AMD's random number generator can't generate a 0?

(board.flatassembler.net)

117 points | by BruceEel 3 hours ago

18 comments

  • jstanley 1 hour ago
    This is not the first RNG bug on Zen 2, I recall after I first got mine that some application or other would quit immediately at startup because rdrand always returned -1, i.e. all 1s. It was fixed with a microcode update.

    Do we now learn that they fixed "always generate all 1s" with "never generate all 0s"??

    EDIT: I've been unable to reproduce the problem on my CPU, FWIW. It's a Ryzen 5 3600.

    EDIT2: OK, update, I can reproduce it with rdrand16, rdrand32 is fine but rdrand16 can never generate all 0s. So my CPU does have this problem!

    • 0x000xca0xfe 23 minutes ago
      I can reproduce it too with rdrand16 on Zen2.

      But it looks like the rdrand16 instruction can produce zeros just fine, it just sets CF=0 erroneously (indicating an error and that the user program should retry).

      So keep that in mind when you try to reproduce it too and use some abstraction that could implement retries internally.

      • dooglius 14 minutes ago
        Good observation, that seems like the most likely explanation. Do you ever see "true" CF=0 (with nonzero arg) or did they just take the lazy approach?
    • yk 1 hour ago

          return 4 # Determined by fair dice roll.
    • peri-cl 37 minutes ago
      Zen 4 reporting in. I'm unable to reproduce it (7840U).

         $ ./a.out | rg '\b\-?\d\b' | sort -n | uniq -c
         15281 -2
         15192 -1
         15273 0
         15243 1
         15269 2
      
      I used the GCC intrinsic ( _rdrand16_step ),

          #include <immintrin.h>
          
          short rdrand16() {     // gcc -mrdrnd
              short ret;
              while (1 != _rdrand16_step(&ret)) { }    
              return ret;
          }
    • RandomOnyx 1 hour ago
      Does rdrand32 and then taking the lowest 16 bits of its result yield any zeroes?

      Basically I'm wondering if it's a bug in the version of the instruction that writes to a 16-bit reg, or a bug in the underlying RNG

      • jstanley 1 hour ago
        Yes it does. rdrand32()%65535 was my first attempt, and generated zeroes at about the expected rate, that's why I initially erroneously thought my CPU did not have this problem.
        • goalieca 1 hour ago
          You should be using &0xFFFF for masking. Your mod is off by 1 too.
          • jstanley 37 minutes ago
            You're right, the code was correct but my comment above is wrong.
        • RandomOnyx 1 hour ago
          How about* rdrand32()%65536? Taking the remainder by 65535 doesn't take the lowest 16 bits after all

          *: missed a word the first time around

    • JdeBP 1 hour ago
    • rbanffy 59 minutes ago
      Even if you reproduce the issue, it is not a proof it can't generate a zero - just that it's very unlikely.

      To prove it, we'd need to examine the chip and its microcode.

  • strenholme 1 hour ago
    This is why I use, in security critical contents of my software (where the numbers have to be computationally infeasible to produce), a type of random number generator called an XOF (extendable-output function).

    It takes entropy from multiple different sources, makes it all input to the XOF, then the XOF uses cryptography to output a stream that has as much entropy as the combined entropy of all of its sources of randomness. So if an XOF, for example, takes 100 runs of rdrand16, along with the system time in microseconds and the number of milliseconds between receiving 100 packets over the network, the XOF will output a completely random stream without artifacts like never returning 0x0000, even if rdrand16 never outputs 0x0000.

    • stingraycharles 58 minutes ago
      Isn’t this effectively what systems like /dev/(u)rand do? Pool multiple random sources together to hedge against these things?

      I fail to see why one should either rely on a single random source nor roll their own.

