Re: [Haskell-beginners] How to improve the accuracy of floating point calculation?

We just figured it out over here. The short version is that the 0.1 and 0.2 just happen to round pretty well (to what looks perfect to a human) while other figures don't. This can be easily represented by forcing a [Float] type for lower precision as opposed to the default double precision: *Main> [0.1, 0.2 .. 1] :: [Float] [0.1,0.2,0.3,0.40000004,0.50000006,0.6000001,0.7000001,0.80000013,0.90000015,1.0000002] Above we can easily see that with a float the errors compound so much that the numbers no longer round to a human-friendly representation. This can also be shown with doubles: the 0.1 was displayed as 0.1 because it just happened to be fairly precise. Start compounding errors early enough and it no longer rounds so nicely: [-0.5,-0.4,-0.30000000000000004,-0.20000000000000007,-0.10000000000000009,-1.1102230246251565e-16,9.999999999999987e-2,0.19999999999999984,0.2999999999999998,0.3999999999999998,0.4999999999999998] You can see just how much precision we lost at mid-point by looking at what should be 0. Thanks On 06/02/13 00:47, Mateusz Kowalczyk wrote:

I thought that KC was asking why is 0.5 not displayed as 0.5 when it's supported by the IEEE754. Even if that wasn't question, it is my question now.

IEEE754 is capable of representing 0.5 with perfect precision; in fact, it would be `00111111000000000000000000000000'.

My question is why [0.1, 0.2 .. 1.0] comes up with an imprecise midpoint: [0.1,0.2,0.30000000000000004,0.4000000000000001,0.5000000000000001,0.6000000000000001,0.7000000000000001,0.8,0.9,1.0]

My initial guess is that the expansion is done by working out the difference between the first two elements and using that to generate the rest. This also makes me question why 0.1 is shown properly when it can't be represented precisely in IEEE754. Quickly rolling own list generation function:

*Main> let f x y = x : f y (y + y - x) *Main> take 10 $ f 0.1 0.2 [0.1,0.2,0.30000000000000004,0.4000000000000001,0.5000000000000001,0.6000000000000001,0.7000000000000001,0.8,0.9,1.0]

reveals the same result as the short-hand which would imply that it's exactly what it's doing (for the simple case). My question for why can 0.1 be shown properly still stands in this case.

On 06/02/13 00:12, Darren Grant wrote:

...

I'm not sure how CReal implements its values, but IEEE754 also supports decimal formats preferred for accuracy in many applications. Take a look:

http://en.wikipedia.org/wiki/IEEE_floating_point

Cheers, d

On Tue, Feb 5, 2013 at 3:24 PM, Patrick Mylund Nielsen mailto:haskell@patrickmylund.com> wrote:

http://floating-point-gui.de/ http://floating-point-gui.de/formats/fp/

http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems

On Wed, Feb 6, 2013 at 12:08 AM, KC mailto:kc1956@gmail.com> wrote:

0.1 cannot be represented exactly in floating point.

0.5 can be represented exactly. Why?

On Tue, Feb 5, 2013 at 2:41 PM, yi lu mailto:zhiwudazhanjiangshi@gmail.com> wrote: > Hi, > > I found that in ghci, I input > [0.1,0.2..2] > and run, I get a result of > > [0.1,0.2,0.30000000000000004,0.4000000000000001,0.5000000000000001,0.6000000000000001,0.7000000000000001,0.8,0.9,1.0,1.1,1.2000000000000002,1.3000000000000003,1.4000000000000004,1.5000000000000004,1.6000000000000005,1.7000000000000006,1.8000000000000007,1.9000000000000008,2.000000000000001] > > But, as you know, it is not the exact answer. > > So, I wonder if there is something I can do to achieve a better performance > and get [0.1,0.2,0.3,0.4..] as the result. > > Thanks. > > _______________________________________________ > Beginners mailing list > Beginners@haskell.org mailto:Beginners@haskell.org > http://www.haskell.org/mailman/listinfo/beginners >

-- -- Regards, KC

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