19

As otus suggests in the comments, it's better to first calculate the frequency of each letter in the decrypted message, and then compare the frequency distribution to what would be expected for English text. For the comparison, you can use chi-squared ($\chi^2$) testing. (Actually, for just comparing the likelihoods of different decryptions, you don't even ...


10

I'll assume that the plaintext consists entirely of capital ASCII letters as in the example. This implies the high 3 bits of each byte of plaintext are 010. It is useful to visualize how 3 consecutive bytes of plaintext map to 4 consecutive Base64 characters. 1. Frequency analysis of the last character of 4-char blocks in ciphertext We see there is a ...


7

Classical ciphers operate on letters. If we consider the frequency attack on classical ciphers it considers the frequency of the letters. Modern ciphers, if we consider only block ciphers, operates on blocks - 64, 128, or more bit blocks. Let see how you can perform a frequency attack on modern ciphers to infer some data. Example attack on Databases: Let ...


6

If one can prove that a large amount of image and audio data doesn't exhibit a frequency patter, only then can we consider frequency analysis as a non-viable attack. For a modified simple example, let us say each guitar chord is encoded as a byte of audio data. If you analyse about 70+ songs, you will see that 4 chords are the most frequently used (as ...


6

My humble contribution to this thread: in order to give some plaintext an "English-score", I found out that the Bhattacharyya coefficient is quite usefull: The Bhattacharyya coefficient is an approximate measurement of the amount of overlap between two statistical samples. The coefficient can be used to determine the relative closeness of the two ...


6

The site also misrepresents RSA encryption. If you are using textbook RSA (i.e. this scheme, with no padding), you don't independently encrypt each letter. Rather, you take the entire message, treat it as a number, and encrypt that. This doesn't work if your primes are 3 and 11, but if they're each, say, 8 bytes long, then you could encrypt a 16-byte ...


5

Regarding the first part of the question, I will just link to another answer I wrote in the past: Why don't homophones hide multiple-letter patterns? Summary: If you adjust the frequencies so that every single symbol is equally likely, then bigrams can be used for frequency analysis because they won't be uniform distributed. The structure of language is ...


5

You are correct that in this case simple frequency analysis would be possible since textbook RSA encryption is deterministic. One can get around this by using RSA with random padding. Here are a few references: Why is padding used for RSA encryption given that it is not a block cipher? OAEP Why RSA encryption padding is critical In practice, we never ...


5

There is no security difference. Of course, purely random characters with entropy rate $\log M$ where $M$ is the size of the alphabet should be independently generated and used for the OTP, whatever the size $M$ of the alphabet.


5

The feasibility depends a lot on the length of the corpus. The more statistics, the better guesses an attacker would be able to make. He'll try to use statistical attacks to fit the frequency curve of the tokens to the frequency curve of English words. This will allow him to guess the preimages of more frequent words with high confidence, but will he be ...


5

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: http://...


4

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


4

This sounds like a classic codebook or nomenclator. Even if we assume a perfect random oracle that generates a completely random codeword for each word of English text, I agree with otus that frequency attacks and N-grams would likely be able to decode the most-frequently-used words. Also, a known-plaintext attack (or worse, a chosen-plaintext attack) would ...


4

The encryption is weak This encryption is more susceptible to frequency analysis than original "substitution ciphers" because the frequency tables should be much more Non-uniform. In my opinion, it should be less secure than substitution cipher although the key space is much much bigger (compare $64!$ to $26!$). Some evidences of the weakness If I assume ...


4

The accepted answer is correct, but I wanted to add an update since the sample set of frequencies given in the accepted answer doesn't always work so well for the Matasano Crypto Challenges. In particular, it is nice to have the frequency of the space character (0x20). For example, the following array has the experimental frequencies for all 26 letters ...


4

All the Quagmire ciphers (see e.g. here for definitions) are combinations of a Vigenère shift cipher and a keyword-based simple substitution cipher, where the substitution cipher is used to scramble the alphabet before and/or after the Vigenère encryption. In particular, the Quagmire III cipher in effect scrambles the plaintext alphabet before doing a ...


