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() % 20000;
>
> 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=10000 (10000 and 30000), but only one output value from rand()
> resulting in x=15000 (15000).
>
> For any case where the statistical quality of the pseudo-random stream
> is important, such as simulation, using the built-in rand() function
> is not a good idea. Use a pseudo-random algorithm with much longer
> period and better properties, such as Mersenne Twister.
>
> But if you are using rand, it is usually recommended to use the high
> order bits rather than the low order bits. Many implementations of
> rand() have cycles in the low bits which are much shorter than the
> period of the generator. He is one way to generate unbiased values of
> quality as good as the generator can provide:
>
> // Return pseudo-random integer in the range 0..n-1
> int randRange(int n)
> {
> int result, div = RANDMAX / n;
> do {
> result = rand() / div;
> } while(result >= n);
> return result;
> }
>
> Don
>
> On Feb 26, 10:10 am, karthikeya s <karthikeya.a...@gmail.com> wrote:
>> RAND() func returns value between 1 to INTMAX, as we know. But when
>> smone tries to find out value between 1 to N he takes remainder of o/p
>> of RAND() with N and adds one......but isn't it wrong coz RAND() will
>> generate numbers with equal probability between 1 and INTMAX but
>> taking remainder can alter the prob. of generating numbers.....
>> e.g.
>>
>> INTMAX=50
>> N=30
>> RAND(50) gives numbers 1 to 30, then prob. will remain same but if it
>> gives numbers 31 to 50, they'll be mapped to the numbers 1 to 20,
>> which means probability of getting numbers 1 to 20 is more than the
>> probability for 21 to 30.
>
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