This is the problem I am dealing with:

Given a set P of n points in general position, let a graph G be defined as follows:

The vertex set is P. Two vertices, a and b, are joined by an edge provided there exists an axis parallel square S with a and b on the boundary and no other point of P in the interior of S.

I need to prove or disprove that G is a near-triangulation where every face except the outerface is a triangle and the outerface is a cycle.

Thanks in advance

  • $\begingroup$ Could you define exactly what a near triangulation is? $\endgroup$ – Tassle Dec 9 '19 at 22:07
  • $\begingroup$ By triangulating any polygon, all the faces except the outer face becomes a triangle. This is called Near-Triangulation. Now suppose the polygon itself is a triangle, then by triangulation, all the faces including the outer face is a triangle. This is perfect triangulation. $\endgroup$ – AlgorithmUser785 Dec 10 '19 at 1:36
  • $\begingroup$ I think a picture will worth a thousand words $\endgroup$ – HEKTO Dec 10 '19 at 4:38

An easy contradicting example is three points forming an obtuse triangle. Note that any Axis-parallel square containing $A$ and $B$ on its boundaries, contains $C$ in its interior and hence, the outer face is not bounded by a cycle since $AB$ is not an edge in the graph.

Obtuse triangle


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