The Complexity of Playing Durak

Transcription

1 The Complexty of Playng Durak Édouard Bonnet Insttute for Computer Scence and Control, Hungaran Academy of Scences (MTA SZTAKI), Budapest, Hungary, Abstract Durak s a Russan card game n whch players try to get rd of all ther cards va a partcular attack/defense mechansm. The last player standng wth cards loses. We show that, even restrcted to the perfect nformaton two-player game, fndng optmal moves s a hard problem. More precsely, we prove that, gven a generalzed durak poston, t s PSPACE-complete to decde f a player has a wnnng strategy. We also show that decdng f an attack can be answered s NP-hard. Introducton The computatonal complexty of games s a frutful research topc whch started to formalze n the late seventes [Schaefer, ]. From an AI perspectve, t offers an nsght nto what may and may not be computed effcently n the process of solvng a game. The complexty of games has been and s stll extensvely studed, gvng rse to a few tractablty results, such as solvng n polynomal tme NIM [Bouton, 0] and SHANNON EDGE SWITCHING GAME [Bruno and Wenberg, 0], and a seres of ntractablty results. For nstance, HEX [Resch, ], OTHELLO [Iwata and Kasa, ], AMAZONS [Furtak et al., 00; Hearn and Demane, 00], and HAVANNAH [Bonnet et al., 0a] are PSPACEcomplete, whle CHESS (wthout ffty-move rule) [Fraenkel and Lchtensten, ], GO (wth Japanese ko rules) [Robson, ], and CHECKERS [Robson, ], are EXPTIMEcomplete. That lst suggests that the computatonal complexty of board games s relatvely well understood. The man motvaton of ths paper s to go towards a smlar understandng for card games. Indeed, although card games are arguably as popular as board games, far less s known concernng ther complexty. We only know of a handful of results mostly on trck-takng card games. Brdge (or whst) wth two hands and a sngle sut, or wth two hands and mrror The author s supported by the ERC grant PARAMTIGHT: Parameterzed complexty and the search for tght complexty results, no. 0. A sut s sad mrror whenever both players have the same number of cards n t. suts can be solved n polynomal tme [Kahn et al., ; Wästlund, 00a; 00b]. Some generalzatons of brdge wth more hands were proven PSPACE-complete [Bonnet et al., 0b]. Fnally, the complexty of problems lnked to the games of UNO [Demane et al., 00] and SET [Lamps and Mtsou, 0] has been studed. Here, we wsh to pursue ths lne of works by nvestgatng the complexty of durak whose game mechansm s not based on takng trcks. Durak s a two to sx-player card game ntensvely played n Russa and East European countres. Durak s the Russan word for fool whch desgnates the loser. There s no wnner n durak, there s just a loser: the last player standng wth cards. We sketch a smplfed verson of the rules for two players and wthout trumps. The game s played wth cards, by keepng the cards from the sxes (lowest cards) to the aces (hghest cards) n a standard -card deck. Both players, let us call them P and O, are dealt a hand of sx cards and ther goal s to empty ther hand before the opponent does. The remanng cards form the ple. The game s made of rounds. A desgnated player, say P, leads the frst round by playng any card c of hs hand. In ths round, P s the attacker, O s the defender, and c s the frst attackng card. The defender can skp, at any tme. In that case, the defender pcks up all the cards played durng the round (by both players) and puts them nto hs hand; then, the attacker remans the attacker for the next round. The defender can also defend the current attackng card by playng a hgher card n the same sut. Each tme hs opponent defends, the attacker can (but s not forced to) play an addtonal attackng card (up to a lmt of sx cards) provded t has the same rank as a card already played durng the round (by ether hmself or hs opponent). If the defender does defend all the attackng cards played by the attacker, all the cards played durng the round are dscarded and the defender leads the next round, thereby becomng the new attacker. After each round, any player wth less than sx cards, draws cards n the ple untl he reaches the total of sx. In fact, we wll consder that the ple s empty and that the two players have perfect nformaton. Why do we make those assumptons? In durak, one does not wn but has to avod losng. Whle the ple s not empty, or whle there are For a full descrpton of the rules of Durak, see pagat.com/beatng/podkdnoy_durak.html

