Any regular polygon you choose (with $n>3$ sides) as a "house", the highest number $(r)$ of "roofs" that you can drawd (without lift the pen or pass two times on the same segment,inscribing a coross inside the polygon) is: $ \ \ r=n-1 \ \ $
Can it be mathematically proven?
I made this sort of "conjecture" when I was a child... Me and my friends used to play a game that consisted in drawing a house whith a cross inside (without lift the pen or pass two times on the same segment)
The maximum number of roofs is indeed $n-1$, and rests on graph theory.
The drawing is an example of an Eulerian path. Suppose we have an $n$-sided polygon with $n$ roofs. Without the inside cross, all the vertices have even degree and an Eulerian cycle/path is possible. With the cross, however, four vertices become odd-degree and no Eulerian path is possible.
If the cross is drawn such that two of the vertices it touches are adjacent along the polygon, removing the roof touching those two adjacent vertices makes those vertices of even degree, causing the whole diagram to have an Eulerian path (but not cycle) again. Furthermore, the path's endpoints will be the other two vertices of the cross, which remain of odd degree.