THE DOMINATION NUMBER OF GRIDS *

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1 SIAM J. DISCRETE MATH. Vol. 2, No. 3, pp Society for Industrial and Applied Mathematics THE DOMINATION NUMBER OF GRIDS * DANIEL GONÇALVES, ALEXANDRE PINLOU, MICHAËL RAO, AND STÉPHAN THOMASSÉ Abstract. In this paper, we conclude the calculation of the domination number of all ðn; mþ grid graphs. Indeed, we prove Chang s conjecture saying that for every 16 n m, γðg n;m Þ¼b ðnþ2þðmþ2þ c 4. Key words. graph, domination, grid AMS subject classification. 0C69 DOI / Introduction. A dominating set in a graph G is a subset of vertices S such that every vertex in V ðgþ \ S is a neighbor of some vertex of S. The domination number of G is the minimum size of a dominating set of G; we denote it by γðgþ. This paper is devoted to the calculation of the domination number of complete grids. The notation ½iŠ denotes the set f1; 2; :::;ig.ifwis a word in the alphabet A, w½iš is the ith letter of w, and for every a in A, jwj a denotes the number of occurrences of a in w (i.e., jfi f1; :::;jwjg w½iš¼agj). For a vertex v, N ½vŠ denotes the closed neighborhood of v (i.e., the set of neighbors of v and v itself). For a subset of vertices S of a vertex set V of a graph, we denote by N½SŠ ¼ S v S N½vŠ. Note that D is a dominating set of G if and only if N½DŠ ¼V ðgþ. Let G n;m be the complete ðn; mþ grid; i.e., the vertex set of G n;m is V n;m ½nŠ ½mŠ, and two vertices ði; jþ and ðk; lþ are adjacent if jk ijþjl jj ¼1. The couple ð1; 1Þ denotes the bottom-left vertex of the grid and the couple ði; jþ denotes the vertex of the ith column and the jth row. We will always assume in this paper that n m. Let us illustrate our purpose by an example of a dominating set of the complete grid G 24;24 on Figure 1. The first results on the domination number of grids were obtained about 30 years ago with the exact values of γðg 2;n Þ, γðg 3;n Þ, and γðg 4;n Þ found by Jacobson and Kinch [8] in In 1993, Chang and Clark [3] found those of γðg ;n Þ and γðg 6;n Þ. These results were obtained analytically. Chang [2] devoted his Ph.D. thesis to study the domination number of grids; he conjectured that this invariant behaves well provided that n is large enough. Specifically, Chang conjectured the following. CONJECTURE 1 (Chang [2]). For every 16 n m, ðn þ 2Þðm þ 2Þ γðg n;m Þ¼ 4: *Received by the editors February 24, 2011; accepted for publication (in revised form) June 3, 2011; published electronically September 22, This work has been partially supported by the ANR Project GRATOS ANR-JCJC Université Montpellier, 2, CNRS, LIRMM 161 rue Ada, 3409 Montpellier Cedex, France (daniel. goncalves@lirmm.fr, alexandre.pinlou@lirmm.fr, stephan.thomasse@lirmm.fr). Département Mathématiques et Informatique Appliqués, Université Paul Valéry, Montpellier 3, Route de Mende, Montpellier Cedex, France. CNRS, Laboratoire J.-V. Poncelet, Moscow, Russia. Université Bordeaux CNRS, LABRI, 31, cours de la Libération 3340 Talence, France (michael.rao@labri.fr). 1443