2 three players or more stll n the game, the rsk of quckly losng s relatvely weak. Ths s one motvaton for focusng on the two-player game wth an empty ple. Now, from hs hand and the cards played and dscarded so far, a player can nfer the hand of hs opponent, yeldng perfect nformaton. More mportantly, we almost exclusvely prove negatve results, and our hardness proofs do not requre more than two players, nor a non empty ple, nor trumps. After precsng the notatons, the vocabulary and the rules of durak n Secton, we show that decdng f one player can defend any attack s NP-hard, n Secton. The man result of the paper s the PSPACE-hardness of two-player perfect nformaton durak and s presented n Secton. Our reducton (from -TQBF) requres the ntroducton of several notons: weaknesses, well-covered weaknesses, and strong suts. We beleve that those notons can be of mportance n desgnng good artfcal players for durak. Prelmnares For any ntegers x y, [x, y] := {x, x +,, y, y} and [x] := [, x]. A card s defned by a sut symbol s j and an nteger called rank, and s denoted by (s j, ). A hand s a set of cards. Example. h = {(s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, )} s a hand. Card (s, ) has rank n sut s. Defnton. A durak poston P = h(p ), h(o), L, y s gven by two hands h(p ) and h(o) of P and O, an ndcator L {P, O} of who leads the next round (equvalently, whose turn t s) and a threshold y, that s the maxmum number of attackng cards allowed n a round. Rules. Relaton defnes a partal order over the cards by: for any sut s j and any, [r], (s j, ) (s j, ) ff. If c c and c c, we wrte c c. A game from an ntal poston P = h(p ), h(o), L, y s composed of rounds that are themselves composed of moves. If h(p ) = or h(o) = the game ends, the player stll havng cards loses, and hs opponent wns. We assume that P s the current attackng player (.e., L = P ). If c, c, c p s the lst of attackng cards played by P, so far, and d, d, d p the lst of defendng cards played by O then p y, and for each [p ], c d and c + has the same rank as at least one card n {c, d,, c, d }. O can skp. In that case, we say that O takes the cards. P can add extra attackng cards c p+,, c q (wth p + q y) provded that they are of the same rank as a card n {c, d,, c p, d p, c p }. The next poston s h(p ) \ {c,, c q }, h(o) {c,, c q }, P, y. O can also try to defend by playng a card d p such that c p d p. In that case, P can contnue the attack (f p < y) or skps. If the attacker P skps, the next poston s h(p ) \ {c,, c p }, h(o) \ {d,, d p }, O, y. The cards played durng the round are dscarded, O has defended untl the end, and O takes the lead. When a player plays a seres of attackng cards that cannot be defended by the opponent, we say that he gves those cards to hs opponent. A draw occurs f h(p ) = h(o) = s s s s s Fgure : The geometrc representaton of poston {(s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, ), (s, )}, h, P,. Generalzed durak. In generalzed durak, there are s suts and the ranks range from to r. The threshold poses some questons. It seems sound that, n a generalzaton of the game wth an unbounded number of suts and ranks, the number of moves wthn a round s not lmted by a constant. Therefore, as a part of the nstance, the threshold should be allowed to grow. Besdes, t does not make sense to mpose that r, s, and y satsfy a constrant that s satsfed by r =, s =, y = snce there s no canoncal such constrant. In case y rs, the threshold cannot come nto play, and we denote ts value as. Algebrac notaton. We wrte fragments of game, called varatons or contnuatons n the followng way. A move s a card, the defensve skp, or the attackng skp. Pars of an attackng card and ts defensve card are