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digitalmars.D.learn - Random double

reply "qznc" <qznc go.to> writes:
I want to generate a random "double" value, excluding wierdos 
like NaN and Infinity. However, std.random.uniform seems to be 
useless. I tried things like

   std.random.uniform( double.min, double.max);
   std.random.uniform(-double.max, double.max);
   std.random.uniform(0.0, double.max);

However, I just get Inf values. :(

I assume this is due to floating point computation within 
uniform, which easily becomes Inf, if you come near the 
double.max boundary. Should that be considered a bug? 
Nevertheless, any ideas how to work around that issue?
Apr 23 2013
next sibling parent =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 04/23/2013 07:43 AM, qznc wrote:


 I want to generate a random "double" value, excluding wierdos like NaN
 and Infinity. However, std.random.uniform seems to be useless. I tried
 things like

    std.random.uniform( double.min, double.max);
    std.random.uniform(-double.max, double.max);
    std.random.uniform(0.0, double.max);


Since double.max is a valid double value, you may want to call uniform 
with a closed range:

   uniform!"[]"(-double.max, double.max)

However, that still produces double.inf. :)


 Should that be considered a bug?


I would say yes, it is a bug.


 Nevertheless, any ideas how to work around that issue?


Floating point numbers have this interesting property where the number 
of representable values in the range [double.min_normal, 1) is equal to 
the number of representable values in the range [1, double.max]. The 
number line on this article should help with what I am trying to say:

   http://dlang.org/d-floating-point.html

So, to workaround this problem I would suggest simply using the 
following range:

         uniform(-1.0, 1.0);

One benefit is, now the range is open ended: You don't need to provide 
!"[)" to leave 1 out, because it is the default. On the other hand, the 
width of the range is now 2.0 so you must keep that in mind when you 
scale the value:

Additionally, note that the random numbers in the range between 
[-double.min_normal, double.min_normal] and outside of those may have a 
different distribution. Test before using. :) (I have no experience with 
floating point random numbers.)

Ali
Apr 23 2013
prev sibling next sibling parent reply qznc <qznc web.de> writes:
Tue, 23 Apr 2013 16:43:14 +0200: qznc wrote


 I want to generate a random "double" value, excluding wierdos like NaN
 and Infinity. However, std.random.uniform seems to be useless. I tried
 things like
 
    std.random.uniform( double.min, double.max);
    std.random.uniform(-double.max, double.max);
    std.random.uniform(0.0, double.max);
 
 However, I just get Inf values. :(
 
 I assume this is due to floating point computation within uniform, which
 easily becomes Inf, if you come near the double.max boundary. Should
 that be considered a bug? Nevertheless, any ideas how to work around
 that issue?


Using a union seems to be a good workaround:

  union foo { ulong input; double output; }
  foo val = void;
  do {
  	val.input = uniform(ulong.min, ulong.max);
  } while (val.output == double.infinity
        || val.output == -double.infinity
        || val.output != val.output);
  return val.output;

Maybe the implementation of uniform should use a similar trick?
Apr 23 2013
parent reply =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 04/23/2013 11:55 AM, qznc wrote:> Tue, 23 Apr 2013 16:43:14 +0200: 
qznc wrote

 I want to generate a random "double" value, excluding wierdos like NaN
 and Infinity. However, std.random.uniform seems to be useless. I tried
 things like

     std.random.uniform( double.min, double.max);
     std.random.uniform(-double.max, double.max);
     std.random.uniform(0.0, double.max);

 However, I just get Inf values. :(

 I assume this is due to floating point computation within uniform, which
 easily becomes Inf, if you come near the double.max boundary. Should
 that be considered a bug? Nevertheless, any ideas how to work around
 that issue?


 Using a union seems to be a good workaround:

    union foo { ulong input; double output; }
    foo val = void;
    do {
    	val.input = uniform(ulong.min, ulong.max);
    } while (val.output == double.infinity
          || val.output == -double.infinity
          || val.output != val.output);
    return val.output;

 Maybe the implementation of uniform should use a similar trick?


