Zero Forcing Number of Random Regular Graphs
Deepak Bal, Patrick Bennett, Sean English, Calum MacRury, Paweł Prałat
ZZero Forcing Number of Random Regular Graphs
Deepak Bal ∗ Patrick Bennett † Sean English ‡ Calum MacRury § Pawe(cid:32)l Pra(cid:32)lat ¶ December 18, 2018
Abstract
The zero forcing process is an iterative graph colouring process in which at each time step acoloured vertex with a single uncoloured neighbour can force this neighbour to become coloured.A zero forcing set of a graph is an initial set of coloured vertices that can eventually force theentire graph to be coloured. The zero forcing number is the size of the smallest zero forcing set.We explore the zero forcing number for random regular graphs, improving on bounds given byKalinowski, Kam˘cev and Sudakov [12]. We also propose and analyze a degree-greedy algorithmfor finding small zero forcing sets using the differential equations method.
Let G = ( V, E ) be any simple, undirected graph. The zero forcing process on a graph G is definedas follows. We start with a subset S of vertices and colour them black; all other vertices are colouredwhite. At each time step of the process, a black vertex, v , with exactly one white neighbour, w , willforce its white neighbour to become black. In this case we say that v forced w . The process stopswhen no more white vertices can be forced to become black. A chronological list of forces for aset of black vertices S is a sequence of pairs ( v , w ) , . . . , ( v k , w k ) such that initially v can force w ,and iteratively at the i th time step, assuming the that v j has forced w j for all 1 ≤ j < i , v i can force w i . It is easy to see that, given an initial set S of black vertices, the set of black vertices obtained byapplying the colour-change rule until no more changes are possible is unique, regardless of the orderthat the forcing is done. Indeed, it follows immediately from the property that once a white vertex isready to become black it stays at this state until it actually becomes black. Therefore, the followingdefinition is natural. The set S is said to be a zero forcing set if at the end of the zero forcingprocess all vertices of V become black. The zero forcing number of G is the minimum cardinalityof a zero forcing set in G , and is denoted by Z ( G ). This graph parameter is well defined as, trivially, V is a zero forcing set and so Z ( G ) ≤ n .The concept of zero forcing was first proposed in [1] as a bound for the minimum rank problem.Given a n × n symmetric matrix A = [ a i,j ] over some field F , the graph of A , denoted G ( A ) is thegraph on vertex set V = { v , . . . , v n } with edge set E = { v i v j | i (cid:54) = j, a i,j (cid:54) = 0 } . Note that the graphof a matrix does not depend on the main diagonal of the matrix. The minimum rank of a graph G ,denoted mr F ( G ) is the minimum rank over all symmetric matrices A over F such that G ( A ) = G . Theminimum rank problem for graphs has been extensively studied both from the lens of linear algebra ∗ Department of Mathematical Sciences, Montclair State University, Montclair, NJ, USA, email: [email protected] † Department of Mathematics, Western Michigan University, Kalamazoo, MI, USA, e-mail:
[email protected] ‡ Department of Mathematics, Ryerson University, Toronto, ON, Canada, e-mail:
[email protected] § Department of Computer Science, University of Toronto, Toronto, ON, Canada, email : [email protected] ¶ Department of Mathematics, Ryerson University, Toronto, ON, Canada, e-mail: [email protected] a r X i v : . [ m a t h . C O ] D ec nd combinatorics. For a survey on the subject, see [9]. In [1], it was shown that Z ( G ) ≥ n − mr F ( G )for any field F . Zero forcing numbers have provided very good bounds for the minimum rank problem,and so have recently been a topic of extensive study.In addition to the application to the minimum rank problem, zero forcing sets have been of interestin their own right. Many facets and generalizations of zero forcing have been studied. Some of note arethe propagation time, which is defined to be the minimum amount of time for a minimum zero forcingset to force the entire graph, where at each time step all possible forces happen simultaneously [11],positive semi-definite zero forcing, a variant which bounds the minimum rank problem when theminimum is taken only over positive semi-definite matrices with graph G [8], and the zero forcingpolynomial, which algebraically encodes information about the number of zero forcing sets of a givensize in a graph [5].In this paper, we will explore zero forcing for random regular graphs, or more precisely, we considerthe probability space of random d -regular graphs with uniform probability distribution. This spaceis denoted G n,d , and asymptotics are for n → ∞ with d ≥ n even if d is odd. We say thatan event in a probability space holds asymptotically almost surely (or a.a.s. ) if the probabilitythat it holds tends to 1 as n goes to infinity. Since we aim for results that hold a.a.s., we will alwaysassume that n is large enough. All logarithms which follow, unless otherwise stated, are naturallogarithms.The case of binomial random graphs was studied by Kalinowski, Kam˘cev and Sudakov [12], whoproved that the zero forcing number of G ( n, p ) is a.a.s. n − (2 + √ o (1)) · log npp for a wide rangeof p . In addition, this group gave bounds involving the spectral properties of regular graphs. Theseresults applied to the random regular graph implies the following theorem. Theorem 1.1 ([12]) . Let d ≥ be fixed. Then a.a.s. n (cid:18) −
40 log dd (cid:19) ≤ Z ( G n,d ) ≤ n (cid:18) − log d d (cid:19) . Here we improve both the upper and lower bound when d is large. Theorem 1.2.
