How to prove Tait's theorem about planar cubic bridgeless graph being 3-edge-colorable?












2














How can be proved, that



The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.



I can't figure out or find any prove of this theorem.



Thanks.






graph-theory coloring







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    2














    How can be proved, that



    The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.



    I can't figure out or find any prove of this theorem.



    Thanks.






    graph-theory coloring







    share|cite|improve this question

























      2












      2








      2


      1





      How can be proved, that



      The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.



      I can't figure out or find any prove of this theorem.



      Thanks.






      graph-theory coloring







      share|cite|improve this question













      How can be proved, that



      The four-color theorem is equivalent to the claim that every planar cubic bridgeless graph is 3-edge-colorable.



      I can't figure out or find any prove of this theorem.



      Thanks.






      graph-theory coloring




      graph-theory coloring






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 1 at 14:19









      Přemysl Šťastný

      1595




      1595






















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          The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.



          A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.



          Since the four color problem has been already proved, also the three edge coloring is true.






          share|cite|improve this answer





























            1














            I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:



            $$(a, b) equiv (c,d) equiv 1$$
            $$(a, c) equiv (b,d) equiv 2$$
            $$(a, d) equiv (b,c) equiv 3$$



            In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.






            share|cite|improve this answer








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              2 Answers
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              2 Answers
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              active

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              active

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              active

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              1














              The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.



              A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.



              Since the four color problem has been already proved, also the three edge coloring is true.






              share|cite|improve this answer


























                1














                The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.



                A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.



                Since the four color problem has been already proved, also the three edge coloring is true.






                share|cite|improve this answer
























                  1












                  1








                  1






                  The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.



                  A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.



                  Since the four color problem has been already proved, also the three edge coloring is true.






                  share|cite|improve this answer












                  The two problems of three-edge-coloring and four-face-coloring for the same map are equivalent.



                  A proof of this equivalency can be round here: http://www.mathpuzzle.com/4Dec2001.htm. Search for "material added 19 November 2001" within the page.



                  Since the four color problem has been already proved, also the three edge coloring is true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 23:59









                  Mario Stefanutti

                  607613




                  607613























                      1














                      I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:



                      $$(a, b) equiv (c,d) equiv 1$$
                      $$(a, c) equiv (b,d) equiv 2$$
                      $$(a, d) equiv (b,c) equiv 3$$



                      In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.






                      share|cite|improve this answer








                      New contributor




                      Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.























                        1














                        I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:



                        $$(a, b) equiv (c,d) equiv 1$$
                        $$(a, c) equiv (b,d) equiv 2$$
                        $$(a, d) equiv (b,c) equiv 3$$



                        In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.






                        share|cite|improve this answer








                        New contributor




                        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.





















                          1












                          1








                          1






                          I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:



                          $$(a, b) equiv (c,d) equiv 1$$
                          $$(a, c) equiv (b,d) equiv 2$$
                          $$(a, d) equiv (b,c) equiv 3$$



                          In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.






                          share|cite|improve this answer








                          New contributor




                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          I'd wondered about this before, but never worked it out until now. If $G$ is a cubic bridge-less graph, it is polyhedral and its dual is a maximal planar graph. Consider a maximal planar dual, $D(G)$ colored by the colors $(a, b, c, d$). Now color each edge, $e_{ij}$, according to $(i,j)$ like so:



                          $$(a, b) equiv (c,d) equiv 1$$
                          $$(a, c) equiv (b,d) equiv 2$$
                          $$(a, d) equiv (b,c) equiv 3$$



                          In each triangle, no two edges can possible have the same color, because that require all four vertex colors. Thus, in $G$, the edges all have different colors around every vertex. Since the edge coloring is valid around every vertex, it is valid for the whole graph $G$.







                          share|cite|improve this answer








                          New contributor




                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|cite|improve this answer



                          share|cite|improve this answer






                          New contributor




                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered 2 days ago









                          Zachary Hunter

                          6019




                          6019




                          New contributor




                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          New contributor





                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






























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