class Eq a where
(==), (/=) :: a -> a -> Bool
x /= y = not (x == y)instance Eq Bool where
False == False = True
True == True = True
_ == _ = FalseBool is an instance of the Eq class, with (==) defined as above.
Only types that are defined using the data or newtype mechanisms can be made instances of classes.
Classes can also be extended to form new classes. For example, the class Ord of types whose values are totally ordered is declared as an extension of Eq class.
class Eq a => Ord a where
(<), (<=), (>), (>=) :: a -> a -> Bool
min, max :: a -> a -> a
min x y | x <= y = x
| otherwise = y
max x y | x <= y = y
| otherwise = x
instance Ord Bool where
False < True = True
_ < _ = False
b <= c = (b < c) || (b == c)
b > c = c < b
b >= c = c <= b
When new types are declared, it is usually appropriate to make them instances of a number of built-in classes using the "deriving" clause.
data Bool = False | True deriving (Eq, Ord, Show, Read)False == False
False < True
show False
read "False" :: Bool
The use of :: Bool in the above example is necessary as the read function needs to know what type it needs to convert the string "False" into.
For the purposes of deriving instances of the Ord type, the ordering of the constructors of a type is determined by the position in the declaration. So, False would be "less-than" True
In the case of constructors that have arguments, the types of these arguments should also be instance of any derived classes.
For example, in
data Shape = Circle Float | Rect Float Floatto derive Shape as an equality type requires Float also as deriving equality type.
or in
data Maybe a = Nothing | Just ato derive Maybe as an equality type, the type argument a should also derive equality.
data Shape = Circle Float | Rectangle Float Float deriving (Eq, Ord, Show)
Rectangle 1.0 4.0 == Rectangle 2.0 3.0
Rectangle 1.0 4.0 > Rectangle 2.0 3.0
-- Note this example works in interactive ghci on terminal; has some issues in jupyter lab
Circle 3.6 == Circle 5.5
show (Circle 4.4)
Consider a language of propositions built up from basic values (False, True) and variables (A,B,...,Z) using "not", "and", "implication" and parentheses. For example the following are propositions:
A and not A
(A and B) -> A
A -> (A and B)
(A and (A -> B)) -> BRecall from your Discrete Math class the meaning of such propositions (truth tables)
data Prop = Const Bool
| Var Char
| Not Prop
| And Prop Prop
| Imply Prop Prop deriving Show
Note that we do not need to define parentheses as we can use the inbuilt Haskell parentheses. Using this type, we can instantiate above propositions as follows:
p1 :: Prop
p1 = And (Var 'A') (Not (Var 'A'))
p2 :: Prop
p2 = Imply (And (Var 'A') (Var 'B')) (Var 'A')
p3 :: Prop
p3 = Imply (Var 'A') (And (Var 'A') (Var 'B'))
p4 :: Prop
p4 = Imply (And (Var 'A') (Imply (Var 'A') (Var 'B'))) (Var 'B')
show p4
type Assoc k v = [(k,v)]
find :: Eq k => k -> Assoc k v -> v
find k t = head [v | (k',v) <- t, k == k']
type Subst = Assoc Char Bool
for example, the substitution [('A',False),('B',True)] assigns boolean values to variables. We can now define the following function, eval, which given a substitution and a proposition, evaluates the proposition based on the substitution to result in a boolean value.
eval :: Subst -> Prop -> Bool
eval _ (Const b) = b
eval s (Var x) = find x s
eval s (Not p) = not (eval s p)
eval s (And p q) = eval s p && eval s q
eval s (Imply p q) = eval s p <= eval s q
eval [('A',False),('B',True)] p2
eval [('A',True),('B',False)] p3
-- function to return all variables in a proposition
vars :: Prop -> [Char]
vars (Const _) = []
vars (Var x) = [x]
vars (Not p) = vars p
vars (And p q) = vars p ++ vars q
vars (Imply p q) = vars p ++ vars q
vars p1
vars p2
vars p2
vars p4
Note: We will get rid of duplicates later
-- function to generate all combinations of False and True for n variables
-- method: think of each combination as an integer in binary notation with 0=False,1=True
type Bit = Int
int2bin :: Int -> [Bit]
int2bin 0 = []
int2bin n = n `mod` 2 : int2bin (n `div` 2)
bools1 :: Int -> [[Bool]]
bools1 n = map (reverse . map conv . make n . int2bin) range
where
range = [0..(2^n)-1]
make n bs = take n (bs ++ repeat 0)
conv 0 = False
conv 1 = True
bools1 3
-- A simpler and recursive definition of bools
bools :: Int -> [[Bool]]
bools 0 = [[]]
bools n = map (False:) bss ++ map (True:) bss
where bss = bools (n-1)
bools 3
-- now we generate all possible substitions for variables
rmdups :: Eq a => [a] -> [a]
rmdups [] = []
rmdups (x:xs) = x : filter (/= x) (rmdups xs)
substs :: Prop -> [Subst]
substs p = map (zip vs) (bools (length vs))
where vs = rmdups (vars p)
substs p1
substs p2
isTaut :: Prop -> Bool
isTaut p = and [eval s p | s <- substs p]
isTaut p1
isTaut p2
isTaut p3
isTaut p4
data Expr = Val Int | Add Expr Expr
value :: Expr -> Int
value (Val n) = n
value (Add x y) = value x + value y
e1 = Add (Add (Val 2) (Val 3)) (Val 4)
value e1
value (Add (Add (Val 2) (Val 3)) (Val 4))
= value (Add (Val 2) (Val 3)) + value (Val 4)
= value (Val 2) + value (Val 3) + value (Val 4)
= (2 + value (Val 3)) + value (Val 4)
= (2 + 3) + value (Val 4)
= 5 + value (Val 4)
= 5 + 4
= 9Note that the definition of value does not specify the order of operations; So, Haskell selects its own order; in this case left to right.
We introduce a method that uses a control stack containing operations that introduce a particular order of operation.
type Cont = [Op]
data Op = EVAL Expr | ADD Int
eval :: Expr -> Cont -> Int
eval (Val n) c = exec c n
eval (Add x y) c = eval x (EVAL y : c)
exec :: Cont -> Int -> Int
exec [] n = n
exec (EVAL y : c) n = eval y (ADD n : c)
exec (ADD n : c) m = exec c (n + m)
value1 :: Expr -> Int
value1 e = eval e []
value1 e1
value1 (Add (Add (Val 2) (Val 3)) (Val 4))
= eval (Add (Add (Val 2) (Val 3)) (Val 4)) []
= eval (Add (Val 2) (Val 3)) [EVAL (Val 4)]
= eval (Val 2) [EVAL (Val 3), EVAL (Val 4)]
= exec [EVAL (Val 3), EVAL (Val 4)] 2
= eval (Val 3) [ADD 2, EVAL (Val 4)]
= exec [ADD 2, EVAL (Val 4)] 3
= exec [EVAL (Val 4)] 5
= eval (Val 4) [ADD 5]
= exec [ADD 5] 4
= exec [] 9
= 9