rapsoulCecil - 1 year ago 72

C++ Question

I'm new to C++. I'm trying to statistically determine the value of Pi based on Ernesto Cesaro's Theorem using the computer system's random number generator. But what I have done now can input a seed number and generate 100 pseudo random numbers and then estimate the value of pi. The generator can generate different group of Pseudo random numbers. However the confusion is I always get the estimate of pi for 2.8284 with no change. Here is the code:

`#include <iostream>`

#include <math.h>

using namespace std;

int main()

{

int seed;

cout << "input a seed number: " << endl;

cin >> seed;

srand(seed);

int i, a[100];

for (i = 0; i < 100; i++)

a[i] = rand() % 100 + 1;

cout << "The generated random numbers are: " << endl;

for (i = 0; i < 100; i++)

cout << a[i] << "\t";

int m, n, j, r;

int sum = 0;

for (j = 0; j < 100; j++)

{

m = a[j];

n = a[j + 1];

j = j + 2;

do

{

r = m%n;

m = n;

n = r;

} while (r != 0);

if (n = 1)

sum = sum + 1;

}

double Pi, p;

p = 300 / sum;

Pi = sqrt(p);

cout << "The estimate value of Pi is: " << Pi << endl;

system("pause");

return 0;

}

Note that Cesaro's theorem states that given two random integers, x and y, the probability that gcd(x, y) = 1 is 6/(Pi^2). And PRNG used affects how close the resulting estimate is to Pi(3.1416).

Answer Source

There are several issues with your code.

**Issue #1:**

```
if (n = 1)
```

It should be `if (n == 1)`

or else you are assigning `1`

to `n`

and always evaluating to true.

**Issue #2:**

```
n = r;
} while (r != 0);
if (n == 1)
```

If you think about it, the loop will end only when `r`

is `0`

, but then `n`

will also be `0`

because of the last line of the loop. So `n`

will never be equal to `1`

. You probably want `if (m == 1)`

.

**Issue #3:**

```
for (j = 0; j < 100; j++)
{
...
j = j + 2;
```

You are incrementing the `j`

in the `for`

line and in the loop body. You just need one.

```
for (j = 0; j < 100; j += 2)
{
//no j increment
```

**Issue #4:**

```
p = 300 / sum;
```

That is an integer division, because both numbers are integers. You want foating point: `p = 300.0 / sum;`

.

With those changes I'm getting about `3.16`

.