ART GALLERIES WITH INTERIOR WALLS. March 1998

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1 ART GALLERIES WITH INTERIOR WALLS Andé Kündgen Mach 1998 Abstact. Conside an at galley fomed by a polygon on n vetices with m pais of vetices joined by inteio diagonals, the inteio walls. Each inteio wall has an abitaily placed, abitaily small dooway. We will show that the minimum numbe of guads that suffice to guad all at galleies with n vetices and m inteio walls is min{ (2n )/, (2n+m 2)/, (2m+n)/ }. If we estict ouselves to galleies with convex ooms of size at least, the answe impoves to min {m, (n + m)/ }. The poofs lead to linea time guad placement algoithms in most cases. The oiginal at galley poblem, posed by Klee and solved by Chvátal [6], is to find the smallest numbe of guads necessay to cove any simple polygon, the at galley, not necessaily convex, on n vetices. Hee a coveing by g guads means that one can find g points in the inteio of the polygon such that evey point in the inteio is coveed by some guad, that is fo each point in the inteio the line segment between it and some guad does not intesect the polygon. The comb polygons in figue 1 show that n/ guads ae sometimes necessay if n is not divisible by simply take a comb on n/ vetices and subdivide one o two of its edges. Chvátal also showed that n/ guads always suffice. Fo moe infomation on the histoy of this poblem and elated poblems, see [15] and [17]. Figue 1: Comb polygons Hutchinson [11] genealized the basic at galley poblem by allowing inteio walls. Thoughout this pape an at galley (with inteio walls) will be a simple polygon on n vetices with some pais of vetices joined by non-intesecting inteio diagonals, the inteio walls. Also suppose that in the inteio of each of the walls Typeset by AMS-TEX 1

2 2 ANDRÉ KÜNDGEN thee is an abitaily placed, abitaily small opening, the dooway. Figue 2 is an example of an at galley on n = 15 vetices and 8 inteio walls that equies 9 guads. Hutchinson now asked to find the minimum numbe of guads that suffice to cove any such at galley on n vetices. To motivate ou poofs we will give Fisk s [9] elegant poof of Chvátal s esult: Fist tiangulate the polygon. The esulting plane gaph, with vetex set the cone points of the polygon, has all its vetices on the outside face. Gaphs that can be embedded in the plane in such a way ae called outeplana. It is well known that outeplana gaphs ae -coloable, which can be easily seen by cutting along a chod and applying induction. Since each tiangle in the tiangulation must have vetices of all thee colos, putting a guad at each vetex in the smallest colo class poduces a coveing set of n/ guads. We will now answe Hutchinson s question with an agument in the spiit of Fisk s poof (see also [1]). Lemma 1. (2n )/ guads suffice to cove any at galley on n vetices and thee ae galleies with 2n/ 2 inteio walls whee this many guads ae equied. Poof. We may assume that the inteio walls tiangulate the at galley, since adding exta inteio walls cannot make it easie to guad the galley. This outeplana gaph can now be -coloed. Fom such a coloing, we get a labeling of the edges of the gaph by assigning to each edge the colo not used on its endpoints. Now each tiangle has each colo appeaing on one of its incident edges. Placing a guad into the dooway, fo inteio walls, o just next to the wall, fo exteio walls, we can see that each set of labels coesponds to a set of guads that coves the whole at galley. Since n-vetex outeplana tiangulations have 2n edges, which can be seen by induction o by Eule s fomula, taking the least fequent colo suffices. Figue 2: At galley with inteio walls Figue : V At galleies of the type in figue 2 achieve this bound. They ae obtained by stating with a small galley, and attaching k = n/ 1 of the V -shaped galleies in figue. If we attach a V with the leftmost wall to an aleady existing at galley and put the dooway exactly in the cente we incease the numbe of vetices by, the numbe of inteio walls by 2 and the numbe of guads needed by 2. Note that even if a guad fom the smalle galley can be placed in the dooway of the inteio wall connecting the V to the smalle galley that guad still can not see the othe banch of the V o the tiangle, so that those will equie one additional guad each. When n = k+ ou stating galley is a tiangle (as in figue 2), when n = k+ any quadilateal and when n = k + 5 a V. This poduces at galleies with