For any fixed (cid:15) > , for sufficiently large fixed d = d ( (cid:15) ) , we have that a.a.s. n (cid:18) − (1 + (cid:15) ) 4 log dd (cid:19) ≤ Z ( G n,d ) ≤ n (cid:18) − (1 − (cid:15) ) log d d (cid:19) . The proof of the upper bound can be found in Section 3, while the lower bound is proved inSection 4. In addition to these improvements for large d , we also give (numerical) upper and lowerbounds for some small d . In Section 2 using the differential equations method, we propose and analyzea degree-greedy algorithm that finds a zero forcing set. In particular, the analysis of the performanceof this algorithm provides the upper bounds presented in Table 1. Most of the numerical bounds arecalculated in Section 2.3, but the best bound for d = 3 is found in Section 2.4, where we provide aslightly improved algorithm and analysis. The lower bounds are discussed at the end of Section 4.Of particular interest is the bound for cubic graphs, or 3-regular graphs. Denote the independencenumber of a graph G by α ( G ). We say a cubic graph G is claw-free if there is no induced copy of K , (the star on 4 vertices) in G . Davila and Henning [7] showed that for all connected, claw-free,cubic graphs, we have that Z ( G ) ≤ α ( G ) + 1. It has been conjectured that this holds for all cubicgraphs. Since all cubic graphs G have α ( G ) ≥ n/
4, this bound along with our numerical bound for d = 3 implies that the conjecture is true for almost all cubic graphs. Instead of working directly in the uniform probability space of random regular graphs on n vertices G n,d , we use the pairing model or configuration model of random regular graphs, first introducedby Bollob´as [3], which is described next. Suppose that dn is even, as in the case of random regular2 lower bound upper bound3 0.06992 0.170574 0.14508 0.253295 0.21137 0.314956 0.26783 0.364377 0.31581 0.405388 0.35689 0.440219 0.39244 0.4703210 0.42351 0.4968911 0.45092 0.5200112 0.47534 0.5408713 0.49726 0.5596514 0.51706 0.57668Table 1: Numerical upper and lower bounds on Z ( G n,d ) /n that hold a.a.s. for d = 3 . . . dn configuration points partitioned into n labelled buckets v , v , . . . , v n of d points each. A pairing of these points is a perfect matching into dn/ P ,we may construct a multigraph G ( P ), with loops allowed, as follows: the vertices are the buckets v , v , . . . , v n , and a pair { x, y } in P corresponds to an edge v i v j in G ( P ) if x and y are containedin the buckets v i and v j , respectively. It is an easy fact that the probability of a random pairingcorresponding to a given simple graph G is independent of the graph, hence the restriction of theprobability space of random pairings to simple graphs is precisely G n,d . Moreover, it is well knownthat a random pairing generates a simple graph with probability asymptotic to e − ( d − / depending on d . When d is constant, e − ( d − / > G n,d . For this reason, asymptoticresults over random pairings suffice for our purposes. One of the advantages of using this model isthat the pairs may be chosen sequentially so that the next pair is chosen uniformly at random overthe remaining (unchosen) points. For more information on this model, see [15]. It can be easily shown that the zero forcing number for random 2-regular graphs is a.a.s. (1+ o (1)) log n .We provide the details here for the sake of completeness.Let Y n = Y n ( G ) be the total number of cycles in any 2-regular graph G on n vertices (not necessarilyrandom). Since exactly two adjacent vertices are needed to be present in a forcing set, we get that Z ( G ) = 2 Y n .We know that the random 2-regular graph is a.a.s. disconnected; by simple calculations one canshow that the probability of having a Hamiltonian cycle is asymptotic to e / √ πn − / (see, for ex-ample, [15]). We also know that the total number of cycles Y n is sharply concentrated near (1 /
2) log n .It is not difficult to see this by generating the random graph sequentially using the pairing model.The probability of forming a cycle in step i is exactly 1 / (2 n − i + 1), so the expected number ofcycles is (1 /
2) log n + O (1). The variance can be calculated in a similar way. So we get that a.a.s. thezero forcing number of a random 2-regular graph is (1 + o (1)) log n . d Let us start with a few observations that will allow us to propose an alternative definition of thezero forcing number and then, based on that, to design a heuristic, greedy algorithm. This algorithm3ill be used to obtain small zero forcing sets for random d -regular graphs and so to get strong upperbounds for Z ( G ) that holds a.a.s. (both numerical and explicit).Let G = ( V, E ) be any graph on n vertices. For convenience, assume that V = [ n ] := { , , . . . , n } .For any v ∈ V , we will use N ( v ) for the (open) neighbourhood of v , that is, the set of neighboursof v ; N [ v ] will be used for the closed neighbourhood , that is, N [ v ] := N ( v ) ∪ { v } . The degree of v is defined as deg( v ) := | N ( v ) | . Finally, for any S ⊆ V , N ( S ) := (cid:32) (cid:91) v ∈ S N ( v ) (cid:33) \ S, and N [ S ] := (cid:91) v ∈ S N [ v ] = N ( S ) ∪ S. The main idea behind our reformulation is that the zero forcing set does not need to be fixedin advance. Instead, one can simply start the zero forcing process and try to decide which vertexshould be the next one to force, and adjust the zero forcing set accordingly so that this operationis permissible. The idea is quite simple and natural but, unfortunately, the formal definitions areslightly technical.In order to build our zero forcing set, we will build a
Z-sequence . Given a graph G , a sequence S = ( v , . . . , v k ), where v i ∈ V for 1 ≤ i ≤ k is a Z-sequence if for each 1 ≤ i ≤ k , we have N ( v i ) \ i − (cid:91) j =1 N [ v j ] (cid:54) = ∅ . A maximal Z-sequence is called a
Z-Grundy dominating sequence , and the length of a longestZ-Grundy dominating sequence of G is known as the Z-Grundy domination number , denoted γ Zgr ( G ). The following result of Breˇsar et al. [6] shows that the zero forcing number of a graph isdetermined by the Z-Grundy domination number. Theorem 2.1 ([6]) . If G = ( V, E ) is a graph without isolated vertices, then γ Zgr ( G ) + Z ( G ) = | V | . We provide here an alternative proof of Theorem 2.1 for the sake of completeness, and also becausethe proof will be relevant to how our algorithm works. First though we provide a few definitions neededfor the proof. Let S = ( v , . . . , v k ) be a Z -sequence. We will say a vertex w is a witness for v i if w ∈ N ( v i ) \ (cid:83) i − j =1 N [ v j ]. Note that for each Z -sequence S = ( v , . . . , v k ), there exists at least onesequence W = ( w , . . . , w k ) such that w i is a witness for v i for all 1 ≤ i ≤ k . We will call such asequence W a witness sequence for S . In a slight abuse of notation, we will sometimes use S and W to refer to the set of vertices in the sequences S or W respectively. It is worth noting that W and S can have non-empty intersection. We are now ready to present an alternative proof of Theorem 2.1. Proof of Theorem 2.1.
