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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 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, ...


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 ...


4

You need to split up your key into eight 7 bit pieces, and put these 7 bits into a byte each. The parity is in the least significant bit on most platforms, so the 7 bits need to go into the most significant bits. Of course, as the key is probably in bytes, you need to shift and combine the values in the bytes to retrieve the 7 bits. It's possible the ...


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

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

There are a couple of things going on: First of all, the DES key FF FF FF FF FF FF FF FF happens to be a "DES weak key"; by that, we mean that if you send a block through the cipher twice, it'll end up with the original value; that is: $$X = DES_{weak}( DES_{weak} ( X ))$$ You are obviously encrypting in CBC mode with a zero IV. So, let us look at what ...


3

You have a ciphertext (or maybe multiple), a list of possible plaintexts, but no key. Therefore, your process would be Generate random decryption key Decrypt ciphertext with that key (base64 decode CT first) See if result appears in your list of possible plaintexts If it does, return that plaintext; otherwise goto 1 This is a basic brute force attack and ...


3

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 ...


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

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.


2

Your problem is which part of the algorithm has to be transmitted (and you are missing an actual IV). Your encryption algorithm of using $TDEA(m)=E_{K_1}(D_{K_2}(E_{K_1}(m)))$ is fine for encrypting a single block, and it can be used in CBC. However, in CBC with a message of 4 blocks $m=(m_1,m_2,m_3,m_4)$, you calculate this: Choose random $iv$. ...


2

See the Wikipedia article on 3DES: http://en.wikipedia.org/wiki/Triple_DES#Algorithm First of all, you are only using 2 keys, so you may want to follow a hybrid technique like this: encrypted = eK1( dK2( plainText ) ) plainText = eK2( dK1( encrypted ) ) However, because I'm not familiar with the DES algorithm, I can't guarentee that this approach is even ...


2

After base64 decoding we get (hex) 5d f8 be 87 82 2b ea 5e f3 5c 23 fe 37 81 0f bd which has a size of two blocks. Of your small word-list, only marketing has so many letters that it needs two blocks: m a r k e t i n as the first, g 07 07 07 07 07 07 07 as the second (or another padding, but this is a common one), and so can correspond to this ciphertext. ...


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: ...


1

Note: you should also take into consideration the Expansion table, as it glues together with the Pbox. The simplest thing you would like to want from a Pbox is to provide a good diffusion on the inter-sbox level. That is, a single sbox should have an effect on many sboxes in the next round. This is not sufficient, for example you could have some indepent ...


1

You're missing a component : a padding convention. Yes, if you're trying to reduce a block size, it will reduce the cipher strength. That's why the less-sized blocks are padded/filled to fit the exact size. What to do : pad or fill or both - that is a question. First you need to understand, that the more predictible the message, the less secure the ...


1

For 1) you need to show that $S(X)$ is linear with respect to xor. We can define $S(X)$ as: $S(X) = (x_{7}, x_{0}, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})$ And define $A \oplus B$ as: $A \oplus B = (a_0 \oplus b_0, a_1 \oplus b_1,a_2 \oplus b_2,a_3 \oplus b_3,a_4 \oplus b_4,a_5 \oplus b_5,a_6 \oplus b_6,a_7 \oplus b_7,)$ So we have: $S(A \oplus B) = ...


1

It is not at all secure if you fix the key and IV in the code, no matter what language you use. Ideally you should generate the key from a password based key derivative function like PBKDF2 or SCRYPT or provision the key from an external key management server. You also need to chose a encryption mode along with the scheme. The modes are picked based on the ...


1

In general, the key length and number of rounds are the dominant factors in deciding cipher strength. But you need to consider how the rounds are constructed and how the key is used. Substitution and permutation are the bread and butter of DES. That's literally all it is - substitution, permutation, and XOR. Here is a diagram of the DES fiestel function ...



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