A New Variation of Hat Guessing Games

A New V ariation of Hat Guessing Games T engyu Ma ∗ Xiaoming Sun † Huac heng Y u ‡ Institute for Theoretical Computer Sc ience Tsingh ua Univ ersit y , Beijing, China Abstract Several va riations of hat gu essing games hav e been p opularly discussed in recreational mathematics. In a typical hat guessing game, after initially coord in ating a strategy , each of n pla yers is assigned a hat from a given color set. Simultaneously , eac h p laye r tries to gu ess the color of h is/her o wn hat by lo oking at colors of hats worn by other play ers. In this pap er, w e consider a new v ariation of this game, in whic h w e require at least k correct guesses and no wrong guess for the play ers to win the game, b u t t hey can choose to “pass”. A strategy is called p erfe ct if it can a chiev e the simple up p er b ound n n + k of the wi nning probabilit y . W e present sufficient and necessary condition on the parameters n and k f or t h e existence of p erfect strategy in the hat guessing games. In fact for any fixed parameter k , the existence of p erfect strategy can b e determined for every sufficiently large n . In our construction we in tro d uce a new notion: ( d 1 , d 2 )-regular partition of the b o olean hypercub e, whic h is w orth to study in its own right. F or example, it is related to the k - dominating set of the hypercub e. It also migh t b e interesting in co ding theory . The existence of ( d 1 , d 2 )-regular partition is explored in the pap er and th e existence of p erfect k -dominating set follo ws as a corollary . Keywords: Hat guessing game; perfect strategy; hypercub e; k-dominating set; p erfect code 1 In tro duction Several different hat guessing games hav e been studied in recent years [1, 2, 3, 4, 5, 6, 7]. In this pap er we inv estigate a v ariatio n where play ers ca n either give a guess o r pass. It was first prop osed by T o dd Ebe r t in [3]. In a standa rd s e tting there are n play ers sitting around a table, who ar e allow ed to co or dinate a strateg y b efore the game b egins. Each pla yer is assigned a hat who se colo r is chosen randomly and indepe ndently with probability 1 / 2 from tw o p ossible colo rs, red a nd blue. Each play er is allowed to see all the hats but his own. Sim ultaneously , each play er guesses its own hat color or pas ses, according to their pre-co o rdinated stra tegy . If at least one player gues ses correctly and no player guesses wrong , the play ers win the ga me. Their goa l is to desig n a strateg y to max imum their winning probability . By a simple c o unt ing arg umen t there is an upper b o und o f the maximum winning probabilit y , n/ ( n + 1). It is known that this upp er b ound can b e achiev ed if and only if n has the form 2 t − 1 [3]. It turns out that the existence of s uch p erfect stra tegy that achieves the upp er b o und cor resp onds to the existence o f p erfect 1-bit er ror-co rrecting co de in { 0 , 1 } n . ∗ Email: mat engyu1989@gmail.com † Email: x iaomings@tsinghua.edu.cn ‡ Email: y uhch12 3@gmail.com 1 In this pap er, we present a natura l g e neralization of Eb ert’s hat g uessing problem: The s etting is the same as in the original problem, every player can see all other hats except his own, and is allow ed to guess o r pass. How ev er, the req uir ement for them to win the game is g eneralized to b e that at leas t k play ers fro m them should guess cor rectly , a nd no play er guesses wrong (1 ≤ k ≤ n ). Note that when k = 1 , it is exac tly the o riginal problem. W e denote by P n,k the maxim um winning