public interface Distance
Two distinct members of the same family, d = f(x) and
d' = f(x - x0) + y0, must have exactly one intersection point (for
each dimension):
|{ x : f(x) = f(x -x0) + y0 }| = 1
This interface is used in DistanceTransform:
D( p ) = min_q f(q) + d(p,q) where p,q are points on a grid/image.
| Modifier and Type | Method and Description |
|---|---|
double |
evaluate(double x,
double xShift,
double yShift,
int dim)
Evaluate function with family parameters
xShift and
yShift at position x in dimension dim. |
double |
intersect(double xShift1,
double yShift1,
double xShift2,
double yShift2,
int dim)
Determine the intersection point in dimension dim of two members of the
function family.
|
double evaluate(double x,
double xShift,
double yShift,
int dim)
xShift and
yShift at position x in dimension dim.x - position on x-axis at which function is evaluatedxShift - shift along the x-axisyShift - shift along y-axisdim - dimension in which to evaluatexdouble intersect(double xShift1,
double yShift1,
double xShift2,
double yShift2,
int dim)
xShift1,
yShift1, xShift2, yShift2, with xShift1 >
xShift2.xShift1 - shift along the x-axisyShift1 - shift along the y-axisxShift2 - shift along the x-axisyShift2 - shift along the y-axisdim - dimension in which to evaluatedimCopyright © 2015–2022 ImgLib2. All rights reserved.