Lower Bound for the Communication Complexity of the Russian Cards Problem

Lo wer Bound for the Co mmunication Comple xity of the Russ ian Cards Problem Aiswarya Cyriac, K. Murali Krishnan, Departmen t of Co mputer Scien ce an d En gineerin g National Institute of T echnology , Calicut 6 7360 1, India. Email: aiswaryan itc@gmail.com, kmu ralinitc@gmail.co m Abstract —In thi s paper it is shown that n o p ublic announce- ment scheme th at can b e modeled in Dynamic Ep istemic Logic (DEL) can solve the Russian Cards Problem (RCP) in one announcement. Since DEL is a general model fo r any public announcement scheme [11], [3], [6], [21], [12] we conclude that there exist n o sin gle announcement solution to the RCP . The proof demonstrates the utili ty of DEL i n proving lower boun ds fo r communication protocols. It is also shown that a general version of RCP has no two announcement solution wh en the adversar y has su fficiently large nu mber of cards. K ey words: Russian cards problem, Dynamic Epistemic Logic, Communication complexity , Lower boun d. I . I N T RO D U C T I O N In the Russian cards pro blem (RCP), there a re th ree p layers and se ven cards. The cards are randomly distributed amo ng the players such that two players g et three cards each and the third player gets one card. The problem is to find a sequence of public an nounce ments by which play ers with th ree cards eac h are able to acquir e comp lete inf ormation abo ut all the ca rds, without the third player knowing about any of their cards. The solution to th e pr oblem will imply a m ethod to com mu- nicate inf ormation amon g parties in a d istributed co mputing setting securely witho ut usin g any encryptio n [15], [23], [ 20]. The analo gy is that, the commu nicating agents and a dversaries are modeled as p layers an d th e info rmation to be com muni- cated as the ownership o f cards. It is gen erally b eliev ed th at the ab ove gam e gives un condition al security [11], [15], [23], [20]. V arious solutions to the a bove p roblem can be f ound in th e literature [9], [10], [22]. T hey all requ ire at least two public annou ncements. Here we address the pro blem of formally establishing that no public announ cement scheme can s olve the problem in fewer an nounce ments. The fra mew o rk of Dynamic Epistemic Log ic (DEL) is u sed to establish the lower bou nd. Similar boun ds using o ther m odels for related pr oblems c an be f ound in [1], [ 2], [8], [15], [16]. The f ollowing sections briefly discuss dy namic epistemic logic, mod eling of RCP in DEL and a proo f for the lower bound . Fin ally a generalization of the RCP is pr esented and it is shown that tw o ann ouncem ents are not sufficient to solve the general ca se when the adversary has sufficiently large n umber of c ards. I I . R U S S I A N C A R D S P RO B L E M ( R C P ) The problem was po sed in 20 00 [ 11] as the following: F r om a pack of seven known car d s two pla yers each draw three cards and the thir d player gets th e remaining ca r d. How can the players with three cards open ly(publicly) info rm each oth er ab out th eir cards, withou t the third player learning fr om any of their car ds who h olds it? Let us call th e cards 0, 1 , ..., 6 . The players are Anne, Bill and Cath. Anne and Bill have three cards each and Cath has one. No secret commun ication is possible. On ly annou ncements are allowed. The anno uncemen ts are assumed to be truthf ul and p ublic. Throu gh a sequen ce of su ch an- nounc ements Anne and Bill have to learn the actual deal o f the cards. i.e., for each card fro m th e a bove p ack, An ne an d Bill shou ld be able to say to wh om that car d belo ngs. Also for any ca rd from the pac k o ther than the one Cath is h olding, she should not be able