# Tag Info

5

How on earth did you arrive at that formula? You can break a Caesar cipher by calculating the result of applying all of the $n-1$ (i.e., 25) possible shifts to the ciphertext and picking the one that makes sense. The computational complexity is just $\mathcal{O}(n)$. If you want to automate the process based on frequency analysis, the correlation step ...

5

I think it's better to define the keyspace of an unkeyed function as having one element. Some advantages: Computing the key size as $log_2 1$ correctly tells you it's a 0 bit key For encryption you pick one key from the set. You can pick an element of a single element set, but can't pick from an empty set Functions with multiple inputs are often defined ...

3

In 2013, I visited the National Cryptologic Museum and researched/photographed the donated materials on the Chaocipher from the Byrne family. As it pertains to your question regarding the security of the cipher, it is true that while the original algorithm has been now been known for several years, to date no known cipher-text only decryption has yet been ...

2

It's not elegant, but the brute force method is to write a program that creates a table of 25x25 digraphs (assuming i=j), yielding 625 rows. I'd also add a column that lists the relative frequency of each digraph (given enough ciphertext you can use that to identify frequent substitutions, as you already have done). You start off with 625! possible ...

2

It depends on the time you want to spend. But most likely, there is nothing with reasonable efficiency. For arithmetic operations, humans are really bad compared to computers, and the difference is at least a factor of $10.000.000$ (very very rough guess, probably even 1+ additional zeros there). So, since you have to assume that the attacker has access to ...

2

Every classical cipher can be used without a computer's assistance; while simple mechanical ciphers can fall into the "classical cipher" category, in general classical ciphers are pen-and-paper ciphers, almost all of which are more secure than your "press the key to the right of the real one." Vigenere, for instance, has flaws; however, it is much more ...

2

How secure is this modified Bazeries Cylinder? That depends on what you expect us to compare it to. I doubt you’re looking for an answer along the lines of “it’s more secure than Scytale, but less secure than AES”. Therefore, I will try to narrow it down a bit by saying that it is safe to assume it can not – in any way – provide the same security levels ...

1

Assume the length is $n$. If the cipher text is $c_0, c_1, c_2, \ldots, c_N$ then consider the sub-text consisting of the characters $c_0, c_n, c_{2n},\ldots$. These have all been encrypted with the same Caesar, and you can break it by frequency analysis (the shifted 'e' should be the most common in standard English texts, or else maybe the 't' etc.). ...

1

If the ciphers are different, with independent keys, you can say that it is at least as strong as the first cipher. If the ciphers commute, like with stream ciphers, you can even say that it is at least as strong as the strongest. See Cascade Ciphers: The Importance of Being First. That's really all you can say in general. In practice, the combinations you ...

1

In general, all you can say is it can be as weak as the weakest encryption layer, if you're lucky. Edit: It can also be even weaker, for badly chosen components that cancel out some mathematically desirable properties, as pointed out in the comment.

1

There are several algorithms available which can attack a playfair cipher. Hill climbing might be one option. Basically it starts with a random key (assuming it's the best one) and decrypts the cipher. The resulting clear text is scored using a fitness function. Then small changes are applied to the key and if the resulting clear text of the modified key ...

1

The "logic" of the Enigma machine and the development of the Polish solution, in principle, are well described in David Kahn's "Seizing The Enigma". There may be better descriptions that have come out since, but I found this very clear and continue to recommend it. In addition to the nuts and bolts of the machine itself, Kahn describes the history from ...

1

If you consider arbitrary permutations, you have $\frac{n(n+1)}{2}$ possibilities. That means, $O(n^2)$ is the correct complexity in big-O notation, but I don't understand why you need that at all, if you can provide the result as exact formula. Caesar cipher contains just a subset of $n$ possibilities, and therefore obviously $O(n)$. Anyway, this doesn't ...

1

It depends how you look at it. If you regard it as a Caesar cipher then the key is 13 (out of a key space of 26 for uppercase ASCII, although key 0 is a very weak key, resulting in the identify function). If you consider that 13 is part of the ROT13 cipher then it indeed has no key. Of course having a static key or no key does not make a difference in ...

1

There was a telegraphic code that mapped the comparatively frequently used 10000 words (that's very much more than for common daily use, e.g. newspapers) of Chinese to [0,9999]. There was once also a patented mechanical device that effected a poly-alphabetical substitution of [0,9] with a set of disks containing certain scrambled alphabet of [0,9], which the ...

1

My understanding is given that $E_K(x)=(x+K)~mod~26$ is the character by character encryption of a shift cipher, you want to extend it to words as the function $E'$ given by $$E'_K(x_1,\ldots,x_m)=(E_K(x_1),E_K(x_m)).$$ If you intended the key $K$ to change per character you can modify.

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