[Pixman] [PATCH 01/13] demos/scale: Compute filter size using boundary of xformed ellipse, not rectangle

[Oded Gabbay]

On Mon, Jan 4, 2016 at 5:12 AM,  <spitzak at gmail.com> wrote:
> From: Bill Spitzak <spitzak at gmail.com>
>
> This is much more accurate and less blurry. In particular the filtering does
> not change as the image is rotated.

You forgot to sign-off this patch.

> ---
>  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
> --
> 1.9.1
>
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Add the sign-off and my:

Reviewed-by: Oded Gabbay <oded.gabbay at gmail.com>