Can someone explain to me the significance of $e leq 3v-6$ in graph theory?











up vote
-1
down vote

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I'm studying for a final and my textbook often uses the equation
$$
e le 3v-6
$$
(seems to be a theory or corollary) for some of the graph theory proofs, but I can't find anywhere as to where this equation is derived from therefore making it hard for me to attempt to use it as part of a solution.




Could someone please tell me how it's derived and why it's significant?




Much Thanks!






combinatorics graph-theory







share|cite|improve this question




























    up vote
    -1
    down vote

    favorite












    I'm studying for a final and my textbook often uses the equation
    $$
    e le 3v-6
    $$
    (seems to be a theory or corollary) for some of the graph theory proofs, but I can't find anywhere as to where this equation is derived from therefore making it hard for me to attempt to use it as part of a solution.




    Could someone please tell me how it's derived and why it's significant?




    Much Thanks!






    combinatorics graph-theory







    share|cite|improve this question


























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I'm studying for a final and my textbook often uses the equation
      $$
      e le 3v-6
      $$
      (seems to be a theory or corollary) for some of the graph theory proofs, but I can't find anywhere as to where this equation is derived from therefore making it hard for me to attempt to use it as part of a solution.




      Could someone please tell me how it's derived and why it's significant?




      Much Thanks!






      combinatorics graph-theory







      share|cite|improve this question















      I'm studying for a final and my textbook often uses the equation
      $$
      e le 3v-6
      $$
      (seems to be a theory or corollary) for some of the graph theory proofs, but I can't find anywhere as to where this equation is derived from therefore making it hard for me to attempt to use it as part of a solution.




      Could someone please tell me how it's derived and why it's significant?




      Much Thanks!






      combinatorics graph-theory




      combinatorics graph-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 at 18:46









      user376343

      2,7782822




      2,7782822










      asked Dec 5 at 18:26









      DevAllanPer

      1296




      1296






















          5 Answers
          5






          active

          oldest

          votes

















          up vote
          3
          down vote













          For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
          $e le 3 v - 6$. See Wikipedia.






          share|cite|improve this answer





















          • does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58


















          up vote
          2
          down vote













          This is only applicable to planar graphs, for instance $K_5$ has $5$ vertices but $10 > 15 - 6$ edges. This answer will give you a proof to this common theorem.






          share|cite|improve this answer





















          • does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58










          • You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
            – Larry B.
            Dec 5 at 20:06


















          up vote
          1
          down vote













          This formula is valid specifically for planar graphs. If a graph with $3$ or more vertices is planar, and $e$ is the number of edges and $v$ the number of vertices of that graph, then it is true that
          $$
          e leq 3v-6
          $$






          share|cite|improve this answer





















          • does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58










          • @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
            – Arthur
            Dec 5 at 19:56




















          up vote
          1
          down vote













          Let $G$ be a planar graph and let $F$ denote the number of faces and $E$ denote the number of edges, and $V$ denote the number of vertices. Euler's formula says that $V-E+F=2$ which means $F=2+E-V$.
          On the other hand, the length of perimeter surrounding a face is at least $3$. So then if we sum up the length of each face, which is twice number of edges, we get:
          $6+3E-3V=3(2+E-V)=F*3 leq sum_{f in face(G)}{len(f)}=2*E$. which means $E leq 3V-6$.






          share|cite|improve this answer




























            up vote
            0
            down vote













            That's the number of edges in a planar graph with all faces triangles, and thus the highest possible number of edges in a planar graph.
            begin{align*}V+F &= E+2\
            V+frac23 E &= E+2\
            V-2 &= frac13E\
            3V-6 &= Eend{align*}






            share|cite|improve this answer





















            • So does it change if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:48










            • Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
              – jmerry
              Dec 5 at 19:42











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            5 Answers
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            active

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            5 Answers
            5






            active

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            active

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            active

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            up vote
            3
            down vote













            For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
            $e le 3 v - 6$. See Wikipedia.






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58















            up vote
            3
            down vote













            For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
            $e le 3 v - 6$. See Wikipedia.