3 ART GALLERIES WITH INTERIOR WALLS 2n/ 2 (that is 2k, 2k and 2k + 1 espectively) inteio walls. The galleies also equie exactly one moe guad than they have inteio walls. This settles the poblem when the numbe of inteio walls is unspecified. But what happens when we have a specified numbe of inteio walls, say m? This question was suggested by J.Giggs. If m 2n/ 2, then Lemma 1 shows that the answe is still (2n )/, since adding additional inteio walls in the at galleies povided does not make guading any easie. Theoem 2. The minimum numbe of guads that suffice to cove all at galleies with n vetices and m inteio walls, g(n, m), is { 2n min, 2m + n, 2n + m 2 }, o moe pecisely: 2n, fo m 2 n 2 g(n, m) = 2m+n, fo m < 2 5 n 2n+m 2, othewise. Lemma 1 poves the fist pat of the statement, and the othe two pats will be Lemmas and. Lemma. Always g(n, m) (2m+n)/, and fo m < 2n/5 thee ae galleies with m inteio walls whee this many guads ae equied. Poof. The bound can be easily established by induction, but thee is also a Fisk type agument. Befoe tiangulating the at galley, assign each vetex v a weight of d(v) 1, whee d(v) is the degee of v in the outeplana gaph detemined by the galley i.e. the numbe of walls meeting at v. Fo example in the at galley without inteio walls each vetex has weight 1. Now tiangulate the galley, also using the inteio walls that ae aleady pesent, -colo the tiangulation, and find the colo class of smallest total weight W. It will suffice to find a guad set with W guads, since the total weight on all vetices is v V (G) (d(v) 1) = 2 E(G) n = 2m + n. To do this simply put d(v) 1 guads at each vetex v in the colo class of total weight W by putting one guad on evey inteio angula bisecto of walls that meet v, as shown in figue, close to v. Note that when we place the guads we ignoe the chods that ae not walls and wee intoduced in the tiangulation step. Since evey tiangle has a vetex of ou chosen colo, and each tiangle is coveed by a guad at the vetex associated with it, we ae done.

4 ANDRÉ KÜNDGEN Exteio Figue : Guad placement Call the at galley in figue 5 an E. Notice that an E has 7 vetices, 1 inteio wall and equies guads since no guad can cove moe than one of the alcoves. If we attach an E with the vetical wall to an aleady existing at galley and put the dooway exactly in the cente we incease the numbe of vetices by 5, the numbe of inteio walls by 2 and the numbe of guads necessay by. We will put the dooway in the middle of the vetical wall, so that a guad thee can t cove eithe alcove. Figue 5: E To achieve the bound, let m = 2k + ɛ with ɛ {0, 1}. Stat the constuction with a comb on n 5k ɛ vetices. Since n 5k ɛ +ɛ, such a comb exists. If m is odd, attach one tiangula oom to a wall paallel to the long hoizontal wall in the comb. This adds 1 vetex, 1 inteio wall and 1 guad, since the guad that is needed to cove the tiangle cannot cove any pong of the comb. (If n 5k 1 {, 5}, then stat with a V and subdivide any wall if necessay). In eithe case add k E s, stating fom an end of the comb. The esulting at galley has n vetices and m inteio walls. The numbe of guads equied is fo m = 2k: n 5k n + k 2m + n + k = =. fo m = 2k + 1: n 5k 1 n + k + 2 2m + n k = =. Lemma. Always g(n, m) (2n + m 2)/, and fo (2n )/5 m (2n 5)/ thee ae galleies with m inteio walls whee this many guads ae equied.

