Use Mersenne Twister to generate 32-bit integers and do something like
this:
long long x = MT.gen();
x = (x32) + MT.gen();
Don
On Feb 27, 5:58 pm, Prakash D cegprak...@gmail.com wrote:
i've another doubt. what to do when I need to generate a random long long?
On Mon, Feb 27, 2012 at 9:07
For instance, if RANDMAX= 32768, then
x = rand() % 2;
is twice as likely to result in the value 10,000 as the value 15,000.
This is because there are two output values from rand() which result
in x=1 (1 and 3), but only one output value from rand()
resulting in x=15000 (15000).
i've another doubt. what to do when I need to generate a random long long?
On Mon, Feb 27, 2012 at 9:07 PM, Don dondod...@gmail.com wrote:
For instance, if RANDMAX= 32768, then
x = rand() % 2;
is twice as likely to result in the value 10,000 as the value 15,000.
This is because there
with equal probability
On Tue, Feb 28, 2012 at 5:28 AM, Prakash D cegprak...@gmail.com wrote:
i've another doubt. what to do when I need to generate a random long long?
On Mon, Feb 27, 2012 at 9:07 PM, Don dondod...@gmail.com wrote:
For instance, if RANDMAX= 32768, then
x = rand() % 2;
@Karthikeya: Doesn't rand() actually return numbers in the range 0 to
RANDMAX? Proceeding as if that is the case:
If N is much smaller than RANDMAX, the process probably works well
enough for most applications, but if you want numbers as good as the
numbers rand() generates, do the following:
int m = (RANDMAX / N) * N
isn't m= RANDMAX simplyit couldn't understand the what is the
logic here.
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@Karthikeya: Integer division truncates. So m is the largest multiple
of N that is less than or equal to RANDMAX. E.g., in your example, m =
(50 / 30) * 30 = 1 * 30 = 30, since 50/30 truncates to 1.
Dave
On Feb 26, 12:33 pm, karthikeya s karthikeya.a...@gmail.com wrote:
int m = (RANDMAX / N) *
oh my bad.really nice concept+1 dave.
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@dave +1.. :)
On Mon, Feb 27, 2012 at 10:55 AM, karthikeya s karthikeya.a...@gmail.comwrote:
oh my bad.really nice concept+1 dave.
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