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 230, 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 10001st 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.

Pages: 1 2

11 Responses to “Minimum Standard Random Number Generator”

1. Paul said

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

from __future__ import division

import time

def minst(seed=None):
M = 2147483647
if seed is None:
seed = (int(time.clock() * 1e15)) % M
while 1:
seed = (16807 * seed) % M
yield seed / M

2. Felipe Ceotto said

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!

3. programmingpraxis said

1622650073 is correct. See A096550.

4. Daniel said

I believe that if x0 = 1, then x10000 is 1043618065. It currently says that x10001 is 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.

function nextInt1(x, a = 16807, m = 2147483647)
(a * x) % m;
end

function nextInt2(x, a = 16807, m = 2147483647)
q = div(m, a);
r = m % a;
lo = x % q;
hi = div(x, q);
x = a*lo - r*hi;
if (x <= 0)
x += m;
end
x;
end

function nextInt3(x, a = 16807, m = 2147483647)
ax = a*x;
q = ax & (1 << 31 - 1);
p = ax >> 31;
s = p + q;
if (s >= m)
s -= m;
end
s;
end

5. treeowl said

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.

6. treeowl said

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.

7. treeowl said

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

module Main where
import           Data.Bits
import           Data.Int

a32=7^5::Int32

a64::Int64
a64=7^5

m32=2^31-1::Int32

m64::Int64
m64= 2^31-1

(q32,r32)=m32 `divMod` a32

getRandom1:: Int64 -> Int64
getRandom1 xold = (a64*xold) `mod` m64

low31=0x7FFFFFFF::Int64

getRandom3 :: Int64 -> Int64
getRandom3 xold = if newmodded Int32

getRandom2 xold = if res >0 then res else res +m32
where
(hi,lo)=xold `divMod` q32
res = a32*lo-r32*hi

randoms1 = iterate getRandom1

randoms2 = iterate getRandom2

randoms3 = iterate getRandom3

numtest = 1000::Int

main :: IO ()
main = do
let ones = take numtest . randoms1
let twos = take numtest . map (fromInteger.fromIntegral) . randoms2
let threes = take numtest . randoms3

print (ones 1 == threes 1)
print (twos 1==threes 1)

8. treeowl said

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.

{-# START_FILE Lincon.hs #-}
module Lincon (getRandom1,getRandom2,getRandom3,randoms1,randoms2,randoms3) where
import           Data.Bits
import           Data.Int

a::Integral x => x
a=7^5
m::Integral x => x
m=2^31-1

a32=a::Int32
a64=a::Int64
m32=m::Int32
m64=m::Int64

getRandom1:: Int64 -> Int64
getRandom1 xold = (a64*xold) `mod` m64

getRandom3 :: Int64 -> Int64
getRandom3 xold = if newmodded Int32
getRandom2 xold = if res >0 then res else res +m32
where
(q,r)=m `divMod` a
(hi,lo)=xold `divMod` q
res = a32*lo-r*hi

randoms1 = iterate getRandom1
randoms2 = iterate getRandom2
randoms3 = iterate getRandom3

{-# START_FILE Main.hs #-}
module Main (main) where
import Lincon

numtest = 100000::Int

main :: IO ()
main = do
let ones = take numtest . randoms1
let twos = take numtest . map fromIntegral . randoms2
let threes = take numtest . randoms3
print (ones 1 == threes 1)
print (twos 1==threes 1)

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