OpenEuphoria

mathmatical question.. - Reply

• Posted by "BABOR, JIRI" <J.Babor at GNS.CRI.NZ> May 29, 1997
• 777 views

Lee woo seob wrote:

> Can anyone give me nice answer over the problem below;
> Assumes the polygon with arbitrary numbers of points is drawn on the
> pixel-graphics screen, for example;

> polygon(RED,1,{{100,200},{200,200},......,}

> Problem is:
> How can I know whether the mouse pointer is INSIDE the polygon or not,
> at the time the mouse button is pressed?


I am not sure I can give you a nice answer, but I thought it was interesting
enough problem. I am assuming you already know how to get the mouse pointer
coordinates - if not, please ask again.

There are several possible approaches to this problem, I simply coded the first
one that occurred to me: imagine a straight line through the point. In a
non-trivial case, the line will intersect with one or more sides of the
polygon. Now, travel from the given point in one or the other direction and
count the intersections. If the total is odd, you obviously started inside the
polygon and vice versa. - Incidentally, I chose a horizontal line in the
function inside() below, sequence p is x,y coordinates of the point, and P is
the polygon sequence. Jiri

Sorry, this is almost as long as some of Ralf's messages...


-- inpoly.ex -------------------------------------------------------------------
-- j.babor at gns.cri.nz
--------------------------------------------------------------------------------

include graphics.e
include get.e

object junk
sequence s,p1,p2,p3
integer key

function inside(sequence p, sequence P)
   atom x
   integer count,n
   n=length(P)
   P=append(P,P[1])
   count=0
   for i=1 to n do
      if (P[i][2]<p[2] and P[i+1][2]>p[2]) or
         (P[i][2]>p[2] and P[i+1][2]<p[2]) then
         x=P[i][1]+(p[2]-P[i][2])*(P[i+1][1]-P[i][1])/(P[i+1][2]-P[i][2])
         if x<p[1] then count=count+1 end if
      end if
   end for
   return remainder(count,2)  -- return 1 (true) if count is odd
end function

-- main ------------------------------------------------------------------------

junk=graphics_mode(18)

p1={30,250}
p2={80,250}
p3={200,250}

polygon(12,1,s)
pixel(15,p1)
pixel(14,p2)
pixel(9,p3)
print(1,{inside(p1,s),inside(p2,s),inside(p3,s)})
--printf(1,"%d",inside(p1,s))
key = wait_key()
junk=graphics_mode(-1)