to say to whom that card belon gs. V arious solu tions to the above prob lem can be seen in the literature [9], [10], [2 2]. All these so lutions require two annou ncements. I I I . D Y N A M I C E P I S T E M I C L O G I C ( D E L ) In this pap er we express the RCP in the fr amew ork of Dynamic Epistemic L ogic (DEL). This sectio n briefly pr esents the syn tax and semantics o f Dynam ic Epistemic Lo gic. More detailed discussion of th e DEL, examples and its app lications can b e fo und in [3], [5], [13], [17], [18]. Kripke models and action mo dels are semantical structur es of dynamic epistemic log ic. Given a set of agents (players) and b asic pr opositions a Kripke m odel co nsists of the set of possible states and accessibility (or in distinguishab ility ) relation betwee n the states for every agent. The knowledge of th e players abo ut th e state o f the g ame in imper fect informa tion g ames 1 can b e mod eled by viewing g ame states as K ripke states and play ers as agen ts. Action models model the actions of the playe rs that will alter the knowledge of the play ers. Given an initial Kripke model modeling th e k nowledge of the play ers, the action models can be sequentially executed in the Kripke model, resulting in a new K ripke mod el that models the knowledge of th e players af ter th e action . 1 In imperfect information games, players do not have complete information about other players’ moves. Epistemic Lang uage: Epistemic log ic can be used to model knowledge in g ames o f imp erfect inf ormation [4], [ 5], [6], [7], [19]. Let N be a finite set of agents 2 and P be a finite set of p roposition al atoms. T he Ep istemic langu age L P,N is the smallest closed set for which the following hold s: • p ∈ P ⇒ p ∈ L P,N • φ, ψ ∈ L P,N ⇒ ¬ φ, ( φ ∧ ψ ) ∈ L P,N • φ ∈ L P,N and n ∈ N ⇒ K n φ ∈ L P,N The sentence K 1 φ is read as: agent 1 knows φ . ( φ ∨ ψ ) , φ → ψ , an d φ ↔ ψ and are abb reviations for ¬ ( ¬ φ ∧ ¬ ψ ) , ¬ φ ∨ ψ , an d ( φ ∧ ψ ) ∨ ( ¬ φ ∧ ¬ ψ ) respectively . The notation ⊤ stands for ¬ ( p ∧ ¬ p ) fo r some p ∈ P . Let a fin ite set of agen ts N and a finite set of proposition al atoms P be giv en . A Kripke model [24], [5], [4], [ 13] is a tuple ( W , R, V ) wher e: • The set W is non empty set of states { w 1 , . . . w | W | } • The accessibility function R : N → 2 W × W assigns for each agen t n ∈ N a set of or dered pairs of states. ∀ n ∈ N , R ( n ) ⊆ W × W is a n eq uiv alence relation. • The valuation functio n V : W → 2 P assign to each state a set of p ropo sitional atoms. ∀ w ∈ W , V ( w ) ⊆ P . ( w, w ′ ) ∈ R ( n ) is in terpreted as state w ′ is accessible ( or indistinguishab le) from state w for the agent n . The set of propo sitional atom s assigne d to a state by V is the ato ms which hold in that state. A Krip ke world is a pair c onsisting of a Kr ipke model M = ( W , R, V ) and a state w ∈ W and is de noted by ( M , w ) . Semantics of Epistemic Logic:: Let a Krip ke model M = ( W , R, V ) and the e pistemic lan guage L P,N be g iv en . • M , w | = p ⇔ p ∈ V ( w ) • M , w | = ¬ φ ⇔ M , w 6| = φ • M , w | = φ ∧ φ ⇔ M , w | = φ and M , w | = ψ • M , w | = K n φ ⇔ For all w ′ such that ( w , w ′ ) ∈ R ( n ) , M , w ′ | = φ A Player x k nows the fact φ in th e state w only if φ holds in all the states in distinguishab le from w . Also if φ is true in all th e states ind istinguishable from w for Player x , she can infer φ . The a ction mo dels are used to upda te Kr ipke Mod els. An action model c onsist of a set of actio ns, an accessibility (indistingu ishability) relation between the action s fo r every agent, and a p recond ition function for each action. Let a set of ag ents N an d the epistemic lang uage L P,N be giv en. An Action m odel µ is a tuple ( A, R ∗ , Π) : • The set A is the n onemp ty set of action s { a 1 , . . . , a | A | } • The accessibility fun ction R ∗ : N → 2 A × A is a fun ction which assigns to each agent a set of ordered pairs of actions. ∀ n ∈ N , R ∗ ( n ) ⊆ A × A is an equiv alence relation. 