separated by commas. The extra attackng cards played after the defender skps are wrtten to the rght of symbol. Rounds are separated by semcolons. Geometrc representaton. Each card (s j, ) h(p ) s represented by a black dsk n (, j); each card (s j, ) h(o) s represented by a crcle n (, j) (see Fgure ). In the followng sectons, the suts are ndexed by symbols rather than ntegers and the columns are dsplayed n a convenent order. Observe that permutng the columns of the representaton preserves the poston. Example. P has a wnnng strategy n the poston of Fgure. He can play (s, ) (s, ); and after both (s, ) ; or (s, )(s, ), (s, ) ; P gves all hs cards but (s, ) by ncreasng ranks and fnsh wth (s, ). Ths process wll be generalzed n Lemma. On Defendng an Attack Defendng untl the end f possble, and takng the frst attackng card otherwse, consttutes a decent heurstc for the defender. Unfortunately, we show that decdng f a defense s possble s already a hard problem. Theorem. Gven a durak poston P, decdng f P can defend any attack of O untl the end s NP-hard. Proof. We reduce from -SAT. Let C = {C,, C m } be any nstance of -SAT, where each C s a -clause over the

3 0 s x s x s x s x s x s x s C s C s C s C s C s C s C s C Fgure : The durak poston P for the nstance {x x x, x x x, x x x, x x x, x x x, x x x, x x x, x x x }. set of varables X = {x,, x n }. We construct a durak poston P = h(p ), h(o), O, wth n + m suts, n + ranks, and n + m cards n total (O has n + m cards and P has n + m cards) such that C s satsfable ff P can defend untl the end any attack of O. Let r : {x, x, x, x,, x n, x n } [, n + ] such that r(x ) = and r( x ) = + for all [n], and l : [, n + ] {x, x, x, x,, x n, x n } be the nverse functon. For each varable x ( [n]), we devote a sut s x where O has the card (s x, ) and P has the two cards (s x, ) and (s x, +). For each clause C j = l l l, we devote a sut s Cj where O has the three cards (s Cj, r(l )), (s Cj, r(l )), and (s Cj, r(l )), whle P has the two cards (s Cj, n + ) and (s Cj, n + ). Ths ends the constructon (see Fgure and Fgure ). Frst, we may observe that f O starts the attack wth a card (s Cj, u), the defense s easy snce P can follow ths famly of varatons: (s Cj, u)(s Cj, n + ), (s Ck, u)(s Ck, n + ), (s Ck, u)(s Ck, n + ), (s Ckh, u)(s Ckh, n + ), where each k ( [h]) s the ndex of a clause where lteral l(u) appears. The only remanng attempt for O s to start attackng wth a card (s x, ), for some [n]. If C s satsfable, we fx a satsfyng assgnment a : X {, }. Symbol (respectvely ) s nterpreted as settng the varable to true (respectvely false). P can defend the attack n the followng way. On each attackng card (s x, ) ( [n]), P plays (s x, + ) f a(x ) = and plays (s x, ) f a(x ) =. Now, n each sut s Cj, O can attack wth at most two cards, and P can defend wth (s Cj, n + ) and (s Cj, n + ). Indeed, f there s a sut s Cj where O can play hs three cards of rank, say, u, u, and u, then no lteral among l(u ), l(u ), and l(u ) would be set to true by assgnment a, so the clause C j would not be satsfed. If C s not satsfable, no assgnment a : X {, } satsfes every clauses. In partcular, after O attacks wth all the cards (s x, ) ( [n]) and P has to defend wth (s x, u ) 0 0 sx sx sx sx sx sxscscscscscscscsc sx sx sx sx sx sxscscscscscscscsc Fgure : After the contnuaton (s x, ) (s x, ), (s x, ) (s x, ), (s x, ) (s x, ), (s x, ) (s x, ), (s x, ) (s x, ), (s x, ) (s x, ), correspondng to the truth assgnment x, x, x, x, x, x, P can defend untl the end. (u {, + }), the assgnment defned by a(x ) = f u = + and a(x ) = f u =, does not satsfy some clause C j. Thus, P has played cards of rank r(l ), r(l ), and r(l ) where C j = l l l. Hence, O can attack wth the three cards (s Cj, r(l )), (s Cj, r(l )), and (s Cj, r(l )), and P can not defend, snce he has only two cards n the sut s Cj. On Playng Optmally Proposton. Gven a durak poston P, decdng f P has a wnnng strategy s n PSPACE. Proof. We have to show that the length of a game s polynomally bounded by the sze of the nstance, or equvalently by the total number n of cards