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


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 ...


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 ...


1

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.


1

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 ...



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