float paraheight() {
if (type!=Paragraph && type!=Block)
printf("Requested paraheight from non-Paragraph!\n");
if (memparaheight==NULL)
memparaheight=p2newfloat(furthest(lineori().perp()));
return *memparaheight;
}
float parawidth() {
if (type!=Paragraph && type!=Block)
printf("Requested parawidth from non-Paragraph!\n");
if (memparawidth==NULL)
memparawidth=p2newfloat(furthest(lineori()));
return *memparawidth;
}
float width() {
if (type==Paragraph || type==Block)
printf("Requested width from Paragraph!\n");
if (memwidth==NULL)
memwidth=p2newfloat(furthest(ori()));
return *memwidth;
}
float height() {
if (type==Paragraph || type==Block)
printf("Requested height from Paragraph!\n");
if (memheight==NULL) {
memheight=p2newfloat(furthest(ori().perp()));
}
return *memheight;
}
// returns (a, (b,c)), three points which define the narrowest diameter of the hull as the pair of
// lines going through b,c, and through a, parallel to b,c TODO: This can be made linear time by
// moving point tc incrementally from the previous value (it can only move in one direction). It
// is currently n*O(furthest)
double ConvexHull::narrowest_diameter(Point &a, Point &b, Point &c) {
Point tb = boundary.back();
double d = INFINITY;
for(unsigned i = 0; i < boundary.size(); i++) {
Point tc = boundary[i];
Point n = -rot90(tb-tc);
Point ta = *furthest(n);
double td = dot(n, ta-tb)/dot(n,n);
if(td < d) {
a = ta;
b = tb;
c = tc;
d = td;
}
tb = tc;
}
return d;
}