The main idea behind our proof will be to show that the complement of awitness set is a zero forcing set, and vice versa. Since witness sequences and their correspondingZ-sequences are of the same size, this will complete the proof that γ Zgr ( G ) + Z ( G ) = | V | . In particular,longer Z-Grundy dominating sequences will yield smaller zero forcing sets, and vice versa.Let S = ( v , . . . , v k ) be a Z-sequence. Let W = ( w , . . . , w k ) be a witness sequence for S . We claimthat B := V \ W is a zero forcing set. Indeed, note that initially v can force w since N [ v ] \{ w } ⊆ B (note that, by definition, for any 2 ≤ i ≤ k we have w i / ∈ N [ v ] and so all vertices but w are in B ).Then v can force w , and in general once the vertices v , . . . , v i − have forced w , . . . , w i − , v i canforce w i since (cid:16) N [ v i ] ∪ (cid:83) i − j =1 N [ v j ] (cid:17) \ { w i } must have all started black or been turned black. Thus,all the vertices in W will become black, so B is a zero forcing set.4ow, let B be a zero forcing set of size | V | − k for some integer k . For some chronological listof forces for B , let v i and w i denote the i th vertex that forced and was forced, respectively, for1 ≤ i ≤ k . We claim that S = ( v , . . . , v k ) is a Z-sequence (not necessarily maximal) with witnesssequence W = ( w , . . . , w k ). Indeed, we have for each 1 ≤ i ≤ k , w i ∈ N ( v i ) \ (cid:83) i − j =1 N [ v j ] since afterthe ( i − (cid:83) i − j =1 N [ v j ] must all have been turned black, while w i remains white. Thus, S is a Z -sequence with witness sequence W . This completes the proof. In light of Theorem 2.1, our goal is to build a long Z-Grundy dominating sequence. Indeed, given agraph G = ( V, E ) on n vertices, if we can build such a sequence of length γ ∗ ( G ) ≤ γ Zgr ( G ), then wehave the upper bound on the forcing number, Z ( G ) = n − γ Zgr ( G ) ≤ n − γ ∗ ( G ) . We attempt to build our Z-Grundy dominating sequence greedily. Throughout the algorithm, wemaintain a Z -sequence S and we also track the set of vertices that S dominates, T := (cid:83) v ∈ S N [ v ], aswell as its complement U := V \ T . We say that a vertex v ∈ T is of type r if | N ( v ) ∩ U | = r . Notethat the sequence S can be extended by any vertex v which has a neighbour in U (that is, any vertexof type r for some r ≥ G is connected and U (cid:54) = ∅ , then we can always find a vertexfrom T which extends S . We can think of U as a “reservoir” of vertices which shrinks by r whenevera vertex from T of type r is added to the Z-sequence. We would like to extend our Z-sequence foras long as possible and so it is natural to always choose a vertex which decreases the size of U by aslittle as possible. Thus we consider the following greedy algorithm (see Algorithm 1). Input :
Connected graph G = ( V, E ) Output:
Z-Grundy dominating sequence, S Initialization:
Let v be a vertex of minimum degree, S = ( v ), T = N [ v ], U = V \ T ; while U (cid:54) = ∅ do Let v ∈ T be a vertex of minimum positive type;Move vertices of N ( v ) ∩ U from U to T ;Append v to the end of S ; end Algorithm 1: Degree greedy algorithmNote that the first step of the while loop is always possible since we assume the input graph isconnected. An iteration of the loop is said to be a step of type r if the chosen vertex v is of type r . G n,d for Small d In this section, our goal is to analyze how many iterations Algorithm 1 lasts. To this end, we make useof the so-called differential equations method. See [17] for an extensive survey of the general method.We will assume throughout this section that d ≥
3, that dn is even, and that we are running ouralgorithm on G ∈ G n,d . Note that the degree greedy algorithm only produces a Z-Grundy dominatingsequence if the graph is connected. Otherwise the algorithm finds a set that is Z-Grundy dominatingin just one component. In this case we can say the algorithm fails, but it is very unlikely. Indeed, itwas proven independently in [4, 16] that for constant d ≥ G is disconnected with probability o (1)(this also holds when d is growing with n , as shown in [13]). However, let us stress the fact that wedo not condition on the fact that G is connected; we simply work with the (unconditional) pairingmodel allowing the algorithm to finish prematurely if G is disconnected.5et T = T ( t ) and U = U ( t ) = V \ T ( t ) be the two sets of vertices at time t as defined in ouralgorithm. In order to analyze the algorithm, we will use the pairing model and only reveal partialinformation about G n,d at each step. More precisely, all edges within T ( t ) will be revealed, but nomore. If v of type r is being processed, then we reveal the r neighbours of v in U ( t ), the edges from T ( t ) to N ( v ) ∩ U ( t ), and the edges within N ( v ) ∩ U ( t ), but keep the edges from T ( t + 1) = T ( t ) ∪ N ( v )to U ( t + 1) = U ( t ) \ N ( v ) hidden.We need to track how many vertices of each type are in T ( t ). For any 0 ≤ i ≤ d , let T i ( t ) denotethe number of vertices of type i in T ( t ) after t steps of the algorithm. Observe that by the structureof our algorithm, T d ( t ) = 0 in every step of the process. Let f i,j (( t − /n, T ( t − /n, T ( t − /n, . . . , T d − ( t − /n ) := E [ T i ( t ) − T i ( t − | G [ T ( t − t is of type j ] . In order to simplify the notation, we will write f i,j ( t −
1) in place of f i,j (( t − /n, T ( t − /n, T ( t − /n, . . . , T d − ( t − /n ). Given a statement A , we will denote by δ A , the Kronecker delta function δ A = (cid:40) A is true,0 otherwise.Further, let us denote by U ( t ) = | U ( t ) | = n − (cid:80) d − (cid:96) =0 T (cid:96) ( t ). Note that we do not need to track U ( t )since it can be determined based on the other variables we are tracking. Then we have f i,j ( t −
1) = j · (cid:18) d − i (cid:19) · (cid:32) (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t − d · U ( t − (cid:33) d − i − (cid:32) − (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t − d · U ( t − (cid:33) i + j · ( d − · (cid:18) ( i + 1) T i +1 ( t − − iT i ( t − d · U ( t − (cid:19) − δ i = j + δ i =0 + O (cid:18) n (cid:19) . (1)The explanation of the preceding equality is as follows. The first term is the expected contributionof the j vertices from U ( t −
1) that are adjacent to the vertex v of type j chosen in the algorithm.Indeed, once we expose one neighbour of v in U ( t − i if precisely d − − i outof d − T ( t − T ( t −
1) have already been revealed, we get that there are exactly (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t − U ( t −
1) matched to configuration points in T ( t − (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t − / ( d · U ( t − U ( t −
1) ismatched to a configuration point in T ( t − O (1)) edge, so for the entire step t the probability that a point in U ( t − T ( t −
