JOSEPH O ' R O U R K E

G A L L E R I E S N E E D F E W E R M O B I L E G U A R D S :

A V A R I A T I O N O N C H V f i ~ T A L ' S T H E O R E M

1. INTRODUCTION

In 1975 Chv~ital proved that In/31 guards are always sufficient and occasional- ly necessary to see every point in the interior of an n-wall room, settling a question posed by Klee [1], [5]. The guards are assumed to be stationary points who can 'see' any other point that can connect to them with a line segment within the room, and the room is taken to be the region bounded by a simple polygon. Chvfital's proof of his 'Art Gallery Theorem' is an inductive combinatorial argument, starting with an arbitrary triangulation of the polygon, and partitioning off a small piece for the induction step. Fisk later offered a simpler proof based upon a 3-coloring of the triangulation graph [2], [6]. This proof idea was subsequently used by Kahn et al. to Qbtain an interesting specialization of Ch%tal's result: In/4] guards are necessary and sufficient to cover an n-wall rectilinear room, one whose walls are orthogonal [7],[8].

In this paper we pursue a suggestion of Toussaint by conside, ring mobile guards [9]. In particular, we show that if the guards are permitted to patrol fixed interior line segments of a simple polygon, then In/4] guards are always sufficient and sometimes necessary for n 1> 4. The necessity follows from the generic example of Toussaint shown in Figure 1 : each 4-edge lobe requires its own mobile guard. The proof of sufficiency follows the model of Chvfital's original inductive proof rather than that of Fisk. It remains an interesting challenge to determine whether the same result can be obtained by a coloring argument.

2. G U A R D DEFINITIONS

We start with formal definitions of the above visibility concepts. A simple polygon P is a sequence of n straight edges in the plane joined at n vertices forming a closed path that intersects itself only in the identity of the first and last vertex. A polygonal region R is a closed subset of the plane whose bound- ary is a simple polygon P. A guard g is a subset of a polygonal region R : g _~ R. A point x ~ R is said to be visible to or seen by a guard g !- R if there exists a point y~g such that 2~, the line segment with endpoints x and y, is a subset of R : x ~ _ R. A collection of guards C = {g~,... ,gk}, gi -~- R, is said to cover their polygonal region R if every point x ~ R can be seen by some guard gl ~ C.

Several special guard types may be defined. Let P be a simple polygon and R its associated polygonal region. A vertex guard is any single vertex of P.

Geometriae Dedicata 14 (1983) 273 283. 0046 5755/83/0143-0273501.65. ~) 1983 by D. Reidel Publishing Company.

Fig. 1. A polygon that requires In/4] edge, diagonal, line, or triangle guards.

274 JOSEPH O 'ROURKE

An edge guard is an edge of P, including the endpoints. A diagonal guard is an edge or internal diagonal between vertices of P, again including the endpoints. A line guard is any line segment wholly contained in R, and a triangle guard is a triangular subset of R. It is the main goal of this paper to prove that, for polygonal regions R with n ~> 4 vertices, [n/4J diagonal, line, or triangle guards are always sufficient and occasionally necessary to cover R. The proof, like Chv/ttaI's, is mainly combinatoric; we next define the combinatorial counterparts of the above geometric notions.

A triangulation graph G of an n vertex simple polygon P is a graph obtained by triangulating P with internal diagonals between vertices. The nodes of G correspond to the n vertices of P, and the arcs correspond to the n edges and n - 3 diagonals. G has n - 2 triangular faces. There are usually several triangulation graphs for a given polygon.

Define a guard in a triangulation graph G of a polygon P to be a subset of the nodes of G. Then a vertex guard in G is a single node of G, an edge guard in G is a pair of nodes adjacent across an arc corresponding to an edge of P, and a diagonal guard in G is a pair of nodes adjacent across any arc of G. Finally, the analog of covering is domination: a collection of guards C -- {g,,... ,gk} is said to dominate G if every triangular face of G has at least one of its three nodes in some gle C. Guards in a graph will be called combina- torial guards to distinguish them from the geometric guards introduced earlier.

3. SUFFICIENCY PROOF

The reason for introducing triangulation graphs is the following lemma.

