Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.












2












$begingroup$



Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.




Attempt:



With handshaking lemma, I get this:



$2e = 26 implies e=13$



Then with Euler's formula, I get:



$6-13+f=2 implies f = 9$



However, since $e leq 3v - 6$ for a simple, connected, planar graph I would get:



$13 leq 3(6)-6$



$13 leq 12$



I can't figure out how I could get 9 faces for my graph. I can only get 8 faces as shown here:



enter image description here



This is the best attempt I had on this problem with no success.






combinatorics graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Euler's formula counts the external face. Your graph has 9.
    $endgroup$
    – Misha Lavrov
    Dec 9 '18 at 2:50










  • $begingroup$
    3 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 9 '18 at 10:54










  • $begingroup$
    I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
    $endgroup$
    – bof
    Dec 9 '18 at 10:59
















2












$begingroup$



Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.




Attempt:



With handshaking lemma, I get this:



$2e = 26 implies e=13$



Then with Euler's formula, I get:



$6-13+f=2 implies f = 9$



However, since $e leq 3v - 6$ for a simple, connected, planar graph I would get:



$13 leq 3(6)-6$



$13 leq 12$



I can't figure out how I could get 9 faces for my graph. I can only get 8 faces as shown here:



enter image description here



This is the best attempt I had on this problem with no success.






combinatorics graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Euler's formula counts the external face. Your graph has 9.
    $endgroup$
    – Misha Lavrov
    Dec 9 '18 at 2:50










  • $begingroup$
    3 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 9 '18 at 10:54










  • $begingroup$
    I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
    $endgroup$
    – bof
    Dec 9 '18 at 10:59














2












2








2





$begingroup$



Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.




Attempt:



With handshaking lemma, I get this:



$2e = 26 implies e=13$



Then with Euler's formula, I get:



$6-13+f=2 implies f = 9$



However, since $e leq 3v - 6$ for a simple, connected, planar graph I would get:



$13 leq 3(6)-6$



$13 leq 12$



I can't figure out how I could get 9 faces for my graph. I can only get 8 faces as shown here:



enter image description here



This is the best attempt I had on this problem with no success.






combinatorics graph-theory







share|cite|improve this question











$endgroup$





Draw a planar graph with two vertices of degree 3 and four vertices of degree 5, if possible.




Attempt:



With handshaking lemma, I get this:



$2e = 26 implies e=13$



Then with Euler's formula, I get:



$6-13+f=2 implies f = 9$



However, since $e leq 3v - 6$ for a simple, connected, planar graph I would get:



$13 leq 3(6)-6$



$13 leq 12$



I can't figure out how I could get 9 faces for my graph. I can only get 8 faces as shown here:



enter image description here



This is the best attempt I had on this problem with no success.






combinatorics graph-theory




combinatorics graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 9:32









Maria Mazur

47.4k1260120




47.4k1260120










asked Dec 9 '18 at 0:36









cosmicbrowniecosmicbrownie

1016




1016












  • $begingroup$
    Euler's formula counts the external face. Your graph has 9.
    $endgroup$
    – Misha Lavrov
    Dec 9 '18 at 2:50










  • $begingroup$
    3 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 9 '18 at 10:54










  • $begingroup$
    I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
    $endgroup$
    – bof
    Dec 9 '18 at 10:59


















  • $begingroup$
    Euler's formula counts the external face. Your graph has 9.
    $endgroup$
    – Misha Lavrov
    Dec 9 '18 at 2:50










  • $begingroup$
    3 minutes. $ $ $ $
    $endgroup$
    – Did
    Dec 9 '18 at 10:54










  • $begingroup$
    I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
    $endgroup$
    – bof
    Dec 9 '18 at 10:59
















$begingroup$
Euler's formula counts the external face. Your graph has 9.
$endgroup$
– Misha Lavrov
Dec 9 '18 at 2:50




$begingroup$
Euler's formula counts the external face. Your graph has 9.
$endgroup$
– Misha Lavrov
Dec 9 '18 at 2:50












$begingroup$
3 minutes. $ $ $ $
$endgroup$
– Did
Dec 9 '18 at 10:54




$begingroup$
3 minutes. $ $ $ $
$endgroup$
– Did
Dec 9 '18 at 10:54












$begingroup$
I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
$endgroup$
– bof
Dec 9 '18 at 10:59




$begingroup$
I had no trouble drawing a simple connected planar graph with two vertices of degree $3$ and four vertices of degree $5$. It is a tree of order $22$. (Was there some other condition you didn't mention?)
$endgroup$
– bof
Dec 9 '18 at 10:59










2 Answers
2






active

oldest

votes


















3












$begingroup$

Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @greedoid 5. As is mentioned in my post.
    $endgroup$
    – Did
    Dec 9 '18 at 11:03












  • $begingroup$
    I don't understand. What have I proved then?
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 11:05










  • $begingroup$
    @greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
    $endgroup$
    – Did
    Dec 9 '18 at 11:15





















2












$begingroup$

From formula you wrote $eleq 3v-6$ you can see that such a (planar) graph does not exist.



Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.




Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also



$$ 4,4,4,2,2implies 3,3,1,1implies 2,0,0$$
but last one clearly it is not graphicaly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
    $endgroup$
    – cosmicbrownie
    Dec 9 '18 at 0:47










  • $begingroup$
    Yes it is, but your graph is clearly connected, it has to many edges to not be connected
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 0:48











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @greedoid 5. As is mentioned in my post.
    $endgroup$
    – Did
    Dec 9 '18 at 11:03












  • $begingroup$
    I don't understand. What have I proved then?
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 11:05










  • $begingroup$
    @greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
    $endgroup$
    – Did
    Dec 9 '18 at 11:15


















3












$begingroup$

Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    @greedoid 5. As is mentioned in my post.
    $endgroup$
    – Did
    Dec 9 '18 at 11:03












  • $begingroup$
    I don't understand. What have I proved then?
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 11:05










  • $begingroup$
    @greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
    $endgroup$
    – Did
    Dec 9 '18 at 11:15
















3












3








3





$begingroup$

Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.



enter image description here






share|cite|improve this answer











$endgroup$



Here is a planar example. The vertices A and B are linked by two simple edges. The vertices A, B, E and F have degree 5. The vertices C and D have degree 3.



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 10:57

























answered Dec 9 '18 at 10:35









DidDid

248k23225464




248k23225464








  • 1




    $begingroup$
    @greedoid 5. As is mentioned in my post.
    $endgroup$
    – Did
    Dec 9 '18 at 11:03












  • $begingroup$
    I don't understand. What have I proved then?
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 11:05










  • $begingroup$
    @greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
    $endgroup$
    – Did
    Dec 9 '18 at 11:15
















  • 1




    $begingroup$
    @greedoid 5. As is mentioned in my post.
    $endgroup$
    – Did
    Dec 9 '18 at 11:03












  • $begingroup$
    I don't understand. What have I proved then?
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 11:05










  • $begingroup$
    @greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
    $endgroup$
    – Did
    Dec 9 '18 at 11:15










1




1




$begingroup$
@greedoid 5. As is mentioned in my post.
$endgroup$
– Did
Dec 9 '18 at 11:03






$begingroup$
@greedoid 5. As is mentioned in my post.
$endgroup$
– Did
Dec 9 '18 at 11:03














$begingroup$
I don't understand. What have I proved then?
$endgroup$
– Maria Mazur
Dec 9 '18 at 11:05




$begingroup$
I don't understand. What have I proved then?
$endgroup$
– Maria Mazur
Dec 9 '18 at 11:05












$begingroup$
@greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
$endgroup$
– Did
Dec 9 '18 at 11:15






$begingroup$
@greedoid That no graph such that, between two given vertices, there are either zero or one edge, solves the question. And I provided a graph solving the question such that, between two given vertices, there are either zero or one or two edges (when one admits more than one edge, these are often called multi-edged graphs). No mathematical mystery here. (But the reason why the OP, who is obviously considering graphs with multi-edges, chose to instantly accept an answer dealing only with graphs without multi-edge, is a true mystery, yes.)
$endgroup$
– Did
Dec 9 '18 at 11:15













2












$begingroup$

From formula you wrote $eleq 3v-6$ you can see that such a (planar) graph does not exist.



Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.




Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also



$$ 4,4,4,2,2implies 3,3,1,1implies 2,0,0$$
but last one clearly it is not graphicaly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
    $endgroup$
    – cosmicbrownie
    Dec 9 '18 at 0:47










  • $begingroup$
    Yes it is, but your graph is clearly connected, it has to many edges to not be connected
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 0:48
















2












$begingroup$

From formula you wrote $eleq 3v-6$ you can see that such a (planar) graph does not exist.



Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.




Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also



$$ 4,4,4,2,2implies 3,3,1,1implies 2,0,0$$
but last one clearly it is not graphicaly.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
    $endgroup$
    – cosmicbrownie
    Dec 9 '18 at 0:47










  • $begingroup$
    Yes it is, but your graph is clearly connected, it has to many edges to not be connected
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 0:48














2












2








2





$begingroup$

From formula you wrote $eleq 3v-6$ you can see that such a (planar) graph does not exist.



Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.




Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also



$$ 4,4,4,2,2implies 3,3,1,1implies 2,0,0$$
but last one clearly it is not graphicaly.






share|cite|improve this answer











$endgroup$



From formula you wrote $eleq 3v-6$ you can see that such a (planar) graph does not exist.



Also you could note that this graph contains $K_{3,3}$ so again it can not be planar.




Edit: Actualy this graph doesn't even exist since the sequence $5,5,5,5,3,3$ is not graphicaly. If it is, then following would be also



$$ 4,4,4,2,2implies 3,3,1,1implies 2,0,0$$
but last one clearly it is not graphicaly.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 0:51

























answered Dec 9 '18 at 0:44









Maria MazurMaria Mazur

47.4k1260120




47.4k1260120












  • $begingroup$
    Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
    $endgroup$
    – cosmicbrownie
    Dec 9 '18 at 0:47










  • $begingroup$
    Yes it is, but your graph is clearly connected, it has to many edges to not be connected
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 0:48


















  • $begingroup$
    Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
    $endgroup$
    – cosmicbrownie
    Dec 9 '18 at 0:47










  • $begingroup$
    Yes it is, but your graph is clearly connected, it has to many edges to not be connected
    $endgroup$
    – Maria Mazur
    Dec 9 '18 at 0:48
















$begingroup$
Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
$endgroup$
– cosmicbrownie
Dec 9 '18 at 0:47




$begingroup$
Can you still get a planar graph even if it's not a simple, connected one based on the formula? I thought that formula was just for simple, connected planar graphs.
$endgroup$
– cosmicbrownie
Dec 9 '18 at 0:47












$begingroup$
Yes it is, but your graph is clearly connected, it has to many edges to not be connected
$endgroup$
– Maria Mazur
Dec 9 '18 at 0:48




$begingroup$
Yes it is, but your graph is clearly connected, it has to many edges to not be connected
$endgroup$
– Maria Mazur
Dec 9 '18 at 0:48


















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