Gap test- 
assume we have 3 numbers and the sequences is
1 2 3 2 1 3 1 3 2 1 2 3 2 1 3.

for the 1s gaps are 3 1 2 3 

In theory for gap of length 3 is
p(no 1)p(no 1)p(no 1)p(1)= 2/3*2/3*2/3*1/3

So a probability for n different types of things of gap k is
(n-1/n)(k-1) (1/n)

So probability of gaps less than k is

sum( from i=1 to k-1) (n-1/n)(i-1) (1/n) = 1 - ((n-1)/n)^(k+1)

Algorithm
(1) Specify the theoretical frequency- for dice set n=6
(2) Arrange the observed sample of gaps  in a cumulative distribution
(3) Find maximum deviation from theory
Look at Table




How do we generate numbers for other distributions:

For exponential distribution , the pdf is

f(x) =  a* exp(-a *x)  for x>0 and 0 otherwise

where 1/a is the mean interarrival time.


The CDF is the integral of the pdf and for x>0
it is  F = 1- exp(-a*x)     for the exponential.

To generate random numbers it is best if  one can find the inverse of the CDF

Let  R =1- exp(-a*x) , a random number in the range of (0 to 1)

Then
     exp(-a*x) = 1-R
     
       -ax = ln(1-R)
         x= -1/a ln(1-R)

Therefore generate a random number R, and then find x the random number in the distribution from the formula  

x= -1/a ln(1-R)

Since in a uniform distribution the probability of R and 1-R  are the same , we usually use

x = -1/a ln (R)


Let's set a=1 for simplicity. Pick R(1), a random number in the uniform distribution. It corresponds to X(1) =- ln(1-R(1)).

P( x<X(1))= p(r<F(X(1))= F(X(1))

Let's try it:

#include <iostream.h> 
#include <math.h>
#include <stdlib.h>

int main()
{
  int save;
  int i, place;
	double r, x;
  int count[11]={0};
 
  for ( i=0;i<1000;i++)
  {
	  r=1.0*rand()/RAND_MAX;
	  x=-log(r);
	  
	  place=x*10;
	  cout <<r<<" "<<x<<" "<<place<< endl;
	  count[place]++;

  }
  for (i=0;i<10;i++)
	  cout <<i<<"     "<<count[i]<<endl;
  return 0;
}
0     84
1     85
2     83
3     80
4     79
5     61
6     49
7     43
8     37
9     43
10     32
11     27
12     25
13     18
14     18
15     19
16     25
17     18
18     15
19     19
20     11
21     10
22     13
23     7
24     6
25     11
26     8
27     9
28     7
29     9
	  
Is this the right log function?

Weibull distribution :


f(x)= b/a^b x^(b-1)exp( -(x/a)*b)

F(x) = 1-exp(-(x/a)b)

Therefor use
x =a(-ln(1-R))1/b

Triangular distribution

f(x) = x for 0 <=x <=1
f(x) = 2-x for 1<x<=2, 
0 everywhere else
Then

F(x) = x*x/2  for x<1

F(x) = 1 - (2-x)*(2-x)/2   for 1<x<=2

F(x) = 1  for x>2


Given R

set  z  = sqrt(2R)   for 0 <R < 1/2

set z=   2- sqrt(2(1-R))     for 1/2 <R <1