n P. Then, we can conclude by dong a depth-frst mnmax search. A player cannot have the lead on n consecutve rounds. Indeed, when a player keeps the lead, at least one card s transferred, at each round, from hs hand to the hand of hs opponent. So, f a player keeps the lead for n consecutve rounds, he wns. When the lead goes from a player to hs opponent, at least two cards are dscarded. Thus, a game cannot contan more than (n )n rounds. A round lasts at most n + moves, so the game length s bounded by (n )n(n + )/ = O(n ). We now need some extra defntons and observatons. Defnton. A weakness for player P s a rank [r] satsfyng the two followng condtons: () h(p ) contans at least one card of rank, and () for each sut s j wth (s j, ) h(p ), there s a rank > such that (s j, ) h(o). Informally, P has each of hs cards of rank domnated by a card of O. The set of cards of rank n h(p ) s also called weakness and each card of the set s called weakness card. A rank whch s not a weakness for P, or the set of cards of rank n h(p ) s called a non weakness (for P ). Assumng that the threshold y s greater than the total number of cards of rank, for any [r], (we wll refer to ths assumpton as (H) n what follows) we may observe that player P, at hs turn, can gve all hs cards of rank to O, provded that s a non weakness for P. Indeed, by defnton, there s

4 a sut s j such that (s j, ) h(p ) and no card c h(o) satsfes (s j, ) c. Thus, O cannot defend ths attack. Therefore, we can show the followng. Lemma. Under (H), f P, at hs turn, has only one weakness, then he has a wnnng strategy. Proof. If s a non weakness for P, and P gves to O all hs (non weakness) cards of rank smaller than, then s stll a non weakness for P n the resultng poston. So, P wns by gvng all hs non weaknesses to O by ncreasng ranks and fnally plays all hs cards of rank. Defnton. A strong sut for player P s a sut s j where he has at least one card and O has none. We observe that the rank of any card n a strong sut of P s a non weakness for P. We say that P can wn by attackng only f he has a wnnng strategy such that O can never take the lead. Example. Let P = h(p ) = {(s, ), (s, )}, h(o) = {(s, ), (s, ), (s, )}, P,. P can wn by attackng only due to the varatons: (a)(s, )(s, ), (s, )(s, ), ; (b)(s, )(s, ), (s, ) ; (c)(s, ) ; (s, )(s, ), ; and (d)(s, ) ; (s, ) ;. Note that f O had the lead n P, then he would wn by Lemma snce he only has as a weakness. The followng lemma s very useful to reduce the number of potentally good frst attackng card. Intutvely, t says that f you cannot wn by attackng only, t s useless (and possbly harmful) to gve cards to your opponent that he wll be able to gve you back when he wll have the lead. Lemma. Under (H), f P has a wnnng strategy but cannot wn by attackng only, O has a card (s j, ) n a strong sut s j, and s a non weakness for P, then P has a wnnng strategy that does not start the round wth cards of rank. Proof. O can accept to take the set S of cards of rank played by P. O wll eventually get the lead back. By defnton, P has no card n the strong sut s j of O. It mples that O has not been attacked n s j, so he has exactly the same cards n s j as n the ntal poston. In partcular, (s j, ) h(o) and O can gve S back to P all hs cards of rank, makng the frst attack of P useless. There s qute a lot of condtons n Lemma, and checkng that P cannot wn by attackng only, to know f the lemma apples, may be problematc. Therefore, we gve a suffcent condton mplyng that a player cannot wn by attackng only. Defnton. A well-covered weakness for P s a weakness such that for each (s j, ) h(p ), there s a hgher card (s j, ) h(o) and P has no card of rank. Intutvely, f P attacks wth a well-covered weakness, O can defend so that P cannot play any other attackng card at ths round. Lemma. If P has two well-covered weaknesses, O can prevent P from wnnng by attackng only. Proof. Let be the two well-covered weaknesses for P. Frst, we remark that whle P gves cards to O whch are not of rank or, they reman well-covered weaknesses. (a) The gadget. (b) x true. (c) x