Prove how to brute-force the double encryption with complexity 2^57 rather than 2^112?

Question

We know that DES has key length that is too short to defend brute-force attack. One idea to improve DES was to encrypt a message using DES twice with two different keys. Let E be the encryption algorithm of DES and k1, k2 be two DES keys. Such a double DES encryption performs like c = E(k2; E(k1; m)) We might hope now the total key length is 112 bits, enough for defending brute-force attack which should take 2112 steps. It turns out, such a system doesn’t provide that level of brute-force security. Given one pair of plaintext-ciphertext pair, show how to brute-force the double encryption with complexity 257 rather than 2112. This attack demonstrates that double symmetric ciphering doesn’t increase the security much. In fact, To get 112 bits security, Triple-DES (a.k.a 3-DES) was introduced and standarlised. You will see that in many security protocols, 3-DES remains an option.