CLEANING OUT BOXES: THE BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS

Transcription

1 CLEANING OUT BOXES: THE BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS ANTHONY BONATO AND MARGARET-ELLEN MESSINGER Abstract. We consider a graph cleaning model and study its associated parameter called the brush number. Although several bounds and algorithmic results are known on the brush number for general graphs, the brush number of product graphs has remained largely uninvestigated. We study the brush number of Cartesian products of graphs. A conjecture is introduced on bounds for the brush number of Cartesian product graphs, and the conjecture is proven for paths and partially for cycles. Brush numbers are computed exactly for Cartesian products of paths with cycles, and cliques with paths and cycles. 1. Introduction Graph products are powerful tools in graph theory, and are concerned with taking two or more graphs and generating new ones. Products often lead to more complex graphs, and yet they are good instruments for probing the structure of graphs. Cartesian products are one of the most studied graph products. The Cartesian or box product of G and H, written G H, has vertex set V G V H. Vertices a, b and c, d are joined if a = c and bd EH, or ab EG and b = d. See [7] for further background on the Cartesian and other graph products. Several unresolved and central conjectures focus on graph products, such as those of Vizing on the domination number of Cartesian products [, 15], and Hedetniemi s conjecture on the chromatic number of categorical products [5, 1]. In the present work, we consider a graph cleaning problem on Cartesian products. Regenerating contaminants, such as algae or zebra mussels, can accumulate in man-made water systems, restricting water flow even in large diameter piping. Mechanical brush cleaning technology is often used to remove biofouling from these water systems as routine maintenance [6, 8]. The cleaning model introduced in [9, 11] considers a network that must be cleaned periodically of a regenerating contaminant. It follows the decontamination metaphor for searching a network and is related to both the edge-searching problem and the chip-firing game. We only consider undirected connected graphs. Brushes are assigned to a set of vertices and initially, all edges are considered contaminated, or dirty. At each time-step, a vertex is cleaned, whereupon it sends one brush along each dirty incident edge, cleaning those edges. Brushes may only traverse dirty edges and a vertex may only be cleaned if it contains at least as many brushes as dirty incident edges. Thus, a network has been cleaned once every edge has been cleaned. The brush number of a graph G, denoted by bg, is the minimum number of brushes needed to clean G. Figure 1 illustrates the cleaning process for a graph G where there are initially two brushes at vertex a Mathematics Subject Classification. 05C75, 05C76. Key words and phrases. Cartesian products, brush number, graph cleaning, paths, cycles, cliques. The authors gratefully acknowledge support from NSERC and MITACS. 1

2 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER b a b a b a b a b a c d c d c d c d c d brushes at a 1 brush at b 1 brush at c brushes at c 1 brush at c 1 brush at d 1 brush at c 1 brush at d Figure 1. A cleaning process on a given graph. The cleaning process is considering to be on-going: once a network has been cleaned, it is regarded as contaminated again, and the brushes reclean the network. It was shown in [11] that using a fixed number of brushes, once a network G has been cleaned, the final configuration of brushes can always act an initial configuration of brushes to reclean G when it is recontaminated. Computing bg is a non-trivial problem for general graphs. For example, it was shown in [] that determining whether bg k is NP-complete and the problem remains NPcomplete for bipartite graphs of maximum degree 6, planar graphs of maximum degree, and 5-regular graphs. Some bounds on the brush number have been determined, as well as exact values for trees, cycles, complete graphs, and complete multipartite graphs; see [10, 11]. Asymptotic bounds and concentration results for the brush number of binomial random [13] and d-regular random graphs were presented in [13] and [1, 1], respectively. Bounds on bg are generally difficult to determine, since by adding or removing an edge from a graph G, the brush number may increase, decrease, or remain the same. See Figure. bg = bg + e 1 = 3 bg + e = bg + e 3 = 5 e 1 e 3 e Figure. The addition of an edge may increase, decrease, or not affect the brush number The Box Cleaning Conjecture. We know of no formulas for bg H in terms of the brush numbers of the factors G and H. Instead, we focus on bounds for the brush number of the Cartesian product. The following upper bound was proved in [11]: bg H V H bg + V G bh,

