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13

Decrypt the ciphertext with every possible key and store the result: $2^{56}$ decryptions. Now encrypt the (known) plaintext of the ciphertext with every possible key: $2^{56}$ encryptions. Now you have to check every entry, which is in both lists and try it with another plaintext-ciphertext pair. If you can successfully decrypt that, you are very likely to ...


9

Re-using their design might be no good idea - there are cheaper designs for sure. This new DES cracker would just need to try every possible key - like the one of the EFF already did. DES was a big standard for encryption, so some people did build such machines, right? Of course did they: COPACOBANA is able to break DES in under 9 days and costs under ...


7

In computer science, and implementation of crypto, ROTL stands for ROTate Left. ROTL is also noted ROL, or RLNC for Rotate Left No Carry. On a $w$-bit word with bits numbered from $0$, bit number $j$ of the input of ROTL with a shift count of $n$ goes to bit $j+n\bmod w$ of the result; $n=1$ unless otherwise specified (and is the only value available on ...


7

DES is slow in software because it was designed back in the early 70's even before the 8086 processor existed, and uses several bit oriented operations that are just not implemented efficiently in a processor with a word oriented instruction set. Its intended product was ASIC hardware designs, in which DES runs quickly. DES hardware processors are quite ...


6

DES with 2 rounds is broken. It is trivial to find a way to get the key with much less work than for the full DES (and even that is broken). DES is a Feistel cipher, so we have two halves, the left and the right half. For every round, we do something with the one half and a subkey, and then XOR it with the other half. After that we switch both halves, ...


6

No, that is impossible. The reason is simple: How would you decrypt this? If you input the ciphertext and the key into the decryption function, than you have to get exactly one output, not two. How would you decide which output is the correct one? The DES encryption and decryption functions are bijective under one given key. This means that for every ...


6

DES is slow compared to AES including in hardware because for comparable security we must use 3DES, which triples the number of rounds per block, to 48 for 3DES versus 10, 12, or 14 for AES; DES's block size is 64 bits, half of AES's 128 bit; so when encrypting a sizable block of data, 3DES does more rounds that AES by a factor of 96/10, 96/12, or 96/14; ...


5

Given a function $F: A \rightarrow B$ and functions $R_1, R_2, \dots, R_k:B \rightarrow A$, we can create a chain of length $k$ from a starting point $a_0$ to an end point $a_k$ using $a_i = R_i(F(a_{i-1}))$. A rainbow table for $(F, R_1, \dots, R_k, k)$ is a collection of chains with end points $(a_0, a_k)$ organized so that searching for chains ending at ...


5

I add my whitebox AES implementation on GitHub in: C++ Java C++ version implements both Chow's (mixing bijections, input/output encodings, external encodings) and Karroumi's (dual AES in each column) whitebox AES scheme plus Billet's key recovery attack on both schemes. Java implements Chow's scheme only. PS: Due to low reputation I post links to ...


4

One simple approach is to truncate the output to 56 bits. I believe this was considered in Hellman's original paper on time-space tradeoffs. Sometimes people get all excited by rainbow tables (partly because it has a cool name, maybe) but forget about Hellman's original paper on the time-space attack. Hellman's paper is very much worth reading, especially ...


4

Yes, it would be more secure if they were used correctly. But as it would require a substantially different algorithm, you really would not be talking about DES anymore. Brute forcing usually scales exponentially with the size of the key. However, if the algorithm is substantially altered then it is required to analyze the algorithm again. Note that AES is ...


4

If you mean DES as block cipher without mode of operation then no, this is impossible. DES is a block cipher, and block ciphers are Pseudorandom Permutations (PRP). As permutations in turn are bijective functions of $\{0,1\}^n$ to $\{0,1\}^n$ there is always a one to one relationship between plaintext and ciphertext. If this wasn't the case then you would ...


3

It looks like there's an error in the test vector. The text of Appendix B.1 states: P1 = “The quic” = 5468652071756663 ... which is incorrect. The hex encoding of The quic is actually 5468652071756963 (note the transposition of the i/69 to an f/66 in the encoding. e.g. encrypting the test vector as intended: $ echo -n 'The quick brown fox jump' | ...


3

Yes, it can; within the DES round function, two different 'right side' inputs can, after the sboxes, come up with the same value to xor into the 'left side'. This was a deliberate decision by the DES designers, who thought that this was an important property. I don't know their reasoning about why they thought it was important.


3

Actually, because of DESX works, the meet-in-the-middle attack can be optimized to take $2^{119}$ $DES$ (and no $DES^{-1}$ operations), and no additional storage. Here is how it works: let us assume that we have three known plaintext/ciphertext pairs $(P_1, C_1)$, $(P_2, C_2)$ and $(P_3, C_3)$. We know that: $$C_1 = K_2 \oplus DES( K, K_1 \oplus P_1 )$$ ...


