# Generating 2 independent random numbers

Discussion in 'C++' started by bilgekhan, May 28, 2008.

1. ### bilgekhanGuest

What is the correct method for generating 2 independent random numbers?
They will be compared whether they are equal.

srand(time(0));
int r1 = rand();
srand(rand());
int r2 = rand();
bool f = r1 == r2;

bilgekhan, May 28, 2008

2. ### bilgekhanGuest

In my prev. posting I forgot to mention that the
random numbers shall be in the range 0 to 99.
So here my updated question:

What is the correct method for generating
2 independent random numbers in the range 0 to 99 ?
They will immediately be compared for equality.

srand(time(0));
int r1 = rand() % 100;
srand(rand());
int r2 = rand() % 100;
bool f = r1 == r2;

bilgekhan, May 28, 2008

3. ### Pascal J. BourguignonGuest

"bilgekhan" <> writes:

> In my prev. posting I forgot to mention that the
> random numbers shall be in the range 0 to 99.
> So here my updated question:
>
> What is the correct method for generating
> 2 independent random numbers in the range 0 to 99 ?
> They will immediately be compared for equality.

What Sam said.

>
> srand(time(0));
> int r1 = rand() % 100;
> srand(rand());
> int r2 = rand() % 100;
> bool f = r1 == r2;

The rand() function returns a pseudo-random integer between 0 and RAND_MAX.

Therefore, assuming it's excluding RAND_MAX, and including 0, and
assuming each value is equiprobable (which is not specified by the man
page of rand(3) on Linux), if RAND_MAX % 100 is not 0, then your
randoms r1 and r2 won't be equiprobable.

But this is already assuming too much, as further reading of the
rand(3) man page.

If you care so little about your randomness, you could as well write

bool f=((rand()%100)==(rand()%100));

--
__Pascal Bourguignon__

Pascal J. Bourguignon, May 28, 2008
4. ### Kai-Uwe BuxGuest

bilgekhan wrote:

> In my prev. posting I forgot to mention that the
> random numbers shall be in the range 0 to 99.
> So here my updated question:
>
> What is the correct method for generating
> 2 independent random numbers in the range 0 to 99 ?
> They will immediately be compared for equality.
>
>
> srand(time(0));
> int r1 = rand() % 100;
> srand(rand());
> int r2 = rand() % 100;
> bool f = r1 == r2;

Not good: the reseeding of the RNG just has unknown effects. If the RNG used
by rand() is good, the reseeding very likely will make it worse. At the
very least, you loose all the good things that the theory inspiring the RNG
guarantees.

a) How to generate independently distributed random numbers?

b) How to generate random numbers in [0,100)?

As for (a), any RNG is supposed to yield pseudo random numbers that look
independent. Therefore, just using a decent RNG will take care of the
independence requirement.

As for (b), the %-trick is probably good enough for your purposes. However,
beware that unless RAND_MAX+1 is divisible by 100.

As for the immediate comparison part: if you discard r1 and r2 immediately
after the comparison, and only keep f, then it suffices to generate one
number r in [0,100) and do:

bool f = ( r == 0 );

because that will yield true with probability 0.01, too.

Best

Kai-Uwe Bux

Kai-Uwe Bux, May 28, 2008
5. ### Lionel BGuest

On Wed, 28 May 2008 13:08:17 +0200, bilgekhan wrote:

> In my prev. posting I forgot to mention that the random numbers shall be
> in the range 0 to 99. So here my updated question:
>
> What is the correct method for generating 2 independent random numbers
> in the range 0 to 99 ? They will immediately be compared for equality.

As others have mentioned, most random number generators (and almost
certainly your `rand()' call) generate *pseudo*-random sequences of
numbers. "Real" random number sequences (and it's by no means trivial to
say what that even means) are hard to come by and will generally involve
some hardware-based solution. I'll assume that pseudo random is good

But be aware that there is a huge variation in the "randomness" of
sequences generated by pseudo random generators. The system-supplied
generators (like `rand()') are often astonishingly non-random, and the
(current) C++ standard allows them to be as cruddy as they please.
Thankfully, the forthcoming C++ standard will include some higher quality
generators such as the Mersenne Twister (recommended, Google it). But
even that would not be good enough e.g. for cryptographic purposes.

>
> srand(time(0));

This is ok as long as you don't call this routine again so quickly that
time(0) returns the same value as for the last call.

