Prove that the graph dual to Eulerian planar graph is bipartite.












5












$begingroup$


How would I go about doing this proof I am not very knowledgeable about graph theory I know the definitions of planar and bipartite and dual but how do you make these connection










share|cite|improve this question











$endgroup$












  • $begingroup$
    It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:00
















5












$begingroup$


How would I go about doing this proof I am not very knowledgeable about graph theory I know the definitions of planar and bipartite and dual but how do you make these connection










share|cite|improve this question











$endgroup$












  • $begingroup$
    It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:00














5












5








5


1



$begingroup$


How would I go about doing this proof I am not very knowledgeable about graph theory I know the definitions of planar and bipartite and dual but how do you make these connection










share|cite|improve this question











$endgroup$




How would I go about doing this proof I am not very knowledgeable about graph theory I know the definitions of planar and bipartite and dual but how do you make these connection







graph-theory planar-graph eulerian-path






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 6 '15 at 6:40









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jul 5 '15 at 20:00









Fernando MartinezFernando Martinez

3,332104281




3,332104281












  • $begingroup$
    It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:00


















  • $begingroup$
    It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:00
















$begingroup$
It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
$endgroup$
– Jorge Fernández
Jul 5 '15 at 22:00




$begingroup$
It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
$endgroup$
– Jorge Fernández
Jul 5 '15 at 22:00










2 Answers
2






active

oldest

votes


















3












$begingroup$

So $G$ is planar and eulerian. We must prove $G'$ is bipartite. Asume $G'$ is not bipartite. Now I want you to forget about the fact that $G'$ is the dual of $G$. Just think of $G'$ as a normal graph in which the vertices of $G'$ are drawn as vertices and not as the faces of $G$.



Since $G'$ is not bipartite it has an odd cycle, one of the faces inside that odd cycle must therefore have an odd number of edges. That face is a vertex of odd degree in $G''$,so $G''$ is not eulerian. Now,$Gcong G''$ so $G$ is not eulerian, a contradiction. The contradiction comes from assuming $G'$ is not bipartite.



A key step is the fact $Gcong G''$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:30



















0












$begingroup$

Let $G$ be a planar Eulerian graph, and consider a particular planar diagram of $G$. There is an Eulerian circuit that does not pass "through" vertices, but instead always takes a hard left or right turn.



Acceptable turns in the Eulerian circuit



While there is probably some argument involving depth-first traversals, one can also construct such an Eulerian circuit by starting with any Eulerian circuit then performing the following kind of local modification to the circuit whenever the circuit crosses over itself (which happens exactly when the circuit fails to make a hard turn).



Circuit surgery



If we think of the circuit as being a curve in the plane, we can push it slightly away from the vertices to get an embedded closed curve. By the Jordan Curve Theorem, this curve separates the plane into two components. The dual graph's vertices are partitioned into two sets depending on which side of the circuit the vertex lies on, and each edge of the dual graph crosses the circuit transversely in one point. Hence, the dual graph is a planar bipartite graph.



Here is a worked example of the dual of on octahedral graph, with the blue curve being the pushed-off embedded Eulerian circuit, and with the cyan and green vertices representing the two partitions of the bipartite graph:



Bipartite dual of an octahedral graph



For the converse, you can take small blue circles around each green vertex to get a multicurve. Then around each vertex in the dual graph you can join them in such a way to get an Eulerian circuit of the dual. Or just note that each face in a planar bipartite graph has an even number of sides.



(It seems there should be a way to think about all of this in terms of (co)homology with $mathbb{Z}/2mathbb{Z}$ coefficients and Poincaré duality.)





A possible simplification to the above argument could be that Eulerian implies each vertex has even degree. Then, one can replace each vertex of degree $2n$ with $n$ vertices of degree $2$ in a "starburst" pattern. This is a planar graph composed entirely of circles. By the Jordan curve theorem you can $2$-color the regions between the circles, meaning the regions on either side of a circle have opposite colors. The coloring partitions the vertices of the dual graph into two parts, and again edges cross the circles, so the dual is bipartite.