LEMMA 1. Let P, R, and G be a simple polygon, its polygonal region, and one of its triangulation graphs, respectieely. I f G can be dominated by k combinatori- al vertex guards ( = nodes), then R ean be covered by k geometric vertex guards.

Proof Since G is dominated, each triangle has at least one combinatorial

G A L L E R I E S N E E D F E W E R M O B I L E G U A R D S 275

guard at one of its nodes. Placing geometric guards at the corresponding vertices of P ensures that each triangular region is covered. []

The implication of this 1emma is that a proof of the sufficiency of a certain number of combinatorial guards in a triangulation graph establishes the sufficiency of the same number of geometric guards in a polygonal region. We will take this approach by proving the sufficiency of In/4] combinatorial diagonal guards.

The proof is by induction and follows the main outlines of Chvfital's inductive proof (and Honsberger's exposition [5]). Before commencing the proof, it will be convenient to establish certain facts that will be used in various cases of the proof. The most important of these concerns 'edge contractions'. Let P be a simple polygon and G a triangulation graph for P, and let e be an edge of P, and u and v the two nodes of G corresponding to the endpoints of e. The contraction of e is a transformation that alters G by removing nodes u and v and replacing them with a new node x adjacent to every node to which u or v was adjacent. 1 Compare Figures 2a and 2d. Note than an edge contraction is a graph transformation, not a polygon transforma- tion: the geometric equivalent ('squashing' the polygon edge) could result in nonsimple polygons. Edge contractions are nevertheless useful because of the following lemma.

LEMMA 2. Let G be a triangulation graph of a simple polygon P, and G' the graph resulting from an edge contraction of G. Then G' is a triangulation graph of some simple polygon P'.

Proof We construct a figure with curved edges corresponding to G', then straighten the edges to obtain P'.

Let Pt be the planar figure corresponding to the triangulation G, and let e be the edge contracted and u and v its two endpoints in Pt. Let the vertices to which u and v are connected by diagonals and edges be Y0, "",Yi and

Z 3

Y4

Fig. 2a. A triangulation graph.

Harary calls this transformation an elementary contraction [4].

276 J O S E P H O ' R O U R K E

z 0 . . . . . zj. respectively, with Yo = v and z o = u, and remainder labeled accord- ing to their sorted angular order. See Figure 2a. Note that Yl = zt is the apex of the triangle supported by e.

Now introduce a new vertex x on the interior of e, and connect the y and z vertices to x by the following procedure. Connect Yt to x; this can be done without crossing any diagonals because Yl is the apex of a triangle on whose base x lies. Remove the diagonal (u, yl). Connect Y2 to x within the region bounded by (x, y~, Y2, u); the line may need to be curved but again no cross- ings are necessary. Remove the diagonal (u, Y2)' Continue in this manner (see Figure 2b) until all the y's have been connected to x. Then apply a similar

Fig. 2b.

Y,~

Several diagonals connected to u have been connected to x instead.

Fig. 2c.

z3

Y4

All the arcs incident to u and v have been made adjacent to x.

Fig. 2d.

7 3

Y4

A straight line figure with the same graph structure as Figure 2c.

G A L L E R I E S NEED F E W E R M O B I L E G U A R D S 277

procedure to the z vertices. The result is a planar figure whose connections are the same as those of G'. See Figure 2c.

Finally, apply F~iry's theorem [3] : for any planar graph drawn iin the plane, perhaps with curved lines, there is a homeomorphism in the plane onto a straight-line graph such that vertices are mapped to vertices and edges to edges. Applying such a homeomorphism to the figure constructed above yields P', a polygon that has G' as one of its triangulations. See Figure 2d. []

The main use of this contraction result is the following.

LEMMA 3. Suppose that f(n) combinatorial diagonal guards are always sufficient to dominate any n-node triangulation graph. Then if G is an arbitrary triangulation graph of polygon P with one vertex guard placed ai ~. any one of its n nodes, then an additional f ( n - 1) diagonal guards are sufficient to dominate G.

