Since the word 'intersect' is used, I assume you mean the boundary of the circle/sphere (and not the area), so that a line segment which is completely inside is not intersecting. One point at the boundary means intersecting, I assume. So the segment is not intersecting if and only if it is completely inside or completely outside the circle/sphere.
If the line segment is given by the endpoints A and B, the pseudo code is rather straightforward: ((A-E)<R)*.((B-E)<R) -...@+. ((A-E)>R)*.((B-E)>R) so ones gets D=: +&.*:/"1 (A,:B)-"1 E (D<R) -...@+.&(*./) D>R (untested) R.E. Boss > -----Oorspronkelijk bericht----- > Van: [email protected] [mailto:programming- > [email protected]] Namens Henry Rich > Verzonden: woensdag 6 oktober 2010 22:54 > Aan: Programming forum > Onderwerp: [Jprogramming] Puzzle: line-circle/sphere intersection > > Given circle/sphere with center C and radius R, and a line-segment with > startpoint S and endpoint E, write the J code to tell whether the > line-segment intersects the circle/sphere. > > R is an atom, the rest are lists with 2 or 3 atoms. > > This problem arises in collision detection for games and simulators, or > if you are trying to see whether a path intersects a round obstacle. > > I found a solution whose brevity surprised me. > > Henry Rich > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
