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11

If you combine two affine ciphers, you obtain one affine cipher. Say the first cipher is $e_1(x) = a_1x+b_1$ and the second is $e_2(x) = a_2x+b_2$. Then $e_1(e_2(x)) = a_1(a_2x+b_2)+b_1 = (a_1a_2)x+(a_1b_2+b_1)$. Note that if $a_1$ and $a_2$ are both relatively prime with the modulus, then so is $a_1a_2$, so the new cipher can also be deciphered.


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 ...


4

Wrapping up my comment as an answer: Imagine you’re a Japanese cryptanalyst in the year 1944. There is no such thing yet called “television”, and you’re still decades away from a wordwide network feeding you with all the knowledge you could wish for. In that case there’s only a minimal chance you’ve ever heard or seen a Navajo. So, you’ll be wondering ...


3

The other answer correctly explains how we could make a modern stream cipher, with a public random IV at the beginning of ciphertext used (together with the key) in the setup of a keystream generator, restricted to the range $[0..35]$, and using addition modulo $36$ (Beaufort cipher) to encipher each character. Here is a more detailed description of such a ...


3

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 ...


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

Depending on your video game design you maybe can utilise a many-time pad. Many-time pad is many plaintexts encrypted by XORing them with a common string. One-time pad cannot be broken but many-time pad can, especially if the players can make an educated guess about what the plaintext message might contain. You can also try encrypting numbers with RSA but ...


2

You could obscure the frequency using a block-cipher or stream-cipher or use a code-book substitution, then maybe use Vigenere Cipher on it, but if you really want your scheme to be secure, then your underlying substitution must be secure enough. In that case using secure block or steam encryption itself is enough. The second level Vigenere encryption is ...


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 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 ...


2

1:1 ciphers Many modern ciphers create ciphertexts that include the plaintext encrypted with a 1:1 stream cipher, plus a few more characters that help defend against certain known stream cipher attacks. As poncho and Rup have suggested, Alice and Bob somehow generate the same long stream of characters that is unguessable by the adversary. (If that stream ...


2

As explained by Cort Ammon, they mostly relied on physical, rather than mathematical processes. Here is a video on Youtube titled "How to launch a nuclear missile", taken in the Titan Missile Museum mentioned by Cort Ammon. The message contains a seven character string. The personnel in the shelter have a set of envelopes with the first two letters on the ...


2

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 ...


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

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 ...


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 ...


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

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

The Chinese used digits. So there is an encoding into digit groups, like a code book. This code book could be standard (like the Chinese used for telegraphy and radio as well, which used a publicly available book), or for extra secrecy could be kept a secret. The cipher could then be made to work on digit groups (and there standard techniques like addition ...


1

In the past, they relied far more on human techniques than cryptographic ones. The Trident example you linked is interesting to me, because it is conflicting with my experience receiving a tour in the Titan Missile Museum in Arizona. I won't claim my tour qualifies as a reliable source, but it's what I was given. In the tour, they explained the process ...


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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