On Rall's 1/2 -conjecture on the domination game
aa r X i v : . [ m a t h . C O ] J un On Rall’s 1 / Csilla Bujt´as a , Vesna Irˇsiˇc b , Sandi Klavˇzar a,b,c , Kexiang Xu da Faculty of Mathematics and Physics, University of Ljubljana, Slovenia b Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia c Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia d College of Science, Nanjing University of Aeronautics & Astronautics,Nanjing, Jiangsu 210016, PR China [email protected] , [email protected]@fmf.uni-lj.si , [email protected] June 5, 2020
Abstract
The 1 / G is a traceablegraph, then the game domination number γ g ( G ) of G is at most l n ( G )2 m . Atraceable graph is a 1 / γ g ( G ) = l n ( G )2 m holds. It is proved that theso-called hatted cycles are 1 / / Keywords: domination game; 1 / AMS Math. Subj. Class. (2010) : 05C57, 05C69, 91A43
The domination game on a graph G is played by two players referred to as Dominatorand Staller. If Dominator (resp. Staller) is the one to start the game, we speak ofthe D-game (resp. S-game). The players alternately select vertices such that at eachmove at least one vertex is dominated that has not yet been dominated by the set1f previously selected vertices. As soon as this is not possible, the game is over;at that point the selected vertices form a dominating set of G . Dominator’s goalis to reach the end of the game as soon as possible, while Staller has the oppositegoal. Assuming that both players play optimally, the number of moves played inthe D-game (resp. S-game) is a graph invariant denoted by γ g ( G ) and named gamedomination number (resp. Staller-start game domination number γ ′ g ( G )) of G .The domination game was introduced in [6]. Early influential references on thisgame include [5, 7, 8, 11, 17, 19], while from the very extensive recent developmenton the domination game and its variants we select the papers [2, 3, 9, 14, 16, 21, 22].In this paper we are interested in the following conjecture that has been proposedseveral years ago by D. Rall, the first published source of it is [15, Conjecture 1.1].Recall that a traceable graph is a graph that contains a Hamiltonian path. Conjecture 1.1 If G is a traceable graph, then γ g ( G ) ≤ l n ( G )2 m . We say that a graph G is a 1 / -graph if G is traceable and γ g ( G ) = l n ( G )2 m .In other words, 1 / / / / For a positive integer k we will use the notations [ k ] = { , . . . , k } and [ k ] = { , , . . . , k − } . The order of a graph G will be denoted by n ( G ). A vertex ofa graph dominates itself and its neighbors; a dominating set in a graph G is a set ofvertices of G that dominates all vertices in the graph. The cardinality of a smallestdominating set of G is the domination number γ ( G ) of G . If G is a graph and S ⊆ V ( G ), then a partially dominated graph G | S is a graph together with a decla-ration that the vertices from S are already dominated. We will need the followingfundamental result on the domination game. Lemma 2.1 [17, Lemma 2.1] (Continuation Principle)
Let G be a graph, and let A, B ⊆ V ( G ) . If B ⊆ A , then γ g ( G | A ) ≤ γ g ( G | B ) and γ ′ g ( G | A ) ≤ γ ′ g ( G | B ) . P ′ n = P n +1 | u and P ′′ n = P n +2 |{ u, v } , where u and v are the end-vertices ofthe path in question. We will also need the following result. Lemma 2.2 [20, Lemmas 2.1 and 2.3] If n ≥ , then γ g ( P ′ n ) = γ g ( P ′′ n ) = (cid:26) ⌈ n ⌉ − n ≡ , ⌈ n ⌉ ; otherwise .γ ′ g ( P ′ n ) = γ ′ g ( P ′′ n ) = (cid:26) ⌈ n ⌉ + 1; n ≡ , ⌈ n ⌉ ; otherwise . Define the weighting function w of partially dominated paths P ′ q + r and P ′′ q + r with w( P ′ q + r ) = w( P ′′ q + r ) = 2 q + r = 0 , r = 1 , ; r = 2 , ; r = 3 . Here is another result that will be applied.
