I don't think it would be that hard to add DEPRECATE for instances. A bit fiddly, and what should the syntax be.. but not really hard. I can offer advice if someone wants to undertake it. It's a cost/benefit question: how often will it be used? Simon | -----Original Message----- | From: libraries-bounces@haskell.org [mailto:libraries- | bounces@haskell.org] On Behalf Of Henning Thielemann | Sent: 16 June 2013 10:12 | To: harry | Cc: libraries@haskell.org | Subject: Re: Proposal: Warn when using Enum instance of Float or Double | | | On Sun, 16 Jun 2013, harry wrote: | | > The Enum instances for Float and Double have dubious semantics which | cause | > endless confusion, e.g. | > http://stackoverflow.com/questions/13203471/the-math-behind-1- | 0999999999999999-in-haskell, | > http://stackoverflow.com/questions/9810002/floating-point-list- | generator, | > http://stackoverflow.com/questions/7290438/haskell-ranges-and-floats, | > http://stackoverflow.com/questions/10328435/how-to-solve-floating- | point-number-getting-wrong-in-list-haskell, | > and many more. | > | > I would therefore like to propose that the usage of an Enum instance | of | > Float or Double generate a compiler warning, such as "The Enum | instance of | > Float is subject to rounding errors". Deadline: 2 weeks. | | I would also like to see these instances be removed or deprecated. | Unfortunately, GHC currently does not allow to DEPRECATE an instance. | With | this feature one might also tag intentionally omitted instances, like | Num | instance for (or 'of' :-) functions: | http://hackage.haskell.org/trac/ghc/ticket/7775 | | _______________________________________________ | Libraries mailing list | Libraries@haskell.org | http://www.haskell.org/mailman/listinfo/libraries

Simon Peyton-Jones wrote

I don't think it would be that hard to add DEPRECATE for instances. A bit fiddly, and what should the syntax be.. but not really hard. I can offer advice if someone wants to undertake it.

It's a cost/benefit question: how often will it be used?

I've already proposed deprecation to prime, but they probably aren't going to do it unless GHC can actually implement said deprecation. In response to "how often will it be used", would it assist with changes to the class hierarchy? -- View this message in context: http://haskell.1045720.n5.nabble.com/Proposal-Warn-when-using-Enum-instance-... Sent from the Haskell - Libraries mailing list archive at Nabble.com.

Roman Cheplyaka-2 wrote

Hi,

This is the mailing list for libraries changes proposals.

What's the library you're proposing a change to, and what is the change itself?

In other words, let's say your proposal gets accepted — what's next?

Roman

This is really more of a compiler change, but when I've submitted such proposals through GHC's trac in the past, I was directed to move it to the libraries list. OTOH, one possible implementation is that GHC will add a pragma allowing for warnings to be attached to usages of specific instances, in which case it really is a library change as well. -- View this message in context: http://haskell.1045720.n5.nabble.com/Proposal-Warn-when-using-Enum-instance-... Sent from the Haskell - Libraries mailing list archive at Nabble.com.

Right now, I'd view it as consensus that it is out of scope for the libraries@ process without such a capability in any existing compiler to make it work. On Jun 24, 2013, at 4:18 AM, harry wrote:

Coming to the end of the discussion period, and given that - Deprecation isn't an option, because this is part of the Haskell report - The libraries submission process is the correct place for this can this be considered a consensus in favour?

-- View this message in context: http://haskell.1045720.n5.nabble.com/Proposal-Warn-when-using-Enum-instance-... Sent from the Haskell - Libraries mailing list archive at Nabble.com.

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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:

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 this is 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...

* 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... It therefore suffers from numerical inaccuracies: $ python

...

...

...

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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

-- Felipe.

+1 from me too, as I mentioned on the other thread, I switched to using my own equivalent of enumFromStepTo years ago. On Fri, Jul 5, 2013 at 4:30 PM, Ivan Lazar Miljenovic wrote:

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:

...

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

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

...

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 this is 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...

* 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... It therefore suffers from numerical inaccuracies: $ python

...

...

...

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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

-- Felipe.

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

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

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On Fri, 5 Jul 2013, Twan van Laarhoven wrote:

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 this is close enough to not be a problem.

That is, it will make problems with numerical inaccuracies more seldom, i.e. harder to detect? Why is it important to use the enumFrom* functions, i.e. [x..] syntax? Why not using 'takeWhile (

There is a pretty big potential problem with this. Not every type that is enumerable admits a torsor where you can talk about forward differences. What would the enumFromStepTo function mean for, say, Ordering or for most derived Enum instances? -Edward On Sun, Jul 7, 2013 at 12:17 PM, Edward A Kmett wrote:

+1 from me.

The massive cancellation destroying significant figures of significand in enumFromThenTo has always driven me nuts.

Sent from my iPhone

On Jul 5, 2013, at 8:50 AM, Twan van Laarhoven wrote:

...

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 this is 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...

* 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... It therefore suffers from numerical inaccuracies: $ python

...

...

...

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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

Sadly this technically requires a language extension over Haskell 98/2010: ConstrainedClassMethods http://www.haskell.org/ghc/docs/6.12.2/html/users_guide/type-class-extension... GHC seems to be somewhat lax about requiring that extension on code in practice these days, but it does have implications when it comes to other approaches to implementing a compiler for Haskell. -Edward On Sun, Jul 7, 2013 at 11:18 PM, John Lato wrote:

My understanding was that the proposal required enumFromStepTo as a separate function (i.e. not a class method) with a Num constraint on the types. So the function wouldn't be available for Ordering or derived Enum instances unless they also had a Num instance.

John

On Mon, Jul 8, 2013 at 10:46 AM, Edward Kmett wrote:

...

There is a pretty big potential problem with this.

Not every type that is enumerable admits a torsor where you can talk about forward differences.

What would the enumFromStepTo function mean for, say, Ordering or for most derived Enum instances?

-Edward

On Sun, Jul 7, 2013 at 12:17 PM, Edward A Kmett wrote:

...

+1 from me.

The massive cancellation destroying significant figures of significand in enumFromThenTo has always driven me nuts.

Sent from my iPhone

On Jul 5, 2013, at 8:50 AM, Twan van Laarhoven wrote:

...

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 this is 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...

* 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... It therefore suffers from numerical inaccuracies: $ python

...

...

> 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

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

_______________________________________________ Libraries mailing list Libraries@haskell.org http://www.haskell.org/mailman/listinfo/libraries

+1 from me, and +1 for exposing enumFromStepTo. Like many others, the current behavior has always bothered me. On Fri, Jul 5, 2013 at 9:50 PM, Twan van Laarhoven wrote:

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 this is 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/**Range.cc#l525http://hg.savannah.gnu.org/hgweb/octave/file/787de2f144d9/liboctave/array/Ra...

* 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#L2742https://github.com/numpy/numpy/blob/master/numpy/core/src/multiarray/ctors.c... It therefore suffers from numerical inaccuracies: $ python

...

...

...

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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