Proof Let u be the node at which the one guard is placed, and let v be a node adjacent to u across an arc corresponding to an edge e of P. Edge contract G across e, producing the graph G' of n - 1 nodes. By Lemma 2 G' is a triangulation graph, and so can be dominated byf(n - 1) diagonal guards. Let x be the node of G' that replaced u and v. Suppose that no guard is placed at x in the domination of G'. Then the same guard placements will dominate G, since the given guard at u dominates the triangle supported by e, and the remaining triangles of G have dominated counterparts in G'. Again compare Figures 2a and 2d. If a guard is used at x in the domination of G', then this guard can be assigned to v in G, with the remaining guards maintaining their position. Again every triangle of G is dominated. []

We note in passing that the same lemma holds for other types of guards, but we will only need to use it with diagonal guards. Intuitively, one can view this lemma as saying that one edge can be 'squashed' out for guard coverage calculations if a guard is assigned to either of the edge's endpoints.

The next three lemmas establish special diagonal guard results for small triangulation graphs.

LEMMA 4. Every triangulation graph of a pentagon (n = 5) can be dominated by a single combinatorial diagonal guard with one endpoint at any selected node.

4 4 4 4 4

2 I 2 I 2 I 2 I 2 I

Fig. 3. A pentagon can be dominated by a single diagonal guard (shown dotted) with one end at node 1.

278 J O S E P H O ' R O U R K E

Proof Let G be a triangulation graph of a pentagon, and let the selected node be labeled 1. It is easy to show that there are only five distinct triangula- tions. In each case, a single combinatorial diagonal guard (pair of adjacent nodes), with one end at node 1 can dominate the graph (see Figure 3). []

LEMMA 5. Every triangulation graph of a septagon (n = 7) can be dominated by a single combinatorial diagonal guard.

Proof Let G be a triangulation graph ofa septagon, and let d be an arbitra- ry internal diagonal. This diagonal partitions the seven boundary edges of G according to either 2 + 5 = 7 or 3 + 4 = 7; clearly the partition 1 + 6 = 7 is not possible.

Case 1.2 + 5 = 7. Let d = (1, 3). Then d supports another triangle T, either (1, 3, 4), (1, 3, 5), (1, 3, 6), or (1, 3, 7). Only two of these cases are distinct.

la. T = (1, 3,4). Then (1,4, 5,6, 7) is a pentagon (see Figure 4a), By Lemma 4, this pentagon can be covered with a single diagonal guard with one end of node 1. This guard dominates the entire graph.

lb. T = (1, 3, 5). Choose diagonal (1, 5) for the guard (see Figure 4b). Regardless of how the quadrilateral (1, 5, 6, 7) is triangulated, all of G is dominated.

Case2. 3 + 4--7. Let d--(1, 4). Then both ways of triangulating the quadrilateral (1, 2, 3, 4) lead to situations equivalent to Case la above. []

LEMMA 6. Every triangulation graph of an enneagon (n = 9) can be dominat- ed by two combinatorial diagonal guards such that one of their endpoints coincides with any selected node.

Proof Let G be a triangulation graph of an enneagon, let the selected node be labeled 1, and let d be any internal diagonal with one end at 1. This diagonal partitions the boundary edges of G according to either 2 + 7 = 9, 3 + 6 = 9, o r 4 + 5 = 9 .

Case 1.2 + 7 = 9. Let d = (1, 3). The diagonal d supports another triangle T whose apex is at either 4, 5, 6, 7, 8, or 9. Only three of these cases are distinct.

la. T = (1, 3, 4). Dominate the septagon (1, 4, 5, 6, 7, 8, 9) with one guard by Lemma 5, and use (1, 3) for the second guard (see Figure 5a).

5 5

2 I 2 I

Fig. 4. A septagon can be dominated by a single diagonal guard.

GALLERIES NEED FEWER MOBILE GUARDS 279

Fig. 5.

6 6 6

2 1 2 I 2 I

A enneagon can be dominated by two diagonal guards, with one of their ends at node 1.

Fig. 6.

G

k i l ~ n - I

f

The diagonal d separates G into two pieces, one of which (G1) shares 5 ~ k ~ 8 edges with G.

lb. T = (1, 3, 5). Domina t e the septagon (1, 3, 5, 6, 7, 8, 9) with one guard by L e m m a 5, and use (1, 3) for the second guard (see Figure 5b).

lc. T = (1, 3, 6). D o m i n a t e the hexagon (1, 2, 3, 4, 5, 6) with one guard by L e m m a 5, and domina te the pen tagon (1, 6, 7, 8, 9) with one guard whose endpoint is at 1 by L e m m a 4 (see Figure 5c).