Lemma 2.3 [12, Lemma 3.2] (Union Lemma) If F , . . . , F k are vertex-disjoint pathswhere F i = P ′ n i or F i = P ′′ n i for i ∈ [ k ] and n i ≥ , then γ ′ g k [ i =1 F i ! ≤ & k X i =1 w( F i ) ' . Note that Lemma 2.3 clearly remains true under the weaker condition n i ≥ i ∈ [ k ]. / -graphs In this section we first state which paths and cycles are 1 / / / Paths and cycles
For paths P n ( n ≥
1) and cycles C n ( n ≥
3) the following non-trivial result holds: γ g ( P n ) = γ g ( C n ) = (cid:6) n (cid:7) − n ≡ , (cid:6) n (cid:7) ; otherwise . The only published proof of this theorem can be found in [20]. The result impliesthat each of P n and C n is a 1 / n (mod 4) ∈ { , , } .3 roken ladders If k ≥
1, then the broken ladder BL k is the graph obtained from the Cartesianproduct P (cid:3) K by adding a path of length 4 k + 1 between two adjacent vertices ofdegree 2. See Fig. 1 for BL . Moreover, we set BL = P (cid:3) K .Figure 1: The broken ladder BL . Proposition 3.1 [19, Theorem 3.3] If k ≥ , then γ g ( BL k ) = 2( k + 2) = n ( BL k )2 . Hatted cycles If n ≥
4, then the hatted cycle b C n is obtained from the cycle C n by adding a newvertex and connecting it to two vertices at distance 2 on the cycle; see Fig. 2 for b C .(In [19] these graphs were denoted with C ′ n .) x ′ xy y ′ Figure 2: The graph b C .Using similar reasoning as in the proof of [19, Theorem 2.2], we now prove thathatted cycles of order 4 k + 2 are 1 / Proposition 3.2 If k ≥ , then γ g ( b C k +1 ) = 2 k + 1 = n ( b C k +1 )2 . Proof.
We use the notation from Fig. 2. If Dominator starts the game on d = y ,then at most one of the vertices x and x ′ can be played in the remaining moves.Thus, after d = y is played, the game is the same as if it was played on C k +1 . Itfollows that γ g ( b C k +1 ) ≤ γ g ( C k +1 ) = 2 k + 1.Before describing an optimal strategy of Staller, we define a run to be a maximalsequence of consecutive dominated vertices. Her strategy is to dominate only one4ew vertex in each of her moves. She achieves this by playing on the end of a run,or on x or x ′ if there is only one run and y, y ′ are the end-vertices of this run. Let m denote the number of moves in the D-game on b C k +1 .If m is even, then Staller dominates m vertices and Dominator can dominate atmost 4 + 3( m −
1) vertices. Together, both players dominate at most 2 m + 1 vertices,which must be at least n ( b C k +1 ) = 4 k + 2. Hence, m ≥ k + 1.If m is odd, we use a similar reasoning as in the previous case. Together bothplayers dominate at most m − + 4 + 3( m +12 −
1) = 2 m + 2 vertices. Hence, m ≥ k .But as m is odd, it cannot equal 2 k . Thus m ≥ k + 1 holds also in this case.It follows that γ g ( b C k +1 ) ≥ k + 1 and we conclude the equality. (cid:3) The same reasoning proves that γ g ( b C n ) = l n ( b C n )2 m − n (mod 4) ∈ { , , } .Therefore, every hatted cycle satisfies Conjecture 1.1, but only those which areobtained from C k +1 are 1 / In this section we prove that Conjecture 1.1 holds for all unicyclic traceable graphs.Clearly, cycles are unicyclic traceable graphs. On the other hand, if two nonadja-cent vertices of the cycle of a unicyclic graph G are of degree at least 3, then G is nottraceable. From this fact it is easy to deduce that if G is traceable then G is a cycle,or a graph obtained by attaching a path to a vertex of a cycle, or a graph obtainedby attaching two disjoint paths to adjacent vertices of a cycle. The graphs fromthe latter two families will be called tadpole graphs and two tailed tadpole graphs ,respective. Since we already know that Conjecture 1.1 holds for cycles, to goal ofthis section is thus to prove that the conjecture also holds for tadpole graphs andfor two tailed tadpole graphs.From the preliminaries recall that the Union Lemma uses the weighting func-tion w( P ′ n ). To make computations simpler, we sometimes use this function in thefollowing equivalent form:w( P ′ n ) = w( P ′′ n ) = n ; n ≡ , n + ; n ≡ , , n + ; n ≡ . Tadpole graphs If m ≥ n ≥
1, then the ( m, n ) -tadpole graph T m,n is the graph obtained fromthe disjoint union of a cycle C m and a path P n by joining a vertex of C m with anend-vertex of P n . Clearly, n ( T m,n ) = n + m .5 heorem 4.1 If m ≥ and n ≥ , then γ g ( T m,n ) ≤ (cid:6) m + n (cid:7) . Proof.
