By Kai-Uwe Sattler
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If is a winning m0 -state and is a borderline winning `-state for some 1 Ä ` Ä n 1, then i ` C 3. Induction Step. Let be a borderline n-state with n > 1. Suppose (absurdum hypothesis) that there exists a winning strategy of size n C 2 for . Let ı D Œa0 ; a1 ; : : : ; ae be the first question in such a strategy. 0; a0 ; : : : ; ae 1 /. x0no ; : : : ; xeno / the two possible states resulting from Carole’s answer to the question ı when Paul’s state of knowledge is . Similarly, denote by yes and no , the two possible states resulting from Carole’s answer to the question ı 0 when Paul’s state of knowledge is .
Indeed, we have V2e i j. /D e i X 2e kD0 D e i X 2e kD0 X 2e i j D kD0 i k j i k 2e j i k ! C e j X 2e i k kD0 ! X 2e i j C j ! 2e i k i j kDe i C j 2e e i ! j ! 1. The following theorem yields a lower bound on the size of the smallest winning strategy in a game with e C 1 lies, given the size of the smallest winning strategy in a game with e lies. 3 (Translation Bound). x0 ; : : : ; xe 1 ; xe / with ej D0 xj 3. 0; x0 ; : : : ; xe 1 / is a borderline winning nstate (n > 0), then m n C 3. Proof.
0 / 1 question of type ı 0 D Œ0; : : : ; 0; 1; 0; : : : ; 0; 2ch. / 1 . j D0 j 0 By recursively applying questions of type ı , Paul perfectly gets through to the end. This theorem has an immediate consequence on the existence of perfect strategies for specific instances of the game: Let us fix two integers e; m 0. Let S be a search space of cardinality M D 2m . Then, up to finitely many exceptional m’s, there exists a perfect winning strategy for Paul. Stated differently, Paul can win the game over S with e lies using n questions, with n being the smallest integer satisfying the Volume Bound.
Algorithmen und Datenstrukturen by Kai-Uwe Sattler