Determine if two triangles overlap: Difference between revisions

→‎{{header|Haskell}}: added Haskell solution
(→‎{{header|Haskell}}: added Haskell solution)
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which have only a single corner in contact, if boundary points do not collide
do not overlap</pre>
 
=={{header|Haskell}}==
Uses the solution of the task [[Find_if_a_point_is_within_a_triangle#Haskell]]
 
<lang haskell>isOverlapping :: Triangle Double -> Triangle Double -> Bool
isOverlapping t1 t2 = vertexInside || midLineInside
where
vertexInside =
any (isInside t1) (vertices t2) ||
any (isInside t2) (vertices t1)
 
isInside t = (Outside /=) . overlapping t
midLineInside =
any (\p -> isInside t1 p && isInside t2 p) midPoints
midPoints =
[ intersections l1 l2 | l1 <- midLines t1
, l2 <- midLines t2 ]
 
intersections (a1,b1,c1) (a2,b2,c2) =
( -(-b2*c1+b1*c2)/(a2*b1-a1*b2)
, -(a2*c1-a1*c2)/(a2*b1-a1*b2) )
 
midLines (Triangle a b c) =
[line a b c, line b c a, line c a b]
 
line (x,y) (ax, ay) (bx, by) =
(ay+by-2*y, -ax-bx+2*x, -ay*x-by*x+ax*y+bx*y)
 
test = map (uncurry isOverlapping)
[ (Triangle (0,0) (5,0) (0,5), Triangle (0,0) (5,0) (0,6))
, (Triangle (0,0) (0,5) (5,0), Triangle (0,0) (0,5) (5,0))
, (Triangle (0,0) (5,0) (0,5), Triangle (-10,0) (-5,0) (-1,6))
, (Triangle (0,0) (5,0) (2.5,5), Triangle (0,4) (2.5,-1) (5,4))
, (Triangle (0,0) (1,1) (0,2), Triangle (2,1) (3,0) (3,2))
, (Triangle (0,0) (1,1) (0,2), Triangle (2,1) (3,-2) (3,4))
, (Triangle (0,0) (1,0) (0,1), Triangle (1,0) (2,0) (1,1))]</lang>
 
<pre>λ> test
[True,True,False,True,False,False,True]</pre>
 
=={{header|Java}}==
Anonymous user