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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 ...


Well, DES is weak and since you can guess a plaintext you can use a rainbow table to crack the key and then of course decrypt and re-encrypt the message. Apart from that I don't think you can re-order the blocks to change the first and last line, if you really want to do this specific operation.


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.


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 ...


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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