Can someone explain to me the significance of $e leq 3v-6$ in graph theory?
$begingroup$
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
edited Dec 5 '18 at 18:46
user376343
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
$endgroup$
add a comment |
$begingroup$
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
edited Dec 5 '18 at 18:46
user376343
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
$endgroup$
add a comment |
$begingroup$
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
edited Dec 5 '18 at 18:46
user376343
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
$endgroup$
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
edited Dec 5 '18 at 18:46
user376343
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
edited Dec 5 '18 at 18:46
user376343
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
edited Dec 5 '18 at 18:46
user376343
3,9234829
edited Dec 5 '18 at 18:46
user376343
3,9234829
edited Dec 5 '18 at 18:46
user376343
3,9234829
3,9234829
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
asked Dec 5 '18 at 18:26
DevAllanPerDevAllanPer
1336
1336
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
add a comment |
$begingroup$
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
$$
e leq 3v-6
$$
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
add a comment |
$begingroup$
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 $E leq 3V-6$.
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
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*}
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
$endgroup$
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
add a comment |
$begingroup$
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
add a comment |
$begingroup$
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
$endgroup$
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
answered Dec 5 '18 at 18:34
Robert IsraelRobert Israel
326k23215469
326k23215469
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
add a comment |
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
$endgroup$
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.
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
answered Dec 5 '18 at 18:34
Larry B.Larry B.
2,801828
2,801828
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
add a comment |
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
$begingroup$
You can shape the faces however you want, but $K_5$ will be non-planar no matter how you draw it.
$endgroup$
– Larry B.
Dec 5 '18 at 20:06
add a comment |
$begingroup$
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
$$
e leq 3v-6
$$
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
add a comment |
$begingroup$
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
$$
e leq 3v-6
$$
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
$endgroup$
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
add a comment |
$begingroup$
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
$$
e leq 3v-6
$$
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
$endgroup$
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
$$
e leq 3v-6
$$
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
answered Dec 5 '18 at 18:34
ArthurArthur
116k7116199
116k7116199
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
add a comment |
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
does this change at all if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:58
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
$begingroup$
@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.
$endgroup$
– Arthur
Dec 5 '18 at 19:56
add a comment |
$begingroup$
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 $E leq 3V-6$.
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
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 $E leq 3V-6$.
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
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 $E leq 3V-6$.
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
$endgroup$
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 $E leq 3V-6$.
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
answered Dec 5 '18 at 19:21
nafhgoodnafhgood
1,803422
1,803422
add a comment |
add a comment |
$begingroup$
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*}
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
$endgroup$
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
add a comment |
$begingroup$
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*}
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
$endgroup$
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
add a comment |
$begingroup$
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*}
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
$endgroup$
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*}
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
answered Dec 5 '18 at 18:35
jmerryjmerry
12k1628
12k1628
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
add a comment |
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
So does it change if the faces were per se, a pentagon?
$endgroup$
– DevAllanPer
Dec 5 '18 at 18:48
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
$begingroup$
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.
$endgroup$
– jmerry
Dec 5 '18 at 19:42
add a comment |
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