Timeline for Rabin-Williams, blinding and size of Integer r?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:48 | history | edited | CommunityBot |
replaced http://crypto.stackexchange.com/ with https://crypto.stackexchange.com/
|
|
| Feb 26, 2016 at 13:24 | comment | added | fgrieu♦ | @bayo15: assuming $(e\cdot d\bmod\varphi(n))=1$ or equivalently $e\cdot d\equiv1\pmod{\varphi(n)}\;$, the equality $(r^{e\cdot d}\bmod n)=r$ holds for integers $r$ with $0\le r<n$, while the congruence $r^{e\cdot d}\equiv r\pmod n$ holds for all integers $r$. I can't think of another interpretation of mathematically correct. | |
| Feb 26, 2016 at 12:15 | comment | added | bayo |
It's is obvious that r^ed mod n = r, as e*d mod phi(n) = 1 so this concept works (equals sign means "is congruent to"). However, what are the formal requirements for the random variable r so this is still mathematicaly correct? Thanks.
|
|
| Apr 27, 2015 at 9:38 | history | edited | fgrieu♦ | CC BY-SA 3.0 |
State another advantage of blinding with public exponent
|
| Apr 27, 2015 at 5:55 | history | edited | fgrieu♦ | CC BY-SA 3.0 |
polish
|
| Apr 27, 2015 at 5:49 | history | edited | fgrieu♦ | CC BY-SA 3.0 |
Add discussion about unusually large e, following comment
|
| Apr 27, 2015 at 5:16 | history | edited | fgrieu♦ | CC BY-SA 3.0 |
expand
|
| Apr 27, 2015 at 5:09 | comment | added | user10496 | "...requires computing r^d (mod N), which is just as costly as the m^d (mod N) operation being protected" - right, that's why I thought reducing hamming weight might be helpful. "So all in all there's no appreciable savings..." - perfect, thanks. | |
| Apr 27, 2015 at 5:09 | vote | accept | CommunityBot | moved from User.Id=10496 by developer User.Id=76 | |
| Apr 27, 2015 at 5:07 | history | answered | fgrieu♦ | CC BY-SA 3.0 |