2 Players are modeled as agents. Agents are assumed to be perfect logicians, i.e. the agents know all the consequences of their knowledge . • The p recond ition fu nction Π : A → L P,N assigns to ev e ry action a pr econdition . ∀ a ∈ A , Π( a ) ∈ L P,N Execution: Action models are used to upda te Krip ke model. Thus an action model is an operator on a Kripke Model. The execution of an action in a state results in a new state if and only if the preco ndition of the action holds in that state. Let a Kripke mode l M = ( W, R, V ) and an action m odel µ = ( A, R ∗ , Π) be g iv e n. The execution of action model µ in Kripke mode l M results in a Kripke mo del deno ted by M ⊗ µ . M ⊗ µ = ( W ′ , R ′ , V ′ ) su ch that: • The set of worlds W ′ = { ( w , a ) ∈ W × A | M , w | = Π( a ) } • For e very Player n ∈ N , (( w, a ) , ( w ′ , a ′ )) ∈ R ′ ( n ) iff ( w, w ′ ) ∈ R ( n ) an d ( a, a ′ ) ∈ R ∗ ( n ) • The valuation fu nction V ′ is su ch that V ′ ( w, a ) = V ( w ) . States at any par ticular p oint carr y a tag o f a ll preceding actions. The state ( w, a ) will exist in the fin al Kripke model only if the preco ndition of th e action a is satisfi ed by w in the initial m odel. T wo states ( w , a ) an d ( w ′ , a ′ ) are indistin- guishable f or Player x in the new Kripke mo del if and on ly if the states w and w ′ were ind istinguishable fo r the Player x in the initial Kripke Mo del and the ac tions a and a ′ were indistinguishab le in the ac tion m odel µ . The kn owledge of the p layers abo ut th e state of the gam e at the b eginning o f th e ga me is mo deled b y a Kripke mode l. Th e knowledge ac tions which o ccur durin g the gam e are modeled by actio n mo dels. The knowledge development is modeled by sequential execution of the action mod el in th e Kr ipke model, r esulting in a new up-to- date Kr ipke model mo deling the k nowledge of the players after the knowledge actions. I V . P RO B L E M M O D E L I N G Giv en th e set of players and the basic pr opositions, th e Russian c ards prob lem (RCP) can be m odeled in Dynam ic Epistemic Log ic (DEL) . Let U = { 0 , 1 , 2 , 3 , 4 , 5 , 6 } be the set of cards and N = { a, b, c } (rep resenting Anne, Bill and Cath) be the set of players. The basic p ropositions are ‘card 0 is w ith Ann e, ’ ‘card 3 is with Bill’ and so on. If i x denotes ‘card i is with x ’ the n the set o f basic prop ositions P = { i x | x ∈ N , i ∈ U } . Initially Play er a and Player b have thr ee c ards each and Player c has one c ard. Th e Kripke mod el for the initial ga me state is g i ven by M = h W, R, V i where, W = { ( A, B , C ) | | A | = | B | = 3 , | C | = 1 , A ∪ B ∪ C = U } R ( a ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | A = A ′ } R ( b ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | B = B ′ } R ( c ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | C = C ′ } V  ( A, B , C )  = { i a | i ∈ A } ∪ { i b | i ∈ B } ∪ { i c | i ∈ C } In any state w = ( A, B , C ) the set of card s with Player a is A , th e set o f card s with Player b is B and that with Player c is C . R ( x ) , x ∈ N co ntains the state pair s tha t ar e indistinguishab le f or Play er x . The definition of R ( x ) follows from th e fact that initially Player x k nows only th e card s she is h olding. Henc e all the states in which her han d of cards is the same will be indistingu ishable for her . For ea ch A ⊆ U, | A | = 3 , let T A = { ( A ′ , B ′ , C ′ ) : A ′ = A , | B ′ | = 3 , | C ′ | = 1 and A ′ ∪ B ′ ∪ C ′ = U } . Similarly for