NIST SP 800-57 Part 1 rev 5 section gives following comparison between different encryption types. For example, it shows that 3TDEA, RSA-2048, ECC224 provides security strength of 112 bits.

Does it mean that with computational power of $2^{112}$, chances of breaking 3TDEA, RSA-2048 and ECC224 are equal? or breaking one of these cipher is difficult than other?

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1 Answer 1


It means that you should be able to retrieve a key using $2^{112}$ operations, assuming a normal attack scenario and the most efficient method of attack at this time, assuming that the rest of the system is secure.

Some algorithms such as classic DH (relying on the discrete logarithm problem or DLP, called Finite Field Cryptography or FFC in the table) will allow more effective attacks if the key can get calculated, but note that generally these figures are out of reach.

However, generally being able to retrieve a key doesn't amount to breaking the algorithm. A break of the algorithm would allow an attacker to do it in (significantly) fewer operations than the number of operations mentioned. You'd just get a single key.

Usually we don't care if it is a bit or so off. These are ballpark figures. There is an attack that would allow AES-128 to be broken in about $2^{126.1}$ operations (using a lot of memory), but we generally still indicate that it has 128 bit security.

Note that NIST doesn't recommend a 112 bit key strength anymore. Cryptographers will generally recommend to aim for at least a 128 bit key strength. Lower key strengths may be prone to be broken and should only be used for realtime applications, only when needed and after careful analysis of the situation.

Of course, the above table doesn't take fully operational, capable quantum computers into consideration.

I'd specifically recommend against using DH or ECDHE with a security margin of 112 bits though as it might be possible that an adversary can reach the number of operations mentioned, or if a slightly more efficient attack is found. DH is vulnerable to attacks using precomputation, and classic Elliptic Curve cryptography is relatively vulnerable against attacks using quantum computers.

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    $\begingroup$ "You'd just get a single key" holds for symmetric key algorithms, IFC/RSA, and AFAIK ECC. But for FCC that applies only to some attacks (those that target a low value of $N$, like Pollard's rho and Pollard's kangaroo). As noted in this answer, there are other plausible attacks (those that target a low value of $L$, like index calculus, Function Field Sieve, I think NFS) that compromise all keys using the same group parameters $p$ and $q$. $\endgroup$
    – fgrieu
    Jan 27, 2023 at 9:55
  • $\begingroup$ "Of course, the above table doesn't take fully operational, capable quantum computers into consideration." how many bits of security will be reduced by taking into account quantum computers? $\endgroup$
    – crypt
    Jan 27, 2023 at 13:32
  • $\begingroup$ @crypt: if you have a crypto relevant quantum computer, FFC, IFC and ECC can be solved quickly (in polynomial time). For symmetric key algorithms (e.g. 3DES), it reduces it somewhat (depending on how long the attacker is willing to wait for the answer - if he's willing to wait millions of years, it effectively halves the key strength - it reduces it less if he is more impatient) $\endgroup$
    – poncho
    Jan 27, 2023 at 13:52
  • $\begingroup$ @fgrieu Good point. I've always assumed that such an attack would also translate to ECDH though, but without any mathematical backing. $\endgroup$
    – Maarten Bodewes
    Jan 27, 2023 at 15:53
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    $\begingroup$ @MaartenBodewes: as far as we know, elliptic curves don't have anything analogous to a factor base, hence those sorts of 'do the computation once, reuse the computation to solve a number of instances quickly' attacks do not appear to apply. $\endgroup$
    – poncho
    Jan 27, 2023 at 21:22

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