Let n = 4 k + x + 1 and m = 4 ℓ + y + 3, where x, y ∈ { , , , } . Let v bethe vertex of T m,n of degree 3. The first move d = v of Dominator implies that γ g ( T m,n ) ≤ γ ′ g ( P ′ k + x ∪ P ′′ ℓ + y ) . Since γ ′ g ( P ′ r ) = γ ′ g ( P ′′ r ) holds by Lemma 2.2, it suffices to consider the cases when x ≤ y , that is, ten such cases. Each of them can be handled using the Union Lemma.Suppose first that x = y = 0. Then γ g ( T m,n ) ≤ γ ′ g ( P ′ k ∪ P ′′ ℓ ) ≤ P ′ k ) + w( P ′′ ℓ ) = 1 + 2 k + 2 ℓ , where the second inequality follows by the Union Lemma. Since n ( T m,n ) = 4 k +4 ℓ +4we get that γ g ( T m,n ) ≤ (cid:6) m + n (cid:7) .Suppose next that x = 1 and y = 3. Then γ g ( T m,n ) ≤ γ ′ g ( P ′ k +1 ∪ P ′′ ℓ +3 ) ≤ P ′ k +1 ) + w( P ′′ ℓ +3 )= 1 + (2 k + 1) + (2 ℓ + 7 /
4) = 2 k + 2 ℓ + 15 / . Since n ( T m,n ) = 4 k + 4 ℓ + 8 we get that γ g ( T m,n ) ≤ (cid:6) m + n (cid:7) . x y ⌈ w ( P ′ k + x ) + w ( P ′′ ℓ + y ) ⌉ n ( T m,n ) ⌈ n ( T m,n )2 ⌉ ⌈ k + 2 ℓ ⌉ + 1 = 2 k + 2 ℓ + 1 4 k + 4 ℓ + 4 2 k + 2 ℓ + 20 1 ⌈ k + 2 ℓ ⌉ + 2 = 2 k + 2 ℓ + 2 4 k + 4 ℓ + 5 2 k + 2 ℓ + 30 2 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 3 4 k + 4 ℓ + 6 2 k + 2 ℓ + 30 3 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 3 4 k + 4 ℓ + 7 2 k + 2 ℓ + 41 0 ⌈ k + 2 ℓ ⌉ + 2 = 2 k + 2 ℓ + 2 4 k + 4 ℓ + 5 2 k + 2 ℓ + 31 1 ⌈ k + 2 ℓ ⌉ + 3 = 2 k + 2 ℓ + 3 4 k + 4 ℓ + 6 2 k + 2 ℓ + 31 2 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 4 4 k + 4 ℓ + 7 2 k + 2 ℓ + 41 3 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 4 4 k + 4 ℓ + 8 2 k + 2 ℓ + 42 0 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 3 4 k + 4 ℓ + 6 2 k + 2 ℓ + 32 1 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 4 4 k + 4 ℓ + 7 2 k + 2 ℓ + 42 2 ⌈ k + 2 ℓ ⌉ + 4 = 2 k + 2 ℓ + 4 4 k + 4 ℓ + 8 2 k + 2 ℓ + 42 3 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 5 4 k + 4 ℓ + 9 2 k + 2 ℓ + 53 0 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 3 4 k + 4 ℓ + 7 2 k + 2 ℓ + 43 1 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 4 4 k + 4 ℓ + 8 2 k + 2 ℓ + 43 2 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 5 4 k + 4 ℓ + 9 2 k + 2 ℓ + 53 3 (cid:6) k + 2 ℓ + (cid:7) + 1 = 2 k + 2 ℓ + 5 4 k + 4 ℓ + 10 2 k + 2 ℓ + 5Table 1: The calculations for all different cases.The remaining cases to be considered can be treated along the same lines asthe above two cases. In Table 1 the summary of calculations for all the cases ispresented. (cid:3) wo tailed tadpole graphs If m ≥ n, k ≥
1, then the notation T m,n,k means a graph obtained from thedisjoint union of a cycle C m and paths P n and P k by joining adjacent vertices of C m with end-vertices of P n and P k with two independent edges. Clearly, n ( T m,n,k ) = n + m + k . Theorem 4.2 If m ≥ and n, k ≥ , then γ g ( T m,n,k ) ≤ l n ( T m,n,k )2 m . Proof.