5 ART GALLERIES WITH INTERIOR WALLS 5 Poof. Fo the constuction let k = (2n m 5)/ 0. Stat with the at galley with n 5k vetices and m 2k inteio walls constucted in Lemma 1. It is a staightfowad computation to check that indeed 2(n 5k)/ 2 m 2k (use the fact that fo evey intege x, x y x + 1 > y), so that the constuction fom Lemma 1 applies. Now add k E s. The numbe of guads needed fo this at galley is k + 2(n 5k) = 2n (2n m 5)/ 8n 2n + m = 2n + m 2, unless 2n + m 2 is divisible by. But in that case ou inequality was not shap and we can gain an additional 1/12 fom that tem. We now pove the uppe bound by induction on n, with the added equiement that when 2n + m 2 is divisible by we can place one of the guads abitaily at an exteio wall. The bound holds fo m = 0, since g(n, 0) = n/ (n 1)/2 fo n. This is the only feasible value fo the base case n =. The added equiement also holds when n = : we can place the guad whee we want since tiangles ae convex. Conside an at galley A with n > vetices and m > 0 walls, and let g(a) be the numbe of guads needed to cove A. Cut the at galley along an inteio wall xy (see figue 6). This splits the galley into two pats shaing two vetices. Fo i {1, 2}, let n i and m i be the numbe of vetices and inteio walls in the i-th pat. Hence n 1 + n 2 = n + 2 and m 1 + m 2 = m 1, so that 2n1 + m 1 2 2n2 + m 2 2 g(a) g(n 1, m 1 ) + g(n 2, m 2 ) + 2(n 1 + n 2 ) + (m 1 + m 2 ) = 2n + m 1. Theefoe g(a) (2n + m 2)/, unless equality holds eveywhee, which implies that (2n 1 + m 1 2) and (2n 2 + m 2 2) ae divisible by. Howeve in that case we can invoke the stonge hypothesis and equie guads to be on eithe side of the sepaating wall xy ight next to the dooway. Now eplacing these two guads by a single guad in the dooway yields the claim. n 1 x n 2 m 1 y m 2 n 1 m 1 u v w n 2 m 2 Figue 6 Figue 7

6 6 ANDRÉ KÜNDGEN To show the additional equiement now suppose that 2n+m 2 is divisible by and that a guad must be placed nea exteio wall uv. Tiangulate the galley, also using the inteio walls that ae aleady pesent, and let w be the thid vetex in the tiangle containing uv. Removing the tiangle uvw, which is aleady coveed, splits the galley into two pats (see figue 7), satisfying n 1 +n 2 = n+1 and m 1 +m 2 m. Thus 2n1 + m 1 2 2n2 + m 2 2 g(a) g(n 1, m 1 ) + g(n 2, m 2 ) (n 1 + n 2 ) + (m 1 + m 2 ) 2n + m + 2. This also woks in the degeneate case when one of the n i = 2. So we ae done unless equality holds eveywhee, since 2n + m 2 is divisible by. This implies that (2n 1 + m 1 2) and (2n 2 + m 2 2) ae divisible by. Also m 1 + m 2 = m so that neithe one of the chods uw o vw is an inteio wall. Again we invoke the stonge induction hypothesis and equie guads nea the chods uw and vw close to w. Replacing these two guads by a single guad ight at w finishes the poof. At Galleies with Convex Rooms In [7] Czyzowicz et al. study at galleies that consist of polygons on n vetices that ae subdivided, not necessaily along chods, into k convex egions and show that these can be coveed with 2(n + k)/ guads. This esult is independent of Theoem 2, since ou poblem allows ooms of abitay shape but equies the inteio walls to be chods. It would be a common special case to study at galleies with n vetices and m inteio walls such that all k = m + 1 inteio ooms ae convex. Notice that we ae not equiing the polygon itself to be convex. Howeve doing so does not change the answe, since ou constuction achieving the uppe bound can easily be built to meet this additional equiement. Since it will pose no additional difficulty, we will also equie each oom to have at least walls, with = being the geneal case. Summing the sizes of the ooms yields n+2m, since the inteio walls ae counted twice. Hence fo at galleies such that all ooms have size at least, we have n + 2m (m + 1), o equivalently n m( 2) +. Theoem 5. The minimum numbe of guads that suffice to cove all at galleies with m > 0 inteio walls and n m( 2) + vetices, such that all ooms ae convex with at least walls, g (n, m), is { } n + m min m,, o moe pecisely: g (n, m) = { m, fo m n 1 n+m, fo m > n 1.

