On the complexity and topology of scoring games: of Pirates and Treasure

Transcription

1 On the complexity and topology of scoring games: of Pirates and Treasure P. Árendás, Z. L. Blázsik, B. Bodor, Cs. Szabó 24 May 2015 Abstract The combinatorial game Pirates and Treasure is played between two players, Left and Right, on a finite, simple, undirected weighted graph. The vertices of the graph correspond to islands, and a weight function on the vertices indicate the amount of treasure the island has. Left player has n ships, Right player has m ships, in pre-defined vertices. Each turn, the current player moves one of his ships into an adjacent, non-visited vertex, and the weight of the vertex is added to the current points of the player. If a player in his turn cannot move, the game ends. The player who collects more treasures (points) than his opponent, wins the game. It is shown that it is PSPACE complete to decide whether or not Left has a winning startegy in the Pirates and Tresure game. For a fixed graph G topological properties of weightings are analyzed. Among others it is shown that the winning space of Left is connected, but not convex, and that it is NP-hard to decide whether or not there is a weighting such that Right has a winning strategy. Corresponding author: Csaba Szabó Address: Mathematical Institute, Eötvös Loránd University, Budapest, Pázmány Péter Sétány 1/C Hungary of correspondong author: csaba at cs.elte.hu 1

2 1 Introduction In the past few decades the directions of the examination of combinatorial games has slightly changed. As a great mathematical invention of the 20th century the theory of misère and partizan type games were developed. In these games the last move decides the outcome of the game. The books and monographs (eg. [1], [2]) on this subject became a part of the classical mathematical knowledge. However, there are some other type of games that were not examined so extensively. These are the so-called scoring games in which the players aim to collect the most points. Although the first few papers about scoring games were published in the middle of the last century ([9], [5]), no deep theory was developed for them. After several attempts to classify the scoring games, the two classes of well tempered and dicot games were introduced [4]. A theory of welltempered scoring games was developed in [7]. There, algebraic and topological (ordering) aspects were investigated and applied to several games, like knot games [6]. In [3] a real-valued metric was defined for positional games, and it was proved that a particular class of games is a topological semigroup. Then a separation property was defined that connnects these games to closely related Conway games. Another attempt to generalize the previous works of Berlekamp, Conway and Guy ([1], [2]) was made to theory for scoring play games [11]. Their type of games show nice behavior for the usual game operations, but the theory is fairly restrictive. It was showed that scoring play games do not form a group, there is no non-trivial identity among them, and that the comparison of two games in the usual sense is impossible. However, the paper also states that scoring-play games are ordered under the disjunctive sum, and we can form equivalence classes among them using a canonical form. Unfortunately, this theory does not involve several types of games, as in the case of the Pirates and Treasure. The combinatorial game Pirates and Treasure was introduced by F. Stewart in [10]. The game is played between two players, Left and Right, on a finite, simple, undirected graph, where each vertex has a weight. The vertices of the graph correspond to islands, and a weight function on the vertices indicate the amount of treasure the island has. Left player has n ships, Right player has m ships, in pre-defined vertices. Each turn, the current player moves one of his ships into an adjacent, non-visited vertex, and the weight of the vertex is added to the current points of the player. If a player in his turn 2

3 cannot move any of his ships to an adjacent, non-visited vertex, the game ends. The player who collects more treasures (points) than his opponent, wins the game. In this paper we will always assume that m = n = 1. All our results easily apply for the general case, as well. In [10] it is shown that it is NP-hard to determine which player wins the game. However, it is asked for certain graph classes such as 3-regular graphs and 3-connected graphs whether the same conclusion is true. It is also conjectured that in the general case it is PSPACE-complete to decide which player has a winning strategy. In this paper we prove that determinig whether or not Left has a wining strategy is PSPACE-complete. We reduce the quantified boolean formula (QBF) problem to the Pirates and Treasures game. With a small modification we present a construction where the two players, Left and Right move in different components of the graph, and it is still PSPACE-complete to decide, who wins. Additionally, answering questions from [10] we give the constructions and ideas of the proofs for the NP-hardness of the winning stretegy for k-regular and k-connected graphs. Furthermore, in the fourth section we investigate topological properties, such as openness, connectivity and convexity, of the winning (non-losing) strategies of Left and Right according to the weight function on a fixed graph, with fixed initial positions with a single ship for both players. We show, for example, that for a graph G it is NP-hard to decide whether or not there is a weight function such that Right has a winning startegy. 2 PSPACE-completeness In [10] it was proved that it is NP-hard to decide which player has a winning strategy in the Pirates and Treasure game. The problem was stated as follows. Problem 2.1 (F. Stewart, [10]). Input: A graph G = (V, E), the initial positions of the players Left and Right (denoted by L and R, respectively) and a W : V \ {L, R} Z + weight function. Question: Does Left have a winning strategy in the Pirates and Treasure game corresponding to the graph G and the weight function W? It is also conjectured in [10] that Problem 2.1 is in fact PSPACEcomplete. In this section we prove this conjecture. For convenience through- 3

