8

I think that you missed a pivotal point in the concept, which is the small blocks that are used to compose a secure PRF (or PRP), i.e. when you permute one bit, you actually change the value of the small block of that bit, i.e. the whole small-block is effected and thus prepared to be confused in the next round, this way you will reach a confusion of the ...


7

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 ...


5

I did it for substitution ciphers but am unable to do it for transposition ciphers. When it comes to transposition ciphers, it’s not really surprising frequency analysis doesn’t turn out to be as useful as it is when looking at substitution ciphers. See, one important strength of transposition ciphers is that they are not susceptible to frequency analysis,...


5

For a scheme to be information-theoretically secure, you need that $$\Pr[M=m\mid C=c]=\Pr[M=m\mid C=c^\prime]$$ for all $c,c^\prime$ (that is, any ciphertext has the same probability $M=m$, so the ciphertext doesn't change the probability $M=m$). Let's suppose we have a $c$ and a $c^\prime$. Both of them have the same number of ones and zeroes, because both ...


4

I wrote an answer to a related question earlier showing an example of how to break a columnar transposition cipher by hand. The basic steps are two-fold: Test different key lengths, and look for a length where the letters in each column look like they might plausibly be consecutive. Once you've picked a key length, shuffle the columns around until they ...


4

First determine the permutation domain. If $L_2 > L_1$, then the cipher is broken by $\lceil log_2 (L_2) \rceil$ pairs of plaintext/ciphertect. The security improvement is small, as the attacker only needs to use $\lceil log_2 (L_2) \rceil - \lceil log_2 (L_1) \rceil$ additional pairs to find the key.


4

As a single transformation considered in isolation, a fixed bitwise permutation does not offer any security. If you examine the hamming weight (the quantity of '1' bits) of the data before and after a bitwise permutation, you will notice that it remains unchanged. A bitwise permutation simply shuffles the order of the bits of data. Repeated applications will ...


4

I find the terms "confusion" and "diffusion" to be slightly nebulous and can lead to over-simplifications. Confusion For example, saying that "substitution" is responsible for "confusion" is not necessarily correct: "Substitution" is actually just a function application to the state; The implementation often utilizes a memoized function, but you can ...


4

Measuring the size of the key space in transposition algorithms is not important, because their security is far less than the size of the key space would suggest. Therefore, any measure of effective key length will be misleading and will not give an accurate picture of the true security of the scheme. In general the standard way to compute the effective ...


4

Think about what properties of the plaintext are preserved by the candidate ciphers. You want to look for those patterns. Transposition moves letters around but keeps their identity. Substituting keeps positioning but changes identity. So, you can do a test for whether letter identity is preserved. The easiest is to do a frequency analysis: if your ...


3

What method of transposition does AES use? Transposition is a simple movement of data to an alternative location. There are two relevant steps in AES when it comes to transposition: addRoundKey and shiftRows. addRoundKey The addRoundKey step incorporates a bitwise rotate left. This is a transposition of bits within a single word. shiftRows The shiftRows ...


3

I don't know where you get that the central element can not be $1$, because the Cardan grille does not have any special "requirements" as you're implying. In the original version, a piece of cardboard has several holes cut in it (= the grille) and when it is placed over an innocent looking message, the holes cover all but specific letters spelling out the ...


2

I think you're referring to Fleissner grilles, which are kind of a subset of Cardan grilles. According to wikipedia, to construct one you "make 16 perforations in an 8x8 grid – 4 holes in each quadrant. If the squares in each quadrant are [identically] numbered 1 to 16, all 16 numbers must be used once only. This allows many variations in placing the ...


2

Obviously such a combination of ciphers is stronger than any of these two ciphers alone. The Vigenere cipher provides "confusion", while the transposition cipher provides "diffusion". These are terms used by C. Shannon, refer to http://en.wikipedia.org/wiki/Confusion_and_diffusion. Modern ciphers like AES are making use of diffusion and confusion as well. ...


2

The key is obtained by taking the encryption steps in reverse. For example, let's encrypt the word "United" with that 4 3 1 6 2 5 key. The first step is to take the 4th letter, a "t" and make it the first. In the decryption, this first ciphertext letter must go to the 4th place again. So the fourth step in the decryption, must be to look at what is in the ...


2

This is not very secure. You directly leak the symbol distribution, because only the order of symbols changes. For short enough messages this allows easy decryption – e.g. "dr olllWeoH" is quite clearly "Hello World". Even for long messages or binary values, the fact that you leak e.g. a crucial byte may be enough. You also have not defined how the same key ...


2

I'm assuming that your Columnar Transposition Cipher (CTC) is as described as this site, since I found that the keyword ANNIE works. OTP First of all, the combined cipher (Caesar shift 7 plus CTC) can always be the OTP, as long as we don't require a truly random OTP. The simple reason; x-or the plaintext and ciphertext get the OTP keystream. This is your ...


1

If you know for sure that it is a transposition of some kind you can chek the average frequency with which the various letters appear in the natural language of the plain text and then compare them to the frequencies of letters in the cyphertext. You can easily find this data on google, here is the example for english. This method works only if the ...


1

One method is to simply try and break it. If you get an answer by analyzing the ciphertext assuming it is from one method then you can disregard the other one, especially if you have a large ciphertext such as 10,000 letters. Trying both methods only doubles the cost to the adversary after all. As breaking a classical/simple transposition or substitution ...


1

can a key be applied to half the cipher text Stream ciphers or Block Cipher with CTR mode can do so. how much time would it take to encrypt a 1.6 mb file with an average encryption algorthm. in seconds or miliseconds. like half a second? It depends on hardware (CPU, speed of the driver, etc), software (OS, the programming language, etc) and the ...


1

In a chosen plaintext attack, the expected ciphertext length required to conclude that a decryption is correct was derived by Claude Shannon and is called the unicity distance. It depends on the distributions of the plaintext and the key and is infinite for the one time pad. So if $K$ is the random variable describing the key with a known distribution, and ...


1

What you are describing is called a transposition cipher. First of all, it would be easy to detect which method you are using since the frequency distribution of your cipher text will be the same as the language of the plain text. I think the biggest problem with using only transposition as opposed to a combination of both substitution and transposition ...


1

For Decryption Process, use the reverse of encryption process. I think you are using a wrong table format. The table should be 4*3 instead of 3*4. Now, Write your encrypted text column by column(4 rows and 3 cols). It will be like this. e h l o l _ h t e e r x Now, Apply the decryption key of key 2 1 3. The decryption key will be the same. 2 1 3 | | | ...


1

This is more a permutation cipher rather than a transposition one. About the principle if in your key you have : ENCRYPTION position -- key 1 4 2 3 3 1 4 6 5 2 6 5 So considering a 6 input block : a b c d e f, you would have : a goes to the 4th position. b goes to the 3rd position... ...


1

(Although your question is for a long time ago) Transposition ciphers don't change the frequency of letters in their ciphertexts. So, you can implement a frequency attack. For decryption by this you only need the table of frequency of English letters and your ciphertext. (Your encryption scheme here is a Columnar transposition with a key that all its letters ...


1

It really depends on the key schedule. IE, how is the key operated on to find the permutation to be used? Absent that knowledge, really the only way to know is to launch searches in parallel at different key lengths, and the one that finds the answer first had the right key length. Unfortunately anything more clever than that will utterly depend on how ...


1

Sorry to burst your bubbles but i disagree, transposition ciphers are susceptible to frequency analysis. Transpotion ciphers only change their relative position not their individual symbols, thus their frequency can be compared with other plaintext candidates to find the most probable ones. (Or by using anagram analysis) To elaborate, transposition cipher ...


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