probabilit y of pla yers. Similarly to the k = 1 case, P n,k has a simple upp er bound P n,k ≤ n n + k . W e call a pair ( n, k ) p erfe ct if th is upper b ound can be achiev ed, i.e. P n,k = n n + k . There is a simple necessary co nditio n for a pair ( n, k ) to b e p erfect, and o ur main r e s ult states tha t this condition is a lmost sufficient: Theorem 1. F or any d, k , s ∈ N with s ≥ 2 ⌈ lg k ⌉ , ( d (2 s − k ) , dk ) is p erfe ct, in p articular, (2 s − k , k ) is p erfe ct. There e xists pair ( n, k ) with the nec e ssary condition but not p erfect, se e the r emark in Sectio n 4. Here is the outline of the pro o f: firs t we give a general characterization of the winner pr obability P n,k by using the size of the minimum k - dominating set of the hyper cube . Then we con vert the condition of ( n, k ) p er fect to so me kind o f re gular p artition o f the hyp ercub e (see the definition in Section 2). Our main contribution is that we pr esent a strong sufficient co ndition for the existence of suc h partition, whic h near ly matches the neces sary co ndition. Then w e can transform it into a per fect hat guessing s tr ategy . As a cor ollary of Theorem 1, we also give asymptotic characterization of the v a lue P n,k . F or example, we show that for any fixed k , the max im um winning proba bilit y approa ches 1 as n tends to the infinit y . Related w ork : F eige [4] consider ed some v ariations including the discar ded hat version and the everywhere balanced v ersion. Lenstra and Serous si [6] s tudied the case that n is not of form 2 m − 1, they also considere d the case with mu ltiple colo rs. In [2 ], Butler, Ha jia g hayi, Klein be r g and Leighton considered the worst case of hat placement with sight gra ph G , in which they need to minimize the maximum wrong guesses ov er all hat placements. In [5] F eige studied the case that each player can see o nly some o f other play ers’ hats with resp ect to the sight graph G . In [7], Peterson and Stinson in vestigated the case that each player can s ee hats in fron t of him and they guess one by one. V ery recen tly , Buhler, Butler, Graham and T ressler [1 ] studied the case that ev ery player needs to guess a nd the play ers win the game if either exactly k 1 or k 2 play ers g ue s s corr ectly , they show ed that the simple necessar y condition is als o sufficient in this g ame. The r e s t of the pap er is org anized as follows: Section 2 descr ibe s the definitio ns , notatio ns and mo dels used in the pap er. Then, Section 3 presents the re sult of the existence of ( d 1 , d 2 )-regular partition of hypercub e w hile Section 4 shows the main result of the hat guessing ga me. Finally , we co nclude the pap er in Sectio n 5 with some op en pro blems. 2 Preliminaries W e use Q n to denote the the n dimension b o olea n h yper cube { 0 , 1 } n . Two no des a r e a djacent on Q n if they differ by only one bit. W e enco de the blue and red colo r by 0 and 1. Th us the placement of hats on the n players’ heads can b e repr esented as a no de of Q n . F or any x ∈ Q n , x ( i ) is used to indicate the string obtained by flipping th e i th bit of x . Thr oughout the paper, a ll the op era tions are over F 2 . W e will clarify explicitly if am biguity app ea r s. Here is the mo del of the hat guess ing game we cons ider in this pap er: The num b er of players is denoted by n a nd players are denoted by p 1 , . . . , p n . The co lo rs of play ers’ hats will b e denoted to 2 be h 1 , . . . , h n , which are randomly a ssigned from { 0 , 1 } with eq ual probability . h = ( h 1 , . . . , h n ). Let h − i ∈ Q n − 1 denote the tuple of colo rs ( h 1 , . . . , h i − 1 , h i +1 , . . . , h n ) that play er p i sees on the others’ heads . The strategy of play er p i is a funct ion s i : Q n − 1 → { 0 , 1 , ⊥} , which ma ps the tuple of co lors h − i to p i ’s a ns wer, where ⊥ repr esents p i answers “pass” (if some play er answers pass, his answer is neither correct nor wrong). A stra tegy S is a collection of n functions ( s 1 , . . . , s n ). The players win the game if at le a st k of them guess co rrectly and no one guesses wrong. W e use P n,k to deno te the maximum winning probability of the play ers. The following tw o definitions a r e very useful in characterization P n,k : Definition 2. A subset D ⊆ V is c al le d a k -do minating set of gr aph G = ( V , E ) if for every vertex v ∈ V \ D , v has at le ast k neighb ors in D . Definition 3. A p artition ( V 1 , V 2 ) of hyp er cub e Q n is c al le d a ( d 1 , d 2 )-regular pa rtition if e ach no de in V 1 has exactly d 1 neighb ors in V 2 , and e ach no de in V 2 has exactly d 2 neighb ors in V 1 . F or example, consider the following par tition ( V 1 , V 2 ) of Q 3 : V 1 = { 000 , 11 1 } , and V 2 = Q 3 \ V 1 . F or each vertex in V 1 , there ar e 3 neighbors in V 2 , and for each vertex in V 2 , there is exactly one neighbor in V 1 . Th us ( V 1 , V 2 ) forms a (3 , 1 )- regular partition of Q 3 . 3 ( d 1 , d 2 )-Regular P artition of Q n In this section we study the existence of ( d 1 , d 2 )-regular par tition of Q n . Prop ositio n 4. Supp ose d 1 , d 2 ≤ n , if ther e ex ists a ( d 1 , d 2 ) -r e gular p artition of hyp er cub e Q n , then the p ar ameters d 1 , d 2 , n should satisfy d 1 + d 2 = gcd( d 1 , d 2 )2 s for some s ≤ n . Pr o of. Suppos e the par tition is ( V 1 , V 2 ), we count the total num b er o f vertices | V 1 | + | V 2 | = 2 n , and the n um b er o f edges b et ween tw o pa r ts d 1 | V 1 | = d 2 | V 2 | . By so lving the equatio ns, we obtain | V 1 | = d 2 d 1 + d 2 2 n , | V 2 | = d 1 d 1 + d 2 2 n . Both | V 1 | and | V 2 | should be in tegers, therefor e d 1 + d 2 = gcd( d 1 , d 2 )2 s holds, since gcd( d 1 , d 1 + d 2 ) = gcd( d 2 , d 1 + d 2 ) = g cd( d 1 , d 2 ). Prop ositio n 5. I f ther e exists a ( d 1 , d 2 ) -r e gular p artition of hyp er cub e Q n , then ther e exists a ( d 1 , d 2 ) -r e gular p artition of Q m for every m ≥ n . Pr o of. It suffices to show that the sta temen t ho lds when m = n + 1, since the desire d result follows by induction. Q n +1 can b e treated as the union of tw o co pies o f Q n (for e x ample partition according to the last bit), i.e. Q n +1 = Q (1) n ∪ Q (2) n . Supp ose ( V 1 , V 2 ) is a ( d 1 , d 2 )-regular par tition of Q (1) n . W e ca n duplicate the partition ( V 1 , V 2 ) to g et another partition ( V ′ 1 , V ′ 2 ) of Q (2) n . Then we can see that ( V 1 ∪ V ′ 1 , V 2 ∪ V ′ 2 ) forms a par tition of Q n +1 , in which eac h no de has an edg e to its duplicate thro ugh the last dimensio n. Obser ve that each no de in V 1 ( V ′ 1 ) still has d 1 neighbors in V 2 ( V ′ 2 ) and same fo r V 2 ( V ′ 2 ), and the ne w edges intro duced by the new dimension are a mong V 1 and V ′ 1 , or V 2 and V ′ 2 , whic h do es not contribute to the edges betw een tw o parts of the partition. Therefore we constructed a ( d 1 , d 2 )-regular par tition of Q n +1 . 