> int r1 = rand() % 100;

This can be a bad idea, particularly for so-called "linear congruential"
rngs (and there's a fair chance your `rand()' will be linear
congruential). The reason for this is that mod-ing a number basically
involves using just the lower-order bits, which for many rngs - and
linear congruential ones in particular - are frequently the "least
random".

Safer is something along the lines of:

int r1 = int(100.0*double(rand())/double(RAND_MAX));

that is, you generate a uniform(ish) random double in the range [0,1),
multiply it by 100 and take the integer part. This gives you a uniform
(ish) random int in the range [0,99). It's safer because it relies more
heavily on the higher-order bits of the number returned by your rng.

> srand(rand());

^^^^^^^^^^^^^^

Probably a bad idea to re-seed your rng with a number generated by the
same rng... and pointless in any case. Seed once and generate away...
Leave out that line.

> int r2 = rand() % 100;

Again, as for r1 above.

> bool f = r1 == r2;

--
Lionel B

Lionel B, May 28, 2008
6. ### bilgekhanGuest

"Pascal J. Bourguignon" wrote:
> "bilgekhan" writes:
>
> > In my prev. posting I forgot to mention that the
> > random numbers shall be in the range 0 to 99.
> > So here my updated question:
> >
> > What is the correct method for generating
> > 2 independent random numbers in the range 0 to 99 ?
> > They will immediately be compared for equality.

>
> What Sam said.
>
> >
> > srand(time(0));
> > int r1 = rand() % 100;
> > srand(rand());
> > int r2 = rand() % 100;
> > bool f = r1 == r2;

>
> The rand() function returns a pseudo-random integer between 0 and RAND_MAX.
>
> Therefore, assuming it's excluding RAND_MAX, and including 0, and
> assuming each value is equiprobable (which is not specified by the man
> page of rand(3) on Linux), if RAND_MAX % 100 is not 0, then your
> randoms r1 and r2 won't be equiprobable.
>
> But this is already assuming too much, as further reading of the
> rand(3) man page.
>
>
> If you care so little about your randomness, you could as well write
>
> bool f=((rand()%100)==(rand()%100));

Have you tried this in a loop of say 1 million times and counted the hits?
I think theoretically there should be about 10,000 matches.
What result do you get?
Maybe 0 ?

bilgekhan, May 28, 2008
7. ### Lionel BGuest

On Wed, 28 May 2008 14:26:27 +0200, bilgekhan wrote:

> "Pascal J. Bourguignon" wrote:
>> "bilgekhan" writes:
>>
>> If you care so little about your randomness, you could as well write
>>
>> bool f=((rand()%100)==(rand()%100));

>
> Have you tried this in a loop of say 1 million times and counted the
> hits? I think theoretically there should be about 10,000 matches. What
> result do you get?
> Maybe 0 ?

Why do you think you might get 0 ?

--
Lionel B

Lionel B, May 28, 2008
8. ### bilgekhanGuest

"Kai-Uwe Bux" wrote:
> bilgekhan wrote:
>
> > In my prev. posting I forgot to mention that the
> > random numbers shall be in the range 0 to 99.
> > So here my updated question:
> >
> > What is the correct method for generating
> > 2 independent random numbers in the range 0 to 99 ?
> > They will immediately be compared for equality.
> >
> >
> > srand(time(0));
> > int r1 = rand() % 100;
> > srand(rand());
> > int r2 = rand() % 100;
> > bool f = r1 == r2;

>
> Not good: the reseeding of the RNG just has unknown effects. If the RNG used
> by rand() is good, the reseeding very likely will make it worse. At the
> very least, you loose all the good things that the theory inspiring the RNG
> guarantees.
>
> Your question has two aspects:
>
> a) How to generate independently distributed random numbers?
>
> b) How to generate random numbers in [0,100)?
>
> As for (a), any RNG is supposed to yield pseudo random numbers that look
> independent. Therefore, just using a decent RNG will take care of the
> independence requirement.

I did some experiments with the standard pseudo-RNG
of the compiler and I come to the conclusion that it is
impossible to let 1 RNG _correctly_ generate 2 independent
random numbers.
If someone manages this let me know pls, though
I solved my original problem by the clever method of Kai-Uwe below.

> As for (b), the %-trick is probably good enough for your purposes. However,
> beware that unless RAND_MAX+1 is divisible by 100.