Unfortunately, that will not produce a uniform distribution. The results 
will mostly be in the range [-0.5, 0.5]. The lower and higher values 
will have half the chance of the middle range:

import std.stdio;
import std.random;

double myUniform()
{
     union foo { ulong input; double output; }
     foo val = void;
     do {
         val.input = uniform(ulong.min, ulong.max);
     } while (val.output == double.infinity
              || val.output == -double.infinity
              || val.output != val.output);
     return val.output;
}

void main()
{
     size_t[3] bins;

     foreach (i; 0 .. 1_000_000) {
         size_t binId = 0;
         auto result = myUniform();

         if (result > -0.5) { ++binId; }
         if (result > 0.5)  { ++binId; }

         ++bins[binId];
     }

     writeln(bins);
}

Here is an output of the program:

[250104, 499537, 250359]

The first value is "less than -0.5", the second one is "between -0.5 and 
0.5", and the third one is "higher than 0.5".

Ali
Apr 23 2013
parent reply qznc <qznc web.de> writes:
Tue, 23 Apr 2013 13:49:48 -0700: Ali Çehreli wrote


 On 04/23/2013 11:55 AM, qznc wrote:> Tue, 23 Apr 2013 16:43:14 +0200:
 qznc wrote
  >
  >> I want to generate a random "double" value, excluding wierdos like
  >> NaN and Infinity. However, std.random.uniform seems to be useless. I
  >> tried things like
  >>
  >>     std.random.uniform( double.min, double.max);
  >>     std.random.uniform(-double.max, double.max);
  >>     std.random.uniform(0.0, double.max);
  >>
  >> However, I just get Inf values. :(
  >>
  >> I assume this is due to floating point computation within uniform,
  >> which easily becomes Inf, if you come near the double.max boundary.
  >> Should that be considered a bug? Nevertheless, any ideas how to work
  >> around that issue?
  >
  > Using a union seems to be a good workaround:
  >
  >    union foo { ulong input; double output; }
  >    foo val = void;
  >    do {
  >    	val.input = uniform(ulong.min, ulong.max);
  >    } while (val.output == double.infinity
  >          || val.output == -double.infinity || val.output !=
  >          val.output);
  >    return val.output;
  >
  > Maybe the implementation of uniform should use a similar trick?
 
 Unfortunately, that will not produce a uniform distribution. The results
 will mostly be in the range [-0.5, 0.5]. The lower and higher values
 will have half the chance of the middle range:



Interesting. Why [-0.5,0.5], though? I would have expected [-1.0,1.0] from 
Clugston's article.

http://dlang.org/d-floating-point.html
Apr 23 2013
parent =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 04/23/2013 11:46 PM, qznc wrote:


 Tue, 23 Apr 2013 13:49:48 -0700: Ali Çehreli wrote



 Unfortunately, that will not produce a uniform distribution. The results
 will mostly be in the range [-0.5, 0.5]. The lower and higher values
 will have half the chance of the middle range:



 Interesting. Why [-0.5,0.5], though? I would have expected [-1.0,1.0] 


from

 Clugston's article.

 http://dlang.org/d-floating-point.html


Hmmm... I don't know. Interestingly, I wrote that *before* running the 
test program, which has confirmed it. So, I was correct without knowing 
why. :)

Ali
Apr 25 2013
prev sibling next sibling parent reply "bearophile" <bearophileHUGS lycos.com> writes:
qznc:


 I want to generate a random "double" value, excluding wierdos 
 like NaN and Infinity. However, std.random.uniform seems to be 
 useless.


Can you explain why you need uniform doubles in their whole 
range? I think I have never had to generate them so far. Maybe 
for a unittest?

Also note by their nature doubles are not equally spread across 
the line of Reals, so getting a truly uniform distribution is 
hard or impossible.

Bye,
bearophile
Apr 23 2013
next sibling parent reply qznc <qznc web.de> writes:
Tue, 23 Apr 2013 22:59:41 +0200: bearophile wrote


 qznc:
 

 I want to generate a random "double" value, excluding wierdos like NaN
 and Infinity. However, std.random.uniform seems to be useless.


 
 Can you explain why you need uniform doubles in their whole range? I
 think I have never had to generate them so far. Maybe for a unittest?


Good guess. :)

I want to port QuickCheck. The core problem is: Given a type T, generate a 
random value.

https://bitbucket.org/qznc/d-quickcheck/src/
eca58bb97b24cedd6128cdda77bbebeaf1689956/quickcheck.d?at=master


 Also note by their nature doubles are not equally spread across the line
 of Reals, so getting a truly uniform distribution is hard or impossible.


