Join Log in
Score:1
avatar Crypto

CDH solver algorithm construction

cn flag
user766787

If $A$ is an efficient algorithm that solves the Computational Diffie-Hellman problem for $\frac{1}{2}$ of the inputs and returns a special symbol for the rest, how can I use $A$ to construct another algorithm B which solves $CDH$ with a higher probability ($1 -\frac{1}{2^k}$) ?

55
1 + 5
SEJPM avatar
us flag
SEJPM
Hint: Can you perhaps transform a given CDH challenge into an independent-looking different one from a solution of which you can recover the original answer?
2
Reply
cn flag
user766787
@SEJPM I am not able to understand how. Could you please elaborate a bit more? For any random instance how should B use A as a subroutine ?
0
Reply
cn flag
user766787
I mean, how should B query A to get the correct answer ?
0
Reply
SEJPM avatar
us flag
SEJPM
As A is unreliable, B needs to construct fresh, related CDH challenge inputs, can you think of ways to create $g^x,g^y$ from $g^a,g^b$ s.t. $g^{xy}$ helps you recover $g^{ab}$?
0
Reply
cn flag
user766787
@SEJPM By taking some k multiple of a,b as x,y ?
0
Reply
Score:2
Chain Markov avatar Crypto
de flag
Chain Markov

Suppose $C_k: (g, g^a, g^b) \mapsto g^{ab}$ is your CDH-solver, that solves it with probability $1 - \frac{1}{2^k}$ and the special symbol otherwise. Let's construct them from $C_1$ recursively.

Construction of $C_{k+1}$:

  1. Calculate $C_k(g, g^a, g^b)$. If it returns $g^{ab}$, output it (probability of this is $1 - \frac{1}{2^k}$). Otherwise, go to the second step.
  2. Generate two independent random numbers $x$ and $y$ uniformly distributed from 1 to $p-1$.
  3. Calculate $g^{ax}$ and $g^{by}$. Note, that they are independent from $g^a$ and $g^b$.
  4. Calculate $C_1(g, g^{ax}, g^{by})$. If it returns the special symbol, output it (probability of it is $\frac{1}{2^{k+1}}$). Otherwise it has calculated $g^{abxy}$. Go to the fifth step.
  5. Use the extended Euclidean algorithm to find $z$, such that $xyz \equiv 1 (\mod p-1)$.
  6. Calculate $g^{abxyz} = g^{ab}$ and output it (probability of it is $\frac{1}{2^{k+1}}$).

It is not hard to see, that the total probability of outputting $g^{ab}$ is $1 - \frac{1}{2^k}$ now.

0 + 0
ธงชาติไทย
Elon Musk
I sit in a Tesla and translated this thread with Ai:

mangohost

Post an answer

Most people don’t grasp that asking a lot of questions unlocks learning and improves interpersonal bonding. In Alison’s studies, for example, though people could accurately recall how many questions had been asked in their conversations, they didn’t intuit the link between questions and liking. Across four studies, in which participants were engaged in conversations themselves or read transcripts of others’ conversations, people tended not to realize that question asking would influence—or had influenced—the level of amity between the conversationalists.