# Tag Info

4

PE is a subclass of FE. This (from the other answer) is correct. Also, from my understanding, your analogy is correct. PE returns the plaintext if the predicate evaluates to true. FE, on the other hand, returns a function of the plaintext. We can say that PE is a subclass of FE, since we can use FE to implement PE. Just use the identity function. ...

4

Public key crypto vs. identity-based crypto made short: In traditional public key cryptography, a user $A$ generates a private/public key pair $(sk_A,pk_A)$ and since this key pair has absolutely no indication to which indentity (user $A$) it belongs, it is necessary to certify the public key, i.e., bind the public key $pk_A$ to the user $A$'s identity. ...

2

You can probably prove the security against your game from the security in the IND-ID-CCA game of Boneh and Franklin (see http://courses.cs.vt.edu/cs6204/Privacy-Security/Papers/Crypto/IBE-Weil-Pairing.pdf). The idea is to create an adversary $\mathcal{B}$ against IND-ID-CCA from your adversary $\mathcal{A}$. Essentially $\mathcal{B}$ will play ...

2

What you are describing is an anonymous credential system. There are two different ways to go about making these and two actual systems that use those techniques: Microsoft's U-prove and IBM's Idemix. If you're interested in smart card usage, you'd probably prefer U-prove as it tends to work better with smart cards. It's described by its original ...

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If Bob and Charlie will share a secret, then it information-theoretic privacy might be possible, otherwise the best that can be hoped for is computational anonymity. I haven't come up up with any way to achieve information-theoretic privacy when Bob and Charlie share a secret, so I will only be addressing computational privacy. I assume that Bob can ...

1

Ok, lets look at the operations. Sign: $s = g * r^{f(t,m)} \pmod n$ This is an assignment. You compute $(g * r^{f(t,m)}) \mod n$ and assign the resulting value to $s$. If you have a multiplication $(a \cdot b) \mod n$, this is equal to $((a \mod n)\cdot (b \mod n)) \mod n$. See for instance here. Verification: $s^e = i * t^{f(t, m)} \pmod n$ This is no ...

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