1. Art gallery theorems for guarded guards T.S. Michael, Val Pinciub Computational Geometry 26 (2003) 247–258

2. 2/24 Definition of a guard set Pn denotes a simple closed polygon with n sides, together with its interior. A point x in Pn is visible from the point w provided that line segment wx does not intersect the exterior of Pn. (Every point in Pn is visible from itself.) G is a guard set for Pn provided that for every point x in Pn there exists a point w in G such that x is visible (guarded) from w. Let g(Pn) denote the minimum cardinality of a guard set for Pn. for each integer n ≧ 3, define the function g(n) = max {g(Pn) : Pn is a polygon with n sides.} Thus g(n) equals the minimum number of guards that are sufficient to cover any gallery with n sides. guard set

3. 3/24 Theorem 1 and Theorem 2 In an orthogonal polygon Pn, each interior angle is 90° or 270. An orthogonal polygon must have an even number of sides. For even n ≧ 4 we define g ⊥ (n) = max { g(Pn): Pn is an orthogonal polygon with n sides } Theorem 1 (Art gallery theorem). For n ≧ 3 we have Theorem 2 (Orthogonal art gallery theorem). For n ≧ 4 we have

4. 4/24 Definition of a guarded guard set A set of points G in a polygon Pn is a guarded guard set for Pn provided that (i) G is a guard set for Pn; and (ii) for every point w in G there exists a point v ≠ w in G such that w is visible (guarded) from v. We let gg(Pn) denote the minimum cardinality of a guarded guard set for the polygon Pn. gg(n) = max { gg(Pn): Pn is a polygon with n sides }, gg ⊥ (n) = max { gg(Pn): Pn is an orthogonal polygon with n sides }. The function gg ⊥ (n) is only defined for even n ≧ 4. guard set

5. 5/24 Theorem 3 and Theorem 4 Theorem 3 (Art gallery theorem for guarded guards). For n ≧ 5 we have Theorem 4 (Orthogonal art gallery theorem for guarded guards). For n ≧ 6 we have One easily verifies that gg(3) = gg(4) = 2 and that gg ⊥ (4) = 2 to treat the small values of n not covered by Theorems 3 and 4.

6. 6/24 Correction of the literature gg(P 12 ) = 5, and P 12 is a counterexample to the formula gg(n)= that appeared in the literature - G. Hernández-Peñalver, Controlling guards (extended abstract), in: Proceedings of the 6th Canadian Conference on Computational Geometry (6CCCG), 1994, pp. 387–392. Our Theorem 3 gives the correct formula for gg(n). P 12

7. 7/24 Triangulation and Quadrangulation One technique to solve an art gallery problem is to translate the geometric situation to a combinatorial one by introducing a graph. The triangulation graph of a simple polygon is 3-colorable and it’s dual graph is a tree. The quadrangulation graph of a simple orthogonal polygon is planar, bipartite and has an even number of vertices. It’s dual graph is also tree.

8. 8/24 Triangulation and Quadrangulation Let Gn be a triangulation or quadrangulation graph on n vertices. (1) A vertex subset G is a guard set of Gn provided every bounded face of Gn contains a vertex in G. (2) Every vertex in G occurs in a bounded face with another vertex in G, then G is a guarded guard set for Gn. If each face in Gn is convex,then G is also a guard set for the corresponding polygon Pn. We let g(Gn) and gg(Gn) denote the minimum cardinality of a guard set and guarded guard set, respectively, of the graph Gn. If Gn arises from a triangulation or quadrangulation of a polygon Pn, then g(Pn) ≦ g(Gn) and gg(Pn) ≦ gg(Gn).

9. 9/24 Proof of Theorem 1 Theorem 1 (Art gallery theorem). For n ≧ 3 we have Proof: g(n) = max { g(Pn) : Pn is a polygon with n sides.} (1) For any simple polygon Pn, let Tn be the corresponding triangulation graph and Tn is 3- colorable. By choosing the smallest color class, we have cardinality of the guard set G at most for Tn. Thus,. And. We obtain (2) Construct Pn that require guards.

10. 10/24 Proof of Theorem 3 Proof of Theorem 3 is divided into tow steps: (1) prove. Tn is triangulation graph of any Pn. (2) construct Pn such that Step (1) relies on several preliminary lemmas. Lemma 6. Let Tn be a triangulation of a polygon Pn with n ≧ 10 sides. Then there exists a diagonal of Tn that separates Pn into two triangulated polygons Tm and Tn−m+2, one of which has m sides, where m ∈ {6, 7, 8, 9}.

11. 11/24 Lemma 7 Lemma 7. Let [x, y] be any boundary edge in a triangulation graph Tm with m vertices. (a) If m = 6, then {w,x} or {w,y} is a guarded guard set for Tm for some w. (b) If m = 7, then gg(Tm) = 2. (c) If m = 8, then {v,w,x} or {v,w,y} is a guarded guard set for Tm for some v and w. (d) If m = 9, then gg(Tm) ≦ 3. Proof : Statement (a) is verified by examining a small number of cases. T 6 :

12. 12/24 Proof of Lemma 7 Proof. (b) If m = 7, then gg(Tm) = 2 (c) If m = 8, then {v,w,x} or {v,w,y} is a guarded guard set for Tm for some v and w. (d) If m = 9, then gg(Tm) ≦ 3