1) is (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t −
1) + O (1) d · U ( t −
1) + O (1) = (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t − d · U ( t −
1) + O (cid:18) n (cid:19) , (2)where the error term is O (1 /n ) because the numerator (cid:80) d − (cid:96) =1 (cid:96)T (cid:96) ( t −
1) is O ( n ) and (we will assumethat) the denominator d · U ( t −
1) is Ω( n ). The second term in (1) accounts for the possibility thatvertices that were type i + 1 could become type i if they end up being adjacent to the neighbours of v , the vertex being processed, and also the possibility that a vertex of type i could become type i − O (1 /n ) error similar to line (2)). Finally, the Kronecker deltas accounts forthe fact that when we process a vertex of type j , that vertex becomes type 0.Now, we will consider the algorithm to run in d − d −
1. Duringthe first phase, we expect T ( t ) to consist mainly of vertices of type 0 and type d − d − d − i ≤ d − d −
1, but after some more steps of this type we may produce another vertex of a lesser type. Whenvertices of type d − d − d − k thphase, a mixture of vertices of type at most d − k are processed. More specifically, in the k th phasewe typically have a linear number of vertices of type d − k , but the algorithm keeps the number ofvertices of type less than d − k sublinear.During the k th phase there are, in theory, two possible endings. It can happen that the numberof vertices of type d − k − d − k (in whichcase we move to the next phase). It is also possible that vertices type d − k are getting so rare thatthose of type d − k disappear (in which case the process goes ‘backwards’) and we begin processingvertices of type d − k + 1 again. With various initial conditions, either one could occur. However, thenumerical solutions of the differential equations for small values of d support the hypothesis that thedegree-greedy process we study never goes ‘back’. The details of the following differential equationsmethod have been omitted, but can be found in [14].According to Theorems 1 and 2 from [14], we can approximate our prioritized algorithm witha deprioritized version that still performs different types of steps in the same proportions as theprioritized algorithm. It means that the vertices are chosen to process in a slightly different way, notalways the minimum positive type, but a random mixture of various types. Once a vertex is chosen,it is treated the same way as in the degree-greedy algorithm.Let us now fix a phase, say phase k , and consider the following system of equations with variables τ i,k ( x, ˜ y , ˜ y , . . . , ˜ y d ) = τ i,k ( x, ˜y ), 1 ≤ i ≤ d − k :1 = d − k (cid:88) i =1 τ i,k ( x, ˜y ) , (3)0 = d − k (cid:88) j =1 τ j,k ( x, ˜y ) · f i,j ( x, ˜y ) , for 1 ≤ i ≤ d − k − . (4)Then if we let x = t/n and ˜ y i ( x ) = T i ( t ) /n , the solution τ i,k of the preceding system of equationscan be interpreted as the proportion of steps of type i that occur in phase k . Indeed, in phase k ,vertices of type less than or equal to d − k − (cid:80) dj =1 τ j,k ( x, ˜y ) · f i,j ( x, ˜y ) = 0 for 1 ≤ i ≤ d − k − d − k occur in phase k , we must have (cid:80) d − ki =1 τ i,k ( x, ˜y ) = 1.Using the standard differential equation method, this suggests that ˜ y i should be approximately y i where the y i are deterministic functions satisfying the following system of differential equations: dy i dx = F ( x, y , i, k ) := d − k (cid:88) j =1 τ j,k ( x, y ) · f i,j ( x, y ) . (5)Now, we can say that phase k is the interval [ x k − , x k ], where x = 0 and for all k ≥ x k isdefined to be the infimum of all x > x k − such that τ d − k,k ( x, y ) = 0. Indeed, this indicates thatvertices of type d − k − d − k , and we move to the next phase. We can then use the final values, y i ( x k ) ofphase k as the initial conditions for phase k + 1. Since we initialize the algorithm with a single vertexof type d , we will have our first initial conditions: y i (0) = 0 for all 1 ≤ i ≤ d . The conclusion is thata.a.s., for any 1 ≤ k ≤ d −
1, any 0 ≤ (cid:96) ≤ d , and any x k − n ≤ t ≤ x k n , we have T (cid:96) ( t ) = ny (cid:96) ( t/n ) + o ( n )7umerical upper bounds are presented in Table 1. Below we present some detailed discussion for d = 3 and d = 4. d = 3Let us concentrate on random 3-regular graphs; that is, on d = 3. During the first phase, vertices oftypes 1 and 2 are processed. The proportions of each step type that occur in this phase is shown inFigure 1(b). The solution to the relevant differential equations is shown in Figure 1(a), namely y ( x )and y ( x ) (recall that the number of vertices of type 1 is sublinear). The phase ends at time t ∼ x n with x ≈ . t but, in fact, y ( x ) ≈ . y ( x ) ≈ . y and y (b) τ , and τ , (c) y , y , and y Figure 1: Solution to the differential equations for d = 3; phase 1 (a-b) and phase 2 (c).During the second phase, only vertices of type 1 are processed. The solution to the relevantdifferential equations is shown in Figure 1(c) The phase ends at time t = x n with x ≈ . x > . Z ( G ) ≤ (1 − x ) n ≤ . n for G ∈ G n, . d = 4There are three phases for random 4-regular graphs. During the first phase vertices of types between1 and 3 are processed; vertices of types 0 and 3 are dominant (vertices of types 1 and 2 are presentbut the number of them is sublinear)—see Figure 2(a). This phase ends at time t ∼ x n with x ≈ . y ( x ) ≈ . y ( x ) ≈ . t ∼ x n with x ≈ . y ( x ) ≈ . y ( x ) ≈ . y ( x ) ≈ . t = x n with x ≈ . Z ( G ) ≤ (1 − x ) n ≤ . n for G ∈ G n, . d = 3 In general, we do not expect Algorithm 1 to find the longest possible Z-Grundy dominating sequencein a given connected graph. Here we present a modification to the algorithm that does slightly better.The improvement is negligible for large degree d , but is noticeable for small d , so we only show theresults of the improved algorithm for d = 3. 8a) y and y (b) y , y , and y (c) y , y , y , and y Figure 2: Solution to the differential equations for d = 4; phase 1 (a), phase 2 (b), and phase 3 (c).Let S be a Z-Grundy dominating sequence with witness sequence W . The idea behind the im-provement is that if there is a vertex u ∈ V ( G ) \ S such that N [ u ] ∩ W = ∅ , then u can be inserted into S , making a longer sequence. In the following algorithm, S will be the Z-sequence we build, T (2) willbe a set of witnesses for the Z-sequence S , and T (1) = (cid:0)(cid:83) v ∈ S N [ v ] (cid:1) \ T (2) . At the end of the process, T (1) will be our zero forcing set. and we will have T (2) = V \ T (1) . The set U will play essentially thesame role it did in Algorithm 1, while the set T from Algorithm 1 is split in two here, sets T (1) and T (2) . Input :
Connected graph G Output:
