Mutually disjoint triangles in certain planar graph












1














Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







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  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago
















1














Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







share|cite|improve this question
























  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago














1












1








1







Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory







share|cite|improve this question















Let G be a connected, planar graph for which every vertex has degree 3, except that one vertex has degree 2.



Is it possible to construct an example of such G whose only odd faces are triangles, and for which no two such triangles share a common vertex?



(For my purposes I may assume there are no faces of length 1 or 2, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)






combinatorics discrete-mathematics graph-theory




combinatorics discrete-mathematics graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago

























asked 2 hours ago









Finallysignedup

495




495












  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago


















  • For my own edification.. what is an odd face ?
    – T. Ford
    2 hours ago










  • Face with odd number of sides.
    – Zachary Hunter
    2 hours ago










  • I think "those bounded by an odd length cycle" is a sensible definition.
    – Finallysignedup
    2 hours ago
















For my own edification.. what is an odd face ?
– T. Ford
2 hours ago




For my own edification.. what is an odd face ?
– T. Ford
2 hours ago












Face with odd number of sides.
– Zachary Hunter
2 hours ago




Face with odd number of sides.
– Zachary Hunter
2 hours ago












I think "those bounded by an odd length cycle" is a sensible definition.
– Finallysignedup
2 hours ago




I think "those bounded by an odd length cycle" is a sensible definition.
– Finallysignedup
2 hours ago










2 Answers
2






active

oldest

votes


















2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago



















2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago
















2














Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer





















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago














2












2








2






Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)






share|cite|improve this answer












Here is one such graph - or, more precisely, one such plane embedding of a graph, since the lengths of the faces are not properties of the graph itself. It has two faces of length 3 (including the external face), two faces of length 4, and two faces of length 6.



enter image description here



(Motivation: we start with a triangular pyramid, which has all the required properties except for the degree-2 vertex, and modify it a little to make it work.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Misha Lavrov

43.8k555104




43.8k555104












  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago


















  • Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
    – Finallysignedup
    1 hour ago
















Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
– Finallysignedup
1 hour ago




Thanks. I now wish to only have degree 3 vertices. I've made a new (hopefully final) question.
– Finallysignedup
1 hour ago











2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago
















2














Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago














2












2








2






Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.






share|cite|improve this answer








New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









Take two cycles of the same order 2n+1, $C_a$ and $C_b$ and number each in clockwise fashion.



"Glue together" $v_{1a}$ and $v_{1b}$, and then do the same for $v_{2a}$ and $v_{2b}$, so that there is just one edge between $v_1$ and $v_2$.



Then, for each $i > 2$ create an edge between $v_{ia}$ and $v_{ib}$.



Finally, subdivide the edge between $v_1$ and $v_2$.







share|cite|improve this answer








New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 2 hours ago









Zachary Hunter

1245




1245




New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    1 hour ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    1 hour ago
















Thanks. Sorry that I could not tick both answers.
– Finallysignedup
1 hour ago




Thanks. Sorry that I could not tick both answers.
– Finallysignedup
1 hour ago












Understandable, it is the fault of my laziness in creating diagrams.
– Zachary Hunter
1 hour ago




Understandable, it is the fault of my laziness in creating diagrams.
– Zachary Hunter
1 hour ago


















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