3 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 3 and was used to give bounds on the brush number of hypercubes. We know that for a graph G of order m, 1 = bp m bg bk m =, where the second inequality follows by Theorem 3.1 in [10], and the final equality follows by results from [11]. Given these facts, it is natural to consider bounds on the brush number of the Cartesian product with paths and cliques. Box Cleaning Conjecture: For m 1 a positive integer, a graph G of order m and a graph H, we have that bp m H bg H bk m H. Our main goal is to verify the conjecture when H is a path P n or cycle C n of order n. We prove the following theorems, the first of which settles the conjecture for paths. Theorem 1. For m 1 and n and a graph G, bp m P n bg P n bk m P n. 1 In the case of a cycle we have the following partial verification of the conjecture. Theorem. For n m 1 and a graph G, bg C n bk m C n. The proofs of Theorems 1 and are deferred to Section. While proving these results, we compute the brush numbers bp m H, bk m H, where H is P n or C n. In [11], the following was proven for grids, which are the Cartesian product of paths. Theorem 3. [11] For m, n > 1, bp m P n = m + n. We use the notation δ m,odd to be either 1 if m is odd, and 0 otherwise. Theorem. For m, n 1 integers we have the following. 1 bk m P n = n + δ m,odd. bk m C n = n +. Hence, by Theorems 1,, and we have the following bounds. Corollary 5. For m 1 and n and a graph G, m + n bg P n n + δ m,odd. Corollary 6. For n m 1 and a graph G, bg C n n +.

4 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER 1.. Background and Notation. We recall some notation from [11] that will be helpful in the proofs of Theorems 1,, and. At each time-step t in cleaning a graph G = V, E let ω t v denote the number of brushes at vertex v. Let ω 0 be the initial configuration of brushes; that is, a placement of the brushes on vertices at the beginning of the cleaning process. If the cleaning process takes T 0 steps, let ω T be the final configuration of brushes; that is, the placement of brushes after the graph is cleaned. A cleaning sequence consists of a sequence α = α 1, α,..., α t of vertices α i cleaned sequentially. Observe that for a graph G and initial configuration ω 0, the cleaning process can return different cleaning sequences and final configurations of brushes; consider, for example, an isolated edge uv and ω 0 u = ω 0 v = 1. It was shown in [11], however, that the final set of dirty vertices is determined by G and ω 0. When a graph G is cleaned each edge of G is traversed exactly once and by exactly one brush. Given some initial configuration ω 0 of brushes, suppose G = V, E admits a cleaning sequence α = α 1, α,..., α T which cleans G. As each edge in G is traversed exactly once and by exactly one brush, an orientation of the edges of G is permitted such that for every α i α j EG, α i α j if and only if i < j. The brush path of a brush b is the oriented path formed by the set of edges cleaned by b note that a vertex may not be repeated in a brush path. Observe that G can be decomposed into b α G oriented brush paths note that no brush can stay at its initial vertex in the minimal brush configuration. All graphs we consider are simple, undirected, connected, and finite. Throughout, for all graphs G, V G = m. For further background in graph theory, see [3] and [16].. Proofs of results In this section, we prove Theorems 1,, and from Section 1. We first make the following observations. By decomposing EG into bg paths, every vertex of odd degree must be the endpoint of at least one path. Thus, the brush number is at least half the number of vertices of odd degree. Let d o G denote the number of vertices of odd degree in graph G. It is well known see [16], for example that the minimum number of paths that decompose a graph G is at least d og, which yields Theorem 7. It was shown in [11] that this bound is exact for trees. Theorem 7. [11] Given initial configuration ω 0, suppose graph G can be cleaned yielding final configuration ω m. Then for every vertex v of odd degree, either ω 0 v > 0 or ω m v > 0. In particular, bg d og..1. Proof of Theorem 1. The proof of Theorem 1 extends the idea of Theorem 7 to count both ω 0 v and ω m v for all vertices v V G. Thus, bg P n = ω 0 v + ω m v. 3 v V G P n.1.1. Lower bound. We prove first the lower bound of Theorem 1. Consider the case when d o G = 0. We determine a lower bound for bg P n in the form of Equation 3 by bounding the number of brushes in the initial configuration as well as the number of brushes in the final configuration.