4 out this paper we will assume that the weights are in R + 0 instead of Z +. It is easy to see that this assumption does not make much difference. Theorem 2.2. The following problem is PSPACE-complete. Problem 2.3 (F. Stewart, [10]). Input: A graph G = (V, E), the initial positions of the players Left and Right (denoted by L and R, respectively) and a W : V \ {L, R} R + 0 weight function. Question: Does Left have a winning strategy in the Pirates and Treasure game corresponding to the graph G and the weight function W? Proof. It is obvious that Problem 2.3 is in PSPACE because if we are given a graph G with n vertices, then any game will end in at most n steps, hence we can check every outcome of the corresponding Pirates and Treasure game in polynomial space, from which we can determine who has a winning or a non-losing strategy. In order to prove the PSPACE-hardness of Problem 2.3 we will do a polynomial time reduction from the QBF problem. The QBF problem is the following. Problem 2.4. Input: A formula of the form Ψ(x 1, y 1,..., x n, y n ) = x 1 y 1 x 2 y 2... x n y n Φ(x 1, y 1,..., x n, y n ), where Φ is a conjunctive normal form, i.e it is of the form Φ = k i=1 C i, where for all i = 1, 2,... k the formula C i is of the form C i = t i j=1 l j, where each l j denotes a variable x i or y i or their negations. Question: Is Ψ true? It is well-known that Problem 2.4 is PSPACE-complete, so from the following reduction it follows that Problem 2.3 is also PSPACE-hard, and therefore PSPACE-complete as well. Suppose that we are given a formula Ψ(x 1, y 1,..., x n, y n ) of the above form. Now we will construct from the formula Ψ a graph G and a weight function W on the vertices of G in the following way. Each l {x i, y i } literal will correspond to two vertices of G: one vertex for l, the other one for l. These vertices will be denoted by the corresponing variables or their negations, x i or x i, or y i or ȳ i. Left and Right alternately choose values for variables x 1, y 1,..., x n, y n of Ψ by moving to the corresponding vertices. In 4

5 the first step Left picks a value for x 1 by moving to the vertex x 1 or x 1. Then, Right chooses a value for y 1 by moving to the vertex y 1 or ȳ 1. Then Left proceeds for x 2, and so on. After walking through the literals, Left choses a clause by moving to the corresponding vertex and Right is allowed to choose an unvisited literal from that clause. If such a literal exists in the clause picked by Left, then Right wins. If Left can pick a clause with all literals visited before, then Left wins. Hence if Right has no chance to visit the literals wisely, than Left has a winning strategy. As we play on undirected graphs, both Left and Right might choose to "turn back" or alter their regular directions, hence we have to carefully choose the weight function, and at a few places we introduce gadgets connected to critical vertices. These extra gadgets are constructed on such a way that if any of the two players "deroutes", the other one would have a shortcut, a short path of vertices with large weights, and wins the game after stopping at the end of the path. In general, Left will have shorter paths with larger weights and Right will have longer paths with smaller weights. Hence, if Right makes an illegal move, Left will have a chance to step on a vertex of high value and then stop the game. For the construction let N := n 2 + k 2. First of all let us consider the vertices L 1, L 1, L 2, L 2,... L n, L n, L n+1, x 1, x 1,... x n, x n, and connect the following pairs of vertices: {L i, x i }, {L i, x i }, {L i, x i }, {L i, x i }, {L i, L i+1 }. Let G 1 denote the graph obtained this way. Let us define the graph G 2 similarly with vertices R 0, R 1, R 1, R 2, R 2,... R n, R n, y 1, ȳ 1,... y n, ȳ n. The initial positions of Left and Right will be L := L 0 and R := R 0, respectively. Let the weight of all vertices defined to be N 2 (except for the vertices L and R). These two graphs can be seen on the top of Figure 1. Note that there is one extra vertex for Right at the beginning, and one extra vertex for Left at the end. This will make sure that Right will have to follow the implicit instructions of Left. Now, let us define the graph H 1 as follows. Let us consider the disjoint paths of length 2N 5

6 P 1, P 2, P 3,..., P k. Note that k was the number of clauses in Φ. The path P i will correspond to the clause C i. Let V i denote one of the endpoints of P i for i = 1, 2,..., k. Let us connect the vertices V 1, V 2,..., V k with a new vertex V 0, and add new paths Q i of length i starting from the vertices V i. The paths Q i will play the role of gadgets. The graph obtained this way will be H 1, and can be seen on the bottom left part of Figure 1. When Left passes through the "literal" part, he will choose a clause C i by moving from V 0 to V i. We proceed with a similar construction for Right as above, except that now the length of the paths will be k instead of 2N. Let us call this graph H 2. For convenience denote the corresponding paths by P 1, P 2,..., P k and Q 1, Q 2,..., Q k. Let the length of P i be k, let the length of Q i be i, and for i > 0 let the vertex V i be an endpoint of P i. Then connect each V i, i > 0 to V 0. This graph can be seen on the bottom right part of Figure 1. Let us connect V 0 to L n+1 and V 0 to R n by an edge. Let C i denote the last (kth) vertex of the path P i (in the graph H 2 ) for all i = 1, 2,... k. We will identify these vertices to the corresponding clauses in Ψ. Connect each literal l x i, x i, y i, ȳ i and clause C i by a path of length N if and only if the clause C i contains the negation of the literal l. Later we will refer to this path as S Ci,l. The graph we obtained this way will be denoted by G. The final construction can be seen in Figure 1. Clearly the graph G can be computed from the formula Ψ in polynomial time. We define the weight function W on H 1 H 2 as follows. Let the weight of the vertices V 0 and V 0 be N 3 ; let the weight of the vertices on the paths P 1, P 2,... P k, P 1, P 2,... P k be 2k 2 ; for all i = 1, 2,..., k; and let the weight of the remaining vertices on the paths Q i and Q i be 2k 2 + k + 1 i except for the endpoint of the path Q i, where we set the weight to be 2k 2 + k + 2 i. Let the weight function be identically 0 on the paths S Ci,l. We claim that Left has a winning strategy in the Pirates and Treasure game corresponding to the the graph G and the weight function W if and only if the formula Ψ is false. If we prove this, then this will provide us a polynomial time reduction from the QBF problem to Problem 2.3, which will imply the PSPACE-hardness of Problem 2.3. We show this in the following several steps. First of all, let us call a play of the game on G regular if the players only 6