2 1444 D. GONÇALVES, A. PINLOU, M. RAO, AND S. THOMASSÉ FIG. 1. Example of a set of size 131 dominating the grid G 24;24. Observe that, for instance, this formula would give 131 for the domination number of the grid in Figure 1. To motivate his bound, Chang proposed some constructions of dominating sets achieving the upper bound. LEMMA 1 (Chang [2]). For every 8 n m, ðn þ 2Þðm þ 2Þ γðg n;m Þ 4. In the meantime, some algorithms based on dynamic programming were designed to compute a lower bound of γðg n;m Þ. There were numerous intermediate results found for γðg n;m Þ for small values of n and m (see [4, 9, 10] for details). In 1986, Hare, Hedetniemi, and Hare [9] gave a polynomial time algorithm to compute γðg n;m Þ when n is fixed. Nevertheless, this algorithm is not usable in practice when n is over 20. Fisher [6] developed the idea of searching for periodicity in the dynamic programming algorithms, and using this technique, he found the exact values of γðg n;m Þ for all n 21. We recall these values for the sake of completeness. THEOREM 2 (Fisher [6]). For all n m and n 21, we have

3 8 m 3 mþ1 2 3mþ1 4 THE DOMINATION NUMBER OF GRIDS 144 if n ¼ 1 if n ¼ 2 if n ¼ 3 m þ 1 if n ¼ 4 and m ¼ ; 6; 9 m if n ¼ 4 and m ; 6; 9 6mþ4 1 if n ¼ and m ¼ 7 6mþ4 if n ¼ and m 7 10mþ4 7 if n ¼ 6 mþ1 3 if n ¼ 7 1mþ7 8 if n ¼ 8 23mþ10 11 if n ¼ 9 30mþ1 >< 13 1 if n ¼ 10 and m or m ¼ 13; 16 γðg n;m Þ¼ : 30mþ1 13 if n ¼ 10 and m and m 13; 16 38mþ if n ¼ 11 and m ¼ 11; 18; 20; 22; 33 38mþ22 1 if n ¼ 11 and m 11; 18; 20; 22; 33 80mþ38 29 if n ¼ 12 98mþ if n ¼ 13 and m 33 13; 16; 18; 19 98mþ4 33 if n ¼ 13 and m 33 13; 16; 18; 19 3mþ if n ¼ 14 and m mþ20 11 if n ¼ 14 and m mþ if n ¼ 1 and m 26 44mþ28 13 if n ¼ 1 and m 26 ðnþ2þðmþ2þ >: 4 if n 16 Note that these values are obtained by specific formulas for every 1 n 1 and by the formula of Conjecture 1 for every 16 n 21. This proves Chang s conjecture for all values 16 n 21.

4 1446 D. GONÇALVES, A. PINLOU, M. RAO, AND S. THOMASSÉ In 2004, Conjecture 1 was confirmed up to an additive constant. THEOREM 3 (Guichard [7]). For every 16 n m, ðn þ 2Þðm þ 2Þ γðg n;m Þ 9: In this paper, we prove Chang s conjecture, hence finishing the computation of γðg n;m Þ. We adapt Guichard s ideas to improve the additive constant from 9 to 4 when 24 n m. Cases n ¼ 22 and n ¼ 23 can be proved in a couple of hours using Fisher s method (described in [6]) on a modern computer. They can also be proved by a slight improvement of the technique presented in the next section. 2. Values of γ G n;m when 24 n m. Our method follows the idea of Guichard [7]. A slight improvement is enough to give the exact bound. A vertex of the grid G n;m dominates at most vertices (its 4 neighbors and itself). It is then clear that γðg n;m Þ n m. The previous inequality would become an equality if there would be a dominating set D such that every vertex of G n;m is dominated only once, and all vertices of D are of degree 4 (i.e., dominates exactly vertices); in this case, we would have jdj n m ¼ 0. This is clearly impossible (e.g., to dominate the corners of the