import FunctionValues
import Misc (all_equal, disjoint)
import Point
-import Tetrahedron (
- Tetrahedron(..),
- c,
- b0,
- b1,
- b2,
- b3,
- volume
- )
+import Tetrahedron (Tetrahedron(..), c, volume)
import ThreeDimensional
data Cube = Cube { h :: Double,
-- | The left-side boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
xmin :: Cube -> Double
-xmin cube = (2*i' - 1)*delta / 2
+xmin cube = (i' - 1/2)*delta
where
i' = fromIntegral (i cube) :: Double
delta = h cube
-- | The right-side boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
xmax :: Cube -> Double
-xmax cube = (2*i' + 1)*delta / 2
+xmax cube = (i' + 1/2)*delta
where
i' = fromIntegral (i cube) :: Double
delta = h cube
-- | The front boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
ymin :: Cube -> Double
-ymin cube = (2*j' - 1)*delta / 2
+ymin cube = (j' - 1/2)*delta
where
j' = fromIntegral (j cube) :: Double
delta = h cube
-- | The back boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
ymax :: Cube -> Double
-ymax cube = (2*j' + 1)*delta / 2
+ymax cube = (j' + 1/2)*delta
where
j' = fromIntegral (j cube) :: Double
delta = h cube
-- | The bottom boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
zmin :: Cube -> Double
-zmin cube = (2*k' - 1)*delta / 2
+zmin cube = (k' - 1/2)*delta
where
k' = fromIntegral (k cube) :: Double
delta = h cube
-- | The top boundary of the cube. See Sorokina and Zeilfelder,
-- p. 76.
zmax :: Cube -> Double
-zmax cube = (2*k' + 1)*delta / 2
+zmax cube = (k' + 1/2)*delta
where
k' = fromIntegral (k cube) :: Double
delta = h cube
-- | Since the grid size is necessarily positive, all tetrahedra
--- (which comprise cubes of positive volume) must have positive volume
--- as well.
+-- (which comprise cubes of positive volume) must have positive
+-- volume as well.
prop_all_volumes_positive :: Cube -> Bool
prop_all_volumes_positive cube =
- null nonpositive_volumes
+ all (>= 0) volumes
where
ts = tetrahedra cube
volumes = map volume ts
- nonpositive_volumes = filter (<= 0) volumes
+
-- | In fact, since all of the tetrahedra are identical, we should
-- already know their volumes. There's 24 tetrahedra to a cube, so