false. Fgure : The exstental gadget x. So, O takes any cards of rank / {, } wthout tryng to defend. At some pont, P has to start an attack wth cards of rank or. In both cases, O can defend untl the end, by defnton of a well-covered weakness. The proof of the followng lemma s smlar to the proof of Lemma and therefore omtted. Lemma. Under (H), f P has a wnnng strategy but cannot wn by attackng only, then P has a wnnng strategy that does not start the round wth the hghest card of some sut. Theorem. Gven a durak poston P, decdng f P has a wnnng strategy s PSPACE-complete. Proof. It s n PSPACE by Proposton. We show that t s PSPACE-hard by a reducton from the PSPACE-hard problem QBF whch remans so even f all the varables are quantfed, the quantfers alternate startng wth and endng wth. Ths restrcted problem s sometmes called -TQBF and conssts of decdng whether x x x x n φ(x, x,, x n ) s true or false, where φ s a conjuncton of clauses wth three lterals. We fx a -CNF formula φ wth m clauses C, C,, C m. We wll buld a durak poston P = h(p ), h(o), P, (n + m + ) wth n + ranks, m + n + suts, and m + n + cards such that ψ = x x x x n φ s true ff P has a wnnng strategy from the poston P. For techncal reasons that wll become relevant later, we defne ψ = x 0 x x x x n φ, where φ s the conjuncton of the m clauses x 0 C, x 0 C,, x 0 C m, x 0 C, x 0 C,, x 0 C m. We denote x 0 C by C and x 0 C by C for all [m]. We observe that ψ s true ff ψ s true, and φ s a conjuncton of -clauses. Exstental quantfer gadget. For each odd [n], we encode x by devotng four suts s, s, s, and s where P has four cards: (s, o ), (s, o ), (s, o + ), and (s, o + ) and O has four cards: (s, o + ), (s, o + ), (s, o + ), and (s, o + ). We set o := +. Fgure a dsplays the geometrc representaton of the exstental gadget and the two local outcomes f P decdes to set x to true (Fgure b) or to set x to false (Fgure c). Unversal quantfer gadget. For each even [n] {0}, we encode x by devotng three suts s, s, and s where P has three cards: (s, o ), (s, o + ), and (s, o + ) and O has four cards: (s, o + ), (s, o + ), (s, o + ), and (s, o +) (see Fgure ). Agan, we set o := +. For the By the form of ψ, nteger n s always even.

5 0 0 (a) The gadget. (b) x true. (c) x false. x0 x Fgure : The unversal gadget x. x x x s 0 s 0 s 0 s s s s s s s s s s s s s s s w s m w sc sc sc sc sc sc sc sc so sp s, d s d s d s d Fgure : The ntal durak poston for the nstance x x x x {x x x, x x x, x x, x x x }. The weaknesses are framed by dotted rectangles. P has weaknesses and O has weaknesses. quantfcaton x 0 and only for ths quantfcaton, P s dealt an extra card (s 0, n + ). Clause gadget. We defne the rank r(l) of lteral l as + f l = x or + f l = x. We denote by s(l) the sut wheren O has a card of rank r(l) n the gadget assocated to the quantfed varable x wth l {x, x }. So, f x s unversally quantfed, then s(x ) = s( x ) = s whle f x s exstentally quantfed, then s(x ) = s and s( x ) = s. For each -clause C = l l l l of φ, we devote a sut s C. Player P has the cards (s C, r(l )), (s C, r(l )), (s C, r(l )), and (s C, r(l )) whle O has the cards (s C, ) and (s C, k) for k [n+, n+]. Weaknesses and strong suts. We add a sut s O where player O has the cards (s O, k) for k [, n + ] [n +, n+] and P has none, and a sut s P where player P has the cards (s P, k) for k {} [, n+] [n+, n+] and O has none. We add (n+m) suts s,k d ( k [(n+m)]) where P has (s,k d, ) and O has (s,k d, ), a sut s d where P has (s d, ) and O has (s d, ), a sut s d where P has (s d, ) and O has (s d, ), and a sut s d where P has (s d, ). Fnally, we add m suts s k w ( k [m]), where P has the card (s k w, n + 0) and O has the card (s k w, n + ). The constructon s now fnshed (see Fgure ) and P satsfes assumpton (H). P has weaknesses:,, and n+0; O has weaknesses: and. P has well-covered weaknesses: and n + 0, and O only one:. Before gong nto the detals, we gve an outlne of the proof. P has one weakness more than O and hs only hope s to get rd of two weaknesses ( and n + 0) before O takes the lead. To do so, P should start