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0

I think Recharge card can be hacked but not by some amatures like us. I tried once.Went deep into it and found that usually those using recharge card use LUHU ALGORITHM. Also permutation is used. If you are intrested you can send me email @ [REMOVED EMAIL ADDRESS]


3

As one of the authors of the paper, let me give you an answer. The operation $F$ is indeed applied to both $x$ and $x'$. By stating that $\oplus$ is invariant under rotation, we mean that if you first rotate $x$ and $x'$ and take the difference with $\oplus$, you get the same result as if you first take the difference with $\oplus$ and then rotate the ...


5

Nobody can tell you not to "have fun with it" but I would strongly recommend you to first study attacks on other ciphers. Spritz (Rivest & Schuldt) fortunately mentions a lot of research on its predecessor, RC4. This makes it a rather good starting point in my opinion. It is necessary to understand the linguistics and mathematical constructs that are ...


2

PBKDF2 is defined for an arbitrary PRF, but in practice HMAC is usually used. Either with SHA-1 (original definition), SHA-256 (e.g. in scrypt) or even SHA-512-256 (NaCL). So first you can look at how swapping key and message affects HMAC. HMAC has two cases depending on the length of the key: if it's no longer than the block size of the hash, it is used as ...


4

Assuming you really had broken AES or another frequently used algorithm that is thought to be secure, the first step would be to prove it. Write the code for the attack. Verify that it works on randomly generated data of the kind it requires. If it can break some challenge (e.g. these), do it. Post the results to the challenger or show the results ...


1

Basically, "length" increases the time a brute force or other generic attack takes, while "complexity" makes it more difficult or less likely to find attacks faster than those. However: The length in question varies depending on the algorithm or operation. It can be the length of keys, the block size of a block cipher, the state size of a PRG or the ...


2

Assuming for the moment that your claim is correct, I would suggest caution in revealing the details of your findings. After having your results validated by one or two people with the skills to do so (and whom you trust to keep things confidential), then some sort of general announcement (without specifics) would be best, to give people time (say three ...


10

In complete honesty: if you have to ask this question, it's overwhelmingly unlikely that you have actually succeeded in breaking the security of AES. At best, you may have discovered a well-known attack against misuse of particular block cipher modes; for instance, plaintext recovery with a chosen-ciphertext attack against ECB, or blind manipulation of the ...


-1

Make sure you put your name prominently in the source, and publish the source anywhere. If you're right, everyone else will make you famous. If wrong, someone will point out your error. Send it to Bruce Schneier if you want to start at the top.


3

If you can read the intermediate states of the encryption algorithm you could recover, one by one all the round keys. Given a AES round, all the operation between the two AddRoundKey (at the beginning and the ond of the round) are invertible. Take for example round 1: you get the internal state before AddRoundKey (of round 2), you get back at the beginning ...


1

If the master key is strong (e.g. random 256-bit key), $c=1$ is fine, or you can use HKDF. A high number of iterations is only needed when you derive the key from a password or other low entropy string. If you can store a 4000 element table securely, you could just use random keys instead of derived ones. If you need the derived key to depend on a low ...


1

You have 3 equations and 2 unknowns, so it is solvable, assuming a solution exists. You can plug this into any linear equation solver. If you subtract equation 3 from equation 2, you get $a=-3$, and can solve for $b=75$. This fits equation 2 and 3, but not equation 1. So, no solution exists.


0

For small numbers the easiest ways to break RSA are: Brute force the private (decryption) exponent. I.e. find the smallest value $b$ such that $(x^a)^b \mod n = x$ for all $x$. Factorize the modulus (trial division is simplest and works fine for small numbers), then use the factors to generate the private exponent. Assuming you are supposed to use the ...


5

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 ...


4

This sounds like a classic codebook or nomenclator. Even if we assume a perfect random oracle that generates a completely random codeword for each word of English text, I agree with otus that frequency attacks and N-grams would likely be able to decode the most-frequently-used words. Also, a known-plaintext attack (or worse, a chosen-plaintext attack) would ...


1

Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations: They would need more samples to differentiate between two tokens at the same level of confidence. For two different tokens with similar frequency they would no longer know that the words differ. The first means you are making the ...


5

The feasibility depends a lot on the length of the corpus. The more statistics, the better guesses an attacker would be able to make. He'll try to use statistical attacks to fit the frequency curve of the tokens to the frequency curve of English words. This will allow him to guess the preimages of more frequent words with high confidence, but will he be ...


2

Provided the text was long enough and used an simple codebook substitution cipher, absolutely. English has common bigrams and trigrams as well as words that are typically positioned in certain places in sentences, like The. Also, if punctuation was tokenized in the codebook, it would be incredibly easy to guess or identify . and ,, because those will be ...


4

I thought you were using a block cipher, i.e. a pseudorandom permutation. Instead as per your comment you are only permuting the order of the plaintext bits. This is not secure. For example, you can imagine the bit permutation is an n-by-n square matrix, where each row and column has a single 1 and the rest 0s. The input and output are then vectors of size ...



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