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These are the only cases where the host opens door 3, so the conditional probability of winning by switching given the host opens door 3 is 1/3 / 1/3 + q/3 which simplifies to 1 / 1 + q . Since q can vary between 0 and 1 this conditional probability can vary between 1 / 2 and 1. This means even without constraining the ...
The 3-input majority gate output is 1 if two or more of the inputs of the majority gate are 1; output is 0 if two or more of the majority gate's inputs are 0. Thus, the majority gate is the carry output of a full adder, i.e., the majority gate is a voting machine. [7] The 3-input majority gate can be represented by the following boolean ...
For an arbitrary n there exists a monotone formula for majority of size O(n 5.3). This is proved using probabilistic method. Thus, this formula is non-constructive. [3] Approaches exist for an explicit formula for majority of polynomial size: Take the median from a sorting network, where each compare-and-swap "wire" is simply an OR gate and an ...
They have but one recourse – their heads explode.") Allowing the "exploding head" case gives yet another solution of the puzzle and introduces the possibility of solving the puzzle (modified and original) in just two questions rather than three. In support of a two-question solution to the puzzle, the authors solve a similar simpler puzzle ...
B is pardoned and the warden mentions C to be executed: 1 / 3 of the cases; C is pardoned and the warden mentions B to be executed: 1 / 3 of the cases; With the stipulation that the warden will choose randomly, in the 1 / 3 of the time that A is to be pardoned, there is a 1 / 2 chance he will say B and 1 / 2 ...
The Ages of Three Children puzzle (sometimes referred to as the Census-Taker Problem [1]) is a logical puzzle in number theory which on first inspection seems to have insufficient information to solve. However, with closer examination and persistence by the solver, the question reveals its hidden mathematical clues, especially when the solver ...
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A variant of the 3-satisfiability problem is the one-in-three 3-SAT (also known variously as 1-in-3-SAT and exactly-1 3-SAT). Given a conjunctive normal form with three literals per clause, the problem is to determine whether there exists a truth assignment to the variables so that each clause has exactly one TRUE literal (and thus exactly two ...