3 ms·
There's no "answer" to precalculate, it's drawing from a distribution. Could you calibrate a non-quantum circuit to have a distribution passing the test? Suppos
by dooglius 7y ago
There's no "answer" to precalculate, it's drawing from a distribution. Could you calibrate a non-quantum circuit to have a distribution passing the test? Supposedly, this is not possible even given arbitrary pre-computation time.
- lisper 7y agoI can pre-calculate as many samples from the distribution as you can generate during the coherence time of the QC. Once you lose coherence, you have to re-calibrate, and during that time I can pre-compute another set of samples.
- dooglius 7y agoThat's true, if you pre-sample you can beat the test. Is that the effect you think the calibration is having? > Once you lose coherence, you have to re-calibrate I didn't realize this, where did you see that?
- lisper 7y ago> Is that the effect you think the calibration is having? I have no idea. All I know (or at least I'm pretty sure of) is that you have to re-calibrate every time you lose coherence, otherwise it would not be calibration, it would be a one-time process that would be considered part of the construction of the QC. > where did you see that? Nowhere. I inferred it. It's possible that I'm wrong, but I'll give you long odds against. I have a pretty good understanding of how QCs work.
- Barrin92 7y agoYou can bring the complexity on a classical machine down to linear by allowing for large disk space, that is precisely the claim that IBM makes in response to the argument.