#### Smallest Circle

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/**
* @file
* @brief Get centre and radius of the
* [smallest circle](https://en.wikipedia.org/wiki/Smallest-circle_problem)
* that circumscribes given set of points.
*
* @see [other
* implementation](https://www.nayuki.io/page/smallest-enclosing-circle)
*/
#include <cmath>
#include <iostream>
#include <vector>

/** Define a point */
struct Point {
double x, /**< abscissa */
y;    /**< ordinate */

/** construct a point
* \param [in] a absicca (default = 0.0)
* \param [in] b ordinate (default = 0.0)
*/
explicit Point(double a = 0.f, double b = 0.f) {
x = a;
y = b;
}
};

/** Compute the Euclidian distance between two points \f$A\equiv(x_1,y_1)\f$ and
* \f$B\equiv(x_2,y_2)\f$ using the formula:
* \f[d=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\f]
*
* \param [in] A point A
* \param [in] B point B
* \return ditance
*/
double LenghtLine(const Point &A, const Point &B) {
double dx = B.x - A.x;
double dy = B.y - A.y;
return std::sqrt((dx * dx) + (dy * dy));
}

/**
* Compute the area of triangle formed by three points using [Heron's
* formula](https://en.wikipedia.org/wiki/Heron%27s_formula).
* If the lengths of the sides of the triangle are \f$a,\,b,\,c\f$ and
* \f$s=\displaystyle\frac{a+b+c}{2}\f$ is the semi-perimeter then the area is
* given by \f[A=\sqrt{s(s-a)(s-b)(s-c)}\f]
* \param [in] A vertex A
* \param [in] B vertex B
* \param [in] C vertex C
* \returns area of triangle
*/
double TriangleArea(const Point &A, const Point &B, const Point &C) {
double a = LenghtLine(A, B);
double b = LenghtLine(B, C);
double c = LenghtLine(C, A);
double p = (a + b + c) / 2;
return std::sqrt(p * (p - a) * (p - b) * (p - c));
}

/**
* Check if a set of points lie within given circle. This is true if the
* distance of all the points from the centre of the circle is less than the
* \param [in] P set of points to check
* \param [in] Center coordinates to centre of the circle
* \param [in] R radius of the circle
* \returns True if P lies on or within the circle
* \returns False if P lies outside the circle
*/
bool PointInCircle(const std::vector<Point> &P, const Point &Center, double R) {
for (size_t i = 0; i < P.size(); i++) {
if (LenghtLine(P[i], Center) > R)
return false;
}
return true;
}

/**
* Find the centre and radius of a circle enclosing a set of points.\n
* The function returns the radius of the circle and prints the coordinated of
* the centre of the circle.
* \param [in] P vector of points
* \returns radius of the circle
*/
double circle(const std::vector<Point> &P) {
double minR = INFINITY;
double R;
Point C;
Point minC;

/* This code is invalid and does not give correct result for TEST 3 */
// for each point in the list
for (size_t i = 0; i < P.size() - 2; i++)
// for every subsequent point in the list
for (size_t j = i + 1; j < P.size(); j++)
// for every subsequent point in the list
for (size_t k = j + 1; k < P.size(); k++) {
// here, we now have picked three points from the given set of
// points that we can use
// viz., P[i], P[j] and P[k]
C.x = -0.5 * ((P[i].y * (P[j].x * P[j].x + P[j].y * P[j].y -
P[k].x * P[k].x - P[k].y * P[k].y) +
P[j].y * (P[k].x * P[k].x + P[k].y * P[k].y -
P[i].x * P[i].x - P[i].y * P[i].y) +
P[k].y * (P[i].x * P[i].x + P[i].y * P[i].y -
P[j].x * P[j].x - P[j].y * P[j].y)) /
(P[i].x * (P[j].y - P[k].y) +
P[j].x * (P[k].y - P[i].y) +
P[k].x * (P[i].y - P[j].y)));
C.y = 0.5 * ((P[i].x * (P[j].x * P[j].x + P[j].y * P[j].y -
P[k].x * P[k].x - P[k].y * P[k].y) +
P[j].x * (P[k].x * P[k].x + P[k].y * P[k].y -
P[i].x * P[i].x - P[i].y * P[i].y) +
P[k].x * (P[i].x * P[i].x + P[i].y * P[i].y -
P[j].x * P[j].x - P[j].y * P[j].y)) /
(P[i].x * (P[j].y - P[k].y) +
P[j].x * (P[k].y - P[i].y) +
P[k].x * (P[i].y - P[j].y)));
R = (LenghtLine(P[i], P[j]) * LenghtLine(P[j], P[k]) *
LenghtLine(P[k], P[i])) /
(4 * TriangleArea(P[i], P[j], P[k]));
if (!PointInCircle(P, C, R)) {
continue;
}
if (R <= minR) {
minR = R;
minC = C;
}
}

// for each point in the list
for (size_t i = 0; i < P.size() - 1; i++)
// for every subsequent point in the list
for (size_t j = i + 1; j < P.size(); j++) {
// check for diameterically opposite points
C.x = (P[i].x + P[j].x) / 2;
C.y = (P[i].y + P[j].y) / 2;
R = LenghtLine(C, P[i]);
if (!PointInCircle(P, C, R)) {
continue;
}
if (R <= minR) {
minR = R;
minC = C;
}
}
std::cout << minC.x << " " << minC.y << std::endl;
return minR;
}

/** Test case: result should be:
* \n Circle with
* \n centre at (3.0454545454545454, 1.3181818181818181)
*/
void test() {
std::vector<Point> Pv;
Pv.push_back(Point(0, 0));
Pv.push_back(Point(5, 4));
Pv.push_back(Point(1, 3));
Pv.push_back(Point(4, 1));
Pv.push_back(Point(3, -2));
std::cout << circle(Pv) << std::endl;
}

/** Test case: result should be:
* \n Circle with
* \n centre at (1.0, 1.0)
*/
void test2() {
std::vector<Point> Pv;
Pv.push_back(Point(0, 0));
Pv.push_back(Point(0, 2));
Pv.push_back(Point(2, 2));
Pv.push_back(Point(2, 0));
std::cout << circle(Pv) << std::endl;
}

/** Test case: result should be:
* \n Circle with
* \n centre at (2.142857142857143, 1.7857142857142856)
* @todo This test fails
*/
void test3() {
std::vector<Point> Pv;
Pv.push_back(Point(0.5, 1));
Pv.push_back(Point(3.5, 3));
Pv.push_back(Point(2.5, 0));
Pv.push_back(Point(2, 1.5));
std::cout << circle(Pv) << std::endl;
}

/** Main program */
int main() {
test();
std::cout << std::endl;
test2();
std::cout << std::endl;
test3();
return 0;
}  