import Cube
import FunctionValues (FunctionValues)
import Tests.FunctionValues ()
-import Tetrahedron (v0, volume)
+import Tetrahedron (b0, b1, b2, b3, c,
+ Tetrahedron(Tetrahedron),
+ v0, v1, v2, v3, volume)
instance Arbitrary Cube where
arbitrary = do
prop_tetrahedron23_volumes_positive :: Cube -> Bool
prop_tetrahedron23_volumes_positive c =
volume (tetrahedron23 c) > 0
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0120_identity1 :: Cube -> Bool
+prop_c0120_identity1 cube =
+ c t0 0 1 2 0 ~= (c t0 0 0 2 1 + c t3 0 0 1 2) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0210_identity1 :: Cube -> Bool
+prop_c0210_identity1 cube =
+ c t0 0 2 1 0 ~= (c t0 0 1 1 1 + c t3 0 1 1 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0300_identity1 :: Cube -> Bool
+prop_c0300_identity1 cube =
+ c t0 0 3 0 0 ~= (c t0 0 2 0 1 + c t3 0 2 1 0) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c1110_identity :: Cube -> Bool
+prop_c1110_identity cube =
+ c t0 1 1 1 0 ~= (c t0 1 0 1 1 + c t3 1 0 1 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c1200_identity1 :: Cube -> Bool
+prop_c1200_identity1 cube =
+ c t0 1 2 0 0 ~= (c t0 1 1 0 1 + c t3 1 1 1 0) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t3 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v2> and v3,v3-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c2100_identity1 :: Cube -> Bool
+prop_c2100_identity1 cube =
+ c t0 2 1 0 0 ~= (c t0 2 0 0 1 + c t3 2 0 1 0) / 2
+ where
+ t0 = tetrahedron0 cube
+ t3 = tetrahedron3 cube
+
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0102_identity1 :: Cube -> Bool
+prop_c0102_identity1 cube =
+ c t0 0 1 0 2 ~= (c t0 0 0 1 2 + c t1 0 0 2 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0201_identity1 :: Cube -> Bool
+prop_c0201_identity1 cube =
+ c t0 0 2 0 1 ~= (c t0 0 1 1 1 + c t1 0 1 1 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c0300_identity2 :: Cube -> Bool
+prop_c0300_identity2 cube =
+ c t0 0 3 0 0 ~= (c t0 0 2 1 0 + c t1 0 2 0 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c1101_identity :: Cube -> Bool
+prop_c1101_identity cube =
+ c t0 1 1 0 1 ~= (c t0 1 0 1 1 + c t1 1 0 1 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c1200_identity2 :: Cube -> Bool
+prop_c1200_identity2 cube =
+ c t0 1 2 0 0 ~= (c t0 1 1 1 0 + c t1 1 1 0 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+-- | Given in Sorokina and Zeilfelder, p. 79. Note that the third and
+-- fourth indices of c-t1 have been switched. This is because we
+-- store the triangles oriented such that their volume is
+-- positive. If T and T-tilde share <v0,v1,v3> and v2,v2-tilde point
+-- in opposite directions, one of them has to have negative volume!
+prop_c2100_identity2 :: Cube -> Bool
+prop_c2100_identity2 cube =
+ c t0 2 1 0 0 ~= (c t0 2 0 1 0 + c t1 2 0 0 1) / 2
+ where
+ t0 = tetrahedron0 cube
+ t1 = tetrahedron1 cube
+
+
+
+-- | Given in Sorokina and Zeilfelder, p. 78.
+-- prop_cijk1_identity :: Cube -> Bool
+-- prop_cijk1_identity cube =
+-- and [ c t0 i j k 1 ~=
+-- (c t1 (i+1) j k 0) * ((b0 t0) (v3 t1)) +
+-- (c t1 i (j+1) k 0) * ((b1 t0) (v3 t1)) +
+-- (c t1 i j (k+1) 0) * ((b2 t0) (v3 t1)) +
+-- (c t1 i j k 1) * ((b3 t0) (v3 t1)) | i <- [0..2],
+-- j <- [0..2],
+-- k <- [0..2],
+-- i + j + k == 2]
+-- where
+-- t0 = tetrahedron0 cube
+-- t1 = tetrahedron1 cube
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