== code(c) |
def 4sum(A):

  S = {}

  sums = {}

  for a in A:
    for b in A:

      S[a + b] = (a, b)

      sums[a + b] = (a, b)

  for c in A:
    for d in A:
      if -(c + d) in sums:

        print (S[-(c + d)][0], S[-(c + d)][1], c, d)

        print (sums[-(c + d)][0], sums[-(c + d)][1], c, d)

        return
  print "No solution."

For the pattern p it tries all the possible keys to obtain a set of numbers corresponding Ek(p). Also for the pattern s it uses all the possible keys to decrypt s, Dk(s).
If we find any match in the two sets it means that Eki(p) = Dkj(s) so the secret keys are ki and kj.
The naive brute force algorithm does $2^56^ * 2^56^$ iterations going through all possible values of k1 and k2 while this algorithm uses $2^56^ * 56$ memory to store all Eki(p) and does $2^56^$ work to find a match.

This is quite a bit of space and quite a bit of computation time. But a for a large enough company or country it starts being within the realm of posibility.

Breaking 2DES was basically the DOUBLE problem from the International Olympiad in Informatics 2001.

The problem DOUBLE the International Olympiad in Informatics 2001 was basically asking to break 2DES for keys of size 24 which is quite feasible.

h2. Discrete logarithm

The naive solution goes through all possible values of k and takes $O(n)$ time.
The baby-step, giant-step algorithm solves the problem more efficiently using the meet in the middle trick.

Let's write k = $i([sqrt(n)] + 1) + j$

Let's write k = $i[sqrt(n)] + j$

Notice that $i <= sqrt(n)$ and $j <= sqrt(n)$.

Replacing k in the equality we get $p^(i ([sqrt(n)] + 1) + j)^ = q (mod n)$.
Dividing by $p^j^$ we get $p^(i[sqrt(n)] + 1)^ = qp^-j^ (mod n)$.

Replacing k in the equality we get $p^(i[sqrt(n)] + j)^ = q (mod n)$.
Dividing by $p^j^$ we get $p^i[sqrt(n)]^ = qp^-j^ (mod n)$.

At this point we can brute force through the numbers on each side of the equality and find a colision.
The algorithm takes $O(sqrt(n))$ space and $O(sqrt(n))$ time.

public

protected