(determine direction for polygon described by ordered points)
(minimum 3 points, up to 6 points,may fail for pathological cases)

(return: _dir: == 2 for cw, 3 for ccw)

o<dir> sub
(not_a_subfile)
#<n>  =  #1 (=4 no. of points)
#<x1> =  #2 (minimum 3 points)
#<y1> =  #3
#<x2> =  #4
#<y2> =  #5
#<x3> =  #6
#<y3> =  #7
#<x4> =  #8 (=0 0 if not used)
#<y4> =  #9 (=0)
#<x5> = #10 (=0)
#<y5> = #11 (=0)
#<x6> = #12 (=0)
#<y6> = #13 (=0)

         (print dir:n=#<n> #<x1> #<y1> #<x2> #<y2> #<x3> #<y3> #<x4> #<y4>)
         (compute direction using cross product: _cross:dir, ignore if EQ 0)
         #<dir> = 0
         o<cross> call [#<x1>][#<y1>][#<x2>][#<y2>][#<x3>][#<y3>]
         #<dir> = [#<dir> + #<_cross:dir>]
         o<cross> call [#<x2>][#<y2>][#<x3>][#<y3>][#<x4>][#<y4>]
         #<dir> = [#<dir> + #<_cross:dir>]

o<l10>   if [#<n> EQ 4]
           o<cross> call [#<x3>][#<y3>][#<x4>][#<y4>][#<x1>][#<y1>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x4>][#<y4>][#<x1>][#<y1>][#<x2>][#<y2>]
           #<dir> = [#<dir> + #<_cross:dir>]
o<l10>   endif

o<l20>   if [#<n> EQ 5]
           o<cross> call [#<x3>][#<y3>][#<x4>][#<y4>][#<x5>][#<y5>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x4>][#<y4>][#<x5>][#<y5>][#<x1>][#<y1>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x5>][#<y5>][#<x1>][#<y1>][#<x2>][#<y2>]
           #<dir> = [#<dir> + #<_cross:dir>]
o<l20>   endif

o<l30>   if [#<n> EQ 6]
           o<cross> call [#<x3>][#<y3>][#<x4>][#<y4>][#<x5>][#<y5>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x4>][#<y4>][#<x5>][#<y5>][#<x6>][#<y6>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x5>][#<y5>][#<x6>][#<y6>][#<x1>][#<y1>]
           #<dir> = [#<dir> + #<_cross:dir>]
           o<cross> call [#<x6>][#<y6>][#<x1>][#<y1>][#<x2>][#<y2>]
           #<dir> = [#<dir> + #<_cross:dir>]
o<l30>   endif

o<l40>   if [[#<n> GE 7] or [#<n> LE 0]]
           (print, dir:problem bad n=#<n>)
           (debug, dir:problem bad n=#<n>)
o<l40>   endif

;o<l60>   if [#<dir> eq 0]
;           (print, dir:problem dir=#<dir>)
;           (debug, dir:problem dir=#<dir>)
;           m2
;o<l60>   endif

         (above computes cross product, to give traversal direction, swap)
o<l70>   if [#<dir> GT 0]
           #<_dir:> = 2 (cw)
o<l70>   else
           #<_dir:> = 3 (ccw)
o<l70>   endif
o<l80>   if [#<dir> EQ 0]
           #<_dir:> = 3 (use ccw)
o<l80>   endif
         (print  dir:direction=#<_dir:>)
         (debug  dir:direction=#<_dir:>)

o<dir> endsub

;wikipedia
;convex polygon: any line drawn through the polygon and not tangent to
;                an edge or corner) meets its boundary exactly twice.

;above uses cross product 1-2 X 3-2
;below uses cross product 2-1 X 3-2 (negative of above, opposite test)

;rhttp://debian.fmi.uni-sofia.bg/~sergei/cgsr/docs/clockwise.htm
;If the polygon is known to be convex then one only has to consider the
;cross product between any two adjacent edges. A positive cross product
;means we have a counterclockwise polygon. There are some tests that may
;need to be done if the polygons may not be "clean". In particular two
;vertices must not be coincident and two edges must not be colinear.

;For the more general case where the polygons may be convex, it is
;necessary to consider the sign of the cross product between adjacent
;edges as one moves around the polygon. If there are more positive cross
;products then the overall polygon is ordered counterclockwise. There are
;pathological cases to consider here as well, all the edges cannot be
;colinear, there must be at least 3 vertices, the polygon must be simple,
;that is, it cannot intersect itself or have holes.