On Bichromatic Triangle Game

Transcription

1 On Bichromatic Triangle Game Gordana Manić Daniel M. Martin Miloš Stojakoić Agst 16, 2012 Abstract We stdy a combinatorial game called Bichromatic Triangle Game, defined as follows. Two players R and B constrct a trianglation on a gien planar point set V. Starting from no edges, they take trns drawing one straight edge that connects two points in V and does not cross any of the preiosly drawn edges. Player R ses color red and player B ses color ble. The first player who completes one empty monochromatic triangle is the winner. We show that each of the players can force a tie in the Bichromatic Triangle Game when the points of V are in cone position, and also in the case when there is eactly one inner point in the set V. As a conseqence of those reslts, we obtain that the otcome of the Bichromatic Complete Trianglation Game (a modification of the Bichromatic Triangle Game) is also a tie for the same two cases regarding the set V. Keywords. Combinatorial game, geometric game, trianglation. 1 Introdction Games on trianglations belong to the more general area of combinatorial games, which sally inole two players. We consider games with perfect information (i.e., there is no hidden information, in contrast to card games, like poker) and each of the players plays optimally (i.e., both players do their best to win). Any sch game that cannot end in a draw has two possible otcomes: first player s win, where the first player has a strategy to win no matter how Centro de Matemática, Comptação e Cognição, Uniersidade Federal do ABC, São Palo, Brail. Grant 2008/ from Fndação de Amparo à Pesqisa do Estado de São Palo. Partially spported by Conselho Nacional de Desenolimento Científico e Tecnológico Project / manic.gordana@fabc.ed.br Centro de Matemática, Comptação e Cognição, Uniersidade Federal do ABC, São Palo, Brail. Partially spported by Conselho Nacional de Desenolimento Científico e Tecnológico Project / daniel.martin@fabc.ed.br Department of Mathematics and Informatics, Uniersity of Noi Sad, Serbia. Partly spported by Ministry of Edcation and Science, Repblic of Serbia, and Proincial Secretariat for Science, Proince of Vojodina. milos.stojakoic@dmi.ns.ac.rs 1

2 the other player moes throghot the game, and second player s win, where we hae the same sitation with the roles swaped between the players. On top of those two otcomes, combinatorial games with two players may hae a third otcome, a tie, where both of the players hae a strategy to preent the opponent from winning. For more information on combinatorial game theory we refer the reader to [2], [3] and [4]. Let V R 2 be a set of points in the plane with no three collinear points. A trianglation of V is a simplicial decomposition of its cone hll whose ertices are precisely the points in V. Aichholer et al. [1] consider seeral combinatorial games inoling the ertices, edges (straight line segments), and faces (triangles) of some trianglation. Their ltimate goal for each game is to characterie who wins the game and design efficient algorithms to compte a winning strategy, or alternatiely, show that both players can force a tie and again determine and efficiently compte their defense strategies. We first gie the definition of the Bichromatic Complete Trianglation Game, as introdced in [1]. Two players R and B constrct a trianglation on a gien point set V. Starting from no edges, players R and B play in trns by drawing one straight edge in each moe, with R making the first moe. In each moe, the chosen edge is not allowed to cross any of the preiosly drawn edges. Player R ses color red and player B ses color ble. A triangle formed by the already drawn edges is said to be empty if it contains no points from V in its interior. Each time a player completes one or more empty monochromatic triangles, the player wins the corresponding nmber of points and it is again his trn (he has an etra moe). Once the trianglation is complete, the game stops and the player who owns more points is the winner. Bichromatic Triangle Game, also introdced in [1], starts as Bichromatic Complete Trianglation Game, bt has a different winning condition. Namely, the player who completes an empty monochromatic triangle first is the winner. If the trianglation is complete and no player has won to that point, the game is a draw. Aichholer et al. [1] posed an open problem, to determine the otcomes of the Bichromatic Triangle Game and the Bichromatic Complete Trianglation Game, possibly depending on the configration of points in V. Besides those two games, they analye a whole family of games on trianglations and manage to obtain game otcomes only for some special configrations of points, sggesting that the whole family of problems in fll generality (in terms of point configrations) is qite hard, and for now ot of reach. On top of that, they conjectre that for many of the games een determining the otcome may be NP-hard for general trianglations, that is, when there is no predetermined condition for the points of V apart from not haing any three collinear points. Therefore, they focs their attention to special classes of trianglations, e.g., when points from V are in cone position, and obtain positie reslts. In the present paper we make a step forward in resoling Bichromatic Triangle Game and Bichromatic Complete Trianglation Game, giing the otcomes 2

