Timeline for How to attack Oblivious Transfer from a malicious sender that can deviate from the protocol
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2021 at 18:42 | vote | accept | Gabi G | ||
| Jan 1, 2021 at 14:01 | comment | added | poncho♦ | @GabiG: well, if $b=0$, then $v - x_0 = k^e$, Alice doesn't know what $k$ is, but it will exist, and hence $v - x_0$ will have a $e$th root. In contrast, $v - x_1$ is effectively a random value, and random values have an aproximately $1/e$ probability (if $e$ is prime and $e$ is a factor of only one of $p-1, q-1$; otherwise, it's more complicated) of having an $e$th root, that is, being $k'^e$ for some value $k'$ | |
| Jan 1, 2021 at 11:54 | comment | added | Gabi G | Thanks! Could you clarify why the correct one will be an $e$th root, and the wrong one will likely not be? | |
| Dec 31, 2020 at 14:05 | history | answered | poncho♦ | CC BY-SA 4.0 |