Talk:Kahan summation

Task

The idea of showing the Kahan summation on Rosettacode is good, but the Task is not good yet. I suggest to remove the requirements of using a Decimal module and make it optional. So many languages can use normal floating point values. I also suggest to increase the number of input values and the number of their digits, to better show why the Kahan summation is useful.

This is a D version of the Task, without a Decimal:

<lang d>real kahanSum(T)(in T[] data) pure nothrow @safe @nogc {

   real tot = 0, c = 0;
   foreach (immutable x; data) {
       immutable y = x - c;
       immutable t = tot + y;
       c = (t - tot) - y;
       tot = t;
   }
   return tot;

}

void main() {

   import std.stdio, std.algorithm;

   enum a = 10000.0L, b = 3.14159L, c = 2.71828L;
   writefln("%5.10f", (a + b) + c);
   writefln("%5.10f", [a, b, c].kahanSum);
   writefln("%5.10f", [a, b, c].sum);

}</lang>

It prints (the D+Phobos sum performs a Kahan summation):

10005.8598700000
10005.8598700000
10005.8598700000

-bearophile (talk) 10:28, 11 December 2014 (UTC)

Would you need different numbers for different precision numbers? What numbers to choose as I would prefer some standardisation.

Or should it be written so that the example is forced to show the effect with the constants they use? --Paddy3118 (talk) 10:49, 11 December 2014 (UTC)

I also agree Kahan summation is a good task but have some issues. It seems Kahan summation is meaningful for fixed-precision floating-point numbers. That is, floating point numbers where the number of significant digits (or bits) is limited. This needs to be clear in the task description. If a language has no convenient fixed-precision floating-point representation, people should feel free to omit it.

I think the fixed-precision floating-point type most languages will support is IEEE 754 64-bit float. I suggest we provide task data crafted for this type. The Python example data works for floating point math that can be limited to 6 significant decimal places. It would be fine to retain this as alternative data for languages that can conveniently limit the significant decimal digits as Python can. If other languages come along that can do neither IEEE 754 64-bit floats nor 6-significant decimal digit floats but have some other limited precision floating point representation, we should allow them appropriate test data. —Sonia (talk) 20:29, 11 December 2014 (UTC)

Example added. Bearophile's suggestion of adding more numbers turned out to be good. —Sonia (talk) 20:49, 11 December 2014 (UTC)

I also suggest we explicitly disallow solutions showing the same answer from both a naive method and Kahan and concluding, "my language already does it." A valid solution should show that Kahan provides some advantage over some other method. —Sonia (talk) 20:57, 11 December 2014 (UTC)

All good suggestions.Please join in in improving this. --Paddy3118 (talk) 23:41, 11 December 2014 (UTC)

OK made my task revisions for the night, taking your comments into consideration... --Paddy3118 (talk) 05:40, 12 December 2014 (UTC)

Modified the R example to comply with the new task, still in R there is no difference because you start losing precision in R after the 15 or 16 decimal place. Perhaps the tasks needs a different set of numbers to illustrate the case? --Jmcastagnetto (talk) 16:45, 12 December 2014 (UTC)

Further modification to include the subtask for R. Now, using the precision that the main tasks defines, apparently there is no difference, but if you compare the results of both operations, there is a difference. Added code to show that in the R example. --jmcastagnetto (talk) 17:27, 12 December 2014 (UTC)

Hi I wonder if you might bring this up in your R language community and ask about the perplexing result? Maybe the setting of precision only affects output precision and internally a higher precision is used for values? --Paddy3118 (talk) 18:41, 12 December 2014 (UTC)

Yep. I found this: https://stat.ethz.ch/pipermail/r-help/2010-January/226016.html, which leads me to believe that R is using its full, floating point, for arithmetic and you are only limiting the digits in the output - not even calculating in decimal arithmatic. --Paddy3118 (talk) 18:48, 12 December 2014 (UTC)

Just found this R info: https://www.stat.auckland.ac.nz/~stat782/downloads/04-Numerics.pdf. according to that, R is floating point and the FP task section should be followed. --Paddy3118 (talk) 18:56, 12 December 2014 (UTC)

OK, indeed the options() call just restricts the number of digits being displayed, not the actual number used in calculations. In that respect it is in the same boat as PHP. AFAIK there is a package (not a base component) that can deal with arbitrary precision (http://cran.r-project.org/web/packages/Rmpfr/), but that is not what this particular task is aiming to show. I have redone the operations and results on a Windows 7 64-bit machine from a friend, will check again on my Ubuntu 64-bit box to corroborate the results. --jmcastagnetto (talk) 02:11, 17 December 2014 (UTC)

So far, in J and R

We have decimal arithmetic not being output digits, and built-in equality fuzz. This, I think, is all useful info to know when comparing languages, . The j example manages to follow the spirit of the task and leave a readable and informative entry. The R example could become just as good if it broke down the task in a similar way, but that would need more detailed knowledge of how the language treats numbers.

Indeed the J example is much better that my plain R one, will rework it to make it more understandable --jmcastagnetto (talk) 02:15, 17 December 2014 (UTC)

The task description could always be improved, I personally like the direction it is changed to, but I am starting to wonder if it will end in a morass of floating point fine-point'erry!

Any ideas for a save? --Paddy3118 (talk) 06:48, 13 December 2014 (UTC)

I have the same concerns. How about something like this,

<lang python>def kahan(vals, start = 0):

   tot = start
   c = 0
   for x in vals:
       y = x - c
       t = tot + y
       c = (t - tot) - y
       tot = t
   return tot

def triangle(n):

   return n * (n - 1) / 2

bigVal = 1e20 numSmall = 54321

print("Sequential: ", sum(range(numSmall), bigVal)) print("Kahan ", kahan(range(numSmall), bigVal)) print("Triangle: ", triangle(numSmall))</lang>

Output:

Sequential:  1.0000000000146198e+20
Kahan        1.0000000000147536e+20
Triangle:                1475358360.0

The idea is to sum enough numbers to skip over the fine-point-erry, while still picking numbers where it's easy to see the correct answer. Adding just three numers is always going rely on doing stuff to the last bit. If you add lots of numbers you can get the result we're interested in to span multiple decimal places. Most people will accept that the triangle fomula should give a good anwer, triangle(54321) fits in a 32 bit int. For types other than 754 binary64, it should be easy to pick other parameters bigVal and numSmall that will still show the loss of precision in the sequential sum. —Sonia (talk) 03:26, 14 December 2014 (UTC)

Thanks Sonia. I am still digesting your suggestion, but just on first reading, (and little cogitation I admit), the solution is not "self scaling" in any way, for different precisions. The current task floating point examples might self scale in say Python, but do not in J because of their "fuzzy comparison" (a problem with the task description and not a slur of J by the way). --Paddy3118 (talk) 05:18, 14 December 2014 (UTC)

Task 2

I suggest to rewrite this task from scratch, giving in the task description 7 or 10 64 bit floating point numbers (or more if generated with a portable random generator), and show the results with a basic summation algorithm versus doing it with Kahan. And show in some ways that the second loses less precision. Nothing more. -bearophile (talk) 10:34, 20 December 2014 (UTC)

But what about the issues we have already uncovered in several languages? What about languages that have much better control of number representation and calculations? There is a chance for languages that have these extra capabilities to shine and I would not want to lose that by fixing on one representation. --Paddy3118 (talk) 16:03, 20 December 2014 (UTC)