5 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 5 Express G P n in the form of Figure 3 and label the copies of G in G P n as Copy 1, Copy,..., Copy n. Consider a cleaning of G P n, using the minimum number of brushes and let v i be the last vertex cleaned in an interior Copy i that is, 1 < i < n. For each pair of neighbours v i, v i+1, where 1 < i < i + 1 < n, one is cleaned before the other. Without loss of generality, suppose v i is cleaned before v i+1. Then at least two brushes must remain at v i+1 after it is cleaned it receives one brush from a vertex in Copy i, receives at least two brushes from neighbours within Copy i + 1, and sends at most one brush to a vertex in Copy i + 1. In general, at least n 3 brushes must be located at vertices in the set {v, v 3,..., v n 1 } in the final configuration by pairing up interior copies of G, giving n 3 pairs with two brushes left on each. By the same argument, at least n 3 brushes must be located at vertices in {u, u 3,..., u n 1 } in the initial configuration. As d o G = 0, there are m odd vertices in Copy 1 and m odd vertices in Copy n. From Theorem 7, these m vertices must be the initial and final locations of brushes. Finally, n 3 bg P n + m which implies bg P n m + n δ n,odd = bp m P n δ n,odd. Finally, consider the first vertex cleaned in G P n. Because d o G = 0, regardless of whether the vertex lies in an interior or exterior copy of G, it requires at least one additional brush in the initial configuration. Finally, bg P n m + n + δ n,odd bp m P n. Next, consider the case when d o G. Then d o G P n = m d o G + n d o G, but we note that at least two additional brushes are required in order to clean the first vertex in G P n. Thus, bg P n 1 d og P n + = + 1 m d o G + n d o G n n = + d o G + m d o G = + d o G + m m + n = bp m P n with the second inequality following since d o G and n. The proof of the lower bound of Theorem 1 now follows..1.. Upper bound. For the proof of the upper bound of Theorem 1, we first prove Theorem 1: bk m P n = n + δ m,odd. The proof of Theorem is in the next subsection. Consider the following cleaning sequence of the complete graph K m, using bk m = brushes first given in [11]. To clean K m, the first vertex cleaned requires m 1 brushes, the second vertex cleaned requires m 3 brushes, and so on, until the m th vertex cleaned, which requires either one or two brushes, depending on the parity of m. If m is odd, for

6 6 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER Copy 1 Copy Copy 3 Copy n-1 Copy n of G of G of G of G of G Exterior copy of G The n interior copies of G Exterior copy of G Figure 3. Description of G P n. example, then m 1 + m = m max{0, m j 1} brushes are required in the initial configuration. The initial configuration of brushes for K 7 is indicated in Figure. 6 Figure. An initial configuration and final configuration for K 7. For the lower bound in Theorem 1, suppose that m is odd. Express K m P n in the form given in Figure 3, where the n copies of K m are labeled Copy 1, Copy,..., Copy n. We count the minimum number of brushes which must be located at vertices in each of the n copies of K m in K m P n.

7 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 7 Consider the event that for each i {, 3,..., n 1}, every vertex in Copy i receives one brush from Copy i 1 and one brush from Copy i + 1 and also every vertex in Copy 1 and Copy n receives one brush from a Copy and Copy n 1. Using, the j th vertex cleaned in Copy i will require at least max{0, m j + 1 } brushes in the initial configuration for i {, 3,..., n 1}. Similarly, the j th vertex cleaned in Copy 1 and Copy n will require at least max{0, m j + 1 1} brushes in the initial configuration. This gives a total number of brushes equalling n m { } max 0, m j 1 + m = n = m 1 { } max 0, m j m 1/ n m j 1 + mn m n + 1. m 1/ m j Now we have a lower bound on the brush number K m P n based on the notion of two brushes coming into each vertex in the n copies of K m and one brush coming into each vertex in the two exterior copies of K m. Clearly, however, this number of brushes does not suffice. Label the j th vertex cleaned in Copy i of K m as v i,j for i [n] and j [m]. Let k j be the number of vertices contained in {v,j, v 3,j,..., v n 1,j } that each receive two brushes from neighbouring copies of K m. When a vertex v i,j is cleaned, it has already received j 1 brushes from vertices within Copy i, and has only m j dirty neighbours within Copy i. If j > m+1, then regardless of the cases below, these vertices will require no additional brushes in the initial configuration. Thus, we will consider vertices v i,j for which j m 1 and then we will consider vertices v i,j for which j = m+1. For a fixed value of j, there are three possible cases to consider, depending on the brushes received by v 1,j and v n,j. Case 1: Assume that v 1,j and v n,j both receive a brush from a neighbouring copy of K m. There are at least k j + 1 vertices in {v,j, v 3,j,..., v n 1,j } that send two brushes out to other copies of K m. These vertices will require an additional four brushes each. The remaining vertices in row j that is, the vertices v i,j must each receive a brush from a neighbouring copy of K m and send one brush out to a neighbouring copy of K m : these require an additional two brushes. In total, at least k j n k j k j 1 = n additional brushes are required on each vertex in {v 1,j, v,j,..., v n,j } for j = 1,,..., m 1. Case : Assume that v 1,j and v n,j both send a brush to neighbouring copies of K m. Then each of these two vertices require two additional brushes. Then there are at least k j 1 vertices that send brushes to other copies of K m. The remaining vertices in {v,j, v 3,j,..., v n 1,j } must receive one brush from a neighbouring copy of K m and send one brush out to a neighbouring copy of K m. So at least k j 1 + n k j = n additional brushes are required on vertices in {v 1,j, v,j,..., v n,j } for j = 1,,..., m 1. Case 3: One of v 1,j, v n,j receives one brush from a neighbouring copy of K m, and one sends a brush out to a neighbouring copy of K m one requiring an additional two brushes.