grid, we need vertices of degree at most 3). Therefore, our goal is to find a dominating set D of G n;m such that the difference jdj n m is the smallest. Let S be a subset of V ðg n;m Þ. The loss of S is lðsþ ¼ jsj jn½sšj. PROPOSITION 4. The following properties of the loss function are straightforward: (i) For every S, lðsþ 0 (positivity). (ii) If S 1 S 2 ¼, then lðs 1 S 2 Þ¼lðS 1 ÞþlðS 2 ÞþjN½S 1 Š N½S 2 Šj. (iii) If S 0 S, then lðs 0 Þ lðsþ (increasing function). (iv) If S 1 S 2 ¼, then lðs 1 S 2 Þ lðs 1 ÞþlðS 2 Þ (superadditivity). Let us denote by l n;m the minimum of lðdþ when D is a dominating set of G n;m. LEMMA. γðg n;m Þ¼d n mþl n;m e. Proof. If D is a dominating set of G n;m,thenlðdþ¼ jdj jn½dšj ¼ jdj n m. Hence, by minimality of l n;m and γðg n;m Þ, we have l n;m ¼ γðg n;m Þ n m. Our aim is to get a lower bound for l n;m. As the reader can observe in Figure 1, the loss is concentrated on the border of the grid. We now analyze more carefully the loss generated by the border of thickness 10. We define the border B n;m V n;m of G n;m as the set of vertices ði; jþ where i 10, or j 10, ori>n 10, orj>m 10 to which we add the four vertices ð11; 11Þ, ð11;m 10Þ, ðn 10; 11Þ, ðn 10;m 10Þ. Given a subset S V, let I ðsþ be the internal vertices of S, i.e., I ðsþ ¼fv S N½vŠ Sg. These sets are illustrated in Figure 2. We will compute b n;m ¼ min D lðdþ, where D is a subset of B n;m and dominates I ðb n;m Þ, i.e., D B n;m and I ðb n;m Þ N½DŠ. Observe that this lower bound b n;m is a lower bound of l n;m. Indeed, for every dominating set D of G n;m, the set D 0 D B n;m dominates I ðb n;m Þ; hence b n;m lðd 0 Þ lðdþ. In the remainder, we will compute b n;m,andwe will show that b n;m ¼ l n;m. In the following, we split the border B n;m into four parts, O m 12, P n 12, Q m 12, R n 12, which are defined just below. For p 12, let P p B n;m defined as follows: P p ¼ð½10Š f12gþ ð½11š f11gþ ð½pš ½10ŠÞ. We define the input vertices of P p as ½10Š f12g and the output vertices of P p as fpg ½10Š. The set P p, illustrated for p ¼ 19 in Figure 3, corresponds to the set of black and gray vertices. The input vertices are the gray circles, and the output vertices are the gray squares. Recall that in our drawing conventions, the vertex ð1; 1Þ is the

5 THE DOMINATION NUMBER OF GRIDS 1447 FIG.2. The graph G 30;40. The set B 30;40 is the set of vertices filled in black or in gray. The set I ðb 30;40 Þ is the set of vertices filled in black. bottom-left vertex, and hence the vertex ði; jþ is in the ith column from the left and in the jth row from the bottom. For n; m N, let f n;m ½nŠ ½mŠ ½mŠ ½nŠ be the bijection such that f n;m ði; jþ ¼ ðj; n i þ 1Þ. It is clear that the set B n;m is the disjoint union of the following four sets depicted in Figure 4: P n 12, Q m 12 ¼ f n;m ðp m 12 Þ, R n 12 ¼ f m;n f n;m ðp n 12 Þ, and O m 12 ¼ f 1 n;mðp m 12 Þ. Similar to P n 12, the sets O m 12, Q n 12,andR m 12 have input FIG. 3. The set P 19 (black and gray), the set of input vertices (gray circles), and the set of output vertices (gray squares).