Unfortuately it's not divisable by 100, and
unfortunately on my machine RAND_MAX is only 32767 :-(

> As for the immediate comparison part: if you discard r1 and r2 immediately
> after the comparison, and only keep f, then it suffices to generate one
> number r in [0,100) and do:
>
> bool f = ( r == 0 );
>
> because that will yield true with probability 0.01, too.

That's really a nice solution!
So there is no need for 2 random numbers at all,
and all the other associated problems with it.
Thanks, Kai-Uwe!

bilgekhan, May 28, 2008
9. ### bilgekhanGuest

"Lionel B" wrote:
> On Wed, 28 May 2008 14:26:27 +0200, bilgekhan wrote:
> > "Pascal J. Bourguignon" wrote:
> >>
> >> If you care so little about your randomness, you could as well write
> >> bool f=((rand()%100)==(rand()%100));

> >
> > Have you tried this in a loop of say 1 million times and counted the
> > hits? I think theoretically there should be about 10,000 matches. What
> > result do you get?
> > Maybe 0 ?

>
> Why do you think you might get 0 ?

Because it is the same sequence of the RNG...
A seed sequence of rand() always generates
RAND_MAX different random numbers.
So within a sequence it cannot generate the same number twice.
After generating RAND_MAX numbers the sequence
starts over again to generate again the same sequence of numbers.

Since the program is within the same sequence
and since the two numbers are generated together
it never can happen that these 2 numbers somehow match...

Ie. rand() is predictable: if you generate RAND_MAX numbers
and store them somewhere then you can look there what random
number it will generate next.
But this works only if you don't change the rand sequence
by calling srand() a second time.

I did some experiments and I come to the conclusion
that it is not NOT possible to let 1 RNG _correctly_ generate
2 independent random numbers.
One has to solve it cleverly via just 1 random number only
as was shown by Kai-Uwe Bux (cf. his posting).

bilgekhan, May 28, 2008
10. ### Lionel BGuest

On Wed, 28 May 2008 16:22:43 +0200, bilgekhan wrote:

> "Lionel B" wrote:
>> On Wed, 28 May 2008 14:26:27 +0200, bilgekhan wrote:
>> > "Pascal J. Bourguignon" wrote:
>> >>
>> >> If you care so little about your randomness, you could as well write
>> >> bool f=((rand()%100)==(rand()%100));
>> >
>> > Have you tried this in a loop of say 1 million times and counted the
>> > hits? I think theoretically there should be about 10,000 matches.
>> > What result do you get?
>> > Maybe 0 ?

>>
>> Why do you think you might get 0 ?

>
> Because it is the same sequence of the RNG... A seed sequence of
> rand() always generates RAND_MAX different random numbers.

Really? The documentation for my system doesn't say that. It simply says:
"The rand() function returns a pseudo-random integer between 0 and
RAND_MAX". Maybe your rand() does what you say... there is no requirement
for it to do so.

In any case, that's irrelevant, since x !=y does not imply x%100 != y%
100. Think about it - and try it:

#include <iostream>
#include <cstdlib>

int main()
{
const int N = 1000000;

int hits = 0;
for (int n=0;n<N;++n)
if ((std::rand()%100)==(std::rand()%100)) ++hits;

std::cout << "hits = " << hits << '\n';
}

On my system this outputs:

hits = 10195

--
Lionel B

Lionel B, May 28, 2008
11. ### Owen JacobsonGuest

On May 28, 10:22 am, "bilgekhan" <>
wrote:
> "Lionel B" wrote:
> > On Wed, 28 May 2008 14:26:27 +0200, bilgekhan wrote:
> > > "Pascal J. Bourguignon" wrote:

>
> > >> If you care so little about your randomness, you could as well write
> > >>    bool f=((rand()%100)==(rand()%100));

>
> > > Have you tried this in a loop of say 1 million times and counted the
> > > hits? I think theoretically there should be about 10,000 matches. What
> > > result do you get?
> > > Maybe 0 ?

>
> > Why do you think you might get 0 ?

>
> Because it is the same sequence of the RNG...
> A seed sequence of rand() always generates
> RAND_MAX different random numbers.

No. It generates numbers between 0 and RAND_MAX. RAND_MAX has no
relationship to how long the period of the rand() RNG is.

Furthermore, even if you were right and it could only generate each
value once, mapping the RNG output to values [0, 100) means that many
(RAND_MAX/100) values from the RNG map to one value in the output, so
the expression rand() % 100 could very well generate '5' twice in a
row.