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58













            up vote
            3
            down vote










            up vote
            3
            down vote









            For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
            $e le 3 v - 6$. See Wikipedia.






            share|cite|improve this answer












            For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
            $e le 3 v - 6$. See Wikipedia.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 at 18:34









            Robert Israel

            317k23206457




            317k23206457












            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58


















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58
















            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58




            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58










            up vote
            2
            down vote













            This is only applicable to planar graphs, for instance $K_5$ has $5$ vertices but $10 > 15 - 6$ edges. This answer will give you a proof to this common theorem.






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
              – Larry B.
              Dec 5 at 20:06















            up vote
            2
            down vote













            This is only applicable to planar graphs, for instance $K_5$ has $5$ vertices but $10 > 15 - 6$ edges. This answer will give you a proof to this common theorem.






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
              – Larry B.
              Dec 5 at 20:06













            up vote
            2
            down vote










            up vote
            2
            down vote









            This is only applicable to planar graphs, for instance $K_5$ has $5$ vertices but $10 > 15 - 6$ edges. This answer will give you a proof to this common theorem.






            share|cite|improve this answer












            This is only applicable to planar graphs, for instance $K_5$ has $5$ vertices but $10 > 15 - 6$ edges. This answer will give you a proof to this common theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 at 18:34









            Larry B.

            2,676727




            2,676727












            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
              – Larry B.
              Dec 5 at 20:06


















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
              – Larry B.
              Dec 5 at 20:06
















            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58




            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58












            You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
            – Larry B.
            Dec 5 at 20:06




            You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
            – Larry B.
            Dec 5 at 20:06










            up vote
            1
            down vote













            This formula is valid specifically for planar graphs. If a graph with $3$ or more vertices is planar, and $e$ is the number of edges and $v$ the number of vertices of that graph, then it is true that
            $$
            e leq 3v-6
            $$






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
              – Arthur
              Dec 5 at 19:56

















            up vote
            1
            down vote













            This formula is valid specifically for planar graphs. If a graph with $3$ or more vertices is planar, and $e$ is the number of edges and $v$ the number of vertices of that graph, then it is true that
            $$
            e leq 3v-6
            $$






            share|cite|improve this answer





















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
              – Arthur
              Dec 5 at 19:56















            up vote
            1
            down vote










            up vote
            1
            down vote









            This formula is valid specifically for planar graphs. If a graph with $3$ or more vertices is planar, and $e$ is the number of edges and $v$ the number of vertices of that graph, then it is true that
            $$
            e leq 3v-6
            $$






            share|cite|improve this answer












            This formula is valid specifically for planar graphs. If a graph with $3$ or more vertices is planar, and $e$ is the number of edges and $v$ the number of vertices of that graph, then it is true that
            $$
            e leq 3v-6
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 5 at 18:34









            Arthur

            110k7104186




            110k7104186












            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
              – Arthur
              Dec 5 at 19:56




















            • does this change at all if the faces were per se, a pentagon?
              – DevAllanPer
              Dec 5 at 18:58










            • @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
              – Arthur
              Dec 5 at 19:56


















            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58




            does this change at all if the faces were per se, a pentagon?
            – DevAllanPer
            Dec 5 at 18:58












            @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
            – Arthur
            Dec 5 at 19:56






            @DevAllanPer You won't get equality if there are pentagonal faces (note that each pentagonal face can be made into three triangular faces by adding two edges between the vertices). So it could be strengthened. To what? I don't know. Something like (I'm just guessing wildly here) $eleq v + 3$, perhaps? A closer look with Euler's formula will probably reveal an actual answer.
            – Arthur
            Dec 5 at 19:56