Z-Grundy dominating sequence, S Initialization:Let v be a vertex of minimum degree. S = ( v ), T (1) = { v } , T (2) = {} , U = V \ { v } while U (cid:54) = ∅ do Let v ∈ T (1) ∪ T (2) be a vertex of minimum positive type; if v is an element of T (1) of type ≥ and there exists u ∈ N ( v ) ∩ U such that N ( u ) ⊆ T (1) then Append u to the end of S ;Move v from T (1) to T (2) ;Move u from U to T (1) ; end Arbitrarily choose a vertex w ∈ N ( v ) ∩ U that minimizes the quantity | ( N ( w ) \ N ( v )) ∩ U | ;Append v to the end of S ;Move w from U to T (2) ;Move the vertices of ( N ( v ) \ { w } ) ∩ U from U to T (1) ; end Algorithm 2: Smart degree greedy algorithmThe if checks to see if we can insert a vertex u into the sequence, giving us a longer sequence.Furthermore, this algorithm gives preference to vertices that will end up being a lower type whenchoosing which vertex to add to the witness set W = T (2) . The rationale behind this is that we onlyget the extra savings from this new algorithm when we process a vertex of type ≥ W ,so we prioritize adding vertices of higher type to T to make this happen more often.To analyze this algorithm, we will assume that n is even and we are running it on G = G n, .we need to track a few more variables than Algorithm 1. Here, let T (1) ( t ), T (2) ( t ) and U ( t ) be thevertices in the sets T (1) , T (2) , and U at iteration t of the algorithm. Similarly to before, we will say9 vertex in T (1) ( t ) ∪ T (2) ( t ) is of type r if it has r neighbours in U ( t ). For each 0 ≤ i ≤
2, and k ∈ { , } , let T i,k ( t ) be the number of vertices of type i in T ( k ) ( t ). For ease of notation, we will let T i ( t ) = T i, ( t ) + T i, ( t ) and let U ( t ) = n − T ( t ) − T ( t ) − T ( t ).We can analyze this algorithm using the differential equations method similarly to our analysis inSection 2.3. We will not provide all the details as it is a routine extension of our work in Section 2.3,instead we will simply provide the expected changes in our tracked variables and then provide thenumerical solution for this case. Before we needed only to track the change in vertices of type i whenprocessing a vertex of type j , but here we also need to distinguish between vertices in T (1) and T (2) .Thus, we will let f i,j,k,(cid:96) ( t −
1) track the expected change in T i,k ( t − j in T ( (cid:96) ) ( t − T (1) ( t − ∪ T (2) ( t − t −
1) in our variables T i ( t − T i,j ( t −
1) and U ( t − P := T +2 T · U , so P represents the probability that an exposed edge lands in T (1) ∪ T (2) .For a ≥ b , let P a,b := 2 δ a (cid:54) = b (cid:0) a (cid:1)(cid:0) b (cid:1) P − a − b (1 − P ) a + b . Then P a,b is the probability that when we processa vertex of type 2, the neighbours become types a and b . Then f i,j,k,(cid:96) ( t −
1) =2 j · (cid:18) ( i + 1) T i +1 ,k − iT i,k · U (cid:19) + δ j =1 (cid:18) δ i =0 ,k = (cid:96) − δ i =1 ,k = (cid:96) + δ k =2 (cid:18) i (cid:19) P − i (1 − P ) i (cid:19) (6)+ δ j =2 (cid:88) ≥ a ≥ b ≥ P a,b ( δ i = a,k =1 + δ i = b,k =2 + δ i =0 ,k = (cid:96) − δ i =2 ,k = (cid:96) ) (7)+ (cid:88) ≤ a ≤ P a, (cid:32) δ (cid:96) =1 (cid:18) T , + 2 T , T + 2 T (cid:19) ( δ i = a,k =2 + δ i =0 ,k =1 + δ i =0 ,k =2 − δ i =2 ,k =1 ) (8)+ (cid:32) − δ (cid:96) =1 (cid:18) T , + 2 T , T + 2 T (cid:19) (cid:33) ( δ i = a,k =1 + δ i =0 ,k =2 + δ i =0 ,k = (cid:96) − δ i =2 ,k = (cid:96) ) (cid:33) (9)+ P , δ (cid:96) =1 − (cid:32) − (cid:18) T , + 2 T , T + 2 T (cid:19) (cid:33) ( δ i =0 + δ i =0 ,k =2 − δ i =2 ,k =1 ) (10)+ − δ (cid:96) =1 − (cid:32) − (cid:18) T , + 2 T , T + 2 T (cid:19) (cid:33) ( δ i =0 + δ i =0 ,k = (cid:96) − δ i =2 ,k = (cid:96) ) (11)The explanation for the equation above is as follows. There are two main changes that we need toaccount for in this algorithm that were not in Algorithm 1, namely we need to account for when the if clause is triggered, and in the algorithm right after the if clause ends, we need to make sure thevertex we chose minimizes the quantity | ( N ( w ) \ N ( v )) ∩ U | .In line (6), the term 2 j · (cid:16) ( i +1) T i +1 ,k − iT i,k · U (cid:17) accounts for the situation when a vertex in T (1) ∪ T (2) ends in being neighbours with the neighbours of v . The other term in line (6) accounts for when thevertex v being processed is of type 1. More precisely, in this case, v changes from type 1 to type0. Furthermore, the neighbour of v in U moves from U to T (2) , and is of type i with probability (cid:0) i (cid:1) P − i (1 − P ) i . The remaining lines of the equation account for when we process a vertex of type 2.In line (7) inside the square brackets, we account for when neither neighbour of the vertex beingprocessed ends up type 0 (i.e. in these cases, we know the if clause of the algorithm is not triggered).We have probability P a,b of the neighbours being types a and b , and then in this case, since a ≥ b , wemove the neighbor of v of type a to T (1) and the neighbour of type b to T (2) , and then v goes fromtype 2 to type 0.In lines (8) and (9) we deal with the situation where one of the neighbours of v is type 0 and theis of type 1 or 2. If v is in T (1) , then we need to consider the possibility that the if clause is triggered.10his happens with probability (cid:16) T , +2 T , T +2 T (cid:17) (since we are conditioning on the fact that exactly oneneighbour of v is of type 0), and in this case we move the positive type vertex to T (2) , the type 0vertex to T (1) , and then v becomes type 0 and moves to T (2) . Then in line (9), we deal with whenthe if clause was not triggered, in which case we prioritize adding vertices to T (1) of higher type, sothe positive type neighbour of v is moved to T (1) , the type 0 neighbour is moved to T (2) , and then v becomes type 0.Then lines (10) and (11) account for when v is of type 2 and both neighbours of v become type 0.In this case, if v is in T (1) , then we have probability 1 − (cid:18) − (cid:16) T , +2 T , T +2 T (cid:17) (cid:19) of one of the neighboursof v having all neighbours in T (1) since we are conditioning on both neighbours of v being type 0. Ifthis is the case, then the if clause is triggered, so we add one vertex to T (1) and one to T (2) of type0 (note δ i =0 = δ i =0 ,k =1 + δ i =0 ,k =2 ). Then v becomes type 0 and moves from T (1) to T (2) . Finally, online (11), if neither neighbour of v has all of its neighbours in T (1) , then we add a vertex of type 0 toeach of the sets T (1) and T (2) , and then v goes from type 2 to type 0. This finished the explanationfor the expression for f i,j,k,(cid:96) .We will omit most of the remaining details for the rest of the analysis of Algorithm 2, since it isvery similar to Section 2.3. To summarize the main steps we are omitting, we define τ i,k,m to be thesolution to a system that resembles (3)-(4), with the interpretation that τ i,k,m is the proportion ofsteps of type i in which we process a vertex from T ( k ) in phase m ∈ { , } . It is worth noting thatsince Algorithm 2 does not prioritize processing vertices in T (1) over