# Tag Info

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Cipher details Cipher type The Felix cipher can be broken down into two algorithms: a substitution cipher and a permutation of the character pairs. We obtain the substitution if we read the number pairs in figure 3.3 vertically rather than horizontally. Since the permutation is fixed, it has no cryptographic value. Therefore, we'll only analyze the ...

7

I think I understand what you're asking for. You're trying to learn how we know which algorithm was used, so we know how to attack it. That's a part of what is known as cryptanalysis, the task of breaking ciphers. If you are using a standard computer protocol, the encryption algorithm is defined as a part of the protocol. The computers can't talk unless ...

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Kerckhoffs's principle states, that a cryptographic system shall be secure even if everything about the system, except the key, is known to the attacker. Typically an encryption algorithm has two inputs: a key and the data. In the case of Rot13, there is no key. So if you know the algorithm, there is nothing left to guess. Let's assume the algorithm ...

5

The fact that a given cipher has a key length of 296 bits doesn't mean at all that it provides 296 bits of security or even that a brute force attack would take $2^{296}$ steps. The problem of mono-alphabetic substitution cipher is the ridiculously small block size (in this case, barely $\log 64 = 6$ bits). If absolutely nothing about the plaintext is ...

5

This is a simple substitution cipher, specifically a mixed/deranged alphabet cipher. See wikipedia's description: Substitution of single letters separately—simple substitution—can be demonstrated by writing out the alphabet in some order to represent the substitution. This is termed a substitution alphabet. The cipher alphabet may be shifted or reversed ...

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Frequency analysis would work on average quite poorly on ciphertext of the proposed cipher. How exactly depends a lot on the value of the key: for some weak keys likes 1, 3, 30, 30000.. it works essentially as well as for any mono-alphabetic cipher. For 103, it still works well. For any key, given enough (lots of) ciphertext, it could still distinguish the ...

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When we consider that a Playfair key consists of the alphabet (reduced to 25 letters) spread on a 5x5 square, that's $25!$ keys (another formulation consider any string to be a key; then strings leading to the same square are equivalent keys). The rules of Playfair are such that any rotation of the lines in the square, and any rotation of its columns, lead ...

4

As the page explains, the cipher it describes is a simple variant of the bifid cipher, with the alphabet extended from the traditional 25 to 36 letters. As such, most techniques for breaking the bifid cipher ought to be more or less directly applicable to it. The bifid cipher is nowadays mainly used for crypto puzzles. Like most classical ciphers, it is ...

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First guess the key length(Just try every plausible length, there aren't many). Then for each position where you know both plain- and ciphertext, calculate the key char. If you get a contradiction, the guessed key length was wrong. If the key length is short enough compared to the number of known pairs this will probably give you a large part of the key.

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Your system is essentially: $c_0 = m_0 + k$ $c_i = m_i + c_{i-1}$ for $i>0$ It's absolutely trivial to break this, since the attacker knows the ciphertext, and thus knows both $c_i$ and $c_{i-1}$. Decrypting the first group is hard. But to decrypt any group but the first, simply subtract the previous group from the current group $m_i = c_i - ... 3 yes (guessing ur doing the cypher challenge...?) This link has a really good decoding tool (saves you time), then trial and error keywords. also this tool; can be used to find the length of the keyword; paste the texts you're decoding then the number of the column(s) with the most x's is the length of the keyword 2 This can be broken. The exact nature of the attack will depend what modulus you use for the Hill cipher: are you working modulo a prime number, or working modulo 26? Working modulo a prime$p\$ A simple attack, with no fancy mathematics needed. One simple attack is to start by requesting the encryption of the 26 messages AAAA, BBBB, CCCC, DDDD, ..., ZZZZ. ...

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If you don't know the system, you just check one after the other: frequency analysis of bigrams detects Ceasar and Playfair. Try Caesar first then Playfair. Auto correlation method for Vigenere (for each x: count the number of occurances, where letter at position i and i+x are equal. For the correct codeword length, it will spike) If you have a Hill ...

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given that the permutation is fixed and the key step is independent of the permutation you can reduce this to an ordinary text-substitution cipher if the key is as long as the input you have a weak one-time pad (because the per letter change is limited to 10 instead of 26) however if the key is short then you have a vignere cipher (if you "decode" with ...

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Encrypting the AES key does not actually make a brute force search any harder: an attacker doesn't need to know the encrypted key to decode messages, they only need to know the actual AES key. Thus, the attacker only(!) needs to search the 256 bit AES keyspace, not the roughly 296+256 = 552 bit encrypted keyspace. Besides, even if the attacker did try an ...

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I'm not exactly certain why "the obvious reasons" ROT13 isn't secure wouldn't be considered the appropriate answer; it's not secure (that is, doesn't provide privacy) because anyone can decrypt it trivially (whether they're the intended recipient or not). If you want to get into the details about why it is not secure, well, we need to talk about "security ...

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It's sometimes called a keyword cipher. As dr jimbob notes, it's a particular type of monoalphabetic substitution cipher. Ps. See also this recent question about breaking such ciphers.

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The Caesar cipher (aka Shift cipher) has, as you said, a key space of size 26. To achieve perfect secrecy, it thus can have at most 26 plaintexts and ciphertexts. With a message space of one character (and every key only used once), it would fit the definition of perfect secrecy. For the usual use with messages longer than one character, or multiple ...

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Strange... WNZTNV is repeated... If i'm counting correct on position 1-6 and 22-27. My guess would be the key length is a divider of 21? 7 seems to be consistent with the other information, but my knowledge of foreign languages is close to none, so i don't recognize anything readable in key or plaintext... key DM''DDS of EN''EET (depending on your ...

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Let's say the person who intercepts the message doesn't know what cipher was used. As long as the message is large enough (sar 30 to 40 characters), it can still be broken (I do it all the time playing Cryptograms.org, which uses a simple substitution cipher, a general class of ciphers to which ROT13 belongs). The process is easy, the most common letter in ...

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CodesInChaos nailed it -- the key size is zero. The strength of an algorithm is roughly determined by how many tries someone who didn't know the key would need in order to successfully decrypt the message. In the case of ROT13, the key size is zero, and anyone attempting to decrypt the message would succeed on the first try.

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