Single DES has been cracked. Double DES is not a solution either; briefly explai
ID: 3146026 • Letter: S
Question
Single DES has been cracked. Double DES is not a solution either; briefly explain how an attack in double DES can be done?
What is the output of the first round of the DES algorithm when the plaintext is 0x00000000000000AB and the key is 0x000000000000C2 ? Show all work; i.e. list the steps of the algorithm and give the input/output of the different functional units when going through these steps. Simply providing an answer earns no points.
The DES algorithm provides confusion and diffusion. Define confusion and define diffusion. From your steps in part b, note if confusion, diffusion or neither is provided by the step
Explanation / Answer
A)))))
Double-DES is two successive DES instances.
We do not use 2DES because 2DES does not yield the security increase that you would believe. Namely, 2DES uses 112 key bits (two 56-bit DES keys) but offers a security level of about 257, not 2112, because of a "meet-in-the middle attack". Hence, we use 3DES as it uses 168 key bits, but offers "only" 2112 security (which is quite sufficient in practice).
Hence, we don't use double DES since we aim to use n-DES where n>2.
This can be summarized as: we use n-DES because a simple DES is too weak (a 56-bit key can be brute-forced by a determined attacker), but in order to really improve security, we must go to n 3. Of course, every additional DES implies some computational overhead (simple DES is already quite slow in software, 3DES thrice as much).
B) 64-bit input is X0=00...0 (64 zero). The first permutations have without effect. So L0=00...0 (32 zero) and R0=00...0 (32 zero). Apply the key list which is a permanent variation on the input bits of the type fields the about key K1= (00...0) (48 zero).
The around calculate R1=L0 xor f (R0, K1)
The f task
Primary expand R0 keen on 48 bit lengthy bit string use a set permuted growth law. Since only permutation and repetition are use these resolves field a 48 bit zero string.
The outcome is XOR’s with K1, which produce a 48 bit 0 string.
The 48 bit zero string is alienated into eight 6 bit chunk plus the it chunk is distorted below the rule specify in the Si box. 000000 is map 8 times with each box Si i=1...8 and produce the follow sequence of 4 bit value: 14, 15, 10, 7, 2, 12, 4, 13. In binary we get the follow series 1110 1111 1010 0111 0010 1100 0100 1101
Lastly the bit string is permuted according to the P chart to the subsequent:1101 1000 1101 1000 1101 1011 1011 1100
This is the outcome of f (R0), K1)
The exact half R1=L0 XOR f (R0), K1) is cleanly the output of f (R0),K1) while the L0 is 0. L1 = R0 = (00...0) (32 zeroes). Concatenates both fields the next 64 bit string, which is the output of around 1.
0000 0000 0000 0000 0000 0000 0000 0000 1101 1000 1101 1000 1101 1011 1011 1100