3 Prop ositio n 6. If ther e exists a ( d 1 , d 2 ) -r e gular p artition of Q n , then ther e exists ( td 1 , td 2 ) -r e gular p artition of Q tn , for any p ositive inte ger t . Pr o of. Suppos e ( V 1 , V 2 ) is a ( d 1 , d 2 )-regular pa rtition of Q n . Let x = x 1 x 2 · · · x nt be a no de in Q nt . W e can divide x into n se ctions of length t , and denote the s um of i th section by w i , i.e. w i ( x ) = ti X j = ti − t +1 x j , (1 ≤ i ≤ n ) . Let R ( x ) = w 1 ( x ) w 2 ( x ) . . . w n ( x ) ∈ Q n . Define V ′ i = { x ∈ Q nt | R ( x ) ∈ V i } , ( i = 1 , 2 ) . W e claim that ( V ′ 1 , V ′ 2 ) is a ( td 1 , td 2 )-regular par tition of Q nt . This is b ecause f or any vertex x in V ′ 1 , R ( x ) is in V 1 . So R ( x ) ha s d 1 neighbors in V 2 , and e ach of which corresp onds t neighbor s of x in V ′ 2 , thu s in total td 1 neighbors in V ′ 2 . It is the same fo r vertices in V ′ 2 . By P rop osition 4-6 we only need to consider the exis tence of ( d 1 , d 2 )-regular partition of Q n where gcd( d 1 , d 2 ) = 1 a nd d 1 + d 2 = 2 s (where s ≤ n ), o r equiv ale n tly , the ex is tence of ( d, 2 s − d )- regular partition of Q n , wher e s ≤ n and d is odd. The following Lemma from [1] showed that when n = 2 s − 1 such regula r partition always exis ts. Lemma 7. [1 ] Ther e exists a ( t, 2 s − t ) -r e gular p artition of Q 2 s − 1 , for any inte ger s, t with 0 < t < 2 s . The following theo rem shows ho w to construct the ( t, 2 s − t )-r e gular partition for n = 2 s − r (where r ≤ t ). Theorem 8. Supp ose ther e exists a ( t, 2 s − t ) -r e gular p artition for Q 2 s − r and t > r , then ther e exists a ( t, 2 s − t ) -r e gular p artition for Q 2 s +1 − min { t, 2 r } . Pr o of. F or con venience, let m = 2 s − r , and l = 2 r − min { t, 2 r } ( ≥ 0). Observe that 2 s +1 − min { t, 2 r } = 2 m + l , and if t ≥ 2 r then l = 0. Supp ose tha t ( V 1 , V 2 ) is a ( t, 2 s − t )-regular pa rtition for Q m . W e wan t to construct a ( t, 2 s +1 − t )-r e gular partition fo r Q 2 m + l . The basic idea o f the construction is as follows: W e start from set V 2 . W e construc t a collection of linear equatio n systems, each of which corres p o nds to a no de in V 2 . The v ariables of the linear systems are the (2 m + l ) bits of node x ∈ Q 2 m + l . Let V ′ 2 be the union of solutions of these linear equation systems, and V ′ 1 be the complement o f V ′ 2 . Then ( V ′ 1 , V ′ 2 ) is the ( t, 2 s +1 − t )-regular partition as w e desired. Here is the construction. Since ( V 1 , V 2 ) is a ( t, 2 s − t ) regular partition for Q m , the subgraph induced by V 2 of Q m is a ( t − r )-regula r graph, i.e. for every no de p ∈ V 2 , there ar e ( t − r ) neighbors of p in V 2 . F or each p ∈ V 2 , arbitrar ily choose a subse t N ( p ) ⊆ V 2 of neig h b ors of no de p with size | N ( p ) | = r − l . (here r − l = r − (2 r − min { t, 2 r } ) = min { t, 2 r } − r , so r − l ≤ t − r , a nd r − l > 0 since t > r ) Now for ea ch no de p = ( p 1 , . . . , p m ) ∈ V 2 , we co nstruct a linea r equation s ystem as follo ws:                x 1 + x 2 = p 1 , x 3 + x 4 = p 2 , . . . . . . . . . , x 2 m − 1 + x 2 m = p m , P m j =1 x 2 j − 1 + P j ∈ N ( p ) x 2 j + P 1 ≤ j ≤ l x 2 m + j = 0 . (1) 4 Note that in the last equation the la st term P 1 ≤ j ≤ l x 2 m + j v anishe s if l = 0. Denote by S ( p ) ⊆ Q 2 m + l the solutions of this linear s ystem. F or conv enience, let f : Q 2 m + l → Q m be the o p er ator such that f ( x 1 , . . . , x 2 m + l ) = ( x 1 + x 2 , x 3 + x 4 , . . . , x 2 m − 1 + x 2 m ) . Then in the linea r system (1 ) the first m equations is nothing but f ( x ) = p . Let V ′ 2 = ∪ p ∈ V 2 S ( p ), and V ′ 1 = Q 2 m + l \ V ′ 2 be its complement. W e claim that ( V ′ 1 , V ′ 2 ) is a ( t, 2 s +1 − t )-regular partition of Q 2 m + l . T o begin