Number of triangle regions in a connected planar simple graph
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Suppose a connected planar simple graph G with v vertices such that all its regions are triangles (cycle with 3 edges).
I'm trying to figure out into how many regions does a representation of the planar graph G.
So here's my thinking:
If G is a connected planar simple graph, with e edges and v>=3 , then e<= 3v-6 . Plus if we add all degrees of all regions, the sum must be equal to 2e. Since we know that the regions are triangles, we know that the degree of each region is 3. Also, r <= e – v + 2
In the end, here's what we have :
r <= e – v + 2
r <= 3v –6 - v +2
r <= 2(v+1)
So r = 2(v+1)?
Something must be wrong in my reasoning.
Thanks by advance for your help.
graph-theory
asked Nov 24 at 10:42
Cleola
11
add a comment |
up vote
0
down vote
favorite
Suppose a connected planar simple graph G with v vertices such that all its regions are triangles (cycle with 3 edges).
I'm trying to figure out into how many regions does a representation of the planar graph G.
So here's my thinking:
If G is a connected planar simple graph, with e edges and v>=3 , then e<= 3v-6 . Plus if we add all degrees of all regions, the sum must be equal to 2e. Since we know that the regions are triangles, we know that the degree of each region is 3. Also, r <= e – v + 2
In the end, here's what we have :
r <= e – v + 2
r <= 3v –6 - v +2
r <= 2(v+1)
So r = 2(v+1)?
Something must be wrong in my reasoning.
Thanks by advance for your help.
graph-theory
asked Nov 24 at 10:42
Cleola
11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose a connected planar simple graph G with v vertices such that all its regions are triangles (cycle with 3 edges).
I'm trying to figure out into how many regions does a representation of the planar graph G.
So here's my thinking:
If G is a connected planar simple graph, with e edges and v>=3 , then e<= 3v-6 . Plus if we add all degrees of all regions, the sum must be equal to 2e. Since we know that the regions are triangles, we know that the degree of each region is 3. Also, r <= e – v + 2
In the end, here's what we have :
r <= e – v + 2
r <= 3v –6 - v +2
r <= 2(v+1)
So r = 2(v+1)?
Something must be wrong in my reasoning.
Thanks by advance for your help.
graph-theory
asked Nov 24 at 10:42
Cleola
11
Suppose a connected planar simple graph G with v vertices such that all its regions are triangles (cycle with 3 edges).
I'm trying to figure out into how many regions does a representation of the planar graph G.
So here's my thinking:
If G is a connected planar simple graph, with e edges and v>=3 , then e<= 3v-6 . Plus if we add all degrees of all regions, the sum must be equal to 2e. Since we know that the regions are triangles, we know that the degree of each region is 3. Also, r <= e – v + 2
In the end, here's what we have :
r <= e – v + 2
r <= 3v –6 - v +2
r <= 2(v+1)
So r = 2(v+1)?
Something must be wrong in my reasoning.
Thanks by advance for your help.
graph-theory
graph-theory
asked Nov 24 at 10:42
Cleola
11
asked Nov 24 at 10:42
Cleola
11
asked Nov 24 at 10:42
Cleola
11
asked Nov 24 at 10:42
Cleola
11
asked Nov 24 at 10:42
Cleola
11
11
add a comment |
add a comment |
1 Answer
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Let $n=|V|$, $e=|E|$, and the number of faces be $f$.
In a triangulation (a plane graph where every face boundary is a 3-cycle) $G$, we know that $e=3n-6$.
We substitute this value for $e$ into Euler's Formula for planar surfaces;
Euler's Formula
$n-e+f=2$
$n-(3n-6)+f=2$
$f=2-n+(3n-6)$
$f=2n-4$
answered 2 days ago
Steve Schroeder
1759
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $n=|V|$, $e=|E|$, and the number of faces be $f$.
In a triangulation (a plane graph where every face boundary is a 3-cycle) $G$, we know that $e=3n-6$.
We substitute this value for $e$ into Euler's Formula for planar surfaces;
Euler's Formula
$n-e+f=2$
$n-(3n-6)+f=2$
$f=2-n+(3n-6)$
$f=2n-4$
answered 2 days ago
Steve Schroeder
1759
add a comment |
up vote
0
down vote
Let $n=|V|$, $e=|E|$, and the number of faces be $f$.
In a triangulation (a plane graph where every face boundary is a 3-cycle) $G$, we know that $e=3n-6$.
We substitute this value for $e$ into Euler's Formula for planar surfaces;
Euler's Formula
$n-e+f=2$
$n-(3n-6)+f=2$
$f=2-n+(3n-6)$
$f=2n-4$
answered 2 days ago
Steve Schroeder
1759
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $n=|V|$, $e=|E|$, and the number of faces be $f$.
In a triangulation (a plane graph where every face boundary is a 3-cycle) $G$, we know that $e=3n-6$.
We substitute this value for $e$ into Euler's Formula for planar surfaces;
Euler's Formula
$n-e+f=2$
$n-(3n-6)+f=2$
$f=2-n+(3n-6)$
$f=2n-4$
answered 2 days ago
Steve Schroeder
1759
Let $n=|V|$, $e=|E|$, and the number of faces be $f$.
In a triangulation (a plane graph where every face boundary is a 3-cycle) $G$, we know that $e=3n-6$.
We substitute this value for $e$ into Euler's Formula for planar surfaces;
Euler's Formula
$n-e+f=2$
$n-(3n-6)+f=2$
$f=2-n+(3n-6)$
$f=2n-4$
answered 2 days ago
Steve Schroeder
1759
answered 2 days ago
Steve Schroeder
1759
answered 2 days ago
Steve Schroeder
1759
answered 2 days ago
Steve Schroeder
1759
1759
add a comment |
add a comment |
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