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by Asdfbla
3030 days ago
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How long does a computation step in a quantum computer (which I guess is technically a measurement) actually take? I'm (at a superficial level) aware of Grover's and Shor's algorithms, but only in the sense that they provide asymptotic speedups in the number of steps required for the computation. In classical computers it's easy to envision some model where a step is just some constant number of cycles or whatever, but how long do the measurements take in quantum computers? I guess it depends on the specific experimental setup of your quantum computer somehow, but I'm just curious about the real-world speeds. A hypothetical computer that does few steps but takes a minute for each measurement wouldn't be so useful. |
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With error correction... it depends how many spare qubits you have. All the non-trivial operations are done by "magic state distillation", where you produce special states in one part of the computer then consume them to perform the operation elsewhere. The more qubits you have, the more space you can allocate to magic state factories.
For example, a 32-bit addition can be done with 124 |T> states [1]. If you have N qubits allocated to factories then you can expect to get about 100N magic states per second. So if you have a tiny computer with 20 qubits dedicated to a factory, then a single addition will take over half a second. But if you can spare 2000 qubits for factories (about the same amount of qubits you'd need to store the problem in the first place) then you'd be producing T-states at 200KHz and 32-bit-adding at 1600Hz.
Shor's algorithm is, effectively, a single modular exponentiation. Performing an N-bit modular exponentiation requires something like 4N^3 |T> states. So 1024^3*4/200KHz = 6 hours to a factor a 1024 bit key, assuming 2000 qubits on factories and that the other rough numbers are close.
1: https://arxiv.org/abs/1709.06648