with, obs e rve the following tw o fac ts . Observ ation 1 F or ev ery x ∈ V ′ 2 , w e ha ve f ( x ) ∈ V 2 . It can be seen clearly from the first m equations in eac h equatio n system. Observ ation 2 If k ≤ 2 m , then f ( x (2 k ) ) = f ( x (2 k − 1) ) = ( f ( x )) ( k ) . If k > 2 m , f ( x ( k ) ) = f ( x ). Recall the x ( i ) is the no de obtained by flipping the i th bit of x . The observ ation can be seen fro m the definition of f ( x ). F or any no de x ∈ V ′ 1 , w e show that there are t different wa ys of flipping a bit of x so tha t we can g et a no de in V ′ 2 . There are t w o p ossible cases: Case 1: f ( x ) 6∈ V 2 . In this case if w e flip the i th bit of x for some i > 2 m , then from Observ ation 2, f ( x ( i ) ) = f ( x ), so f ( x ( i ) ) will remain not in V 2 , and therefore x ( i ) will not b e in V ′ 2 , by O bserv ation 1 . So w e can only flip the bit in ( x 1 , . . . , x 2 m ). Suppo se by flipping the i th bit of x we get x ( i ) ∈ V ′ 2 ( i ∈ [2 m ]), from the definition of V ′ 2 we hav e : f ( x ( i ) ) ∈ V 2 , and x ( i ) satisfies the last equation in the equa tion systems corresp o nding to f ( x ( i ) ): m X j =1 x 2 j − 1 + X j ∈ N ( f ( x ( i ) )) x 2 j + X 1 ≤ j ≤ l x 2 m + j = 0 . (2) Since f ( x ) / ∈ V 2 and ( V 1 , V 2 ) is a ( t, 2 s − t )-regular partition of Q m , s o there are exactly t neighbors of f ( x ) in V 2 , which implies there are t bits of f ( x ) b y flipping which we can get a neighbor o f f ( x ) in V 2 . Let { j 1 , . . . , j t } ⊆ [ m ] b e these bits, i.e. f ( x ) ( j 1 ) , . . . , f ( x ) ( j t ) ∈ V 2 , b y Observ ation 2, f ( x (2 j k − 1) ) = f ( x (2 j k ) ) = f ( x ) ( j k ) ∈ V 2 , ( k = 1 , . . . , t ) . But exactly one of { x (2 j k − 1) , x (2 j k ) } satis fie s the equatio n (2) (here we use the fact f ( x ) / ∈ V 2 , note that j k 6∈ N ( f ( x ) ( j k ) ). Thus totally , there are t pos s ible i such that x ( i ) ∈ V ′ 2 . Case 2: f ( x ) ∈ V 2 . Since x / ∈ V ′ 2 , the last linear equation must b e vio lated, i.e. m X j =1 x 2 j − 1 + X j ∈ N ( f ( x ( i ) )) x 2 j + X 1 ≤ j ≤ l x 2 m + j = 1 . (3) W e further consider three cases her e: flip a bit in { x 1 , . . . , x 2 m } \ { x 2 j , x 2 j − 1 : j ∈ N ( f ( x )) } ; flip a bit in { x 2 j , x 2 j − 1 : j ∈ N ( f ( x )) } ; flip a bit in { x 2 m +1 , . . . , x 2 m + l } : a) if i ∈ [ m ] , i / ∈ N ( f ( x )), and f ( x ) ( i ) ∈ V 2 . Since f ( x ) ∈ V 2 , ( V 1 , V 2 ) is a ( t, 2 s − t )-regula r partition of Q m , there are m − (2 s − t ) − | N ( f ( x )) | = (2 s − r ) − (2 s − t ) − ( r − l ) = t − 2 r + l such index i , and x (2 i − 1) is the ex actly the one in { x (2 i − 1) , x (2 i ) } which is in V ′ 2 . (deter mined b y Equation 3 ). Th us in this c a se there a re ( t − 2 r + l ) neighbo rs of x in V ′ 2 . b) if i ∈ N ( f ( x )), then bo th of x (2 i − 1) , x (2 i ) are in V ′ 2 , there are 2 · | N ( f ( x )) | = 2( r − l ) such neighbors. c) if i > 2 m , then ev ery x ( i ) is in V ′ 2 , ( i = 2 m + 1 , . . . , 2 m + l ), there ar e l suc h neighbor s. Hence, totally x has ( t − 2 r + l ) + 2 ( r − l ) + l = t neigh b ors in V ′ 2 . 