Euler's formula and graph duality
$begingroup$
I am confused with this video on YouTube.
In the graph attached, the edge taken by the Randolph (the blue pi creature) forms a spanning tree and the remaining edge (colored in red) is taken by Mortimer (the orange pi creature).
The video state these two points:
- (Number of Randolph's Edges) + 1 = V
- (Number of Mortimer's Edges) + 1 = F
I understand why "(Number of Randolph's Edges) + 1 = V". It's because an edge requires two vertices. Starting from a single vertice, you need another vertice to draw an edge (so we have a single branch of edge=1, and vertices=2). We can expand this tree by connecting a new vertice to any existing vertice (provided that it doesn't form a cycle), and by connecting these two vertices, we can add one vertice and one edge to the total. Each time we add a branch, we keep doing the same thing; the total will always result in one more vertice than edges.
What I am confused about is how do you get "(Number of Mortimer's edge) + 1 = F"? In the video, it says that "the number of edges he gets is one more than the number of vertices of the dual graph, which are faces cut out by the original graph." I'm counting the number of edges, and I get Number of Mortimer's edge = 7, which is the number of faces in the planar graph. So where am I going wrong?
Here is the video link:
Euler's Formula and Graph Duality
duality-theorems eulerian-path
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
$endgroup$
add a comment |
$begingroup$
I am confused with this video on YouTube.
In the graph attached, the edge taken by the Randolph (the blue pi creature) forms a spanning tree and the remaining edge (colored in red) is taken by Mortimer (the orange pi creature).
The video state these two points:
- (Number of Randolph's Edges) + 1 = V
- (Number of Mortimer's Edges) + 1 = F
I understand why "(Number of Randolph's Edges) + 1 = V". It's because an edge requires two vertices. Starting from a single vertice, you need another vertice to draw an edge (so we have a single branch of edge=1, and vertices=2). We can expand this tree by connecting a new vertice to any existing vertice (provided that it doesn't form a cycle), and by connecting these two vertices, we can add one vertice and one edge to the total. Each time we add a branch, we keep doing the same thing; the total will always result in one more vertice than edges.
What I am confused about is how do you get "(Number of Mortimer's edge) + 1 = F"? In the video, it says that "the number of edges he gets is one more than the number of vertices of the dual graph, which are faces cut out by the original graph." I'm counting the number of edges, and I get Number of Mortimer's edge = 7, which is the number of faces in the planar graph. So where am I going wrong?
Here is the video link:
Euler's Formula and Graph Duality
duality-theorems eulerian-path
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
$endgroup$
add a comment |
$begingroup$
I am confused with this video on YouTube.
In the graph attached, the edge taken by the Randolph (the blue pi creature) forms a spanning tree and the remaining edge (colored in red) is taken by Mortimer (the orange pi creature).
The video state these two points:
- (Number of Randolph's Edges) + 1 = V
- (Number of Mortimer's Edges) + 1 = F
I understand why "(Number of Randolph's Edges) + 1 = V". It's because an edge requires two vertices. Starting from a single vertice, you need another vertice to draw an edge (so we have a single branch of edge=1, and vertices=2). We can expand this tree by connecting a new vertice to any existing vertice (provided that it doesn't form a cycle), and by connecting these two vertices, we can add one vertice and one edge to the total. Each time we add a branch, we keep doing the same thing; the total will always result in one more vertice than edges.
What I am confused about is how do you get "(Number of Mortimer's edge) + 1 = F"? In the video, it says that "the number of edges he gets is one more than the number of vertices of the dual graph, which are faces cut out by the original graph." I'm counting the number of edges, and I get Number of Mortimer's edge = 7, which is the number of faces in the planar graph. So where am I going wrong?
Here is the video link:
Euler's Formula and Graph Duality
duality-theorems eulerian-path
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
$endgroup$
I am confused with this video on YouTube.
In the graph attached, the edge taken by the Randolph (the blue pi creature) forms a spanning tree and the remaining edge (colored in red) is taken by Mortimer (the orange pi creature).
The video state these two points:
- (Number of Randolph's Edges) + 1 = V
- (Number of Mortimer's Edges) + 1 = F
I understand why "(Number of Randolph's Edges) + 1 = V". It's because an edge requires two vertices. Starting from a single vertice, you need another vertice to draw an edge (so we have a single branch of edge=1, and vertices=2). We can expand this tree by connecting a new vertice to any existing vertice (provided that it doesn't form a cycle), and by connecting these two vertices, we can add one vertice and one edge to the total. Each time we add a branch, we keep doing the same thing; the total will always result in one more vertice than edges.
What I am confused about is how do you get "(Number of Mortimer's edge) + 1 = F"? In the video, it says that "the number of edges he gets is one more than the number of vertices of the dual graph, which are faces cut out by the original graph." I'm counting the number of edges, and I get Number of Mortimer's edge = 7, which is the number of faces in the planar graph. So where am I going wrong?
Here is the video link:
Euler's Formula and Graph Duality
duality-theorems eulerian-path
duality-theorems eulerian-path
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
asked Dec 20 '18 at 8:34
supmethodssupmethods
72
72
add a comment |
add a comment |
1 Answer
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I suppose you forgot one face - the unbounded "outer" face. Things become clearer when playing on a sphere instead of in the plane
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
$endgroup$
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I suppose you forgot one face - the unbounded "outer" face. Things become clearer when playing on a sphere instead of in the plane
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
$endgroup$
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
add a comment |
$begingroup$
I suppose you forgot one face - the unbounded "outer" face. Things become clearer when playing on a sphere instead of in the plane
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
$endgroup$
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
add a comment |
$begingroup$
I suppose you forgot one face - the unbounded "outer" face. Things become clearer when playing on a sphere instead of in the plane
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
$endgroup$
I suppose you forgot one face - the unbounded "outer" face. Things become clearer when playing on a sphere instead of in the plane
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
answered Dec 20 '18 at 8:54
Hagen von EitzenHagen von Eitzen
1843
1843
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
add a comment |
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
$begingroup$
Rather odd to have an unbounded "outer" face, but I am assuming it has its purpose and applications. Graph theory itself seems quite intensive, I think I'll require a lot more learning prior to tackling these types of videos.
$endgroup$
– supmethods
Dec 20 '18 at 11:42
add a comment |
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