Feedback game on 3 -chromatic Eulerian triangulations of surfaces
aa r X i v : . [ c s . D M ] F e b Feedback game on 3-chromatic Eulerian triangulationsof surfaces
Akihiro Higashitani ∗ , Kazuki Kurimoto † and Naoki Matsumoto ‡ Abstract
Recently, a new impartial game on a connected graph has been introduced, calleda feedback game , which is a variant of the generalized geography. In this paper, westudy the feedback game on 3-chromatic Eulerian triangulations of surfaces. Weprove that the winner of the game on every 3-chromatic Eulerian triangulation of asurface all of whose vertices have degree 0 modulo 4 is always fixed. Moreover, wealso study the case of 3-chromatic Eulerian triangulations of surfaces which have atleast two vertices whose degrees are 2 modulo 4. In addition, as a concrete class ofsuch graphs, we consider the octahedral path, which is obtained from an octahedronby adding octahedra in the same face, and determine the winner of the game onthose graphs.
Keywords:
Feedback game, Octahedron addition, Eulerian triangulation.
AMS 2010 Mathematics Subject Classification:
All graphs considered in this paper are finite simple undirected graphs. We say that agraph G is Eulerian if every vertex of G has even degree. We refer the reader to [3] forthe basic terminologies.Recently, a new impartial game on a connected graph has been introduced, called a feedback game . Definition 1.1 ([8]) . There are two players, Alice and Bob. Alice starts the game. For agiven connected graph G with a starting vertex s , a token is put on s . They alternatelymove the token on a vertex u to another vertex v which is adjacent to u and the edge uv will be deleted after he or she moves the token. The first player who is able to move thetoken back to the starting vertex s or to an isolated vertex (after removing an edge usedby the last move) wins the game. ∗ Department of Pure and Applied Mathematics, Graduate School of Information Science and Tech-nology, Osaka University, Suita, Osaka 565-0871, Japan. E-mail: [email protected] † Department of Mathematics, Graduate School of Science, Kyoto Sangyo University, Kyoto 603-8555,Japan. E-mail: [email protected] ‡ Research Institute for Digital Media and Content, Keio University, Yokohama, Kanagawa 232-0062,Japan. E-mail: [email protected] generalized geography , which is a most famoustwo-player impartial game played on (directed) graphs. It is known that for many variantsof the game to determine the winner of the game is PSPACE-complete (see [5, 7, 10] forexample). Similarly, the decision problem of the feedback game is PSPACE-completeeven if the maximal degree of a given connected graph is 3 or it is 4 and the graph isEulerian [8]. Thus it is an important problem to determine the winner of the game onconcrete classes of graphs.In the recent paper [8], they focus on connected Eulerian graphs, since the token isfinally moved back to the starting vertex if a given connected graph is Eulerian. It is trivialthat Bob wins the game if a given connected Eulerian graph is bipartite, and hence, theydeal with non-bipartite ones in that paper. In particular, they gave sufficient conditionsfor Bob to win the game on two concrete classes of such graphs, triangular grid graphsand toroidal grid graphs, and they also conjectured that those conditions are necessary.One of challenging problems in this study is to determine the winner of the game on3-chromatic graphs. In this paper, we focus on the feedback game on 3-chromatic Euleriantriangulations of surfaces and determine the winner of the game, where a triangulation is a graph of a surface with each face triangular and every two faces sharing at mostone edge. (In this paper, a surface means a closed surface, i.e., a 2-dimensional manifoldwithout boundary.) Let c n,k ( G ) be the number of vertices of a graph G whose degrees arecongruent to k modulo n ; note that G satisfies | V ( G ) | = c , ( G ) + c , ( G ) if G is Eulerian.We begin with the following general result and a remark for the game on Euleriantriangulations. Theorem 1.2.
Let G be a -chromatic Eulerian triangulation of a surface. If c , ( G ) = 0 ,then Bob always wins the game on G with any starting vertex. Remark 1.3.
