I have seen that there are PRNG that can generate a specific number of random-numbers. The Mersenne Twister as an example, can generate 2**19937 (if I'm not wrong) but... can we use a cryptographic hash function to generate infinite random numbers setting any seed to it? So I can set text for the seed and do it like this:

seed set to "myseed"
counter set to 0
first random number generated = first 64 bits of "myseed0" hashed with sha512
secound number = first 64 bits of "myseed1" hashed with sha512

Or we can save the other 448 bits left and use them later. The main idea, is 100% secure to generate PRN's from a hash function? why don't we do it?

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    $\begingroup$ "why dont we do it?" We do. $\endgroup$
    – Maeher
    Commented Dec 12, 2019 at 9:17
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    $\begingroup$ Note that the answers will explain how can generate (pseudo-) random numbers from a seed using a hash. They should not be seen as analysis of your described function - it is too simple and uses too small a seed to be considered one - and analysis of such functions is considered off topic on this site. $\endgroup$
    – Maarten Bodewes
    Commented Dec 12, 2019 at 13:53
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    $\begingroup$ Do you actually expect "to generate infinite random numbers" from a finite machine with finite memory producing results from finitely many algorithms each accepting only finitely many distinct inputs? (I'm not asking about "a lot" or "an unimaginably huge number", but "infinitely many", which is what you seem to be asking about.) $\endgroup$ Commented Dec 13, 2019 at 4:34
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    $\begingroup$ I hate to nitpick, but to mathematicians infinity should not be used in place of "very large number". Since SHA512 is limited to a maximum 2**128 bit long inputs that limits the number of outputs to be a finite number. $\endgroup$ Commented Dec 13, 2019 at 5:33
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    $\begingroup$ Technically, you can make an infinite number of calls to your PRNG; it's just that the numbers will start repeating after you've gone through all possible states. $\endgroup$
    – dan04
    Commented Dec 13, 2019 at 18:02

4 Answers 4


This construction gives you cryptographic-quality pseudorandom output, but it isn't as secure as it can be for a random generator.

With commonly used hash functions $H$ (such as any of the SHA2 and SHA3 family), as far as we know, the bits of $H(\textrm{seed}, n)$ are unpredictable if you only know $n$ and $H(\textrm{seed}, m_i)$ for any number of values $m_i \ne n$, but you don't know $\textrm{seed}$. This makes $D(n) = H(\textrm{seed}, n)$ a good key derivation function: its output is essentially indistinguishable from random.

A good random generator must have the property that a bit in the output is not predictable even if the adversary knows all the other bits of the output, but not the seed. The construction $H(\textrm{seed}, \textrm{counter})$ has this property. But a good random generator also has an additional property: backtracking resistance. Backtracking resistance means that if the adversary compromises the hash state at some point, then they can't recover past outputs. (Of course the adversary will know every future output, at least until the random generator is reseeded.) Your construction does not have this property since the original seed remains a part of the hash state.

A good random generator has a “ratcheting” step, which makes it impossible to recover the previous state from the current state when generating some output. It's easy to build ratcheting with a hash function: you basically just run the hash function on the hash state. Take a hash function with an $n$-bit output. Start with an $n$-bit secret seed; that's the original state of the random generator. To generate up to $n$ pseudorandom bits, calculate $H(0 || \textrm{state})$ and output that; also calculate $H(1 || \textrm{state})$ and use this as the next internal state. In pseudocode:

state = seed
while True:
    output(hash('0' + state))
    state = hash('1' + state)

Hash_DRBG specified in NIST SP 800-90A is a popular pseudorandom generator construction based on this principle.

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    $\begingroup$ The question asks about acquiring an infinite quantity of output from the generator. It would be accurate to note that NIST dictates that Hash_DRBG requires reseeding after $2^{48}$ requests (See Table 2 on page 38), which is not an infinite amount of data. $\endgroup$
    – Ella Rose
    Commented Dec 12, 2019 at 16:14
  • $\begingroup$ @EllaRose The amount of data you can get is as practically infinite with either method. It isn't really a consideration here. The reseed limit is one of several ways Hash_DRBG differs from the basic algorithm I present. $\endgroup$ Commented Dec 12, 2019 at 23:55
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    $\begingroup$ Can we reliably predict the period of this PRNG? See crypto.stackexchange.com/questions/24623/cycles-in-sha-256 $\endgroup$ Commented Dec 13, 2019 at 10:27
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    $\begingroup$ Maybe we could fix everything with Random_number = seed + salt + counter, am I wrong? $\endgroup$ Commented Dec 14, 2019 at 16:33

Your scheme can be re-defined as; take a cryptographically secure hash function $\operatorname{H}$ and generate the sequence as;

  1. Init the seed with $\text{seed}= \text{"myseed"}$
  2. $\text{hash} = \operatorname{H}(seed\mathbin\|counter)$
  3. $\text{counter} = \text{counter} +1 $
  4. output $hash_{|\text{required size}}$ (trimming the output)
  5. return step 2 for more random.