7 ART GALLERIES WITH INTERIOR WALLS 7 Poof. The bounds ae staightfowad. Fo the fist bound, place a guad in each dooway. Since the ooms ae convex evey oom can be coveed by at least one of the guads. Fo the second bound, one can fist povide a labeling of the walls with labels, such that each oom has evey label on one of its walls. We will do so by induction on m with the esult being tivial fo m = 0. Fo m > 0 cut along an inteio wall and apply induction on both pats. To combine both labelings into one it may be necessay to swap labels in one of the pats if they disagee on the sepaating wall. Now placing one guad at each wall suggested by the label used least fequently establishes the bound. Fo the constuction, let D be the at galley with one cental oom and 1 side ooms of size. The side ooms shae a wall only with the cental oom. Figue 8 shows D, D and D 5. Each D has 1 inteio walls, ( 2) + 2 vetices and equies 1 guads. Figue 8: D, D and D 5 If m n 1, ou galley will be simila to a D : Take a cente oom of size n m( 2), and attach m side ooms of size each. This is possible, since when m n we have n m( 2) m, and this galley equies m guads. If on the othe hand m > n 1 1 m( 1) n, then let k = > 0. We will fom ou galley by taking a galley on n = n k( 2) vetices and m = m k 0 inteio walls and then attaching k D s to it. Always attaching the next D with the fee wall of its cente oom to any othe oom we obtain a galley on n vetices and m walls. It can be eadily checked that n m ( 2) + and n m ( 1) so that we can take a galley fom the fist case to stat out with. So we need m + k( 1) = m k = m guads to cove this galley. m( 1) n = m + n Complexity In implementing Fisk s poof, Avis and Toussaint [2] obtained an O(n log n) algoithm to place n/ guads to guad an n-vetex at galley. This appoach