8 8 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER Then there are at least k j vertices that send brushes to other copies of K m. The remaining vertices must have one brush coming in and send one brush out. So at least k j +n k j + = n brushes are required on vertices in row j for i = j,,..., m 1. Finally, we consider the first vertex cleaned in the set {v 1,j, v,j,..., v n,j } for j = m+1. It will require at least one additional brush in the initial configuration. In total, we have a minimum of m 1 n mn m n n m 1 m = = bk m n + 1 brushes. The lower bound for m even is similar and so is omitted. n + 1 For the upper bound of Theorem 1, we provide an initial configuration using bk m n+ δ m,odd brushes. Suppose m is even. Initially place 1, 3, 5,..., m 1 brushes arbitrarily on m vertices in each of the n interior copies of K m. We initially place,, 6,..., m brushes arbitrarily on m vertices in one end copy of K m. Initially place,, 6,..., m brushes arbitrarily on m 1 vertices in the other end copy of K m. The total number of brushes used is m m m + m m n + + = n m = nbk m. The case for m odd is analogous and so omitted. It can easily be seen that this configuration of brushes will suffice to clean K m P n. As an example, an initial configuration using 5bK 6 brushes is illustrated in Figure 5. All the vertices in Copy 1 of K 6 are cleaned first, then all the vertices in Copy of K 6, and so on. The proof of the upper bound of Theorem 1 follows. Copy 1 Copy Copy 3 Copy Copy 5 of K 6 of K 6 of K 6 of K 6 of K Figure 5. An initial configuration for K 6 P 5. We now prove the upper bound of Theorem 1. Let α be a cleaning sequence of G that uses bg brushes. Consider Copy 1 of G. We place bg brushes on the vertices of Copy

9 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 9 1 of G such that sequence α will clean the graph induced by the vertices of Copy 1. For each vertex in α that contains no brushes in both the initial and final configurations of G, we place one additional brush. Now all the vertices of Copy 1 can be cleaned in G P n. For i {, 3,..., n}, we place bg brushes on vertices in Copy i such that the sequence α can be used to clean the subgraph induced by the vertices of Copy i. Then we can certainly clean Copy 1, then Copy, then Copy 3, and so on. Hence, we have the following. Lemma 8. Let ω 0 be an initial configuration that uses bg brushes to clean G. Let α be the cleaning sequence that has the minimum number of vertices with ω 0 = ω m = 0. Then bg P n bgn + sg, where sg is the number of vertices in sequence α with ω 0 = ω m = 0. We also derive the following lemma. Lemma 9. Let G be a connected graph. Then EKm EG bk m bg. Proof. Let G be a connected graph where EG = EK m k for some k N. Then by results in [1, 10], we have that bg EG + m = EK m k + m = mm 1 k + m = m { bk m k = + 1 if m is odd bk m k if m is even. As bg is an integer, bg bk m EK m EG /. k For the upper bound of Theorem 1, we begin with K m and delete arbitrary edges to obtain a connected graph G. Note that sk m = δ m,odd and every edge deleted from K m will increase s by at most. Thus, for a connected graph G on m vertices, We begin by noting that sg EK m EG + δ m,odd. sg δ m,odd EKm EG = EK m EG EKm EG 1 mm 1 EG n. 1 mm 1 EG Using this along with Lemma 9, we have that EKm EG n bk m bg n sg δ m,odd. Hence, nbk m + δ m,odd sg + nbg. 5

