2
$\begingroup$

I came across this problem, while currently reading "Introduction to Modern Cryptography" by Katz and Lindell. I am new to crypto and just trying to go through the book and solve the exercises, unfortunately, got stuck at this one.

Consider the following public-key encryption scheme. The public key is $(G, q, g, h)$ and the private key is $x$, generated exactly as in the El-Gamal encryption scheme. In order to encrypt a bit b, the sender does the following:

  • If $b = 0$ then choose a uniform $y ∈ Z_{q}$ and compute $c_{1} := g^{y}$ and $c_{2} := h^{y}$. The ciphertext is $(c_{1} , c_{2}) $.

  • If $b = 1$ then choose independent uniform $y,z ∈ Z_{q}$ , compute $c_1 := g^{y}$ and $c_{2} := g^{z}$ , and set the ciphertext equal to $(c_{1},c_{2})$.

Show that it is possible to decrypt efficiently given knowledge of $x$.

Would appreciate some help

$\endgroup$
  • $\begingroup$ Hint 1: take $x$-th power of $c_1$. $\endgroup$ – kelalaka Jan 16 at 19:49
  • $\begingroup$ Thanks, that helped for b=0! Still not sure how to approach for the case b=1 though. $\endgroup$ – badoo Jan 16 at 20:14
  • $\begingroup$ So you get equality right? $\endgroup$ – kelalaka Jan 16 at 20:14
  • $\begingroup$ Yes, I do for b=0. $\endgroup$ – badoo Jan 16 at 20:17
  • $\begingroup$ Then if there is no equality it means? $\endgroup$ – kelalaka Jan 16 at 20:17
1
$\begingroup$

To determine the bit plaintext from $(c_1,c_2)$ take $x$-th power of the $c_1$. There are two options that we don't know

  1. $(c_1,c_2) = (g^y,h^y)$ then $(c_1^x,c_2) = (g^{xy},g^{xy})$
  2. $(c_1,c_2) = (g^y,g^z)$ then $(c_1^x,c_2) = (g^{xy},g^{z})$

Now check whether $c_1^x = c_2$ or not.

  • If they are equal then than bit $b=0$
  • If they are not equal than bit $b=1$.

There is a small chance that we will have $g^{xy}=g^{z}$ as false $b=0$. The probability of this event is $1/g$.

The cost of operation: if we use Left-to-right binary method for modular exponentiation than the cost is $\mathcal{O}(\log(x))$, therefore, it is efficient.

$\endgroup$
  • $\begingroup$ Note: the question is solved by OP with the hints. $\endgroup$ – kelalaka Jan 16 at 20:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.