Is there a non-planar, non-hamiltonian and eulerian graph?
0
$begingroup$
I'm trying to find a graph that is non-planar, non-hamiltonian and eulerian but I can't find anyone.
Is this possible?
Thanks
discrete-mathematics graph-theory planar-graph hamiltonian-path
asked Dec 10 '18 at 0:33
emeeemee
43
$endgroup$
add a comment |
0
$begingroup$
I'm trying to find a graph that is non-planar, non-hamiltonian and eulerian but I can't find anyone.
Is this possible?
Thanks
discrete-mathematics graph-theory planar-graph hamiltonian-path
asked Dec 10 '18 at 0:33
emeeemee
43
$endgroup$
add a comment |
0
0
0
1
$begingroup$
I'm trying to find a graph that is non-planar, non-hamiltonian and eulerian but I can't find anyone.
Is this possible?
Thanks
discrete-mathematics graph-theory planar-graph hamiltonian-path
asked Dec 10 '18 at 0:33
emeeemee
43
$endgroup$
I'm trying to find a graph that is non-planar, non-hamiltonian and eulerian but I can't find anyone.
Is this possible?
Thanks
discrete-mathematics graph-theory planar-graph hamiltonian-path
discrete-mathematics graph-theory planar-graph hamiltonian-path
asked Dec 10 '18 at 0:33
emeeemee
43
asked Dec 10 '18 at 0:33
emeeemee
43
asked Dec 10 '18 at 0:33
emeeemee
43
asked Dec 10 '18 at 0:33
emeeemee
43
asked Dec 10 '18 at 0:33
emeeemee
43
43
add a comment |
add a comment |
1 Answer
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Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
$endgroup$
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
$endgroup$
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
add a comment |
4
$begingroup$
Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
$endgroup$
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
add a comment |
4
4
4
$begingroup$
Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
$endgroup$
Take two copies of $K_5$ and identify one of the nodes in one with one of the nodes in the other.
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
answered Dec 10 '18 at 0:44
Henning MakholmHenning Makholm
242k17308550
242k17308550
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
add a comment |
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
$begingroup$
But K5 is 4-regular. Eulerian graphs has even degrees. If I join two vertex, this will have odd degree.
$endgroup$
– emee
Dec 10 '18 at 0:55
1
1
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
@emee: i'm not saying to connect the two vertices with an edge. I'm saying to combine them into one vertex. The combined vertex will have degree 8.
$endgroup$
– Henning Makholm
Dec 10 '18 at 0:57
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
Sorry but I don't undestand. d(v) + d(u) = 2.8 > 5 [Ore's theorem]
$endgroup$
– emee
Dec 12 '18 at 22:24
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
@emee: What is $v$ and $u$? The condition in Ore's theorem talks about pairs of distinct vertices. The graph I describe has eight vertices of degree 4 and one vertex of degree 8, so how do you get $2cdot 8$ for even one pair of vertices?
$endgroup$
– Henning Makholm
Dec 12 '18 at 22:44
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
$begingroup$
I get it. Thanks so much. I was drawing another graph
$endgroup$
– emee
Dec 12 '18 at 22:47
add a comment |
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