Thread: Consecutive one's problem

Hey everyone, I am having some difficult trying to calculate the probability of a sequence of consecutive ones being generated in an array when the position is chosen randomly.

For example: I want to create an array of length K and which entry will be initialized with 0. Then I will select M number of locations that will be picked randomly, once I chose those position I will set it to 1. Now if I try to this problem N number of times, what is the chance I will get a sequence of 1's?

Like this 1's in an array of 10: 0001111110

this is what I did to pick a random number for the position:

Code:

 
long discrete_uniform (long a, long b)
{
  if (b > a) {
    double x = uniform ();
    long c = ( a + (long) floor((b-a+1)*x) );
    return c;
  }
  else if (a == b) return a;
  else { printf("ERROR in uniform.discrete_uniform(): a=%2ld b=%2ld\n",a,b);return 0;}
}

and then I did a loop to set all the positions to 0. After that I called the discrete_uniform function and set the 1's in the array. But I am having trouble trying to do the code to see if I repeat the experiment N number of times what is the chance that all the 1's will be next to each other.

I am a beginner, can anyone help me?

anyone?

So you have an array(?) of numbers, some of which are zero and some of which are one. You need to think about how to go through the array and see how many of them are in a row.

I thought about it this way:

Code:

double consecutiveOnesProbability (int K, int M, int numTrials)
{
        int randomarray[];

        for (int a=0; a<=K; a++){      /* initialize an array with 0's
                randomarray[a]= 0;
        }
        for (int b=0; b<=M; b++){        /* set 1 in M different positions
                int i = uniform_discrete(0,9);
                randomarray[i]=1;
        } 
        for (int e=1; e=k; e++){    /*find the sequence ones
              int f = randomarray[e];
              while (f==0)
                       e++;
              else  if (f==1)
                   saveposition = e;
                   savecount =1;
                   int l=e+1;
                   int h= (e+M)-1;
                      for ( int g=l; g<=h; g++){
                           if (randomarray[g]==1)
                                 savecount=savecount+1;
                           else
                                printf ("not a sequence"); 
                                printf ( "Sequence size", savecount);
                       }
             }
}

No wrap around in this one

Originally Posted by tabstop

Code:

        int randomarray[];

I shudder to think that you think this is legal.

Code:

        for (int a=0; a<=K; a++){      /* initialize an array with 0's

How many times do you expect this for loop to run?
[/code]
for (int b=0; b<=M; b++){ /* set 1 in M different positions
[/code]
And this one?

Code:

                int i = uniform_discrete(0,9);

Why 0 through 9?

Code:

        for (int e=1; e=k; e++){    /*find the sequence ones

How many times do you expect this for loop to run? And how many elements do you intend to miss?

And after that's done, you need to turn counting into probability.

The user would entry those values, but you can use K=10 and M=3

Originally Posted by iMalc

Are the consecutive ones allowed to involve "wrap around"?
Where K = 4 and M = 2, that makes the difference between 50% and 67%.

I assume you mean 37.5% and 50%?

Originally Posted by Samyx

The user would entry those values, but you can use K=10 and M=3

That's nice but completely irrelevant.

how did u guys come up with those percentages?

Originally Posted by Samyx

how did u guys come up with those percentages?

Counting.