Distribution of Fix numbers

**laserlight** · 10-30-2012

Originally Posted by newbi

well its either 1-4

I suggest that you rewrite your program slightly:

Code:

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
                 
int main (void)
{
    int C_A = 0, C_B = 0, C_C = 0, C_D = 0;
    int loop = 0, i = 0, pre = 0;
    int votes [500];
    srandom( (unsigned) time(NULL) );

    for (i = 0; i < 500; i++)
    {
        votes [i] = random ( ) % 4 + 1;
        printf("\n %d \n", votes[i]);
    }

    if (i == 1)
    {       
        C_A = C_A + 1;
    }
    else if (i == 2)
    {
        C_B = C_B + 1;
    }
    else if (i == 3)
    { 
        C_C = C_C + 1;
    }
    else if (i == 4)
    {
        C_D = C_D + 1;
    }
    else
    {
        printf("DESPITE CLAIMS TO THE CONTRARY, THE VALUE OF i = %d\n", i);
    }

    loop = 0;
    printf("\n %d \n", C_A);
    printf("\n %d \n", C_B);
    printf("\n %d \n", C_C);
    printf("\n %d \n", C_D);

    return (0);
}

Notice my more consistent indentation. Anyway, compile and run your program with my modifications. What do you see?

**newbi** · 10-30-2012

Well apart from the "else" change, its all the same. Still "0" value for the Candidates at the end.

4

4

1

2

2

1

2

Despite Claims i = 500

0

0

0

0

**laserlight** · 10-30-2012

Did you not see this?

Code:

DESPITE CLAIMS TO THE CONTRARY, THE VALUE OF i = 500

EDIT:
Well, okay, in case you missed it among all that other output, compile and run this program:

Code:

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

int main (void)
{
    int C_A = 0, C_B = 0, C_C = 0, C_D = 0;
    int loop = 0, i = 0, pre = 0;
    int votes [500];
    srandom( (unsigned) time(NULL) );

    for (i = 0; i < 500; i++)
    {
        votes [i] = random ( ) % 4 + 1;
    }

    printf("DESPITE CLAIMS TO THE CONTRARY, THE VALUE OF i = %d\n", i);

    return (0);
}

The whole point here is that you are saying that the value of i will be either 1, 2, 3 or 4. We're telling you that that is not the case. The value of i will be 500. As such, your if-else chain to increment the different variables fails.

If you want to do things this way with the array, then for each candidate, you need to loop over the array and count the number of votes the candidate was given. Just a single if-else chain that is not within a loop will not do.

**laserlight** · 10-30-2012

Right, so I repeat:

The whole point here is that you are saying that the value of i will be either 1, 2, 3 or 4. We're telling you that that is not the case. The value of i will be 500. As such, your if-else chain to increment the different variables fails.

If you want to do things this way with the array, then for each candidate, you need to loop over the array and count the number of votes the candidate was given. Just a single if-else chain that is not within a loop will not do.

Of course, in this case you probably do not have to do things this way with the array, but it is really up to you.

**newbi** · 10-30-2012

Alright....I guess i was a bit sleepy b4 i read ur reply. Well i got things working now. Thanks. Now I just have to loop it 5 times for 5 election come out.

**Nominal Animal** · 10-30-2012

Wow, I seriously admire laserlight's patience!

The algorithm she was expecting newbi to arrive at up till post #21 is very simple. In pseudocode:

Code:

Initialize an array containing the number of votes per candidate to zero
Loop for each vote:
    Generate a random integer to match a candidate (number of candidates)
    Increase the number of votes cast for that candidate by one
End loop

After the loop you end up with the number of votes given to each candidate. (Note, the loop simulates a voter, not the results. Like laserlight stated, this approach does not record which voter voted for whom, only the totals.)

In C, I'd definitely use a function for this. A good prototype for such a function would be

Code:

void vote(unsigned long candidate[], const int candidates, const unsigned long votes)

where the first parameter would be an array containing the number of votes per candidate, the second parameter is the number of candidates (entries in the array), and the third parameter the number of votes to cast. When the function returns, you'd know that the sum of entries in candidates[] would be votes.

I believe unsigned long is the best integer type for number of votes, but feel free to use int instead if it freaks you out.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Throwing a curve-ball for advanced programmers who are interested in the subject:

If you have a vote distribution you wish to simulate, then you could use

Code:

void vote(unsigned long candidate[], const unsigned long expected[], const int candidates, const unsigned long votes);

where the expected[] array contain the distribution of votes, by stating expected number of votes per candidate for some unspecified number of total votes. You sum the total expected number of votes, and generate a random number between zero and that number (excluding the upper limit), and pick the candidate that has that range.

For example, if you have three candidates, A, B, and C, with their expected vote distribution being 2, 1, and 3 votes, then your random number range would be

Code:

0 1 2 3 4 5
A A B C C C

and you'd generate a random number between 0 and 6 (excluding the upper limit, so [0, 5], really). If it is 0 or 1, you give the vote to candidate A. If it is 2, you give the vote to candidate B. Otherwise it is 3, 4, or 5, and you give the vote to candidate C.

There are a large number of different ways to implement this.

One robust way is to use an array to hold the sum total of estimates for candidates prior to the current candidate, plus the total sum in an extra final element. In the above case, that would be 0 2 3 6. After generating a random number n = [0,6], you find the index i in the array that fulfils cumulative_i <= n < cumulative_i+1. For a reasonably small number of candidates (say, up to at least a dozen candidates) a linear search will work just fine, but for a much larger number a binary search will usually be more efficient.

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