Case 2. 3 + 6 = 9. Let d = (1, 4). This is equivalent to Case l a above. Case 3 .4 + 5 = 9. Let d = (1, 6). This is equivalent to Case lc above. [ ]

Final ly we must establish the existence of a special d iagonal that will allow us to take the induct ion step.

L E M M A 7. Let P be a simple polygon of n >>- 10 vertices, and G a triangulation graph of P. There exists a diagonal d in G that partitions G into two pieces, one of which contains k = 5, 6, 7, or 8 ares correspondin9 to edges of P.

Proof Choose d to be a diagonal of G that separates off a m i n i m u m n u m b e r of po lygon edges that is at least 5. Let k >/5 be this m i n i m u m number , and label the vertices 0, 1 . . . . ,n - 1 such that d is (0, k). See Figure 6o The diagonal d suppor ts a tr iangle T whose apex is at t, 0 ~< t ~< k. Since k is minimal , t ~< 4 and k - t ~< 4. Adding these two inequalities yields k ~< 8. [ ]

280 JOSEPH O'ROURKE

With the preceding lemmas available, the induction proof is a nearly straight- forward enumeration of cases.

T H E O R E M 1. Every triangulation graph G of a polygon of n >~ 4 vertices can be dominated by In/4] combinatorial diagonal guards. Proof Lemmas 4, 5, and 6 establish the truth of the theorem for 5 ~< n ~< 9, so assume that n ~> 10, and that the theorem holds for all n' < n. Lemma 7 guarantees the existence of a diagonal d that partitions G into two graphs G 1 and G 2 where G 1 contains k boundary edges of G with 4 ~< k ~< 8. Each value of k will be considered in turn.

Case 1. k = 5 or 6. G1 has k + 1 ~<7 boundary edges including d. By Lemma 5, G 1 can be dominated with a single diagonal guard. G 2 has n - k + l ~ < n - 5 + l = n - 4 boundary edges including d, and by the induction hypothesis, it can be dominated with [ ( n - 4 ) / 4 J = in /4] - 1 diagonal guards. Thus G 1 and G z together can be dominated by In/4] diagonal guards.

Case 2. k = 7. The presence of any of the diagonals (0, 6), (0, 5), (1, 7), or (2, 7) would violate the minimality of k. Consequently the triangle T in G 1 that is bounded by d is either (0, 3, 7) or (0, 4, 7); since these are equivalent cases, suppose that T is (0, 3, 7). The quadrilateral (0, 1, 2, 3) has two distinct triangulations. Each will be considered separately.

2a. (1, 3) is included. Dominate the pentagon (3, 4, 5, 6, 7) with one diagonal guard with one end at node 3. This is possible by Lemma 4. This guard dominates all of G 1 . Since G 2 has n - 7 + 1 = n - 6 boundary edges, it can be dominated by [ (n - 6)/4] ~< [n/4] - 1 diagonal guards by the induction hypothesis. This yield a domination of G by [n/4] diagonal guards.

2b. (0, 2) is included. Form graph G O by adjoining the two triangles T = (0, 3, 7) and T' = (0, 2, 3) to G 2 (see Figure 7). G O has n - 7 + 3 = n - 4 edges, and so can be dominated by [ ( n - 4)/4J = [n/4J- 1 diagonal guards by the induction hypothesis. In such a domination,

i I

G O is formed by adding T and T' to G 2 . Fig. 7.

G A L L E R I E S NEED F E W E R M O B I L E G U A R D S 281

at least one of the vertices of T' = (0, 2, 3) must be a diagonal guard endpoint. There are three possibilities:

(0) If node 0 is a guard end, then G o can be extended to include (0, 1, 2) without need of further guards.

(2) If node 2 is a guard end, then G o can again be extended to include (0, 1, 2).

(3) If node 3 is a guard end, then there are three possible locations for the other end of the guard. If the other end is at either node 0 or 2, then we fall into the two cases above. If the other end is at node 7, then replace the diagonal guard (3, 7) with (0, 7). Every triangle that was previously dominated is still dominated, arid again G O can be extended to include (0, 1, 2).

