Re: Proposal: fix Enum Double instance

+1 from me, I always think in terms of start, delta step, end anyways, this is definite a more natural api surface, as well as making the floating point story much saner. On Sat, Jul 6, 2013 at 12:18 AM, Evan Laforge wrote:

+1 from me too, as I mentioned on the other thread, I switched to using my own equivalent of enumFromStepTo years ago.

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Also +1 from me, including exporting enumFromStepTo (I typically want this a lot more than enumFromThenTo, and having to calculate what the "then" value is tends to get rather tedious).

On 6 July 2013 02:14, Felipe Almeida Lessa wrote:

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+1, it can't get better than this.

On Fri, Jul 5, 2013 at 10:50 AM, Twan van Laarhoven wrote:

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Hello All,

I would like to make a counterproposal in the Enum Double debate: Instead of deprecating or removing the instances, how about just fixing them?

A perfect instance for Enum Double is not possible, because arithmetic is inexact. But you can actually get awfully close. I.e. instead of allowing the final value to be at most step/2 past the end, we can allow it to be at most about 2e-16*step past the end. In many practical applications

On Fri, Jul 5, 2013 at 4:30 PM, Ivan Lazar Miljenovic wrote: this is

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close enough to not be a problem.

Now for some more detail on the design:

* First of all, Haskell's enumFromThenTo is stupid, because calculating step=then-from destroys numerical accuracy. So I will focus on implementing a function enumFromStepTo. You can still implement enumFromThenTo on top of it. But it might also make sense to expose enumFromStepTo.

* It is possible to write an exact instance for Rational. It is IMO unacceptable that Haskell currently does not use this correct instance. I will use this Rational instance as a baseline for comparison.

instance EnumFromStepTo Rational where enumFromStepTo f s t | s >= 0 && f <= t = f : enumFromStepTo (f + s) s t | s < 0 && f >= t = f : enumFromStepTo (f + s) s t | otherwise = []

* Writing a loop naively, by recursing with from' = from+step will result in accumulating the error of the addition. On the other hand, Doubles can represent integer numbers exactly up to 2^53, so it is better to use values from+i*step.

* Assume for simplicity that step>0. The stopping condition will then be of the form from+i*step-to > 0. When this quantity is 0 then the final value should be included, otherwise it shouldn't be.

* Now for an error analysis. Assume that we start with f,s,t :: Rational and call let f' = fromRational f let s' = fromRational s let t' = fromRational t enumFromThenTo f' s' t'

The fromRational function rounds a rational number to the nearest representable Double. The relative error in f' is therefore bounded by abs (f-f') ≤ ε * abs f, where ε is the half the maximum distance between two adjacent doubles, which is about 1.11e-16. similarly for s' and t'. the result of an addition or multiplication operation will also be rounded to the nearest representable Double.

So, the total error in our stopping condition is bounded by err(f+i*s-t) ≤ ε * abs f -- representing f + ε * i * abs s -- representing s, amplified by i + ε * i * abs s -- error in calculating (*), i * s + ε * abs (f + i*s) -- error in calculating (+), f + (i*s) + ε * abs t -- representing t + ε * abs (f + i*s - t) -- error in calculating (-)

Note that the last term of the bound is the thing we are bounding. So we can handle that by picking a slightly larger epsilon, eps = ε/(1-ε).

This leads to the following implementation of enumFromStepTo:

enumFromStepToEps eps !f !s !t = go 0 where go i | s >= 0 && x <= t = x : go (i+1) | s < 0 && x >= t = x : go (i+1) | abs (x - t) < eps * bound = [x] | otherwise = [] where x = f + i * s bound = abs f + 2 * abs (i * s) + abs t + abs x

with eps = 1.12e-16 :: Double or eps = 5.97e-8 :: Float

I have done extensive experiments with this implementation and many variants, and I believe it will work in all cases.

Now for enumFromTo, i.e. step=1 we could make a special case, since we know that step is exactly equal to 1, so there is no representation error. We could get rid of the multiplication, and replace i*s by 0 in the error calculation.

Another issue that remains is enumFromThenTo. If we take step = then - from then the relative error in step is bounded by |step' - step| ≤ ε * (|then| + |from| + |then - from|). This error gets multiplied by i, so it can unfortunately become quite large.

I think it would be best if we use the more general

enumFromStepToEps' !eps !f !s !sErr !t = go 0 where go i | s >= 0 && x <= t = x : go (i+1) | s < 0 && x >= t = x : go (i+1) | abs (x - t) < eps * bound = [x] | otherwise = [] where x = f + i * s bound = abs f + abs (i * s) + abs (i * sErr) + abs t + abs x

enumFromStepTo f s t = enumFromStepToEps' eps f s s t enumFromTo f t = enumFromStepToEps' eps f 1 0 t enumFromThenTo f h t = enumFromStepToEps' eps f (h - f) (abs h + abs f + abs (h - f)) t

I have also looked at what other language implementations do. So far I have found that: * Octave uses a similar method, with the bound 3 * double_epsilon * max (abs x) (abs t), which is less tight than my bound. see

http://hg.savannah.gnu.org/hgweb/octave/file/787de2f144d9/liboctave/array/Ra...

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* numpy just uses an array with length ceil((to-from)/step) see

https://github.com/numpy/numpy/blob/master/numpy/core/src/multiarray/ctors.c...

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It therefore suffers from numerical inaccuracies: $ python

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> from numpy import arange > notone = 9/7. - 1/7. - 1/7. > notone 1.0000000000000002 > len(arange(1,1)) 0 > len(arange(1,notone)) 1

In summary: * change Enum Rational to the always correct implementation * change Enum Double and Enum Float to the almost-always correct implementation propose above. * (optional) expose the enumFromStepTo function.

Twan

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-- Felipe.

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-- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com http://IvanMiljenovic.wordpress.com

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