            up vote
            1
            down vote













            Let $G$ be a planar graph and let $F$ denote the number of faces and $E$ denote the number of edges, and $V$ denote the number of vertices. Euler's formula says that $V-E+F=2$ which means $F=2+E-V$.
            On the other hand, the length of perimeter surrounding a face is at least $3$. So then if we sum up the length of each face, which is twice number of edges, we get:
            $6+3E-3V=3(2+E-V)=F*3 leq sum_{f in face(G)}{len(f)}=2*E$. which means $E leq 3V-6$.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Let $G$ be a planar graph and let $F$ denote the number of faces and $E$ denote the number of edges, and $V$ denote the number of vertices. Euler's formula says that $V-E+F=2$ which means $F=2+E-V$.
              On the other hand, the length of perimeter surrounding a face is at least $3$. So then if we sum up the length of each face, which is twice number of edges, we get:
              $6+3E-3V=3(2+E-V)=F*3 leq sum_{f in face(G)}{len(f)}=2*E$. which means $E leq 3V-6$.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Let $G$ be a planar graph and let $F$ denote the number of faces and $E$ denote the number of edges, and $V$ denote the number of vertices. Euler's formula says that $V-E+F=2$ which means $F=2+E-V$.
                On the other hand, the length of perimeter surrounding a face is at least $3$. So then if we sum up the length of each face, which is twice number of edges, we get:
                $6+3E-3V=3(2+E-V)=F*3 leq sum_{f in face(G)}{len(f)}=2*E$. which means $E leq 3V-6$.






                share|cite|improve this answer












                Let $G$ be a planar graph and let $F$ denote the number of faces and $E$ denote the number of edges, and $V$ denote the number of vertices. Euler's formula says that $V-E+F=2$ which means $F=2+E-V$.
                On the other hand, the length of perimeter surrounding a face is at least $3$. So then if we sum up the length of each face, which is twice number of edges, we get:
                $6+3E-3V=3(2+E-V)=F*3 leq sum_{f in face(G)}{len(f)}=2*E$. which means $E leq 3V-6$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 5 at 19:21









                mathnoob

                1,763322




                1,763322






















                    up vote
                    0
                    down vote













                    That's the number of edges in a planar graph with all faces triangles, and thus the highest possible number of edges in a planar graph.
                    begin{align*}V+F &= E+2\
                    V+frac23 E &= E+2\
                    V-2 &= frac13E\
                    3V-6 &= Eend{align*}






                    share|cite|improve this answer





















                    • So does it change if the faces were per se, a pentagon?
                      – DevAllanPer
                      Dec 5 at 18:48










                    • Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                      – jmerry
                      Dec 5 at 19:42















                    up vote
                    0
                    down vote













                    That's the number of edges in a planar graph with all faces triangles, and thus the highest possible number of edges in a planar graph.
                    begin{align*}V+F &= E+2\
                    V+frac23 E &= E+2\
                    V-2 &= frac13E\
                    3V-6 &= Eend{align*}






                    share|cite|improve this answer





















                    • So does it change if the faces were per se, a pentagon?
                      – DevAllanPer
                      Dec 5 at 18:48










                    • Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                      – jmerry
                      Dec 5 at 19:42













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    That's the number of edges in a planar graph with all faces triangles, and thus the highest possible number of edges in a planar graph.
                    begin{align*}V+F &= E+2\
                    V+frac23 E &= E+2\
                    V-2 &= frac13E\
                    3V-6 &= Eend{align*}






                    share|cite|improve this answer












                    That's the number of edges in a planar graph with all faces triangles, and thus the highest possible number of edges in a planar graph.
                    begin{align*}V+F &= E+2\
                    V+frac23 E &= E+2\
                    V-2 &= frac13E\
                    3V-6 &= Eend{align*}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 5 at 18:35









                    jmerry

                    6915




                    6915












                    • So does it change if the faces were per se, a pentagon?
                      – DevAllanPer
                      Dec 5 at 18:48










                    • Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                      – jmerry
                      Dec 5 at 19:42


















                    • So does it change if the faces were per se, a pentagon?
                      – DevAllanPer
                      Dec 5 at 18:48










                    • Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                      – jmerry
                      Dec 5 at 19:42
















                    So does it change if the faces were per se, a pentagon?
                    – DevAllanPer
                    Dec 5 at 18:48




                    So does it change if the faces were per se, a pentagon?
                    – DevAllanPer
                    Dec 5 at 18:48












                    Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                    – jmerry
                    Dec 5 at 19:42




                    Faces with more edges lead to fewer total edges. Of course, we could always draw in new edges in each face, cutting them up until only triangles remain.
                    – jmerry
                    Dec 5 at 19:42


















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                    Cabo Verde

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                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                    Gyllenstierna

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