Algorithms_Part17 - S. Dasgupta, C.H. Papadimitriou, and...

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Unformatted text preview: S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 321 Let’s make this more precise. Lemma Suppose s independent samples are drawn uniformly from , M k , 2 M k , . . . , ( k- 1) M k . Then with probability at least 1- k/ 2 s , the greatest common divisor of these samples is M/k . Proof. The only way this can fail is if all the samples are multiples of j · M/k , where j is some integer greater than 1 . So, fix any integer j ≥ 2 . The chance that a particular sample is a multiple of jM/k is at most 1 /j ≤ 1 / 2 ; and thus the chance that all the samples are multiples of jM/k is at most 1 / 2 s . So far we have been thinking about a particular number j ; the probability that this bad event will happen for some j ≤ k is at most equal to the sum of these probabilities over the different values of j , which is no more than k/ 2 s . We can make the failure probability as small as we like by taking s to be an appropriate multiple of log M . 10.5 Quantum circuits So quantum computers can carry out a Fourier transform exponentially faster than classical computers. But what do these computers actually look like? What is a quantum circuit made up of, and exactly how does it compute Fourier transforms so quickly? 10.5.1 Elementary quantum gates An elementary quantum operation is analogous to an elementary gate like the AND or NOT gate in a classical circuit. It operates upon either a single qubit or two qubits. One of the most important examples is the Hadamard gate, denoted by H, which operates on a single qubit. On input , it outputs H ( ) = 1 √ 2 + 1 √ 2 1 . And for input 1 , H ( 1 ) = 1 √ 2- 1 √ 2 1 . In pictures: 1 √ 2 + 1 √ 2 1 H 1 H 1 √ 2- 1 √ 2 1 Notice that in either case, measuring the resulting qubit yields with probability 1 / 2 and 1 with probability 1 / 2 . But what happens if the input to the Hadamard gate is an arbitrary superposition α + α 1 1 ? The answer, dictated by the linearity of quantum physics, is the superposition α H ( ) + α 1 H ( 1 ) = α + α 1 √ 2 + α- α 1 √ 2 1 . And so, if we apply the Hadamard gate to the output of a Hadamard gate, it restores the qubit to its original state! Another basic gate is the controlled- NOT , or CNOT . It operates upon two qubits, with the first acting as a control qubit and the second as the target qubit. The CNOT gate flips the second bit if and only if the first qubit is a 1 . Thus CNOT ( 00 ) = 00 and CNOT ( 10 ) = 11 : 322 Algorithms 00 00 10 11 Yet another basic gate, the controlled phase gate, is described below in the subsection describing the quantum circuit for the QFT. Now let us consider the following question: Suppose we have a quantum state on n qubits, α = ∑ x ∈{ , 1 } n α x x . How many of these 2 n amplitudes change if we apply the Hadamard gate to only the first qubit? The surprising answer is—all of them! The new superposition becomes β = ∑ x ∈{ , 1 } n β x x , where β y = α y + α 1 y √ 2 and β 1 y = α y- α 1 y √ 2 . Looking at the...
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Algorithms_Part17 - S. Dasgupta, C.H. Papadimitriou, and...

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