Tags: linear_algebra
Rating:
A group of sheaves of grain stood on end in a field!
The description makes this challenge sound complicated, with abstract algebra and algebraic geometry but the challenge is actually quite simple
The cipher is basically doing c=k1/m+k2 mod b where these are polynomials instead of integers, however most normal operations works for polynomials.
We are also given a lot of m,c pairs, though only 2 was needed
Furthermore everything is done in GF(2), which simplifies a lot, addition and subtraction are also the same thing which is quite convenient
Using 2 pairs of m,c, we get(under mod b):
c1 = k1/m1 + k2
c2 = k1/m2 + k2
c1m1 = k1 + k2m1
c2m2 = k1 + k2m2
k2 = (c1m1+c2m2)/(m1+m2)
k1 = m1*(c1+k2)
Now using these keys, we can easily retrive the flag
Flag :
CCTF{GF2_F1nI73_Crc13_f1elds}