This is rehashing a proof that the dual of a planar graph with vertices of only even degree can be $2$-colored. For example the shadow of a knot diagram.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1350552%2fprove-that-the-graph-dual-to-eulerian-planar-graph-is-bipartite%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    So $G$ is planar and eulerian. We must prove $G'$ is bipartite. Asume $G'$ is not bipartite. Now I want you to forget about the fact that $G'$ is the dual of $G$. Just think of $G'$ as a normal graph in which the vertices of $G'$ are drawn as vertices and not as the faces of $G$.



    Since $G'$ is not bipartite it has an odd cycle, one of the faces inside that odd cycle must therefore have an odd number of edges. That face is a vertex of odd degree in $G''$,so $G''$ is not eulerian. Now,$Gcong G''$ so $G$ is not eulerian, a contradiction. The contradiction comes from assuming $G'$ is not bipartite.



    A key step is the fact $Gcong G''$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
      $endgroup$
      – Jorge Fernández
      Jul 5 '15 at 22:30
















    3












    $begingroup$

    So $G$ is planar and eulerian. We must prove $G'$ is bipartite. Asume $G'$ is not bipartite. Now I want you to forget about the fact that $G'$ is the dual of $G$. Just think of $G'$ as a normal graph in which the vertices of $G'$ are drawn as vertices and not as the faces of $G$.



    Since $G'$ is not bipartite it has an odd cycle, one of the faces inside that odd cycle must therefore have an odd number of edges. That face is a vertex of odd degree in $G''$,so $G''$ is not eulerian. Now,$Gcong G''$ so $G$ is not eulerian, a contradiction. The contradiction comes from assuming $G'$ is not bipartite.



    A key step is the fact $Gcong G''$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
      $endgroup$
      – Jorge Fernández
      Jul 5 '15 at 22:30














    3












    3








    3





    $begingroup$

    So $G$ is planar and eulerian. We must prove $G'$ is bipartite. Asume $G'$ is not bipartite. Now I want you to forget about the fact that $G'$ is the dual of $G$. Just think of $G'$ as a normal graph in which the vertices of $G'$ are drawn as vertices and not as the faces of $G$.



    Since $G'$ is not bipartite it has an odd cycle, one of the faces inside that odd cycle must therefore have an odd number of edges. That face is a vertex of odd degree in $G''$,so $G''$ is not eulerian. Now,$Gcong G''$ so $G$ is not eulerian, a contradiction. The contradiction comes from assuming $G'$ is not bipartite.



    A key step is the fact $Gcong G''$.






    share|cite|improve this answer











    $endgroup$



    So $G$ is planar and eulerian. We must prove $G'$ is bipartite. Asume $G'$ is not bipartite. Now I want you to forget about the fact that $G'$ is the dual of $G$. Just think of $G'$ as a normal graph in which the vertices of $G'$ are drawn as vertices and not as the faces of $G$.



    Since $G'$ is not bipartite it has an odd cycle, one of the faces inside that odd cycle must therefore have an odd number of edges. That face is a vertex of odd degree in $G''$,so $G''$ is not eulerian. Now,$Gcong G''$ so $G$ is not eulerian, a contradiction. The contradiction comes from assuming $G'$ is not bipartite.



    A key step is the fact $Gcong G''$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 7 '15 at 10:38

























    answered Jul 5 '15 at 22:27









    Jorge FernándezJorge Fernández

    75.2k1190192




    75.2k1190192












    • $begingroup$
      $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
      $endgroup$
      – Jorge Fernández
      Jul 5 '15 at 22:30


















    • $begingroup$
      $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
      $endgroup$
      – Jorge Fernández
      Jul 5 '15 at 22:30
















    $begingroup$
    $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:30




    $begingroup$
    $G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
    $endgroup$
    – Jorge Fernández
    Jul 5 '15 at 22:30











    0












    $begingroup$

    Let $G$ be a planar Eulerian graph, and consider a particular planar diagram of $G$. There is an Eulerian circuit that does not pass "through" vertices, but instead always takes a hard left or right turn.



    Acceptable turns in the Eulerian circuit



    While there is probably some argument involving depth-first traversals, one can also construct such an Eulerian circuit by starting with any Eulerian circuit then performing the following kind of local modification to the circuit whenever the circuit crosses over itself (which happens exactly when the circuit fails to make a hard turn).



