37

The key space of a cryptographic algorithm whose key length is $n$ is given by $2^n$ No. There is confusion between: keyspace (or key space) $\mathcal K$, which is the set of possible keys. keyspace size (or size of the keyspace) $\|\mathcal K\|$, which is the number of possible keys (an integer). key length (or key size) in bit, which can be defined as ...


28

If the substitution ciphers belong to the same family, then their composition will also (typically, assuming that the family is closed under composition) belong to the same family. Thus, breaking the combined cipher will be no harder than breaking an arbitrary cipher in the family. For a simple example, combining two Caesar shift ciphers with shifts ...


25

He is talking about the original version of the Caesar Cipher where the substitution was just a +3: A -> D B -> E C -> F D -> G E -> H F -> I G -> K H -> L ... X -> A Y -> B Z -> C Because the shift is fixed, it does not have a key (but you could say it is a substitution cipher with a key equal to +3). However it is common ...


17

DES actually demonstrated that a Feistel structure was not a guarantee against attacks. In "academic" terms, DES is broken by both differential and linear cryptanalysis, because they require, respectively, $2^{47}$ chosen plaintexts and $2^{43}$ known plaintexts, whereas the DES key is (effectively) 56 bits. Of course, for practical attacks, we would brute ...


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 ...


10

The composition of any number of substitution ciphers is still a substitution cipher, hence no.


9

In the substitution cipher, the answer lies in the permutations, a key is one of the all possible permutations of the alphabet (keyspace), i.e. each letter is substituted with another. Therefore, for a key of an alphabet with 26 characters, the first letter can have the 26, the second can have 25, etc. and the last letter can get only one letter for ...


7

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://...


7

Since this is homework, let me just give you a hint: consider the two-character messages $m_1 = \text{"aa"}$ and $m_2 = \text{"ab"}$. Given a ciphertext $c$ encrypted with a monoalphabetic substitution cipher, can you tell which of $m_1$ or $m_2$ it corresponds to, even without knowing the key? Why (not)? What does this imply about perfect secrecy? Ps. ...


7

It's called a keyword cipher. See this question for some ways to break it.


7

Is this something that exists and could be plausible? Yes, things like that already exist and have even been used by well-known serial killers! (So much for creating a dramatic intro – lol) Monoalphabetic Substitution Cipher What you are referring to, could be categorized as a classical “pigpen cipher”; a monoalphabetic substitution cipher where ...


7

Actually, we have a four-way (that is, four words that will can be converted into any of the others with the right shift). These words are: ax, by, he, if Other two letter words are: am <-> my at <-> pi do <-> it hi <-> no We also have the <-> max I didn't do a systematic search for words over two letters; there certainly ...


7

A monoalphabetic substitution cipher uses a fixed permutation of an alphabet $A$ namely $$\pi:A\rightarrow A,$$ to encrypt a plaintext $P=(P_1,\ldots P_n)\in A^n$ of length $n$ into the ciphertext $C=(C_1,\ldots,C_n) \in A^n$ via $$C_i=\pi(P_i),\quad i=1,\ldots,n.$$ For a natural language, the letter frequencies are non-uniform. Thus, given a long enough ...


7

any other considerations? Yes. In many common use cases the mapping table needs to be retained. That map changes each time a number is added; that's a backup / continuity of service headache. The map is security-sensitive: it contains all the clear phone numbers, and information which (combined with other information) allows getting back to users. The map ...


7

CBC in this context means that each letter of the ciphertext is used to help encrypt the next letter. Let's say, making up some numbers, that a certain ciphertext letter is C. The next plaintext letter is T. So to encrypt it, we will add C (or 2) to T, giving V. Then we do the substitution step, which could be anything and happens to be J. The key ...


6

Yes, we can almost certainly break this, given enough ciphertext. One approach would be to use a dictionary and use word patterns. For instance, if the ciphertext word is qddxfozogf, then the plaintext word was probably ammunition. Notice how the 2nd and 3rd letters are the same; and the 5th and 10th letters, and the 6th and 8th letters? The word ...