7 L R x 1 x 1 R 1 L 1 y 1 ȳ 1 L 2 R 1 L n L n+1 R n V 0 V 0 V 1 V 2... V k V 1 V 2... V k Q 1 Q 2 Q k Q 1 Q 2 Q k P 1 P 2 P k P 1 P 2 P k C 1 C 2... C k S C1,l (connected back to vertex l) Figure 1: Construction for the proof of Theorem 2.2. make the following types of moves. 1. Left moves to the vertices l 1, L 1, L 2, l 2, L 2,... l n, L n, L n+1 in his first 3n steps, respectively, where l i = x i or l i = x i for all i. 2. Right moves to vertices R 1, l 1, R 1, R 2, l 2, R 2,... R n, l n, R n in his first 3n steps, respectively, where l i = y i or l i = ȳ i for all i. 3. Left moves to V 0 is his 3nth step, and then starts moving on the path P i for some i (we require that he takes at least two steps on it) 7

8 4. Right moves to V 0, and then if Left s (3n + 2)nd move was to V i for some i, then he starts moving on the path P i (we require that he takes at least two steps on it) Sometimes we will refer to these kind of moves as regular moves. A move is called irregular if it is not regular. In the following lemmas we prove that both players have to move regularly (in the above sense) during the game, because if someone makes an irregular move, then the other player would have a winning strategy. Lemma 2.5. Suppose that one of the players makes an irregular move in the first 6n steps of the game. Then the player making the first irregular move loses. Proof. Let A be the player who makes the first irregular move. This means that after this move he either starts moving on a path S Ci,l or moves back to some vertex corresponding a literal l, and starts moving on a path S Ci,l in his next step. In this case the other player, B, can win as follows: after A s first irregular step he moves regularly until his (3n + 1)st step, then he gets himself stuck in the path Q 1 or Q 1. This is possible because the paths S Ci,l are long enough. By doing this B will collect at least N 3 points since he will collect the treasure on the vertex V 0 or V 0, while A can collect at most 3(n + 1)N 2 altogether. Therefore B wins if n is large enough. Lemma 2.6. Suppose that both players play regularly in the first 6n steps of the game, but later one of them makes an irregular move. Then the player making the first irregular move loses. Proof. Suppose that the first 6n steps were regular. Then Left is forced to move to V 0 and then Right is forced to move to V 0. Up to this point both players collected exactly 3nN 2 + N 3 points. For this reason we will ignore these points in the remaining part of the proof and we will only count with the points collected after this step. In the next ((6n + 3)rd) step Left will move to the vertex V i for some i. Now there are 3 possible cases. Case 1: Right makes an irregular move at his next ((6n + 4)th) step by moving to the vertex V j for some j i. We claim that in this case Left wins by taking the path Q i. 8

9 Subcase 1/1: Right starts moving on the path P j. In this case, both players will make i + 1 moves in which Left will collect 2k 2 +(2k 2 +k+1 i)i points and Right will collect 2k 2 (i+1) points. Therefore Left wins. Subcase 1/2: j > i and Left starts moving on the path Q j. In this case, both players will make i + 1 moves again in which Left will collect 2k 2 +(2k 2 +k+1 i)i points and Right will collect 2k 2 +(2k 2 +k+1 j)i points. Then Left wins again since j > i. Subcase 1/3: j < i and Left starts moving on the path Q j. In this case, Left will make j + 2 and Right will make j + 1 moves. In these steps Left will collect 2k 2 +(2k 2 +k +1 i)(j +1) and Right will collect 2k 2 + (2k 2 + k + 1 j)j + 1 points. Then 2k 2 +(2k 2 +k+1 i)(j+1) > 2k 2 (j+2) = 2k 2 j+2k 2 > 2k 2 +(2k 2 +(k+1 j)j+1, since j, k + 1 j k. Therefore Left wins again. Case 2: Right moves to V i in the (6n + 4)th step, but in the next step Left makes an irregular move, i.e. starts moving on the path Q i. We claim that in this case Right wins if he starts moving on the corresponding path Q i. Indeed, in this case both players will make i + 1 moves, in which Left will collect 2k 2 + (2k 2 + k i)i and Right will collect 2k 2 + (2k 2 + k i)i + 1 points. Hence Right wins. Case 3: Right moves to V i in the (6n + 4)th step, and in the next step Left starts moving on the path P i, but after that in the (6n + 6)th step Right makes an irregular move, i.e. starts moving on the path Q i. In this case Left will make i + 2 and Right will make i + 1 moves. By taking these steps, Left will collect 2k 2 (i + 2) points, while Right will collect 2k 2 + 2(k 2 + k i)i + 1 points altogether. We have already seen (in Subcase 1/3) that the inequality 2k 2 (i + 2) > 2k 2 + 2(k 2 + k i)i + 1 holds. This implies that Left wins in this case as well. By definition, if the first 6n + 6 steps of the game are regular, then it must be a regular play of the game. 9

10 From now on we will always assume that the players play regularly. By Lemmas 2.5. and 2.6. we can assume this. In this case the first 6n + 6 steps of the game can be considered as if they are setting values of the Boolean variables x 1, y 1,..., x n, y n alternately (by assigning the true literal with their moves), and then start moving on the paths P i and P i (of the same index). We claim that after setting the values of the variables x 1, y 1,..., x n, y n Left can win if and only if the formula Φ(x 1, y 1,..., x n, y n ) is false. If the players play regularly, then they will collect the same amount of treasure in the first 6n + 6 steps, hence we can ignore these points as in the proof of Lemma 2.6. and only count with the points collected afterwards. At first, assume that Φ is false. This implies that C i must be false for some 1 i k. Then Left must take the path P i. In this case Left and Right will move alternately until Right will arrive to C i. Now in his next step Right has to take a path S Ci,l for some l. Since C i is false, the value of this l must be true which means that somebody has already moved to l earlier. In this case Left will make k +N steps and Right will make k +N 1 steps, and they will collect 2(k + N)k 2 and 2k 3 points, respectively. Hence left wins. Now assume that Φ is true. Then it does not matter which path Left takes, Right can win with the following strategy. Suppose that Right has already arrived to C i for some 1 i k. This must be true since Φ is true. This implies that C i contains a literal l which is true. We claim that Right wins, if he takes the path S Ci, l. Indeed, in this case Right will eventually arrive to the vertex l and collect at least N 2 points, while Left can collect at most (k + N + 1)k 2 points, which is less than N 2 if n is large enough. Finally, the reduction works as follows. It can be seen from our previous arguments that Left has winning strategy if and only if he can choose the value of x 1 such that for any choice of y 1 he can chose the value of x 2 such that etc., and finally he can chose the value of x n such that for any choice of y n the formula Φ(x 1, y 1,..., x n, y n ) will be false. This holds if and only if the formula Ψ is false. Therefore by this construction we reduced the QBF problem to Problem 2.3 which finishes the proof of the theorem. One can notice that in our proof of Theorem 2.2 the constructed graph G is almost disconnected, therefore one can ask what happens if consider the game Pirates and Treasure only for those graphs in which the initial positions of the players are in different components. Now by slightly modifying the 10