Let G = T m,n,k and let the vertices of G be denoted by v , . . . , v n + m + k suchthat v . . . v n + m + k is the Hamiltonian path and v n +1 v n + m is the extra edge. Duringthe game, an antirun is a component of the subgraph induced by the undominatedvertices.Suppose that d = v n +1 . After this move, we have three antiruns: X = v . . . v n − , Y = v n +3 . . . v n + m − , and Z = v n + m +1 . . . v n + m + k . Together with theneighboring dominated vertices, we may consider the antiruns X , Y , Z as a P ′ n − ,a P ′′ m − , and a P ′ k , respectively. By the Continuation Principle, we can considerantiruns Y and Z as one, and get γ ′ g ( G | N [ d ]) ≤ γ ′ g ( P ′ n − ∪ P ′ m + k − ). Denote n = 4 n ′ + x + 1, k = 4 k ′ + z , m = 4 m ′ + y + 2, where x, y, z ∈ { , , , } and m ′ , n ′ , k ′ are integers. Hence the Union Lemma yields γ g ( G ) ≤ γ ′ g ( P ′ n − ∪ P ′ m + k − ) ≤ (cid:6) w( P ′ n ′ + x ) + w( P ′ m ′ + k ′ )+ y + z ) (cid:7) . Repeating similar calculations as in the proof of Theorem 4.1 for all 64 differentvalues of ( x, y, z ), we see that for most cases γ g ( G ) ≤ l n ( G )2 m holds. The exceptionalcases, after transforming from ( x, y, z ) to ( n, m, k ) ( mod 4), are gathered in Table 2. n (mod 4) 2 2 2 2 3 3 3 3 3 3 3 3 0 0 0 0 m (mod 4) 2 3 0 1 2 2 3 3 0 0 1 1 2 3 0 1 k (mod 4) 2 1 0 3 1 3 0 2 1 3 0 2 2 1 0 3Table 2: Exceptional cases.Most of those cases can be omitted by symmetry (assuming d = v n + m and thenrepeating the calculations using the Union Lemma). The only problematic cases leftare: • m ≡ n ≡
2, and k ≡ • m ≡ n ≡
2, and k ≡ • m ≡ n ≡
3, and k ≡ • m ≡ n ≡
3, and k ≡ m ≡ n ≡
3, and k ≡ • m ≡ n ≡
0, and k ≡ m ≡ n ≡
2, and k ≡ d = v n . His second move d is such that after s and d , allthree vertices in { v n + m , v n + m +1 , v n + m +2 } are dominated. By using the ContinuationPrinciple, the orders of the antiruns are n − ≡ m − ≡
0, and k − ≡ γ g ( G ) ≤ (cid:24) n −
22 + m −
22 + k − (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . For all the remaining cases, we assume again that d = v n +1 is the first moveof Dominator. His second move d will be specified depending on the move s ofStaller.At the beginning, there is only one antirun. After the move d = v n +1 thatdominates four vertices, we have the following three antiruns: X = v . . . v n − , Y = v n +3 . . . v n + m − , and Z = v n + m +1 . . . v n + m + k . Together with the neighboringdominated vertices, we may consider the antiruns X , Y , Z as a P ′ n − , a P ′′ m − , anda P ′ k , respectively. Note that X or Y might be a path of order 0 (i.e., an emptygraph), but we always assume that n, m, k are positive integers. After the move d ,at any point in the game, each antirun will be a path P ′ j or P ′′ j for an appropriate j .In all the remaining cases n ( G ) is even. Suppose that Staller’s move s increasesthe number of antiruns. Then, we have four antiruns X , . . . , X and, as such a move s increases the number of dominated vertices by 3, we have | X | + | X | + | X | + | X | = n + m + k −
7. We consider three cases. If there is an antirun containing at leastthree vertices, Dominator can choose d such that the number of antiruns is notincreased and the move dominates three new vertices. Applying the Union Lemmafor this graph with antiruns X ∗ , . . . , X ∗ , we obtain the following estimation on thetotal number of moves t : t ≤ γ ′ g [ i =1 X ∗ i ! ≤ & X i =1 w( X i ) ' ≤ & X i =1 (cid:18) | X ∗ i | (cid:19)' = 3 + (cid:24) n + m + k −
102 + 4 · (cid:25) = 3 + (cid:24) n ( G )2 − (cid:25) = n ( G )2 . Note that the last step uses the fact that n ( G ) is even. In the second case, afterStaller’s move s , there is no antirun of order at least three but there is an antirunwith | X i | = 2. Then, Dominator may play a vertex d that dominates the entire X i X ∗ , X ∗ , and X ∗ , we have t ≤ γ ′ g [ i =1 X ∗ i ! ≤ & X i =1 (cid:18) | X ∗ i | (cid:19)' = 3 + (cid:24) n + m + k −
92 + 3 · (cid:25) = n ( G )2 . In the third case, every antirun consists of one vertex after Staller’s move s and,therefore, n ( G ) = 11. It can be checked by hand (we also checked by computer)that the game can be finished in 6 moves, thus γ g ( G ) ≤ ⌈ n ( G ) / ⌉ holds.From now on, we may assume that Staller’s move s does not increase the numberof antiruns. Let X ′ ⊆ X , Y ′ ⊆ Y , and Z ′ ⊆ Z be the antiruns after the move s .Remark that n ( G ) = n + m + k is even for each of the following cases. • Case 1. m ≡ n ≡