3

It seems to me you can do everything as when calculating a rainbow table for a hash function, except that choosing a good reduction function is very easy. For example, define a chain starting from $k$ as: $$c_k(0) = T(E_k(0))$$ $$c_k(i) = T(E_{c_k(i-1) \oplus i}(0)),$$ where $T$ truncates its input to 56 bits. Now you can create a rainbow table with $n$ ...


3

If the plaintext format is indeed as you describe, then you're out of luck: the insertion of the newlines and the consequent shifting of the plaintext records is enough to disrupt any structure in the ciphertext. If the plaintext were longer, say, 8 records, then it could work, but with just 7 records there's no way to switch the first and last record ...


3

You cannot encrypt 720 bits plaintext using just AES-128. AES is a 128 bit block cipher. Such a block cipher has an input of 128 bits of plaintext and an output of 128 bits ciphertext; and that's it. You need some kind of construction to make block ciphers encrypt larger or smaller plaintext. Such constructions are known as (block cipher) modes of ...


2

According to the following link (Slide 5) and to what I studied last semester, http://www.ee.ic.ac.uk/pcheung/teaching/ee4_network_security/L02DESIDESAES.pdf During the final round (Round 16) before the inverse permutation, the left and right halves of the bits will be swapped then the inverse permutation will be applied.


2

This is known as the "key complementation" property of DES; I had thought that it actually predated Biham and Shamir's work. In any case, your questions: Does this hold for only that particular combination of s box or it will be same for any S-box combination It'd remain even if you change the sbox's arbitrarily. The reason for this is that it is not ...


2

Put another way, you can say that the key is whatever information the recipient possesses which allows him to decrypt the message, and which must be kept secret from everybody else. Thus, "algorithm" and "key" are not mutually exclusive: if knowledge of the algorithm allows one to decrypt a message, then the algorithm is the key.


2

Picking up what has been said in the comments: to simplify: symmetric ciphers are like mathematical operations with 2 operands and 1 result. There is The plaintext message $m$ and $k$ as the key and they result in the ciphertext $c$. In your example, the algorithm can be cut down to a addition and modulo: $c = (m + k) \mod k_{max}$ And of course there is ...


2

The key space for DES is far too small (56 bits). Therefore, any use of DES is not secure. It doesn't matter what mode you use. If the attacker has one plaintext, ciphertext pair, they can brute force the key space and recover the key in a feasible amount of time (24 hours using the cloud). But most importantly, how it could be made secure? Will change ...


2

I understand that if a block cipher has $k$-bit keys and $n$-bit input/output blocks, then if $k>n$, we can expect one message-ciphertext pair to narrow us down (I think?) to $2^{k−n}$ possible keys, right? That is approximately correct (if the block cipher with the wrong key acts like a random permutation; this is generally a safe assumption); if ...


2

That's an optimization for the attack. It would work without it, but slower. To do a Meet-in-the-middle attack, we need to encrypt a known plaintext with every possible key and save the resulting text (with the used key) in a list. Now we decrypt a known ciphertext with every possible key and look if we got the resulting text in our list. We want to know ...


2

Feistel networks were broken in DES but not triple DES. Some final AES candidates not approved also used Feistel networks $2^{36}$ plain text attacks. Reduction of $2^{16}$ possible keys for single DES: $4^{48/6} = 4^{8} = 2^{16}$. First for a one round Feistel network: $R_0$ and $f (R_O, k_1) = R_1 \oplus L_0$, $k_1$ becomes known. For two round Fiestel: ...


2

In short, yes. The complementation property of DES states that if $DES_K(P) = C$, then $DES_{\overline{K}}(\overline{P}) = \overline{C}$, where $\overline{X}$ is the complement of a string $X$. ECB with DES takes a message $M_1M_2\cdots M_\ell$ and computes $C_1C_2\cdots C_\ell$, where $C_i = DES_K(M_i)$, for $1\le i\le\ell$. Therefore, if you encrypt ...


2

According to Wikipedia, GCM is defined for block ciphers with a block size of 128 bits. So no, you can't use GCM with 3DES or DES, because of the 64-bit block size. You could use something similar to GCM, but it wouldn't be GCM.


1

Including Parity bits absolutely does not increase the cryptographic strength of DES (or 2-key TDES, or 3-key TDES). The DES feistel network itself only accepts eight 7-bit blocks as input, no more and no less. You can add as many parity as you want, but that doesn't mean any more bits are being processed by the algorithm. Parity bits are either used ...


1

First, note that $192=3\cdot64$, so the real key length of 3DES is $192$ bits. However, since $8$ bits in each subkey are parity bits, this reduces to $3\cdot56=168$ bits of non-redundant key material. Now, the reason that 3DES' effective key length is usually classified as $2\cdot56=112$ bits is that 3DES is susceptible to a meet-in-the-middle attack: When ...



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