An improvement of the upper bound for GKS communication game
aa r X i v : . [ c s . CC ] J a n An improvement of the upper bound for GKScommunication game
Ivan Petrenko ∗ Moscow State University
Abstract
The GKS game was formulated by Justin Gilmer, Michal Kouck´y,and Michael Saks in their research of the sensitivity conjecture. MarioSzegedy invented a protocol for the game with the cost of O ( n . ).Then a protocol with the cost of O ( n . ) was obtained by DeVon In-gram who used a bipartite matching. We propose a slight improvementof Ingram’s method and design a protocol with cost of O ( n . ). The GKS game is a three-player game where two players, Alice and Bob,play as a team against their opponent Merlin. The game has two naturalparameters n, k . In the first phase of the game Merlin and Alice fill aninitially empty string x of length n with bits in the following fashion: FirstMerlin arbitrarily picks a permutation of the set { , , . . . , n } . Then hesends indices 1 , , . . . , n one-by-one in the chosen order to Alice. Once Alicereceives an index i , she immediately has to pick a bit to fill the i th position inthe string, that is she chooses x i . This goes on until only one empty positionis left, that position is filled by Merlin, not Alice. We will commonly referto this bit as the last bit. The now-filled string is sent to Bob, whose goalis to determine the position filled by Merlin. To this end Bob chooses a set S of k natural numbers. Alice and Bob win if S contains the position filledby Merlin, otherwise Merlin wins. Alice and Bob can communicate beforethe game starts but cannot communicate during the play. ∗ The article was in part funded by RFBR according to the research project 19-01-00563. k, n )-game is called a ( k, n ) -strategy . The minimal k for which there is a ( k, n )-strategy is called the costof the n -game .The following proposition is very important for the study of the strategiesfor the GKS game: Proposition 1.
If there are a ( k, n ) -strategy and a ( k ′ , n ′ ) -strategy, thenthere exists a ( kk ′ , nn ′ ) -strategy. The proof of this proposition can be found in [2]. This result allows to use( k, n )-strategies constructed for small values of n to obtain upper boundsfor the cost of the n -game for arbitrary n . More precisely, if a ( K, N )-strategy exists, then the cost of the n -game is O ( n log N K ), where the constanthidden in O-notation depends on K, N . Szegedy proved a O ( n . ) upperbound using (5 , O ( n . ) bound is derived froma (11 , Theorem 2.
There exists a (9 , -strategy.Proof. The proof relies heavily on a table that is presented in AppendixA and was found by a computer search. This table consists of 220 binarystrings of length 12. We call those strings codewords . In every codewordthree positions are underlined, all those positions contain ones. An im-portant property of the table is that every possible combination of threeunderlined positions out of (cid:0) (cid:1) = 220 combinations appears exactly once.Another property that is important for us is the following:
Proposition 3.
Let x and y be different codewords. Let x ′ and y ′ be stringsobtained from x and y , respectively, by flipping at most one non-underlinedbit. Then x ′ = y ′ . Informally this property can be understood as follows: the code correctsone error provided it appears in a non-underlined position. This is where ourconstruction differs from that proposed by Ingram [3], who used Hammingcodes that correct one error (in any position). Unfortunately, it is ratherdifficult to verify this property by hand and we cannot provide any proof ofthis fact. We verified this property by a computer program. The table itselfwas also found by that program. 2e will now construct a (9 , differs from theassigned codeword.Observe that every block completed by Alice alone is a codeword withone error in a non-underlined position. From Proposition 3 we can concludethat every such block is different from each codeword. So Bob can receivetwo kinds of strings:1. Either the received string has a block which is a codeword. In thatcase Bob knows that this block contains the last filled position and thatthat position is not underlined. Thus, he forms the set of 9 non-underlinedpositions in that block.2. Or, there is no such block in the received string, that is, all the blocksare codewords with one error, and that error occurs in a non-underlined po-sition. Then, due to Proposition 3, for each block Bob can find the assignedcodeword and the position of error, that position was filled last. The set ofthose positions has size 9 and contains the last bit.In both cases Bob can find a set of 9 positions containing the last bit. Corollary 4.