CareerCup

Chi Square test, distribution test

This is harder than it sounds. Just generating numbers and binning them will not work, because then this would be a valid random number generator:

unsigned int randVal = 0;

float rand() {
  randVal++;
  return randVal / (float)INT_MAX;
}

.johndcook.com /Beautiful_Testing_ch10.pdf

monte carlo simulation.

print the numbers into a file and try to compress using e.g. Zip/huffman. Random set should be impossible to compress

I think none of the solution is efficient.
I was wondering if this can be solved by following various criteria:
1) Test How Georgi Kalchev suggested
2) Do a Chi Square Test
3) Try XORing all numbers?

I think none of the solution is efficient.
I was wondering if this can be solved by following various criteria:
1) Test How Georgi Kalchev suggested
2) Do a Chi Square Test
3) Try XORing all numbers?

take 100000 entries
assume that the range in which the random number are being generated is between minimum and maximum (can be assumed for large entries)
Now, for each position, count the number of times the next number appearing after it is more than it or not. Take this ratio. This ratio should be proportioinal to the (Max - E)/(Max-Min)
where E is the current value
Now, plotting these values for all positions, we take the variance. If the variance is small, then the random number generator can be assumed good

Hi ,

What if we use time to generate the randon number. Get the current time in milliseconds and divide it by 1000.

class TimeTest{
		public static void main(Strings[] args){
			system.out.println(System.currentTimeMilliseconds()/1000);
		}
	}

This should generate unique number definetly.


Regards,
Vijay Kalkundri

Generate random numbers in large range and plot the random numbers in a graph which should be a uniform distribution for all the numbers in the range.

1.) we can check how the elements distribute over a given range
2.) we can check how many times one number is hit and what is the difference between the most hit number and the least hit one.
3.) *Visually check for patterns of each number hits by looking (in the code bellow I would look) the "map".

point 1.) sample Java code:

public static double randPercent() {
  int N = 1000;
  int[] map = new int[N];
  int cnt = 0;

  int rand;
  for (int i = 0; i < N; i++) {
   rand = (int) (Math.random() * N);
   map[rand]++;
 
   if (map[rand] == 0)
    cnt++;
  }
 
  //System.out.format("Result is %s (different numbers) out of %s", cnt, N);
 
  return (double) cnt / N;
 }

None of the solution above really works.
Think about this fake random number generator.

static n = 0;
void fakerand()
{
n++;
if(n%2==0)
return rand()*0.5;
else
return 1-(rand()*0.5);
}

We need to generate a really long sequence. And try to proof, as much as we can, that there is no pattern in this sequence.

srand((unsigned)time(NULL)); //seed with time function
printf("%d\n",rand());