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Let $G$ be a finite group of prime order $p$, and $g$ a generator of $G$. The standard DDH is hard to distinguish two distributions $$ \{ (g, g^a, g^b, g^{ab}) : a, b \leftarrow \mathbb{Z}_p\} \text{ and } \{ (g, g^a,g^{b}, g^r) : a, r \leftarrow \mathbb{Z}_p\}. $$

Is still secure DDH with multiple instances? That is, is hard to distinguish two following distributions? $$ \{ (g, g^a, g^{b_i}, g^{ab_i}) : a, b_i \leftarrow \mathbb{Z}_p\} \text{ and } \{ (g, g^a,g^{b_i}, g^r) : a, r_i \leftarrow \mathbb{Z}_p\}. $$ We also suppose that the cardinality of the set, $|\{b_i\}|$, is much smaller than $p$ to avoid easy cases.

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    $\begingroup$ Is it naturally true due to the self-reducibility of DDH? $\endgroup$ Aug 23 at 13:07
  • $\begingroup$ Is this homework? If it is, you should clarify it. $\endgroup$ Aug 23 at 14:05
  • $\begingroup$ @GeoffroyCouteau No. Not homework. just curious things $\endgroup$ Aug 23 at 14:26
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This can be solved via a standard hybrid argument. I won't give you all the details. However, note that given a single tuple $(g,h_1,h_2,h_3)$ you can generate a tuple of the form $(g,g^a,g^{b_i},g^{ab_i})$ by choosing $b_i$ and forming $(g,h_1,g^{b_i},h_1^{b_i})$ and you can generate a tuple of the form $(g,g^a,g^{b_i},g^r)$ by choosing $b_i$ and forming $(g,h_1,g^{b_i},g^r)$. This suffices for building hybrid distributions as needed for a hybrid argument.

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  • $\begingroup$ Thank you for your hopeful comments! $\endgroup$ Aug 24 at 13:59

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