Galactic Algorithm

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

by jonbaer • 7 years ago

382 71 7 years ago

en.wikipedia.org

_5659• 7 years ago

Related discussion: https://news.ycombinator.com/item?id=21151646

cornstalks• 7 years ago

1729

Why is this hyperlinked? I was hoping the link would help explain why 1729 and not some other number, but it’s just trivia...

One immediate practical effect would be to earn the discoverer a million dollar prize from the Clay Mathematics Institute.

I mean, I can’t argue with the practicality of that.

ehsankia• 7 years ago

Probably because it's a pretty famous number, being the Hardy–Ramanujan number, and having a pretty famous story. There's quite a few videos on it [0] [1].

[0] https://www.youtube.com/watch?v=_o0cIpLQApk

[1] https://www.youtube.com/watch?v=LzjaDKVC4iY

namanyayg2• 7 years ago

Yeah not sure why it's hyperlinked, it's a rather dull number.

zaroth• 7 years ago

“But it’s not dull at all,” Ramanujan replies, “it’s actually the first number that can be written as the sum of two cubes, two different ways!”

edanm• 7 years ago

I also wonder why that number. But as others said, it's a pretty famous number all by itself so it makes sense for there to be some notice paid to that- if there were no hyperlink I would've thought it's weird and maybe someone made a mistake.

nabla9• 7 years ago

1729 has no special meaning for the multiplication algorithm. In the paper Harvey writes:

In Section 5 we establish (1.3) with the explicit constant K= 1728, and in Section 5.4 we list some optimisations that improve it to K= 8.
In Section 5, we will simply take d:= 1729 (any constant larger than K would do)

zeristor• 7 years ago

No mention of the number of dimensions for Fourier transform multiplication though: https://en.wikipedia.org/wiki/1729_(number)

saagarjha3• 7 years ago

I'd argue that it shouldn't be linked, because it's being used in the wrong context. It would be like linking an music album titled "Outer space" to the Wikipedia article on outer space.

hcs• 7 years ago

I can't explain it, but it's the algorithm discussed here, if you want to read up on it: https://news.ycombinator.com/item?id=19474280

BlueGh0st• 7 years ago

The 1729 link has since been removed, likely for the reasons user saagarjha stated -- wrong context.

saagarjha3• 7 years ago

Interestingly, the change that added it was also quite recent: https://en.wikipedia.org/w/index.php?title=Galactic_algorith...

lidHanteyk• 7 years ago

Because 1728 is an incredibly surprising number, showing up most prominently at the center of elliptic curves, modular forms, and L-functions: https://en.wikipedia.org/wiki/J-invariant

And 1729 is 1728 + 1. Not even kidding, that is a thing that happens.

cperciva2• 7 years ago

In addition to galactic algorithms, there are also galactically proven algorithms. One example is the algorithm for matching with mismatches which I presented in the first chapter of my doctoral thesis; I proved that it was faster than other algorithms for inputs of at least ~10^30 bytes.

In practice, it wins starting at around 10^4 bytes.

Gehinnn6• 7 years ago

How do you prove that something is faster than everything else? Was it already running in linear-time?

There aren't many non-trivial lower bounds out there, and I think they all use one of three ideas (diagonalization, crossing sequences or oracle reconstruction)!

cperciva2• 7 years ago

Sorry, I was unclear. I proved that my algorithm was faster than other known algorithms; I didn't prove that it was faster than all other possible algorithms.

gmoot• 7 years ago

Ah. I still remember my computational geometry professor telling us about an algorithm to do triangulation (composing a polygon into triangles) in linear time: "It's very complex. I don't think anyone has actually implemented it".

It shocked me at the time that there were algorithms like this.

https://en.wikipedia.org/wiki/Polygon_triangulation#Computat...

alanbernstein• 7 years ago

What was shocking about this? Seems like something you could always solve by hand, even if you couldn't quite enumerate the steps you need to take to do it. I guess that's sort of my baseline for whether I expect an algorithm to exist.

tripzilch3• 7 years ago

The shocking thing was that it is a known algorithm but (presumably) nobody ever implemented it.

Sharlin• 7 years ago

In linear time was the key part.

mfabbri77• 7 years ago

Fast and robust polygon triangulation was the hardest challenge in the period when we developed the OpenGL version of our 2D vector graphics engine (www.amanithvg.com), it was in the early 2000.

I clearly remember all such research papers. After studying all of them and after three complete rewrite of the engine, we finally implemented a classic 2 step algo: sweepline polygon decomposition to monotone polygons, and then monotone polygons to triangles.

