Timeline for Matrix Trapdoor AB+BA
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 24, 2017 at 3:09 | review | Close votes | |||
| Sep 24, 2017 at 12:49 | |||||
| Sep 20, 2017 at 1:21 | vote | accept | DannyNiu | ||
| Sep 19, 2017 at 15:54 | comment | added | poncho | Is the trapdoor the knowledge of $D$, where $a = P \cdot D \cdot P^{-1}$? If so, then $D = P^{-1} \cdot a \cdot P$ is easy to recover... | |
| Sep 19, 2017 at 13:49 | answer | added | poncho | timeline score: 2 | |
| Sep 19, 2017 at 10:43 | history | edited | DannyNiu | CC BY-SA 3.0 |
edited title
|
| Sep 19, 2017 at 10:34 | history | edited | DannyNiu | CC BY-SA 3.0 |
added 7 characters in body
|
| Sep 19, 2017 at 10:24 | comment | added | poncho | If the dimension of $R$ is $n \times n$, isn't the operation $a \cdot b + b \cdot a$ for public $a$ linear in terms of the $n^2$ elements that make up the matrix $b$? Can't you then express $u = a \cdot b + b \cdot a$ as $n^2$ linear equations, and solve it in $O(n^6)$ time? | |
| Sep 19, 2017 at 8:36 | history | edited | DannyNiu | CC BY-SA 3.0 |
Restructure.
|
| Sep 19, 2017 at 6:04 | history | edited | DannyNiu | CC BY-SA 3.0 |
added 427 characters in body
|
| Sep 19, 2017 at 1:33 | history | edited | DannyNiu | CC BY-SA 3.0 |
de-confuse.
|
| Sep 19, 2017 at 1:23 | history | edited | DannyNiu | CC BY-SA 3.0 |
Clarifying what the trapdoor is.
|
| Sep 19, 2017 at 1:11 | history | edited | DannyNiu | CC BY-SA 3.0 |
Rewording to remove "generator".
|
| Sep 19, 2017 at 1:02 | history | edited | DannyNiu | CC BY-SA 3.0 |
details clarification.
|
| Sep 19, 2017 at 1:01 | comment | added | DannyNiu | $x=PDP^{-1}$ isn't necessarily a diagonal matrix, it's just some matrix $x \in S$. $R$ is a matrix ring. | |
| Sep 18, 2017 at 18:25 | comment | added | Aleph | It is not clear to me if $R$ is supposed to be a matrix ring, or just any commutative ring. Over what ring is the matrix $D$? I don't think you're using the term generator in the usual meaning (i.e. $S$ is actually cyclic), so please clarify. | |
| Sep 18, 2017 at 12:15 | review | Close votes | |||
| Sep 18, 2017 at 12:45 | |||||
| Sep 18, 2017 at 11:56 | comment | added | tylo | First, $S$ is commutative and diagonal matrices are commutative, then $P D P^{-1} = D$, so you defined $S$ as the subgroup of diagonal matrices. Secondly: Why do you think it's hard? What is the trapdoor here? And why is the problem much easier when knowing the trapdoor? In the current state, the question might be better suited for math-SE, there isn't much cryptography so far. | |
| Sep 18, 2017 at 11:47 | comment | added | DannyNiu | @fgrieu yes, now that you've made it explicit. | |
| Sep 18, 2017 at 10:40 | comment | added | fgrieu♦ | I guess it is restricted to finite ring, as common in crypto. | |
| Sep 18, 2017 at 9:06 | history | asked | DannyNiu | CC BY-SA 3.0 |