6 1448 D. GONÇALVES, A. PINLOU, M. RAO, AND S. THOMASSÉ R n 12 Q m 12 V Gn m Bn m O m 12 P n 12 FIG. 4. The sets O m 12, P n 12, Q m 12, and R n 12. and output vertices. For instance, the output vertices of Q m 12 correspond in Figure 3 to the white squares. Every set playing a symmetric role, we now focus on P n 12. Given a subset S of V n;m, let the labeling ϕ S V n;m f0; 1; 2g be such that 8 < 0 if ði; jþ S; ϕ S ði; jþ ¼ 1 if ði; jþ N½SŠ \ S; : 2 otherwise: Note that ϕ S is such that any two adjacent vertices in G n;m cannot be labeled 0 and 2. Given p 12 and a set S P p, the input word (respectively, output word) ofs for P p, denoted by w in ðsþ (respectively, w out p ðsþ), is the 10-letter word in the alphabet f0; 1; 2g obtained by reading ϕ S on the input vertices (respectively, output vertices) of P p. More precisely, its ith letter is ϕ S ði; 12Þ (respectively, ϕ S ðp; iþ). This notion extends to O p, Q p, and R p in a natural way: For example, the output word of S O p for O p is w out p ðf n;m ðsþþ. According to the definition of ϕ, the input and output words belong to the set W of 10-letter words in f0; 1; 2g, which avoid 02 and 20. The number of k-digit trinary numbers without 02 or 20 is given by the following formula [6]: ð1þ The size of W is therefore jw j¼8119. p ð1 þ ffiffiffi p 2 Þ kþ1 þð1 ffiffi 2 Þ kþ1. 2

7 THE DOMINATION NUMBER OF GRIDS 1449 Given two words w; w 0 W, we define D w;w 0 p as the family of subsets D of P p such that D dominates I ðp p Þ, w is the input word w in ðdþ, w 0 is the output word w out p ðdþ. To complete our calculation, we will need to know, for two given words w; w 0 W, the minimum loss over all losses lðdþ, where D D w;w 0 p. For this purpose, we introduce the ð8119; 8119Þ square matrix C p. For w; w 0 W, let C p ½w; w 0 Š¼min D D w;w 0 lðdþ, p where the minimum of the empty set is þ. Let w; w 0 W be two given words. Because of the symmetry of P 12 with respect to the first diagonal (bottom-left to top-right) of the grid, if a vertex set D belongs to D w;w 0 12, then D 0 ¼fðj; iþjði; jþ Dg belongs to D w 0 ;w 12. Moreover, it is clear that, again because of the symmetry, lðdþ ¼lðD 0 Þ. Therefore, we have C 12 ½w; w 0 Š¼C 12 ½w 0 ;wš, and thus C 12 is a symmetric matrix. Despite the size of C 12 and the size of P 12 (141 vertices), it is possible to compute C 12 in less than one hour by computer. Indeed, we choose a sequence of subsets X 0 ¼ ;X 1 ; :::;X 141 ¼ P 12 such that for every i f1; :::;141g, X i X iþ1 and X iþ1 \ X i ¼fx iþ1 g. Moreover, we choose the sequence such that for every i, X i \ I ðx i Þ is at most 21. This can be done, for example, by taking x iþ1 ¼ minfðx; yþ ðx; yþ P 12 \ X i g, where the order is the lexical order. For i f0; :::; 141g, we compute for every labeling f F i, where F i is the set of functions ðx i \ I ðx i ÞÞ f0; 1; 2g, the minimal loss l i;f of a set S X i that dominates I ðx i Þ and such that ϕ S ðvþ ¼fðvÞ for every v X i \ I ðx i Þ. Note that not every labeling is possible since two adjacent vertices cannot be labeled 0 and 2. The number of possible labelings can be computed using formula (1), and since jx i \ I ðx i Þj can be covered by a path of at most 23 vertices, this gives, in the worst case, that this number is less than 10 9 and can then be processed by a computer. We compute inductively the sequence ðl i;f Þ f F i from the sequence ðl i 1;f Þ f F i 1 by dynamical programming, and the matrix C is easily deduced from ðl 141;f Þ f F 141. In the following, our aim is to glue P n 12, Q m 12, R n 12, and O m 12 together. For two consecutive parts of the border, say, P n 12 and Q m 12, the output word of Q m 12 should be compatible with the input word of P n 12. Two words, w and w 0 of W, are compatible if the sum of their corresponding letters is at most 2, i.e., w½išþw 0 ½iŠ 2 for all i ½9Š. Note that w½10šþw 0 ½10Š could be greater than 2 since the corresponding vertices can be dominated by some vertices of V n;m \ B n;m. Given two words w; w 0 W, let lðw; w 0 Þ¼jfi ½10Š w½iš 2andw 0 ½iŠ ¼0gj þ jfi ½10Š w 0 ½iŠ 2andw½iŠ ¼0gj: LEMMA 6. Let D be a dominating set of G n;m, and let us denote D P ¼ D P n 12 and D Q ¼ D Q m 12. Then lðd ðp n 12 Q m 12 ÞÞ ¼ lðd P ÞþlðD Q Þþlðw; w 0 Þ, where w ¼ w in ðd P Þ and w 0 ¼ w out m 12ðf 1 n;mðd Q ÞÞ. Moreover, w and w 0 are compatible. Proof. By Proposition 4(ii), lðd ðp n 12 Q m 12 ÞÞ ¼ lðd P ÞþlðD Q Þþ jn½d P Š N½D Q Šj. It suffices then to note that lðw; w 0 Þ¼jN½D P Š N½D Q Šj to get lðd ðp n 12 Q m 12 ÞÞ ¼ lðd P ÞþlðD Q Þþlðw; w 0 Þ. In what remains, we prove that w and w 0 are compatible. If those two words were not compatible, there would exist an index i ½9Š such that w out 1 m 12ðf n;mðd Q ÞÞ½iŠþ w in ðd P Þ½iŠ > 2. Thus at least one of these two letters should be a 2, and the other one should not be 0. Suppose that w out 1 m 12ðf n;mðd Q ÞÞ½iŠ ¼2, and note that this means that the vertex ði; 13Þ is not dominated by a vertex in D Q. Since D is a dominating set of G n;m, every

8 140 D. GONÇALVES, A. PINLOU, M. RAO, AND S. THOMASSÉ output vertex of Q m 12 except ð10; 13Þ (and every input vertex of P n 12 except ð10; 12Þ) is dominated by a vertex of D Q or by a vertex of D P. Thus ði; 13Þ should be dominated by its unique neighbor in P n 12, namely, ði; 12Þ. This would imply that ði; 12Þ D, contradicting the fact that w in ðd P Þ½iŠ 0. Similarly, if w in ðd P Þ½iŠ ¼2, the vertex ði; 12Þ is not dominated by a vertex in D P ; thus ði; 12Þ must be dominated by the vertex ði; 13Þ D, contradicting the fact that w out 1 m 12ðf n;mðd Q ÞÞ½iŠ 0. Lemma 6 is designed for the two consecutive parts P n 12 and Q m 12 of the border of G n;m. It is easy to see that this extends to any pair of consecutive parts of the border, i.e., Q m 12 and R n 12, R n 12 and O m 12, O m 12 and P n 12. We define the matrix ð8119; 8119Þ square matrix L, which contains, for every pair of words w; w 0 W, the value