3 of the two games for a few special configrations of point configrations. In particlar, for the Bichromatic Triangle Game, we show that B can force a tie when the points in V are in cone position, and also when there is eactly one inner point in V (we say that V is an inner point of V if belongs to the interior of the cone hll of V ). We also show that B cannot win the game, so the otcome has to be a draw. As a conseqence of the reslts we obtained for the Bichromatic Triangle Game, we immediately get that the otcome of the Bichromatic Complete Trianglation Game is also a draw if the points of V are in cone position, and in the case when there is eactly one inner point in the set V. 1.1 Notation Let V R 2 be a set of points in the plane. We say that the points of V are in cone position if they are the ertices of some cone polygon. A point V is referred to as inner, if it is strictly inside the cone hll of V. For, V, denote by the line segment with endpoints and, which we will sometimes call an edge. We denote the set of all sch line segments with ( V 2). A configration in a Bichromatic Triangle Game is the triple (V, E R, E B ), where V R 2, and E R, E B ( V 2) are two disjoint sets of edges, drawn by R and B, respectiely, dring the corse of a (possibly nfinished) game. So, no two edges in E R E B are allowed to cross. A free edge with respect to a configration (V, E R, E B ) is an edge in ( V 2) \ (ER E B ) that does not cross any of the edges in E R E B. Gien a configration C = (V, E R, E B ) and a set of points W V, we define the indced configration C [W ] = (W, E R, E B ), ) ER and E B = ( W 2 ) EB. Two indced configrations are said where E R = ( W 2 to be independent if they share precisely two points 1 and 2, and they both contain the edge 1 2 which is taken by the same player in both configrations. Gien non-collinear points, y, R 2 so that the seqence (, y, ) is in clockwise order, we denote by y the open set of points w R 2 sch that the seqences (, y, w) and (w, y, ) are both in clockwise order and both noncollinear. For the seqence (, y, ) in conter-clockwise order, we say y := ẑy. Let W be a finite planar set of points in cone position, denoted 0,..., n 1 in clockwise order. For = i, y = j, by y we denote the collection of points { i, i+1,..., j }, where inde addition is taken modlo n. Moreoer, for, y W, if = i and y = i+1, we say that and y are consectie in W. 2 Or Reslts Lemma 1 Sppose C = (V, E R, E B ) is a configration of the Bichromatic Triangle Game where V is a set of points in cone position and each of the drawn edges, i.e., edges in E R E B, are between consectie points in V. If E R < 2, then B can force a tie in C regardless of which player is net to make a moe. 3

4 Proof The proof is by indction on V. If V 3, the statement triially holds. Hence, sppose V = n > 3 and sppose the statement is tre for V < n. We denote by r the cardinality of E R, so r is either 0 or 1. Case 1. If B is net to play he can choose to draw any free edge that does not connect two consectie points from V, diiding C into two independent configrations C 1 and C 2. By the indction hypothesis B can force a tie both in C 1 and in C 2. Hence, B can force a tie in C. Case 2. If R is net to play, let be the edge that he draws, and let C be the reslting configration, i.e., C = (V, E R {}, E B ). This moe diided C into two independent configrations C 1 = C [ ] and C 2 = C [ ], with belonging to both configrations. Clearly, the remainder of the game goes on independently in these two configrations. In order for B to force a tie in C, he mst force a tie as second player in both C 1 and C 2. (i) If r = 1 and and are any two (different) points, we may assme, withot loss of generality, that C 1 has two red edges, and that C 2 has one red edge (inclding the edge in both cases). Note that a tie in C 1 can be easily forced when the configration C 1 has at most three points (by playing arbitrarily), so from now on we assme that it has at least for points. Now B picks a free edge y in C 1 with C 1 [ y] and C 1 [ y] each containing a single red edge. The eistence of sch an edge is garanteed by the properties of C 1, i.e., it has at least for points, its ertices are in cone position, and its drawn edges are between consectie points (note that the two red edges may hae a common erte). After B draws y we obtain a configration C 1 which is diided into two independent configrations C 1 [ y] and C 1 [ y]. By the indction hypothesis, B can force a tie in both C 1 [ y] and C 1 [ y] and hence B can force a tie in C 1. As for C 2, it is either triial, i.e., a single edge, or B can force a tie in it by the indction hypothesis. (ii) If, howeer, r = 0 and and are not consectie in V, then both C 1 and C 2 hae one red edge, and by the indction hypothesis B can force a tie in C 1, and in C 2. (iii) Finally, if r = 0 and and are consectie in V, then we set C = C and apply the argment from Case 1. The preios lemma helps s to determine the otcome of the game for points in cone position. Theorem 1 Player B can force a tie in the Bichromatic Triangle Game when the points in V are in cone position. Proof The first edge that R draws diides the initial configration C into two 4

