> 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 17 minutes 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 19 minutes 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.
topaz0 7 minutes 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.
sdenton4 13 minutes 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.
klodolph 14 minutes 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.
9 minutes ago [-]
danbruc 2 hours ago [-]
All the primes above 2^32 are out, but that accounts for only two point something percent.
PaulHoule 2 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 18 minutes 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.
adgjlsfhk1 2 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 41 minutes 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.
pants2 2 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 29 minutes 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 23 minutes ago [-]
Enough of this divided binary world, we are all one
jihadjihad 1 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!
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 60 minutes ago [-]
There should be a law!
kleiba2 1 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 1 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.
da_chicken 1 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.
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 1 hours ago [-]
There are about 18.446 quintillion more 64-bit integers than 32-bit integers.
moefh 43 minutes ago [-]
I think they meant to write "There are about 4 billion TIMES more 64 bit integers than 32 bit integers".
henry2023 14 minutes ago [-]
Indeed, edited the mistake
adrian_b 31 minutes 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.
58 minutes ago [-]
layer8 1 hours ago [-]
The chance of a random 64-bit integer matching some pair of 32-bit integers is a 100%, though.
brookst 1 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 54 minutes ago [-]
Wonder what the limit is as you add more 32 bit integers to the product. Just the primes over 32 bit?
thaumasiotes 37 minutes 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.
6 minutes ago [-]
2 hours ago [-]
crest 30 minutes ago [-]
So you're better of using a 8x8->16 widening multiplication SIMD instruction or even just a multi register TBL/TBX instruction?
MarkusQ 1 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 34 minutes ago [-]
where is the graph and the theorem for integers of n bits, with n going to infinity?
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.
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.
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.
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.
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.
0: https://en.wikipedia.org/wiki/Indiana_pi_bill
Most 1s won't go towards 1.5, but sometimes you're lucky.
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.
[0]: https://en.wikipedia.org/wiki/Benford%27s_law
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
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.
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.
https://arxiv.org/pdf/1908.04251