Question

Assume that Alice uses Bob's Elgammal public key (e1 = 2 and e2 = 8) to send two messages P = 17 and P' = 37 using the same random integer r = 9. Eve intercepts the ciphertext and somehow she finds the value of P = 17. Show how Eve use a known-plaintext attack to find the value of P'

Answer 1

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Answer 2

Answer:

Concept:

Public-key cryptography is used in ElGamal encryption. It encrypts the message and employs asymmetric key encryption for two-party communication.

This cryptosystem is based on the fact that it is quite challenging to compute $g^{ak}$ even when we know $g^{k}$ and $g^{a}$, a discrete logarithm in a cyclic group.

Explanation:

Given:

Public key $\alpha=2, y_{n}=8$

Two message $P=17 and P^{\prime}=37$

integer r = 9

Find:

Show how Eve uses a known-plaintext attack to find the value of P'.

Solution:

Assume parameter = 11

let's assume that working in$z_{p}^{*-1}$you have a generator g Public key is $L=g^{x}\left(c_{1}, c_{2}, p\right)$

$c_{1}=2, c_{2}=8, p=11$

Both are public parameters and g be a generator of $Z_{p}^{*}$. Alice selected $P=17 which \alpha \in Z_{p} and the send \alpha= ga \bmod \ p$ Bob. Bob selects a random $Y_{n}=8$send $P^{\prime}=17, Y_{n} \in Z_{p}$ and mod p to Alice. The shared key is $g^{ab}$  Mod p,  Alice $=y^{a}$mod $p$ , and gab $\mathrm{mod} \ p$, message $m \in z^{*}$ Bob $=x^{b}$ mod p.

A message $m \in z_p^{*}$ may be encrypted using the key as C = P

so, $g^{ab}$ mod p. Both ciphertexts share the randomness k for $P^{\prime}$ from P.

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