Art gallery theorems for guarded guards

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

2. 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.) • Gis a guard setfor Pn provided that for every point x in Pn there exists a point w in Gsuch 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 functiong(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. 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 defineg⊥(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. Definition of a guarded guard set • A set of points Gin a polygon Pn is a guarded guard set for Pn provided that(i) Gis a guard set for Pn; and(ii) for every point w in Gthere exists a point v ≠ w in Gsuch 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. 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. Correction of the literature • gg(P12) = 5, and P12 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 3gives the correct formula for gg(n). P12

7. 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 andit’s dual graph is a tree. • The quadrangulation graphof a simple orthogonalpolygon is planar, bipartite and has an even number of vertices. It’s dual graph is also tree.

8. Triangulation and Quadrangulation • Let Gn be a triangulation or quadrangulation graph on n vertices. (1) A vertex subset Gis 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. 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. 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 triangulationof a polygon Pn with n≧ 10 sides. Then there exists a diagonal of Tn that separatesPn into two triangulated polygonsTmandTn−m+2, one of which has m sides, where m∈ {6, 7, 8, 9}.

11. 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. T6:

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

13. 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(T7) ≦Φ(n −5)+2≦Φ(n). A similar argument holds for m=9.

14. Proof of Theorem 8 For m=6 Let G* and G be guarded guards for T*n-5 and T6 , 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 T8 , respectively. G=G*∪{y,w,v} => |G|≦|G*|+3 ≦ Φ(n −7)+3≦Φ(n).

15. 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. 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 x1x2 with a suitable orientation. Fig 5.

17. 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 P10’ , excluding segment x1x2) Thus, |Gn | =|Gn-7 | + |G’ | (2) Gn = {x3,x1,x’7,x’4,x’3}Gn-7 = {x3,x1}G’ = {x’7,x’4,x’3 }

18. 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. 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. 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 polygonPn.(2) construct Pn such that • Theorem 11. If Qn is a quadrangulation graph on n vertices (n ≧ 6), thenProof.We construct a set Gof vertices in Qn that satisfies:(i)(ii) every quadrilateral of Qn contains a vertex of G ;(iii) every vertex in Gis guarded in a quadrilateral with another vertex in G.

21. Proof of Theorem 11

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

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

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