      • strenholme 35 minutes ago
        Yes, /dev/(u)random is supposed to do that, but what if there’s a bug in the kernel which causes /dev/(u)ramdom to be less than secure? There’s also issues where, for example, it may no longer be possible to read /dev/(u)random after putting the process in a chroot() sandbox (chroot() isn’t defined in POSIX so its behavior is not guaranteed to be consistent across multiple operating systems).

        getrandom() is often times suggested, but alas isn’t a standardized function, i.e. it’s not part of the POSIX specification. Considering how the C23 changes to the C specification caused a lot of perfectly good C code to no longer compile, I’m very anal about sticking to specs; I use '-std=C99' for my code these days (even though it can compile as C23 code) and stick to POSIX functions (except chroot() and setgroups(), but both of those predate POSIX, and even here I have a compile-time option to compile my code without those non-POSIX syscalls).

        The code using a secure XOF (the algorithm was developed by the same team which later on made SHA-3, and includes people who helped make AES) has been around for nearly two decades (the code where I roll my own RNG to make secure random numbers has been around for over 25 years, but used AES before XOFs existed) and not one security problem has found with the RNG code has ever been found. [1] “Don’t roll your own RNG” is a suggestion, but it is possible to do so securely if one knows what they are doing (i.e. they have read Applied Cryptography and keep current with cryptographic developments).

        For anything vibe coded (my code is 100% human written, for the record), rolling one’s own RNG is a really bad idea.

        [1] There was a theoretical issue with cache timing attacks over two decades ago, so I put mitigations in place, and then chose to use an XOF for newer code.

        • boltzmann64 3 minutes ago
          i remember some linux kernel dev got ousted by the community because he/she wanted to not implement a backdoor that would compromise the results of /dev/urandom.
        • sltkr 2 minutes ago
          > getrandom() is often times suggested, but alas isn’t a standardized function

          The POSIX standard function is getentropy(), which internally calls getrandom() on Linux.

          > what if there’s a bug in the kernel which causes /dev/(u)ramdom to be less than secure?

          It's often the other way around: the Linux kernel contains thousands of workarounds for buggy hardware, while the buggy hardware itself doesn't always get patched. Linux developers take this stuff very seriously. As a result it's often safer to rely on kernel APIs than to access the hardware directly.

          The kernel code involving random number generation receives an exceptionally high amount of scrutiny because of its security implications, so I'd trust it to do the right thing over a naked call to RDRAND which nobody knows how exactly it's implemented in proprietary hardware or a handrolled solution to mix the RDRAND output with other entropy sources.

          Remember the Debian openssl disaster from 2008? That happened exactly because someone had handrolled their entropy mixing solution, then someone else broke it.

        • NooneAtAll3 12 minutes ago
          > but what if there’s a bug in the kernel which causes /dev/(u)ramdom to be less than secure?

          so instead you suggest trusting your own untested unlooked at implementation more?

          • strenholme 4 minutes ago
            Black-and-white thinking like this is always inaccurate.

            >untested

            The automated tests includes tests that make sure the XOF is correctly implemented. [1]

            >unlooked at

            People have been looking at my code for security holes for well over 20 years, and I have been getting multiple AI assisted security reports over the last year, things like “there’s a buffer overflow in this code which is nay to impossible to exploit, using code which hasn’t even been able to compile since 2022”.

            [1] https://github.com/samboy/MaraDNS/tree/master/deadwood-githu... and https://github.com/samboy/MaraDNS/tree/master/deadwood-githu...

          • UnlockedSecrets 6 minutes ago
            No you see what we do, Is we ask Claude to make no mistakes in implementing the CSPRNG. This way we ensure there are no mistakes in the implementation or mathematics.

            https://xkcd.com/221/

  • 349ru3h4f03 1 hour ago
    • peri-cl 1 hour ago
      The OP says they discovered this on a Zen 2, which is not covered by that bulletin (?)
  • CodesInChaos 2 hours ago
    Embarrassing, but probably little practical impact, since these hardware random numbers are typically not used directly and instead seed a CSPRNG.
    • leonidasrup 57 minutes ago
      According to Theodore Ts there was pressure from Intel engineers to let /dev/random rely only on the RDRAND instruction.

      " I am so glad I resisted pressure from Intel engineers to let /dev/random rely only on the RDRAND instruction. To quote from the article below:

      "By this year, the Sigint Enabling Project had found ways inside some of the encryption chips that scramble information for businesses and governments, either by working with chipmakers to insert back doors...."

