# Tag Info

7

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 ...

6

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 ...

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 ...

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

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 ...

4

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.

3

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://...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

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

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 $char) {$...

1

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 ...

1

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 ...

1

Provided the text was long enough and used an simple codebook substitution cipher, absolutely. English has common bigrams and trigrams as well as words that are typically positioned in certain places in sentences, like The. Also, if punctuation was tokenized in the codebook, it would be incredibly easy to guess or identify . and ,, because those will be ...

1

Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations: They would need more samples to differentiate between two tokens at the same level of confidence. For two different tokens with similar frequency they would no longer know that the words differ. The first means you are making the ...

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