10 10 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER For n we therefore have that bk m P n = bk m n + δ m,odd bgn + sg bg P n, where the equality follows by Theorem 1, the first inequality follows by 5, and the second inequality follows by Lemma 8... Proof of Theorem. We begin by proving Theorem : bk m C n = n +. We express K m C n in the form given in Figure 6, where the n copies of K m are labeled Copy 1, Copy,..., Copy n. Let m be odd. Consider the event that for all i, every vertex in Copy i received one brush from Copy i 1 and one brush from Copy i + 1. Then the j th vertex cleaned in Copy i will require at least max{0, m j + 1 } brushes in the initial configuration. This gives a total of n m { } max 0, m j 1 = n m 1/ m j 1 = n = nbk m mn + n m 1 mn + n. Now we have a lower bound on the brush number of K m C n based on the idea of two brushes coming into each vertex in the n copies of K m. Clearly, however, this number does not suffice. Let v i,j denote the j th vertex cleaned in Copy i of K m, and let N j = {v 1,j, v,j,..., v n,j } that is, N j is the jth row. When a vertex v i,j is cleaned, it has already received j 1 brushes from vertices within Copy i and has only m j dirty neighbours within Copy i. If j > m+1, then the vertices will require no additional brushes in the initial configuration. Thus, we consider the case when j m 1. For each j m 1, let k j denote the number of vertices in N j that each receive two brushes from neighbouring copies of K m. Then for a fixed j, there must be exactly k j vertices which also send two brushes out to neighbouring copies of K m. These vertices will require an additional brushes each in the initial configuration. The remaining N j k j vertices each must receive one brush from a neighbouring copy and send one brush out to a neighbouring copy: these vertices each require an additional brushes. In total, at least m 1/ k j + n k j = mn n brushes. Now we consider the first vertex cleaned in N m+1/. This vertex receives m 1 brushes from previously cleaned vertices within its copy of K m and has m 1 dirty neighbours within its copy of K m. Thus, as it has two edges to neighbouring copies of K m, it will require an additional two brushes in the initial configuration.

11 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 11 Finally, we have a minimum of nbk m mn + n + mn n + = nbk m + brushes required in the initial configuration. The lower bound for m even is similar and has been omitted. It remains to be shown this number of brushes suffices to clean K m C n. We now describe an initial configuration using bk m n + many brushes. Suppose that m is even. We initially place 1, 3, 5,..., m + 1 brushes on m+ vertices in one copy of K m. In n copies of K m, we initially place 1, 3, 5,..., m 1 brushes on m vertices. In the remaining copy of K m, we initially place 1, 3, 5,..., m 3 brushes on m vertices. This is a total of n m + = nbk m + brushes. The case when m is odd is analogous and so is omitted. It can easily be seen that this configuration of brushes will suffice to clean K m C n. An initial configuration using 5bK 6 + brushes is illustrated for K 6 C 5 in Figure 6. All the vertices in Copy 1 of K 6 are cleaned first, then all the vertices in Copy of K 6, and so on. This concludes the proof of Theorem. We now prove the upper bound of Theorem. Note that our hypothesis is that n m. Express G C n in the form given in Figure 6, where the n copies of K m are labeled Copy 1, Copy,..., Copy n. Copy 1 Copy Copy 3 Copy Copy 5 of K 6 of K 6 of K 6 of K 6 of K Figure 6. An initial configuration for K 6 C 5. Let ω 0 be an initial configuration of brushes that cleans G using bg brushes. Let s be the number of vertices with ω 0 = 0 = ω m, let s 1 be the number of vertices with ω 0 = 1, and let s 1 be the number of vertices with ω m = 1. We now prove that bg C n nbg + s + s 1 + s 1. 6 To prove 6, the idea is to clean Copy 1, then Copy, and so on. So the vertices in Copy will all have to send two brushes out to neighbouring copies of G. To clean all of Copy 1 of G, we will require bg brushes plus an additional brushes for each vertex v with ω 0 v > 0; an additional brushes for each vertex v with ω 0 v = 0 = ω m v; plus one