the attack wth the lowest card n the gadget encodng the frst quantfed varable (namely, hs weakness card of rank ). O has to defend, and they slowly clmb up from rank to rank n + 0 passng through each quantfer gadget. In unversal gadgets x, O has two ways of defendng: wth a card of rank r(x ) or r( x ). In exstental gadgets x, P has two suts s and s to contnue the attack, but due to the threshold lmt, he has to choose only one. So, P and O act as the exstental and the unversal player n QBF seen as a two-player game. At the next round, O has to get rd of hs weakness of rank and wns ff one clause of φ s not satsfed by ther jont assgnment. P has to start the attack wth (s 0, ). We frst show that P has to start the attack wth (s 0, ) wth the dea of gettng rd of the two weaknesses and n + 0 n the same round. By Lemmas and, the three other optons are to start the attack wth a card of rank,, or n + 0. In case of the second or thrd opton, O can defend untl the end: (s d, )(s d, ), ; or (s w, n + 0)(s w, n + ), (s m w, n + 0)(s m w, n + ), ; and then O wns wth the followng strategy, whch we denote by S. Player O starts the next round wth all hs cards (s C, ) for each clause C of φ. P has to defend, snce otherwse O leads the next round wth a sngle weakness, so O wns by Lemma. In partcular, P should defend the card, ). The only way to do so s to play, r(l)) where l appears n C. If l = x 0 the wnnng contnuaton for O s, ), r(x 0 )), (s O, r(x 0 )) (s(x 0 ), r(x 0 )), ), ); whereas, f l x m 0 the varaton s, ), r(l)), (s(l), r(l)), ), ); and n both cases O leads the next m round wth only one weakness. Fnally, startng an attack wth a card of rank cannot help P ; O would just skp. Indeed, let S be the set of cards of rank played by P and taken by O. Ether S {(s P, )}, and O can gve all those cards back to P the next tme he takes the lead; or S = {(s P, )}, but P could gve ths card to O any tme he s the attacker. P cannot play cards of s P. We show that durng the frst round startng wth (s 0, ), P loses f he plays a card (s P, ). Assume P does. O has to take all the cards played durng the round, n partcular (s 0, ). s a new weakness for O, but P has also as a new weakness because he played (s P, ). So, each player has weaknesses and P has stll the lead. However, O wns n the followng way. Whle P starts attacks wth non weaknesses, O skps. Agan, we observe that, as the poston s, ths step cannot create weaknesses for O. When P starts an attack wth a weakness, O defends untl the end. Ths s possble snce and n + 0 are well-covered

6 weaknesses, and P has (m + n) + cards of rank, so he would be allowed to add at most extra attackng card, due to the threshold (m + n + ). One can check that O can always defend ths card of rank. Thus, O wll take the lead at least twce. The frst tme O takes the lead, he attacks wth (s 0, ). P has to defend, otherwse O wns thanks to strategy S. The second tme, O s left wth weaknesses and, and wns wth S. O should defend untl the end. We show that f P does not play a card of the sut s P durng ths frst round, then O should defend untl the end. Suppose O skps at some pont. Player O takes n hs hand the cards played durng the round; n partcular, the card (s 0, ) whch s now a weakness card for O, snce P has the card (s 0, n + ) that cannot have been played n the prevous attack of P, for t s the only card wth rank n +. P can wn by playng (s d, ) n the next round. O has to defend, snce otherwse P s left wth only one weakness n + 0 and wns by Lemma. So, the contnuaton s (s d, )(s d, ), ;. Now, O leads the round and has weaknesses:, and. Cards (s 0, ) and (s d, ) are well-covered weakness cards for O. P can skp on all the attacks of O untl one of these cards s played. Then, he defends and wns by Lemma, snce O cannot gve cards to P that would consttute weaknesses for P. P and O smulates QBF. If P does not play all hs cards of rank n + 0 durng the frst round, and O defends untl the end, then O wns. O starts the next round wth (s d, ). P has to defend: (s d, )(s d, ), ; otherwse O wns by Lemma. Then, P has the