2, and k ≡ X , we may assume by the Con-tinuation Principle that | X ′ | = | X | − n − | X ′ | ≡ d = v n + m +2 which dominates threevertices from Z ′ = Z . The order of the antiruns are n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
22 + m −
32 + 12 + k − (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . If Staller dominates at least one vertex from Y , we may apply the ContinuationPrinciple again. Then, Dominator plays d = v n + m +2 which dominates threevertices. The antiruns are of the following orders: n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
12 + 12 + m −
42 + 12 + k − (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . In the third case, Staller does not dominate any vertices from X ∪ Y andconsequently, she has to dominate at least two vertices from Z by playingeither v m + n +1 , v m + n +2 , v m + n + k − , or v m + n + k . By the Continuation Principle,we may assume that | Z ′ | = | Z | −
2. In the next move, Dominator plays d = v n +2 and dominates two vertices from Y ′ = Y . This creates antirunswith the following orders: n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
12 + 12 + m −
52 + k −
22 + 12 (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . This finishes the proof for Case 1, since it follows that γ g ( G ) ≤ t ≤ n ( G )2 .9 Case 2. m ≡ n ≡
3, and k ≡ s . If Staller dominates at least one vertex from X , Dominator replieswith d = v n + m +2 . After this move, the orders of the antiruns are n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
22 + 12 + m −
32 + 14 + k − (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2by the Union Lemma. If Staller dominates at least one vertex from Y , Dom-inator plays d = v n + m +2 again and we have antiruns satisfying n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
12 + 12 + m −
42 + 12 + k − (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . If Staller does not dominate any vertices from X ∪ Y , then she dominates atleast two vertices from Z . We may assume by the Continuation Principle that | Z ′ | = | Z | −
2. Then, Dominator plays d = v n +3 and dominates three verticesfrom Y ′ = Y . After this move we have the following antiruns: n − ≡ m − ≡ k − ≡ t ≤ (cid:24) n −
12 + 12 + m −
62 + k −
22 + 12 (cid:25) = 3 + (cid:24) n + m + k − (cid:25) = n ( G )2 . • Case 3. m ≡ n ≡
3, and k ≡ X , Dominator plays d = v n + m +2 and then, the antirunsare n − ≡ m − ≡ k − ≡ Y , Dominator plays d = v n + m +2 again. Then, we haveantiruns of order n − ≡ m − ≡ k − ≡ Z , Dominator replies by dominating at least two verticesfrom X . The antiruns are: n − ≡ m − ≡ k − ≡ t ≤ n ( G ) / • Case 4. m ≡ n ≡
0, and k ≡ s from X , Dominator selects d = v n + m +2 and then, we havethe antiruns n − ≡ m − ≡
2, and k − ≡ s dominatesat least one vertex from Y , Dominator plays d = v n + m +2 and then, theantiruns are n − ≡ m − ≡
1, and k − ≡ s and dominates at least two vertices from Z , Dominator can10lay d = v . Since n > n ≡ n − ≡ m − ≡
2, and k − ≡ t ≤ n ( G ) / • Case 5. m ≡ n ≡
0, and k ≡ n ≥ k . (Otherwise, Dominatorwould play v m + n instead of v n +1 .) If Staller’s move dominates a vertex from X and X ′ still contains at least three undominated vertices, then Dominatorplays in X ′ and leaves | X | − n − ≡ m − ≡
1, and k ≡ n ( G ) /
2. IfDominator cannot do this, then n − Y ′ if possible. In this case, the orders of the antiruns are n − m − ≡ k ≡ X ′ nor Y ′ contains three undominated vertices, then n = 4, m = 4 and, by theassumption n ≥ k > k ≡ k = 4 follows. It can be checkedby hand (or computer), that the game domination number of this graph is6 = n ( G ) /
2. For the remaining cases Dominator’s startegy is the usual. IfStaller dominates a vertex from Y , then Dominator dominates three verticesfrom X ′ . Note that this is possible as the antirun contains n − ≡ Z ,Dominator can reply by dominating at least three vertices from X ′ .Since for each possible case, we have shown a strategy of Dominator which ensuresthat the game is finished in at most ⌈ n ( G ) / ⌉ moves, the desired upper bound γ g ( G ) ≤ ⌈ n ( G ) / ⌉ follows. (cid:3) Applying a relationship between the domination game and minimal edge cuts, non-trivial families of 1 / Cycles with a chord
A graph obtained from a cycle C n by adding an edge between two nonadjacent ver-tices is clearly traceable. These graphs form our next family for which Conjecture 1.1holds. 11 roposition 5.1 If G is a graph obtained from a cycle C n by connecting two non-adjacent vertices of the cycle, then γ g ( G ) ≤ l n ( G )2 m . Proof.