T (2) , or vice versa in the firststep of the while loop, this system has one degree of freedom. For our analysis, in phase 1, we willassume in the while loop, whenever we are processing a vertex of type 2, we choose if we process avertex from T (1) or T (2) randomly with probability proportional to the size of T (1) and T (2) . Thiscorresponds to adding in the equation τ , , ( x, ˜y ) · T , ( x, ˜y ) = τ , , ( x, ˜y ) · T , ( x, ˜y ).Then we let vector-valued function y be the solution to a differential equation similar to (5).Analogously to the analysis of Algorithm 1 for d = 3 there are two phases. Numerical solutions tothe system we have described give a numerical bound of 0 . . T (1) and T (2) can affect the performanceof the algorithm. For example, if we always prioritize processing vertices from T (1) over vertices of T (2) , Algorithm 2 actually performs identically to Algorithm 1 as in this case the vertices of positivetype in T (1) do not accumulate, and so the number of vertices we find that can be inserted into our Z -sequence is negligible. It may be possible to improve further on our analysis by changing how thealgorithm chooses between sets T (1) and T (2) in the first step of the while loop. d In this section, we provide explicit bounds on the forcing number of d -regular graphs. These boundsare derived from tools based in spectral graph theory; in particular, the expander mixing lemma isused frequently. In what follows, we assume that G is a n -vertex d -regular graph, with eigenvalues ofits adjacency matrix listed in decreasing order as λ ( G ) ≥ . . . ≥ λ n ( G ). We refer to G as a ( n, d, λ )graph, provided | λ i | ≤ λ for each i = 2 , . . . , n , where λ ≥ Lemma 3.1. If G = ( V, E ) is a ( n, d, λ ) graph, then the following statements hold:1. λ ( G ) = d .2. If λ < d , then G is connected.3. λ ≥ √ d − d √ n . emma 3.2 (Expander Mixing Lemma) . If G = ( V, E ) is a ( n, d, λ ) graph, and U, W ⊆ V ( G ) , then (cid:12)(cid:12)(cid:12)(cid:12) d | U || W | n − e ( U, W ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ λ (cid:115) | U || W | (cid:18) − | U | n (cid:19) (cid:18) − | W | n (cid:19) , where e ( U, W ) := |{ ( u, w ) ∈ U × W : { u, w } ∈ E }| (edges with both ends in U ∩ W are counted twice). We first observe that Lemma 3.2 guarantees the existence of edges between subsets of sufficientlylarge size.
Proposition 3.3.
For any
U, W ⊆ V ( G ) , if u := min {| U | , | W |} > λd + λ n, then e ( U, W ) > .Proof. Indeed, it follows immediately from the Expander Mixing Lemma that e ( U, W ) ≥ d | U || W | n − λ (cid:115) | U || W | (cid:18) − | U | n (cid:19) (cid:18) − | W | n (cid:19) = | U || W | n (cid:34) d − λ (cid:115)(cid:18) n | U | − (cid:19) (cid:18) n | W | − (cid:19)(cid:35) ≥ u n (cid:104) d − λ (cid:16) nu − (cid:17)(cid:105) > . The result holds.We are now ready to analyze Algorithm 1 for general d . This will provide us with upper boundson the forcing number of arbitrary d -regular graphs. That being said, it will be convenient to assumethat G is connected.Recall that Algorithm 1 works by building a Z -sequence S = ( v , . . . , v k ) of the input graph G = ( V, E ). For each 1 < t ≤ k , let us denote by B t the set (cid:83) t − i =1 N [ v i ], and W t = V ( G ) \ B t . Wemay then denote B := ∅ and W := V ( G ). We will call the vertices in B t black and the vertices in W t white.For each vertex v ∈ V ( G ), we define its type at time 1 ≤ t ≤ k , denoted type t ( v ) (or type( v )when clear), to be the number of neighbours v has in W t , namely deg W t ( v ). We may then partition B t into the sets B t , . . . , B dt , where for 0 ≤ i ≤ d , B it is defined to be the vertices of B t with type i .Similarly, we may partition W t into the sets W t , . . . , W dt . It will be convenient to denote B + t and W + t as (cid:83) di =1 B it and (cid:83) di =1 W it , respectively.We initialize Algorithm 1 by first selecting the vertex v arbitrarily. For t >
1, it then chooses v t among the vertices of B t whose type is positive and nonzero; that is, we choose v t from the membersof B + t . In particular, it looks to select a vertex of this kind whose type is minimal. The algorithmfinishes executing when no such vertex exists. Observe that since the graph G is connected, once thevertex v k is processed, all the vertices of the graph must be coloured black. In particular, we knowthat | W k | ≤ d .Let us first consider an essential time in the execution of the algorithm. Namely, define t as thesmallest 1 < t ≤ k , for which | W t | ≤ λd + λ n. It is clear from the above discussion that t exists. We may therefore think of phase one as those t ≥
1, such that 1 ≤ t < t (note that the phases referred to in this section are different than thephases referred to in Section 2). Similarly, we may define the remaining time increments as the secondand final phase. It turns out that in every iteration of the algorithm in phase one, we are guaranteedto find a black vertex of type proportional to | W t | .12 emma 3.4. Let G = ( V, E ) be a ( n, d, λ ) graph, and assume that it is passed as input to Algorithm 1.For each < t < t , there exists a vertex u ∈ B t , such that ≤ type t ( u ) ≤ ( d + λ ) | W t | n + λ. Proof.
Observe that by Lemma 3.2, we are guaranteed a lower bound on twice the number of edgeswithin B + t . In particular, e ( B + t , B + t ) ≥ | B + t | (cid:18) d + λn | B + t | − λ (cid:19) . As a result, there exists a vertex u ∈ B + t , for whichdeg B + t ( u ) ≥ d + λn | B + t | − λ, where deg B + t ( u ) is the number of neighbours u has in B + t . Thus, we have thattype t ( u ) = d − deg B t ( u )= d − deg B + t ( u ) − deg B t ( u ) ≤ d − d + λn | B + t | + λ. However, | B + t | = n − | W t | − | B t | , so after simplification,type t ( u ) ≤ d + λn | W t | + d + λn | B t | . It remains to bound the size of | B t | . Recall that we are in phase 1, which implies that | W t | > λd + λ n .On the other hand, we know that e ( B t , W t ) = 0 by definition of B t . Thus, by Proposition 3.3, wehave that | B t | ≤ λd + λ n . Applying this observation to the above equation, we get thattype t ( u ) ≤ d + λn | W t | + λ, thereby proving the lemma, as type t ( u ) ≥ < t < t , the algorithm selects a vertex v t ∈ B + t , for which type( v t ) ≤ ( d + λ ) | W t | n + λ . Sincetype( v t ) many vertices are removed from W t when v t is processed, for each 1 < t < t we have | W t +1 | ≥ | W t | − ( d + λ ) | W t | n − λ. (12)Let us now define w t := | W t | for each t ≥
1. We shall use the above recursion to lower boundthe length of phase one, namely t . As the length of the Z-sequence we are generating is preciselythe number of iterations the algorithm performs, this will give us a lower bound on the length of the Z -sequence, and as a consequence of Theorem 2.1, an upper bound on the forcing number of G . Proposition 3.5.
