frings

901211

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}

share|improve this answer

answered 3 hours ago

Christoph FringsChristoph Frings

901211

answered 3 hours ago

Christoph FringsChristoph Frings

901211

answered 3 hours ago

Christoph FringsChristoph Frings

901211