> So within a sequence it cannot generate the same number twice.

False, or at best buggy. In fact, this is a dangerous property for an
RNG, and no respectable PRNG has it. If your system's rand() RNG
can't generate a value twice in the same period, report it to the
appropriate vendor's bug list and get it fixed.

> Ie. rand() is predictable: if you generate RAND_MAX numbers
> and store them somewhere then you can look there what random
> number it will generate next.

This is true: iff you know the period of your rng (the number of
samples it produces before repeating its list). However, the period
of rand() is allowed to be much larger than RAND_MAX -- in which case
it must, by simple counting logic, produce at least one value more
than once.

> But this works only if you don't change the rand sequence
> by calling srand() a second time.
>
> I did some experiments and I come to the conclusion
> that it is not NOT possible to let 1 RNG _correctly_ generate
> 2 independent random numbers.
> One has to solve it cleverly via just 1 random number only
> as was shown by Kai-Uwe Bux  (cf. his posting).

Indeed. It's better to understand the probability distribution you
want -- often you can achieve it with only a single RNG, even for more
complicated distributions. Multiple RNG samples aren't any "more
random" than a single sample.

Owen Jacobson, May 28, 2008
12. ### Kai-Uwe BuxGuest

Pete Becker wrote:

> On 2008-05-28 10:22:43 -0400, "bilgekhan"

[snip]
>> I did some experiments and I come to the conclusion
>> that it is not NOT possible to let 1 RNG _correctly_ generate
>> 2 independent random numbers.

If that is true, then the implementation of the RNG you are using sucks. A
good pseudo random number generator will pass tests for statistical
independence of subsequent returns.

> Define "independent". <g>

As far as I can see, the OP just uses the ordinary concept of independent
random events/variables: knowing one does not help predicting the other.
Why do you ask? do you have any reason to suspect otherwise?

Best

Kai-Uwe Bux

Kai-Uwe Bux, May 28, 2008
13. ### Lionel BGuest

On Wed, 28 May 2008 10:58:51 -0400, Pete Becker wrote:

> On 2008-05-28 10:22:43 -0400, "bilgekhan"
> <> said:
>
>> "Lionel B" wrote:
>>> On Wed, 28 May 2008 14:26:27 +0200, bilgekhan wrote:
>>>> "Pascal J. Bourguignon" wrote:
>>>>>

>>
>> I did some experiments and I come to the conclusion that it is not NOT
>> possible to let 1 RNG _correctly_ generate 2 independent random
>> numbers.

>
> Define "independent". <g>

I assume the OP intends "independent" in the (perfectly well-defined)
probability-theoretic sense.

I guess you could argue that it doesn't make much sense to talk about
independence of *pseudo* random numbers. In practice, however, you can
perfectly well look for statistics such as serial (sample) correlation
between successive random deviates generated by some PRNG - this is one
intuitive notion of "how independent" successive deviates are. This would
certainly be one (of many) tests routinely applied to gauge the quality
of a PRNG.

The whole question of "how random is a sequence of numbers" leads into
astonishingly deep and murky mathematical/philosophical waters (of which
I have little knowledge).

--
Lionel B

Lionel B, May 28, 2008
14. ### Pascal J. BourguignonGuest

"bilgekhan" <> writes:

> "Pascal J. Bourguignon" wrote:
>> "bilgekhan" writes:
>>
>> > In my prev. posting I forgot to mention that the
>> > random numbers shall be in the range 0 to 99.
>> > So here my updated question:
>> >
>> > What is the correct method for generating
>> > 2 independent random numbers in the range 0 to 99 ?
>> > They will immediately be compared for equality.

>>
>> What Sam said.
>>
>> >
>> > srand(time(0));
>> > int r1 = rand() % 100;
>> > srand(rand());
>> > int r2 = rand() % 100;
>> > bool f = r1 == r2;

>>
>> The rand() function returns a pseudo-random integer between 0 and RAND_MAX.
>>
>> Therefore, assuming it's excluding RAND_MAX, and including 0, and
>> assuming each value is equiprobable (which is not specified by the man
>> page of rand(3) on Linux), if RAND_MAX % 100 is not 0, then your
>> randoms r1 and r2 won't be equiprobable.
>>
>> But this is already assuming too much, as further reading of the
>> rand(3) man page.
>>
>>
>> If you care so little about your randomness, you could as well write
>>
>> bool f=((rand()%100)==(rand()%100));

>
> Have you tried this in a loop of say 1 million times and counted the hits?
> I think theoretically there should be about 10,000 matches.
> What result do you get?
> Maybe 0 ?