11 proof of Theorem 2.2, we can show that Problem 2.3 is PSPACE-complete in this case, as well. Theorem 2.7. The following problem is PSPACE-complete. Problem 2.8. Input: A graph G = (V, E), the initial positions of the players Left and Right (denoted by L and R, respectively) which are assumed to be in different components of G and a W : V \ {L, R} R + 0 weight function. Question: Does Left have a winning strategy in the Pirates and Treasure game corresponding to the graph G and the weight function W? Proof. We will modify the construction in the proof of Theorem 2.2 in such a way that the vertices L and R will be in different components of G. Let us consider a formula Φ and the corresponding construction G as in the proof of Theorem 2.2. We will make the following modifications. 1. We delete the edges (L i, x i ), (L i, x i ), (R i 1, R i ) and put the graph in Figure 2 in place of them. We will denote the new neighbours of R i 1 by x i and x i (x i is the left neighbour in the figure). The numbers on the vertices in Figure 2 denote their weight. We chose the value of ε in such a way that the sum of weights of the newly added vertices will be less than The endpoint of the paths S Ci,x i and S Ci, x i will be x i and x i, respectively (instead of x i and x i ). L i R i 1 13ε 12ε 11ε 11ε 12ε x i x i 14ε 12ε 11ε 11ε 13ε 10ε 10ε 10ε 10ε 10ε 10ε 10ε 10ε x i x i R i Figure 2: Construction for the proof of Theorem

12 Let G denote the graph obtained this way. This graph can be computed from the formula Ψ in polynomial time again. We claim that Left has a winning strategy in the Pirates and Treasure game on the graph G if and only if the formula Ψ is false. We define the concept of regular play of a game again: let us call a play of the game on G regular if both players make only the following types of moves. 1. The players make a regular move on the edges of G which existed before the modification according to the definition in the proof of Theorem Left walks from L i to l in his (6(i 1)+1)st, (6(i 1)+2)nd, (6(i 1)+ 3)rd and (6(i 1) + 4)th step, Right moves l in his (6(i 1) + 1)st and walks to R i in his (6(i 1) + 2)nd, (6(i 1) + 3)rd and (6(i 1) + 4)th step, where l = x i or l = x i. (After Left s (6(i 1) + 1)st move it turns out whether he walks towards x i or x i ). Now we prove again that both players have to move regularly during the game. Lemma 2.9. Suppose that one of the players makes an irregular move in the first 12n steps of the game. Then the player making the first irregular move loses. Proof. Let A be the player who makes the first irregular move. Then we distinguish 4 cases according to when the first irregular move happened. Case 1: Player A starts moving on a path S Ci,l (maybe after a few other moves). As in the proof of Theorem 2.2, the other player B wins by moving regularly until his 3n + 1st step, then getting himself stuck in the path Q 1 or Q 1. This is possible because the paths S Ci,l are long enough. Now the same argument proves that B wins. Case 2: Right chooses the wrong path. If the players moved regularly up to their 6(i 1)th step, then they collected the same amount of points, thus we can ignore these points (as we did in earlier proofs). Here we will distinguish two subcases. 12

13 Subcase 2/1: Left moves to the left neighbour of L i, but Right chooses the right one ( x i). In this case Left moves to the 13ε vertex in his next step, then Right has one more move in which he collects at most 11ε points. Then Left collects 25ε and Right collect at most 22ε points, thus Left wins. Subcase 2/2: Left moves to the right neighbour of L i, but Right chooses the left one (x i). In the following two steps Left moves to the vertices that worth 11ε and 12ε. Then he will collect 34ε points. If Right moves to the 14ε worth vertex, then he will collect 26ε points, and if he moves to the 10ε worth vertex, then he will collect 30ε points. Either way, Right loses. Case 3: Left starts moving on the short path in his (6(i 1) + 2)nd step. In this case Right wins by taking the short path as well, because in both cases Right will collect ε points more then Left. Case 4: Right starts moving on the short path in his (6(i 1) + 2)nd step. We will distinguish two subcases again. Subcase 4/1: Both players have chosen the left path in their (6(i 1)+1)st step. Then Left will collect 32ε, and Right will collect 26ε points, hence Left wins. Subcase 4/2: Both players have chosen the right path in their (6(i 1) + 1)st step. Then Left will collect 30ε+N 2 > 30ε+1 and Right will collect 35ε points, hence Left wins again. Lemma Suppose that both players play regularly in the first 12n steps of the game, but later one of them makes an irregular move. Then the player making the first irregular move loses. 13

