ref: c918b54f25340dcf1a6bf57f747f545ac1f30568
parent: a255125fe482772feade4334635fc8a3967199b9
author: Anuj Verma <[email protected]>
date: Tue Aug 18 06:28:16 EDT 2020
[sdf] Add function to resolve corner distances. * src/sdf/ftsdf.c (resolve_corner): New function.
--- a/ChangeLog
+++ b/ChangeLog
@@ -1,5 +1,11 @@
2020-08-18 Anuj Verma <[email protected]>
+ [sdf] Add function to resolve corner distances.
+
+ * src/sdf/ftsdf.c (resolve_corner): New function.
+
+2020-08-18 Anuj Verma <[email protected]>
+
[sdf] Add essential math functions.
* src/sdf/ftsdf.c (cube_root, arc_cos) [!USE_NEWTON_FOR_CONIC]: New
--- a/src/sdf/ftsdf.c
+++ b/src/sdf/ftsdf.c
@@ -1556,4 +1556,108 @@
#endif /* !USE_NEWTON_FOR_CONIC */
+ /*************************************************************************/
+ /*************************************************************************/
+ /** **/
+ /** RASTERIZER **/
+ /** **/
+ /*************************************************************************/
+ /*************************************************************************/
+
+ /**************************************************************************
+ *
+ * @Function:
+ * resolve_corner
+ *
+ * @Description:
+ * At some places on the grid two edges can give opposite directions;
+ * this happens when the closest point is on one of the endpoint. In
+ * that case we need to check the proper sign.
+ *
+ * This can be visualized by an example:
+ *
+ * ```
+ * x
+ *
+ * o
+ * ^ \
+ * / \
+ * / \
+ * (a) / \ (b)
+ * / \
+ * / \
+ * / v
+ * ```
+ *
+ * Suppose `x` is the point whose shortest distance from an arbitrary
+ * contour we want to find out. It is clear that `o` is the nearest
+ * point on the contour. Now to determine the sign we do a cross
+ * product of the shortest distance vector and the edge direction, i.e.,
+ *
+ * ```
+ * => sign = cross(x - o, direction(a))
+ * ```
+ *
+ * Using the right hand thumb rule we can see that the sign will be
+ * positive.
+ *
+ * If we use `b', however, we have
+ *
+ * ```
+ * => sign = cross(x - o, direction(b))
+ * ```
+ *
+ * In this case the sign will be negative. To determine the correct
+ * sign we thus divide the plane in two halves and check which plane the
+ * point lies in.
+ *
+ * ```
+ * |
+ * x |
+ * |
+ * o
+ * ^|\
+ * / | \
+ * / | \
+ * (a) / | \ (b)
+ * / | \
+ * / \
+ * / v
+ * ```
+ *
+ * We can see that `x` lies in the plane of `a`, so we take the sign
+ * determined by `a`. This test can be easily done by calculating the
+ * orthogonality and taking the greater one.
+ *
+ * The orthogonality is simply the sinus of the two vectors (i.e.,
+ * x - o) and the corresponding direction. We efficiently pre-compute
+ * the orthogonality with the corresponding `get_min_distance_`
+ * functions.
+ *
+ * @Input:
+ * sdf1 ::
+ * First signed distance (can be any of `a` or `b`).
+ *
+ * sdf1 ::
+ * Second signed distance (can be any of `a` or `b`).
+ *
+ * @Return:
+ * The correct signed distance, which is computed by using the above
+ * algorithm.
+ *
+ * @Note:
+ * The function does not care about the actual distance, it simply
+ * returns the signed distance which has a larger cross product. As a
+ * consequence, this function should not be used if the two distances
+ * are fairly apart. In that case simply use the signed distance with
+ * a shorter absolute distance.
+ *
+ */
+ static SDF_Signed_Distance
+ resolve_corner( SDF_Signed_Distance sdf1,
+ SDF_Signed_Distance sdf2 )
+ {+ return FT_ABS( sdf1.cross ) > FT_ABS( sdf2.cross ) ? sdf1 : sdf2;
+ }
+
/* END */