5 independent configrations (one possibly triial). By Lemma 1, B can force a tie in both of these configrations, and hence, B can force a tie in C. Theorem 2 Player B can force a tie in the Bichromatic Triangle Game when the set of points V has eactly one inner point. Proof First we gie an otline of the proof. Roghly speaking, we analye eery possible moe of R and determine what the response of player B shold be. Eery time player B draws an edge in response to a moe of R, we diide the problem of forcing a tie into independent sbgames. This process may be repeated more than once ntil each sbgame falls into one of the following three types, in which player B can force a tie. The first type of sbgame has a configration with the set of points in cone position, with all drawn edges being between consectie points and with at most one red edge. To show that B can force a tie in sch a sbgame, we apply Lemma 1. The second type of sbgame has a configration on for points, possibly with some edges already drawn and at most one of them red. As sch sbgames are of finite sie, they are easy to analye checking throgh all possible configrations and possibilities we get that player B can force a tie in each of them. Finally, the third configration type is more comple, and we are going to analye it separately. Throghot the game, it is part of B s defense strategy to make his moe in the sbgame in which player R played his last moe. Eceptionally, if player R draws the last free edge in a sbgame, B will respond in another sbgame which still has free edges (if there is no more free edges, the game is finished). Clearly, this etra moe does not harm player B, so ignoring those moes does not affect or analysis. Let be the niqe inner point of V. Initially, there are no edges drawn and it is R s trn, so we hae the configration (V,, ). Regarding the first moe of R, we distingish two cases. Case (i). R draws an edge, for some. Let r be the line passing throgh that is perpendiclar to the segment, and let s be the line containing the segment, see Figre 1. Note that r diides the plane into two half-planes. There mst eist at least one point in the halfplane not containing, otherwise wold not be an inner point of V. Among all possible choices, we pick as close as possible to the line s. Now B draws and we are left with sbgames ( {}, {}, {}) and ( {}, {}, {}). One of these sbgames is cone, so B can force a tie by Lemma 1. The other one will be handled in the following claim. Note that is a free edge that belongs to both sbgames. Howeer, the color of does not inflence the otcome of the non-cone sbgame. Hence, if edge is eer played dring the game we consider it to be a moe in the cone sbgame. Claim 1 (Special sbgame) Player B can force a tie playing on the config- 5

6 s r Figre 1: ration (W, {}, {}), where is the inner point of W, = {, } (i.e., and are consectie in W ), and no configration points are contained in. Proof of the claim. We look at the points of W in clockwise order, and let w be the first point after. When player R makes another moe in the configration (W, {}, {}), if player B can draw w, he does so and we are left with a triial sbgame on the points,,, w, and one or more cone sbgames (depending on the moe of player R), which can be taken care of by Lemma 1, see Figre 2a. Otherwise, R has made a moe that preents B from playing w. Hence, R has drawn either w or or, for some w, with, w. If R drew w, then B draws w, splitting the problem into a triial sbgame on the points,,, w, and a cone sbgame, see Figre 2b. If, howeer, R drew, then B draws, and prodces two cone sbgames (one of them possibly triial - a single edge), and a sbgame on the points,,,, see Figre 2c. Finally, if R drew, then again B draws and we are left with two sbgames (,, {}) and ( {}, {}, {}), see Figre 2d. While the first sbgame is always cone (or triial - a single edge), the second sbgame may or may not be cone. Howeer, if ( {}, {}, {}) is not cone, then it is a smaller instance of this special sbgame, which can be handled recrsiely. Hence, the claim is proed. Case (ii). R draws an edge, for,. We assme, withot loss of generality, that is in the interior of the cone hll of, and we proceed as follows. Sbcase (ii.a). Sppose first that there eists an edge y (distinct from ) leaing on one side and on the other side. In this case, we assme that y and are chosen so that there are no other points in y and y. (We may choose and y according to the following simple algorithm. Starting from, consider the points of V in conterclockwise order and pick y as the last point that is on the same side as with respect to the line. Now choose as the last point in clockwise order, starting from, that is on the same side as with respect to y. Note that y may be eqal to and can be eqal to, bt not both simltaneosly. This follows from the fact that y is distinct 6

