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## Re: Mersenne Twister range in rand.oct and randn.oct

 From: Paul Kienzle Subject: Re: Mersenne Twister range in rand.oct and randn.oct Date: Sun, 4 Dec 2005 10:22:56 -0500

```
On Dec 4, 2005, at 2:28 AM, Mike Miller wrote:

```
```On Sat, 3 Dec 2005, Paul Kienzle wrote:

```
```On Dec 1, 2005, at 9:58 AM, Mike Miller wrote:

```
```On Thu, 1 Dec 2005, Mike Miller wrote:
```
```In one Mersenne Twister FAQ I see this:
```
2002-version mt19937ar.c has versions [0,1],[0,1),[0,1),(0,1) (32-bit
```  precision), and [0,1) with 53-bit precision.
```
Which am I getting when I use rand.oct in Octave forge? What I would really like to know are the largest and smallest values that can be produced both by rand.oct and by randn.oct.
```I should say that it's really one .oct file:  rand.oct.
randn is derived from rand.oct.
```
I took a look at Octave Forge's rand.cc but I can't find in it what I need to know (because of my incompetence, obviously).
```
```
The answer is in Makefile rather than rand.cc. This sets the ALL_BITS flag for building randmtzig.c. Here's the relevant code:
```
/* generates a random number on (0,1) with 53-bit resolution */
USE_INLINE double randu53(void)
{
const uint32_t a=randi32()>>5;
const uint32_t b=randi32()>>6;
return(a*67108864.0+b+0.4) * (1.0/9007199254740992.0);
}
```
```

```
OK. Thanks. I guess "(0,1)" means that neither 0 nor 1 will ever be generated as a random uniform, but does that last line of the code tell me the largest and smallest possible values for a random uniform? I can't figure out what the largest and smallest numbers would be.
```
It does.  Expressed in powers of two the expression is:

( [0,2^27-1] * 2^26 + [0,2^26-1] + 0.4 ) / 2^53

So the min is 0.4/2^53 and the max is (2^53-1 + 0.4)/2^53 = 1 - 0.6/2^53

```
Is it difficult to determine the range of possible values for the normal deviates (randn with mean 0 and variance 1 in Octave Forge)?
```
Again, looking at the code:

#define ZIGGURAT_NOR_R 3.6541528853610088
#define ZIGGURAT_NOR_INV_R 0.27366123732975828
...
/* As stated in Marsaglia and Tsang
*
* For the normal tail, the method of Marsaglia[5] provides:
* generate x = -ln(U_1)/r, y = -ln(U_2), until y+y > x*x,
* then return r+x. Except that r+x is always in the positive
* tail!!!! Any thing random might be used to determine the
* sign, but as we already have r we might as well use it
*
* [PAK] but not the bottom 8 bits, since they are all 0 here!
*/
double xx, yy;
do
{
xx = - ZIGGURAT_NOR_INV_R * log (RANDU);
yy = - log (RANDU);
}
while ( yy+yy <= xx*xx);
return (rabs&0x100 ? -ZIGGURAT_NOR_R-xx : ZIGGURAT_NOR_R+xx);

```
Given x^2 < 2y and the largest possible value of 2y is -2 *log(min(RANDU)), that means |randn| is at most
```
sqrt( -2*log(min(RANDU)) ) + 3.6541528853610088

```
Plugging in 0.4/2^53 for min(RANDU), randn is in (-12.332,12.332). Not quite (-Inf,Inf) like it should be :-(
```
- Paul

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