14 Proof. This proof is the same as that of Lemma 2.6. in the proof of Theorem 2.2. Now we can notice that if the players move regularly in the above sense, then for all 1 i n during the game they will visit the vertex x i if and only if they will visit the vertex x i, and they will visit the vertex x i if and only if they will visit vertex x i. Using this, the same argument as in the proof of Theorem 2.2 shows that the formula Ψ is false if and only if Left has a winning strategy. This gives us a polynomial time reduction from the QBF problem to Problem 2.3, which proves the theorem. 3 NP-hardness in certain graph families First, we recite the proof of NP-hardness of Problem 2.3 from [10]. Theorem 3.1 ([10]). Problem 2.3 is NP-hard. Proof. The theorem is proven by a reduction to the problem of finding a Hamiltonian path in a graph. Determining whether an input graph has a Hamiltonian path is NP-hard [8]. First, pick a vertex in a simple graph G(V, E), label it as L. This vertex will be the starting position for Left. For all remaining vertices v V, let w(v) = 1. In the next step, add a path P of length P = V to the graph G so that the vertex L is one of the endpoints of P. The vertex adjacent of L on the path P is labeled R, this will be the starting position of Right. For the remaining vertices on the path P, we set the weights to 1. The construction obtained is shown in Figure??. In this game, Right will move through the path P, visiting V 2 vertices, then getting stuck. Therefore the only possible winning strategy for Left is to visit all vertices in G, yielding V 1 points. This is viable if and only if G contains a Hamiltonian path starting from L. In [10] it was conjectured that the game is still NP-hard if the graph is 3-connected or 3-regular. Because a single path is neither 3-connected nor 3-regular, the construction from the proof of Theorem 3.1 doe snot work 14

15 G P L R Figure 3: Proving NP-hardness in [10]. here. In this section, we give the basic idea of proving NP-hardness for k-connected and k-regular graphs. The idea for the construction in the general case was to put the two starting nodes L and R into neighbouring vertices. The two players blocked each other, and were forced to take steps in their respective subgraphs. For the k-connected and k-regular cases, we take a reversed approach: let us put the two starting positions sufficiently far from each other in the graph, separate them with a large subgraph of zero-valued vertices, and put the valuable vertices in a certain pattern near the two starting vertices L and R. Let us call the latter the value subgraphs of Left and Right, respectively. Sufficiently far means here that a decision from either player to travel to the rival s value subgraph would always result in a loss. Aside of this change, the concept of deciding the winner will be the same as in the original proof: if and only if there is a Hamiltonian path in a certain subgraph, has Left a winning strategy. The two constructions to achieve these goals are the following. We start with a k-connected or k-regular graph G, label one of its vertices as L, and for the remaining vertices of G, we set the weights to 1. The vertex L will be the starting position of Left. The decision problem of the existence of a Hamiltonian path will correspond to this G graph. If there is a Hamiltonian path in G, then suppose that Left could collect n 1 points in n 1 steps. To place the two starting positions sufficiently far away, we need a long bridge between the starting positions of Left and Right. This can be built as a chain of subgraphs on k vertices each. In the k-connected case, this is only a chain of complete bipartite graphs K k,k. This chain is denoted by B 1,..., B n in Figure 4. In the k-regular case this is a bit more complicated, because we must preserve the regularity in each step. A solution here is that between B i and B i+1, for odd i we connect the vertices pairwise, and for even 15

16 i we take the complement of the pairwise edge selection. We set the weights of the vertices in the bridge to zero. In this paper, we omit the technical steps required to connect the two ends of the bridge to the value subgraphs of Left and Right, while preserving k-connectedness or k-regularity. Now we want to give such a starting position and graph structure for Right, that Left would have a winning strategy if and only if there is a Hamiltonian path in G. If we suppose that Left takes n 1 points while he moves along the Hamiltonian path in n 1 steps, this can be achieved if Right can take n 2 points in n 2 steps, after which Right gets stuck, and loses in the next step. In the k-connected case, we can take a complete bipartite graph K n 1 2, n 1 2, where k n 1 2 (denoted by I 1 and I 2 in Figure 4). We place the starting vertex R such that when Right moves along all vertices of this subgraph, he would get stuck, then we set the weights of the remaining vertices in this subgraph to 1. In the k-regular case, we set a complete graph as the value subgraph of Right, where all weights are 1, and with a deletion of an edge, connect it to a single vertex. The bridge connects to this single vertex. We place the starting vertex R into the bridge such that in order to win, Right must travel to this single vertex, then take all points from the initially complete graph, then trapping himself. G, G = n L B 1, B 1 = k B 2, B 2 = k B n 1, B n 1 = k B n, B n = k I 1, I 1 = n 1 2 R, if n 1 is even I 2, I 2 = n 1 2 R, if n 1 is odd Figure 4: The construction in the k-connected case 16

17 4 About connectivity and convexity at a fixed graph In this section we examine what happens if we fix a graph and also fix the initial positions of the ships, but we do not fix the weight function on the vertices of the graph. The main question is how do the outcome and the strategies of the game depend on our particular choice of the weights. Throughout this section we will always assume that Left moves first. We know that in Pirates and Treasure either Left or Right has a winning strategy or they both have a non-losing strategy. Definition 4.1. Let G = (V, E) a graph, V = {L, R, v 1,..., v n }, where L and R denote the initial position of player Left and Right, respectively. Then let L G := {(x 1, x 2,..., x n ) i(x i 0), and Left has a winning strategy if the weight on vertex v i is x i for all i}; R G := {(x 1, x 2,..., x n ) i(x i 0), and Right has a winning strategy if the weight on vertex v i is x i for all i}; T G := {(x 1, x 2,..., x n ) i(x i 0), i(x i > 0), and both Left and Right have a non-losing strategy if the weight on vertex v i is x i for all i}. Sometimes we omit the subscripts if the graph G is clear from the context. Because of our previous remark this definitions are meaningful and give a partition of R n 0 \ 0. We would like to investigate how certain properties of these sets L G, R G, T G depend on the graph G. It is obvious that these sets are cone-like subsets of R n 0 \ 0, i.e. if x L (similarly for R and T), then λx L for all λ > 0. Proposition 4.2. For all graph G the sets L G, R G are open and T G, L G T G, R G T G are closed subsets of R n 0 \ 0. Proof. Let s define V (G) = {L, R, v 1, v 2,..., v n } as above, and suppose that x = (x 1, x 2,..., x n ) L. Then Left has a winning strategy for the weighting corresponding to the point x. Now let δ > 0 be such that 2nδ n i=1 ε i x i for any choice of ε i {0, 1, 1} if this sum is not 0. One can find such δ because there are only finitely many choice for such a sum. Then we claim that if for all 1 i n, x i x i δ (and x i 0), then x = (x 1, x 2,..., x n) L. In fact we will show that if Left plays as his winning strategy in the game where the weights correspond to the coordinates of x, then we he will also win the game where the weights correspond to the coordinates of x. For 17