It is well known that every Eulerian triangulation of the sphere is 3-chromatic. Moreover, we can see by the Handshaking Lemma that we have c , ( G ) ≥ G of a surface with c , ( G ) = 0 and that if c , ( G ) = 2, then the two such vertices belong to the same partite set of G .By the above theorem and remark, we mainly deal with Eulerian triangulations G ofsurfaces with c , ( G ) ≥
2. It is relatively easy to see that Bob wins the game on everyEulerian double wheel , which is an Eulerian triangulation on the sphere with n ≥ , , . . . , , n − , n − Theorem 1.4.
For any surface F and any positive integer m ≥ , there exists a -chromatic Eulerian triangulation G of F with c , ( G ) = m and a starting vertex suchthat Alice wins the game on G . Thus we study one of reasonable concrete classes of Eulerian triangulations, called an octahedral path , denoted by E n . (See Figure 1.) This graph is obtained from the octahe-dron (i.e., E ) by repeatedly adding a new octahedron in the same face, which way lookslike a path. By the construction, the degree sequence of E n is (4 , , , , , , , , . . . , | {z } n − ).2ote that every Eulerian triangulation of the sphere can be constructed from E by twooperations, one of which is the octahedron addition. (The octahedron addition is to addthree vertices a , a , a to a face u , u , u of an Eulerian triangulation and edges so that a i u j u k and a i a j u k are faces of the resulting graph for { i, j, k } = { , , } . See [2, Theorem3] for more detail.) u u u u n v w v v v n w n w w Figure 1: Octahedral path E n The feedback game on the graph E n with its starting vertex u p or v p or w p is denotedby E ( n, p ), where n ≥ ≤ p ≤ n . Notice that the choice of u p , v p , w p does notmatter. We give the complete characterization whether Alice or Bob wins the game on E ( n, p ): Theorem 1.5.
For any integers n and p with n ≥ and ≤ p ≤ n , the winner of thegame E ( n, p ) is determined as follows: n ≡ n ≡ n ≡ p ≡ Alice Bob Alice p ≡ Alice Bob Bob p ≡ Alice Alice AliceHere, all “ ≡ ” mean that “congruent modulo ”. In the next section, we introduce an important concept for the feedback game, calledan even kernel , and we prove Theorems 1.2 and 1.4 and Proposition 2.5. In Section 3, weshall give a proof of Theorem 1.5. -chromatic Eulerian triangulations We first introduce the key notion, called an even kernel , which is first introduced in [5].3 efinition 2.1 (Even kernel [5]) . Let G be a connected graph with a starting vertex s .An even kernel for G is a nonempty subset S ⊂ V ( G ) such that • s ∈ S , • no two vertices in S are adjacent, and • every vertex not in S is adjacent to an even number (possibly 0) of vertices in S . Definition 2.2 (Even kernel graph [8]) . Let S be an even kernel of a connected Euleriangraph G with a starting vertex. An even kernel graph with respect to S is a bipartitesubgraph H S with the bipartition V ( H S ) = S ∪ R and E ( H S ) = E G ( S, R ), where R isa superset of the set N G ( S ) = { v ∈ V ( G ) \ S : v is adjacent to a vertex u ∈ S } and E G ( A, B ) denotes the set of edges between A and B . Remark 2.3.
It is easy to see that if a connected graph G with some starting vertexhas an even kernel, then Bob wins the feedback game on G since Bob can always move atoken from a vertex not in S back to a vertex in S [5, 8]. As a corollary, Bob also wins thegame on every connected Eulerian bipartite graph. It is NP-complete in general to findan even kernel of a given graph [4]. Furthermore, there exist infinitely many connectedEulerian graphs without an even kernel (for a specified starting vertex) on which Bobwins the game [8].Now we shall prove Theorems 1.2 and 1.4. Proof of Theorem 1.2.