each B ⊆ U, | B | = 3 , let S B = { ( A ′ , B ′ , C ′ ) : B ′ = B , | A ′ | = 3 , | C ′ | = 1 and A ′ ∪ B ′ ∪ C ′ = U } and for each C ⊆ U, | C | = 1 , let Q C = { ( A ′ , B ′ , C ′ ) : C ′ = C , | A ′ | = | B ′ | = 3 an d A ′ ∪ B ′ ∪ C ′ = U } . Hence we have: R ( a ) = S A ⊆ U, | A | =3 T A × T A and A 6 = A ′ ⇒ T A ∩ T A ′ = ∅ R ( b ) = S B ⊆ U, | b | =3 S B × S B and B 6 = B ′ ⇒ S B ∩ S B ′ = ∅ R ( c ) = S C ⊆ U, | C | =1 Q C × Q C and C 6 = C ′ ⇒ Q C ∩ Q C ′ = ∅ R ( a ) is a partitio n of W and T A will b e called a compone nt of R ( a ) . Her e W = S A ⊆ U, | A | =3 T A and the union is disjoint. R ( a ) has  7 3  = 35 c ompon ents each h aving  4 3  ×  1 1  = 4 states. All fou r states in any compo nent T A are indistinguish- able f or Play er a by the definition of R ( a ) . Similarly R ( b ) is a partition of W and S B is called a compon ent of R ( b ) . W = S B ⊆ U, | B | =3 S B and the u nion is d isjoint. R ( b ) has  7 3  = 35 com ponen ts each having  4 3  ×  1 1  = 4 states a nd all four states in any comp onent S B are indistingu ishable for Player b . R ( c ) is a p artition of W an d Q C is ca lled a co mpone nt of R ( c ) . W = S C ⊆ U, | C | =1 Q C and the un ion is d isjoint. R ( c ) has  7 1  = 7 c ompon ents each h aving  6 3  ×  3 3  = 20 states. The twenty states in a compo nent Q C are indistinguishab le for Player c . As an example for A = { 0 , 1 , 2 } , T A = { ( { 0 , 1 , 2 } , { 3 , 4 , 5 } , { 6 } ) , ( { 0 , 1 , 2 } , { 3 , 4 , 6 } , { 5 } ) , ( { 0 , 1 , 2 } { 3 , 5 , 6 } , { 4 } ) , ( { 0 , 1 , 2 } , { 4 , 5 , 6 } , { 3 } ) } . Player a canno t distingu ish between these fou r states because in all the fo ur states Player a ’ s h and is { 0 , 1 , 2 } . W ithou t loss of g enerality let us assum e that Player a is having cards { 0 , 1 , 2 } , Player b is having { 3 , 4 , 5 } and Player c is having { 6 } initially . W e den ote this state b y w ∗ . Thus w ∗ = ( { 0 , 1 , 2 } , { 3 , 4 , 5 } , { 6 } ) . Suppose that a sing le annou ncement scheme so lves the RCP . In the final model we claim th at ther e will be at least one c ompon ent T A 3 and S B of the partitions ge nerated by R ( a ) and R ( b ) re spectiv e ly such that T A = S B = { w ∗ } . This is because if one m ore state was present in the com ponen t, the Players can not d istinguish between those states. Also there will be at least one component Q C of the p artition gen erated b y R ( c ) such that w ∗ ∈ Q C and | Q C | > 1 . Otherwise Pl ayer c will be able to find out the actual state. we present the claim form ally: Lemma 1 : Assume th at RCP is solved in Kripke mod el M = ( W , R , V ) an d for all A , B and C such tha t | A | = | B | = 3 an d | C | = 1 , T A , S B and Q C are c ompon ents of R ( a ) , R ( b ) and R ( c ) respectively . The n the following statements 4 hold: 3 Let the initial state be M = ( W, R, V ) and the action model be µ = ( A, R ∗ , π ) . The states in the final model M ⊗ µ will be a subset of W × A . But for notation al con venience , we ignore the action-tags in the states. Hence the set of states in the final model is seen as a s ubset of W . 4 These statements are necessary but not suffic ient. 1) ∃ A, ∃ B suc h that T A = S B = { w ∗ } 5 2) ∀ C , w ∗ ∈ Q C ⇒ | Q C | > 1 . Pr oof: Let w ∗ ∈ T A . If possible let T A contain an other state — say w 1 such that w ∗ 6 = w 1 . Let w ∗ = ( A, B ∗ , C ∗ ) and w 1 = ( A, B 1 , C 1 ) . B ∗ 6 = B 1 and C ∗ 6 = C 1 , other wise w ∗ = w 1 . Player a can not distinguish whe ther Playe r b is having B ∗ or B 1 . Also she can not d istinguish wheth er Play er c is having C ∗ or C 1 . This m eans the RCP is n ot y et solved contradictin g the hyp othesis. Th erefore T A = { w ∗ } . Similar argument shows S B = { w ∗ } . ∀ C , w ∗ ∈ Q C ⇒ | Q C | > 1 . If | Q C | = 1 , then c w ill understan d which state she is in . There does not exist e ven a single card that belongs to Player a in all the states of Q C . If such a card exists, say 4, Player c will und erstand that 4 is with Player a . (Recall th e semantics for K n φ is discussed in sectio n II I ). Similar ly , there doe s not exist e ven a sing le card that belo ngs to Player b in all th e states o f Q C . The ab ove claim is tuned to o ur req uiremen t of solving the RCP in on e an nounc ement. It is easily seen th at th e claim holds in any fina l mo del reached by any seque nce of annou ncements. V . L O W E R B O U N D F O R R C P Theor e m 1: Th ere exist no single an nouncem ent s o lution to the RCP . Pr oof: For the sake of con tradiction assum e ther e exists a single anno uncemen t solution to RCP . Without lo ss of generality it can be assum ed that Player a is m aking the annou ncement. Let th e action m odel for the anno uncemen t be µ = ( A, R ∗ , Π) . In sectio n III we have seen that R ( a ) partitio ns W into 35 compo nents. Each compo nent will have 4 states . Sup pose w 1 , w 2 , w 3 and w 4 belong to one co mponen t — say T A . Let w 1 be the actual state. Since Player a will be determin istically making the an noun cement, she will make the same anno unce- ment, say a 1 for all the e lements in T A . In section I II we hav e seen that the states ( w 1 , a 1 ) and ( w 2 , a 2 ) will be be indistinguisha ble for Player a if ( w 1 , w 2 ) ∈ R ( a ) an d ( a 1 , a 2 ) ∈ R ∗ ( a ) . So in the final model ( w 1 , a 1 ) , ( w 2 , a 1 ) , ( w 3 , a 1 ) an d ( w 4 , a 1 ) will belon g to th e same compon ent of R ( a ) . This contra dicts Claim 1. Hence ther e cannot b e a single annou ncement solution to th e RCP . V I . A G E N E R A L I Z A T I O N In this sectio n we will consider a natu ral g eneralization of the RCP in whic h Anne and Bill ar e holding k car ds each and Cath is ho lding l cards fro m a pack of 2 k + l cards. W e deno te this version of the RCP as RCP( k ; l ). Hence the orig inal RCP discussed b efore is RCP( 3; 1 ) in the n ew notation. It can be easily seen that f or any k ≥ 1 an d l ≥ 1 there does n ot exist a o ne an nounce ment solution for RCP( k ; l ) as the Theo rem 1 an d the proof extends to th is case as well . W e 5 Actuall y it should be T A = S B = { ( w ∗ , t ) } , where t is the list of actio ns that led to the final state, but for the ease of writing we have omitte d the action tag list t . will now examine the impossibility o f a two an nounce ment solution for RCP( k ; l ) usin g similar stra tegies. The set of card s U = { 1 , 2 , . . . , 2 k + l } . The initial Kr ipke model M = h W, R, V i wh ere, W = { ( A, B , C ) | | A | = | B | = k , | C | = l , A ∪ B ∪ C = U } R ( a ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | A = A ′ } R ( b ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | B = B ′ } R ( c ) = { (( A, B , C ) , ( A ′ , B ′ , C ′ )) | C = C ′ } V  ( A, B , C )  = { i a | i ∈ A } ∪ { i b | i ∈ B } ∪ { i c | i ∈ C } T A , S B and Q C are also defined similar ly as in Section IV. It follows that R ( a ) and R ( b ) will hav e  2 k + l k  compon ents each with  k + l k  ×  l l  =  k + l k  elements each. Suppose Play er a is makin g an anno uncemen t α for a set of compon ents T α . Since Player c sh ould n ot lear n ab out a single card other than his own han d, we get the fo llowing lemm a. Lemma 2 : [ T A ∈T α A = U (1) \ T A ∈T α A = ∅ (2) Pr oof: Assume that this was n ot the case. i.e., S T A ∈T α A = Q , Q ⊂ U . Player c can infe r that the c ards in U \ Q are n ot with Player a . Hence U \ Q ha s to be ∅ . Similarly it can be seen that T T A ∈T α A = ∅ since if T T A ∈T α A = Q , Q ⊆ U , Q 6 = ∅ , then Player c can in fer that the set of cards Q is with Player a . Before proving th e impossibility of a two a nnoun cement solution for R CP( k ; l ) we need to prove the following technical lemma. Lemma 3 : For k ≥ 2 , l ≥ 2 k 2 ln k ⌈ 2 k + l k ⌉ ×  k + l k  >  2 k + l k  (3) Pr oof: Enou gh to have 2 k + l k ×  k + l k  >  2 k + l k  i.e., 2 k + l k × ( k + l )! k ! × l ! > (2 k + l )! k ! × ( k + l )! 