Random Number selection

Wed May 26, 2010 9:32 pm   

Hi All,

Can somebody help me with a better approach to perform the random
number selection from a set of numbers than the one mentoned below.

If there are set of numbers like & i would like to get one number out
of this:

0,1,2,3,256,257.258,259,512,513,514,515,768,769,770,771

int ii;
int number;

ii = {$random(seed)} % 2;

number = 0 + {$random(seed)} % 4 + (0 + ($urandom_range(259,256)) *
ii) + (0 + ($urandom_range(515,512)) * ii) + (0 +
($urandom_range(771,768)) * ii)


Thanks for your time..

Wed May 26, 2010 9:32 pm   

On Wed, 26 May 2010 11:32:43 -0700 (PDT), Deepu wrote:


Depends how much SystemVerilog you would like to try.

Answer #1 (easy in any Verilog dialect):
Populate an array with your numbers. Randomly pick an index.

int values[16] = '{ 0, 1, 2, 3,
256,257.258,259,
512,513,514,515,
768,769,770,771 };
number = values[$dist_uniform(seed, 0, 15)];

Answer #2 (SystemVerilog testbench licence required):

int number;
std::randomize(number) with { number inside
{[0:3], [256:259], [512:515], [768:771]};};

Answer #3
(easy, but depends on special properties of the set of numbers):
Observe that the numbers are all of the form 256*A+B, where
A and B are both in the range 0..3. Choose random values
for A and B, then construct the number.

An Observation:
~~~~~~~~~~~~~~~
Of the three methods I offered above, just one (#2) depends
on detailed knowledge of (System)Verilog. The other two
are based solely on simple arithmetical or logical insight.
Since it's a fair bet that you are younger and therefore
faster-thinking than I am, I wonder why it needed me to
point them out?
--
Jonathan Bromley

Wed May 26, 2010 11:46 pm   

Thanks! Jonathan 3rd version was more comfortable for me. Thanks for
all the other approach, would have liked to check the 2nd approach but
was not able because of the license..