It also raises the question what uniform means in the context of floating 
point. Uniform over the numbers or uniform over the bit patterns?
Apr 23 2013
parent reply "Ivan Kazmenko" <gassa mail.ru> writes:
On Wednesday, 24 April 2013 at 06:37:50 UTC, qznc wrote:

 It also raises the question what uniform means in the context 
 of floating point. Uniform over the numbers or uniform over
 the bit patterns?


I'd like to mention that there's no such mathematical object as 
"uniform distribution on [0..+infinity)".  On the other hand, a 
(discrete) uniform distribution of the bit pattern, or restricted 
bit pattern (that is, no special values), is of course valid.  A 
"random positive double" as a bit pattern most closely resembles 
exponential distribution I think.
Apr 23 2013
next sibling parent reply "Andrea Fontana" <nospam example.com> writes:
On Wednesday, 24 April 2013 at 06:56:44 UTC, Ivan Kazmenko wrote:

 On Wednesday, 24 April 2013 at 06:37:50 UTC, qznc wrote:

 It also raises the question what uniform means in the context 
 of floating point. Uniform over the numbers or uniform over
 the bit patterns?


 I'd like to mention that there's no such mathematical object as 
 "uniform distribution on [0..+infinity)".


... you neither can choose a random real number in any interval 
...
Apr 24 2013
parent reply "Ivan Kazmenko" <gassa mail.ru> writes:
On Wednesday, 24 April 2013 at 10:26:19 UTC, Andrea Fontana wrote:

 I'd like to mention that there's no such mathematical object 
 as "uniform distribution on [0..+infinity)".


 ... you neither can choose a random real number in any interval 
 ...


... but that is at least valid mathematically, albeit achievable 
only approximately on a computer.  On the other hand, an infinite 
case, even if it would be possible, won't be practical anyway 
since with probability 1, the result would require more bits to 
store than available on any modern hardware.
Apr 24 2013
parent reply "Andrea Fontana" <nospam example.com> writes:
On Wednesday, 24 April 2013 at 10:33:49 UTC, Ivan Kazmenko wrote:

 On Wednesday, 24 April 2013 at 10:26:19 UTC, Andrea Fontana 
 wrote:

 I'd like to mention that there's no such mathematical object 
 as "uniform distribution on [0..+infinity)".


 ... you neither can choose a random real number in any 
 interval ...


 ... but that is at least valid mathematically, albeit 
 achievable only approximately on a computer.  On the other 
 hand, an infinite case, even if it would be possible, won't be 
 practical anyway since with probability 1, the result would 
 require more bits to store than available on any modern 
 hardware.


I mean that a random real number is not valid mathematically too. 
In any given real interval there are infinite numbers, how you 
can choose a number in an infinite (and non-numerable!) interval? 
I think you always need some sampling.

What's the probability to guess a precise number in [0..1]? I 
think is 0 as long as you have infinite numbers.

What's the probability to guess a interval in [0..1]? I think 
it's the interval size.

Am I wrong?
Apr 24 2013
parent "Ivan Kazmenko" <gassa mail.ru> writes:
On Wednesday, 24 April 2013 at 10:46:57 UTC, Andrea Fontana wrote:

 What's the probability to guess a precise number in [0..1]? I 
 think is 0 as long as you have infinite numbers.


Right.


 What's the probability to guess a interval in [0..1]? I think 
 it's the interval size.


Right again.


 I mean that a random real number is not valid mathematically 
 too. In any given real interval there are infinite numbers, how 
 you can choose a number in an infinite (and non-numerable!) 
 interval? I think you always need some sampling.

 Am I wrong?


In a sense, yes.  A continuous probability distribution is 
well-defined.  In short, as you pointed out, you can coherently 
define the probabilities to hit each possible segment and get a 
useful mathematical object, though the probability to hit each 
single point is zero.  It's no less strict than a typical high 
school definition of an integral.  See the link for more info: 
http://en.wikipedia.org/wiki/Probability_distribution#Continuous_prob
bility_distribution 
.
Apr 24 2013
prev sibling parent "qznc" <qznc go.to> writes:
On Wednesday, 24 April 2013 at 06:56:44 UTC, Ivan Kazmenko wrote:

 On Wednesday, 24 April 2013 at 06:37:50 UTC, qznc wrote:

 It also raises the question what uniform means in the context 
 of floating point. Uniform over the numbers or uniform over
 the bit patterns?