12 1 ANTHONY BONATO AND MARGARET-ELLEN MESSINGER additional brush for each vertex v with ω m v = 1. For Copy, Copy 3, and so on to Copy n 1, each copy will require bg brushes. Note that each vertex in Copy n has already received two brushes. Thus, Copy n will require bg subtracting two brushes from each vertex v with ω 0 v and subtracting one brush from each vertex v with ω 0 v = 1. This gives us a total of nbg + s + s 1 + s 1 brushes. We show that bk m bg 1 n s+s 1 +s 1. First, note that for any graph G, s m and s + s 1 + s 1 m. As n m, if bk m bg, then bk m bg 1 n m 1 n s + s 1 + s 1. Next, consider the case where bk m bg = 1. By Theorem. b of [10], bg = bk m \ e for some e EK m. This implies s + s 1 + s 1 m. As s {1, } depending on the parity of m, bk m bg = 1 1 n m 1 n m + s 1 n s + s 1 + s 1. The proof of Theorem follows. 3. Further Work The Box Cleaning Conjecture remains open for all graphs H except paths. lower bound bp m C n bg C n Even the remains unproven, and the upper bound is open if m n. We mention that if m n, then it can be shown that the upper bound in Theorem holds if either bk m bg m 1, n or if bk m bg m 1. We can also show that bp m C n = m + n the proof is analogous to the proof of Theorem and so is omitted. We do not know the exact value of bc m C n for all m and n. The categorical product of graphs G and H, written G H, has vertex set V G V H. Vertices a, b and c, d are joined if ac EG and bd EH. The brush number Figure 7. An initial configuration for the categorical product P C 5 using bp C 5 brushes. of categorical product is not well understood, and there are only scattered results. For

13 BRUSH NUMBER OF CARTESIAN PRODUCT GRAPHS 13 example, we can prove that n if m > n bp m C n = m 1 if m n and n m 1 if m n and n. See Figure 7 for the case m = and n = 5. Further, { m if m and m n; bc m C n = min{m, n} if m and n, and bp m P n = min{m 1, n 1}. References [1] N. Alon, P. Pra lat, N. Wormald, Cleaning d-regular graphs with brushes, SIAM Journal on Discrete Mathematics 3 008, [] T. Biedl, T. Chan, Y. Ganjali, M. Hajiaghayo, D. Wood, Balanced vertex orderings of graphs, Discrete Applied Mathematics [3] R. Diestel, Graph theory, Springer-Verlag, New York, 000. [] B. Hartnell, D.F. Rall, Domination in Cartesian products: Vizing s conjecture, In: Domination in Graphs Advanced Topics Ed. T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. New York: Dekker, [5] S. Hedetniemi, Homomorphisms of graphs and automata, Technical Report T, University of Michigan, [6] B. Hobbs, J. Kahabka, Underwater Cleaning Technique Used for Removal of Zebra Mussels at the Fitzpatrick Nuclear Power Plant, In: Proceedings of The Fifth International Zebra Mussel and Other Aquatic Nuisance Organisms Conference, [7] W. Imrich, S. Klavzar, Product Graphs-Structure and Recognition, Wiley-Interscience series in discrete mathematics and optimization, Wiley-Interscience, New York, 000. [8] S. R. Kotler, E. C. Mallen, K. M. Tamms, Robotic Removal of Zebra Mussel Accumulations in a Nuclear Power Plant Screenhouse, Proceedings of The Fifth International Zebra Mussel and Other Aquatic Nuisance Organisms Conference, Toronto, Canada, February [9] S. McKeil, Chip Firing Cleaning Processes, MSc Thesis, Dalhousie University, 007. [10] M. E. Messinger, Methods of Decontaminating a Network, PhD Thesis, Dalhousie University, 008. [11] M. E. Messinger, R. J. Nowakowski, P. Pra lat, Cleaning a Network with Brushes, Theoretical Computer Science [1] M. E. Messinger, R. J. Nowakowski, P. Pra lat, N. Wormald, Cleaning random d-regular graphs with brushes using a degree-greedy algorithm, Proceedings of the th Workshop on Combinatorial and Algorithmic Aspects of Networking CAAN007, Lecture Notes in Computer Science, Springer, 007, [13] P. Pra lat, Cleaning random graphs with brushes, Australasian Journal of Combinatorics 3 009, [1] N. Sauer, N. Hedetniemi s conjecture: a survey, Discrete Mathematics [15] V.G. Vizing, The Cartesian product of graphs, Vycisl. Sistemy [16] D.B. West, Introduction to Graph Theory, nd edition, Prentice Hall, 001. Department of Mathematics, Ryerson University, Toronto, ON, Canada, M5B K3 address: abonato@ryerson.ca Department of Mathematics, Ryerson University, Toronto, ON, Canada, M5B K3 address: messinger@ryerson.ca