## Minimum Standard Random Number Generator

### January 14, 2014

Here is a basic version of the minimum standard random number generator, which updates a global variable `seed`

and returns the new `seed`

.

`(define (minstd0) (set! seed (modulo (* 16807 seed) 2147483647)) seed)`

A better design hides the seed inside a closure and returns a random fraction between zero and one, exclusive. It is also better to set the seed randomly, then allow the user to reset the seed if it is desired to repeat the same random sequence. A common way to set the seed uses the system clock:

`(define minstd1`

(let ((a 16807) (m 2147483647) (seed (time-second (current-time))))

(lambda args (if (pair? args) (set! seed (modulo (car args) m)))

(set! seed (modulo (* a seed) m)) (/ seed m))))

The expression `(time-second (current-time))`

gets the number of seconds since the Unix epoch; it works in Chez Scheme, but other Scheme systems have other conventions for getting the time. When called as `(minstd1 `

*seed*`)`

, this resets the seed before returning a random fraction.

Using Schrage multiplication makes the function only a little bit more complicated:

`(define minstd2`

(let ((a 16807) (m 2147483647) (q 127773) (r 2836)

(seed (time-second (current-time))))

(lambda args (if (pair? args) (set! seed (modulo (car args) m)))

(let* ((lo (modulo seed q)) (hi (quotient seed q))

(test (- (* a lo) (* r hi))))

(set! seed (if (positive? test) test (+ test m)))

(/ seed m)))))

Although the point of Schrage multiplication is to prevent intermediate overflow, this *doesn’t* prevent intermediate overflow in Scheme. Most Scheme systems use the two least-significant bits of any value as tag bits, so the largest integer that can be encoded natively is 2^{30}, which is too small. This will work in other languages, however.

Implementation of Carta’s version of the function uses the bit operations from the Standard Prelude. An optimized version of this function is available at http://www.firstpr.com.au/dsp/rand31/.

`(define minstd3`

(let ((a 16807) (m 2147483647) (seed (time-second (current-time))))

(lambda args (if (pair? args) (set! seed (modulo (car args) m)))

(let* ((lo (* a (logand seed #xFFFF)))

(hi (* a (ash seed -16)))

(lo (+ lo (ash (logand hi #x7FFF) 16)))

(lo (+ lo (ash hi -15)))

(lo (if (>= lo #x7FFFFFFF) (- lo #x7FFFFFFF) lo)))

(set! seed lo) (/ seed m)))))

This actually looks better in C, assuming `seed`

is a global unsigned long integer:

`long unsigned int minstd()`

{

long unsigned int lo, hi;

```
``` lo = 16807 * (seed & 0xFFFF);

hi = 16807 * (seed >> 16);

lo += (hi & 0x7FFF) <> 15;

if (lo >= 0x7FFFFFFF)

lo -= 0x7FFFFFFF;

` return ( seed = (long) lo );`

}

We can test the four implementations by computing the 10001^{st} value starting from 1; we stop at 9999 because one iteration is consumed by the initial setting of the seed and one iteration is performed in the final call after the loop completes:

`> (define seed 1)`

> (minstd0)

16807

> (do ((n 1 (+ n 1))) ((= 9999 n) (minstd0)) (minstd0))

1043618065

> (minstd1 1)

16807/2147483647

> (do ((n 1 (+ n 1))) ((= 9999 n) (minstd1)) (minstd1))

1043618065/2147483647

> (minstd2 1)

16807/2147483647

> (do ((n 1 (+ n 1))) ((= 9999 n) (minstd2)) (minstd2))

1043618065/2147483647

> (minstd3 1)

16807/2147483647

> (do ((n 1 (+ n 1))) ((= 9999 n) (minstd3)) (minstd3))

1043618065/2147483647

You can run the program at http://programmingpraxis.codepad.org/vqL1b0fL.

In Python. As in Python integer overflow cannot happen, I only implemented the simple version.

I’m having trouble with the results of the suggested solution: when I run minstd0 three times I get results 16807, 282475249 and 1622650073 but when I do manual calculations, the third result for me is: 1622652946. And I can verify that by getting (M * 221) + 1622652946 = (4745938856997 + 1622652946) = 4747561509943 = (16807 * 282475249).

So, which one is correct? Am I missing something in the calculations? Obviously, due to this difference I can’t get the same X(10001) result as suggested in the post.

Thanks!

1622650073 is correct. See A096550.

I believe that if x0 = 1, then x10000 is 1043618065. It currently says that

x10001is 1043618065.Here’s a solution using Julia. I’m not too confident that I did nextInt3 as intended, but it got the right answer in my tests.

Don’t you mean “ensuring that r<a” rather than “r<q”? The former is the result of the usual division algorithm; the latter, it seems, may force r to be negative.

I just checked the paper, and it does seem that r<a is intended. More significantly, you swapped the definitions of lo and hi, so the math doesn’t work out right.

It’s not terribly pretty, but here it is. Prettified on codepad

Since that was such a mess, I’ll post a cleaned up version. I’m hoping PP will delete the first one. To actually compile this, you will have to split it into two files with appropriate names.

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