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a day ago by rjmunro

I still think the best way to deal with this stuff is to use interval arithmetic. Calculate the numbers with rounding set to round up, and separately to round down and compare the results. See PyInterval https://pyinterval.readthedocs.io/ for an implementation.

>>> from interval import interval

>>> data = (interval(1_000_000_000.1), interval(1.1)) * 50_000

>>> sum(data) / len(data)

interval([500000000.5971994, 500000000.601347])

The result is definitely between the 2 values. If you try another method to do the calculation, (e.g. sorting first) you will know which one was better without the answer having to be obvious as it is in this case.

You might find one method is better at the lower bound, the other is better at the higher bound, which means you can narrow the value further than using either individually. It's a really powerful system that I think should be much better known.

a day ago by nestorD

You should be okay with a sum but interval arithmetic can blow up in a way that is not correlated with the numerical accuracy of the algorithm. Newton's method, for example, is strongly convergent but produces diverging intervals.

a day ago by uyt

Why does that happen? Is it because the errors are in opposite directions and will cancel each other out?

a day ago by nestorD

That (and overall the unability to take correlation between error into account) and the fact that intervals propagated are upper bounds for some things like the division.

2 days ago by MaxBarraclough

Can't we just use Kahan summation [0] then divide by n?

edit Someone on the reddit thread [1] had the same thought.

[0] https://en.wikipedia.org/wiki/Kahan_summation_algorithm

[1] https://old.reddit.com/r/perl/comments/2zneat/how_you_averag...

2 days ago by SuchAnonMuchWow

Kahan summation is better than naive summation, but not perfect. The stable_mean algorithm in the article is not perfect either. It just has more precision than the naive one, but it more or less has the same precision as twice-as-precise floats.

>>> data = (1_000_000.1, 1.1, -1_000_000.1, -1.1) * 25_000

>>> sum(data) / len(data) == -2.3283153183228934e-16

>>> sum(sorted(data)) / len(data) == 2.7122441679239272e-11

>>> stable_mean(data) == -2.571225228716577e-13

>>> float(sum(numpy.array(data, dtype=numpy.float128)) / len(data)) == 2.2648549702353195e-19

Note that in python there is a math.fsum function which return a perfectly rounded summation of floats. There are multiple different techniques for that, but they are more complex than Kahan summation.

>>> math.fsum(data) / len(data) == 0.0

a day ago by vitus

> Note that in python there is a math.fsum function which return a perfectly rounded summation of floats.

Quirks of the underlying data can yield surprising results.

I was playing around with the "well-known" floating point error result of 0.3 - 0.2 - 0.1 = some small number (instead of 0), and found:

    >>> sum((0.3, -0.2, -0.1) * 1_000_000) / 3e6
    >>> math.fsum((0.3, -0.2, -0.1) * 1_000_000) / 3e6
    >>> stable_mean((0.3, -0.2, -0.1) * 1_000_000)
Namely, the "naive" mean yielded the smallest error (due to cancellation lining up just right such that it's capped after two rounds), fsum yielded a nominally larger one, and stable_mean yielded the biggest error.

Of course, if you sort the values (such that we're accumulating -0.2 + ... -0.2 + -0.1 + ... + -0.1 + 0.3 + ... + 0.3, then sum performs much worse, stable_mean is a little worse, and fsum performs exactly the same.

    >>> sum(sorted((0.3, -0.2, -0.1) * 1_000_000)) / 3e6
    >>> math.fsum(sorted((0.3, -0.2, -0.1) * 1_000_000)) / 3e6
    >>> stable_mean(sorted((0.3, -0.2, -0.1) * 1_000_000))
edit: Naturally, if you want accurate decimal calculations, you should consider using decimal.Decimal or fractions.Fraction (albeit at the cost of significant performance). Using numpy.float128 will help, but it's still subject to rounding errors due to imprecise representation.
a day ago by MauranKilom

Some of these results are indeed very surprising. The "correct" result of (double(0.3) - double(0.2) - double(0.1)) is -1 * 2^-54 = -5.5511151231257827e-17, and dividing by three comes out to -1.8503717077085942e-17.

Hence, as far as those functions are concerned, the naive mean yielded the worst error in both tests, but stable_mean coincidentally happens to be the best in the first test. I'm slightly surprised that none of them got the correct result.