2 k + l k k Y i =1 ( l + i ) > k Y i =1 ( l + k + i ) i.e., 2 k + l k > k Y i =1  1 + k l + i  . Since (1 + x ) ≤ e x RH S ≤ e k P l + k i = l +1 1 i . Bounding the summation by integral we g et, RH S ≤ e k. ln ( l + k l ) =  l + k l  k . Enoug h to have 2 k + l k >  l + k l  k . Let l = 2 k 2 ln k . Enoug h to have 2 + 2 k ln k >  1 + ln k 2 k  k since (1 + x ) ≤ e x we get  1 + ln k 2 k  k ≤ e ln k 2 = √ k ∴ 2 + 2 k ln k > √ k . W e c an see th at the above equatio n holds for all k ≥ 2 . Now we will p rove tha t f or sufficiently large l , the re exist at least two states in S T A ∈T α T A which are indisting uishable for player b for any anno uncemen t α satisfyin g Lem ma 2 . Lemma 4 : For k ≥ 2 , l > 2 k 2 ln k for any announce ment α satisfying Lemm a 2 ∃ s 1 , s 2 ∈ S T A ∈T α T A such th at s 1 6 = s 2 and ( s 1 , s 2 ) ∈ R ( b ) Pr oof: Assume that th ere d o n ot exist two indistingu ish- able states fo r Player b in S T A ∈T α T A . From L emma 2 we will get |T α | ≥ ⌈ 2 k + l k ⌉ since | A | = k and | U | = 2 k + l (as the 2 k + l elemen ts ne ed to be distributed among sets of size k ). Now the n umber of different hands possible for Player b shou ld be a t least ⌈ 2 k + l k ⌉ ×  k + l k  . But the d ifferent number of co mbination s po ssible is  2 k + l k  . By Le mma 3 ⌈ 2 k + l k ⌉ ×  k + l k  >  2 k + l k  for l ≥ 2 k 2 ln k and k ≥ 2 . Now we will prove tha t there doe s not exist a two- annou ncement solution for RCP( k ; l ) if k ≥ 2 and l ≥ 2 k 2 ln k . Theor e m 2: For k ≥ 2 , l ≥ 2 k 2 ln k , th ere exists no two annou ncement solution to th e RCP( k ; l ). Pr oof: The in itial Kripke model M 1 = h W 1 , R 1 , V 1 i as d escribed above. L et Player a make th e first anno unce- ment α . Th e action model for the first annou ncement is µ 1 = h A 1 , R ∗ 1 , Π 1 i . α ∈ A 1 . Let s 1 , s 2 ∈ W 1 be the two states wh ich are indisting uishable for Playe r b in S T A ∈T α A . ( s 1 , s 2 ) ∈ R 1 ( b ) . Let the re sulting Krip ke mo del be M 2 = h W 2 , R 2 , V 2 i . Th e states ( s 1 , α ) , ( s 2 , α ) ∈ W 2 will be in- distinguishab le for player b since ( s 1 , s 2 ) ∈ R 1 ( b ) an d ( α, α ) ∈ R ∗ 1 ( b ) . Thus (( s 1 , α ) , ( s 2 , α )) ∈ R 2 ( b ) . Hence Player b will make the same anno uncemen t say β for both these states. Let µ 2 = h A 2 , R ∗ 2 , Π 2 i be th e action mo del for the seco nd announ cement w ith β ∈ A 2 . Let the r esult- ing Krip ke mode l be M 3 = h W 3 , R 3 , V 3 i . Now the states ( s 1 , α, β ) , ( s 2 , α, β ) ∈ W 3 will still be indistinguishable f or Player b since (( s 1 , α ) , ( s 2 , α )) ∈ R 2 ( b ) and ( β , β ) ∈ R ∗ 2 ( b ) . i.e., (( s 1 , α, β ) , ( s 2 , α, β )) ∈ R 3 ( b ) . Hen ce two announce- ments ar e not sufficient to so lve the RCP( k ; l ) if l ≥ 2 k 2 ln k . V I I . C O N C L U S I O N W e have an alyzed the Ru ssian Cards Problem and the g en- eralization RCP( k ; l ) in the fra mew o rk of Dyn amic Ep istemic Logic. It is shown th at the re d oes no t exist a sing le an - nounc ement s olution for the Russian Card s Problem within the framework of Dyn amic E pistemic Lo gic. Since the fram ew or k is co nsidered sufficiently general [11], [3], [6 ], [21], [12] we claim that ther e can be no one-an noun cement solution to RCP in gener al and no two anno uncemen t solution to RCP( k ; l ) for l ≥ 2 k 2 ln k , k ≥ 2 . 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