Let G = ( V, E ) be a connected ( n, d, λ ) -graph, for which ≤ d ≤ √ n . In this case,we have that Z ( G ) ≤ n − log (cid:18) λd + λ (cid:19) (cid:18) log (cid:18) − d + λn (cid:19)(cid:19) − = n − (1 + o (1)) log (cid:18) d + λ λ (cid:19) nd + λ . roof. As above, let us assume that we pass G to Algorithm 1, which returns the Z -sequence S =( v , . . . , v k ). We denote t as the first time 1 ≤ t ≤ k for which w t ≤ λd + λ n . Our goal will be to lowerbound the size of t .By inequality (12), we know that for t ≥ w t ≥ (cid:0) − d + λn (cid:1) w t − − λ . Let us consider the sequence( a t ) t ≥ , where a := w = n and a t := (cid:0) − d + λn (cid:1) a t − − λ for each t ≥
1. This sequence has an exactsolution, in which a t = n (cid:18) λd + λ (cid:19) (cid:18) − d + λn (cid:19) t − − λnd + λ for each t ≥
1. In particular, observe that a = n − d − λ . On the other hand, we know that v and all of its neighbours are in B , so we have that w = n − d −
1. Moreover, the assumption that3 ≤ d ≤ √ n , together with Lemma 3.1 imply that λ ≥ . We may therefore conclude that w ≥ a .It follows that w t ≥ a t for all t ≥
1, as the function f ( x ) = (cid:0) − d + λn (cid:1) x − λ is monotone increasing.We may therefore analyze the sequence ( a t ) t ≥ to lower bound how long the first phase lasts. Inparticular, observe that for all 1 ≤ t ≤ log (cid:16) λd + λ (cid:17) (cid:0) log (cid:0) − d + λn (cid:1)(cid:1) − , w t ≥ a t > λnd + λ . Then the Z -Grundy dominating sequence S returned by the algorithm has length k ≥ t ≥ log (cid:16) λd + λ (cid:17) (cid:0) log (cid:0) − d + λn (cid:1)(cid:1) − . By Theorem 2.1, we may conclude that Z ( G ) ≤ n − log (cid:18) λd + λ (cid:19) (cid:18) log (cid:18) − d + λn (cid:19)(cid:19) − . Now, since d + λn →
0, we have 1 − d + λn = (1 + o (1)) e − ( d + λ ) /n , and thus (cid:18) log (cid:18) − d + λn (cid:19)(cid:19) − = − nd + λ (1 + o (1))thereby completing the proof.To conclude the section, we consider the case when for each n ≥
1, we are given a random d -regulargraph G n,d . The value of λ for random d -regular graphs has been studied extensively. A major resultdue to Friedman [10] is the following: For every fixed (cid:15) > G ∈ G n,d , a.a.s. λ ( G ) ≤ √ d − (cid:15). As a result, the above proposition implies that for each fixed d a.a.s, Z ( G n,d ) ≤ n − (1 + o (1)) log (cid:18) d + λ λ (cid:19) nd + λ , where λ = 2 √ d . If we consider large but constant d , then this implies the main result of the section. Theorem 3.6.
For any fixed (cid:15) > and for sufficiently large fixed d = d ( (cid:15) ) , it holds that a.a.s. G n,d satisfies Z ( G n,d ) ≤ n (cid:18) − (1 − (cid:15) ) log d d (cid:19) . Lower bound and bipartite holes
In this section we improve on a comment made in [12], where they observed that if a graph G doesnot contain two disjoint sets of q vertices each, with no edges from one set to the other (we call sucha substructure a q -bipartite hole), then Z ( G ) ≥ n − q . The reason for this is that if S = ( v , . . . , v q )is a Z -sequence with corresponding witness sequence W = ( w , . . . , w q ), the sets { v , . . . , v q } and { w q +1 , w q +2 , . . . , w q } form a q -bipartite hole. In [12] they use a known (but loose) result stating thata.a.s. G n,d has no n log dd -bipartite hole. We improve this and show the following: Theorem 4.1.
For any fixed (cid:15) > and for sufficiently large d = d ( (cid:15) ) , it holds that a.a.s. G n,d has no (1 + (cid:15) ) n log dd -bipartite hole. As a result, a.a.s. Z ( G n,d ) ≥ n (cid:18) − (1 + (cid:15) ) 4 log dd (cid:19) . Proof.
We use the first moment method. Fix some real numbers a, b and let X a,b be the expectednumber of induced subgraphs consisting of disjoint sets A , A of an vertices each, such that there areno edges between A and A and there are exactly bn edges within A (the number of edges within A is unspecified here). For our first moment calculation, the expected number of bipartite holes with an vertices on each side is (cid:80) b X a,b . Since the sum is over a linear number of terms, it suffices to choose a such that X a,b is exponentially small for all b .Using the pairing model described in Section 1.1, let us estimate X a,b for given parameters a and b . First we choose our sets A , A so we have (cid:18) nan an n − an (cid:19) = n !( an )!( an )!( n − an )!choices. Next we will choose 2 bn configuration points from among the dan points in A , so we have (cid:18) dan bn (cid:19) = ( dan )!(2 bn )!( dan − bn )!choices. Now we put a matching on these 2 bn points so we have(2 bn )!! = (2 bn )!2 bn ( bn )!choices. Now for the rest of the dan − bn configuration points in A , they just need to be matchedto any points outside of A ∪ A , so there are( dn − dan ) ( dan − bn ) = ( dn − dan )!( dn − dan + 2 bn )!choices. At this point there are dn − dan + 2 bn many configuration points left unmatched and anyof them can match to each other, so we have( dn − dan + 2 bn )!! = ( dn − dan + 2 bn )!2 dn − dan +2 bn (cid:0) dn − dan +2 bn (cid:1) !choices. Now we will multiply the previous several lines, and divide by the number of possible match-ings, that is, ( dn )!! = ( dn )!2 dn (cid:0) dn (cid:1) !and after a little cancellation we get X a,b = n !( dan )!( dn − dan )!( dn − dan + 2 bn )! (cid:0) dn (cid:1) !( an )!( an )!( n − an )!( dan − bn )!( bn )!( dn − dan + 2 bn )! (cid:0) dn − dan +2 bn (cid:1) !