Now the first case, if the size of the counter if fixed then there will be a periodic sequence. The counter will be set back to 0 like in CTR mode.

If the let the counter run as long as it requires (not fixed size), then it won't have a periodic property. Somehow, if it has (that we don't expect), you will find a weakness for the underlying hash function. Of course, after running $2^{\ell/2}$ times, where $\ell$ is the hash output size, you will start to see collisions more frequently due to the birthday-paradox. The collisions are inevitable. If you use this construction for IV generation that requires lower bits than the output size, the collision probabilities will be much lower than the actual hash function itself like $2^{64}$ for 128-bit IV.

Can we use a cryptographic hash function to generate infinite random numbers?

Infinite is not a good measure here. One can define an infinite sequence but predictable 1010010010001... We want then Cryptographically secure pseudo Random number generator (CSPRNG). Usually, we require a fixed size random for example 128-bit random IV, 128,192,256-bit random key, or larger sizes as in RSA key-gen and as in RSA signatures.

Your construction is similar to HASH_DRBG NIST SP 800-90A which is seem secure if the hash function behaves like a random oracle.

why don't we do it?

We have it. HASH_DRBG and HMAC_DRGB are examples in better design. However, the NIST suggests that after $2^{48}$ requests the HASH_DRBG needs re-seeding page 38. Therefore, we cannot run them infinitely.

note: as point out Gilles's comment and answer your construction lacks backtracking resistance.

In conclusion, your scheme can go infinity but lacks backtracking resistance.

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    $\begingroup$ This construction lacks backtracking resistance, which is a useful property for a PRNG used as a component of a RNG, and mandated by security standards such as SOG-IS. $\endgroup$ Commented Dec 12, 2019 at 12:30
  • $\begingroup$ @Maarten-reinstateMonica Is it better know? $\endgroup$
    – kelalaka
    Commented Dec 12, 2019 at 15:27
  • $\begingroup$ Not sure, I mean, if I have a PRNG I generally can require any amount of random data from it, without specifying a size in advance. Cycles are usually very large, to the point that commonly no limit is even implemented. I still get the feeling that only allowing a relatively small amount of bits from a PRNG is a requirement from the answer. $\endgroup$
    – Maarten Bodewes
    Commented Dec 12, 2019 at 16:17

Can we use a Cryptographic hash function to generate ...

Yes we can. Your code example shows that we can if the hash function is secure. That means a currently non invertible function. A slightly modified form is one of the older cryptographic Java RNGs, called SHA1PRNG.

$$ \left\{ \begin{alignat}{7} & \texttt{State}_{0} && = \texttt{SHA}_{1} \left(\texttt{SEED} \right) \\ \\ & \texttt{Output}_{i} && = \texttt{SHA}_{1} \left(\texttt{State}_{i-1} \right) \\ & \texttt{State}_{i} && = \texttt{State}_{i-1} + \texttt{Output}_{i} + 1 \, \operatorname{mod} \, {2}^{160} \end{alignat} \right. $$

Notice the modification: The output is fed back into the state. The 'counter' is simply a +1 in this case. Similar constructions are also key components of stream ciphers.

SHA-512 is fine but perhaps a bit over the top if you're looking only for 64 bit outputs. And sticking with SHA-1 and SHA-256 allows you to leverage Intel's hardware SHA extensions to make your RNG run faster.


Well not absolutely. All RNGs have a finite internal state. As you've already said. the common Twister's is of size $2^{19,937}$. A similar limit will apply to a whatever state variable you use for yours. When the limit is reached, the numbers will roll over and repeat. Good enough for practical purposes, but not entirely infinite.

...random numbers?

And it's important to call these 'pseudo-random' numbers rather than just random numbers, as in the world of random number generation there is an alternative called 'true' random numbers.


In short, no.

Or rather, yes, but you don't want to do that.

Note, by the way, that "can we do that" and "is it 100% secure" in your question are different, antipodal things. Much like "MT" and "secure" are antipodal. All generated pseudorandom numbers (including those coming from secure pseudorandom generators and those coming out of cryptographic hash functions are deterministic, so they cannot be 100% secure anyway). Now, MT, which you gave as an example, is not in any way secure, it is very easily exploited. If you meant "secure" in a way of "are we confident that the bits will look random", then that's a different question. But in that case, you most likely want to use a generator that is orders of magnitude faster than a cryptographic hash.