      Relying solely on the hardware random number generator which is using an implementation sealed inside a chip which is impossible to audit is a BAD idea. "

      https://web.archive.org/web/20180611180213/https://plus.goog...

      Putting a backdoor into CSPRNG is a favored way to break crypto, for example Dual_EC_DRBG.

      "

      Weaknesses in the cryptographic security of the algorithm were known and publicly criticised well before the algorithm became part of a formal standard endorsed by the ANSI, ISO, and formerly by the National Institute of Standards and Technology (NIST). One of the weaknesses publicly identified was the potential of the algorithm to harbour a cryptographic backdoor advantageous to those who know about it—the United States government's National Security Agency (NSA)—and no one else. In 2013, The New York Times reported that documents in their possession but never released to the public "appear to confirm" that the backdoor was real, and had been deliberately inserted by the NSA as part of its Bullrun decryption program. In December 2013, a Reuters news article alleged that in 2004, before NIST standardized Dual_EC_DRBG, NSA paid RSA Security $10 million in a secret deal to use Dual_EC_DRBG as the default in the RSA BSAFE cryptography library, which resulted in RSA Security becoming the most important distributor of the insecure algorithm. RSA responded that they "categorically deny" that they had ever knowingly colluded with the NSA to adopt an algorithm that was known to be flawed, but also stated, "We have never kept this relationship [with the NSA] a secret and in fact have openly publicized it."

      "

      https://en.wikipedia.org/wiki/Dual_EC_DRBG

  • matja 2 hours ago
    I'm getting 16-bit zeros on my Zen 3 chip (+1:3821, 0:3893, -1:3895), I will wait to get some statistically significant samples for the 32-bit values and update the forum thread. Maybe it was fixed after Zen 2?
    • rbanffy 46 minutes ago
      Does anyone have access to an HPC cluster with thousands of Zen2 chips? We might want to check 64-bit ones with that - should take just a couple years depending on the size of the machine.

      Anyone from the High-Performance Computing Center Stuttgart willing to play on the 720,320 Zen2 cores?

  • 20k 1 hour ago
    I always wonder how hardware bugs like this happen with the sheer amount of hardware validation that's done. It'd be fascinating to know how it slipped through the cracks, though I know almost nothing about this side of the industry sadly
  • hnacobsxph 1 hour ago
    Chased a similar bug in a KDF once and only caught it by histogramming the 16 bit draws, statistical suites never flagged it.
  • rbanffy 49 minutes ago
    I have a couple questions:

    Looks like they tried 16-bit numbers. Does the odd behavior happen also on 32 and 64 (might take a long time to check - I'd start scratching my head after a couple hundred years of no zeroes) ones? Is the zero masking as some other fixed number, increasing its output count? Is RDRAND implemented as multiple reads of an internal state so that a larger random number takes longer?

  • deadbabe 26 minutes ago
    I would be very concerned if an RNG simply produced a natural 0.
    • flippingheck 2 minutes ago
      I would be very concerned if an RNG simply produced a natural 1.
  • Plainharbor21 1 hour ago
    [dead]
  • Ledgermellow 1 hour ago
    [dead]
  • dark-star 1 hour ago
    Usually you do "rdrand % <some-number>" anyways, and in that case you will still get zeroes. True, your result might be skewed by 1/(maxint/some-number) but I guess that's not a big problem in practice
  • throwawayffffas 1 hour ago
    So what? The point is to be non predictable not to pick all the numbers in the range with exactly the same probability. Would it be a problem if it never generated 16542?
    • gnfargbl 1 hour ago
      Consider an 8-bit RNG.

      By your argument, it would not be a problem if the RNG never generated 0. So, it must follow that it would also not be a problem if it never generated {1, 2, 3, ..., 253}.

      That means that our RNG now only generates the values 254 and 255. Which of the values is generated is unpredictable on any given call. However, 7 of the 8 output bits are now always fixed and so completely predictable. Can you imagine how an attacker could exploit that?

      Failing to generate only the number 0 is a weaker version of the same class of flaw.

      • brookst 56 minutes ago
        This is the “what’s the big deal if I lost $100k in a casino, it’s really the same thing as if I had lost $5” argument.