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

In a Feistel networks (from the German IBM cryptographer Horst Feistel), the input is divided into two blocks ($L_0$ and $R_0$) which interact with each other. Main example is DES. basic construction: In a SPN (Substitution Permutation Network), the input is divided into multiple small blocks, applied to a S-box (substitution), then the bits positions are ...


6

Your construction is completely insecure: a single known plaintext / ciphertext block pair is sufficient to decrypt all blocks encrypted with the same key. Specifically, let me write your block encryption function $E_K$ as $$c = E_K(p) = P^{(n)}(S^{(n)}(p \oplus K_1 \oplus K_2 \oplus \dots \oplus K_n)),$$ where $p$ is the plaintext block, $c$ is the ...


6

The table lookup implementations usually combine the SubBytes and ShiftRows steps with the MixColumns step. Different implementations/hardware/etc make general statements impossible, but this paper gives some benchmarks and information on a table-lookup implementation. It won't necessarily be quicker across all systems. As per this paper: It must be noted,...


5

Based on your sample code I do not consider the scheme secure enough for implementation. Additionally you will run into a few problems if you actually try to implement this to generate DNA strands with encrypted messages in them (ala some kind of futuristic scifi thriller). As the other answer suggests, it would be best to think about using the DNA sequence ...


5

As a page at ibm.com indicates, there could have been a bit of a "contra" attitude against Feistel ciphers thanks to DES having seen the first breaks in it's security etc. Down with the Feistel structure! In most ciphers, the round transformation has the well-known Feistel structure. In this structure typically part of the bits of the intermediate ...


5

There's two missing pieces. First, the ring setting changes the output letter, it doesn't rotate the whole exit pattern. Second, the rotors are advanced before the letter is encrypted. If your rotor (Enigma I Rotor I) is set up like this with the ring at A. abcdefghijklmnopqrstuvwxyz ekmflgdqvzntowyhxuspaibrcj Then if you advance the ring to B all the ...


5

This is, in fact, not a Vigenère cipher. One clue to this is the fact that the ciphertext (which, conveniently, includes unenciphered word breaks) contains lots of repeated words like UTL, VCI, V, UB and QVRY that would be very unlikely to occur by chance in the output of a polyalphabetic cipher like Vigenère. Another clue can be obtained by examining the ...


5

I don't have enough space to expand on yyyyyyy's answer in a comment so I am making this an answer in and of itself. TruthSerum is correct, but it seems like an explanation is wanted, so here goes. Imagine you have a regular (all the sides have the same length, all the angles are the same) n-gon. That sounds complex, but trust me it isn't. A 4-gon is ...


5

This is called homophonic encryption, and has been around for a long time. In terms of cryptanalysis of such ciphers, there is a nice thesis from SJSU on this topic which is available here. The attacks tested in that cipher were based on hill climbing and local optimization techniques. The conclusion states: We designed and implemented an efficient ...


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

The formula reads: C: ciphertext of a character E(P): encryption using Caesar of plaintext character = (Pi + 3) mod 26: index of character in alphabet, plus 3 (the key) and then modulus 26, the size of the alphabet So basically for A you encode the value to 0 and add 3, which gets you the value 3, which decodes back to D. For Y you use 24, add 3 and get ...


5

You're right. For a substitution cipher the huge size of the keyspace is quite irrelevant in terms of security: it does not change the frequencies of letters, just maps them to other letters. So you you can go from the most common letter and map that, then the 2. most common, etc. And after maybe the 10 most common letters you can use a dictionary with ...


5

I suspect it was a semi-deliberate feature. That is, while it probably wasn't a design goal in and of itself, it neatly solved a mechanical issue that would otherwise have required a more complicated and failure-prone solution. What was the issue? Simply, it was making the third wheel only advance one step at a time, rather than 26 steps in a row. That's ...


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