3

Any course material or book with a title like introduction to cryptography (or similar) will address this topic and provide examples. Alternatives are: Frequency analysis and substitution cipher. The basic steps are: Decide on what you want to count: Usually first you look at single symbols first. And then bigrams. But there are other possibilities (...


3

You probably mean the VIC cipher. It can be set up in a way such that the initial substitution that also achieves fractionation through the use of a so-called straddling checkerboard is followed by a transposition or substitution step. These steps make the distribution of symbols in the cipher text more uniform than in the plain text. An optional ...


3

I would have expected Vigenère and Playfair ciphers to reduce the skew in the letter distribution, so I think you can rule them out for the time being. Here are some other things you could consider: Perhaps one of the letters corresponds to a space: hello world → itssgxvgksr Perhaps there is no letter "e" in the plaintext A Void by G. Adair &...


3

Yes, there are many. Here are the few that I know: Brute-Force: Simply trying all possible keys. Kasiski examination: Exploiting the repeated words in the ciphertext to figure out the key length in Vigenere Cipher. Friedman test aka Kappa Test: Measures the unevenness of the cipher letter frequencies to break the Vigenere cipher. Key elimination: Eleminate ...


3

Here's a simple PHP script that calculates bytes frequencies from 10 random Wikipedia articles: <?php $frequencies = array(); $total = 0; for ($i=0; $i<256; $i++) $frequencies[$i] = 0; for ($i=0; $i<10; $i++) { $src = file_get_contents('https://en.wikipedia.org/wiki/Special:Random'); foreach (str_split($src) as $char) { $...


3

This is a monoalphabetic substitution cipher - the same letter in the plaintext will always be represented as a identical letter in the ciphertext (for example the plaintext-letter "E" is always a "V" in the ciphertext). Frequency analysis works better the longer the text is. For very short texts (as yours) it can be very difficult to ...


3

One approach you might try is applying additive smoothing via pseudocounts. At its most basic, this can be as simple as adding 1 to the number of times each byte (or $n$-tuple, or whatever) occurs in your corpus before calculating the frequencies. This allows you to calculate a non-zero (although possibly very small) likelihood for any decrypted byte ...


3

I was wondering, would calculating the entropy of the resulting plaintext not serve the same purpose, if not to a better degree seeing as it's not confined to the english language? First note that a cipher is susceptible to frequency analysis if it doesn't change the frequency at which characters occur. That is the "probability" of each substituted letter ...


2

It probably refers to the Index of coincidence, or more accurately the un-normalized index of coincidence, referred to in the Wikipedia article as "kappa-plaintext".


2

You should not just "pick the highest frequency character and assume it should be E" because it will probably fail most of the time, except if your ciphertext is really long. If your alphabet is small enough (usually either 26 or 255), it would be wiser to try all the possibilities for each group, and to check if the output looks like real english (...


2

You can compute a statistical distance measure between the observed letter frequencies in your candidate plaintext and the typical letter frequencies in English text. There are a number of different ways to measure the similarity between two statistical distributions, and I'm not aware of any theoretical or empirical studies on which of them would be ...


2

This will make it more clear: A B C D E F G H I J ... 0 1 2 3 4 5 6 7 8 9 ... The Caesar cipher starts with the letter and goes through the alphabet using the key, wrapping around after 'Z'. So if 'I' is the most common letter then this is probably the enciphered letter 'E'. And since index('I') - index('E') = 8 - 4 = 4, that's the key that was used. Note ...


2

Substitution cipher (e.g., Caesar) and polyalphabetic substitution cipher (e.g., Vigenere) are deterministic encryption schemes. The same plaintext will always result in the same ciphertext. So, they are not secure in the sense of indistinguishability under chosen plaintext attack (IND-CPA). It may be possible that "all the letters of the alphabet are ...


2

The Caesar cipher, Vigenere, monoalphabetic substitution, the autokey cipher, columnar transposition, the Playfair cipher, the Rail Fence cipher, disrupted transposition, the ADFGVX cipher, Quagmire III, etc., are all interesting and good to understand, but compared to modern cryptographic systems they are almost always utterly worthless for providing real-...


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