Now, after I fixed the errors, the main problem of this program is speed. I mean it works, but terribly slow.
I have an array of points and i need to find squares here, so in order to do this i need to check all possible combinations of 4 points. So I used three nested loops. In general it would take about n^4 operations.If i had up to 100 elements of an array it would be more or less normal, but I have 500 elements, so 500^4 it`s quite a lot. Is here any faster way to do this?
/*j, o, k, l are numbers of each point. This loop will check all possible combinations
(except 2 or more points have the same number(it means also the same coordinades) or they already are part of some figure*/
for (int j=0; j < i-1; j )
{
if (Point[j].p != true)
{
for (int o = 1; o < i - 2; o )
{
if ((o != j) && (Point[o].p != true))
{
for (int k = 2; k < i - 3; k )
{
if ((k!= j) && (k != o) && (Point[k].p != true))
{
for (int l = 3; l < i - 4; l )
{
if ((l != k) && (Point[l].p != true) && (l != o) && (l!=j))
{
vx1 = abs(Point[o].x - Point[j].x); //vectors coordinates
vx2 = abs(Point[k].x - Point[o].x);
vy1 = abs(Point[o].y - Point[j].y);
vy2 = abs(Point[k].y - Point[o].y);
vx3 = abs(Point[l].x - Point[k].x);
vy3 = abs(Point[l].y - Point[k].y);
vx4 = abs(Point[j].x - Point[l].x);
vy4 = abs(Point[j].y - Point[l].y);
dx1 = abs(Point[k].x - Point[j].x); //diagonals coordinates
dy1 = abs(Point[k].y - Point[j].y);
dx2 = abs(Point[o].x - Point[l].x);
dy2 = abs(Point[o].y - Point[l].y);
v1 = sqrt(vx1 * vx1 vy1 * vy1); // sides length
v2 = sqrt(vx2 * vx2 vy2 * vy2);
v3 = sqrt(vx3 * vx3 vy3 * vy3);
v4 = sqrt(vx4 * vx4 vy4 * vy4);
d1 = sqrt(dx1 *dx1 dy1 * dy1); //diagonals length
d2 = sqrt(dx2 * dx2 dy2 * dy2);
if (
((abs(v1-v2)<=0.5) && (abs(v3-v4)<=0.5) && (abs(v3-v2)<=0.5) && (v1<d1)) /*cheks all sides are equal and if the diagonal is bigger than side*/
&& (Point[k].p != true && Point[o].p != true && Point[j].p != true)/*checks if the points aren`t the part of any figure yet*/
&&(abs(d1 - d2)<=0.5)/*checks if the diagonals are equal*/)
{
q ;
Point[j].p = true; // it means that point with this number is already a part of some figure, so it will be skipped in next operations
Point[o].p = true;
Point[k].p = true;
Point[l].p = true;
// the output
out << "Figure " << q << ":" << "x1=" << Point[j].x << " y1=" << Point[j].y << " x2="
<< Point[o].x << " y2=" << Point[o].y <<
" x3=" << Point[k].x << " y3=" << Point[k].y <<
" x4=" << Point[l].x << " y4=" << Point[l].y << endl;
}
}
}
}
}
}
}
}
CodePudding user response：
You will have to test for any combination of 2 initial points. Let us call them A and B. But as you only want each possible square once, you can require the third point, let us call it C, to verify (AC) is (AB) turned with 90° in trigonometric sense. If a possible candidate could be obtained by turning in the opposite sense, you would find the final square from a different initial pair.
But here the condition is damned simple:
- yC - yA = xB - xA and
- xC -xA = - (yB - yA)
And when you have a possible candidate for point C, point D will have to verify (CD) = (AB) meaning:
- xD - xC = xB - xA and
- yD - yC = yB - yA
But if you use floating point values for your coordinates, you could fall in an accuracy problem. We all know that floating point arithmetics is broken inaccurate. So you should define an epsilon value (generally 10-6 is an acceptable value) and replace all equality tests with code close to:
if (abs(x - y) < epsilon) ...
I cannot currently write and test code (no access to a acceptable compiler) but this algorithm should slightly reduce time: operations are much simpler and as you only search the 4th point if you have found an acceptable 3rd one the complexity should be closer to n3.
CodePudding user response：
Since you mentioned it's a "computer graphics coordinate system(without negative coordinates)" I'm also assuming integer values which makes it rather straight forward.
- Create a
std::unordered_setofPoint - For each combination of two points,
AandB, calculate the position of a third point,C, rotated 90° in relation to the first two,A -> B. - If such a third point exists in the set, calculate the position of the last point,
D, in relation toB -> C. - If a match is found, remove the four points from the set.
- Continue until you've gone through all points.
Example:
First, a Point class template:
template <class T = unsigned long long> // or some other integer type
struct Point {
bool operator==(const Point& o) const { return x == o.x && y == o.y; }
bool operator!=(const Point& o) const { return !(*this == o); }
Point& operator-=(const Point& o) {
x -= o.x;
y -= o.y;
return *this;
}
T x, y;
};
template <class T>
Point<T> operator-(const Point<T>& lhs, const Point<T>& rhs) {
Point rv = lhs;
rv -= rhs;
return rv;
}
template <class T>
std::ostream& operator<<(std::ostream& os, const Point<T>& p) {
return os << '{' << p.x << ',' << p.y << '}';
}
template <class T>
struct std::hash<Point<T>> {
std::size_t operator()(const Point<T>& p) const {
std::hash<T> h;
// boost::hash_combine:
std::size_t rv = h(p.x);
rv ^= h(p.y) 0x9e3779b9 (rv << 6) (rv >> 2);
return rv;
}
};
This will let us do B - A to get a Point with the x and y difference and the specialized std::hash will let us store it in containers like std::unordered_set that requires the Key to be hashable. The operator<< overload is just to make printing Point<> instances easier.
Next, a function template to get a Point<> from rotating 90° in relation to two other Point<>s:
template<class T>
Point<T> rot90(const Point<T>& A, const Point<T>& B) {
// C.x = B.x (B-A).y
// C.y = B.y - (B-A).x
auto BA = B - A;
return {B.x BA.y, B.y - BA.x};
}
Then doing the actual matching could look like this:
int main() {
std::unordered_set<Point<>> points{ /* ... */ };
// first, second, third and last iterator:
// fit(A), sit(B), tit(C), lit(D)
for (auto fit = points.begin(); fit != points.end();) {
bool found = false;
// temporarily extract the node from the set to loop over the rest
// of the nodes:
auto next = std::next(fit);
auto node = points.extract(fit);
// loop over the rest of the nodes:
auto sit = points.begin();
decltype(sit) tit, lit;
for(; sit != points.end(); sit) {
// calculate where C should be:
auto candidate = rot90(node.value(), *sit);
if(tit = points.find(candidate); tit != points.end()) {
// third Point found - try for the last too:
candidate = rot90(*sit, candidate);
if(lit = points.find(candidate); lit != points.end()) {
found = true;
break;
}
}
}
if(found) {
std::cout << "FOUND: " << node.value() << ' ' << *sit << ' '
<< *tit << ' ' << *lit << '\n';
// erase the points from the set
if(next == sit) next; // next being erased, step next
points.erase(sit);
if(next == tit) next; // next being erased, step next
points.erase(tit);
if(next == lit) next; // next being erased, step next
points.erase(lit);
} else {
// reinsert the first Point node since no square was found
points.insert(fit, std::move(node));
}
fit = next; // try next
}
}