7 w w (a) (b) w w (c) (d) Figre 2: (a) The dashed red edges are eamples of possible moes of R that do not preent B from playing w. (d) If one of the sbgames is not cone, then it is a smaller instance of the special sbgame, which can be handled by indction. from.) Then B draws y, and the problem is diided into three smaller sbgames ( y, {}, {y}), (, {}, ) and ( y {},, {y}), see Figre 3. Note that the first two sbgames are cone (some of them possibly degenerate). For the third sbgame we show that B can force a tie in the following paragraph. Note that there can be no configration points in ŷ or in ẑy. Consider the points of V in clockwise order, let b be the point immediately after, and let a be the point preceding y. After R has played in the configration ( y {},, {y}), if B can draw either a or yb, then B plays sch a moe, and again, we are left with a triial sbgame on the points a,, y,, or b,, y,, and one or two cone sbgames (depending on the moe of player R), see Figre 4a, so B can force a tie on them by Lemma 1. Otherwise, R has jst made a moe that preents B from playing a and yb, which means that R drew either a or yb or w, for some w y, with w, y. If R has drawn a (or yb), then B plays a (respectiely b), splitting the problem into a triial sbgame on the points a,, y, (b,, y, respectiely), and a cone sbgame, see Figre 4b. Assme now that R has drawn w for some w y, with w, y. Then B draws, forcing R to play y, and finally B plays yw (see Figre 4c). At this 7

8 y Figre 3: point we obtain two more cone sbgames, namely ( w {}, {w}, {}) and ( wy,, {wy}) (one of them may be triial). a a b y b y (a) w a (b) b y (c) Figre 4: (a) The dashed red edges are eamples of possible moes of R that do not preent B from playing a. Sbcase (ii.b). If, on the other hand, there eists no edge y which has on one side and on the other side, there can be no points in or û. Consider the points of V in clockwise order and let y be the point preceding. Now B draws y, and we are left with a triial sbgame on the points,,, y, and two cone sbgames ( y,, {y}) and (, {}, ) (one of them may be degenerate), see Figre 5 below, resoled by Lemma 1. This completes the proof of Theorem 2. Now we show that player R can force a tie in the Bichromatic Triangle Game 8

9 y Figre 5: when the points in V are in cone position, and when V has eactly one inner point. Lemma 2 Sppose C = (V, E R, E B ) is a configration of the Bichromatic Triangle Game where V is a set of points in cone position and all drawn edges, i.e., edges in E R E B, are between consectie points in V. If E B < 2, then R can force a tie in C regardless of who is net to play. Proof The proof is symmetric to the proof of Lemma 1 this time R plays the role of B, and ice ersa. Theorem 3 Player R can force a tie in the Bichromatic Triangle Game when the points in V are in cone position. Proof Player R is the first to make a moe, ths he can draw any edge between two consectie points in V. Now, the statement follows immediately from Lemma 2. Theorem 4 Player R can force a tie in the Bichromatic Triangle Game when the set of points V has eactly one inner point. Proof First, player R draws any edge y with the property that there are no points in ŷ or ẑy, and sch that is in the interior of y. Sch an edge may be obtained as follows. Start with any edge y, with y,, sch that is in the interior of y. If y has the desired property, we are done. Otherwise, set = and = y and se the same simple algorithm as in Case (ii) of the proof of Theorem 2 to find the desired edge y. By drawing y, player R splits the problem into two smaller sbgames, namely ( y, {y}, ) and ( y {}, {y}, ). Player R can force a tie in the first sbgame by Lemma 2. To show that he can also force a tie in the second sbgame, we proceed analogosly as in the second paragraph of Case (ii) of the proof of Theorem 2, bt this time R plays the role of B, and ice ersa. We also se Lemma 2 instead of Lemma 1. 9

10 We presented each of the players with a strategy to preent his opponent from claiming a monochromatic triangle in the Bichromatic Triangle Game. Hence, we immediately obtain the following reslt for the Bichromatic Complete Trianglation Game. Corollary 1 If the points in V are in cone position, or V has eactly one inner point, then the otcome of the Bichromatic Complete Trianglation Game played on V is a tie, i.e., each of the players has a strategy to force at least a tie. 3 Smmary and Remarks We consider the Bichromatic Triangle and the Bichromatic Complete Trianglation Games and show that either player can force a tie in these games when the gien points are in cone position or when there is eactly one inner point. Natral open qestions that arise are to design polynomial algorithms to compte winning strategies or to determine which player can force a tie in these games in the general case, i.e., when there is no predetermined conditions for the gien set of points, or else to show that the problem of characteriing the otcome of any of these games in the general case is NP-hard. Acknowledgments We wold like to thank the anonymos referees for helpfl comments. References [1] O. Aichholer, D. Bremner, E.D. Demaine, F. Hrtado, E. Kranakis, H. Krasser, S. Ramaswami, S. Sethia, J. Urrtia, Games on trianglations, Theor. Compt. Sci. 343 (2005), [2] E.R. Berlekamp, J.H. Conway, R.K. Gy, Winning Ways of yor mathematical plays, Academic Press, [3] J.H. Conway, On Nmbers and Games, Academic Press, [4] E.D. Demaine, Playing games with algorithms: Algorithmic combinatorial game theory, Proc. 26th Symp. on Math Fond. in Comp. Sci., Lect. Notes in Comp. Sci (2001),