Mixing
Degree of a region in a planar graph
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I have from my notes that they claim in a planar graph,
$$2|E| = text{sum of all degrees of regions}$$
where $|E|$ is the cardinality of the edge-set of the graph
They say this because each edge in the graph contributes twice to the degree of the region.
However, I can't see this in the following example:
The inner region seems to have degree $4$, however the outer region seems to have degree $5$. (from edges AB,BC,CD,DE,EB)
graph-theory
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
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add a comment |
$begingroup$
I have from my notes that they claim in a planar graph,
$$2|E| = text{sum of all degrees of regions}$$
where $|E|$ is the cardinality of the edge-set of the graph
They say this because each edge in the graph contributes twice to the degree of the region.
However, I can't see this in the following example:
The inner region seems to have degree $4$, however the outer region seems to have degree $5$. (from edges AB,BC,CD,DE,EB)
graph-theory
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
$endgroup$
add a comment |
$begingroup$
I have from my notes that they claim in a planar graph,
$$2|E| = text{sum of all degrees of regions}$$
where $|E|$ is the cardinality of the edge-set of the graph
They say this because each edge in the graph contributes twice to the degree of the region.
However, I can't see this in the following example:
The inner region seems to have degree $4$, however the outer region seems to have degree $5$. (from edges AB,BC,CD,DE,EB)
graph-theory
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
$endgroup$
I have from my notes that they claim in a planar graph,
$$2|E| = text{sum of all degrees of regions}$$
where $|E|$ is the cardinality of the edge-set of the graph
They say this because each edge in the graph contributes twice to the degree of the region.
However, I can't see this in the following example:
The inner region seems to have degree $4$, however the outer region seems to have degree $5$. (from edges AB,BC,CD,DE,EB)
graph-theory
graph-theory
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
asked Oct 22 '17 at 4:11
Natash1Natash1
619213
619213
add a comment |
add a comment |
1 Answer
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$begingroup$
For this identity to hold, we need to count the edge $AB$ twice, on the basis that as you go around the boundary of the outer face, you trace that edge twice: once going from $A$ to $B$ and once from $B$ to $A$.
If we follow this convention, then the inner region (blue) has degree $4$ while the outer region (red) has degree $6$, and everything works out.
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
For this identity to hold, we need to count the edge $AB$ twice, on the basis that as you go around the boundary of the outer face, you trace that edge twice: once going from $A$ to $B$ and once from $B$ to $A$.
If we follow this convention, then the inner region (blue) has degree $4$ while the outer region (red) has degree $6$, and everything works out.
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
$endgroup$
add a comment |
$begingroup$
For this identity to hold, we need to count the edge $AB$ twice, on the basis that as you go around the boundary of the outer face, you trace that edge twice: once going from $A$ to $B$ and once from $B$ to $A$.
If we follow this convention, then the inner region (blue) has degree $4$ while the outer region (red) has degree $6$, and everything works out.
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
$endgroup$
add a comment |
$begingroup$
For this identity to hold, we need to count the edge $AB$ twice, on the basis that as you go around the boundary of the outer face, you trace that edge twice: once going from $A$ to $B$ and once from $B$ to $A$.
If we follow this convention, then the inner region (blue) has degree $4$ while the outer region (red) has degree $6$, and everything works out.
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
$endgroup$
For this identity to hold, we need to count the edge $AB$ twice, on the basis that as you go around the boundary of the outer face, you trace that edge twice: once going from $A$ to $B$ and once from $B$ to $A$.
If we follow this convention, then the inner region (blue) has degree $4$ while the outer region (red) has degree $6$, and everything works out.
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
answered Oct 22 '17 at 4:18
Misha LavrovMisha Lavrov
47.5k657107
47.5k657107
add a comment |
add a comment |
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