5 The res t thing is to sho w that ev ery no de x ∈ V ′ 2 has (2 s +1 − t ) neighbors in V ′ 1 . The pro of is similar to the pro of of Ca se 2 ab ove, we cons ider thr ee cas es: a) If i ∈ [ m ], i 6∈ N ( f ( x )), and f ( x ) ( i ) ∈ V 2 . Then exactly one of x (2 k − 1) , x (2 k ) in V ′ 2 , thus there are m − (2 s − t ) − | N ( f ( x )) | = 2 s − r − (2 s − t ) − ( r − l ) = t − 2 r + l suc h neighbors of x in V ′ 2 . b) If i ∈ N ( f ( x )) both x (2 i − 1) , x (2 i ) are not in V ′ 2 . c) If i > 2 m , then every x ( i ) is not in V ′ 2 . Hence totally , x has ( t − 2 r + l ) neighbor s in V ′ 2 , and therefore (2 m + l ) − ( t − 2 r + l ) = 2 s +1 − t neighbors in V ′ 1 . Hence we pr ov e that ( V ′ 1 , V ′ 2 ) is indeed a ( t, 2 s +1 − t )-regular par titio n of Q 2 s +1 − min { t, 2 r } . Theorem 9 . F or any o dd numb er t and any c ≤ t , when s ≥ ⌈ lg t ⌉ + ⌈ lg c ⌉ , ther e exists a ( t, 2 s − t ) - r e gular p art ition of Q 2 s − c . Pr o of. Let s 0 = ⌈ lg t ⌉ . By Lemma 7, there exists a ( t, 2 s 0 − t )-regular partitio n of Q 2 s 0 − 1 . By repe a tedly using Theo rem 8, we obtain that there ex ists ( t, 2 s 0 +1 − t )-reg ula r par tition o f Q 2 s 0 +1 − 2 , ( t, 2 s 0 +2 − t )-regular partition o f Q 2 s 0 +2 − 2 2 , etc., ( t, 2 s 0 + ⌈ lg c ⌉− 1 − t )-re g ular pa rtition of Q 2 s 0 + ⌈ lg c ⌉− 1 − 2 ⌈ lg c ⌉ − 1 , and ( t, 2 s 0 + ⌈ lg c ⌉ − t )-regula r pa rtition of Q 2 s 0 + ⌈ lg c ⌉ − c . By using P rop osi- tion 5, we get that there exists a ( t, 2 s − t )-r e gular partition of Q 2 s − c , for any s ≥ s 0 + ⌈ lg c ⌉ = ⌈ lg t ⌉ + ⌈ lg c ⌉ . Combining Pr op osition 6 and Theorem 9, w e hav e the following coro llary . Corollary 10. S u pp ose d 1 = dt, d 2 = d (2 s − t ) , n = d ( 2 s − c ) , wher e d, t, s ar e p ositive inte gers with 0 < t < 2 s , c ≤ t and s ≥ ⌈ lg c ⌉ + ⌈ lg t ⌉ , then t her e exists a ( d 1 , d 2 ) -r e gular p artition for Q n . 4 The M axim um Winning Probabilit y P n,k The following lemma characterizes the rela tionship b etw een the maximum winner proba bility P n,k and the minim um k -domina ting set of Q n . The same res ult was showed in [5] for k = 1. Lemma 11 . Supp ose D is a k -dominating set of Q n with minimum numb er of vertic es. Then P n,k = 1 − | D | 2 n . Pr o of. Given a k -domina ting set D of Q n , the follo wing strategy will have winning pr o bability at least 1 − | D | 2 n : F or a ny certain placemen t of hats, each play er can see all hats but his own, so play er p i knows that curre nt placement h is one o f tw o adjacent nodes { x, x ( i ) } of Q n . If x ∈ D (or x ( i ) ∈ D ), he guesses that the cur rent pla cement is x ( i ) (or x ), otherwise he pass es. W e claim that by using this strategy , players win the game when the placement is a no de which is not in D . Observe that since D is a k -do minating set, for an y node y / ∈ D , y has l neigh b ors y ( i 1 ) , y ( i 2 ) , . . . , y ( i l ) that are in D , where l ≥ k . According to the s trategy desrib ed, play ers p i 1 , . . . , p i l would g ue s s correctly and a ll other players will pass. This sho ws the winning probability is at leas t 1 − | D | 2 n . Next we show that P n,k ≤ 1 − | D | 2 n . Suppos e we hav e a strategy with winning probabilit y P n,k . W e prove that there exists a k -dominating se t D 0 , such that | D 0 | = 2 n (1 − P n,k ). The constr uction is straightf orward: Let D 0 = { h ∈ Q n : h is no t a winning placement } . Th us | D 0 | = N (1 − P n,k ). F or ev ery winning placement h / ∈ D 0 , s uppo s e play ers p i 1 , . . . , p i l will guess correctly ( l ≥ k ), consider the placement h ( i 1 ) , which differs from h only at play er p i 1 ’s hat, so play er p i 1 will guess incorr ectly in this cas e, thus h ( i 1 ) ∈ D 0 . Similarly h ( i 2 ) , . . . , h ( i l ) ∈ D 0 , therefore D 0 is a k -dominating s et. W e ha ve | D | ≤ | D 0 | = 2 n (1 − P n,k ) , 6 which implies P n,k ≤ 1 − | D | 2 n . Combining these t w o results, w e hav e P n,k = 1 − | D | 2 n as desire d. Prop ositio n 12. The fol lowing pr op erties hold: (a) If n 1 < n 