Yes, that's what I'd expect. And that's why the OP could expect too,
"If [he] care so little about [his] randomness" so as not to read
the rand(3) man page.

--
__Pascal Bourguignon__

Pascal J. Bourguignon, May 28, 2008
15. ### bilgekhanGuest

"Pete Becker" wrote:
> On 2008-05-28 10:22:29 -0400, "bilgekhan" said:
> >
> > Unfortuately it's not divisable by 100, and
> > unfortunately on my machine RAND_MAX is only 32767 :-(
> >

>
> The usual solution here is to compute the largest multiple of 100
> that's less than RAND_MAX, and ignore any generated values that are
> equal to or greater than that value:
>
> int rand100()
> {
> const int max = (RAND_MAX / 100) * 100;
> int tmp = rand();
> while (max <= tmp)
> tmp = rand();
> return tmp % 100;
> }

Hmm... this method has a drawback of slowing the program.
If RAND_MAX is say 32767 then the window in the routine above
is just 0.3% of the full range, so in 99.7% of the calls it will waste
extra calls to rand()...

bilgekhan, May 28, 2008
16. ### Lionel BGuest

On Wed, 28 May 2008 17:36:52 +0200, bilgekhan wrote:

> "Pete Becker" wrote:
>> On 2008-05-28 10:22:29 -0400, "bilgekhan" said:
>> >
>> > Unfortuately it's not divisable by 100, and unfortunately on my
>> > machine RAND_MAX is only 32767 :-(
>> >
>> >

>> The usual solution here is to compute the largest multiple of 100
>> that's less than RAND_MAX, and ignore any generated values that are
>> equal to or greater than that value:
>>
>> int rand100()
>> {
>> const int max = (RAND_MAX / 100) * 100; int tmp = rand();
>> while (max <= tmp)
>> tmp = rand();
>> return tmp % 100;
>> }

>
>
> Hmm... this method has a drawback of slowing the program. If RAND_MAX is
> say 32767 then the window in the routine above is just 0.3% of the full
> range, so in 99.7% of the calls it will waste extra calls to rand()...

No, if RAND_MAX is 32767 then (RAND_MAX / 100) * 100 = 32700, so you
waste only about 0.2% of calls.

--
Lionel B

Lionel B, May 28, 2008
17. ### bilgekhanGuest

"Pete Becker" wrote:
> On 2008-05-28 10:22:43 -0400, "bilgekhan" said:
>
> If you know how long the period of the generator is, you can store that
> many numbers, and then you can "predict" what the next value will be.
> But RAND_MAX is not the period; it's the maximum value that will be
> produced.

Ok, but it should be no problem to get the length of the period...
For example:
Store, say 256 rand() call results in an array.
Then infinetely call rand() and check after how many calls
it repeats the same sequence stored in the array....
and then quit the loop.
A simpler way would be of course looking in the TM of the compiler/library...
(but OTOH who does RTFM ?...

> > I did some experiments and I come to the conclusion
> > that it is NOT possible to let 1 RNG _correctly_ generate
> > 2 independent random numbers.

>
> Define "independent". <g>

Independent as so far as 2 consecutive calls to rand()
also generate twice the same random number as
theoretically would be expected in 10% of 1 million cases
when each time generating 2 numbers between 0 and 99.
That's 1e6 / 100 = 10000 = 10% of 1e6.

For achieving this independence one needs to switch
to another seed sequence. But my experiments show
that then the distribution is flawed as it deviates from
the theoretical expectation. I managed to get it within
+1.16% more matches by using the following ugly kludge:
srand((1U + rand()) * (1U + rand()) * 4U /* - 1 */);
as RAND_MAX on my machine is only 32767.
But then I discarded the idea of wanting to have just one RNG
generate such 2 independent random numbers.

Using just one RNG, and within the same period (srand sequence),
it is not possible to get the theoretically expected _correct_ number
of matches. Ie. one would need two identical but independent RNGs
and then both initialized with different seed values.

bilgekhan, May 28, 2008
18. ### James KanzeGuest

On May 28, 2:17 pm, Lionel B <> wrote:
> On Wed, 28 May 2008 13:08:17 +0200, bilgekhan wrote:

[...]

