Timeline for Does changing the random number selected for each message increase security in Schnorr's scheme?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2017 at 16:52 | comment | added | Ethan Heilman | I figured out what I was doing wrong last night. I am not used to some of the quirks of EC notation. $kG$ is actually a different operation than say $kx$. Finding $G^{-1}$ would break discrete log, finding $k^{-1}$ is trivial. | |
| Jan 16, 2017 at 8:30 | comment | added | DrLecter | @EthanHeilman Note that $(e'-e)$ can easily be inverted as it is in $\mathbb{Z}_q$. The point is that this helps us computing $x$ using the last equation above. Your equation $x=(kG)(G^{-1})$ makes no sense. Is $G$ your basepoint from an elliptic curve? What is $k$ in your equation? Anyways, let us write $G^{-1}$ as $-G$ (as you write the group additively) and what you get is: $kG-G=(k-1)G$. That doesn't help you. | |
| Jan 15, 2017 at 21:05 | comment | added | Ethan Heilman | Doesn't finding $(e′−e)^{−1}$ s.t $x(e′−e)(e′−e)^{−1} \equiv x$ involve finding the discrete log of $x(e′−e)$. For instance if find these inverses is easy why not find $G^{-1}$ so that you can determine the secret key $x = (kG)(G^{-1})$. | |
| Apr 1, 2014 at 18:55 | history | edited | DrLecter | CC BY-SA 3.0 |
deleted 43 characters in body
|
| Feb 1, 2014 at 23:48 | history | edited | DrLecter | CC BY-SA 3.0 |
added 13 characters in body
|
| Feb 1, 2014 at 23:42 | history | edited | DrLecter | CC BY-SA 3.0 |
added 59 characters in body; added 13 characters in body; deleted 12 characters in body
|
| Jan 28, 2014 at 22:24 | history | edited | DrLecter | CC BY-SA 3.0 |
added 47 characters in body
|
| Jan 26, 2014 at 18:57 | history | edited | DrLecter | CC BY-SA 3.0 |
added 24 characters in body
|
| Jan 26, 2014 at 18:48 | history | answered | DrLecter | CC BY-SA 3.0 |