2 then P n 1 ,k ≤ P n 2 ,k . (b) ( n, k ) is p erfe ct iff ther e exists a ( k, n ) -r e gular p artition of Q n . (c) F or any t ∈ N , P nt,kt ≥ P n,k . As a c onse quenc e, if ( n, k ) is p erfe ct, ( nt, k t ) is p erfe ct. Pr o of. F or part (a), suppo se tha t D is a minimum k -do minating set of Q n 1 . W e make 2 n 2 − n 1 copies of Q n 1 , and b y co m bining them we get a Q n 2 , which has dominating s et of size 2 n 2 − n 1 | D | . By Lemma 11, P n 2 ,k ≥ 1 − 2 n 2 − n 1 | D | 2 n 2 = P n 1 ,k . F or part (b), suppose ( U, V ) is a ( k , n )-regular par tition of Q n , note that V is a k -domina ting set of Q n and | V | = k n + k · 2 n , thus V is a minim um k -dominating set of Q n . W e ha ve that P n,k = 1 − | V | 2 n = n n + k , which implies that ( n, k ) is perfect. On the other hand, if ( n, k ) is p erfect, s uppo s e D is the minimum k -do minating set, we have | D | = k n + k · 2 n . It can b e observed that ( Q n \ D, D ) is a ( k , n )-regular partition of Q n . F or part (c), since n n + k = nt nt + kt , once P nt,kt ≥ P n,k holds, it’s an immedia te consequence that the p erfectness of ( n, k ) implies the perfectness of ( nk , nt ). Suppo se for n play ers, w e ha ve a strategy S with probability of winning P n,k . F or nt players, we divide th em into n groups, each o f which has t pla yers. E ach pla cement h = ( h 1 , h 2 , . . . , h nt ) of nt players can b e mapp ed to a placement P ( h ) of n pla yers in the fo llowing w ay: for Group i , suppo se the sum o f colors in the group is w i , i.e. w i ( h ) = ti X j = ti − t +1 h j , (1 ≤ i ≤ n ) . Let P ( h ) = ( w 1 ( h ) , w 2 ( h ) , . . . , w n ( h )) be a plac e ment of n players. Each play er in Group i knows the color o f all players in P ( h ) o ther than Pla yer i , thus he uses Play er i ’s s trategy s i in S to guess the sum o f colors in Group i or passes. Moreover once he knows the sum, his color can be uniquely determined. Note that the pla yers in Gr oup i would guess cor rectly or incor r ectly o r pass , if a nd only if Play er i in the n -player-game w ould do. Since the hat placemen t is unifor mly at random, the probability o f winning using this s trategy is at le a st P n,k , thu s P nt,kt ≥ P n,k . Now we can prov e our main theorem: Theorem 1. F or any d, k , s ∈ N with s ≥ 2 ⌈ lg k ⌉ , ( d (2 s − k ) , dk ) is p erfe ct, in p articular, (2 s − k , k ) is p erfe ct. Pr o of. It’s an immedia te cor ollary of part (b) of P rop osition 12 and Theor em 9. Remark: By Prop os ition 4 and P rop osition 12(b) there is a simple neces sary co ndition for ( n, k ) to be perfect, n + k = g cd( n, k )2 t . Theor e m 1 indica tes that when n + k = gcd( n, k )2 t and n is sufficiently la rge, ( n, k ) is perfect. The nec e ssary condition and sufficient condition nearly match in the sense that for eac h k , there’s only a few n that w e don’t kno w whether ( n, k ) is per fect. Mo reov er, the following prop os itio n shows that the simple necessary condition ca n’t b e 7 sufficient. The first counterexample is (5 , 3), it is not p erfect while it satisfies the simple nece s sary condition. But (13 , 3) is p erfect b y Theorem 1 and more generally for all s ≥ 4, (2 s − 3 , 3 ) is per fect. W e verified by computer pr ogram that (2 4 − 5 , 5) = (11 , 5) is no t p erfect, while by our main theor em (2 6 − 5 , 5) = (59 , 5) is p erfect. But we still don’t know whether the case betw een them, (2 5 − 5 , 5) = (27 , 5), is perfect. Prop ositio n 13. ( n, k ) is not p erfe ct unless 2 k + 1 ≤ n when n ≥ 2 and k < n . Pr o of. Suppos e ( n, k ) is per