Only 17% of all 64-bit Integers are products of two 32-bit integers (lemire.me)
119 points by sebg 4 days ago
_alternator_ 13 minutes ago
> You might be able to come up with a more efficient algorithm.
Challenge accepted. Suppose we want to know the answer to 3 decimal places (so we'd match the headline). And suppose I allow my algorithm to be wrong one in a thousand times ("probably approximately correct").
Then sample some constant number C of random 64 bit integers. Run the following algorithm which separates each random sample into one of three classes: Y (has 32 but factors), N (does not have 32 bit factors), U (unknown).
Check if prime using probabilistic miller rabin. (Error prob goes to zero exponentially fast). If prime, return N. If it's not a prime, then run T steps of pollard rho to determine whether the number has 32 but factors; return Y,N, or U depending of the factors found up to step T.
The key observation is that T can be chosen to make the UNKNOWN class very small (with high probability), and so our estimate should rapidly converge to 17%Y, 83%N, ~0.001%U
For fixed error tolerance, this would run in roughly a constant number of iterations, independent of N.
Dylan16807 4 hours ago
> I find it interesting to consider that if you pick a value at random, it will usually fail! That is, most 64-bit integers cannot be written as the product of two 32-bit integers.
While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already, and considering even a tiny amount of overlapping results gets you there.
What gets interesting is actually trying to quantify the overlapping results.
ot 2 hours ago
Yeah the number sounds a lot less impressive if you say that you only get 2^61.44 integers out of 2^64. In other words, a 4% entropy loss.
Information quantities are more meaningfully expressed in number of bits.
FabHK 2 hours ago
> Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63.
Not sure I understand.
Adding two 32 bit integers takes you to 33 bit integers. (1111 + 1111 = 11110).
Addition doesn't care about order, so you're instantly cutting 2^33 possibilities down to 2^32. Or so is your argument. But in reality you can reach nearly all of those 2^33 numbers.
sdenton4 2 hours ago
Concatenating arbitrary 32 bit ints covers all possible 64 bit ints. So the space of all pairs of 32 bit ints is in bijection with 64 bit ints.
Commutativity introduces a relation on pairs of 32 bit ints (a,b) ~ (b,a), which accounts for one bit of information. Thus, at most 50% of 64bit ints show up as products of 32 bit ints.
FabHK 2 hours ago
Dylan16807 an hour ago
topaz0 2 hours ago
The 2^64 in gps argument comes from the number of pairs of 32 bit numbers, not from the upper bound of multiplying two 32 bit numbers. So for the addition case the symmetry argument is still only good enough to get you down to about 2^63, which doesn't help you at all because you have much stronger information from the upper bound.
klodolph 2 hours ago
Addition in this case is cutting from 2^64 to 2^33-1.
The 2^64 number is the number of inputs. For an operation which is commutative, you expect the outputs to be 2^63+2^32 or smaller, since you’ve introduced symmetry.
danbruc 4 hours ago
All the primes above 2^32 are out, but that accounts for only two point something percent.
topaz0 an hour ago
But also all of their multiples. I suspect that those account for the vast majority.
HarHarVeryFunny 2 hours ago
Why does order matter?
Whether a 64-bit number can be written as the product of two 32-bit ones depends only on the prime factors of the 64-bit number - it's a property of the number itself, and apparently 17% of 64-bit numbers have this property.
orlp an hour ago
The input space is 32 + 32 = 64 bits. The output space is 64 bits. So the best you can do is an 1-to-1 mapping.
However, since a * b = b * a, our input space has a lot of duplicate outputs. So from this alone you can conclude roughly half of the output space must be uncovered by any input pair, simply because there aren't enough input pairs.
PaulHoule 4 hours ago
... or just considering the even numbers almost all of them are 2 x N where N>2^32 and that gets you to within a hair of "most" and if you add in the odd thirds for which the same is true you get a bound of 2/3 - epsilon.
topaz0 2 hours ago
It's a bit more subtle than that -- most n>2^32 are not prime in which case 2 x n has more factorizations you would have to check.
(Just by way of example, for n=2^33, 2n=2^34 but also =2^17*2^17)
adgjlsfhk1 4 hours ago
A lot of the remaining is multiples of 4, which you can either get from having a 2 in both factors or a 4 in one (multiples of 9 are similar).
thaumasiotes 3 hours ago
> While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already
It's much worse than that. It's difficult for a 64-bit product to have the high bit set if the multiplicands are both no larger than 32 bits.
jandrese 30 minutes ago
This just seems like an expansion of prime numbers to includes factors in the 2^33+ range. Basically you're calculating if a number is prime but stopping the check when the factors go above 2^32.
intuitionist 12 minutes ago
Having a prime factor greater than 2^32 accounts for about 80% of the 64-bit integers that can’t be expressed as a product of 32-bit integers. But it’s not the only way; you can also have three prime factors in the range (2^16, 2^32), for instance.
Taek 24 minutes ago
Well, technically yes, but 'stopping the factors at 32 bits' is a plenty interesting constraint because it excludes all 64 bit composite numbers that have at least one factor above 2^32.
You have to redo the math to make the constraint work.
tobz1000 2 hours ago
> the proportion of all 2n-bit values that can be generated by the product of two n-bit values goes to zero as n becomes large. This means that if you have, say, 10000000-bit integers multiplying 10000000-bit integers, you’d expect relatively few 100000000000000-bit integers to be produced.