18 this let us consider a possible outcome of the game assuming Left played as above. At this outcome let W L (x) and W R (x) be the amount of Treasure Left and Right collected respectively, in the case when the weighting corresponds to the vector x, and let W L (x ) and W R (x ) be defined analogously for the vector x. For x we know that W L (x) > W R (x). Since x i x i δ, this implies W L (x) W L (x ) nδ and W R (x) W R (x ) nδ, and so W L (x ) W R (x ) W L (x) W R (x) 2nδ. By definition both W L (x) and W R (x) are of the form i I x i for some index sets I, thus W L (x) W R (x) are of the form n i=1 ε i x i for some ε {0, 1, 1}. Then W L (x) > W R (x) that is W L (x) W R (x) > 0 and so 2nδ < W L (x) W R (x) hence W L (x ) W R (x ) W L (x) W R (x) 2nδ > 0 by the definition of δ. This means that Left can win the game where the weights correspond to the coordinates of x, hence x L. We have just proven that if x L, then a neighbourhood of x is also contained in L, i.e. L is open. A similar proof shows that R is also open. From this the other statements in the Proposition 4.2 are trivial making use of the fact that the sets L, R, T give a partition of R n 0 \ 0. After discussing these essential topological properties of L, R and T one can be interested in the topological nature for example connectivity properties of L, R and T too. In order to prove connectivity for L T we present a nice observation. Let e i denote the vector the ith coordinate of which is 1, and all other coordinates are 0. Lemma 4.3. For every x L there exists an i such that the vertex v i is a neighbour of L in G and the vectors x and e i are in the same connectivity component of L. Proof. By the definition of L Left has a winning strategy for the weighting corresponding to x. Let v i be a possible first move of Left in his winning strategy. Then by increasing the value of the ith coordinate of x it remains in L. Let λ > 0 such that the ith coordinate of x = x + λe i is larger than the sum of all other coordinates of x. Then by decreasing the value of all but the ith coordinate of x to 0 it still remains in L. We found a continuous curve within L connecting the points x and λ e i for some λ > 0, therefore x and e i (recall that L is cone-like) are in the same connectivity component of L. Proposition 4.4. L T is (path) connected. 18

19 Proof. Using Lemma 4.3 we only have to prove that if v i and v j are neighbours of L in G, then e i and e j are in the same connectivity component of L T. For this it is clearly enough to prove that λe i + µe j L T for all λ, µ 0 (not both are 0). If λ µ, then if Left moves first to v i he does not lose. If λ µ, then if Left moves first to v j he does not lose. Hence Left can not lose at the weighting corresponding to the point λe i + µe j. It is natural to ask whether the set L is always connected as well. The answer in no, in fact it is possible that L has as many connectivity components as many neighbours L has. Clearly (by Lemma 4.3) the latter is always an upper estimate for the number of connectivity components of L. Example 4.1. Let V (G) := {L, R, v 1, v 2,..., v n }, E(G) := {(L, v 1 ), (L, v 2 ),..., (L, v n )} {(R, v 1 ), (R, v 2 ),..., (R, v n )}. Then the vectors e i are in pairwise different connectivity components of L, and hence by Lemma 4.3 the class L has exactly n connectivity components. L v 1 v 2 v n 1 v n R Figure 5: There could be n connectivity components. Proof. Since L is an open subset of R n 0, the connectivity components and the path connectivity components of L are the same. Therefore it is enough to prove that if i j, then e i and e j cannot be connected by a continuous curve within L. For the contradiction let us suppose that γ : [0, 1] L is continuous curve for which γ(0) = e i and γ(1) = e j. Let α(x) be the ith coordinate of γ(x), and let β(x) be the maximum of the other coordinates of γ(x). Then α(0) = β(1) = 1, α(1) = β(0) = 0, and the functions α and β are continuous on [0, 1] as well, thus by Bolzano s theorem there exists an x [0, 1] such that α(x) = β(x). This means that the vector γ(x) has at least two maximal coordinates, and it is easy to see that this implies γ(x) T. This contradiction completes the proof. 19