lead, but O wns snce he has only one weakness (), P has two well-covered weaknesses ( and n + 0), and P cannot gve cards to O whch would be new weaknesses for O. Besdes, as P cannot play cards of the sut s P, one can check that O wll be able to defend untl the end (thanks notably to cards (s C, k) h(o) for k [n +, n + ]). So, P has to fnd a way of playng all hs cards of rank n + 0. Therefore, due to the threshold (m + n + ), P can only play one card of rank r(x ) n each exstental gadget x. Thus, the frst round should be of ths form: (s 0, )(s 0, r(σ(x 0 ))), (s 0 0, r(σ(x 0)))(s 0 0, r( x 0) + ), (s, r( x 0 )+)(s, r(σ(x ))), (s +, r(σ(x )))(s +, r( x ) +), (s n, r( x n ))(s n, r(σ(x n ))), (s n n, r(σ(x n ))) (sn n, n + 0), (s w, n + 0)(s w, n + ), (s m w, n + 0)(s m w, n + ), ; where for each even k (resp. odd k), σ(x k ) {x k, x k } corresponds to the lteral that s set to true by O (resp. P ), and k {, } s the matchng ndex. As n Fgure and, we nterpret the card c of rank n {r(x ), r( x )} played by O (and dscarded at the end of the round) as settng x to true f the rank of c s r(x ) and as settng x to false f the rank of c s r( x ). Player O leads the next round. At ths pont, O has stll hs two weaknesses: and ; whle P has only one weakness:. If ψ s false, O wns. We recall that ψ and ψ are equvalent. Let us assume ψ s false. Then, O had a strategy n the frst round ensurng that there s a clause C = x 0 l l l such that (s(l ), r(l )), (s(l ), r(l )), (s(l ), r(l )) are stll n h(o). O plays all hs cards of rank. By Lemma, P has to defend. In partcular, he has to defend on the card, ). To do so, P can ether play, r(x 0 )) or, r(l k )) for some k {,, }. In the former case, the contnuaton s, ), r(x 0 )), (s O, r(x 0 )) and O add as extra attackng cards all hs cards of rank and potentally hs card (s 0, r(x 0 )). In the latter case, the contnuaton s, ), r(l k )), (s(l k ), r(l k )) and agan, O gves all hs cards of rank to P. In both cases, O wns by Lemma. If ψ s true, P wns. Whchever cards O gves to P, P wll not have addtonal weaknesses. Thus, f P can defend an attack of O untl the end, P wns by Lemma (provded that O has stll at least one card left). Ths s equvalent to sayng that f O has a wnnng strategy, he wns by attackng only. Let us show that O cannot wn by attackng only. The last attack of O should be (s d, )(s d, ), ; whle all hs other cards have been prevously gven to P. At some pont, O wll have to play hs weakness cards of rank. If O has already gven (s O, r(x 0 )) and (s O, r( x 0 )) to P, pror to ths attack, then P can defend:, ), r(x 0 )), m, ) m, r(x 0 )),, ), r( x 0)), )(s m C m, r( x 0)), (s 0, r(x 0 )) (s 0, n + ), ; (ths s why we ntroduced the dummy varable x 0 ) and P wns. So, we can assume that (s O, r 0 ) s stll n h(o) when O starts the attack wth cards of rank, wth r 0 {r(x 0 ), r( x 0 )}. As ψ s true, P had a strategy n the frst round such that, for each clause C = x 0 l l l ( [m]), there exsts k {,, } satsfyng (s(lk ), r(lk )) / h(o). Thus, P can defend n the followng way:, ), r(lk )), m, ) m, r(lk m m )),, ), r(l k )),, ) (s m C m, r(lm k m )), and O, to contnue the attack, has to play a card (s O, r(lk )) for some [m]. P takes all the cards played durng ths round. Now, O has a new well-covered weakness r 0 snce (s O, r 0 ) (s O, r(lk )), (s 0, r 0 ) (s 0, n + ), and O has no card of rank r(lk ) nor n +. O has two well-covered weaknesses and r 0. So, by Lemma, O cannot wn by attackng only, and by the prevous remarks, O loses. Perspectves Our proof of PSPACE-hardness for two-player durak reles on a fnte threshold. One could look for a reducton whch does not use the threshold feature. We also leave as an open queston f the seemngly very smple two-player durak wth a sngle sut s solvable n polynomal tme. References [Bonnet et al., 0a] Édouard Bonnet, Floran Jaman, and Abdallah Saffdne. Havannah and TwxT are PSPACEcomplete. In Computers and Games - th Internatonal Conference, CG 0, Yokohama, Japan, August -, 0, Revsed Selected Papers, pages, 0.

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