/// <param name="p">Vector to the start point of p.</param>
/// <param name="p2">Vector to the end point of p.</param>
/// <param name="q">Vector to the start point of q.</param>
/// <param name="q2">Vector to the end point of q.</param>
/// <param name="intersection">The point of intersection, if any.</param>
/// <param name="considerOverlapAsIntersect">Do we consider overlapping lines as intersecting?</param>
/// <returns>True if an intersection point was found.</returns>
public static bool LineSegementsIntersect(Vector p, Vector p2, Vector q, Vector q2, out Vector intersection, bool considerOverlapAsIntersect = false)
{
intersection = new Vector();
var r = p2 - p;
var s = q2 - q;
var rxs = r.Cross(s);
var qpxr = (q - p).Cross(r);
// If r x s = 0 and (q - p) x r = 0, then the two lines are collinear.
if (rxs.IsZero() && qpxr.IsZero())
{
// 1. If either 0 <= (q - p) * r <= r * r or 0 <= (p - q) * s <= * s
// then the two lines are overlapping,
if (considerOverlapAsIntersect)
if ((0 <= (q - p) * r && (q - p) * r <= r * r) || (0 <= (p - q) * s && (p - q) * s <= s * s))
return true;
// 2. If neither 0 <= (q - p) * r ≤ r * r nor 0 <= (p - q) * s <= s * s
// then the two lines are collinear but disjoint.
// No need to implement this expression, as it follows from the expression above.
return false;
}
// 3. If r x s = 0 and (q - p) x r != 0, then the two lines are parallel and non-intersecting.
if (rxs.IsZero() && !qpxr.IsZero())
return false;
// t = (q - p) x s / (r x s)
var t = (q - p).Cross(s) / rxs;
// u = (q - p) x r / (r x s)
var u = (q - p).Cross(r) / rxs;
// 4. If r x s != 0 and 0 <= t <= 1 and 0 <= u <= 1
// the two line segments meet at the point p + t r = q + u s.
if (!rxs.IsZero() && (0 <= t && t <= 1) && (0 <= u && u <= 1))
{
// We can calculate the intersection point using either t or u.
intersection = p + t * r;
// An intersection was found.
return true;
}
// 5. Otherwise, the two line segments are not parallel but do not intersect.
return false;
}
public double Cross(Vector v)
{
return X * v.Y - Y * v.X;
}
