[SCTF 2018 Finals] LCG
Download the problem file from http://research.samsung.com/sctf2018 or https://www.dropbox.com/s/xx6tnhzrgpdxvd8/LCG.py?dl=0 It is quite simple PRNG with the equation (t = 0xdeadbeef): $x_i = (k_1 - t) x_{i-1} + k_1 t x_{i-2} + k_2 (mod\ k_3)$ We can define $y_i$ as $y_i = x_i + t x_{i-1}$, then $y_i = k_1 y_{i-1} + k_2 (mod\ k_3)$. So, it’s just same as the normal LCG. I used the method to break LCG described in this link, and the solver is here....