A cryptographic hash function can be used to generate (pseudo-) random bits of an apparent quality comparable to dedicated random number generators. I say "apparent" because although cryptographic hash functions are designed with some things in mind that are desirable features of random number generators as well (think e.g. avalanche), they are not designed to be random number generators.
So, they kinda work as such, mostly, but it's not their real purpose and you do not have a hard guarantee that they will pass all tests that a specifically designed high-quality random number generator will pass (MT doesn't even pass them all either, by the way, it's comparatively poor).

The fact that you can use a cryptographic hash as a random generator is demonstrated by the fact that for example, the secure random number generator in at least one free open source operating system is implemented in exactly this way.

Then why am I saying "no"?

A hash function (cryptographic or not) can be considered being a sort of entropy extractor.

You input N bits and the function somehow produces M bits from these (and usually N >> M) in an obscure, hard to predict way such that you cannot easily find collisions, etc etc.

The M bits that the function outputs are (pseudo) random, or at least as good as. So you could say that the function extracts M bits of entropy from the message.

That is the exact reason why, for example, DJB recommended that you use a hash function after you did a curve25519 exchange and want to use the result as encryption key for your block cipher. You have some curve point which is not totally random, and it has more bits than you actually need, but also you know that it only has slightly fewer than 128 bits of entropy somewhere inside, and you do not know where. Obviously you want to use all the entropy that you're given. What to do?! Which bits should you use?
Hashing the point extracts that entropy, and ensures you don't throw any of it away.

So, let's think about what happens in our random number generator. We seed it with a certain amount of entropy, and then we keep extracting entropy from it forever. Wait a moment, if we extract some, what about remaining entropy? Yep, you guessed right. Eventually, very soon, we run out of entropy. It's still a random-looking deterministic sequence, of course. However, it is a sequence about which we practically do not know anything (e.g. what is its period lenght?).

Doesn't any random number generator have the entropy problem? Well yes, output is deterministic, and there is a finite number of numbers in an integer, so necessarily, sooner or later, you get the same sequence of numbers again, but this is a known problem and it's something that is explicitly addressed in the design (not so in the design of a cryptographic hash!).
Good generators try to maximise the period length (and some other things).
That's why MT has such a ridiculously large state. This huge state exists only to turn a rather poor generator into one with a very long period with a very large k-distribution (by only ever updating a small part of a huge state independently, and iterating over it).

PCG or xoroshiro variants (which, too, are not cryptographically secure) achieve practical periods (and, except for k-distribution, better properties otherwise!) with much, much smaller state. I say "practical" because one needs to realize that there is absolutely no difference between a 2^256 and a 2^19937 period. Even in massively parallel applications, a 2^256 period which can be subdivided with skip-ahead into 2^128 independent, non-overlapping sequences, is way more than you can use in your lifetime, even with an utopian farm of impossibly fast supercomputers. So, that's "infinite" for all practical purposes, just like 2^19937 is only "infinite", too.

In the case of the previously mentionend secure random generator used in an operating system, running out of entropy isn't very much a problem because it is being re-seeded all the time. So it never (well, never is a lie... let's say rarely, in normal conditions) runs out of entropy.

  • $\begingroup$ "The M bits that the function outputs are (pseudo) random, or at least as good as. So you could say that the function extracts M bits of entropy from the message" is interesting. I'm confused about your definition of true and pseudo random numbers. The bulk of randomness extractors (seeded and deterministic) in TRNGs are nevertheless some form of wacky hash. They seem to work, yet you're suggesting that they may not even produce pseudo randomness? $\endgroup$
    – Paul Uszak
    Commented Dec 13, 2019 at 14:07
  • $\begingroup$ @PaulUszak: Well true random generators are not predictable because the random is... uh... truly random. As in unforeseeable. Such as atmospheric noise, radioactive decay. All others have a (hopefully) reasonable amount of entropy that came from "somewhere" (without knowing where from) to start with, and work deterministically from there, trying to hide patterns as good as they can, and simulate the statistical properties of "real" random as good as they can. But they all (including cryptographically secure generators) are still deterministic, of course. $\endgroup$
    – Damon
    Commented Dec 13, 2019 at 14:23
  • $\begingroup$ Some generators have better distribution and longer periods, and some make it harder for a prospective attacker to determine the internal state from observed output than others (the internal state of MT for example, can be determined almost trivially, on a cryptographically secure generator that shouldn't ever happen with high confidence). But in the end, of course, they're still all deterministic. $\endgroup$
    – Damon
    Commented Dec 13, 2019 at 14:25
  • $\begingroup$ @PaulUszak The only similarity between hashes and randomness extractors is that they both take input and generate output. They are not "some fork of wacky hash" (although standard hashes can be used as randomness extractors). They're as far away from being hashes as can be. $\endgroup$
    – forest
    Commented Dec 14, 2019 at 3:28
  • $\begingroup$ @forest I don't quite get your point, but would it be better to address Damon rather than me? $\endgroup$
    – Paul Uszak
    Commented Dec 15, 2019 at 0:58

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