20 For a graph G let N denote the set of neighbours of L. We will define an equivalence relation on N. Definition 4.5. For u, v V (G) \ {L} let d(u, v) be the distance between the vertices u and v in G \ {L} (it is defined to be if u and v are different components of G \ {L}). Let r N N be the relation as follows: r(u, v) holds if and only if d(u, v) d(r, v). Let be the equivalence relation generated by r, i.e. u v if and only if there exists a sequence of vertices in N, namely u = u 0, u 1,..., u n = v such that for all 0 i n 1 r(u i, u i+1 ) or r(u i+1, u i ) holds. Now we are ready to give a sufficient condition for L to be connected. Theorem 4.6. The set L has at most as many connectivity components as many equivalence classes the relation has. In particular if all vertices in N are -equivalent, then L is connected. Proof. We are going to show that if r(v i, v j ) holds for some v i, v j N, then e i, e j are in the same connectivity component of L. As in the proof of Proposition 4.4 it is enough to prove that λe i + µe j L T for all λ, µ 0 (not both are 0). If λ > µ, then if Left moves first to v i he wins, if λ < µ, then if Left moves first to v j he wins, so we only have to consider the case when λ = µ. We may assume that λ = µ = 1. We have to show that if the weights of the vertices v i and v j are 1, and all other weights are 0, then Left has a winning strategy. The strategy of Left will be the following: Let u 0 = v i, u 1,..., u d = v j a path between v i and v j where their distance is realized (it is less that d(r, v j ) and hence finite). Then in the kth step Left moves to u k 1 if he can, otherwise he can move anywhere. First we claim that if Left follows this strategy, then for all 1 k n + 1 in the kth step either Left can move to u k 1 or the game has already ended at his earlier step. To the contrary assume that for some 0 i n Left stands in u k 1 (u 1 := L), Right has already taken his kth move, but Left is unable to move to u k. We assume that k is the smallest integer for which it is possible. This means that either Left or Right was in u k at some earlier step. The vertices u 0, u 1,..., u d are pairwise distinct (because they form a path), so it must be player Right who was in u k at some earlier step. But this is impossible, since this would imply d(r, u i ) < i implying d(r, v j ) d(r, u k ) + d(u k, v j ) < i + (d i) = d = d(v i, v j ). So if Left follows this strategy then in a game if he moves at most d times he collects Treasure of value 1, and Right collects 20

21 nothing, and if he moves at least d + 1 times he collects Treasure of value 2, and Right collects nothing. In either case Left wins. Now by the definition of it follows easily that v i v j also imply that e i and e j are in the same connectivity component of L, and then the proof is complete by using Lemma 4.3. The following example shows that the condition in Theorem 4.6 is not necessary. Example 4.2. Consider the graph G as in Figure 6. It is clear that v 1 v 2, and it is easy to check that L G is connected for this graph, because every point can be connected with e 3 = (0, 0, 1) by a continuous curve within L. R v 1 v 2 v 3 L Figure 6: L G is connected but not because of Theorem 4.6. Considering the examples we discussed so far, it is easy to see that the sets L T and the components of L were always convex sets themselves, so one could ask whether it is true in general. The answer in negative again as the following example shows. Example 4.3. Consider the graph G as in Figure 7. If x < 1 or 2 < x is the case then Left wins but for the 1 < x < 2 case Right could win. x 2 R L 2 1 Figure 7: Non-convex example. In the construction of Example 4.3 it can be seen that if we fix the weight of each vertex but x, then the sets L, R, T divides the half-line in R n 0 \ 0 (depending on the choice of x) to more than 2 intervals. The following theorem generalizes this fact. 21

22 Theorem 4.7. For every strictly monotone increasing 0 < a k < a k 1... < a 1 sequence of integers (note the unusual way of indexing!) there exists a graph G with a distinguished vertex z V (G) and with fixed weights on the vertices distinct to z such that Left has a winning strategy if and only if w(z) (a i+1, a i ) for some odd i (or w(z) < a k and k is odd), Right has a winning strategy if and only if w(z) (a i+1, a i ) for some even i (or w(z) < a k and k is even) and both have a non-losing strategy if and only if w(z) is equal to some a i. Proof. Consider an arbitrary but fixed strictly monotone increasing 0 < a k < a k 1... < a 1 sequence of integers. We are going to construct a graph with the desired properties. Our graph will consists of two disjoint paths of length 3k, a central vertex c and the vertex z. The ith vertex of the first path will be denoted by f i and the jth vertex of the second path will be denoted by s j. The starting point of Left is f 1 and the starting point of Right is s 1 (L = f 1 and R = s 1 ). The central vertex is connected with every sixth vertex of the first path starting from the fourth one (the vertices f 6i 2 where 1 i k/2) and with every sixth vertex of the second path starting from the first one (the vertices s 6j 5 where 1 j k/2 ). The central vertex is also connected with the vertex z, and there are no more edges in this graph. The weights of the vertices are the following: w(z) will be varied, w(f 3i 1 ) = w(s 3i 1 ) = a i where 1 i k and all the other weights are 0. Figure 8 shows the construction for k = 6. R a a a a a a w(x) 0 a 1 a 2 a 3 L a a a Figure 8: Varying w(z) ruin convexity We will show first that if a player goes to the c central vertex, and his next step is going back into his own path, then he will lose. Because we do not specify which player moves like this, we will denote the vertices of his 22

23 path by v j instead of f j or s j and we will refer to this player as Player A, and the other player as Player B. Suppose that he stepped to c from v 3i+1, and he arrived at v 3i+1+6j when moving away from c. We have two cases: the first case is when he collects the weights on the vertices between v 3i+1 and v 3i+1+6j. In this case he will have 3i+6j +2 steps, so Player B must have at least 3i + 6j + 2 steps too (Player B might have 3i + 6j + 3 if he is the Left). So Player A will collect the first (i + 2j) nonzero weights, but Player B will collect the first (i+2j+1) ones, so Player B will win. The second case is when Player A collects the weights on the vertices from v 3i+1+6j to v 3k+1. In this case Player A will have 3i+2+(3k+1 (3i+1+6j)) = 3k + 2 6j steps, so Player B must have at least 3k + 2 6j steps too. Player A will collect k 2j non-zero vertices while Player B will collect also k 2j = (3k+2 6j) 2 non-zero vertices. But Player B will collect the largest 3 treasures from the vertices, hence he will win. Now, we would like to show that if a player goes to the c central vertex, and in his next step he is going to the other player s path, then he will lose. Again we do not specify which player plays like this, we will denote the vertices of his path by v j instead of f j or s j, the vertices of the other player s path by u j and we will refer to this player as Player A, and the other player as Player B. Suppose that he stepped to c from v 3i+1, and he arrived at u 3i+4+6j when stepping away from c. We have two cases: the first case is when Player A leaves his spot toward Player B. In this case Player B will collect one non-zero vertex more than Player A and Player B will collect the largest treasures whilst Player A not, so in this case Player B will win. The second case is when Player A leaves his spot in the opposite direction, in this case Player B will collect at least as many non-zero vertex as Player A, and Player B will collect the largest treasures whilst Player A not, hence in this case Player B will win. We will examine now that under which circumstances will be feasible for a player to step on the vertex c, then on the vertex z. If a player does that, and he steps to c from the vertex v 3j+1, then he will collect w(z)+ a i and 1 i j the other player will collect a i, so the first player will win if and only 1 i j+1 if w(z) > a j+1 and the game will end with a draw if and only if w(z) = a j+1. So that player will win who has an opportunity to collect the vertex z such that w(z) is greater than the next a j he can collect, and he has this opportunity earlier than the opposing player. If neither player has such opportunity, then both have non-losing strategy, and this happens exactly 23