2 days ago by dataflow
2 days ago by Majromax

In this case, it seems as if stable_mean is less accurate than the naive sum?

a day ago by SuchAnonMuchWow

The two behaves differently, so you can not strictly compare the two. Most of the time stable_mean will give more accurate results, but this is not always the case. For example you can create test cases where the naive approach will return an exact result but stable_mean will have some rounding error.

Kahan summation however is always more precise than naive summation.

2 days ago by gpderetta

Interestingly enough I was given this problem (well, numerically stable online variance, but close enough) a while ago as a take-home job interview and I ended up researching this quite a bit.

It has been a while, but if I remember correctly pairwise sum was in practice as good or better as Kahan summation (and both more stable than Walford's) but faster.

But by far the fastest and still very stable solution was to use the legacy 80 bit x87 fpu as accumulator. Which is not surprising as Kahan explicitly added 80 bit precision to x87 for intermediate computations.

2 days ago by m4r35n357

Or can we sum more accurately by taking the stable_mean and multiplying by n?

a day ago by rhaps0dy

According to Radford Neal, it is possible to sum a sequence of n-bit floating point numbers exactly, i.e. rounding to the n-bit floating point value closest to the exact result.

The algorithm uses 2n bits for the sum. https://radfordneal.wordpress.com/2015/05/21/exact-computati...

a day ago by asciimov

I have a special hatred of floating point numbers. I wish and advocate for a better standard.

My issue is that I've come across errors due to floating point in weird places. One example, searching a list of strings for another string. If the sort function decides those strings look like numbers it may coerce them into doubles (ints being too small to hold them). Then sort wont match correctly due to the least significant bits being rounded off.

a day ago by Kranar

Not sure what this could possibly have to do with floating point numbers, nor am I aware of any function that does what you're describing. Is this some kind of JavaScript quirky behavior because nothing about floating point numbers has anything to do with coercing strings.

a day ago by TacticalCoder

I've definitely been bitten by similar (not identical) stuff. I've seen data serialized and sent over the wire by I-don't-know-what where several parsers can't correctly parse/serialize and end up with the same thing written back. Be it 32 bit float/64 bit float/string parsed as 32 bit float/string parsed as 64 bit float/string parsed as big decimals: I don't care, the fact is there are buggy serializers, buggy parsers, and even sometimes buggy parsers going into infinite loop (the JVM is famous for that: there was a bug where Double.parseDouble could be sent into an infinite loop by feeing it a carefully crafted string).

The fact is: such kind of SNAFU does happen.

16 hours ago by spacechild1

I think the lesson is rather: don't use text based protocols for data serialization if you care about precision. Such things should never happen with a binary protocol.

a day ago by titzer

I've gone back and forth over the years. Having implemented some of the lowest-level floating point routines, taking care to handle all the corner cases, it definitely is a chore. But is there an alternative?

After all of this, I still think that IEEE 754 is pretty darn good. The fact that it is available and fast in pretty much all hardware is a major selling point.

That said, I absolutely hate -0 (minus zero). It's a stupid design wart that I've never seen any good use case for. But it is observable and therefore cannot be ignored, leading to even more icky special cases.

a day ago by jart

How about determining the quadrant of an angle? That's how atan2() uses signed zero and it's the greatest of all libm functions.

a day ago by titzer

After staring at the MatLab[https://www.mathworks.com/help/matlab/ref/atan2.html#buct8h0...] explanation, I have to admit I don't really get that. Minus zero is a encoding artifact. Values of [x,y] with either `x == 0` or `y == 0` don't have, in the mathematical sense, a quadrant. The fact that a convention preserves an encoding artifact doesn't make it useful.

E.g. I've seen it argued that +0 and -0 represent limits of functions that approach zero from the negative side or the positive side. But that makes no sense, because no other floating point number represents a limit going from above or below the number, so why should 0? Also, all this walking on eggshells around a possible -0 (because that represents the limit of 0 coming from below 0) is just lost when someone writes 0. Admittedly, I've not seen the guts of most numerical code, but the only time I have ever seen a minus 0 written into code was to test the minus-zero handling of something.


a day ago by TinkersW

That isn't floating points fault, but whoever wrote the buggy parser..

Floating point is overall pretty good, sometimes I wish there was a version without some many ways to store NAN(for 16 bit floats it is so wasteful), and maybe versions of sqrt/div/rcp/rsqrt that would clamp invalid inputs to 0, but other than that it is hard to see how you make any major improvements given the fixed # of bits.