        I don’t think you can rebut “you only lose one of many values” with “it’s the same as only having one left”.

        • gnfargbl 33 minutes ago
          We're talking about whether a modification of the expected probabilities changes the dynamics of the game. The example I gave was deliberately extreme, because that makes it easier to reason about.

          If you want a casino example, then consider a roulette wheel that always lands on 36 but still pays out as usual. I think you'd want to play on it. Now consider one that always lands somewhere between 30 and 36. Still worth it, right? With careful bets and a good starting float you're still coming away from the table up (with a very high probability).

          In fact for a roulette wheel you only need two dead pockets for the player to get an edge. Bias is exploitable.

      • throwawayffffas 46 minutes ago
        The value space goes from 2^16, 2^32, 2^64 to 2^16 - 1, 2^32 - 1, and 2^64 - 1 respectively.

        The bug has zero practical impact.

        • gnfargbl 38 minutes ago
          It is absolutely untrue that a biased RNG has "zero practical impact." Modern cryptography has plenty of examples of relatively small biases leading to breaks. Check out Bleichenbacher's attack, for instance.

          You could be correct that the very small bias here is not enough to be exploitable. But, given the history around this, it would be wrong to handwave it away as trivial.

    • antiloper 1 hour ago
      What are you talking about? The point is in fact to pick all the numbers in the range with exactly the same probability.

      See section 7.3.17 of the Intel SDM, and how NIST SP800-90A (which the SDM refers to) defines "random number".

    • swader999 1 hour ago
      Betty from accounting will have words.
      • throwawayffffas 45 minutes ago
        What does Betty from accounting care about RNGs?
    • Hugsbox 1 hour ago
      That may well be a problem, yes.
  • ExoticPearTree 2 hours ago
    The probability of generating a zero is incredibly low if you use the normal distribution curve.

    So it is not necessarily that it doesn't generate zero, they did not run enough times to increase the probability of actually generating a zero.

    • blensor 1 hour ago
      From what I can see they were trying to generate 16bit integers, so the probability is 1 in 65536 and they were running the test for 11 hours.

      You definitely would expect a roughly equal number of 0s as any other of those numbers since it's uniformly distributed. And definitely not 0

      • ExoticPearTree 23 minutes ago
        > You definitely would expect a roughly equal number of 0s as any other of those numbers since it's uniformly distributed.

        How would random numbers be uniformly distributed?

        • Hugsbox 2 minutes ago
          Because each number is equally as likely as every other number. If you know you're more likely to get certain numbers, or in this case have no chance of getting certain other numbers, it is by definition _less random_.
        • thinkingQueen 13 minutes ago
          So you think a weighted die is more random than a fair die? A uniform distribution means each outcome has equal probability; it doesn’t mean the outcome is predictable.
    • zygentoma 1 hour ago
      This also seems to happen for 16 and 32 bit numbers, so you should be able to see zeros easily.

      They also write:

      > Running the same programs on an Intel processor, and the 0's are there with no problem.

    • matja 2 hours ago
      Why would it be a normal distribution?
      • throawayonthe 1 hour ago
        should be a discrete uniform distribution right?
  • m_antis89 1 hour ago
    0 is not a number, it's undefined
  • ZiiS 2 hours ago
    It is just possible they decided crypto code that uses it was safer to skip zeros. (Whist mathematically it should be no more likely; it is vastly more likely someone will actually try that key).

    It is also possible that their code was generating too many zeros and the easiest fix was to discard them all.

    • jstanley 1 hour ago
      Can you clarify what you mean by "it is vastly more likely someone will actually try that key"?

      I'm guessing you don't think there are people calling rdrand in a loop and throwing away the output with high probability except when it is 0, but I can't see how else you imagine people would be vastly more likely to use the output when it is 0?

      • ZiiS 54 minutes ago
        In lots of scenarios I know the software used to generate the key; the only unknown is the random numbers used. If I am searching for weaknesses it is highly likely I would try keys with different seeds; zero, one, are going to me much more likely choices here then hoping I can guess the right values.
    • dark-star 1 hour ago
      this is not how crypto works