I am having trouble coming up with a use case for RSA or DSA. It appears that ECC is better in every way.

Is this true?

I am looking for cases where RSA/DSA is superior to ECC, not where it is used for historical reasons.

  • 3
    $\begingroup$ Tempted to close as dupe of this question. Is there anything that is not answered in the answers there? If so, you should edit to clarify. $\endgroup$
    – otus
    Feb 24 '16 at 8:10
  • $\begingroup$ By themselves, elliptic curves are useless. It is when certain algorithms are done over elliptic curves that they become useful. ECC is not a replacement for DSA, but ECDSA is. $\endgroup$
    – Melab
    Feb 26 '16 at 6:53

There are three use cases where RSA beats common ECC algorithms, such as ECDSA:

  1. Signature with verification frequent or/and by low-power devices. The verification cost of $n$-bit RSA with usual public exponents is $O(n^2)$, but the verification cost of ECC-based signatures is $O(n^3)$ (using usual algorithms). Together with simpler math, that's why RSA can be way over 10 times faster for signature verification at usual security levels, even though it must use a larger $n$ for equivalent security level.

  2. Similarly, encryption by low-power devices or/and with decryption comparatively rare.

  3. Need to minimize the size overhead of adding a signature; using signature with message recovery, that can be 34 bytes for RSA (using SHA-256 hash, ISO 9796-2 mode 3 or the deprecated mode 1, for messages at least 222 bytes before signature at the 2048-bit security level), versus 64 bytes for ECDSA for comparable security.

RSA is thus a good choice (and indeed still the dominant one, I believe) for signing public-key certificates; beside inertia, in the internet domain that's mostly for reason 1 (certificates are verified often), but in the Smart Card and payment industry reason 3 adds up.

Additional arguments for RSA (vs ECC) are

  • RSA was first there / is the most time-proven, and became an industry standard.
  • Simplicity. RSA signature verification is much easier to code, and get right, than ECDSA signature verification.
  • RSA is now clearly patent-free.
  • Perhaps, slightly more quantum resilience; that is, at comparable level of resistance versus non-quantum attacks, RSA arguably would fall after ECC if it ever emerged quantum computers usable for cryptanalysis; see this other answer and section 5.4 of its source; note that Koblitz and Menezes are not making any strong statement, rather, their intro is (emphasis mine):

    We next examine some conjectures about the NSA’s motives in its PQC announcement (..)
    The NSA believes that RSA-3072 is much more quantum-resistant than ECC-256 and even ECC-384. (..)

Note: this answer does not touch use cases where ECC is preferable, or its virtues.

  • 3
    $\begingroup$ Case 2 could occur in a sensor grid scenario where the encryption is on embedded systems but decryption is on a much more powerful server. $\endgroup$
    – Demi
    Feb 24 '16 at 18:26
  • $\begingroup$ OTOH, using RSA signature and key generation on smart card devices really can bog things down, and using the more accpted / proliferate RSA PKCS#1 extends the signature size to less acceptable levels. $\endgroup$
    – Maarten Bodewes
    Feb 28 '16 at 20:02
  • 2
    $\begingroup$ I think citing improved "quantum resilience" is a bit irresponsible, since both algorithms should be considered utterly broken in the presence of scalable quantum computing. While there may be a gap, the authors themselves suggest "It is not likely that the gap between quantum cryptanalysis of a 384-bit key and a 3072-bit key will be great enough to serve as a basis for a cryptographic strategy." This should be interpreted as quantum resilience NOT being a reason to use RSA over ECC algorithms. $\endgroup$
    – bkjvbx
    Oct 18 '16 at 9:40

If practical quantum computers become a reality, the larger bitlengths of RSA keys would make them more quantum-resistant than their ECC counterparts. See section 5.4 of this Koblitz & Menezes paper

  • $\begingroup$ What about Ed448? $\endgroup$
    – Zimba
    Sep 21 '20 at 16:57

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