The majority game with an arbitrary majority

Abstract

Abstract. The k-majority game is played with n numbered balls, each coloured with one of two colours. It is given that there are at least k balls of the majority colour, where k is a fixed integer greater than n/2. On each turn the player selects two balls to compare, and it is revealed whether they are of the same colour; the player's aim is to determine a ball of the majority colour. It has been correctly stated by Aigner that the minimum number of comparisons necessary to guarantee success is 2(n − k) − B(n − k), where B(m) is the weight of the binary expansion of m. However his proof contains an error. We give an alternative proof of this result, which generalizes an argument of Saks and Werman.

Searching for knights and spies: A majority/minority game

“…Choosing Y and Z so that |Y | ≥ |Z|, we define the weight of C to be |Y | − |Z|. The multiset of component weights then encodes exactly the same information as the 'state vector' in [1, page 5] or the 'game position' in [5,Section 2], [10, page 384] and [11,Section 3]. The following lemma also follows from any of these papers, and is proved here only for completeness.…”

“…The majority game. The values of K(n, k) were found for all n and k in [4], but many natural questions about the majority game remain open. Given a multiset M of component weights and e ∈ N such that the sum of the weights in M has the same parity as e, let n − V e (M ) be the minimum number of questions that are necessary and sufficient to find a knight starting from the position M , when spies always lie and the excess of knights over spies is at least e. Thus V e (M ) is the number of components in the final position, assuming optimal play.…”

“…The Switching Knight Hunt gives a lower bound in some of the remaining cases. In [4] a family of statistics SW e (M ) were defined, generalizing the statistic Φ(M ) = SW 1 (M ) used in [5]. In [4,Section 5] it was shown that V e (M ) ≤ SW e (M ).…”

“…Aigner also claims a proof, based on Lemma 5.1 in [11], that K L (n, k) ≥ 2(n − k) − B(n − k). A flaw in these proofs was pointed out in [5], and a correct proof was given. It is obvious that K S (n, k) ≥ K L (n, k), so it follows that K S (n, k) = K L (n, k) = 2(n − k) − B(n − k), giving part of Theorem 3.…”

“…Aigner also claims a proof, based on Lemma 5.1 in [6], that K L (n, k) ≥ 2(n − k) − B(n − k). A flaw in these proofs was pointed out in [4], and a correct proof was given. It is obvious that…”