Hot answers tagged classical-cipher
9
Using the book as a key is relatively similar to one-time pad, insofar as the book can be considered as a random stream of characters. But that's true only to some extent: a book consists of words, with meaning, which implies that characters which may appear at position 321:42:35 are not uncorrelated with characters which appear at positions 321:42:34 and ...
8
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 ...
7
I take your question to mean, how both historically and in the modern age one could construct a pen-and-paper cipher using the Chinese language.
As pointed out in the question, Chinese is a logographic langauge and therefore has a far greater number of characters than Phonetic systems. Historically this has cause chinese codes not to be based around the ...
7
As the other poster rightly pointed out, it's a Playfair cipher.
Even without the known plaintext, the program "playn"
here will give the right text in less than a second.
(you can compile it yourself, and it uses the bigram statistics of English)
I ran it, and the result was the following:
IT XT UR NS OU TX TH AT OR IG AM IX IS AB RI LX
LI AN TW AY TO ...
7
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 ...
6
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
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 ...
5
I do not have a solution, but I pursued the cipher long enough to establish it wasn't one of the easy classical ciphers. This approach should get you started.
The first thing you want to do is convert the text into numbers as many classic ciphers are mathematically-based (or at least easy represented mathematically). Using $A=0$, $B=1$, $\ldots$, the ...
5
An obstacle to proving that a book cipher is secure is that the letters in (most) books are not chosen independently at random. Thus, in principle, if two indices are chosen too close to each other, an adversary could deduce some statistical information about how the corresponding plaintext letters may be correlated.
As a toy example, suppose that an ...
5
This is essentially a Vigenère cipher; it's been known for centuries. As for how secure it is, well, it is actually fairly easy to break (unless the key is both as long as the ciphertext, and randomly chosen; however, at that point, if you could remember the key, you could have well just remembered the plaintext). As for your colleague, he's right, and ...
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 ...
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 special case of the affine cipher where $m=26$.
Let's encrypt a single letter using your $E$. Let it be m, say, which is at index 12. So, $$E(12) = (7 \cdot 12 + 10) \mod{26} = 16$$
Now if we try to use the $D$ in your question, we decrypt this as:
$$D(16) = (7 \cdot 16 - 10) \mod{26} = 24$$
which is obviously not right. The issue is that your ...
4
So the one's i'd bet on are either solitaire.
Solitaire by Bruce schneier is probably your best bet. It has a few issues but it will work well for most things. It ends up having a small bias, but it takes about 15 seconds per character after the initial keystream has been generated.
It is not nearly as widely studied a field since most people are assumed ...
4
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
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.
4
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 - ...
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 ...
3
It seems I can't comment on answers because the question is no longer on the Chinese Q&A, but I wanted to support fgrieu's suggestion. Certain web services don't have much care for security, but they want to avoid containing keywords that are blocked by the Chinese Firewall. One I'm familiar with is PIMCloud, a cloud-supported IME, which does exactly ...
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
3
Both the Vigenère and autokey ciphers are classified as polyalphabetic substitution ciphers, so the cipher in your exam is not likely to be either of those. Rather, the phrasing of the question suggests that it belongs to the other branch of classical ciphers, transposition ciphers.
Indeed, looking at the letter frequencies of the ciphertext strongly ...
3
I'll assume that the objective is to assert if the distribution of the $f'_i/n'$ is sufficiently similar with the distribution of the $f_i/n$ to support that a substitution cipher (including Caesar cipher) with the same permutation table and same frequency of plaintext characters could be used in both case.
If $n \gg n'$, $f_i \gg 5$ and ...
3
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.
...
2
hummm... some thoughts about it:
I think that it could be secure depending on what you want to hide. The bigger and the more "real world words" you want to protect, the easier it gets to crack. Why? Because in books, in general, you'll only have letters, few numbers, and that's only. Ok, so I know your transmitting one or more words. By the size of the ...
2
If you save the current page and examine the file with an hex editor, you will likely find that your example Chinese string is represented by the bytes
E6 88 91 E5 9C A8 E5 AD B8 E4 B8 AD E6 96 87
One option, very suitable for implementation by a machine, is to encipher the bytes representing the message in this (or other) format used by the machine. ...
2
I don't know the solution, but since you say you're only asking for hints, here's a few that occurred to me:
If this is a Vigenère cipher, the missing character at the beginning should not matter (much): if you encrypt a message with the key FOOBAR and drop the first letter of the output, you can decrypt the resulting ciphertext with the key OOBARF.
As ...
2
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 ...
2
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.
2
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 ...
2
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 ...
Only top voted, non community-wiki answers of a minimum length are eligible
