From: Bill Spitzak <[email protected]> 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 <[email protected]> Reviewed-by: Oded Gabbay <[email protected]> Reviewed-by: Soren Sandmann <[email protected]> --- demos/scale.c | 102 +++++++++++++++++++++++++++++++++++----------------------- 1 file 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 -- 1.7.11.7 _______________________________________________ Pixman mailing list [email protected] https://lists.freedesktop.org/mailman/listinfo/pixman