( dn )! · dan − bn . n ! = (1 + o (1)) √ πn (cid:0) ne (cid:1) n = (cid:0) ne (cid:1) n exp( o ( n )), so we estimate (cid:0) ne (cid:1) n (cid:0) dane (cid:1) dan (cid:0) dn − dane (cid:1) dn − dan (cid:0) dn − dan +2 bne (cid:1) dn − dan +2 bn (cid:16) dn e (cid:17) dn · dan − bn + o ( n ) (cid:0) ane (cid:1) an (cid:0) n − ane (cid:1) n − an (cid:0) dan − bne (cid:1) dan − bn (cid:0) bne (cid:1) bn (cid:0) dn − dan +2 bne (cid:1) dn − dan +2 bn (cid:16) dn − dan +2 bn e (cid:17) dn − dan +2 bn (cid:0) dne (cid:1) dn and now a large power of ne cancels and we are left with X a,b = ( da ) dan ( d − da ) dn − dan ( d − da + 2 b ) dn − dan +2 bn (cid:0) d (cid:1) dn · dan − bn + o ( n ) a an (1 − a ) n − an ( da − bn ) dan − bn b bn ( d − da + 2 b ) dn − dan +2 bn (cid:0) d − da +2 b (cid:1) dn − dan +2 bn d dn . Letting g ( x ) := x log x , and f ( a, b, d ) := g ( da ) + g ( d − da ) + g ( d − da + 2 b ) + g (cid:18) d (cid:19) − g ( a ) − g (1 − a ) − g ( da − b ) − g ( b ) − g ( d − da + 2 b ) − g (cid:18) d − da + 2 b (cid:19) − g ( d ) + ( da − b ) log 2 (13)we can then write X a,b = exp ( f ( a, b, d ) n + o ( n )) . So now, to prove that a.a.s. the graph has no bipartite hole with an vertices on each side (for some a which may depend on d ), it suffices to show that f ( a, b, d ) is negative for all b . Note that ∂f∂b = 2 ln( d − da + 2 b ) + 2 ln( da − b ) − ln b − d − da + 2 b ) − ln (cid:18) d − da + 2 b (cid:19) − (cid:32) ( d − da + 2 b ) ( da − b ) b ( d − da + 2 b ) (cid:0) d − da +2 b (cid:1) (cid:33) = ln (cid:18) ( d − da + 2 b )( da − b ) b ( d − da + 2 b ) (cid:19) = 0when ( d − da + 2 b )( da − b ) = 2 b ( d − da + 2 b ) . Expanding, and then factoring again, d (1 − a )( d − ad + 2 b )( d a + 4 dab − b − db ) = 0 . Solving this for b , it is clear that the solution we want is from the quadratic factor, b = 2 da − d + (cid:113) (2 da − d ) + 4 d a , (14)in other words we claim that, given fixed a, d , the value of b above maximizes f ( a, b, d ).Now we imagine d → ∞ , and a = K log dd . The value of b becomes b = 2 da − d + √ d − d a + 8 d a d (cid:16) a − (cid:112) − a + 8 a (cid:17) = d (cid:18) a − − a + 8 a ) −
18 ( − a + 8 a ) + O ( a ) (cid:19) = 12 da + O ( da )Since g (cid:48) ( x ) = log x + 1, g (cid:48)(cid:48) ( x ) = x − and g (cid:48)(cid:48)(cid:48) ( x ) = − x − by Taylor’s theorem we have g ( x + ∆ x ) = g ( x ) + (log x + 1)∆ x + (∆ x ) x + O (cid:18) (∆ x ) x (cid:19) | ∆ x | < x . If we plug in a = K log dd and b = da + O ( da ) we get (asymptotics are in d for thefollowing) f ( a, b, d ) = g ( da ) + g ( d − da ) + g ( d − da + 2 b ) + g (cid:18) d (cid:19) − g ( a ) − g (1 − a ) − g ( da − b ) − g ( b ) − g ( d − da + 2 b ) − g (cid:18) d − da + 2 b (cid:19) − g ( d ) + ( da − b ) log 2= g ( da ) + g ( d ) − da (log d + 1) + ( − da ) d + g ( d ) + ( − da + 2 b ) (log d + 1) + ( − da + 2 b ) d + g (cid:18) d (cid:19) − a log a − ( − a ) − [ g ( da ) − b (log da + 1))] − g ( b ) − (cid:20) g ( d ) + ( − da + 2 b ) (log d + 1) + ( − da + 2 b ) d (cid:21) − (cid:20) g (cid:18) d (cid:19) + ( − da + b ) (cid:18) log d (cid:19) + ( − da + b ) d (cid:21) − g ( d ) + ( da − b ) log 2 + o (cid:18) log dd (cid:19) = − da (log d + 1) + 2 b (log da + 1) + ( da − b ) (cid:18) log d (cid:19) − da + 2 a log d − b log b + ( da − b ) log 2 + o (cid:18) log dd (cid:19) = − b (log d + 1) + 2 b (log da + 1) − da + 2 a log d − b log b − b log 2 + o (cid:18) log dd (cid:19) = b − da + 2 a log d + o (cid:18) log dd (cid:19) = K (2 − K ) log dd + o (cid:18) log dd (cid:19) which is negative if K = 2(1 + (cid:15) ) and d is sufficiently large. This completes the proof.Finally we will justify numerical lower bounds for small d using the same method as above. Forfixed d , to show that a.a.s. G n,d has no an -bipartite hole it suffices to show that f ( a, b, d ) < f isdefined in equation (13) and b is taken to be the value in (14). It then follows that Z ( G n,d ) ≥ (1 − a ) n .The table below provides a summary for 3 ≤ d ≤
14 and justifies Table 2. In particular the thirdcolumn below corresponds to the lower bounds in Table 2.
In this paper, we have analyzed the zero forcing number of random d -regular graphs, providing anupper bound by considering a random greedy algorithm which creates a zero forcing set and providinga lower bound by considering the structure of bipartite holes. For any (cid:15) > d , we find that a.a.s., ( − (cid:15) ) log dd ≤ − Z ( G n,d ) n ≤ (4 + (cid:15) ) log dd . Of course, a central open problemis to determine the constant in front of log d/d and it seems reasonable to guess that it ought tobe 2 + √ G ( n, p ). Note that if S = ( v , . . . , v k ) is a Z -sequence with witnesssequence W = ( w , . . . , w k ), then v i ∼ w i for all 1 ≤ i ≤ k and v i (cid:54)∼ w j whenever i < j . The authorsin [12] call such a structure a k -witness and are able to use first and second moment methods todetermine asymptotically, the size of the largest k -witness which appears in G ( n, p ). In the setting ofrandom regular graphs, the calculations become much more complicated as the edges do not appearindependently. For example, the first moment argument given in Section 4 takes several pages, whereasa corresponding argument in G ( n, p ) takes only a line or two (as seen in [12]). We also wonder whethera different algorithm can be used to give a better general upper bound for large d .17 a − a d together with values a such that a.a.s. G n,d has no ( an )-bipartite hole. The numerical results presented in this paper (in particular, upper bounds in Table 1 and the figureswhich appear in the appendix) were obtained using Julia language [2]. We would like to thank Bogumi(cid:32)lKami´nski from SGH Warsaw School of Economics for helping us to implement it.
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