Diffuse reflection diameter in simple polygons

Gill Barequet, Sarah M. Cannon, Eli Fox-Epstein, Benjamin Hescott, Diane L. Souvaine, Csaba D. Tóth, Andrew Winslow
2016 Discrete Applied Mathematics  
We prove a conjecture of Aanjaneya, Bishnu, and Pal that the minimum number of diffuse reflections sufficient to illuminate the interior of any simple polygon with n walls from any interior point light source is n/2 − 1. Light reflecting diffusely leaves a surface in all directions, rather than at an identical angle as with specular reflections. . For every k ∈ N 0 , (1) We have µ k+1 ≥ min(µ k + 1, n); and (2) If all windows of R k are saturated, then µ k+1 ≥ min(µ k + 2, n). Proof. Note that
more » ... f cl(R k ) = P , then R k has at least one window and λ k ≥ 1. Counting Weakly Covered Edges in R k Let P be a simple polygon with n vertices, and let s ∈ int(P ). In this section, we establish the inequality for all k ∈ N 0 , which immediately implies Theorem 1. It is folklore that V 0 (s) weakly covers at least three edges, hence µ 0 ≥ 3. Proposition 3. If s ∈ int(P ), then V 0 (s) weakly covers at least three edges of P . Consequently, µ 0 ≥ 3. Proof. In any triangulation of P , s lies in some triangle whose vertices partition the edges of ∂P into three sets. At least one edge is seen by s in each of the three sets. We prove ( ) for all k ∈ N 0 by induction on k. Recall that R 0 satisfies ( ) by Proposition 3, and µ k strictly monotonically increases until it reaches n by Corollary 2. Consequently, if ( ) fails for some R k+1 , k ∈ N 0 , then R k must satisfy ( ) with equality, and µ k < n. This motivates the following definition. A region R k is called critical if µ k = 2k + 3 and µ k < n. By Lemma 1, it is enough to show that whenever R k is critical, then λ k ≥ 2 or all windows of R k are saturated. For every critical region R k , we will inductively show (Lemma 2(3)) that one of the following two conditions holds: (A) All windows of R k are saturated; or (B) R k has an unsaturated window and λ k ≥ 2. Note that these two conditions are mutually exclusive, that is, a region R k cannot satisfy both. Initialization. We first show that R 0 satisfies one of the two conditions. Proposition 4. Region R 0 satisfies condition (A) or (B). Proof. First, suppose λ 0 ≥ 2. Then either all windows are saturated so (A) holds, or at least one window is unsaturated and (B) holds. Next, suppose λ 0 = 1. If the single window of R 0 is degenerate, then (A) holds. Assume that R 0 has exactly one window that is nondegenerate, denoted by ab as defined in (ii). Recall that R 0 = V 0 (s), and so every point in ∂R 0 is contained in a window or directly visible from s. Consequently, all points in ∂R 0 \ ab are in R 0 . As ab splits P into R 0 and U ab , every chord of P that crosses ab has exactly one endpoint in ∂R 0 \ab and so has one endpoint in R 0 , as desired. It follows that window ab is saturated, and so R 0 satisfies condition (A). 8 4 5 6 7 8
doi:10.1016/j.dam.2015.04.025 fatcat:hxue6rvkafc2xpoyj7tpc3adhy