24 when w(z) is equal to some a j. It is natural to ask similar connectivity questions for R and R T too. In order to examine these sets we define a graph G L in the following way. For a graph G = (V, E), where V = {L, R, v 2, v 3,..., v n } let V (G L ) := {L, R, v 1, v 2,..., v n }, where R = L, v 1 = R and E(G L ) := E(G) {(L, v 1 )}. When we talk about games on the graphs G and G L, we will assume that Left s initial position is L (and L respectively), and Right s initial position is R (and R respectively). Proposition If Left has a winning (non-losing) strategy for given weighting in the graph G, then Right has a winning (non-losing) strategy in the graph G L for the same weighting extended in v 1 as If Right has a winning (non-losing) strategy for a given weighting in the graph G L, then Left has a winning (non-losing) strategy in the graph G for the same weighting resticted to V \ {L, R }. Proof (sketch). 1. If after the first step of Left Right plays according to a winning (nonlosing) strategy of Left in G, he wins (does not lose). 2. If Left plays just like as a winning (non-losing) strategy of Right after the first step of Left in G L, he wins (does not lose). Corollary 4.9. x L G (0, x) R G L, x L G T G (0, x) R G L T G L. For any c R 0 (c, x) R G L x L G, (c, x) R G L T G L x L G T G. Proof. This is just a reformulation of Proposition

25 Using this corollary it is easy to see that the number of connectivity components of R G L is at least the number of connectivity components of L G (in fact they are equal). Then if G is as in Example 4.1 we get that R G L can have any number of connectivity components. Example 4.3 shows that L G need not to be convex, hence (by Corollary 4.9) the same is true for R G L. We saw that for all graphs G the degree of L (the initial position of player Left) is an upper estimate for the number of connectivity components of L G. Although L G T G is always connected, one could ask if the same applies for the sets R G and R G T G in general. Question 4.1. Is it true in general that the number of connectivity components of R G (if there is any) is at most the degree of R (the initial position of player Right)? Question 4.2. Is that true that R G T G nonempty)? is always connected (if it is It is obvious that L G is nonempty if and only if L has some neighbour in G, however it is a meaningful question that how can we describe the graphs for which the sets R G and R G T G are nonempty. For the set R G T G we will show that this is an NP-hard problem for a graph G. First of all it is easy to see that R G T G is nonempty if and only if T G is nonempty. If T G is nonempty then trivially R G T G is also nonempty. For the other direction suppose that T G =. Then the sets L G and R G gives an open composition of R n 0 \ 0 (by Proposition 4.2). But R n 0 \ 0 is connected, so either R G or L G is empty. In the former case we are ready, and L G = (L has no neighbour ) T G = R n 0 \ 0, which is impossible. Theorem The following problem is NP-hard. Input: a graph G, the initial position of Left and Right on G Question: Is R G T G (or T G ) nonempty, or equivalently is there a nontrivial weighting on G, at which Left has no winning strategy for the Pirates and Treasure game? Proof. It is well-known that the following problem (Hamiltonian Path problem) is NP-complete. Input: a graph H. 25

26 Question: Does H contain a Hamiltonian path? We will prove Theorem 4.10 by doing a reduction from Hamiltonian Path problem to our problem. For a graph H let us consider the following graph G (see it in Figure 9). V (G) := {L, R, u 1, u 2, u 3, w 1, w 2, w 3,..., w n+4, y 1, y 2,..., y n+6 } V (H) E(G) := {(R, u 1 ), (u 1, u 2 ), (u 2, u 3 ), (L, u 1 ), (L, u 2 ), (L, u 3 ), (L, y 1 )} {(y i, y i+1 ) : 1 i n+5} {(u 3, v) : v V (H)} {(u 3, w i ) : 1 i n+4} {(w i, v) : 1 i n + 4, v V (H)}, where n is the number of vertices in H. Clearly this graph can be computed from H in polynomial time. Now, we claim that H contains a Hamiltonian path if and only if R T is nonempty for this graph G (as usual L and R denotes the initial positions of Left and Right, respectively). If we prove this, then we are done. H R u 1 u 2 u 3 w 1 w 2 w 3 w n+4 L x 1 x 2 x 3 x n+6 Figure 9: For proving NP-completeness in Theorem Lemma If there is a positive amount of treasure in u 1, u 2, u 3, w 1, w 2,..., w n+4 or in any vertex of H, then Left has a winning strategy. Proof. If the weight on u 1 is positive, then Left moves there and wins. If the weight on u 1 is 0 but the weight on u 2 is positive, then Left moves to u 2 and wins. If the weight on u 1 and u 2 are 0, but the weight on u 3 is positive, then Left moves to u 3 and wins. So we may assume that the weights on u 1, u 2 and u 3 are all 0. Then let v V (H) {w 1,..., w n+4 } a vertex, where there is a positive amount of treasure. It is easy to check that in this case v and u 3 have a common neighbour v. (If v V (H), then any v = w i suffices, and if v = w i for some i, then any v V (H) suffices.) In this case Left has the following winning strategy: First he moves to u 3, then Right is 26