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 ...


21

I fail to see why one would want to use classical or pencil and paper tools for derivation. For anyone attacking your technique it will make no difference. An attacker with a modern computer will only brute force the part you memorized. Any key stretching done on pencil and paper will be a minor nuisance at best; anything done on paper will add no time at ...


16

Your interesting questions deserve to be answered more thoroughly, but here goes: According to a highly classified document 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, these ...


15

Edit: 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 ...


14

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.


14

Cryptography as we know it today dates from the Renaissance, in a certain sense, in a mathematical sense. --Whitfield Diffie If you look at introductory cryptography texts, you will usually see some of the same ciphers, methods, and cryptographic tools covered in a chapter on classical cryptography: The Scytale, a tool to perform a transposition cipher The ...


12

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 ...


11

The VIC cipher and for something not as secure but easier for encryption and decryption, the double transposition cipher. VIC 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 ...


11

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 ...


11

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 ...


11

I will throw tiny encryption algorithm into the mix: https://en.m.wikipedia.org/wiki/Tiny_Encryption_Algorithm 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 ...


9

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 ...


9

Yes. 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 ...


9

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.


9

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.


8

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 ...


8

I think the most simple ciphers that are available are stream ciphers. Of course there are secure and non-secure stream ciphers. But e.g. LFSR's based ciphers are pretty easy to understand, and generally you just have to deal with bitops and basic possibly (modulo) addition. Those operations are generally easy to perform "by hand". Of course, to ...


7

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 ...


7

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 ...


7

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


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

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 ...


7

You are not likely to find such a construction. One problem you will run in to is that of size: In order to be secure against brute force search with Grover's algorithm, you will need to use at least 200 bits of secret material. And that assumes you are content with a security level of $2^{100}$. This is equal to 25 random bytes or a 60 digit long decimal ...


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

If there is no upper bound on the length of the password to be used, the most common suggestion I know to create strong, easily-memorable (for some definition of "easy") password is diceware. The basic idea behind it is that it chooses each word via a roll of 5 d6's (e.g. each word has $6^5= 7765= 2^{12.92}\approx 2^{13}$ options). The entire ...


6

While a known plaintext attack successfully finds the keys, nobody has been able to put forward a general solution to this cipher. Is that possible? You really have to go back in time to learn that ChaoCipher has been subject to some cryptanalysis before the description of the algorithm/device was published. As an example: Here’s one of the oldest papers I ...


6

First of all, you cannot uniquely determine the keyword of a Playfair cipher, or even the key table constructed from it, simply because there are multiple equivalent key tables that will produce the same ciphertext (and multiple keywords that will produce each table). In particular, the following key tables are all equivalent: Original: Row shift: ...


6

There is imprecision in what is stated in your notes. The Kasisky test only works if the corresponding letters in the two segments are separated by a distance that is a multiple of the key length (in other words they are encrypted by the same letter in the key). For example (source Wikipedia): [abcdea]bcdeabcdeabcde[abcdea]bcdeabc [crypto] is short for [...


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