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I think that letter frequency and entropy can give us some tips if the password source is biased. What advantage can these measures give us if the passwords are chosen with a good percentage of randomness? Therefore, people don't toss coins to choose passwords. So, suppose a bad way of choosing passwords, like this: always starting with (03) three numbers.....


2

I was wondering, would calculating the entropy of the resulting plaintext not serve the same purpose, if not to a better degree seeing as it's not confined to the english language? First note that a cipher is susceptible to frequency analysis if it doesn't change the frequency at which characters occur. That is the "probability" of each substituted letter ...


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One tool that tries to do this is untwister. It's almost certainly not the tool you were thinking of, though, as it cannot determine if the output came from OpenSSL specifically. It can determine Glibc's rand(), Mersenne Twister (MT19937), PHP's MT-variant (php_mt_rand), Ruby's MT-variant DEFAULT::rand(), and Java's Random() class, though, and can recover ...


31

A colleague of mine told me about a website that, given a sufficient quantity of output from an PRNG, had been able to deduce which application the PRNG was from. As you correctly identified this would present an immediate and probably devastating attack to any cryptographic PRNG as it indeed would allow you to easily distinguish a random string from a PRNG ...


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In cryptography you usually want to use min-entropy (which is a lower bound on security) instead of shannon-entropy (which can be higher than the security if the attacker is content with breaking only a fraction of the targets). Min-entropy is simply $-\log p$ where p is the probability of the most likely value. In this case this works out to: $$-16 \cdot \...


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First of all the first Entropy calculation is not correct. Recall the definition of (Shannon) entropy: $$H(X) = -\sum_{i=1}^n {\mathrm{P}(x_i) \log_{\,b} \mathrm{P}(x_i)}$$ If uniformly chosen a digits entropy is $$H(X) = -\frac{1}{10} \log_2\left(\frac{1}{10}\right) = 3.3219280...$$ Case $<250$ with 16 digits; \begin{align} H_1(X) &= 16 \cdot - ...


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