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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

Wonderful2019
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1 Answers1

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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

Narek Bojikian
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