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Guard Sets in Polygons: Minimum Requirements and Theorems

This paper presents results on guard sets for polygons, discussing the concepts of guarded guards and their implications in computational geometry. It defines the minimum cardinality of guard sets and introduces key theorems related to both general and orthogonal polygons. The study explores the relationships between triangulation and quadrangulation graphs and their role in determining the structure of guard sets. Theorems address various polygon configurations, offering insights into visibility problems and the optimization of guard placements.

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Guard Sets in Polygons: Minimum Requirements and Theorems

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  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):

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