My question is about the existence of a dynamic RSA accumulator with deletion of an element in O(1) time.

Do you know some practical implementation?

  • 1
    $\begingroup$ When you say O(1) you mean with respect to the number of elements in the accumulator, right? Because the cost obviously increases with increasing modulus size. $\endgroup$ Dec 7, 2015 at 20:59
  • $\begingroup$ Yes, I mean respect to number of elements. $\endgroup$
    – 4nf3rt
    Dec 7, 2015 at 21:08

2 Answers 2


Given the trapdoor, one would delete an element from RSA accumulator in constant time. In particular, produce an inverse to the element with extended Euclid algorithm and power-to accumulator to the inverse. The element in question would cancel-out from accumulator this way.

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    $\begingroup$ But in general, deletion is slower than add an element? $\endgroup$
    – 4nf3rt
    Dec 7, 2015 at 21:15
  • $\begingroup$ Modular exponentiation is expensive. In terms of exponentiations, cost is exactly one for this particular accumulator. $\endgroup$ Dec 7, 2015 at 21:27
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    $\begingroup$ It would be better to include a definition and refer to an implementation, giving a background to the answer above. Yes, accumulated value is $A_i$, equation (2) at Michael T. Goodrich, Roberto Tamassia, Jasminka Hasic, An Efficient Dynamic and Distributed RSA Accumulator (arXiv:0905.1307, 2009). Thanx fgrieu. Please note a fast deletion was suggested with trapdoor access at the answer; without trapdoor one need to re-calculate the accumulator in $n$ exponentiations. $\endgroup$ Dec 7, 2015 at 21:40
  • $\begingroup$ And without the knowledge of the secret ϕ(n), is possible only add elements to accumulator (in a efficient way)? $\endgroup$
    – 4nf3rt
    Dec 8, 2015 at 12:30
  • $\begingroup$ Nope; it is possible, under some circumstances to delete in constant-time from an RSA accumulator without the trapdoor. See my full answer below. $\endgroup$ Feb 5, 2023 at 2:45

Lattice-based accumulators

Yes, there exists an accumulator with $O(1)$ deletion time from lattices by Papamanthou et. al [PSTY13]. Specifically, given an old digest and the deleted element, one can easily compute the new digest. No auxiliary information is needed.

I do not know how its computational performance compares to RSA accumulators; likely more expensive. There was some (partial) exploration of its performance in [CPZ18], which you could investigate.

I do recall that its proof size is $O(\log{n})$, so it will be worse both asymptotically and concretely than RSA accumulator proof sizes.

RSA accumulators

At the same time, note that there is also a way to update an RSA accumulator without the trapdoor.

Specifically, if you have a membership proof for the deleted element, then the updated accumulator is simply that membership proof.

It gets a little trickier if you want to update an RSA accumulator after two or more deletions. This requires a so-called "Shamir trick", which I'll explain below.

Let $\mathsf{BatchDel}$ denote the algorithm that updates the accumulator after two or more deletions.

Specifically, $\mathsf{BatchDel}$ takes an accumulator $A_t$ as input and deletes all the elements $x_i$ from it given a membership proof $\pi_{x_i}^t$ w.r.t. $A_t$ for each $x_i$

Then, $\mathsf{BatchDel}$ works as follows (screenshot from [BBF19]):

As you can see, $\mathsf{BatchDel}$ makes use of Shamir's trick, which in turn relies on computing Bezout coefficients (see screenshot below also from [BBF19]):

enter image description here


[BBF19] Batching Techniques for Accumulators with Applications to IOPs and Stateless Blockchains; by Boneh, Dan and Bünz, Benedikt and Fisch, Ben; in CRYPTO'19; 2019

[CPZ18] Edrax: A Cryptocurrency with Stateless Transaction Validation; by Alexander Chepurnoy and Charalampos Papamanthou and Yupeng Zhang; 2018; https://eprint.iacr.org/2018/968

[PSTY13] Streaming Authenticated Data Structures; by Papamanthou, Charalampos and Shi, Elaine and Tamassia, Roberto and Yi, Ke; in EUROCRYPT 2013; 2013


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