demos/scale: Compute filter size using boundary of xformed ellipse

Instead of using the boundary of xformed rectangle, use the boundary
of xformed ellipse. This is much more accurate and less blurry. In
particular the filtering does not change as the image is rotated.

Signed-off-by: Bill Spitzak <spitzak@gmail.com>
Reviewed-by: Oded Gabbay <oded.gabbay@gmail.com>
Reviewed-by: Soren Sandmann <soren.sandmann@gmail.com>

1 files changed, 61 insertions, 41 deletions

diff --git a/demos/scale.c b/demos/scale.c
index d00307e..0995ad0 100644
--- a/demos/scale.c
+++ b/demos/scale.c
@@ -55,50 +55,70 @@ get_widget (app_t *app, const char *name)
     return widget;
 }
 
-static double
-min4 (double a, double b, double c, double d)
-{
-    double m1, m2;
-
-    m1 = MIN (a, b);
-    m2 = MIN (c, d);
-    return MIN (m1, m2);
-}
-
-static double
-max4 (double a, double b, double c, double d)
-{
-    double m1, m2;
-
-    m1 = MAX (a, b);
-    m2 = MAX (c, d);
-    return MAX (m1, m2);
-}
-
+/* Figure out the boundary of a diameter=1 circle transformed into an ellipse
+ * by trans. Proof that this is the correct calculation:
+ *
+ * Transform x,y to u,v by this matrix calculation:
+ *
+ *  |u|   |a c| |x|
+ *  |v| = |b d|*|y|
+ *
+ * Horizontal component:
+ *
+ *  u = ax+cy (1)
+ *
+ * For each x,y on a radius-1 circle (p is angle to the point):
+ *
+ *  x^2+y^2 = 1
+ *  x = cos(p)
+ *  y = sin(p)
+ *  dx/dp = -sin(p) = -y
+ *  dy/dp = cos(p) = x
+ *
+ * Figure out derivative of (1) relative to p:
+ *
+ *  du/dp = a(dx/dp) + c(dy/dp)
+ *        = -ay + cx
+ *
+ * The min and max u are when du/dp is zero:
+ *
+ *  -ay + cx = 0
+ *  cx = ay
+ *  c = ay/x  (2)
+ *  y = cx/a  (3)
+ *
+ * Substitute (2) into (1) and simplify:
+ *
+ *  u = ax + ay^2/x
+ *    = a(x^2+y^2)/x
+ *    = a/x (because x^2+y^2 = 1)
+ *  x = a/u (4)
+ *
+ * Substitute (4) into (3) and simplify:
+ *
+ *  y = c(a/u)/a
+ *  y = c/u (5)
+ *
+ * Square (4) and (5) and add:
+ *
+ *  x^2+y^2 = (a^2+c^2)/u^2
+ *
+ * But x^2+y^2 is 1:
+ *
+ *  1 = (a^2+c^2)/u^2
+ *  u^2 = a^2+c^2
+ *  u = hypot(a,c)
+ *
+ * Similarily the max/min of v is at:
+ *
+ *  v = hypot(b,d)
+ *
+ */
 static void
 compute_extents (pixman_f_transform_t *trans, double *sx, double *sy)
 {
-    double min_x, max_x, min_y, max_y;
-    pixman_f_vector_t v[4] =
-    {
-	{ { 1, 1, 1 } },
-	{ { -1, 1, 1 } },
-	{ { -1, -1, 1 } },
-	{ { 1, -1, 1 } },
-    };
-
-    pixman_f_transform_point (trans, &v[0]);
-    pixman_f_transform_point (trans, &v[1]);
-    pixman_f_transform_point (trans, &v[2]);
-    pixman_f_transform_point (trans, &v[3]);
-
-    min_x = min4 (v[0].v[0], v[1].v[0], v[2].v[0], v[3].v[0]);
-    max_x = max4 (v[0].v[0], v[1].v[0], v[2].v[0], v[3].v[0]);
-    min_y = min4 (v[0].v[1], v[1].v[1], v[2].v[1], v[3].v[1]);
-    max_y = max4 (v[0].v[1], v[1].v[1], v[2].v[1], v[3].v[1]);
-
-    *sx = (max_x - min_x) / 2.0;
-    *sy = (max_y - min_y) / 2.0;
+    *sx = hypot (trans->m[0][0], trans->m[0][1]) / trans->m[2][2];
+    *sy = hypot (trans->m[1][0], trans->m[1][1]) / trans->m[2][2];
 }
 
 typedef struct