network profile
| bio | website | sites.google.com/site/… |
|---|---|---|
| location | Alexandria, Egypt | |
| age | 22 | |
| visits | member for | 1 year, 1 month |
| seen | Apr 12 at 16:59 | |
| stats | profile views | 6 |
I am a Senior year student at Computer and Systems Engineering Department,
Faculty of Engineering,
Alexandria University, Egypt.
|
Apr 7 |
comment |
Predicting PRNG given some of its previous output I know that the range for (2x + 5) mod p is [0..p-1], but wouldn't that make x < (p-5)/2 that is (2^56-5)/2 ? and the maximum value for p will be (2^56-1), please correct me if i am getting anything wrong. |
|
Apr 7 |
comment |
Predicting PRNG given some of its previous output No, this is a mistake of mine, x and y and chosen randomly in [0..p-1] and the problem stated that p is 56-bits. |
|
Apr 6 |
comment |
Predicting PRNG given some of its previous output Thanks, for the correction. |
|
Apr 6 |
comment |
Predicting PRNG given some of its previous output Thanks @fgrieu, i have two points that i tried to figure out, but i couldn't. First, how did u limit the range of p to [2^55, 2^56] ... Second, also the same for Xj and Yj. |
|
Apr 6 |
comment |
Predicting PRNG given some of its previous output First, thanks a lot for the help. Second, i still have some points that i cant understand ... how did u determine the ranges for Kx and Ky ? ... Also, i don know whether i got the second step or not, so u mean that for all values of X [0, 2^56] and for all values of bit-0 of p {0, 1}, find 3 pairs of (Kx, Ky) that will generate the 2nd, 3rd and 4th outputs. |
|
Apr 1 |
comment |
How does one attack a two-time pad (i.e. one time pad with key reuse)? Thanks, a lot i solved the problem. I started with ' the ' as you said and got 'robab' which i guessed would be 'probably'. I kept doing this with some guesses until i solved the whole message. Thanks again. |
|
Apr 1 |
comment |
How does one attack a two-time pad (i.e. one time pad with key reuse)? Yes, your totally right. I was mistaken with that part, but it is my very first trial in cryptography. I got it now |
