# Tag Info

## Hot answers tagged permutation

16

You have clarified the question as asking about whether replacing ShiftRows with a random byte permutation would strengthen AES against differential attacks. It would not. ShiftRows and MixColumns were carefully selected to work in tandem, such that every byte affects every other byte in the state within just two rounds. MixColumns ensures that every ...

9

I assume that you mean the S-box. The answer is NO! Randomly chosen S-boxes are not good choices for differential and linear cryptanalysis. When Biham and Shamir presented differential attacks on DES, one of the things that they showed was that if you replace the S-boxes in DES with randomly chosen ones, then the differential attack becomes much more ...

7

Security is clearly broken if there is a polynomial-length period with non-negligible probability (where by this I mean if a random point falls in a cycle with a poly-length period with non-negligible probability). In order to find a preimage, just go forward until you get back to the starting point, keeping the previous value each time.

6

Let us first consider the problem without involving Shamir secret-sharing at all. Suppose that $n = 140$ and that the secret $\sigma$ is a 140-byte Twitter message. The space is thus restricted considerably, from all possible $256$ byte values to the printable characters permitted to be used in Twitter messages, and the distribution in this restricted space ...

5

If a message is longer than the block length, how would changing one part of the message affect the encryption of other parts of the message? That really doesn't depend on the block cipher in use, which may be a feistel cipher like DES or a SP network like AES, but on the mode of operation. Now the answer to this really depends on the actual mode ...

4

If you can generate uniform random numbers, you can use a variant of Fisher-Yates. //given an array s with the elements to be permuted for i from n-1 to 1: t = rand(0, i) # inclusive swap(s[i], s[t])

3

I'm not sure I understand your question entirely. If there is only one possible message, then the ciphertext can be trivially decrypted simply by choosing this message. I'll assume instead that the ciphertext contains the shuffled bit pattern of a name chosen from a set of more than one name. The problem with bit shuffling is that the number of set bits ...

2

A one-way permutation is just a one-way function in which the function is a permutation (id est, a bijective function). Every OWP is a OWF, the converse is not true. IND-CPA security is a security notion specifically related to encryption schemes. OWF and OWP are not encryption schemes, hence they cannot be said "IND-CPA secure"; however, one can construct ...

2

1) and 2) Either S0, or S1 is used for each 4-bit long part of block depending on key bits. 16 S-boxes are used in each round (because you split 64-bit-long block into 16 parts of length 4) of the Lucifer cipher, so each round needs a 2 byte long subkey. Example: 16 starting bits of the key (the first subkey): 0 0 1 0 1 1 1 1 0 0 1 0 1 0 0 1 (2 bytes ...

1

A while ago, I spent time playing with modern field ciphers, especially with Card-Chameleon. Card-Chameleon uses a 52 card deck, assigning a letter to each red and each black card and needs two separate full alphabet permutations as a key. As it's a field I tried to find a computer-less, math-less, way to generate permutations from passwords. My solution is ...

1

First this, a nit about the hash function: one can trivially create a collision by appending a 0 value; an appended 0 does not change the hash value, and hence that produces a collision. If we say that collisions are required to have at least one place there two colliding messages differ with at least one having a nonzero value, then we can show it. Here ...

1

Ryan, the permutation IP is fixed so it is a table of only 64 entries, mapping the $i$-th position to the $P(i)$th position. Each entry $i$ (and $P(i)$) of the table is in the range $$0,1,\ldots,63$$ so 6 bits are enough to represent each, but a byte can also be used.

1

A solution is given by Peter Ryan, Crypto Santa. In The New Codebreakers, vol. 9100 of Lecture Notes in Computer Science, pp. 543-549, Springer, 2016. http://dx.doi.org/10.1007/978-3-662-49301-4_33 An earlier version (with Sjouke Mauw and Sasa Radomirovic) can be found at http://hdl.handle.net/10993/25936 (Open access), Security protocols for Secret Santa.

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