Let v , . . . , v n be the vertices of the cycle and let v v i , 3 ≤ i ≤ n −
1, be theadditional edge. Suppose Dominator starts the game by playing d = v . Then itfollows by the Continuation Principle that γ g ( G ) ≤ γ ′ g ( P ′′ n − | v i ) ≤ γ g ( P ′′ n − ) . Applying Lemma 2.2 in each of the following cases, we get: • If n ≡ γ g ( G ) ≤ (cid:24) n − (cid:25) = 1 + n −
22 = n l n m . • If n ≡ γ g ( G ) ≤ (cid:24) n − (cid:25) = 1 + n −
32 = n − ≤ l n m . • If n ≡ γ g ( G ) ≤ (cid:24) n − (cid:25) + 1 = 2 + n −
32 = n + 12 = l n m . (cid:3) Graphs from F ( X ) , where X is traceable Let X be a traceable graph. Then the family F ( X ) consists of all graphs G thatcan be constructed in the following way. G is obtained from the disjoint union of X and a path P n , n ≥
3, with end-vertices y and y ′ , by connecting y to all verticesof X , and by connecting y ′ to the vertices from W ⊆ V ( X ), where W contains atleast one end-vertex of some Hamiltonian path in X . In the example in Fig. 3 wehave X = P , n = 8, and y ′ is adjacent to all vertices of P , that is, W = V ( P ).If X is a traceable graph and G ∈ F ( X ), then it is easy to observe that G istraceable. Hence the following result is of interest to us. Proposition 5.2 If X is a traceable graph and G ∈ F ( X ) , then γ g ( G ) ≤ l n ( G )2 m . y ′ Figure 3: A graph from F ( P ). Proof.
Let X be a traceable graph and G ∈ F ( X ), where the end-vertex y of thebuilding graph P n of G is adjacent to all the vertices of the building graph X of G .Consider the D-game and set d = y . This move dominates all the vertices of X in G , and at most one of the vertices from V ( X ) can be played in the rest of the game.If follows that after the first move d = y is played, the game is the same as if itwould be played on C n +1 . Since cycles fulfill Conjecture 1.1, the same holds for G . (cid:3) Particular Halin graphs A Halin graph is a graph obtained from a plane embedding of a tree T on at leastfour vertices and with no vertex of degree 2, by connecting the leaves of T into acycle in the clock-wise order with respect to the embedding. These graphs wereintroduced in [13] and are of continuing interest, cf. [10]. For us the most importantproperty of these graphs is that they are Hamiltonian [1].Let k ≥ d ≥
3, and d i ≥ i ∈ [ k − H ( k ; d , . . . , d k − ) be theHalin graph obtained from the tree T = T ( k ; d , . . . , d k − ) defined as follows. Let r be the root of T of degree d . For i ∈ [ k − i from r is ofdegree d i + 1. The vertices at distance k from r are the leaves of T . See Fig. 4 for T (3; 4 , ,
3) and H (3; 4 , , Proposition 5.3 If k ≥ , d i ≥ for i ∈ [ k ] , and H = H ( k ; d , . . . , d k − ) , then γ ( H ) < n ( H )4 . Consequently, γ g ( H ) < n ( H )2 − . Proof.
