feat: add approximate_pi.cpp file/algorithm (#1860)

* Add approximate_pi.cpp file

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update math/approximate_pi.cpp

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Update approximate_pi.cpp

* Update math/approximate_pi.cpp

Co-authored-by: David Leal <halfpacho@gmail.com>

* Apply suggestions from code review

Co-authored-by: David Leal <halfpacho@gmail.com>

math/approximate_pi.cpp

@@ -0,0 +1,78 @@
+/**
+ * @file
+ * @brief Implementation to calculate an estimate of the [number π (Pi)](https://en.wikipedia.org/wiki/File:Pi_30K.gif).
+ *
+ * @details
+ * We take a random point P with coordinates (x, y) such that 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. If x² + y² ≤ 1, then the 
+ * point is inside the quarter disk of radius 1, otherwise the point is outside.
+ * We know that the probability of the point being inside the quarter disk is equal to π/4
+ * double approx(vector<Point> &pts) which will use the points pts (drawn at random) to 
+ * return an estimate of the number π
+ * \note This implementation is better than naive recursive or iterative
+ * approach.
+ *
+ * @author [Qannaf AL-SAHMI](https://github.com/Qannaf)
+ */
+
+#include <iostream>  /// for IO operations
+#include <vector>    /// for std::vector
+#include <cstdlib>  /// for std::rand
+
+/**
+ * @namespace math
+ * @brief Mathematical algorithms
+ */
+namespace math {
+
+    /**
+     * structure of points containing two numbers, respectively x and y such that 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. 
+    */
+    typedef struct {
+    double x;
+    double y;
+    } Point;
+
+    double  approximate_pi(const std::vector<Point> &pts) {
+    /**
+     * This function use the points pts (drawn at random) to return an estimate of the number π  using the given points
+     * @param pts Each item of pts contains a point. A point is represented by a structure containing exactly 
+     * two numbers, respectively x and y such that 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. 
+     * pts always contains at least one item
+     * @return  an estimate of the number π
+     */
+        {
+            int count =0;    // Points in cercle
+            for(Point p:pts)
+                if(p.x * p.x + p.y*p.y <= 1)
+                    ++count;
+
+            return 4.0*count/pts.size();
+        }
+    }
+}  // namespace math
+
+/**
+ * @brief Self-test implementations
+ * @returns void
+ */
+static void test() {
+        std::vector<math::Point> rands;
+    for (std::size_t i = 0; i < 100000; i++) {
+        math::Point p;
+        p.x = rand() / (double)RAND_MAX; // 0 <= x <= 1
+        p.y = rand() / (double)RAND_MAX; // 0 <= y <= 1
+        rands.push_back(p);
+    }
+    std::cout << math::approximate_pi(rands) << std::endl;          // ~3.14
+}
+
+/**
+ * @brief Main function
+ * @param argc commandline argument count (ignored)
+ * @param argv commandline array of arguments (ignored)
+ * @returns 0 on exit
+ */
+int main(int argc, char *argv[]) {
+    test();  // run self-test implementations
+    return 0;
+}