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Optimize get_closest_point_to_segment*.

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By combining all scalar factors we can get rid of a scalar * vector
multiplication and a square root operation, since the resulting formula
only uses the squared length.
This commit is contained in:
Simon Puchert 2019-07-06 17:41:13 +02:00
parent 0b6b49a897
commit 4b78e17b15
1 changed files with 14 additions and 18 deletions
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32
core/math/geometry.h
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@ -455,16 +455,15 @@ public:
Vector3 p = p_point - p_segment[0];
Vector3 n = p_segment[1] - p_segment[0];
real_t l = n.length();
if (l < 1e-10)
real_t l2 = n.length_squared();
if (l2 < 1e-20)
return p_segment[0]; // both points are the same, just give any
n /= l;
real_t d = n.dot(p);
real_t d = n.dot(p) / l2;
if (d <= 0.0)
return p_segment[0]; // before first point
else if (d >= l)
else if (d >= 1.0)
return p_segment[1]; // after first point
else
return p_segment[0] + n * d; // inside
@ -474,12 +473,11 @@ public:
Vector3 p = p_point - p_segment[0];
Vector3 n = p_segment[1] - p_segment[0];
real_t l = n.length();
if (l < 1e-10)
real_t l2 = n.length_squared();
if (l2 < 1e-20)
return p_segment[0]; // both points are the same, just give any
n /= l;
real_t d = n.dot(p);
real_t d = n.dot(p) / l2;
return p_segment[0] + n * d; // inside
}
@ -488,16 +486,15 @@ public:
Vector2 p = p_point - p_segment[0];
Vector2 n = p_segment[1] - p_segment[0];
real_t l = n.length();
if (l < 1e-10)
real_t l2 = n.length_squared();
if (l2 < 1e-20)
return p_segment[0]; // both points are the same, just give any
n /= l;
real_t d = n.dot(p);
real_t d = n.dot(p) / l2;
if (d <= 0.0)
return p_segment[0]; // before first point
else if (d >= l)
else if (d >= 1.0)
return p_segment[1]; // after first point
else
return p_segment[0] + n * d; // inside
@ -521,12 +518,11 @@ public:
Vector2 p = p_point - p_segment[0];
Vector2 n = p_segment[1] - p_segment[0];
real_t l = n.length();
if (l < 1e-10)
real_t l2 = n.length_squared();
if (l2 < 1e-20)
return p_segment[0]; // both points are the same, just give any
n /= l;
real_t d = n.dot(p);
real_t d = n.dot(p) / l2;
return p_segment[0] + n * d; // inside
}