Let G be a 3-chromatic Eulerian triangulation on a surface andlet ( S , S , S ) be the tripartite set of G , i.e., V ( G ) = S i =1 S i and S i ∩ S j = ∅ for i = j .Without loss of generality, the starting vertex s is in S . Since the degree of each vertexof G is zero modulo 4, each vertex in S i is adjacent to an even number of vertices in S j for any i, j ∈ { , , } with i = j . (Note that the subgraph induced by the neighbors ofeach vertex in G is an even cycle of length 4 k for some k ≥ S is clearlyan even kernel. Proof of Theorem 1.4. (The first step) : It suffices to prove the theorem for the sphere,since we can construct the desired Eulerian triangulation on a given surface F by “past-ing” a face of the desired one on the sphere (constructed below) and that of any 3-chromatic Eulerian triangulation on F with only vertices of degree 4 k with k ≥ Remark 2.4.
Let G be a 3-chromatic Eulerian triangulation on F and let uvw and uvw ′ be two faces of G . A of an edge uv of G is to replace uv with a path uabv and to add edges aw, bw, aw ′ , bw ′ . Note that w and w ′ belong to the same partite set of G and that the remainder of their degrees modulo 4 is changed after this operation. Theoctahedron addition also changes the remainder modulo 4 of degrees of three vertices onthe corresponding face boundary. Thus applying octahedron additions and 2-subdivisionssuitably, we can construct a 3-chromatic Eulerian triangulation on F with only verticesof degree 4 k with k ≥ m ≡ m ≥
3, the octahedral path E n with n ≥ m ≥ m ≡ , (The second step) : Let G be the Eulerian triangulation on the sphere shown in the leftof Figure 2. Note that exactly two vertices u and b are of degree 6 and other verticeshave degree 4 k for some k ≥ c , ( G ) = 2). We will show that Alice can win thegame on G with the starting vertex s . s su u u v v v b at t t Figure 2: An Eulerian triangulation on the sphere with the starting vertex s where Alicewins the gameAlice first moves the token to u . After that, unless Bob moves the token to a or u ,Alice moves it to a white vertex shown in the right of Figure 2 corresponding to Bob’smove. As in the argument of an even kernel (see Remark 2.3), Bob finally has to movethe token to a or u , and hence, Alice wins the game. (The third step) : Observe that four vertices t , t , t and b are not used for the abovewinning strategy of Alice in the second step. So, by repeatedly applying an octahedronaddition to a face consisting three vertices of degree 4 in the interior of t t t , we canincrease the number of vertices of degree 6 by 3 r for r ≥ m ≥ m ≡ m ≥ m ≡ bu t and bt t respectively. The resultinggraph has four vertices b, u , u and t of degree 4 k + 2 for some k ≥ m ≡ bt t ).We conclude this section with the following result. Proposition 2.5.
Bob wins the game on every Eulerian double wheel graph. roof. Let G be an Eulerian double wheel graph of n + 2 vertices with n ≥ n ≡ C n = v v . . . v n − be the rim of G and let x, y be two vertices of degree n .If n ≡ x or y is thestarting vertex of G , then Bob wins the game since { x, y } is an even kernel. Thus we mayassume that n ≡ v is the starting vertex by symmetry.If Alice first moves the token from v to v by symmetry, then Bob moves it to v .After that, since Alice can move the token to neither x nor y , she has to move it from v i to v i +1 and then Bob moves it to v i +2 , and hence, Bob can finally move the token backto v .So Alice first moves the token from v to x by symmetry. In this case, Bob first movesthe token to v , and then Alice moves it to v and Bob moves it to v . After that, Bobcan win the game as follows: If Alice moves the token to x , then Bob moves it to v . Sincetwo edges v v and v v are already removed, Alice must move the token to y and thenBob can move it back to v . Otherwise, similarly to the previous case, Bob can finallymove the token back to v . This section is devoted to proving Theorem 1.5. We use the same labeling of the verticesof E n as drawn in Figure 1. We can see that E n can be also drawn as Figure 3. u v u w v Figure 3: Another drawing of the octahedral path
Lemma 3.1.
