Timeline for Combining multiple PRGs together
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 29, 2016 at 18:09 | comment | added | Jjang | Please? 2 weeks have passed and I have no clue. | |
| Apr 9, 2016 at 17:11 | comment | added | Jjang | Why does G2(0^|s|) imply that e,f,g are secure PRGs? Constant or not, I showed G1 such that G is always 1... how can G be safe? And also, got any idea about h,i? | |
| Apr 7, 2016 at 22:50 | comment | added | Suphanat Chunhapanya | I thought, for c, it should be $G_2(s) = G_1(s \oplus 1^{|s|}) \oplus 1$. For d, it should be $G_2(s) = \overline{G_1(s \oplus 1^{|s|})} \oplus 1$. For e,f,g, they are secure PRGs, since $G_2(0^{|s|})$ is just a constant. Finally, you should be careful with this sentence "XOR with s must also be random". | |
| Apr 7, 2016 at 19:49 | comment | added | Jjang | Regarding i, I got a little idea that might show that it IS a PRG. Since G is PRG, then the first n bits of G are also random, XOR with s must also be random, so we got random seed as an input to the outer G which is PRG, so G1 is PRG. | |
| Apr 7, 2016 at 19:42 | comment | added | Jjang | Thanks alot for the direction. I'd like to verify my answer with you. So b can be easily solved using the above method. in c-d we use the same idea but we pick G2(s XOR 1) = G1(s XOR 1) XOR 1, so we get G = 1. In e we choose G1 = 𝐺2(0^|𝑠|) XOR 1 so G = 1. f is the same, we pass the above G1 but with NOT upperline (so it cancels each other). In g, the solution is like e. Am I right? P.S. no clue regarding h and i. | |
| Apr 6, 2016 at 22:34 | review | First posts | |||
| Apr 6, 2016 at 23:32 | |||||
| Apr 6, 2016 at 22:32 | history | answered | Suphanat Chunhapanya | CC BY-SA 3.0 |