> > srand(time(0));

> This is ok as long as you don't call this routine again so
> quickly that time(0) returns the same value as for the last
> call.

It's somewhat implicit in what you just said, but you
*shouldn't* use this if you need independent random sequences on
different machines---the programs could be started within the
same second (and in a lot of applications, typically will be).

I generally use /dev/random if it's available, and a mixture of
time(), pid() and the machine's MAC address (or failing that,

> > int r1 = rand() % 100;

> This can be a bad idea, particularly for so-called "linear
> congruential" rngs (and there's a fair chance your `rand()'
> will be linear congruential). The reason for this is that
> mod-ing a number basically involves using just the lower-order
> bits, which for many rngs - and linear congruential ones in
> particular - are frequently the "least random".

> Safer is something along the lines of:

> int r1 = int(100.0*double(rand())/double(RAND_MAX));

> that is, you generate a uniform(ish) random double in the
> range [0,1), multiply it by 100 and take the integer part.
> This gives you a uniform (ish) random int in the range [0,99).
> It's safer because it relies more heavily on the higher-order
> bits of the number returned by your rng.

Actually, just he reverse is true. While some linear congruent
generators aren't very random in the low order bits, the good
ones are, and the high order bits are never really very random:
a very small number is always followed by a very large one.

--
James Kanze (GABI Software) email:
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34

James Kanze, May 28, 2008
19. ### Jerry CoffinGuest

In article <g1jpsm\$h5v\$>,
says...

[ ... ]

> Because it is the same sequence of the RNG...
> A seed sequence of rand() always generates
> RAND_MAX different random numbers.

False. It usually _will_, but there's no guarantee of it.

> So within a sequence it cannot generate the same number twice.

Even more thoroughly false. I've cut out a bunch more statements, but
they all reflect the same misconception.

You're assuming that RAND_MAX reflects not only the range, but also the
period of the PRNG. While the standard's definition is loose enough to
_allow_ that, it's most assuredly not required. I'd add that, IMO, such
an implementation is _far_ from ideal.

From a practical viewpoint, having the range and the period of the
generator equal appears to be fairly unusual. For example, consider the
following code:

#include <stdlib.h>
#include <iostream>

int main() {
srand(1);

int i;

for (i=0; i<9; i++)
std::cout << rand() << "\t";
std::cout << "\n";

for (; i<RAND_MAX; i++)
rand();

for (i=0; i<9; i++)
std::cout << rand() << "\t";
std::cout << "\n";
return 0;
}

According to your statements, the two lines should be identical (with a
possible offset). With the compilers I have handy, that's not the case.
Here's the output from VC++ 7.1:

41 18467 6334 26500 19169 15724 11478 29358 26962
29246 17262 12374 18983 5450 31630 32102 30438 30332

With g++ (4.3.1), these are the results:

1481765933 1085377743 1270216262 1191391529
812669700553475508 445349752 1344887256 730417256

1039805031 1083326686 1331336834 716584787
2033177353934031309 1079589791 1155192414 1372256943

In neither case does repetition appear after producing RAND_MAX numbers.
For most things, I also test with Comeau, but it normally uses the back-
end compiler's implementation of rand(), so it would produce results
identical to the back-end compiler's.

--
Later,
Jerry.

The universe is a figment of its own imagination.

Jerry Coffin, May 29, 2008
20. ### Jerry CoffinGuest

[ ... ]

> While some linear congruent
> generators aren't very random in the low order bits, the good
> ones are, and the high order bits are never really very random:
> a very small number is always followed by a very large one.

Hmmm...neither of the compilers I have handy seems to suffer from either
of these problems. For example, consider the following two sequences
generated by VC++ 7.1:

41 18467 6334 26500 19169 15724 11478 29358 26962
29246 17262 12374 18983 5450 31630 32102 30438 30332

I see no particular pattern of a particularly large number following a
small one. To help make possible problems in the low bits more apparent,
here's the same sequence printed in hex:

29 4823 18be 6784 4ae1 3d6c 2cd6 72ae 6952
723e 436e 3056 4a27 154a 7b8e 7d66 76e6 767c

At least right off, I don't see any obvious patterns on the low bits
either (though I'll openly admit I haven't tried to do any sort of
extensive analysis).

--
Later,
Jerry.

The universe is a figment of its own imagination.

Jerry Coffin, May 29, 2008