+(defvar image-transform-right-angle-fudge 0.0001
+ "Snap distance to a multiple of a right angle.
+There's no deep theory behind the default value, it should just
+be somewhat larger than ImageMagick's MagickEpsilon.")
+
+(defsubst image-transform-width (width height)
+ "Return the bounding box width of a rotated WIDTH x HEIGHT rectangle.
+The rotation angle is the value of `image-transform-rotation' in degrees."
+ (let ((angle (degrees-to-radians image-transform-rotation)))
+ ;; Assume, w.l.o.g., that the vertices of the rectangle have the
+ ;; coordinates (+-w/2, +-h/2) and that (0, 0) is the center of the
+ ;; rotation by the angle A. The projections onto the first axis
+ ;; of the vertices of the rotated rectangle are +- (w/2) cos A +-
+ ;; (h/2) sin A, and the difference between the largest and the
+ ;; smallest of the four values is the expression below.
+ (+ (* width (abs (cos angle))) (* height (abs (sin angle))))))
+
+;; The following comment and code snippet are from
+;; ImageMagick-6.7.4-4/magick/distort.c
+
+;; /* Set the output image geometry to calculated 'best fit'.
+;; Yes this tends to 'over do' the file image size, ON PURPOSE!
+;; Do not do this for DePolar which needs to be exact for virtual tiling.
+;; */
+;; if ( fix_bounds ) {
+;; geometry.x = (ssize_t) floor(min.x-0.5);
+;; geometry.y = (ssize_t) floor(min.y-0.5);
+;; geometry.width=(size_t) ceil(max.x-geometry.x+0.5);
+;; geometry.height=(size_t) ceil(max.y-geometry.y+0.5);
+;; }
+
+;; Other parts of the same file show that here the origin is in the
+;; left lower corner of the image rectangle, the center of the
+;; rotation is the center of the rectangle and min.x and max.x
+;; (resp. min.y and max.y) are the smallest and the largest of the
+;; projections of the vertices onto the first (resp. second) axis.
+
+(defun image-transform-fit-width (width height length)
+ "Return (w . h) so that a rotated w x h image has exactly width LENGTH.
+The rotation angle is the value of `image-transform-rotation'.
+Write W for WIDTH and H for HEIGHT. Then the w x h rectangle is
+an \"approximately uniformly\" scaled W x H rectangle, which
+currently means that w is one of floor(s W) + {0, 1, -1} and h is
+floor(s H), where s can be recovered as the value of `image-transform-scale'.
+The value of `image-transform-rotation' may be replaced by
+a slightly different angle. Currently this is done for values
+close to a multiple of 90, see `image-transform-right-angle-fudge'."
+ (cond ((< (abs (- (mod (+ image-transform-rotation 90) 180) 90))
+ image-transform-right-angle-fudge)
+ (cl-assert (not (zerop width)) t)
+ (setq image-transform-rotation
+ (float (round image-transform-rotation))
+ image-transform-scale (/ (float length) width))
+ (cons length nil))
+ ((< (abs (- (mod (+ image-transform-rotation 45) 90) 45))
+ image-transform-right-angle-fudge)
+ (cl-assert (not (zerop height)) t)
+ (setq image-transform-rotation
+ (float (round image-transform-rotation))
+ image-transform-scale (/ (float length) height))
+ (cons nil length))
+ (t
+ (cl-assert (not (and (zerop width) (zerop height))) t)
+ (setq image-transform-scale
+ (/ (float (1- length)) (image-transform-width width height)))
+ ;; Assume we have a w x h image and an angle A, and let l =
+ ;; l(w, h) = w |cos A| + h |sin A|, which is the actual width
+ ;; of the bounding box of the rotated image, as calculated by
+ ;; `image-transform-width'. The code snippet quoted above
+ ;; means that ImageMagick puts the rotated image in
+ ;; a bounding box of width L = 2 ceil((w+l+1)/2) - w.
+ ;; Elementary considerations show that this is equivalent to
+ ;; L - w being even and L-3 < l(w, h) <= L-1. In our case, L is
+ ;; the given `length' parameter and our job is to determine
+ ;; reasonable values for w and h which satisfy these
+ ;; conditions.
+ (let ((w (floor (* image-transform-scale width)))
+ (h (floor (* image-transform-scale height))))
+ ;; Let w and h as bound above. Then l(w, h) <= l(s W, s H)
+ ;; = L-1 < l(w+1, h+1) = l(w, h) + l(1, 1) <= l(w, h) + 2,
+ ;; hence l(w, h) > (L-1) - 2 = L-3.
+ (cons
+ (cond ((= (mod w 2) (mod length 2))
+ w)
+ ;; l(w+1, h) >= l(w, h) > L-3, but does l(w+1, h) <=
+ ;; L-1 hold?
+ ((<= (image-transform-width (1+ w) h) (1- length))
+ (1+ w))
+ ;; No, it doesn't, but this implies that l(w-1, h) =
+ ;; l(w+1, h) - l(2, 0) >= l(w+1, h) - 2 > (L-1) -
+ ;; 2 = L-3. Clearly, l(w-1, h) <= l(w, h) <= L-1.
+ (t
+ (1- w)))
+ h)))))
+
+(defun image-transform-check-size ()
+ "Check that the image exactly fits the width/height of the window.
+
+Do this for an image of type `imagemagick' to make sure that the
+elisp code matches the way ImageMagick computes the bounding box
+of a rotated image."
+ (when (and (not (numberp image-transform-resize))
+ (boundp 'image-type)
+ (eq image-type 'imagemagick))
+ (let ((size (image-display-size (image-get-display-property) t)))
+ (cond ((eq image-transform-resize 'fit-width)
+ (cl-assert (= (car size)
+ (- (nth 2 (window-inside-pixel-edges))
+ (nth 0 (window-inside-pixel-edges))))
+ t))
+ ((eq image-transform-resize 'fit-height)
+ (cl-assert (= (cdr size)
+ (- (nth 3 (window-inside-pixel-edges))
+ (nth 1 (window-inside-pixel-edges))))
+ t))))))
+