Timeline for Hardness of problem related to bilinear pairings
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2015 at 15:33 | vote | accept | cygnusv | ||
| Jan 8, 2015 at 10:36 | comment | added | DrLecter | @cygnusv Yes, GPI is not harder than FAPI. If $GPI \implies FAPI$, then the contraposition is $\neg FAPI \implies \neg GPI$. So if $FAPI$ does not hold if follows that $GPI$ does not hold (if you have an oracle for FAPI you can solve GPI). | |
| Jan 8, 2015 at 10:09 | comment | added | cygnusv | OK, I think it is a problem of nomenclature. I thought that if you use an oracle P to solve problem Q, then "P implies Q". I have seen lots of references following this nomenclature, but also the other way around... At least, we agree on that GPI is not harder than FAPI, right? | |
| Jan 8, 2015 at 10:00 | comment | added | DrLecter | @gygnusv Nope, you get an GPI instance $z$, sample one element $h$ from $G$ randomly and call the FAPI oracle (which returns $f$) and output $z,h,f$. | |
| Jan 8, 2015 at 9:55 | comment | added | cygnusv | Thanks for the links! Now I know my problem is the Generalised Pairing Inversion. One comment: I think that FAPI implies GPI, and not the other way around, since FAPI fixes ones of the arguments to the pairing. | |
| Jan 8, 2015 at 9:46 | history | answered | DrLecter | CC BY-SA 3.0 |