That should be "relatively few 20000000-bit integers", right?
tgv an hour ago
Perhaps it's binary.
henry2023 3 hours ago
There are about 4 billion 64 bit integers for each 32 bit integer.
The chance of a random 64 bit integer being a 32 bit integer is 0.0000000233 %
The chance of a random 64 bit integer being a product of two 32 bit integers is 17%
Nice
HWR_14 3 hours ago
There are about 18.446 quintillion more 64-bit integers than 32-bit integers.
adrian_b 2 hours ago
True, but there are as many 64-bit integers as pairs of 32-bit integers.
Therefore the fact that relatively few 64-bit numbers are products of 32-bit integers means that a lot of pairs of 32-bit integers give by multiplication the same product.
moefh 3 hours ago
I think they meant to write "There are about 4 billion TIMES more 64 bit integers than 32 bit integers".
henry2023 2 hours ago
bo1024 an hour ago
There are about 2^64 more 64-bit integers than 32-bit integers.
layer8 3 hours ago
The chance of a random 64-bit integer matching some pair of 32-bit integers is a 100%, though.
brookst 3 hours ago
Or, the odds of a random 64-bit integer being a 32-bit integer are the same as you or me guessing a random 32 bit integer.
rao-v 3 hours ago
Wonder what the limit is as you add more 32 bit integers to the product. Just the primes over 32 bit?
thaumasiotes 2 hours ago
If you're allowed to multiply as many 32-bit numbers as you want, the only numbers you won't be able to achieve by so doing are those with any prime factor larger than 2^32.
This is more than just the prime numbers. For example, a 41-bit prime can be multiplied by 16 and it will still fit into 64 bits.
nyeah 2 hours ago
fogleman 16 minutes ago
Does it actually matter for hash uniformity, though?
pants2 4 hours ago
I dream of a future where all 64-bit integers are products of 32-bit integers. Together, we can change math for the better.
jerf 2 hours ago
Indeed, but justice requires that we recursively continue all the way to the base case, until all 32-bit integers are products of 16-bit integers, all 16-bit integers are products of 8-bit integers, all 8-bit integers are products of 4-bit integers, all 4-bit integers are products of 2-bit integers, and all 2-bit integers are products of 1-bit integers. Only when we have reach all the way down that list to the very, very smallest of the numbers around us and brought justice to them will the future be able to arrive. I literally can not wait for that day.
order-matters 2 hours ago
Enough of this divided binary world, we are all one
jihadjihad 3 hours ago
Why stop there? We can dream of a future where math is bent to our will [0] for the betterment of all mankind!
dvh 3 hours ago
1 + 1 = 3 (for sufficiently large values of 1)
Taek 21 minutes ago
I thought you were making a joke but if we're assuming that the 1's are being rounded or truncated before the final value cake is produced I guess you are right.
sshine 3 hours ago
It helps if you take the limit of 1 going towards 1.5.
Most 1s won't go towards 1.5, but sometimes you're lucky.
brookst 3 hours ago
There should be a law!
kleiba2 3 hours ago
I upvoted you, not because I think your joke is particularly great, but I hate that HN has this tendency to downvote comments that are clearly meant as a humorous contribution. And I get it, no-one wants HN to turn into Reddit. I also understand that not every joke lands. But I just think it's unnecessary to downvote, you could simply ignore.
zamadatix 3 hours ago
"Ignore" is one of those things that sounds like it's a neutral choice but really isn't in practice - it's still just saying "can only ever be positively pressured". IMO people shouldn't go as far as flag though, at the very least, and if it's already at the bottom of the sort there is no sense dumping on it further.
My current comment itself, for instance, also doesn't really add anything to the discussion about the article and I'd have no expectation people leave it from going negative. Maybe the will, maybe they won't, but there is no reason to expect they should in principle of me loving tangents :D.
kingstnap 2 hours ago
This is something I had thought about some time back where I was thinking about the feasibility of somehow using the upper and lower registers inside a multiplier as general purpose storage for fun / seeing if you could make them more compact.
Anyway here is a fun pattern you get when you multiply 8 bit unsigned integers. Not all pairs of (upper bits, lower bits) are reachable, and it has a lot of distinct patterns.
https://i.imgur.com/Gb3HDR0.png
(Should I host the image on GitHub Gists so it doesn't vanish?)
cobbzilla an hour ago
I must be missing something. Aren’t ~50% of 64-bit integers the product of the number 2 and another 32-bit integer?
slazaro an hour ago
Going from 32 bits to 64 bits doesn't double the range (that would be adding 1 bit), it squares the range.
mplanchard an hour ago
I don’t think so, because that only gets you up to 2x2^32, which is nowhere near halfway to 2^64
jlarocco an hour ago
No. 50% of them are the product of 2 and a 63-bit integer.
da_chicken 3 hours ago
This feels like a underlying property that contributes to of Benford's Law[0]. That is, most numbers we measure and record are the results of various independent (addition) and dependent (multiplication) factors stacking together, and we observe this property in the distribution of them.
PunchyHamster an hour ago
Well, that is entirely not surprising. Pretty sure people writing not terrible hash functions figured it decades ago
crest 2 hours ago
So you're better of using a 8x8->16 widening multiplication SIMD instruction or even just a multi register TBL/TBX instruction?
MarkusQ 3 hours ago
If this seems counterintuitive, consider that only about a third of the two-digit numbers ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 63, 64, 72, 81}) can be written as the product of two one-digit numbers.
childintime 2 hours ago
where is the graph and the theorem for integers of n bits, with n going to infinity?
mswphd 2 hours ago
The math was linked in the article