Thus all but the pentagon (3, 4, 5, 6, 7) can be dominated with In/4] - 1 diagonal guards, and the pentagon only requires a single diagonal guard by Lemma 4, resulting in a total of [n/4J diagonal guards for all of G.

Case 3. k = 8. G 1 has k + 1 = 9 boundary edges, and so by Lernma 6, it can be dominated with two diagonal guards, one of whose endpoints is at node 0. Now G 2 has n - k + 1 = n - 7 boundary edges. By Lemma 3, the one guard at node 0 permits the remainder of G 2 to be dominated b y f ( n - 7 1) - f ( n - 8) diagonal guards, where the function f(n') specifies a number of diagonal guards that are always sufficient to dominate a triangulation graph of n' nodes. By the induction hypothesis, f (n ')= [n'/4J. Therefore [ ( n - 8)/4] = [ n / 4 J - 2 diagonal guards suffice to dominate G 2. Together with the two allocated to G 1 , all of G is dominated by In/4] diagonal guards.

[]

COROLLARY 1. Any polygon P of n >~ 4 edges can be covered by [n/4J geometric diagonal, line, or triangle guards.

Proof The diagonal guard result follows immediately from the theorem and Lemma 1. Since diagonal guards are special cases of line ',and triangle guards, the same number of these more powerful guards clearly suffice. []

5

2 I

Fig. 8. A polygon of seven edges that requires two edge guards.

282 J O S E P H O ' R O U R K E

Fig. 9.

16 17 15~ 18

II 15. 14~, ~ " ~ 1 9 .20 Z2

3 2

A triangulation graph that requires two edge guards per seven edges. The central area may be triangulated arbitrarily.

4. D I S C U S S I O N

The above proof depends on the fortunate identity between the number of combinatorial and geometric diagonal guards necessary and sufficient to dominate and cover triangulation graphs and polygons, respectively. This identity is not known to hold for edge guards, however. No polygons are known to need more than [(n + 1)/4J 9eometric edge guards (see Figure 8), but triangulation graphs exist that require [2n/7] = [n/3.5] combinatorial edge guards (see Figure 9). Thus it appears that a different proof technique is required in this case.

A C K N O W L E D G E M E N T S

Godfried Toussaint of McGill University suggested the problem addressed in this paper and provided the example that establishes the necessity of [n/4J line guards (Figure 1). The polygon in Figure 8 was discovered by Chris Paige of McGill. Finally I would like to thank Toussaint and David Avis of McGill for clarifying discussions on the proof presented in this paper.

R E F E R E N C E S

1. Chvfital, V. : 'A Combinatorial Theorem in Plane Geometry' , J. Comb. Theory B 18 (1975), 39-41.

G A L L E R I E S NEED F E W E R M O B I L E G U A R D S 283

2. Fisk, S. : 'A Short Proof of Chv~ital's Watchman Theorem', J. Comb. Theory B 24 (1978), 374. 3. Giblin, P. L. : Graphs, Surfaces, and Homology, Chapman and Hall, London, 1977, pp. 41-45. 4. Harary, F. : Graph Theory, Addison-Wesley, Reading, 1976, pp. 112-113. 5. Honsberger, R.: Mathematical Games I1, Mathematical Association of America, 1976,

pp. 104-110. 6. Honsberger, R. : 'Games, Graphs, and Galleries', in The Mathematical Gardner (ed. David

A. Klarner), Prindle, Weber & Schmidt, Boston, 1981, pp. 274-284. 7. Kahn, J., Klawe, M. and Kleitman, D. : 'Traditional Galleries Require Fewer Watchman' ,

S lAM J. on Algebraic and Discrete Methods, June 1983. 8. O'Rourke, J. : 'An Alternate Proof of the Rectilinear Art Gallery Theorem', Johns Hopkins

University Technical Report 82 15, Dec. 1982. 9. Toussaint, G., Personal Communication, Jan. 1982.

A u t h o r ' s curren t address :

Joseph O'Rourke, Department of Electrical Engineering

and Computer Science, Johns Hopkins University, Baltimore, Maryland 21218, U.S.A.

(Received November 11, 1981 ; revised version : March 31, 1982)