When trying to break an unknown cipher, one first needs to figure out what kind of cipher one it is. Generally, a good starting point would be to start with the most common and well known classical ciphers, eliminate those that obviously don't fit, and try the remaining ones to see if any of them might work.
An obvious first step is to look at the ...
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 ...
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 ...
I think the edit to the question makes it as vigenere cipher; which invalidates my answer below. @galvatron answer gives the suitable answer why vigenere is not secure.
The old answer below ( applies only to substitution)
Baiscally this is a simple substitution cipher, where each letter is mapped to another letter (i.e. the shift). The answers for ...
If you combine two affine ciphers, you obtain one affine cipher. Say the first cipher is $e_1(x) = a_1x+b_1$ and the second is $e_2(x) = a_2x+b_2$. Then $e_1(e_2(x)) = a_1(a_2x+b_2)+b_1 = (a_1a_2)x+(a_1b_2+b_1)$.
Note that if $a_1$ and $a_2$ are both relatively prime with the modulus, then so is $a_1a_2$, so the new cipher can also be deciphered.
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 ...
Since this is an historical question, I am going to digress and make some historical corrections. In science, we give credit for important inventions to the people who published. If it turns out that someone else invented it earlier and didn't publish, they don't get credit. Obviously, they should be mentioned in passing or a footnote in the interests of ...
You'd be trying each possible displacement (offset).
Suppose the ciphertext is CXEKCWCOZKUCAYZEKW. Here's displacement 1:
At displacement 1, there are no matches (nothing where the a letter in the top line is equal to the letter immediately below it).
Here's displacement 2:
The VIC cipher and for something not as secure but easier for encryption and decryption, the double transposition cipher.
The VIC (short for VICTOR) was used by the Soviet spy Reino Häyhänen - a pencil paper cipher. To quote the wikipedia page:
Although certainly not as complex or secure as modern computer
operated stream ciphers or block ...
Wrapping up my comment as an answer:
Imagine you’re a Japanese cryptanalyst in the year 1944. There is no such thing yet called “television”, and you’re still decades away from a wordwide network feeding you with all the knowledge you could wish for.
In that case there’s only a minimal chance you’ve ever heard or seen a Navajo. So, you’ll be wondering ...
Encryption is naïvely viewed as a way to send messages from A to B that cannot be deciphered (at least in practice) by an adversary. Sure, encryption does do that, but modern ciphers do so much more...
A common attack scenario is the known plaintext attack (KPA). Of course, if the adversary already knows the entire plaintext, there's not much to be gained ...
Your interesting questions deserve to be answered more thoroughly, but here goes:
According to a highly classified document that was written in 1947 and finally declassified in 2013, the Germans started using a one-time pad system for diplomatic traffic in 1925. This system (GEE) used a one-time pad of digits to encrypt codes by modulo addition. To be clear,...
I would say MiMC is the simplest block cipher with plausible security. The idea is to cube the state, add a random constant, and repeat. This is typically done in a large prime field, but it is trivial to implement field arithmetic in any language with big integer support. Here's a Python implementation:
def mimc(x, p, k, constants):
x = (x + k) % p
Some additions to the other answer: any given letter can only correspond to a fairly limited number of ciphertext letters: only the ones in the same column or row, and never to itself. So a highly frequent letter like E will still stick out in longer texts and then we will also find its row and column mates, which helps in reconstructing the square. There ...
One option would be to get them to select a one-time MAC of the form:
$mac(m,k_0, k_1) = (k_0 \times m + k_1) \mod p$
You would select $p$ to be something like 29. $k_0$ and $k_1$ would be chosen at random from the values 0-29. $k_0$ has the additional restriction that it can't be 0.
You can aid the computation by giving them a 29x29 matrix of all ...
If the message is shorter than the key, then the Vigenere cipher is essentially the one-time pad, which is unbreakable for a random key. If the key is not random, then you may get some information on the plaintext.
I will throw tiny encryption algorithm into the mix:
It's a very respectable block cipher. It really works as a block cipher with convenient block size of 64 bits and key size of 128 bits. So it behaves much like a DES or AES as in how you use it securely.
It's a Feistel network which anyone starting ...
If it is a simple substitution cipher, there are a few standard techniques:
Frequency analysis. Count how many times each letter appears in the ciphertext. The most common ciphertext-letters probably correspond to the most-common letters in English. The most common letters in English are ETAOINSHRDLU... (in decreasing order of prevalence). Therefore, ...
First, you should start by guessing which symbols in the ciphertext are actually enciphered, and which are simply written in plain. (Don't worry if you guess wrong, you can always make several guesses.) For a Vigenère cipher, one also needs to guess whether any non-encrypted characters should advance the key position or not (usually they do not).
For your ...
The one time pad technically meets all your criteria and I think it's the simplest. It gets used all the time within encryption schemes where it's usually called blinding.
Otherwise I would look into small block ciphers. For example, RC5 and skip32. These are probably the simplest beside the OTP.
A substitution cipher consist of a mapping from letters in the alphabet to letters in the alphabet (not necessarily the same alphabet, but probably is in this case). There are many forms that a key can take on. Ones I've seen in practice are:
The key is the mapping (i.e. a->m, b->x, c->q,...).
The key represents a shift. A key of 5 would mean the ...
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 ...
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 ...
This looks like a sliding window approach to calculating the index of coincidence. So you would have something like:
ABCDE FGHIJ KLMNO
OACBD EFGHI JKLMN
Given enough cyphertext, you'll discover a length at which the IC is high; this is a candidate keylength for the cyphertext, because you've shifted the two texts by one keylength. Multiples of this size ...
This is highly insecure, for the same reason that ECB mode and simple substitution ciphers are. Every time you use the word the in your message, it will be encrypted the same way. The same goes for other, lower-frequency (but still fairly common) words -- like as or with or will (or any of hundreds of other examples).
This is a humongous clue to ...
Your scheme would make a nice puzzle for amateur codebreakers. That's about the best that can be said for it.
It does not meet the generally accepted standards for a modern encryption scheme; in particular, it is not semantically secure. In fact, the security of your scheme would be seriously compromised if an attacker obtained even a small amount of ...
Remember that, in a Vigenère cipher, the $n$-th ciphertext letter is calculated by adding the $n$-th plaintext letter and the $n$-th key letter (where the key is repeated as many times as necessary to make it as long as the plaintext) modulo 26 (for the standard English alphabet), i.e.:
$$c_n \equiv p_n + k_n \mod 26 \tag1$$
(Here, I'll assume the ...
what if you were to incorporate a Block Cipher Mode into a hand cipher
That line is a bit misleading and hints at a potential misunderstanding. A "mode of operation" is more something you wrap around a block cipher… not something you incorporate or embed into a cipher algorithm.
if you have a big enough key space, a small enough cipher text, can a very ...