Triangular grid with 4 edges per vertex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:54:49Z http://mathoverflow.net/feeds/question/105157 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105157/triangular-grid-with-4-edges-per-vertex Triangular grid with 4 edges per vertex halvor 2012-08-21T13:18:44Z 2012-08-21T19:16:48Z <p>I am trying to create a triangular grid/mesh for a rectangular domain in $\mathbb{R}^2$ with the property that each vertex is shared by (at most) four edges. Is this possible to accomplish?</p> http://mathoverflow.net/questions/105157/triangular-grid-with-4-edges-per-vertex/105175#105175 Answer by Kevin Walker for Triangular grid with 4 edges per vertex Kevin Walker 2012-08-21T16:23:57Z 2012-08-21T16:23:57Z <p>This question is too elementary for MO, but here's a hint. (See the FAQ for alternative sites to ask your question.)</p> <ol> <li><p>Familiarize yourself with the notion of Euler characteristic</p></li> <li><p>Convince yourself that the Euler characteristic of your triangulation of a rectangle is 1.</p></li> <li><p>Deduce some inequalities on the numbers of vertices, edges and faces, assuming all faces are triangles and all vertex valences are less than or equal to 4.</p></li> </ol> http://mathoverflow.net/questions/105157/triangular-grid-with-4-edges-per-vertex/105182#105182 Answer by Johan Wästlund for Triangular grid with 4 edges per vertex Johan Wästlund 2012-08-21T18:45:12Z 2012-08-21T19:16:48Z <p>Since there is no requirement that the outer region must be a triangle, the question is not quite as trivial as indicated in earlier comments. The rectangle $[0,n] \times [0,1]$ can be triangulated by dividing it into unit squares and then inserting the SW-NE diagonal in each square. Still, this might not be the kind of grid/mesh one wants. To see the problem, it might be easier to think in terms of angles than to use Euler's polyhedron formula: If there are interior points in the triangulation, then the angles at those points have to be at least $90^\circ$ on average, while the average angle in a triangle is only $60^\circ$. It follows that most vertices of the triangulation have to be on the boundary of the region. </p>