Let V i , i ∈ [ k + 1] , be the set of vertices of H at distance i from the root r .13 r Figure 4: The graphs T (3; 4 , ,
3) (left) and H (3; 4 , ,
3) (right).In particular, V = { r } . Let D = V ∪ S k/ i =1 V i − ; k ≡ ,V ∪ S ( k − / i =1 V i ; k ≡ , S ( k +1) / i =1 V i − ; k ≡ . Since for every i ∈ [ k − V i dominates V i − ∪ V i ∪ V i +1 , it readily followsthat D is a dominating set of H . Moreover, as d i ≥
3, we also have that | V i | < | V i − ∪ V i ∪ V i +1 | /
4, from which the first assertion of the proposition follows.Since for every graph G we have γ g ( G ) ≤ γ ( G ) − H can be bounded as follows: γ g ( H ) ≤ γ ( H ) − < · n ( H )4 − n ( H )2 − (cid:3) Note that the only requirement that the proof of Proposition 5.3 works is that | V i | < | V i − ∪ V i ∪ V i +1 | / i for which V i ⊆ D . This is achievedby only requiring that d i ≥ i . A possible approach to prove (or disprove) Rall’s conjecture is the following. Let G be an arbitrary traceable graph and let P be a Hamiltonian path in G . Thenwe know that the conjecture holds when the game is played on P . Adding edgesto P one by one, until G is reached, while keeping the game domination numberbelow ⌈ n ( G ) / ⌉ , would yield the conjecture. An obstruction with this approach isthe fact [4, Proposition 2.4] which asserts that for any ℓ ≥ G with an edge e such that γ g ( G ) = ℓ and γ g ( G − e ) = ℓ −
2. Thus, going from G − e to G , the result implies that adding an edge to a graph, the game domination14umber can increase by up to 2. But is the situation different for traceable graphs?Moreover, we still have all possible Hamiltonian paths to start with, as well as allpossible orders of the edges not on P to be put back into G . In this section wereport our computer experiments on this approach.Our first partial support for Rall’s conjecture was obtained by computer. Proposition 6.1 If ≤ n ≤ and G is a path P n with two additional edges, then γ g ( G ) ≤ ⌈ n ⌉ . If ≤ n ≤ , the same holds for a path P n with three additionaledges. As mentioned above, proving that in the case of traceable graphs adding edgesto the graph does not increase the game domination number would suffice to proveRall’s conjecture. However, the following example shows that this is not true.
Example 6.2
It holds that γ g ( P ) = 5 , and the same value is achieved for allpossibilities of adding one or two edges to the path P . But when three edges areadded, it can happen that the game domination number increases to ⌈ ⌉ . Let V ( P ) = [11] with naturally defined edges. Configurations that result in the gamedomination number are obtained by adding the following edges to P : ∼ , ∼ , ∼ or ∼ , ∼ , ∼ (see Fig. 5). Denote these graphs with R and R ′ , respectively. Figure 5: The graphs R and R ′ , both have game domination number 6.By adding the same structure of edges as in the Example 6.2 to the paths P n +3 with vertex set [4 n + 3] for n ∈ { , . . . , } , the same phenomena occurs: the valueon a path with three additional edges is larger than on the path. It is possible thatthe same happens for longer paths as well.Let n ≥ R n +3 be the graph with the vertex set [4 n + 3] obtainedfrom the path P n +3 by adding the edges 0 ∼
4, 5 ∼
8, and 1 ∼
7. Note that γ g ( P n +3 ) = 2 n + 1. Proposition 6.3 If n ≥ , then γ g ( R n +3 ) ≤ n + 2 . Proof.
Consider the following subgraphs of R n +3 : the graph R induced by vertices[11] , and n − P induced by vertices { i − , i, i + 1 , i + 2 } for15 ∈ { , . . . , n } . We have checked by computer that γ g ( R ) = 6 and that everyvertex from [11] is an optimal first move for Dominator.To prove the upper bound, we describe an appropriate strategy of Dominator.His first move is d = 10. If Staller plays on R and this subgraph is not yetdominated, then Dominator selects his optimal move in R . If Staller plays on R and this subgraph becomes dominated, then Dominator plays his optimal move inone of the subgraphs P , that is, a vertex of P that dominates three vertices of it. IfStaller plays on some P and it is not yet dominated after her move, then Dominatorreplies optimally on the same copy of P after which all the vertices of the P aredominated. Otherwise, Dominator plays his optimal move on some other P . Bythe above observation, at most 6 moves are played on R . As γ g ( P ) = γ ′ g ( P ) = 2,at most 2 moves are played on each P . Hence, the number of moves is at most6 + 2( n −
2) = 2 n + 2. (cid:3) We believe that in Proposition 6.3 the equality actually holds.Using similar approach, we have also investigated cycles with some additionaledges. The following result was obtained by a computer, but no example when theaddition of edges would increase the game domination number of a Hamiltoniangraph was obtained.