13. 13/24 Proof of Theorem 8 Theorem 8. If Tn is a triangulation graph on n vertices (n ≧ 5), then Proof. We induct on n. (1) The result is easily verified for n = 5, and holds for 6 ≦ n ≦ 9 by Lemma 7. (2) Let Tn be a triangulation graph on n vertices (n ≧ 10). By Lemma 6 there exists an edge [x, y] that separates Tn into two triangulation graphs Tm and Tn−m+2,where m ∈ {6, 7, 8, 9}.Clearly, gg(Tn) ≦ gg(Tn−m+2)+ gg(Tm). (1) Let For m = 7, the inductive hypothesis, Lemma 7(b), and (1) imply that gg(Tn) ≦ gg(Tn−5)+gg(T 7 ) ≦ Φ(n −5)+2 ≦ Φ(n). A similar argument holds for m=9.

14. 14/24 Proof of Theorem 8 For m=6 Let G* and G be guarded guards for T*n-5 and T 6, respectively. G=G* ∪ {y,w} => | G | ≦ | G* |+2 ≦ Φ(n −5)+2 ≦ Φ(n). For m=8 Let G* and G be guarded guards for T*n-7 and T 8, respectively. G=G* ∪ {y,w,v} => | G | ≦ | G* |+3 ≦ Φ(n −7)+3 ≦ Φ(n).

15. 15/24 Construct Pn The cases n ≡ 1, 3, 5 (mod 7) are the critical values for which Φ(n) > Φ(n − 1); we may always add one or two vertices to our polygons to deal with n ≡ 0, 2, 4, 6 (mod 7). Ex: Φ(5)=Φ(6)=Φ(7)=2, Φ(8)=Φ(9)=3, Φ(10)=Φ(11)=4. Construct gg(Pn)=Φ(n) for n =5, 8, 10. Add one or two vertices to Pn to deal with n=6,7, 9, 11 Fig. 4

16. 16/24 Construct Pn For n ≧ 12, we construct Pn from Pn−7 by adjoining a special decagon P’ 10 with vertices x’ 0, x’ 1,..., x’ 7, x’ 1, x’ 2 on side x 1 x 2 with a suitable orientation. Fig 5.

17. 17/24 Proof of Lemma 9 Lemma 9. Any guarded guard set for the polygon Pn defined inductively in Figs. 4 and 5 has cardinality at least for n ≧ 5. Proof: We induct on n. (1) We have base cases with 5 ≦ n ≦ 11 (2) Let Gn be a guarded guard set of Pn and Gn = Gn-7 ∪ G’ Gn-7 = Gn ∩Pn (The points of Gn in the closed polygon Pn-7) G’ = Gn - Gn-7 ( the points of Gn in the decagon P 10 ’, excluding segment x 1 x 2 ) Thus, | Gn | =| Gn-7 | + | G’ | (2) Gn = {x 3,x 1,x’ 7,x’ 4,x’ 3 } Gn-7 = {x 3,x 1 } G’ = {x’ 7,x’ 4,x’ 3 }

18. 18/24 Proof of Lemma 9 Claim 1: | G’| ≧ 3. Claim 2: | Gn-7 | ≧ Φ(n-7)-1. Make Gn-7 ∪ { z } to be a guarded guard set for Pn-7. if x in Pn-7 is visible from a point in P10’, then x is also visible from both x3 and x1. Segment x3x4 that are near x3 must be visible from some point w in Gn-7. Hence, if x3 Gn-7, set z=x3, if x3 ∈ Gn-7, set z=x1, By the induction hypothesis | Gn-7 |+| {z} | ≧ | Gn-7 ∪ {z} | ≧ Φ(n − 7), which establishes the claim.

19. 19/24 Proof of Lemma 9 If | G’ | ≧ 4, by | Gn-7 | ≧ Φ(n-7)-1 and | Gn | =| Gn-7 | + | G’ | we obtain | Gn | ≧ Φ(n) If | G’ |=3, Points on segment x3x2 near x3 are not visible from any point in G’, and hence there is a point w ∈ Gn-7 from which such points are visible. Now the set Gn-7 − {w} ∪ {z} is a guarded guard set for Pn-7, where z is defined in Claim 2. Thus | Gn-7 | = | Gn-7 − {w} |+ | {z} | ≧ | Gn-7 − {w} ∪ {z}| ≧ Φ(n-7), and | Gn | ≧ Φ(n) follows from | Gn | =| Gn-7 | + | G’ |.

20. 20/24 Proof of Theorem 4 Proof of Theorem 4 for n ≧ 6 is divided into tow steps: (1) prove. for the quadrangulation graph Qn of any orthogonal polygon Pn. (2) construct Pn such that Theorem 11. If Qn is a quadrangulation graph on n vertices (n ≧ 6), then Proof. We construct a set G of vertices in Qn that satisfies: (i) (ii) every quadrilateral of Qn contains a vertex of G ; (iii) every vertex in G is guarded in a quadrilateral with another vertex in G.

21. 21/24 Proof of Theorem 11

22. 22/24 Proof of Theorem 11 (i) (ii) every quadrilateral of Qn contains a vertex of G ;

23. 23/24 Proof of Theorem 11 (iii) every vertex in G is guarded in a quadrilateral with another vertex in G.

24. 24/24 Construct Pn construct orthogonal polygon Pn such that for n ≧ 6 and n is even. Consider n = 0, 2, 4 (mod 6):