Mutually disjoint triangles in certain planar graph












2














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 days ago










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










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
















2














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 days ago










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










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














2












2








2







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 days ago

























asked 2 days ago









Finallysignedup

666




666












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










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










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


















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










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










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
















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




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












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




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












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




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










2 Answers
2






active

oldest

votes


















4














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
    2 days 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
    2 days ago










  • Understandable, it is the fault of my laziness in creating diagrams.
    – Zachary Hunter
    2 days 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









4














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
    2 days ago
















4














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
    2 days ago














4












4








4






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








edited 2 days ago

























answered 2 days ago









Misha Lavrov

43.9k555104




43.9k555104












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


















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
















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




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










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










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









Zachary Hunter

3568




3568




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
    2 days ago










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


















  • Thanks. Sorry that I could not tick both answers.
    – Finallysignedup
    2 days ago










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
















Thanks. Sorry that I could not tick both answers.
– Finallysignedup
2 days ago




Thanks. Sorry that I could not tick both answers.
– Finallysignedup
2 days ago












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




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


















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