    Circuit surgery



    If we think of the circuit as being a curve in the plane, we can push it slightly away from the vertices to get an embedded closed curve. By the Jordan Curve Theorem, this curve separates the plane into two components. The dual graph's vertices are partitioned into two sets depending on which side of the circuit the vertex lies on, and each edge of the dual graph crosses the circuit transversely in one point. Hence, the dual graph is a planar bipartite graph.



    Here is a worked example of the dual of on octahedral graph, with the blue curve being the pushed-off embedded Eulerian circuit, and with the cyan and green vertices representing the two partitions of the bipartite graph:



    Bipartite dual of an octahedral graph



    For the converse, you can take small blue circles around each green vertex to get a multicurve. Then around each vertex in the dual graph you can join them in such a way to get an Eulerian circuit of the dual. Or just note that each face in a planar bipartite graph has an even number of sides.



    (It seems there should be a way to think about all of this in terms of (co)homology with $mathbb{Z}/2mathbb{Z}$ coefficients and Poincaré duality.)





    A possible simplification to the above argument could be that Eulerian implies each vertex has even degree. Then, one can replace each vertex of degree $2n$ with $n$ vertices of degree $2$ in a "starburst" pattern. This is a planar graph composed entirely of circles. By the Jordan curve theorem you can $2$-color the regions between the circles, meaning the regions on either side of a circle have opposite colors. The coloring partitions the vertices of the dual graph into two parts, and again edges cross the circles, so the dual is bipartite.



    This is rehashing a proof that the dual of a planar graph with vertices of only even degree can be $2$-colored. For example the shadow of a knot diagram.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $G$ be a planar Eulerian graph, and consider a particular planar diagram of $G$. There is an Eulerian circuit that does not pass "through" vertices, but instead always takes a hard left or right turn.



      Acceptable turns in the Eulerian circuit



      While there is probably some argument involving depth-first traversals, one can also construct such an Eulerian circuit by starting with any Eulerian circuit then performing the following kind of local modification to the circuit whenever the circuit crosses over itself (which happens exactly when the circuit fails to make a hard turn).



      Circuit surgery



      If we think of the circuit as being a curve in the plane, we can push it slightly away from the vertices to get an embedded closed curve. By the Jordan Curve Theorem, this curve separates the plane into two components. The dual graph's vertices are partitioned into two sets depending on which side of the circuit the vertex lies on, and each edge of the dual graph crosses the circuit transversely in one point. Hence, the dual graph is a planar bipartite graph.



      Here is a worked example of the dual of on octahedral graph, with the blue curve being the pushed-off embedded Eulerian circuit, and with the cyan and green vertices representing the two partitions of the bipartite graph:



      Bipartite dual of an octahedral graph



      For the converse, you can take small blue circles around each green vertex to get a multicurve. Then around each vertex in the dual graph you can join them in such a way to get an Eulerian circuit of the dual. Or just note that each face in a planar bipartite graph has an even number of sides.



      (It seems there should be a way to think about all of this in terms of (co)homology with $mathbb{Z}/2mathbb{Z}$ coefficients and Poincaré duality.)





      A possible simplification to the above argument could be that Eulerian implies each vertex has even degree. Then, one can replace each vertex of degree $2n$ with $n$ vertices of degree $2$ in a "starburst" pattern. This is a planar graph composed entirely of circles. By the Jordan curve theorem you can $2$-color the regions between the circles, meaning the regions on either side of a circle have opposite colors. The coloring partitions the vertices of the dual graph into two parts, and again edges cross the circles, so the dual is bipartite.



      This is rehashing a proof that the dual of a planar graph with vertices of only even degree can be $2$-colored. For example the shadow of a knot diagram.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $G$ be a planar Eulerian graph, and consider a particular planar diagram of $G$. There is an Eulerian circuit that does not pass "through" vertices, but instead always takes a hard left or right turn.



