Differential D981 Diff 6792 source/simulation2/helpers/Geometry.cpp

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source/simulation2/helpers/Geometry.cpp

Show First 20 Lines • Show All 180 Lines • ▼ Show 20 Lines	if (dv < fixed::Zero()) // F, H
corner -= v.Multiply(hh);		corner -= v.Multiply(hh);
else // A, C		else // A, C
corner += v.Multiply(hh);		corner += v.Multiply(hh);

return corner;		return corner;
}		}
}		}

		fixed Geometry::DistanceSquareToSquare(const CFixedVector2D& relativePos, const CFixedVector2D& u1, const CFixedVector2D& v1, const CFixedVector2D& halfSize1, const CFixedVector2D& u2, const CFixedVector2D& v2, const CFixedVector2D& halfSize2)
		{
		/*
		* The shortest distance between two non colliding squares equals the distance between a corner
		* and other square. Thus calculating all 8 those distances and taking the smallest.
		templeUnsubmitted Not Done Inline Actions Unless they're overlapping in a weird way (two rectangles in a cross), but that's unlikely to happen. temple: Unless they're overlapping in a weird way (two rectangles in a cross), but that's unlikely to…
		bbUnsubmitted Not Done Inline Actions True, but do notice this is already counted as 0, since at least 1 corner will be inside the other square (and we set the last argument to true) bb: True, but do notice this is already counted as 0, since at least 1 corner will be inside the…
		templeUnsubmitted Not Done Inline Actions I'm saying if you have two long rectangles overlapping in a cross, like +, then they only overlap in the center, so none of the corners of one rectangle will be inside the other rectangle. It should be distance = 0 but instead will be something > 0. We won't often/ever have rectangles overlapping like this so it's not really a concern. (Units are circles, so rectangle vs rectangle would have to be something like fort vs fort, and they'll never overlap.) temple: I'm saying if you have two long rectangles overlapping in a cross, like +, then they only…
		bbUnsubmitted Not Done Inline Actions How could I miss that, we can simply test for TestSquareSquare avoiding this issue bb: How could I miss that, we can simply test for TestSquareSquare avoiding this issue
		* For colliding squares we simply return 0. When one of the points is inside the other square
		* we depend on DistanceToSquare's countInsideAsZero. When no point is inside the other square,
		* it is enough to check that two adjacent edges of one square does not collide with the other square.
		*/
		fixed hw1 = halfSize1.X;
		fixed hh1 = halfSize1.Y;
		fixed hw2 = halfSize2.X;
		fixed hh2 = halfSize2.Y;
		if (TestRaySquare(relativePos + u1.Multiply(hw1) + v1.Multiply(hh1), relativePos - u1.Multiply(hw1) + v1.Multiply(hh1), u2, v2, halfSize2) \|\|
		TestRaySquare(relativePos + u1.Multiply(hw1) + v1.Multiply(hh1), relativePos + u1.Multiply(hw1) - v1.Multiply(hh1), u2, v2, halfSize2))
		return fixed::Zero();

		return std::min(std::min(std::min(
		DistanceToSquare(relativePos + u1.Multiply(hw1) + v1.Multiply(hh1), u2, v2, halfSize2, true),
		DistanceToSquare(relativePos + u1.Multiply(hw1) - v1.Multiply(hh1), u2, v2, halfSize2, true)),
		std::min(
		DistanceToSquare(relativePos - u1.Multiply(hw1) + v1.Multiply(hh1), u2, v2, halfSize2, true),
		DistanceToSquare(relativePos - u1.Multiply(hw1) - v1.Multiply(hh1), u2, v2, halfSize2, true))),
		templeUnsubmitted Not Done Inline Actions minus u2 temple: minus u2
		bbUnsubmitted Not Done Inline Actions Woot woot bb: Woot woot
		std::min(std::min(
		DistanceToSquare(relativePos + u2.Multiply(hw2) + v2.Multiply(hh2), u1, v1, halfSize1, true),
		DistanceToSquare(relativePos + u2.Multiply(hw2) - v2.Multiply(hh2), u1, v1, halfSize1, true)),
		std::min(
		DistanceToSquare(relativePos - u2.Multiply(hw2) + v2.Multiply(hh2), u1, v1, halfSize1, true),
		DistanceToSquare(relativePos - u2.Multiply(hw2) - v2.Multiply(hh2), u1, v1, halfSize1, true))));
		}

		fixed Geometry::MaxDistanceToSquare(const CFixedVector2D& point, const CFixedVector2D& u, const CFixedVector2D& v, const CFixedVector2D& halfSize, bool countInsideAsZero)
		{
		fixed hw = halfSize.X;
		fixed hh = halfSize.Y;

		if (point.Dot(u).Absolute() < hw && point.Dot(v).Absolute() < hh && countInsideAsZero)
		return fixed::Zero();

		/*
		* The maximum distance from a point to an edge of a square equals the greatest distance
		* from the point to the a corner. Thus calculating all and taking the greatest.
		*/
		return std::max(std::max(
		(point + u.Multiply(hw) + v.Multiply(hh)).Length(),
		(point + u.Multiply(hw) - v.Multiply(hh)).Length()),
		std::max(
		(point - u.Multiply(hw) + v.Multiply(hh)).Length(),
		(point - u.Multiply(hw) - v.Multiply(hh)).Length()));
		}

		fixed Geometry::MaxDistanceSquareToSquare(const CFixedVector2D& relativePos, const CFixedVector2D& u1, const CFixedVector2D& v1, const CFixedVector2D& halfSize1, const CFixedVector2D& u2, const CFixedVector2D& v2, const CFixedVector2D& halfSize2)
		{
		/*
		* The maximum distance from an edge of a square to the edge of another square
		* equals the greatest distance from the any of the 16 corner corner distances.
		*/
		bbUnsubmitted Not Done Inline Actions indentation bb: indentation
		fixed hw1 = halfSize1.X;
		fixed hh1 = halfSize1.Y;

		return std::max(std::max(
		MaxDistanceToSquare(relativePos + u1.Multiply(hw1) + v1.Multiply(hh1), u2, v2, halfSize2, true),
		MaxDistanceToSquare(relativePos + u1.Multiply(hw1) - v1.Multiply(hh1), u2, v2, halfSize2, true)),
		std::max(MaxDistanceToSquare(relativePos - u1.Multiply(hw1) + v1.Multiply(hh1), u2, v2, halfSize2, true),
		MaxDistanceToSquare(relativePos - u1.Multiply(hw1) - v1.Multiply(hh1), u2, v2, halfSize2, true)));
		}

bool Geometry::TestRaySquare(const CFixedVector2D& a, const CFixedVector2D& b, const CFixedVector2D& u, const CFixedVector2D& v, const CFixedVector2D& halfSize)		bool Geometry::TestRaySquare(const CFixedVector2D& a, const CFixedVector2D& b, const CFixedVector2D& u, const CFixedVector2D& v, const CFixedVector2D& halfSize)
{		{
/*		/*
* We only consider collisions to be when the ray goes from outside to inside the shape (and possibly out again).		* We only consider collisions to be when the ray goes from outside to inside the shape (and possibly out again).
* Various cases to consider:		* Various cases to consider:
* 'a' inside, 'b' inside -> no collision		* 'a' inside, 'b' inside -> no collision
* 'a' inside, 'b' outside -> no collision		* 'a' inside, 'b' outside -> no collision
* 'a' outside, 'b' inside -> collision		* 'a' outside, 'b' inside -> collision
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