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11

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.


6

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 ...


5

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 ...


4

Block cipher modes of operation don't complicate the crypt-analysis of the underlying cipher (much). They are required to create a CPA secure cipher out of a block cipher (using a random IV, and making sure that the rest of the ciphertext is chained to this IV somehow). Sometimes error propagation is also used for very specific purposes. In other words, the ...


4

To be honest, I think all of them do NOT make much difference, since Caesar Cipher is a weak cipher. You can see Breaking the cipher on Wikipedia. It uses a method called frequency analysis, which ignores key strength. Also, the three methods you mentioned all have their weaknesses. You should use a secure random generator.


3

Depending on your video game design you maybe can utilise a many-time pad. Many-time pad is many plaintexts encrypted by XORing them with a common string. One-time pad cannot be broken but many-time pad can, especially if the players can make an educated guess about what the plaintext message might contain. You can also try encrypting numbers with RSA but ...


3

The other answer correctly explains how we could make a modern stream cipher, with a public random IV at the beginning of ciphertext used (together with the key) in the setup of a keystream generator, restricted to the range $[0..35]$, and using addition modulo $36$ (Beaufort cipher) to encipher each character. Here is a more detailed description of such a ...


3

How secure is this modified Bazeries Cylinder? That depends on what you expect us to compare it to. I doubt you’re looking for an answer along the lines of “it’s more secure than Scytale, but less secure than AES”. Therefore, I will try to narrow it down a bit by saying that it is safe to assume it can not – in any way – provide the same security levels ...


3

It's not elegant, but the brute force method is to write a program that creates a table of 25x25 digraphs (assuming i=j), yielding 625 rows. I'd also add a column that lists the relative frequency of each digraph (given enough ciphertext you can use that to identify frequent substitutions, as you already have done). You start off with 625! possible ...


2

1:1 ciphers Many modern ciphers create ciphertexts that include the plaintext encrypted with a 1:1 stream cipher, plus a few more characters that help defend against certain known stream cipher attacks. As poncho and Rup have suggested, Alice and Bob somehow generate the same long stream of characters that is unguessable by the adversary. (If that stream ...


2

As explained by Cort Ammon, they mostly relied on physical, rather than mathematical processes. Here is a video on Youtube titled "How to launch a nuclear missile", taken in the Titan Missile Museum mentioned by Cort Ammon. The message contains a seven character string. The personnel in the shelter have a set of envelopes with the first two letters on the ...


2

It depends on the time you want to spend. But most likely, there is nothing with reasonable efficiency. For arithmetic operations, humans are really bad compared to computers, and the difference is at least a factor of $10.000.000$ (very very rough guess, probably even 1+ additional zeros there). So, since you have to assume that the attacker has access to ...


2

There was a telegraphic code that mapped the comparatively frequently used 10000 words (that's very much more than for common daily use, e.g. newspapers) of Chinese to [0,9999]. There was once also a patented mechanical device that effected a poly-alphabetical substitution of [0,9] with a set of disks containing certain scrambled alphabet of [0,9], which the ...


2

I have tried with some lengths but it worked so that the product was Vigenère with length of LCM. Actually, the statement really isn't "is encrypting with cipher $\mathcal{V}_1$ and the cipher $\mathcal{V}_2$ always equivalent to some cipher $\mathcal{V}_3$. Instead, it's "are the set of transforms that you can express by first encrypting by some ...


2

There are several algorithms available which can attack a playfair cipher. Hill climbing might be one option. Basically it starts with a random key (assuming it's the best one) and decrypts the cipher. The resulting clear text is scored using a fitness function. Then small changes are applied to the key and if the resulting clear text of the modified key ...


2

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 ...


2

Pre requisite : understand how Vigenère cipher works (basically it is a Cesar cipher where the key change at each character). Breaking a Vigenère Cipher text is decomposed into 3 steps : Find the key length possibles lengths by observing occurrences. Frequency analysis to retrieve the key. Breaking the cipher text. Find the key length possibles ...


1

My impression is that it would at most take a longer ciphertext to be decrypted. Imagine first the scenario where the different keys have the same length: $K_1=hello$ $K_2=world$ In that scenario, the keys add up and the result of applying $E_{K_2}(E_{K_1}(m))$ can be expressed as one single operation: $E_{K*}(m)$, where $K*=dscwr$. Therefore, not a lot ...


1

No, nothing is ever undisputed. There is a cipher that provides perfect security (OTP) but that doesn't mean it is very practical if just for key distribution. So saying it's "the best" can be and should be disputed. However, the answer the quiz master is looking for is probably the Vigenère cipher. I presume however that there must have been at least one ...


1

A few things come to mind: Something based on XOR. I'm imagining an endlessly (and randomly) streaming bridge of tiles that can either be in the left position or the right position (ie the message) and you can only cross the bridge if they all line up. Straight bridge: ---------- You know that the bridge started straight, but has been XOR'd with ...


1

You could obscure the frequency using a block-cipher or stream-cipher or use a code-book substitution, then maybe use Vigenere Cipher on it, but if you really want your scheme to be secure, then your underlying substitution must be secure enough. In that case using secure block or steam encryption itself is enough. The second level Vigenere encryption is ...


1

In a Vigenère cipher, the $i$-th ciphertext letter $c_i$ is calculated as $p_i + k_{i \bmod \ell}$, where $p_i$ is the $i$-th plaintext letter, $k_i$ is the $i$-th key letter, and $\ell$ is the length of the key. Thus, the difference between two ciphertext letters at positions $i$ and $j$ is $$c_i - c_j = p_i - p_j + c_{i \bmod \ell} - c_{j \bmod \ell}.$$ ...


1

Yes, you can perform rudimentary asymmetric encryption or signing by hand, but no it won't be secure. Textbook RSA with small enough numbers is easy to do by hand. Diffie-Hellman to exchange a symmetric key is even easier. However, the kinds of numbers for which this is feasible are far from those needed for real security. There are a lot of other ...


1

Assume the length is $n$. If the cipher text is $c_0, c_1, c_2, \ldots, c_N$ then consider the sub-text consisting of the characters $c_0, c_n, c_{2n},\ldots$. These have all been encrypted with the same Caesar, and you can break it by frequency analysis (the shifted 'e' should be the most common in standard English texts, or else maybe the 't' etc.). ...


1

If the ciphers are different, with independent keys, you can say that it is at least as strong as the first cipher. If the ciphers commute, like with stream ciphers, you can even say that it is at least as strong as the strongest. See Cascade Ciphers: The Importance of Being First. That's really all you can say in general. In practice, the combinations you ...


1

In general, all you can say is it can be as weak as the weakest encryption layer, if you're lucky. Edit: It can also be even weaker, for badly chosen components that cancel out some mathematically desirable properties, as pointed out in the comment.


1

The Chinese used digits. So there is an encoding into digit groups, like a code book. This code book could be standard (like the Chinese used for telegraphy and radio as well, which used a publicly available book), or for extra secrecy could be kept a secret. The cipher could then be made to work on digit groups (and there standard techniques like addition ...


1

In the past, they relied far more on human techniques than cryptographic ones. The Trident example you linked is interesting to me, because it is conflicting with my experience receiving a tour in the Titan Missile Museum in Arizona. I won't claim my tour qualifies as a reliable source, but it's what I was given. In the tour, they explained the process ...


1

My understanding is given that $E_K(x)=(x+K)~mod~26$ is the character by character encryption of a shift cipher, you want to extend it to words as the function $E'$ given by $$E'_K(x_1,\ldots,x_m)=(E_K(x_1),E_K(x_m)).$$ If you intended the key $K$ to change per character you can modify.



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