# transformRect method

Returns a rect that bounds the result of applying the given matrix as a perspective transform to the given rect.

This function assumes the given rect is in the plane with z equals 0.0. The transformed rect is then projected back into the plane with z equals 0.0 before computing its bounding rect.

## Implementation

```
static Rect transformRect(Matrix4 transform, Rect rect) {
final Float64List storage = transform.storage;
final double x = rect.left;
final double y = rect.top;
final double w = rect.right - x;
final double h = rect.bottom - y;
// We want to avoid turning a finite rect into an infinite one if we can.
if (!w.isFinite || !h.isFinite) {
return _safeTransformRect(transform, rect);
}
// Transforming the 4 corners of a rectangle the straightforward way
// incurs the cost of transforming 4 points using vector math which
// involves 48 multiplications and 48 adds and then normalizing
// the points using 4 inversions of the homogeneous weight factor
// and then 12 multiplies. Once we have transformed all of the points
// we then need to turn them into a bounding box using 4 min/max
// operations each on 4 values yielding 12 total comparisons.
//
// On top of all of those operations, using the vector_math package to
// do the work for us involves allocating several objects in order to
// communicate the values back and forth - 4 allocating getters to extract
// the [Offset] objects for the corners of the [Rect], 4 conversions to
// a [Vector3] to use [Matrix4.perspectiveTransform()], and then 4 new
// [Offset] objects allocated to hold those results, yielding 8 [Offset]
// and 4 [Vector3] object allocations per rectangle transformed.
//
// But the math we really need to get our answer is actually much less
// than that.
//
// First, consider that a full point transform using the vector math
// package involves expanding it out into a vector3 with a Z coordinate
// of 0.0 and then performing 3 multiplies and 3 adds per coordinate:
// ```
// xt = x*m00 + y*m10 + z*m20 + m30;
// yt = x*m01 + y*m11 + z*m21 + m31;
// zt = x*m02 + y*m12 + z*m22 + m32;
// wt = x*m03 + y*m13 + z*m23 + m33;
// ```
// Immediately we see that we can get rid of the 3rd column of multiplies
// since we know that Z=0.0. We can also get rid of the 3rd row because
// we ignore the resulting Z coordinate. Finally we can get rid of the
// last row if we don't have a perspective transform since we can verify
// that the results are 1.0 for all points. This gets us down to 16
// multiplies and 16 adds in the non-perspective case and 24 of each for
// the perspective case. (Plus the 12 comparisons to turn them back into
// a bounding box.)
//
// But we can do even better than that.
//
// Under optimal conditions of no perspective transformation,
// which is actually a very common condition, we can transform
// a rectangle in as little as 3 operations:
//
// (rx,ry) = transform of upper left corner of rectangle
// (wx,wy) = delta transform of the (w, 0) width relative vector
// (hx,hy) = delta transform of the (0, h) height relative vector
//
// A delta transform is a transform of all elements of the matrix except
// for the translation components. The translation components are added
// in at the end of each transform computation so they represent a
// constant offset for each point transformed. A delta transform of
// a horizontal or vertical vector involves a single multiplication due
// to the fact that it only has one non-zero coordinate and no addition
// of the translation component.
//
// In the absence of a perspective transform, the transformed
// rectangle will be mapped into a parallelogram with corners at:
// corner1 = (rx, ry)
// corner2 = corner1 + dTransformed width vector = (rx+wx, ry+wy)
// corner3 = corner1 + dTransformed height vector = (rx+hx, ry+hy)
// corner4 = corner1 + both dTransformed vectors = (rx+wx+hx, ry+wy+hy)
// In all, this method of transforming the rectangle requires only
// 8 multiplies and 12 additions (which we can reduce to 8 additions if
// we only need a bounding box, see below).
//
// In the presence of a perspective transform, the above conditions
// continue to hold with respect to the non-normalized coordinates so
// we can still save a lot of multiplications by computing the 4
// non-normalized coordinates using relative additions before we normalize
// them and they lose their "pseudo-parallelogram" relationships. We still
// have to do the normalization divisions and min/max all 4 points to
// get the resulting transformed bounding box, but we save a lot of
// calculations over blindly transforming all 4 coordinates independently.
// In all, we need 12 multiplies and 22 additions to construct the
