Timeline for A question about the definition of PRFs
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2018 at 7:58 | history | edited | Blanco | CC BY-SA 4.0 |
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| Mar 4, 2018 at 16:40 | history | edited | Blanco | CC BY-SA 3.0 |
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| Mar 4, 2018 at 7:36 | history | edited | Blanco | CC BY-SA 3.0 |
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| Mar 3, 2018 at 7:34 | history | edited | Blanco | CC BY-SA 3.0 |
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| Mar 3, 2018 at 7:33 | comment | added | Blanco | @D.W. You are right, I wrote the correct definition at the first time, and I decided to simplify it. And I did not realize that it is wrong. | |
| Mar 3, 2018 at 7:29 | vote | accept | Blanco | ||
| Mar 3, 2018 at 7:27 | history | edited | Blanco | CC BY-SA 3.0 |
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| Feb 27, 2018 at 16:33 | comment | added | D.W.♦ | There's something wrong with your question setup. A PRF takes two inputs: a value in $\{0,1\}^n$, and a key. Your function $f_n$ takes only one input. So, syntactically, it can't be a PRF. It's also not clear to me what you mean by "the distribution of functions $f_n$". You might want to check your understanding of the definitions. Once you fix that up, you'll find that a simple counterexample is $f(k,x) = k \oplus x$. | |
| Feb 27, 2018 at 8:37 | answer | added | Yuval Filmus | timeline score: 2 | |
| Feb 27, 2018 at 7:20 | history | asked | Blanco | CC BY-SA 3.0 |