lðw; w 0 Þ: þ if w and w L½w; w 0 Š¼ 0 are not compatible; lðw; w 0 Þ otherwise: Note that L is symmetric since lðw; w 0 Þ¼lðw 0 ;wþ. Let be matrix multiplication in ðmin; þþ algebra; i.e., C ¼ A B is the matrix where for all i, j, C½i; jš ¼min k A½i; kšþb½k; jš. Let M p ¼ L C p for p 12. By construction, M n 12 ½w; w 0 Š corresponds to the minimum possible loss lðd P n 12 Þ of a dominating set D V n;m that dominates I ðp n 12 Þ and such that w is the output word of Q m 12 and w 0 is the output word of P n 12. LEMMA 7. For all 24 n m, we have b n;m min M n 12½w 1 ;w 2 ŠþM m 12 ½w 2 ;w 3 ŠþM n 12 ½w 3 ;w 4 ŠþM m 12 ½w 4 ;w 1 Š: w 1 ;w 2 ;w 3 ;w 4 W Proof. Consider a set D B n;m that dominates I ðb n;m Þ and achieves lðdþ ¼b n;m. Let D P ¼ D P n 12, D Q ¼ D Q m 12, D R ¼ D R n 12, and D O ¼ D O m 12. Let w P (w Q, w R, and w O, respectively) be the input word of P n 12 (Q m 12, R n 12, and O m 12 ), and w P 0 (w Q 0, w R 0,andw O 0 ) be the output word of P n 12 (Q m 12, R n 12,and O m 12 ). By definition of C p, the loss of D P is at least C n 12 ½w P ;w P 0 Š. Similarly, we have lðd Q Þ C m 12 ½w Q ;w Q 0 Š, lðd RÞ C n 12 ½w R ;w R 0 Š, and lðd OÞ C m 12 ½w O ;w O 0 Š. By the definition of the loss, lðdþ ¼b n;m ¼ jdj jn½dšj ¼ lðd O ÞþlðD P ÞþlðD Q ÞþlðD R ÞþL½w O 0 ;w PŠþL½w P 0 ;w QŠ þ L½w Q 0 ;w RŠþL½w R 0 ;w OŠ by Lemma 6 and since N½D P Š N½D R Š¼N½D Q Š N½D O Š¼ C m 12 ½w O ;w O 0 ŠþC n 12½w P ;w P 0 ŠþC m 12½w Q ;w Q 0 ŠþC n 12½w R ;w R 0 Š þ L½w O 0 ;w PŠþL½w P 0 ;w QŠþL½w Q 0 ;w RŠþL½w R 0 ;w OŠ M m 12 ½w O ;w P ŠþM n 12 ½w P ;w Q ŠþM m 12 ½w Q ;w R ŠþM n 12 ½w R ;w O Š since w O 0 and w P ðrespectively; w P 0 and w Q;w Q 0 and w R;w R 0 and w OÞ are compatibles:

9 According to Lemma 7, to bound b n;m it would be thus interesting to know M p for p>12. It is why we introduce the following ð8119; 8119Þ square matrix T. LEMMA 8. There exists a matrix T such that C pþ1 ¼ C p T for all p 12. This matrix is defined as follows: w out p 8 þ if i ½10Š s:t: w ½iŠ ¼0 and w 0 ½iŠ ¼2 þ if i ½9Š s:t: w½iš ¼2 and w 0 ½iŠ 0 þ if i f2; :::;9gs:t: w >< 0 ½iŠ ¼1;w½iŠ 0;w 0 ½i 1Š 0 T½w; w 0 and w Š¼ 0 ½i þ 1Š 0 þ if w 0 ½1Š ¼1;w½1Š 0 and w 0 ½2Š 0 þ if w 0 ½10Š ¼1;w½10Š 0 and w 0 ½9Š 0 3 jw >: 0 j 0 jwj 2 jw 0 j 1 þjwj 0 1 if w 0 ½10Š ¼0 3 jw 0 j 0 jwj 2 jw 0 j 1 þjwj 0 otherwise: Proof. Consider a set S 0 P pþ1 dominating I ðp pþ1 Þ, and let S ¼ S 0 P p. Let w ¼ ðsþ and w 0 ¼ w out pþ1 ðs 0 Þ. Let ΔðS;S 0 Þ¼lðS 0 Þ lðsþ. By definition of the loss, ΔðS;S 0 Þ¼ js 0 \ Sj jn½s 0 Š \ N½SŠj. Let us compute ΔðS;S 0 Þ in terms of the number of occurrences of 0 s, 1 s, and 2 s in the words w and w 0. The set S 0 \ S corresponds to the vertices fðp þ 1;iÞ i ½10Š;w 0 ½iŠ ¼0g. The set N½S 0 Š \ N½SŠ corresponds to the vertices dominated by S 0 but not dominated by S; these vertices clearly belong to the columns p, p þ 1, and p þ 2. Since S 0 dominates I ðp pþ1 Þ, those in the column p are the vertices fðp; iþ i ½10Š;w½iŠ ¼2g. Those