8 8 ANDRÉ KÜNDGEN can be impoved to obtain a linea time algoithm, since Chazelle ([], []) showed that an n-vetex polygon can be tiangulated, at least theoetically, in time O(n). Fist tiangulate the polygon in time O(n). Since outeplana tiangulations ae chodal gaphs, one can find a vetex elimination scheme (see fo example [16]) and then use this to obtain a -coloing of the vetices ([10]), both in O(n+) = O(n) time. Now placing the guads just equies O(n) time and the algoithm is linea. Fom hee it is easy to see how the uppe bound aguments in Lemma 1 and can be used to find linea time algoithms fo these poblems too. The only poblem could be that the existence of inteio walls does not necessaily make tiangulation easie. Howeve we can tiangulate each oom sepaately. This is still possible in linea time, since they have a total of n + 2m < n vetices. Finding the ooms and theefoe also the weak dual can be done in linea time. See [5] o [12] to find the otation scheme fom which this can be done. A staightfowad implementation of Lemma esults in an O(n 2 ) algoithm, due to the stonge statement, even fo m = 0. Howeve this case can be implemented in linea time even with the stonge statement. Fo n = just add the guad whee equested. Fo n apply the basic algoithm and just add in the exta guad at the equied place if necessay. This will wok since n/ + ɛ (n 1)/2, fo n with ɛ = 1 when n is odd and 0 othewise. This makes a faste algoithm plausible, howeve it is still an open question whethe a linea time algoithm can be obtained in this case. A Fisk type poof fo the uppe bound in Lemma would cetainly yield a fast algoithm. In the case of convex ooms the situation is easie. The fist bound tivially leads to an O(m) algoithm. The second bound can be implemented in time O(m) = O(n) as well, since the labeling can be found in linea time: Find the weak dual of the at galley, then stating at any vetex conduct a Beadth-Fist Seach on this dual tee to detemine the ode in which the ooms will be labeled. In the fist oom label the edges using each one of the labels at least once. On each consecutive oom one wall is aleady labeled, so label the emaining walls accodingly to assue that evey label is being used. It is impotant to note that although these algoithms give fast algoithms fo guading given classes of at galleies efficiently they do not necessaily give the best possible answe fo a specific at galley. This poblem is known to be NP-had even when the at galley has no inteio walls ([1], [1]). Acknowledgments The autho would like to thank Douglas B. West fo binging this poblem to his attention, assigning Hutchinson s poblem as homewok in one of his classes and fo helpful suggestions on the manuscipt. Refeences [1] A. Aggawal, The at galley theoem: its vaiations, applications and algoithmic aspects, PhD Thesis, The Johns Hopkins Univesity, Baltimoe, 198. [2] D. Avis, G.T. Toussaint, An efficient algoithm fo decomposing a polygon into sta-shaped polygons, Patten Recognition 1 (1981), no. 6, [] B. Chazelle, Tiangulating a simple polygon in linea time, Discete Comput. Geom. 6 (1991), no. 5, [], Tiangulating a simple polygon in linea time, Poc. 1st Symp. Foundations of Compute Science I (1990),

9 ART GALLERIES WITH INTERIOR WALLS 9 [5] N. Chiba, T. Nishizeki, S. Abe, T. Ozawa, A linea algoithm fo embedding plana gaphs using P Q-tees, J. Comput. System Sci 0 (1985), no. 1, [6] V. Chvátal, A combinatoial theoem in plane geomety, J. Combinatoial Theoy Se. B 18 (1975), 9 1. [7] J. Czyzowicz, E. Rivea-Campo, N. Santoo, J. Uutia, J. Zaks, Tight bounds fo the ectangula at galley poblem, Lectue Notes in Comput. Sci. 570 (1992), [8], Guading ectangula at galleies, Discete Appl. Math. 50 (199), no. 2, [9] S. Fisk, A shot poof of Chvátal s watchman theoem, J. Combin. Theoy Se. B 2 (1978), no., 7. [10] F. Gavil, Algoithms fo minimum coloing, maximum clique, minimum coveing by cliques, and maximum independent set of a chodal gaph, SIAM J. Comput. 1 (1972), no. 2, [11] J. Hutchinson, At Galleies with Walls, Poblem #1078, Ame. Math. Monthly 102 (1995), no. 8, 76. [12] R. Jayakuma, K. Thulasiaman, M. Swamy, Plana embedding: linea-time algoithms fo vetex placement and edge odeing, IEEE Tans. Cicuits and Systems 5 (1988), no.,. [1] A. Kündgen, At Galleies with Walls, Solution to Poblem #1078, Ame. Math. Monthly 105 (1998), no., [1] D. T. Lee, A. K. Lin, Computational complexity of at galley poblems, IEEE Tans. Infom. Theoy 2 (1986), no. 2, [15] J. O Rouke, At galley theoems and algoithms, The Intenational Seies of Monogaphs on Compute Science, The Claendon Pess, Oxfod Univesity Pess, New Yok, [16] D.J. Rose, R.E. Tajan, G.S. Lueke, Algoithmic aspects of vetex elimination on gaphs., SIAM J. Comput. 5 (1976), no. 2, [17] T. Sheme, Recent esults in at galleies, Poceedings IEEE 80(9) (Septembe 1992), Depatment of Mathematics, Univesity of Illinois, Ubana, Illinois addess: kundgen@math.uiuc.edu