We had really heavy headaches dealing with intersections, autointersections, splitting edges, comparing and merging original vertexes with vertexes created by splitting edges, managing edges connectivity and topology.

After some try, we totally discarded the idea to write it in floating point arithmetic, preferring fixed point integers instead (and this was a real turning point for the robustness!).

Anyway, good times to remember!

carlmr• 7 years ago

They mention O(n log* n) being quasi linear time, I had to look up log * n because I never saw this before.

pfdietz2• 7 years ago

And then there's O(n alpha(n)), where alpha(n) is an inverse the Ackermann function, and which grows even more slowly than log* n. The famous union-find algorithm has this complexity.

MaxBarraclough• 7 years ago

Wikipedia link: https://en.wikipedia.org/wiki/Log-star

IshKebab• 7 years ago

Yeah I discovered a fast algorithm for the travelling salesman problem with neighbourhoods (i.e. cities are shapes not points). Emailed the author to inquire about code, turns out they never wrote any.

jakeogh• 7 years ago

https://rjlipton.wordpress.com/2010/10/23/galactic-algorithm...

jonbaer• 7 years ago

"The famous quantum factoring algorithm of Peter Shor may or may not be a galactic algorithm. It is of course one of the great results in theory, ever. It has sparked funding for research centers on quantum computation that have promoted many other advances.

If and when practical quantum computers are built Peter’s algorithm will be one of the first algorithms run on them. Right now it is a galactic algorithm. But, perhaps it is the best example of the importance of galactic algorithms."

leeoniya• 7 years ago

is Graham's Number a galactic number?

https://en.m.wikipedia.org/wiki/Graham's_number

saagarjha3• 7 years ago

You could say so:

As with these, it is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number—and so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe.

function_seven• 7 years ago

for a number of times far exceeding the total number of Planck volumes in the observable universe.

Just so I'm clear. They're saying that not only can I not fit Graham's Number in all the Planck volumes of the universe; and I can't even count the digits of GN and write that in the Planck volume of the universe (and so on), but the number of "indirections" is itself so large as to not fit in the universe?

Like:

    1. GN (can't fit).  
    2. Number of digits in GN (can't fit).  
    3. Number of digits in #2 (can't fit).  
    4. Number of digits in #3 (can't fit).  
    ...  
    N. <-- The numbered list item itself won't fit.

Am I understanding that right?

theemathas• 7 years ago

But have you heard of Rayo's number?

thewarrior• 7 years ago

Can’t wait to be told to implement some of these for my next whiteboard interview

hrgiger• 7 years ago

Perfect candidate for primality testing! https://www.mersenne.org/various/math.php#lucas-lehmer

asdaddasdad5• 7 years ago

I've never understood the "number of the atoms in the universe" argument. The number of states the universe can be in doesn't seem to be equal to the number of atoms. For example, just two atoms could encode lots of numbers simply by using their distance. Quantum physics would affect it, but I mean in principle: we are not switching atoms on and off to encode state.

Retric2• 7 years ago

It’s 2^1729 digits, (vastly) more digits than there are atoms in the universe.

The vastly bit is kind of an understatement.

Each stable isotope is indistinguishable from every other carbon isotope. So, you can probably encode a few bits per atom assuming you can somehow read this data. However they are talking vastly larger numbers of digits here. Where there are only ~10^80 digits worth of atoms in the visible universe, but hey bump that to 10^90 it does not help.

Sure you might encode 10^6 or hell I will give you 10^100 bits of data per atom or something but that’s not even close to helpful. It’s still on the order of K atoms in the universe and you want to encode 10^400+ * K bits of data. So each atom needs to encode 10^400+ bits and remember each stable isotope is indistinguishable from every other atom of that stable isotope.

nine_k• 7 years ago

You forgot about ordering.

8 binary bits in order can encode 256 states, as if every bit was capable of encoding 32 states.

comboy• 7 years ago

It's just there for a reference, maybe a better one would be Kurzweil's ultimate laptop (made from the whole universe). But the interesting part is that I don't think the ultimate laptop takes into account interactions and state between atoms. AFAIR it assumes computation is done on every available dimension (like electron spin) but not on how atoms move and interact with each other. Intuitively it shouldn't make much difference though.

marcosdumay2• 7 years ago

The number of digits of the possible interactions between the atoms is still proportional to the number of atoms.

lonelappde• 7 years ago

It's not states, it's logarithmic of states (# of dimensions of finite size) . Think about writing down a number. One atom per digit is a pretty natural heuristic for the optimal spatial cost of information. Yes you can get clever, but there's no point, were already at astronomical levels of imprecision.

asdaddasdad5• 7 years ago

As I said, one atom per digit is not really natural. Just two atoms could encode an infinite number of numbers, by measuring their distance. Well except quantum physics might get in the way, not allowing us to measure with arbitrary precision. But that would be another argument.

