Add UE5 A.1
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code/Angabe5.hs
163
code/Angabe5.hs
@ -52,121 +52,132 @@ data PH_ElemTyp a b c d e = A a | B b | C c | D d | E e deriving (Eq,Show)
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data PH_ElemTyp' q r s = Q q | R r | S s deriving (Eq,Show)
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-- Simple helper functions
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roman_to_nat :: Zahlraum_0_10 -> Nat0
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roman_to_nat N = 0
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roman_to_nat I = 1
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roman_to_nat II = 2
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roman_to_nat III = 3
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roman_to_nat IV = 4
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roman_to_nat V = 5
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roman_to_nat VI = 6
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roman_to_nat VII = 7
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roman_to_nat VIII = 8
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roman_to_nat IX = 9
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roman_to_nat X = 10
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roman_to_nat F = -1
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nat_to_roman :: Nat0 -> Zahlraum_0_10
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nat_to_roman 0 = N
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nat_to_roman 1 = I
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nat_to_roman 2 = II
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nat_to_roman 3 = III
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nat_to_roman 4 = IV
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nat_to_roman 5 = V
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nat_to_roman 6 = VI
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nat_to_roman 7 = VII
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nat_to_roman 8 = VIII
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nat_to_roman 9 = IX
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nat_to_roman 10 = X
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nat_to_roman n
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| n > 10 = F
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| n < 0 = F
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| otherwise = error "Could not recognize input" -- this should not happen
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is_correct_num :: Zahlraum_0_10 -> Bool
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is_correct_num n =
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case n of
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F -> False
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_ -> True
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-- Aufgabe A.1
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instance Num Zahlraum_0_10 where
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...
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{- Knapp, aber gut nachvollziehbar geht die Instanzbildung fuer Num folgendermassen vor:
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...
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-}
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(+) n n' = add_r n n'
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(-) n n' = diff_r n n'
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(*) n n' = mult_r n n'
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fromInteger n = nat_to_roman (fromInteger n)
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add_r :: Zahlraum_0_10 -> Zahlraum_0_10 -> Zahlraum_0_10
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add_r n n'
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| is_correct_num n == False || is_correct_num n' == False = F
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| otherwise = nat_to_roman sum_r
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where
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a = roman_to_nat n
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b = roman_to_nat n'
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sum_r = a + b
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diff_r :: Zahlraum_0_10 -> Zahlraum_0_10 -> Zahlraum_0_10
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diff_r n n'
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| is_correct_num n == False || is_correct_num n' == False = F
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| otherwise = nat_to_roman dif_r
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where
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a = roman_to_nat n
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b = roman_to_nat n'
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dif_r = a - b
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mult_r :: Zahlraum_0_10 -> Zahlraum_0_10 -> Zahlraum_0_10
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mult_r n n'
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| is_correct_num n == False || is_correct_num n' == False = F
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| otherwise = nat_to_roman prod_r
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where
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a = roman_to_nat n
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b = roman_to_nat n'
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prod_r = a * b
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-- Aufgabe A.2
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instance Eq Funktion where
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...
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instance Show Funktion where
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...
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{- Knapp, aber gut nachvollziehbar gehen die beiden Instanzbildungen fuer
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Eq und Show folgendermassen vor:
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...
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-}
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-- instance Eq Funktion where
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-- ...
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-- instance Show Funktion where
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-- ...
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-- Aufgabe A.3
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instance Menge_von Int where
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...
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instance Menge_von Zahlraum_0_10 where
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...
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instance Menge_von Funktion where
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...
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{- Knapp, aber gut nachvollziehbar gehen die drei Instanzbildungen fuer
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Menge_von folgendermassen vor:
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...
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-}
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-- instance Menge_von Int where
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-- ...
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-- instance Menge_von Zahlraum_0_10 where
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-- ...
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-- instance Menge_von Funktion where
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-- ...
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-- Aufgabe A.4
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instance (Eq a,Eq b) => Menge_von (Paar a b) where
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...
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instance Eq a => Menge_von (Baum a) where
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...
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{- Knapp, aber gut nachvollziehbar gehen die beiden Instanzbildungen fuer
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Menge_von folgendermassen vor:
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...
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-}
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-- instance (Eq a,Eq b) => Menge_von (Paar a b) where
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-- ...
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-- instance Eq a => Menge_von (Baum a) where
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-- ...
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-- Aufgabe A.5
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instance Eq a => Eq (ElemTyp a) where
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...
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instance Show a => Show (ElemTyp a) where
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...
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{- Knapp, aber gut nachvollziehbar gehen die beiden Instanzbildungen fuer
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Eq und Show folgendermassen vor:
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...
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-}
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-- instance Eq a => Eq (ElemTyp a) where
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-- ...
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-- instance Show a => Show (ElemTyp a) where
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-- ...
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-- Aufgabe A.6
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instance Eq a => Menge_von (ElemTyp a) where
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...
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{- Knapp, aber gut nachvollziehbar geht die Instanzbildung fuer
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Menge_von folgendermassen vor:
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...
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-}
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-- instance Eq a => Menge_von (ElemTyp a) where
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-- ...
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-- Aufgabe A.7
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instance (Eq a,Eq b,Eq c,Eq d,Eq e) => Menge_von (PH_ElemTyp a b c d e) where
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...
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{- Knapp, aber gut nachvollziehbar geht die Instanzbildung fuer
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Menge_von folgendermassen vor:
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...
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-}
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-- instance (Eq a,Eq b,Eq c,Eq d,Eq e) => Menge_von (PH_ElemTyp a b c d e) where
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-- ...
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-- Aufgabe A.8
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instance (Eq p,Eq q,Eq r) => Menge_von (PH_ElemTyp' p q r) where ...
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{- Knapp, aber gut nachvollziehbar geht die Instanzbildung fuer
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Menge_von folgendermassen vor:
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...
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-}
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-- instance (Eq p,Eq q,Eq r) => Menge_von (PH_ElemTyp' p q r) where ...
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