 I'd like to mention that there's no such mathematical object as 
 "uniform distribution on [0..+infinity)".  On the other hand, a 
 (discrete) uniform distribution of the bit pattern, or 
 restricted bit pattern (that is, no special values), is of 
 course valid.  A "random positive double" as a bit pattern most 
 closely resembles exponential distribution I think.


Of course not, since infinity is not a number.

However, double.max is not infinity, but 0x1p+1024. See 
http://dlang.org/d-floating-point.html
Apr 24 2013
prev sibling parent Joseph Rushton Wakeling <joseph.wakeling webdrake.net> writes:
On 04/23/2013 10:59 PM, bearophile wrote:

 Also note by their nature doubles are not equally spread across the line of
 Reals, so getting a truly uniform distribution is hard or impossible.


More than that -- the number of unique values generated by the underlying RNG
should (for Mersenne Twister) be much smaller than the number of unique double
values in [-double.max, double.max], because the type used is uint32.

I think the infinities probably come from the fact that the returned random
value is generated by the following expression:

       _a + (_b - _a) * cast(NumberType) (urng.front - urng.min)
       / (urng.max - urng.min);

So, you've got a double.max + (double.max - double.min) in there which evaluates
to inf.
Apr 24 2013
prev sibling parent reply "Ivan Kazmenko" <gassa mail.ru> writes:
On Tuesday, 23 April 2013 at 14:43:15 UTC, qznc wrote:

 I want to generate a random "double" value, excluding wierdos 
 like NaN and Infinity. However, std.random.uniform seems to be 
 useless. I tried things like

   std.random.uniform( double.min, double.max);
   std.random.uniform(-double.max, double.max);
   std.random.uniform(0.0, double.max);

 However, I just get Inf values. :(


I've just tried to reproduce that. The first line gives a 
deprecation warning, the second gives infinities, but the third 
line does give me non-infinities.  I've checked that under DMD 
2.062 on Windows, and the hardware is Intel Xeon E5450.

Here is an example:

-----
import std.random;
import std.stdio;

void main ()
{
	foreach (i; 0..5)
	{
		double x = uniform (-double.max, +double.max);
		writefln ("%25a %30.20g", x, x);
	}

	foreach (i; 0..5)
	{
		double x = uniform (          0, +double.max);
		writefln ("%25a %30.20g", x, x);
	}
}
-----

And an example output is:

-----
                       inf                            inf
                       inf                            inf
                       inf                            inf
                       inf                            inf
                       inf                            inf
   0x1.9d6fad0f9d6f9p+1023     1.4516239851099787345e+308
   0x1.b53e4c11b53e3p+1020      1.919017005286872499e+307
   0x1.5dc7e6d15dc7dp+1020     1.5351529811186840176e+307
   0x1.bd608187bd606p+1023     1.5637717442069522658e+308
   0x1.e38d5871e38d4p+1023     1.6978092693521789428e+308
-----

Perhaps "std.random.uniform(-double.max, double.max);" call does 
indeed cause overflow.  If the example does not work for you, try 
"double.max / 2" or "double.max / 4" boundary instead.

That said, note that the values generated this way will have 
exponent close to the maximal possible (1024), and that may be 
not the only case you wish to cover.  Your suggestion to generate 
64-bit patterns instead of real numbers, probably excluding 
special values (all-0 or all-1), sounds like the way to go.  I'd 
include some important cases (like +-0 and +-1) manually too.

Ivan Kazmenko.
Apr 24 2013
parent "Ivan Kazmenko" <gassa mail.ru> writes:
On Wednesday, 24 April 2013 at 13:30:41 UTC, Ivan Kazmenko wrote:

 probably excluding special values (all-0 or all-1),


I meant exponent bits being all 0 or all 1, sorry.  That's where 
the special values reside, at least according to IEEE-754-1985 
here: http://en.wikipedia.org/wiki/IEEE_754-1985
Apr 24 2013