Actually one other possible improvement would be a way to request the hardware to dither when rounding down to the fixed bit count, although I'm not sure how it could do anything besides pure white noise, but better than nothing..

a day ago by dolmen

Don't blame the floating point numbers.

Just use a programming language that doesn't coerce strings into numbers. Or use a comparison function/operator that explicitely compares arguments as strings and not as numbers.

a day ago by worik

If you are looking for such accuracy (where 50_000_000.00006 is different from 50_000_000.06) then you need a specialised library and do very careful coding.

If you are averaging bazillions of numbers and they range from weeny to huge and you thing 50_000_000.000006 != 50_000_000.06 for your purposes then it is hard for me to see that you are doing anything sensible.

This is, except in very specialised cases that I cannot imagine, a non issue.

Just add them up and divide, (and check for overflow, that could ruin your day!)

a day ago by jfrunyon


If you do, in fact, need to worry about floating point error, then there is not some magical "it will never happen if you have this much precision and use this neat algorithm". Heck, even fixed/decimal/rational numbers aren't perfect solutions, they just make it easier to tell if you're staying within your precision.

Without infinite storage, you can't retain infinite precision, so the only real question is how much error you are willing to tolerate. And that depends very much on what you're doing - I can tolerate a lot more error in rolls on a MMO than in my bank account balance...

a day ago by johnbcoughlin

> the only real question is how much error you are willing to tolerate

So, you are, in fact, gonna need an algorithm for summation that controls that error, if you're not willing to tolerate very much.

a day ago by jfrunyon

Yes. Similarly, if you design something with total disregard to optimization, it will likely need optimization.

YAGNI is a prediction, not a rule. Just like premature optimization, alternatives to floating point - or using longer floating point types - have real costs, which must be balanced against the needed precision.

a day ago by commandlinefan

Well, his final example suggests otherwise, but I kind of suspect that there must be something else going on there (that I'm too lazy to try to track down). Still, while fixing most floating point errors seem like rearranging the deck chairs on the titanic, if there's one algorithm that's 99.999% accurate and another that's only 99.9% accurate, might as well use and be aware of the more accurate one.

a day ago by jfrunyon

The error in the 'Variance calculated using Statistics::Descriptive' is ~0.0002% of the mean (and of each value).

As the OP says: "One saving grace of the real world is the fact that a given variable is unlikely to contain values with such an extreme range"

My point is that 99.999% algorithm is going to have a cost over the 99.9% algorithm (more complex code, more resource usage), and depending on what you're doing, you may have to consider that cost.

a day ago by jart

Not in this case. Consider naive sum:

    double fsum(const double *p, size_t n) {
      size_t i;
      double s;
      for (s = i = 0; i < n; ++i) s += p[i];
      return s;
Add one line of code and it not only becomes numerically stable but benchmarks faster too.

    double fsum(const double *p, size_t n) {
      size_t i;
      double s;
      if (n > 8) return fsum(p, n / 2) + fsum(p + n / 2, n - n / 2);
      for (s = i = 0; i < n; ++i) s += p[i];
      return s;
Losers always whine about how their laziness and ignorance is a calculated tradeoff.

What we need to consider is that sometimes there is no tradeoff.

a day ago by _0ffh

I suppose if you're not bothered with wasting lots of time, do the following for a precise sum: Select the two smallest numbers from the list, remove them and insert their sum. Repeat until only one number is left.

[Ed. smaller meaning closer to zero]

a day ago by dolmen

This is exactly what happens when you sort the number ascending and compute the sum starting from the smallest. Like in the article.

Except this doesn't work if you must handle negative numbers. So what matters is not the smallest numbers but the number nearer from zero.

a day ago by MauranKilom

No, that's not the same.

Summing [4, 5, 6, 7] would yield the series (4, )9, 15, 22 with "sort then add", but what the previous poster described would first sum 4+5 (leaving the list [9, 6, 7]), then 6+7 (leaving the list [9, 13]) and finally arriving at 22.

a day ago by dolmen


That algorithm could be improved (for accuracy, not speed) by replacing sums of the smallest number with multiplications as I suppose that multiplication of more than 3 identical numbers or more is loosing less precision than multiple additions.

For example:

[4, 3, 7, 7, 8] could be calculated as:

  (3+4) => [7, 7, 7, 8]
  (7*3) => [21, 8]
  (8+21) => 29
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