@Jitesh: I did. But here is another try:
We construct a fraction p' by using the random number generator to
select the bits. So the number is 0.f()f()f()f()..., where each f() is
a zero- or a one-bit. We compare p to p'. If, no matter what remaining
bits we choose for p' we will have p p', then
@Dave: Very nice.
Don
On Sep 12, 10:51 pm, Dave dave_and_da...@juno.com wrote:
Here's another way, using a rejection technique on the bits of the
mantissa of p. Each iteration of the do-while loop exposes another
high-order bit of p, and the do-while loop iterates as long as the
random bits
@Dave: Thanks a lot.. :)
On Tue, Sep 13, 2011 at 7:48 PM, Don dondod...@gmail.com wrote:
@Dave: Very nice.
Don
On Sep 12, 10:51 pm, Dave dave_and_da...@juno.com wrote:
Here's another way, using a rejection technique on the bits of the
mantissa of p. Each iteration of the do-while loop
Here is an extension:
Given a pseudo-random generator f() which return unsigned integers
uniformly distributed over the range 0..2^32-1, create a generator
which produces integer values in the range 0..N-1 with each occurring
with a probability {p0, p1, p2, ..., pn-1} such that all the
For particular values of p we might be able to do better, but for
unknown values of p, I can't think of anything better than this:
int g(double p)
{
int n = 0;
for(int i = 0; i 30; ++i)
n += n+f();
return n (int)(p*1073741824.0);
}
On Sep 12, 9:55 am, JITESH KUMAR jkhas...@gmail.com
Don, how did you come up with 1073741824.0 ?
On Mon, Sep 12, 2011 at 8:49 PM, Don dondod...@gmail.com wrote:
For particular values of p we might be able to do better, but for
unknown values of p, I can't think of anything better than this:
int g(double p)
{
int n = 0;
for(int i = 0; i
On 12 September 2011 20:49, Don dondod...@gmail.com wrote:
For particular values of p we might be able to do better, but for
unknown values of p, I can't think of anything better than this:
int g(double p)
{
int n = 0;
for(int i = 0; i 30; ++i)
n += n+f();
return n
It uses 30 random bits from f() to build a 30-digit number which will
be in the range 0..2^30. 2^30 is 1,073,741,824. p*2^32 is the number
of values in that range which should produce the result 0.
Don
On Sep 12, 10:48 am, siddharth srivastava akssps...@gmail.com wrote:
On 12 September 2011
Here's another way, using a rejection technique on the bits of the
mantissa of p. Each iteration of the do-while loop exposes another
high-order bit of p, and the do-while loop iterates as long as the
random bits produced by f match the high order bit sequence of p. This
most likely will use fewer
Hi Dave,
Can you please explain your approach?
On Tue, Sep 13, 2011 at 9:21 AM, Dave dave_and_da...@juno.com wrote:
Here's another way, using a rejection technique on the bits of the
mantissa of p. Each iteration of the do-while loop exposes another
high-order bit of p, and the do-while loop
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