# Tag Info

6

You'd be trying each possible displacement (offset). Suppose the ciphertext is CXEKCWCOZKUCAYZEKW. Here's displacement 1: CXEKCWCOZKUCAYZEKW CXEKCWCOZKUCAYZEKW At displacement 1, there are no matches (nothing where the a letter in the top line is equal to the letter immediately below it). Here's displacement 2: CXEKCWCOZKUCAYZEKW CXEKCWCOZKUCAYZEKW ...

5

This looks like a sliding window approach to calculating the index of coincidence. So you would have something like: ABCDE FGHIJ KLMNO OACBD EFGHI JKLMN Given enough cyphertext, you'll discover a length at which the IC is high; this is a candidate keylength for the cyphertext, because you've shifted the two texts by one keylength. Multiples of this size ...

4

First guess the key length(Just try every plausible length, there aren't many). Then for each position where you know both plain- and ciphertext, calculate the key char. If you get a contradiction, the guessed key length was wrong. If the key length is short enough compared to the number of known pairs this will probably give you a large part of the key.

3

This method isn't deterministic. That is, just because a keyword has the lowest chi-square value, that doesn't mean it's the keyword. All it means that there is a good chance that it is the keyword. It's not a guarantee. You should look at the first several likely options to see if the keyword lies there. If you want to automate this, you could compare the ...

3

Yes. Remember that, in a Vigenère cipher, the $n$-th ciphertext letter is calculated by adding the $n$-th plaintext letter and the $n$-th key letter (where the key is repeated as many times as necessary to make it as long as the plaintext) modulo 26 (for the standard English alphabet), i.e.: $$c_n \equiv p_n + k_n \mod 26 \tag1$$ (Here, I'll assume the ...

2

For chosen plaintext and a classic Vigenère cipher, you need only as many characters of plaintext as the length of the key to completely recover the key. It's a trivial reversal. (If the cipher uses 26 scrambled alphabets, it will take more.) if you don't know the length of the key, you will spot the repeating sequence after the second repeat. For an ...

2

I suspect, when the IC was invented, it was typically used together with human judgement, so there might not be any standard rules to address this issue. Anyway, it seems like there are two plausible solutions: It might be easy to add a special case for this situation: choose the keylength with the highest IC, except that if it has a divisor whose IC is ...

2

If you have known plaintext, namely one input file that is known in its entirety, this is trivial to break. So I'll explore methods that might lead to a break, if you don't know what's in the input file that was compressed. I suggest that you start by analyzing the DEFLATE stream format carefully (see also these handy notes). This will probably help you ...

2

It was never determined how to derive the keyword PALIMPSEST from the sculpture or its text. It was, as you stated, found by brute-force hill-climbing algorithms. One theory is that crib-dragging the morse phrase SHADOW FORCES reveals IMPS at SHAD, and working form there retrieving the entire keyword, but I find this far-fetched and implausible. The ...

2

Strange... WNZTNV is repeated... If i'm counting correct on position 1-6 and 22-27. My guess would be the key length is a divider of 21? 7 seems to be consistent with the other information, but my knowledge of foreign languages is close to none, so i don't recognize anything readable in key or plaintext... key DM''DDS of EN''EET (depending on your ...

2

Sticking to monoalphabetic ciphers, Vigenere can be combined with a secret random-like substitution of the plaintext or/and ciphertext alphabet, making it significantly more resistant. If Vigenere encryption is $$x_j\mapsto (x_j+K[j\bmod k])\bmod{26}$$ where array $K$ is the key or length $k$, I'm discussing $$x_j\mapsto (S[x_j]+K[j\bmod k])\bmod{26}$$ ...

2

For Vigenère specifically, you can make it harder to break by increasing the size of the key and by making the key truly random. If the key is truly random, longer than the plaintext and never reused, then Vigenère becomes equivalent to a One Time Pad. Even without going that far, it is possible to strengthen many, though not all, cyphers by lengthening ...

2

I don't know the solution, but since you say you're only asking for hints, here's a few that occurred to me: If this is a Vigenère cipher, the missing character at the beginning should not matter (much): if you encrypt a message with the key FOOBAR and drop the first letter of the output, you can decrypt the resulting ciphertext with the key OOBARF. As ...

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