Alice (resp., Bob) can win the game on E ( n, p ) if and only if Alice (resp.,Bob) can win the game on E ( n, n − p ) .Proof. It is trivial by the symmetry of the vertices u p , v p , w p and u n − p , v n − p , w n − p on E n ,respectively. 6or the proof of the lemmas below, we describe the neighbors of each vertex: N ( u i ) = { v i , w i , u i +1 , w i +1 } if i = 0 , { v i , w i , u i − , v i − } if i = n, { v i , w i , u i +1 , w i +1 , u i − , v i − } otherwise ,N ( v i ) = { u i , w i , v i +1 , u i +1 } if i = 0 , { u i , w i , v i − , w i − } if i = n, { u i , w i , v i +1 , u i +1 , v i − , w i − } otherwise ,N ( w i ) = { u i , v i , w i +1 , v i +1 } if i = 0 , { u i , v i , w i − , u i − } if i = n, { u i , v i , w i +1 , v i +1 , w i − , u i − } otherwise . (1) Lemma 3.2.
Bob can win the game on E (3 m + 1 , k ) and E (3 m + 1 , k + 1) for any m, k ∈ Z ≥ .Proof. Let S = { u i , v i +1 : i = 0 , , . . . , m } . For proving the statement, it is enough to show that S is an even kernel. By (1), we caneasily see the following: N ( v i ) ∩ S = N ( w i ) ∩ S = { u i , v i +1 } for i = 0 , , . . . , m ; N ( u i +1 ) ∩ S = N ( w i +1 ) ∩ S = { u i , v i +1 } for i = 0 , , . . . , m ; N ( u i +2 ) ∩ S = N ( v i +2 ) ∩ S = { v i +1 , u i +3 } for i = 0 , , . . . , m − N ( w i +2 ) ∩ S = ∅ . Lemma 3.3.
Bob can win the game on E (3 m + 2 , k + 1) for any m, k ∈ Z ≥ .Proof. Let us consider S = S ∪ S , where S = { u i , v i +1 : i = 0 , , . . . , k } and S = { v k +1+3 j , w k +2+3 j : j = 0 , , . . . , m − k } . Consider the bipartite graph H as depicted in Figure 4:Note that there are two even kernel graphs H and H with respect to S and S ,respectively, although S is not an even kernel since u k +1 and w k +1 are adjacent to u k , v k +1 and w k +2 .If Alice moves the token to the vertex x or y (resp., z or y ), then Bob may move thetoken to u k (resp., w k +2 ). After that, by following the even kernel graph H (resp., H ),we see that Bob can move the token to the vertex u k (resp., w k +2 ). Hence, Bob canfinally win the game. Lemma 3.4.
Alice can win the game on E ( n, k + 2) for any n, k ∈ Z ≥ . k +1 w k +2 y zxu k H H Figure 4: “Almost” even kernel graph Hv k +1 u k s = v k +2 u k +2 even kernel graphFigure 5: The even kernel graph with respect to S in Lemma 3.4 Proof.
Let us consider the same subset as in Lemma 3.2, i.e., S = { u i , v i +1 : i = 0 , , . . . , k } , and let s = v k +2 be the starting vertex (see Figure 5).First, Alice moves the token to the vertex v k +1 ∈ S . Then by following the evenkernel graph with respect to S , we see that Alice can move the token back to v k +1 again.Finally, Bob must move the token to u k +2 from v k +1 . Therefore, Alice can win thegame. Lemma 3.5.