// non-normalized vectors and then 8 divisions (or 4 inversions and 8
// multiplies) for normalization (plus the standard set of 12 comparisons
// for the min/max bounds operations).
//
// Back to the non-perspective case, the optimization that lets us get
// away with fewer additions if we only need a bounding box comes from
// analyzing the impact of the relative vectors on expanding the
// bounding box of the parallelogram. First, the bounding box always
// contains the transformed upper-left corner of the rectangle. Next,
// each relative vector either pushes on the left or right side of the
// bounding box and also either the top or bottom side, depending on
// whether it is positive or negative. Finally, you can consider the
// impact of each vector on the bounding box independently. If, say,
// wx and hx have the same sign, then the limiting point in the bounding
// box will be the one that involves adding both of them to the origin
// point. If they have opposite signs, then one will push one wall one
// way and the other will push the opposite wall the other way and when
// you combine both of them, the resulting "opposite corner" will
// actually be between the limits they established by pushing the walls
// away from each other, as below:
// ```
// +---------(originx,originy)--------------+
// | -----^---- |
// | ----- ---- |
// | ----- ---- |
// (+hx,+hy)< ---- |
// | ---- ---- |
// | ---- >(+wx,+wy)
// | ---- ----- |
// | ---- ----- |
// | ---- ----- |
// | v |
// +---------------(+wx+hx,+wy+hy)----------+
// ```
// In this diagram, consider that:
// ```
// wx would be a positive number
// hx would be a negative number
// wy and hy would both be positive numbers
// ```
// As a result, wx pushes out the right wall, hx pushes out the left wall,
// and both wy and hy push down the bottom wall of the bounding box. The
// wx,hx pair (of opposite signs) worked on opposite walls and the final
// opposite corner had an X coordinate between the limits they established.
// The wy,hy pair (of the same sign) both worked together to push the
// bottom wall down by their sum.
//
// This relationship allows us to simply start with the point computed by
// transforming the upper left corner of the rectangle, and then
// conditionally adding wx, wy, hx, and hy to either the left or top
// or right or bottom of the bounding box independently depending on sign.
// In that case we only need 4 comparisons and 4 additions total to
// compute the bounding box, combined with the 8 multiplications and
// 4 additions to compute the transformed point and relative vectors
// for a total of 8 multiplies, 8 adds, and 4 comparisons.
//
// An astute observer will note that we do need to do 2 subtractions at
// the top of the method to compute the width and height. Add those to
// all of the relative solutions listed above. The test for perspective
// also adds 3 compares to the affine case and up to 3 compares to the
// perspective case (depending on which test fails, the rest are omitted).
//
// The final tally:
// basic method = 60 mul + 48 add + 12 compare
// optimized perspective = 12 mul + 22 add + 15 compare + 2 sub
// optimized affine = 8 mul + 8 add + 7 compare + 2 sub
//
// Since compares are essentially subtractions and subtractions are
// the same cost as adds, we end up with:
// basic method = 60 mul + 60 add/sub/compare
// optimized perspective = 12 mul + 39 add/sub/compare
// optimized affine = 8 mul + 17 add/sub/compare
final double wx = storage[0] * w;
final double hx = storage[4] * h;
final double rx = storage[0] * x + storage[4] * y + storage[12];
final double wy = storage[1] * w;
final double hy = storage[5] * h;
final double ry = storage[1] * x + storage[5] * y + storage[13];
if (storage[3] == 0.0 && storage[7] == 0.0 && storage[15] == 1.0) {
double left = rx;
double right = rx;
if (wx < 0) {
left += wx;
} else {
right += wx;
}
if (hx < 0) {
left += hx;
} else {
right += hx;
}
double top = ry;
double bottom = ry;
if (wy < 0) {
top += wy;
} else {
bottom += wy;
}
if (hy < 0) {
top += hy;
} else {
bottom += hy;
}
return Rect.fromLTRB(left, top, right, bottom);
} else {
final double ww = storage[3] * w;
final double hw = storage[7] * h;
final double rw = storage[3] * x + storage[7] * y + storage[15];
final double ulx = rx / rw;
final double uly = ry / rw;
final double urx = (rx + wx) / (rw + ww);
final double ury = (ry + wy) / (rw + ww);
final double llx = (rx + hx) / (rw + hw);
final double lly = (ry + hy) / (rw + hw);
final double lrx = (rx + wx + hx) / (rw + ww + hw);
final double lry = (ry + wy + hy) / (rw + ww + hw);
return Rect.fromLTRB(
_min4(ulx, urx, llx, lrx),
_min4(uly, ury, lly, lry),
_max4(ulx, urx, llx, lrx),
_max4(uly, ury, lly, lry),
);
}
}
```