in the column p þ 1 are the vertices fðp þ 1;iÞ i ½10Š;w 0 ½iŠ 2;w½iŠ 0g and possibly the vertex ðp þ 1; 11Þ when w 0 ½10Š ¼0. Finally, those in the column p þ 2 are the vertices fðp þ 2;iÞ i ½10Š;w 0 ½iŠ ¼0g. We then get ΔðS;S 0 Þ¼ 3 jw 0 j 0 jwj 2 jw 0 j 1 þjwj 0 1 if w 0 ½10Š ¼0; 3 jw 0 j 0 jwj 2 jw 0 j 1 þjwj 0 otherwise; where jwj n denotes the number of occurrences of the letter n in the word w. Thus ΔðS;S 0 Þ depends only on the output words of S and S 0, and we can denote this value by Δðw; w 0 Þ. Note, however, that there exist pairs of words ðw; w 0 Þ that could not be the output words of S and S 0 ; there are three cases: Case 1. w½iš ¼0 and w 0 ½iŠ ¼2 since the vertex ðp þ 1;iÞ would be dominated by ðp; iþ contradicting its label 2. Case 2. w½iš ¼2 and w 0 ½iŠ 0 for i ½9Š since ðp; iþ would not be dominated, contradicting the fact that S 0 dominates I ðp pþ1 Þ. Case 3. w 0 ½iŠ ¼1 and w 0 ½i 1Š 0, w 0 ½i þ 1Š 0, w½iš 0 since ðp þ 1;iÞ would be dominated according to its label but none of its neighbors belong to S 0. For these forbidden cases, we set Δðw; w 0 Þ¼þ. By definition, C pþ1 ½w i ;w 0 Š is the minimum loss lðs 0 Þ of a set S 0 P pþ1 that dominates I ðp pþ1 Þ, with w i as input word and w 0 as output word. It is clear that S ¼ S 0 P p dominates I ðp p Þ and has w i as input word. Let w be its output word and note that C pþ1 ½w i ;wš¼lðs 0 Þ¼lðSÞþΔðw; w 0 Þ. The minimality of lðs 0 Þ implies the minimality of lðsþ over the sets X D w i;w 0 p. Indeed, any set X D w i;w 0 p could be turned into a set X 0 D w i;w pþ1 by adding vertices of the p þ 1th column accordingly to w 0. Thus which implies that THE DOMINATION NUMBER OF GRIDS 141 C pþ1 ½w i ;wš¼c p ½w i ;w 0 ŠþΔðw; w 0 Þ;

10 142 D. GONÇALVES, A. PINLOU, M. RAO, AND S. THOMASSÉ C pþ1 ½w i ;wš min w C p ½w i ;w 0 ŠþΔðw; w 0 Þ: On the other hand, for every word w o W such that C p ½w i ;w o Š þ and Δðw o ;w 0 Þ þ, there is a set S D w i;w o p,withlðsþ ¼C p ½w i ;w o Š, that can be turned into a set S 0 D w i;w 0 pþ1 with lðs 0 Þ¼C p ½w i ;w o ŠþΔðw o ;w 0 Þ. Thus C pþ1 ½w i ;wš min C p ½w i ;w o ŠþΔðw o ;w 0 Þ: w o This concludes the proof of the lemma. By the definition of M p,wehavealsom pþ1 ¼ M p T. NotethatT is a sparse matrix: About 9.% of its entries are þ. Thus the multiplication by T in the ðmin; þþ algebra can be done in a reasonable amount of time by an algorithm using Oðn 3 Þ additions and comparisons. Note that the best known algorithm for multiplication of the ðn; nþ matrix in the ðmin; þþ algebra is due to Chan [1]: He provided a Oð n3 log n Þ-time algorithm. However, for simplicity reasons, we implemented the algorithm in Oðn 3 Þ. Fact 9. The computations give us that M 126 ¼ M 12 þ 1. Thus, since ða þ cþ B ¼ðA BÞþc for any matrices A, B and any integer c, we have that M 12þk ¼ M 12 þ k for every k N. Let us define M p 0 ¼ min k N ðm pþk kþ. Then, for all q p, M q M p 0 þðq pþ.by Fact 9, M p 0 ¼ min k f0;:::12 pg ðm pþk kþ. Fact 