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Circles of given radius through two points: Difference between revisions
Circles of given radius through two points (view source)
Revision as of 18:07, 9 September 2015
, 8 years agoAdded C++ solution
m (→{{header|REXX}}: changed extra digits and optimized the sqrt function, added a comment.) |
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Case 5)
No circles can be drawn through (0.1234,0.9876)
</pre>
=={{header|C++}}==
{{works with|C++11}}
<lang cpp>
#include <iostream>
#include <cmath>
#include <tuple>
struct point { double x, y; };
bool operator==(const point& lhs, const point& rhs)
{ return std::tie(lhs.x, lhs.y) == std::tie(rhs.x, rhs.y); }
enum result_category { NONE, ONE_COINCEDENT, ONE_DIAMETER, TWO, INFINITE };
using result_t = std::tuple<result_category, point, point>;
double distance(point l, point r)
{ return std::hypot(l.x - r.x, l.y - r.y); }
result_t find_circles(point p1, point p2, double r)
{
point ans1 { 1/0., 1/0.}, ans2 { 1/0., 1/0.};
if (p1 == p2) {
if(r == 0.) return std::make_tuple(ONE_COINCEDENT, p1, p2 );
else return std::make_tuple(INFINITE, ans1, ans2);
}
point center { p1.x/2 + p2.x/2, p1.y/2 + p2.y/2};
double half_distance = distance(center, p1);
if(half_distance > r) return std::make_tuple(NONE, ans1, ans2);
if(half_distance - r == 0) return std::make_tuple(ONE_DIAMETER, center, ans2);
double root = std::hypot(r, half_distance) / distance(p1, p2);
ans1.x = center.x + root * (p1.y - p2.y);
ans1.y = center.y + root * (p2.x - p1.x);
ans2.x = center.x - root * (p1.y - p2.y);
ans2.y = center.y - root * (p2.x - p1.x);
return std::make_tuple(TWO, ans1, ans2);
}
void print(result_t result, std::ostream& out = std::cout)
{
point r1, r2; result_category res;
std::tie(res, r1, r2) = result;
switch(res) {
case NONE:
out << "There are no solutions, points are too far away\n"; break;
case ONE_COINCEDENT: case ONE_DIAMETER:
out << "Only one solution: " << r1.x << ' ' << r1.y << '\n'; break;
case INFINITE:
out << "Infinitely many circles can be drawn\n"; break;
case TWO:
out << "Two solutions: " << r1.x << ' ' << r1.y << " and " << r2.x << ' ' << r2.y << '\n'; break;
}
}
int main()
{
constexpr int size = 5;
const point points[size*2] = {
{0.1234, 0.9876}, {0.8765, 0.2345}, {0.0000, 2.0000}, {0.0000, 0.0000},
{0.1234, 0.9876}, {0.1234, 0.9876}, {0.1234, 0.9876}, {0.8765, 0.2345},
{0.1234, 0.9876}, {0.1234, 0.9876}
};
const double radius[size] = {2., 1., 2., .5, 0.};
for(int i = 0; i < size; ++i)
print(find_circles(points[i*2], points[i*2 + 1], radius[i]));
}</lang>
{{out}}
<pre>
Two solutions: 1.96344 2.07454 and -0.963536 -0.852436
Only one solution: 0 1
Infinitely many circles can be drawn
There are no solutions, points are too far away
Only one solution: 0.1234 0.9876
</pre>
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