Can someone explain to me the significance of $e leq 3v-6$ in graph theory?
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
edited Dec 5 at 18:46
user376343
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
add a comment |
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
edited Dec 5 at 18:46
user376343
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
add a comment |
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
edited Dec 5 at 18:46
user376343
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
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 at 18:46
user376343
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
edited Dec 5 at 18:46
user376343
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
edited Dec 5 at 18:46
user376343
2,7782822
edited Dec 5 at 18:46
user376343
2,7782822
edited Dec 5 at 18:46
user376343
2,7782822
2,7782822
asked Dec 5 at 18:26
DevAllanPer
1296
asked Dec 5 at 18:26
DevAllanPer
1296
asked Dec 5 at 18:26
DevAllanPer
1296
1296
add a comment |
add a comment |
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.
answered Dec 5 at 18:34
Robert Israel
317k23206457
does this change at all if the faces were per se, a pentagon?
– DevAllanPer
Dec 5 at 18:58
add a comment |
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.
answered Dec 5 at 18:34
Larry B.
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
add a comment |
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
$$
e leq 3v-6
$$
answered Dec 5 at 18:34
Arthur
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
add a comment |
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 $E leq 3V-6$.
answered Dec 5 at 19:21
mathnoob
1,763322
add a comment |
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*}
answered Dec 5 at 18:35
jmerry
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
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
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.
answered Dec 5 at 18:34
Robert Israel
317k23206457
does this change at all if the faces were per se, a pentagon?
– DevAllanPer
Dec 5 at 18:58
add a comment |
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.
answered Dec 5 at 18:34
Robert Israel
317k23206457
does this change at all if the faces were per se, a pentagon?
– DevAllanPer
Dec 5 at 18:58
add a comment |
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.
answered Dec 5 at 18:34
Robert Israel
317k23206457
For a simple, connected, planar graph with $v ge 3$ vertices and $e$ edges,
$e le 3 v - 6$. See Wikipedia.
answered Dec 5 at 18:34
Robert Israel
317k23206457
answered Dec 5 at 18:34
Robert Israel
317k23206457
answered Dec 5 at 18:34
Robert Israel
317k23206457
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
add a comment |
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
add a comment |
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.
answered Dec 5 at 18:34
Larry B.
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
add a comment |
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.
answered Dec 5 at 18:34
Larry B.
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
add a comment |
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.
answered Dec 5 at 18:34
Larry B.
2,676727
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 at 18:34
Larry B.
2,676727
answered Dec 5 at 18:34
Larry B.
2,676727
answered Dec 5 at 18:34
Larry B.
2,676727
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
add a comment |
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
add a comment |
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
$$
e leq 3v-6
$$
answered Dec 5 at 18:34
Arthur
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
add a comment |
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
$$
e leq 3v-6
$$
answered Dec 5 at 18:34
Arthur
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
add a comment |
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
$$
e leq 3v-6
$$
answered Dec 5 at 18:34
Arthur
110k7104186
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 at 18:34
Arthur
110k7104186
answered Dec 5 at 18:34
Arthur
110k7104186
answered Dec 5 at 18:34
Arthur
110k7104186
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
add a comment |
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
add a comment |
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 $E leq 3V-6$.
answered Dec 5 at 19:21
mathnoob
1,763322
add a comment |
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 $E leq 3V-6$.
answered Dec 5 at 19:21
mathnoob
1,763322
add a comment |
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 $E leq 3V-6$.
answered Dec 5 at 19:21
mathnoob
1,763322
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 at 19:21
mathnoob
1,763322
answered Dec 5 at 19:21
mathnoob
1,763322
answered Dec 5 at 19:21
mathnoob
1,763322
answered Dec 5 at 19:21
mathnoob
1,763322
1,763322
add a comment |
add a comment |
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*}
answered Dec 5 at 18:35
jmerry
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
add a comment |
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*}
answered Dec 5 at 18:35
jmerry
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
add a comment |
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*}
answered Dec 5 at 18:35
jmerry
6915
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 at 18:35
jmerry
6915
answered Dec 5 at 18:35
jmerry
6915
answered Dec 5 at 18:35
jmerry
6915
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
add a comment |
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
add a comment |
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