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Timeline for How does quantum encryption work?

Current License: CC BY-SA 4.0

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Mar 14, 2022 at 15:42 history edited J.A.K. CC BY-SA 4.0
Added some nuance to the impact
S Feb 16, 2021 at 0:08 history suggested R1w CC BY-SA 4.0
Fixed grammar.
Feb 15, 2021 at 20:37 review Suggested edits
S Feb 16, 2021 at 0:08
Jan 29, 2017 at 18:37 comment added e-sushi A bit aside, but nevertheless related to the quantum cryptanalysis section: What does a “real” quantum computer need for cryptanalysis and/or cryptographic attack purposes? That Q&A goes well beyond the hyped D-Wave and might therefore represent a more valid reference for your “As far as we know.”
Jan 29, 2017 at 18:18 history edited J.A.K. CC BY-SA 3.0
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Jan 29, 2017 at 15:32 vote accept PallavBakshi
Jan 27, 2017 at 21:00 comment added floor cat It's worth saying that there isn't consensus yet on whether D-Wave really is a quantum computer. IBM has granted open access to their 5 qbit quantum processor though, so I think that stands as a better example of future threats. MIT is working on breaking RSA with a 5 qbit processor. Also worth noting, QKD can be compromised by spoofing distributions to match the Bell test.
Jan 27, 2017 at 16:30 history edited J.A.K. CC BY-SA 3.0
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Jan 27, 2017 at 16:25 history edited J.A.K. CC BY-SA 3.0
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Jan 27, 2017 at 16:18 history edited Biv CC BY-SA 3.0
use the $\latex$ formula ;)
Jan 27, 2017 at 16:15 history edited J.A.K. CC BY-SA 3.0
updated correction by commenter
Jan 27, 2017 at 16:03 history edited J.A.K. CC BY-SA 3.0
updated correction by commenter
Jan 27, 2017 at 15:00 history edited J.A.K. CC BY-SA 3.0
edited body. Spelling.
Jan 27, 2017 at 14:58 comment added J.A.K. I didn't know that, thanks so much :). My post was mainly to counterbalance the other post, which only went into QKD.
Jan 27, 2017 at 5:11 comment added Biv Quantum algorithms have about an impact of $O(n/2)$ on symmetryc crypto This is incorrect. We do not compare the key size in such a way via complexity. While the idea of the effective size of the key is divided by 2, the Grover algorithm allows a speed up of $O(\sqrt{n})$. Which, given a key size of $n$ bits, thus a search size of $2^n$ an effective search of $\sqrt{2^n} = 2^{n/2}$. The effective size of the key (if you were to compare to a pre-quantum size) is divided by 2, but the complexity of the search is $O(\sqrt{n})$ where $n$ is the size of the search space.
Jan 27, 2017 at 3:47 history answered J.A.K. CC BY-SA 3.0