# Tag Info

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By taking enough samples, it is possible to ensure that the majority value is the correct one with high probability. To prove this, you use Chernoff's bound. Now, the more samples that you take, the higher the probability that the majority value is the correct one (when plugging into Chernoff). So, this is a parameter that you can choose however you want. We ...

4

Note: In this answer, I stick to a definition of the One Time Pad where the random pad is used only One Time; at least, I've the name of it as support! Otherwise, it is well known that the OTP encryption scheme consisting of XOR with a repeated key is insecure by even the weakest standard (unknown plaintext with redundancy). INDistinguishability under ...

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Here's the next step in the iteration, which should be easy to understand: Let's call the oracle on 2P and 4P: Answer (even,even) means, that $P<N/4$ (this is still easy: Otherwise either 2P or 4P would be greater than N). Answer (even,odd) means $N/4<P<N/2$. (odd,even) means $N/2<P<3N/4$ and (odd,odd) means $3/4N<P<N$. Actually, ...

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We have a Padding Oracle if there is a different response from the server gives us an indication of the correctness of the pad (say if this needs proving). We can establish this by playing a game where we send badly padded cipher-text and random strings to the server, finally submitting some at random and seeing if we can get a non-negligible Advantage is ...

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This represents the probability over all $K$ that $A$ given an oracle access to $F_K(\cdot)$ outputs 1. You usually compare that to the probability of A ouptputting 1 while having oracle access to a random function and the difference tells you how well $A$ does at telling $F_K$ from random

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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 ...

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The approach with which I solved the problem is indeed as @tylo suggested. Initially we know that the target plaintext $P$ is within the bounds $[0,N]$ where the lower bound $LB=0$ and the upper bound $UB=N$. Now we iterate the following algorithm $log_{2}N$ times to find P from the original intercepted ciphertext $C$ $C' = (2^{e}\mod N) * C$ if (Oracle(C'...

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In the standard definition of security for public key encryption schemes there exists only one public key. Therefore the decryption oracle will always decrypt with the secret key that corresponds to the public key given to the attacker. It does not matter how the $c_1$ in your question is computed, it can be computed using the real public key, a different ...

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It is called circular security. It is problematic because it is not captured by the usual security definitions. I.e., even if an encryption scheme is proven secure by some regular definition, it is usually not a given to be circular secure. To see why consider, for example, the usual definition of semantic security for a public-key encryption scheme \$\Pi = ...

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So this feels like a homework question, as such I"m not going to give you the full answer, but yes, yes you can. https://en.wikipedia.org/wiki/Homomorphic_encryption#Unpadded_RSA Is the best starting point I can give without giving away the barn, but essentially rsa is homomorphic, and you can exploit that and repeated calls to the oracle to do what you ...

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Prove? Why does the attacker need to "prove" it? For example, the attacker can check whether there is an oracle by looking at the code and seeing whether such an attack is possible. Or, the attacker can guess that such an attack might be possible and then try the attack. If the attack succeeds, the attacker knows the system is vulnerable. There might ...

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