10. By computing M 12 0, and A 0 ¼ M 12 0 M 12 0, we obtain that min w 1 ;w 3 ða 0 þ A 0T Þ½w 1 ;w 3 Š¼76 (where A T is the transpose of A). This implies that minðmin w 1 ;w 3 w 2 M 0 12 ½w 1;w 2 ŠþM 0 12 ½w 2;w 3 ŠÞ þ ðmin w 4 M 0 12 ½w 3;w 4 ŠþM 0 12 ½w 4;w 1 ŠÞ ¼ 76; min w 1 ;w 2 ;w 3 ;w 4 M 0 12 ½w 1;w 2 ŠþM 0 12 ½w 2;w 3 ŠþM 0 12 ½w 3;w 4 ŠþM 0 12 ½w 4;w 1 Š¼76: THEOREM 11. If 24 n m, then ðn þ 2Þðm þ 2Þ γðg n;m Þ¼ 4: Proof. By Chang s construction [2], γðg n;m Þ b ðnþ2þðmþ2þ c 4. Let us now compute a lower bound for the loss of a dominating set of G n;m. l n;m b n;m min M n 12 ½w 1 ;w 2 ŠþM m 12 ½w 2 ;w 3 ŠþM n 12 ½w 3 ;w 4 ŠþM m 12 ½w 4 ;w 1 Š w 1 ;w 2 ;w 3 ;w 4 by Lemma 7 min M w 1 ;w 2 ;w 3 ;w 12 0 ½w 1;w 2 Šþðn 12 12ÞþM 12 0 ½w 2;w 3 Šþðm 12 12Þ 4 þ M 12 0 ½w 3;w 4 Šþðn 12 12ÞþM 12 0 ½w 4;w 1 Šþðm 12 12Þ 2ðn þ m 48Þþ min M w 1 ;w 2 ;w 3 ;w 12 0 ½w 1;w 2 ŠþM 12 0 ½w 2;w 3 Š 4 þ M 12 0 ½w 3;w 4 ŠþM 12 0 ½w 4;w 1 Š 2ðn þ m 48Þþ76 2ðn þ mþ 20:

11 THE DOMINATION NUMBER OF GRIDS 143 Thus by Lemma, we have n:m þ 2ðn þ mþ 20 γðg n;m Þ ðn þ 2Þðm þ 2Þ 4 4 ðn þ 2Þðm þ 2Þ 4: Our proofs and computations in this paper led us to get a lower bound on the loss b n;m, namely, b n;m 2ðn þ mþ 20. This lower bound appeared to be the best possible and allowed us to conclude the proof of our main theorem. This lower bound is highly related to the shape of the border B n;m (thickness of 10 with four extra vertices in the corners). This shape was obtained after several unsuccessful attempts using other different shapes. Indeed, a border of thickness less than 10 or a border of thickness 10 without the extra four vertices in the corner did not give a big enough lower bound to conclude. REFERENCES [1] T. M. CHAN, All-pairs shortest paths with real weights in Oð n3 log nþ time, Algorithmica, 0 (2007), pp [2] T. Y. CHANG, Domination Numbers of Grid Graphs, Ph.D. thesis, University of South Florida, Tampa, FL, [3] T. Y. CHANG AND W. E. CLARK, The domination numbers of the n and 6 n grid graphs, J. Graph Theory, 17 (1993), pp [4] T. Y. CHANG, W. E. CLARK, AND E. O. HARE, Domination numbers of complete grids I, Ars Combin., 38 (1994), pp [] E. J. COCKAYNE, E. O. HARE, S. T. HEDETNIEMI, AND T. V. WIMER, Bounds for the domination number of grid graphs, Congr. Numer., 47 (198), pp [6] D. C. FISHER, The Domination Number of Complete Grid Graphs, manuscript, [7] D. R. GUICHARD, A lower bound for the domination number of complete grid graphs, J. Combin. Math. Combin. Comput., 49 (2004), pp [8] M. S. JACOBSON AND L. F. KINCH, On the domination number of products of a graph I, Ars Combin., 10 (1983), pp [9] E. O. HARE, S. T. HEDETNIEMI, AND W. R. HARE, Algorithms for computing the domination number of K N complete grid graphs, Congr. Numer., (1986), pp [10] H. G. SINGH AND R. P. PARGAS, A parallel implementation for the domination number of grid graphs, Congr. Numer. 9 (1987), pp

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