        Acceptable turns in the Eulerian circuit



        While there is probably some argument involving depth-first traversals, one can also construct such an Eulerian circuit by starting with any Eulerian circuit then performing the following kind of local modification to the circuit whenever the circuit crosses over itself (which happens exactly when the circuit fails to make a hard turn).



        Circuit surgery



        If we think of the circuit as being a curve in the plane, we can push it slightly away from the vertices to get an embedded closed curve. By the Jordan Curve Theorem, this curve separates the plane into two components. The dual graph's vertices are partitioned into two sets depending on which side of the circuit the vertex lies on, and each edge of the dual graph crosses the circuit transversely in one point. Hence, the dual graph is a planar bipartite graph.



        Here is a worked example of the dual of on octahedral graph, with the blue curve being the pushed-off embedded Eulerian circuit, and with the cyan and green vertices representing the two partitions of the bipartite graph:



        Bipartite dual of an octahedral graph



        For the converse, you can take small blue circles around each green vertex to get a multicurve. Then around each vertex in the dual graph you can join them in such a way to get an Eulerian circuit of the dual. Or just note that each face in a planar bipartite graph has an even number of sides.



        (It seems there should be a way to think about all of this in terms of (co)homology with $mathbb{Z}/2mathbb{Z}$ coefficients and Poincaré duality.)





        A possible simplification to the above argument could be that Eulerian implies each vertex has even degree. Then, one can replace each vertex of degree $2n$ with $n$ vertices of degree $2$ in a "starburst" pattern. This is a planar graph composed entirely of circles. By the Jordan curve theorem you can $2$-color the regions between the circles, meaning the regions on either side of a circle have opposite colors. The coloring partitions the vertices of the dual graph into two parts, and again edges cross the circles, so the dual is bipartite.



        This is rehashing a proof that the dual of a planar graph with vertices of only even degree can be $2$-colored. For example the shadow of a knot diagram.






        share|cite|improve this answer











        $endgroup$



        Let $G$ be a planar Eulerian graph, and consider a particular planar diagram of $G$. There is an Eulerian circuit that does not pass "through" vertices, but instead always takes a hard left or right turn.



        Acceptable turns in the Eulerian circuit



        While there is probably some argument involving depth-first traversals, one can also construct such an Eulerian circuit by starting with any Eulerian circuit then performing the following kind of local modification to the circuit whenever the circuit crosses over itself (which happens exactly when the circuit fails to make a hard turn).



        Circuit surgery



        If we think of the circuit as being a curve in the plane, we can push it slightly away from the vertices to get an embedded closed curve. By the Jordan Curve Theorem, this curve separates the plane into two components. The dual graph's vertices are partitioned into two sets depending on which side of the circuit the vertex lies on, and each edge of the dual graph crosses the circuit transversely in one point. Hence, the dual graph is a planar bipartite graph.



        Here is a worked example of the dual of on octahedral graph, with the blue curve being the pushed-off embedded Eulerian circuit, and with the cyan and green vertices representing the two partitions of the bipartite graph:



        Bipartite dual of an octahedral graph



        For the converse, you can take small blue circles around each green vertex to get a multicurve. Then around each vertex in the dual graph you can join them in such a way to get an Eulerian circuit of the dual. Or just note that each face in a planar bipartite graph has an even number of sides.



        (It seems there should be a way to think about all of this in terms of (co)homology with $mathbb{Z}/2mathbb{Z}$ coefficients and Poincaré duality.)





        A possible simplification to the above argument could be that Eulerian implies each vertex has even degree. Then, one can replace each vertex of degree $2n$ with $n$ vertices of degree $2$ in a "starburst" pattern. This is a planar graph composed entirely of circles. By the Jordan curve theorem you can $2$-color the regions between the circles, meaning the regions on either side of a circle have opposite colors. The coloring partitions the vertices of the dual graph into two parts, and again edges cross the circles, so the dual is bipartite.



        This is rehashing a proof that the dual of a planar graph with vertices of only even degree can be $2$-colored. For example the shadow of a knot diagram.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 22:28

























        answered Dec 3 '18 at 22:18









        Kyle MillerKyle Miller

        8,462928




        8,462928






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1350552%2fprove-that-the-graph-dual-to-eulerian-planar-graph-is-bipartite%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix