  
   

Say you have a 2D broken line you move along, but only some directions are allowed (I give you the angles relative to the usual cartesian plane):


1. (Up-Left): $]\pi, \dfrac{\pi}{2}[$
2. (Down-Left): $]-\pi, \dfrac{-\pi}{2}[$
3. (Down-Right): $ ]\dfrac{-\pi}{2}, 0[$


An additional rule is that you cannot go *Up-left if you just went Down-right*, and vice-versa.   

The goal is to prove that if you want to make a cycle out of this line, you always end up with a line that cross it-self.
  
  

**Context**   

I'm working on unit-square graphs with a clique number of 2. The directions correspond to which corner of some squares lies inside the other. I'm trying to define some order onto the squares.   

This is the last point of a proof that would allow the use of some representation of a graph, ending up with very easy to prove and useful small results.



Broken line that can go in specific directions: can it end up on its starting point?
There has to be a self-intersection; the path cannot be a simple polygon.


Suppose the path were indeed simple. A vertex with maximal $y$ coordinate must connect edges that go up and then down. According to the rules, the only way this can happen is first up-left, then down-left. [For a simple polygon](https://en.wikipedia.org/wiki/Curve_orientation#Orientation_of_a_simple_polygon), this implies the path is oriented counterclockwise. On the other hand a vertex with minimal $y$-coordinate must connect edges that go down-left, then up-left, implying the path is oriented clockwise: a contradiction.


