Abstract
The number of two-qubit gates required to transform deterministically a three-qubit pure quantum state into another is discussed. We show that any state can be prepared from a product state using at most three CNOT gates, and that, starting from the GHZ state, only two suffice. As a consequence, any three-qubit state can be transformed into any other using at most four CNOT gates. Generalizations to other two-qubit gates are also discussed.
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