﻿ Solving for The Illumination Problem - OOQQ

# Solving for the Illumination Problem 🔦

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The illumination problem is a resolved mathematical problem first posed by Ernst Straus in the 1950s. Straus asked if a room with mirrored walls can always be illuminated by a single point light source, allowing for repeated reflection of light off the mirrored walls. Is there a shape possible such that there is a point where it is impossible to illuminate, assuming the reflections are infinite?.
Wikipedia

### Solving for areas of non-polygonal rooms

In 1958 Roger Penrose used the properties of the ellipse to describe a room with curved walls that would always have dark regions, regardless of the position of the source. Penrose unilluminable room (unilluminated regions are indicated in gray). My own solution; Let's assume you have a circle where the light source lies at the geometrical centre of the circle. As a consequence, the rays from the source will irradiate outwards, hit the wall, and bounce back with a reflection angle of 0º.

Let's create another circle, that is tangentially external to the first one.

Since the room must be connected, let's join the two circles at the tangent point and make them a single shape. (open points are indicated with a hollow spot).

As long as the light source remains at one of the two geometric centers, the entire area of the other circle will always be in complete darkness, except for the single ray crossing the figure at the tangent point. Since the circle nor the ellipse aren't regular polygons, the two demonstrations accomplish the same: That for a non-polygonal figure it may could be spots and regions that will not be illuminated.

### The Tokarsky Paper

When dealing with regular polygons in the Illumination Problem, the general rules are:

• The angle of incidence equals the angle of reflection.
• Rays of light always have length (nonzero).
• A ray reaching a vertice does not reflect.

George W. Tokarsky in a 1995 paper titled Polygonal Rooms Not Illuminable from Every Point demonstrated there exists an unilluminable spot on a polygonal 26-sided room. A dark spot which is not illuminable from the light soure in the room, even allowing for repeated reflections. ### Solving for points of polygonal rooms

Now I will explain in layman terms the main points (heh) in Tokarsky demonstration: In a right triangle, if the rules are to be followed, there's no such ray that will origin at vertice A that can bounce back to vertice A. Therefore It's easy to see that if there's a orthographic grid where any point A is always surrounded by B points. As long as all surrounding B points are vertices, then there's no possible connection between any two points A.

As Tokarsky explains, any triangle cannot form a path that emanates from point A and came back to point A if:

• The angle at the origin is an integer (a rational angle).
• The other two vertices B and C are different of 180º.

Therefore any triangle that follows this rules, if mirrored along the oposite side of the light source (BC Segment) will create a new delta shape. On this delta shape, light coming from vertice A cannot reach vertice D (or in the orthographic projection, another point A). This delta construction will come in handy for the next step of this demonstration. ### Solving for areas of polygonal rooms

Lemma: If in a polygon there's a point unreacheable for the light source, and such point is the only connection between two areas, then the entire area beyond the unreacheable point will also be unreacheable.

Proof: At this point I'd like to re-introduce my two circles demonstration. But this time using the previous delta shape instead. A new shape can be created using two of these deltas joined together at the open vertice D.

And I'm sure that figured out already what you'll get: So if the Tokarsky demonstration of a dark spot is to be trusted, then this demonstration proves that there's at least a special case where a polygon can posess infinite dark spots and therefore dark areas. Which was previously considered to be impossible. (Lelièvre, Monteil & Weiss. 2016)

Q.E.D.

As a sidenote, Tokarsky questions what could be the polygon with the smallest number of sides that contains a dark spot not placed on a vertice. This demonstration brings down the previous polygon sides from 24 (David Castro, 1997) to 8. But on my example the most distant vertice from the light source can also be removed, so the minimum number of sides in a polygon that contains a dark spot (or area) is 7.