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Is there any restriction that the members to be accumulated in bilinear map (multiplicative) accumulator need to be relatively prime to $P$, where $P$ is the order of the group.

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  • $\begingroup$ I guess you mean the t-SDH based accumulator? Here $P$ is prime and your accumulation domain is $1...P-1$. So this holds for any value in the domain anyways and you do not need to care about that. $\endgroup$
    – DrLecter
    Jun 25, 2015 at 6:20
  • $\begingroup$ Is there any restriction on the size of P ? Can I use a 3092 bit prime no as P and calculate the witness using polynomial evaluation (with out accumulator secret key)? will it very badly effect the computational complexity? $\endgroup$
    – manju
    Jun 25, 2015 at 6:41
  • $\begingroup$ Why do you want to use such a large prime $P$? The size of $P$ is determined by your security parameter and using for instance BN curves for your pairing setting this will be 256 bit for good security. If you want to enlarge the domain of your accumulator you can always use a collision resistant hash function to map arbitrary size strings to elements in $\mathbb{Z}_P^*$. $\endgroup$
    – DrLecter
    Jun 25, 2015 at 6:44
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    $\begingroup$ thank you for the wonderful paper 'Revisiting cryptographic accumulators ..." $\endgroup$
    – manju
    Jun 25, 2015 at 9:47
  • $\begingroup$ Nice to hear that you enjoy it! :) $\endgroup$
    – DrLecter
    Jun 27, 2015 at 12:42

1 Answer 1

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If the group order $P$ is prime, then you can accumulate any integer in $[0,P)$.

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