# 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 ...

4

Since you encrypt just a single letter, there are $26^2$ combinations of $p$ and $c$ where $c=E(p)$. This is because there are $26$ possible shift keys in the key space, an therefore each $p$ can be mapped to one of $26$ letters in the code space. Now, assuming that the key is distributed uniformly in the key space, each of those combinations of $(p,c)$ has ...

3

A Vigenère cypher can be unbreakable if, and only if: the key is random, the key is at least as long as the message being encrypted the key is used only once. In that case a Vigenère cypher is mathematically equivalent to a One Time Pad, and the same mathematical proof of unbreakability applies. You proposal is closer to being unbreakable than standard ...

3

You can solve it at http://www.quipqiup.com/index.php in about 5 seconds. contrariwise continued tweedle dee if it was so it might be and if it were so it would be but as it isnt it aint thats logic It's an excerpt from Through the Looking-Glass by Lewis Carroll Information on how quipqiup works is available at http://www.quipqiup.com/howwork.php

3

Vigenere Cryptosystem is as follow: You chose a key $(K_0,...,K_{m-1})$ consisting of elements in $Z_{26}$. Then a ciphertext for the message $(M_0,...,M_{n-1})$ is $$(M_i+K_{i\mod m}\mod 26)_{i \in [0..n-1]}$$ It is easy to see that you can generate a ciphertext for the message $(M_0+1,...,M_{n-1}+1)$ by adding 1 to each letter. It is therefore by ...

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 ...

2

There are different approaches to crack a substitution cipher. A human would use a different strategy than a computer. But as the word boundaries are not preserved it will be rather challenging for a human solving this cipher. The quipqiuq tool mentioned by John is using word lists, but there are other methods as well. Resources: ...

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

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 ...

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

The key to a mono-alphabetic substitution cipher is a substitution table. Thus you already have (most of) the key for that cipher, it is a b c d e f g h i j ? D E F G H I ? K L k l m n o p q r s t M N O P Q R S B T U u v w x y z V W X Y C ? where ...

1

As explained on the link you posted, the Vigenere cipher with a key on length $n$ encrypts every $n$-th symbol with the same key under the Caesar cipher. So to calculate the IC you should take all the $n$ sub-sequences separately: $\{1, 1+n, \dots, 1+kn, \dots\}$, $\{2, 2+n, \dots, 2+kn, \dots\}$ and so on and compute the IC for every sub-sequence.

Only top voted, non community-wiki answers of a minimum length are eligible