Proposition 6.4 If ≤ n ≤ and G is a cycle C n with two additional edges, then γ g ( G ) ≤ γ g ( C n ) ≤ ⌈ n ⌉ . If ≤ n ≤ , the same holds for a cycle C n with threeadditional edges. Acknowledgements
We are grateful to Gaˇsper Koˇsmrlj for providing us with his software that com-putes game domination invariants. We acknowledge the financial support from theSlovenian Research Agency (bilateral grant BI-CN-18-20-008, research core fundingP1-0297, projects J1-9109, J1-1693, N1-0095, N1-0108). Kexiang Xu is also sup-ported by NNSF of China (grant No. 11671202) and China-Slovene bilateral grant12-9.
References [1] J.A. Bondy, L. Lov´asz, Lengths of cycles in Halin graphs, J. Graph Theory 9(1985) 397–410.[2] M. Borowiecki, A. Fiedorowicz, E. Sidorowicz, Connected domination game,Appl. Anal. Discrete Math. 13 (2019) 261–289.163] B. Breˇsar, Cs. Bujt´as, T. Gologranc, S. Klavˇzar, G. Koˇsmrlj, T. Marc, B. Patk´osZs. Tuza, M. Vizer, The variety of domination games, Aequationes Math. 93(2019) 1085–1109.[4] B. Breˇsar, P. Dorbec, S. Klavˇzar, G. Koˇsmrlj, Domination game: effect of edge-and vertex-removal, Discrete Math. 330 (2014) 1–10.[5] B. Breˇsar, S. Klavˇzar, G. Koˇsmrlj, D.F. Rall, Domination game: extremalfamilies of graphs for 3 / ppendix just for reviewers In Table 3 the complete calculations that were used to produce Table 2 in the proofof Theorem 4.2 are listed. x y z ⌈ w( P ′ n ′ + x ) + w( P ′ m ′ + k ′ )+ y + z ) ⌉ ⌈ n ( G )2 ⌉ ≤ ?0 0 0 2 k ′ + 2 m ′ + 2 n ′ + 1 2 k ′ + 2 m ′ + 2 n ′ + 2 True0 0 1 2 k ′ + 2 m ′ + 2 n ′ + 2 2 k ′ + 2 m ′ + 2 n ′ + 2 True0 0 2 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 0 3 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 1 0 2 k ′ + 2 m ′ + 2 n ′ + 2 2 k ′ + 2 m ′ + 2 n ′ + 2 True0 1 1 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 1 2 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 1 3 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 4 True0 2 0 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 2 1 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 2 2 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 4 True0 2 3 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True0 3 0 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True0 3 1 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 4 True0 3 2 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True0 3 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True1 0 0 2 k ′ + 2 m ′ + 2 n ′ + 2 2 k ′ + 2 m ′ + 2 n ′ + 2 True1 0 1 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True1 0 2 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 3 False1 0 3 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 1 0 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True1 1 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 3 False1 1 2 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 1 3 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 2 0 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 3 False1 2 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 2 2 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 2 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True1 3 0 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 3 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True1 3 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True1 3 3 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 5 False2 0 0 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True2 0 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 3 False2 0 2 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True2 0 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False19 1 0 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 3 False2 1 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True2 1 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False2 1 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True2 2 0 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True2 2 1 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False2 2 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True2 2 3 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 5 False2 3 0 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False2 3 1 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True2 3 2 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 5 False2 3 3 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 6 True3 0 0 2 k ′ + 2 m ′ + 2 n ′ + 3 2 k ′ + 2 m ′ + 2 n ′ + 3 True3 0 1 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True3 0 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False3 0 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 1 0 2 k ′ + 2 m ′ + 2 n ′ + 4 2 k ′ + 2 m ′ + 2 n ′ + 4 True3 1 1 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False3 1 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 1 3 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 2 0 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 4 False3 2 1 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 2 2 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 2 3 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 6 True3 3 0 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 3 1 2 k ′ + 2 m ′ + 2 n ′ + 5 2 k ′ + 2 m ′ + 2 n ′ + 5 True3 3 2 2 k ′ + 2 m ′ + 2 n ′ + 6 2 k ′ + 2 m ′ + 2 n ′ + 6 True3 3 3 2 k ′ + 2 m ′ + 2 n ′ + 7 2 k ′ + 2 m ′ + 2 n ′ + 6 FalseTable 3: For all possible values of x, y, z we check if1 + ⌈ w( P ′ n ′ + x ) + w( P ′ m ′ + k ′ )+ y + z ) ⌉ ≤ ⌈ n ( G )2 ⌉⌉