fect. According to par t (b) of Prop os ition 12, w e can find ( U, V ), a ( k , n )-r egular partition of Q n . Supp ose x is s o me node in U , and y is some neighbor o f x which is also in U , y has k neig hbors in V . They a ll differ from x at exactly 2 bits and one of them is what y differs fr om x at, i.e. each of them “domina tes ” 2 neighbor s o f x , one of them is y . So x has to tally k + 1 neighbors “dominated” b y k of nodes in V . Since all no des in V ar e pairwise nonadjacent, these k + 1 nodes must b e in U . Now we ha ve k + 1 neigh b ors of x a re in U and k neighbors a re in V , it has totally n neighbors. W e m ust hav e 2 k + 1 ≤ n . F or ea ch o dd num b er k , let s ( k ) b e the s mallest num ber such that (2 s ( k ) − k , k ) is perfect. W e know that s ( k ) ∈ [ ⌈ lg k ⌉ , 2 ⌈ lg k ⌉ ]. The fo llowing pro p o sition indicates that all s ≥ s ( k ), (2 s − k , k ) is als o p er fect. Prop ositio n 14. If (2 s − k, k ) is p erfe ct, (2 s +1 − k, k ) is p erfe ct. Pr o of. If (2 s − k , k ) is p e rfect, by Pro po sition 12(b) ther e is a ( k , 2 s − k )-re g ular partition of Q 2 s − k . Thu s by Prop os ition 5, we hav e a ( k , 2 s − k )-reg ular partition of Q 2 s − k +1 . Combine this par tition and Theorem 8, w e ge t a ( k , 2 s +1 − k )-r egular partition of Q 2 s +1 − k . Therefo r e (2 s +1 − k , k ) is per fect. Using Theorem 1 we can give a general lo wer b ound for the winning proba bility P n,k . Recall that there’s upper bound P n,k ≤ 1 − k n + k . Lemma 15 . P n,k > 1 − 2 k n + k , when n ≥ 2 2 ⌈ lg k ⌉ − k . Pr o of. Let n ′ be the large s t integer of for m 2 t − k whic h is no more than n . By T heo rem 1, ( n ′ , k ) is per fect, i.e. P n ′ ,k = 1 − k n ′ + k . B y part (a) of P rop osition 12, P n,k ≥ P n ′ ,k . O n the other hand we have n + k < 2 t +1 , so we hav e P n,k ≥ 1 − k n ′ + k = 1 − 2 k 2 t +1 > 1 − 2 k n + k . Corollary 16. F or any inte ger k > 0 , lim n →∞ P n,k = 1 . 5 Conclusion In this pap er we in v estiga ted the existence of regular partition for bo o lean hypercub e, and its applications in finding p erfect stra tegies of a new hat guessing games. W e sho wed a sufficient condition for ( n, k ) to b e p erfect, which near ly matches the necessary condition. Sev era l problems remain op en: for exa mple, determine the minimum v alue of s ( k ) such that (2 s ( k ) − k , k ) is p erfect, and determine the exa ct v alue of P n,k . It is a lso v ery in teresting to consider the case when there are mor e than tw o color s in the game. 8 References [1] Jo e Buhler , Steve Butler, Ron Graham, and E ric T re ssler. Hyp ercub e orientations with only t wo in-degr ees. http://a rxiv.or g/abs/1 007.2311 , 2010 . [2] Steve Butler, Mohamma d T. Ha jiagha yi, Rob ert D. Kleinberg , and T o m Leighton. Hat guessing games. SIAM J. Discr ete Ma th. , 22(2):592– 6 05, 2008 . [3] T o dd T. Eb ert. Applic ations of r e cursive op er ators to r andomness and c omplexity . PhD thesis, Univ ers ity of California at Santa Barba r a, 19 98. [4] Uriel F eige. Y ou can leav e your hat on (if you guess its color). T e chnic al R ep ort MCS04-03 , Computer Scienc e and Appli e d Mathematics, The Weizmann Institute of Scienc e , 20 04. [5] Uriel F e ige. On optimal strategies for a hat game on graphs. SIAM Journal of Discr ete Mathematics , 201 0. [6] Hendrik W. Lenstra and Gadiel Serouss i. On hats and o ther cov ers. http://arxiv.o r g/abs/cs/0 509045 , 2005. [7] Maur a B . Peterson and Douglas R. Stinson. Y et ano ther hatt g ame. the ele ctr onic jo urnal of c ombinatorics , 17(1), 2010 . 9