Alice can win the game on E (3 m, k ) for any m, k ∈ Z ≥ . roof. Let us consider the same subsets as in Lemma 3.3, i.e., S = { u i , v i +1 : i = 0 , , . . . , k − } and S = { w k +2+3 j , u k +2+3 j +1 : j = 0 , , . . . , m − k − } , and let s = v k be the starting vertex (see Figure 6). u k v k +1 u k +3 u k − w k − w k +1 u k +1 w k +2 v k − v k − u k − w k s = v k e e Figure 6: The even kernel graphs with respect to S and S in Lemma 3.5First, Alice moves the token to w k . Then Bob should move the token to u k − or w k +1 , otherwise Bob will lose the game immediately. Consider the case u k − (resp., w k +1 ). Then Alice moves the token to v k − (resp., w k +2 ). By following the even kernelgraph with respect to S (resp., S ), we see that Alice can move the token back to v k − (resp., w k +2 ) again. Finally, Bob must move the token to v k − (resp., u k +1 ) from v k − (resp., w k +2 ). Therefore, Alice can win the game.We can now prove Theorem 1.5. Proof of Theorem 1.5.
All of the cases directly follow from Lemmas 3.1–3.5 as follows(where each “ ≡ ” means that “congruent modulo 3”): n ≡ n ≡ n ≡ p ≡ p ≡ p ≡ In this paper, focusing on 3-chromatic Eulerian triangulations of surfaces, we completelydetermine the winner of the feedback game on several classes of the graphs. In general,for any surface F which is not the sphere, there are infinitely many non-3-colorableEulerian triangulations of F . Since Theorem 1.2 strongly depends on the 3-colorability9f the graphs, it is not clear which player wins the game on a non-3-colorable Euleriantriangulation G even if c , ( G ) = 0. In fact, as shown in Figure 7, there exists a 5-chromatic Eulerian triangulation G of the projective plane with c , ( G ) = 0 such thatAlice wins the game on G with a starting vertex s ; she first moves the token on s to v ,and after that, since s is adjacent to all other vertices, Alice can move the token back to s regardless of Bob’s move. sv vv ′ v ′ v ′′ v ′′ Figure 7: The 5-chromatic Eulerian triangulation G of the projective plane with c , ( G ) =0 On the other hand, the proof of Theorem 1.4 does not strongly depend on the 3-colorability so much, since we can obtain a similar statement as Theorem 1.4 by showingthat for a fixed surface F , there exists a non-3-colorable Eulerian triangulation G of F with c , ( G ) = 0. Thus we guess that Theorem 1.4 also holds for non-3-colorable Euleriantriangulations.As in the octahedral path, determining the winner of the game on a 3-chromaticEulerian triangulation G with c , ( G ) > Problem 4.1.
Completely determine the winner of the feedback game on 3-chromatic6-regular triangulations of the torus or Klein bottle.
Acknowledgement
The first author is supported by JSPS Grant-in-Aid for Young Scientists (B) 17K14177.The third author is supported by JSPS Grant-in-Aid for Early-Career Scientists 19K14583.
References [1] A. Altshuler, Construction and enumeration of regular maps on the torus,
DiscreteMath. , (1973), 201–217. 102] V. Batagelj, Inductive definition of two restricted classes of triangulations, DiscreteMath. , (1984), 113–121.[3] R. Diestel, “Graph Theory” (fifth edition), Graduate Texts in Mathematics ,Springer, 2016.[4] A.S. Fraenkel, Even kernels, Electron. J. Combin. , (1994), Theo-ret. Comput. Sci. , (1993), 371–381.[6] M. Kasai, N. Matsumoto, A. Nakamoto, T. Nozawa, H. Seno and Y. Takiguchi, Noteon 6-regular graphs on the Klein bottle, Theory and Applications of Graphs , (2017),Article 5.[7] D. Lichtenstein and M. Sipser, GO is Polynomial-Space Hard, J. ACM , (1980),393–401.[8] N. Matsumoto and A. Nagao, Feedback game on Eulerian graphs, submitted.arXiv:2002.09570[9] S. Negami, Classification of 6-regular Klein-bottlal graphs, Res. Rep. Inf. Sci. T.I.T. , A-96 (1984).[10] T.J. Schaefer, On the complexity of some two-person perfect-information games,
J.Comput. System Sci. ,16