Another argument would perhaps be the energy required to do the computation. Maybe that relates more directly to the number of atoms in the universe, via Einstein's equation?

marzell• 7 years ago

How would you read that? (Distance between two distinct atoms) doesn't seem practical

asdaddasdad5• 7 years ago

Doing computations with single atoms also doesn't sound very practical.

Measuring distances, I suppose there could be numerous ways, like measuring the gravity or electric pull (not sure what it is called in English). I think only Quantum theory says we can not measure to arbitrary precision, or at least if we do, there are other issue. Still, that would be another argument than simply pointing at the number of atoms.

bkanber• 7 years ago

It's really just a phrase used nearly-rhetorically in order to convey the vast scope of a thing. Everyone understands that the number of atoms in the universe is a huge number, so it's a good benchmark.

When people use this phrase they're not trying to claim that the number of atoms in the universe is directly related to the problem at hand; they're just trying to convey scale.

pfdietz2• 7 years ago

This sort of thing highlights the difference between theoretical and practical algorithms.

In practical algorithms, one does not ignore constants, one takes into account actual hardware (particularly cache effects), and one does not ignore real world distributions of inputs.

Gehinnn6• 7 years ago

The godfather of a galatic algorithm is the known (!) algorithm that solves SAT in polytime.

However, its runtime is only guaranteed to be in polytime, if someone guarantees that P = NP.

codeflo2• 7 years ago

Can you give a reference, or a hint on where to read more? This sounds really interesting, but Google has failed me.

Gehinnn6• 7 years ago

Actually, I don't know anymore where I read about this idea - if you find out, please let me know! I've added a description of the algorithm.

Gehinnn6• 7 years ago

Before an unbeliever comments (as I would do myself if I wouldn't know the algorithm): This algorithm involves enumerating f(M, t) where M is the Gödel number of a turing machine T, t an integer and f the output of T after t steps on the original input. This output is interpreted as variable assignment. If it satisfies the formula, the SAT instance is satisfiable. Otherwise the enumeration continues.

If P=NP there is a turing machine that always outputs a satisfying assignment if one exists with t being polynomial bounded by the input size. As this turing machine is constant, the algorithm runs in poly-time.

codeflo2• 7 years ago

Interesting, I don’t remember this proof (maybe I learned it, but it’s been a couple of years since university). Do you remember the “stop condition”? At some point the algorithm would have to “give up” and declare the problem unsatisfiable.

GuB-42• 7 years ago

So, in simpler terms (for me):

- start with a file size (let's say 4kb), and a duration (let's say 1 hour).

- generate all possible files of the given size (there are 2^32768 of them), mark them as executable and run them. If they don't finish after the given duration, kill them.

- check the output of each program that didn't crash. If one matches the solution, OK. If it doesn't, try again with a longer duration and larger size.

It doesn't just solve SAT. At galactic scales, it will either be the optimal solution to any problem, or be as fast as checking the answer, whatever is slower. If all that code generation thing doesn't depend on the input size, it is constant time. That the constant is many times the age of the universe doesn't change the complexity class.

mormegil• 7 years ago

Just yesterday, I read the comments on the Physical Polynomial-Time complexity class on Complexity Zoo: https://complexityzoo.uwaterloo.ca/Complexity_Zoo:P#php

elil17• 7 years ago

This reminds me of the spaghetti sort algorithm: https://en.m.wikipedia.org/wiki/Spaghetti_sort

imglorp• 7 years ago

Same thing, but a little more practical for computing in case your computer doesn't have a hand and a table: https://rosettacode.org/wiki/Sorting_algorithms/Sleep_sort

angel_j• 7 years ago

If all that is required is a super large constant (as in some cases), why not assume the large constant, do the calculation, then factor out the large constant?

nabdab• 7 years ago

Because the “faster” algorithm is faster than the “slow” algorithm in the range where this constant is added... but not faster than just applying the “slow” algorithm directly. So the only thing you would achieve is to slow down the calculation overall.

Galactic Algorithm

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