在Haskell中生成逻辑表达式的真值表

在Haskell中生成逻辑表达式的真值表,haskell,logic,boolean-logic,boolean-expression,Haskell,Logic,Boolean Logic,Boolean Expression,第一部分是具有以下类型签名的求值函数: evaluate :: Logic Expr -> [(Variable, Bool)] -> Bool 它将逻辑表达式和赋值对列表作为输入,并根据提供的布尔赋值返回表达式的值。赋值列表是一个不同的对列表,其中每对包含一个变量及其布尔赋值。也就是说,如果将表达式A传递给函数∧ 当赋值A=1和B=0时,函数必须返回0(这来自数字逻辑设计,0对应于false,1对应于true) 到目前为止,我做到了这一点: type Variable = Ch

第一部分是具有以下类型签名的求值函数:

evaluate :: Logic Expr -> [(Variable, Bool)] -> Bool
它将逻辑表达式和赋值对列表作为输入,并根据提供的布尔赋值返回表达式的值。赋值列表是一个不同的对列表,其中每对包含一个变量及其布尔赋值。也就是说,如果将表达式A传递给函数∧ 当赋值A=1和B=0时,函数必须返回0(这来自数字逻辑设计,0对应于false,1对应于true)

到目前为止,我做到了这一点:

type Variable =  Char

data LogicExpr = V Variable
                 | Negation  LogicExpr
                 | Conjunction LogicExpr LogicExpr
                 | Disjunction  LogicExpr LogicExpr 
                 | Implication  LogicExpr LogicExpr 


evaluate :: LogicExpr -> [(Variable,Bool)] -> Bool

evaluate (V a) ((x1,x2):xs) | a==x1 = x2
                            | otherwise = (evaluate(V a)xs)

evaluate (Negation a) l | (evaluate a l)==True = False
                        | otherwise = True

evaluate (Conjunction a b) l = (evaluate a l)&&(evaluate b l)

evaluate (Disjunction a b) l = (evaluate a l)||(evaluate b l)

evaluate (Implication a b) l
    | (((evaluate b l)==False)&&((evaluate a l)==True)) = False
    | otherwise = True
下一部分是定义
generateTrustTable
,该函数以逻辑表达式作为输入,并以赋值对列表的形式返回表达式的真值表。也就是说,如果将表达式E=A传递给函数∧ B、 函数必须返回A=0,B=0,E=0 | A=0,B=1,E=0 | A=1,B=0,E=0 | A=1,B=1,E=1


我不太熟悉语法,所以不知道如何返回列表。

标准库函数,代码重用。此外,括号的用法和间距也确实有问题

evaluate (V a) l =
    case lookup a l
      of Just x -> x
         Nothing -> error $ "Unbound variable: " ++ show a
-- same as
evaluate (V a) l = maybe (error $ "Unbound variable: " ++ show a) id $ lookup a l

evaluate (Negation a) l = not $ evaluate a l

evaluate (Implication a b) l = evaluate (Negation a `Disjunction` b) l
现在,您想要一个
generateTrustTable
?这很简单,只需获取布尔变量的所有可能状态,并将求值表达式固定到每个变量的末尾

generateTruthTable :: [Variable] -> LogicExpr -> [[(Variable, Bool)]]
generateTruthTable vs e = [l ++ [('E', evaluate e l)] | l <- allPossible vs]
根据我的直觉,这感觉应该是一种变形。毕竟,它确实需要查看列表中的所有内容,但返回的是不同结构的内容,并且它可能可以用一种简单的方式进行分解,因为这是一个介绍级CS类。(我不在乎课程号是多少,这是介绍性的东西。)

现在,
foldr::(a->b->b)->b->[a]->b
,因此前两个参数必须是
step::a->b->b
initial::b
。现在,
所有可能的::[Variable]->[[(Variable,Bool)]]=foldr步骤初始::[a]->b
。嗯,这一定意味着
a=Variable
b=[[(Variable,Bool)]]
。这对于
步骤
初始
意味着什么

    step :: Variable -> [[(Variable, Bool)]] -> [[(Variable, Bool)]]
    initial :: [[(Variable, Bool)]]
有趣。不知何故,需要有一种方法从变量状态列表中
step
并向其中添加单个变量,以及一些
初始
列表中完全没有变量

如果您的大脑已经成功地“点击”到函数式编程范式中,那么这就足够了。如果没有,那么在作业到期的几小时内,不管你在这里收到了什么指示,你都会被搞砸。祝你好运,如果作业到期后你还没做完,你应该问你的教授,或者在这里问一个非紧急问题


如果您对该语言存在基本的可用性问题(“语法是什么”、“运行时语义是什么”、“xxx是否已有功能”等):

  • 是对基础语言和库的免费、规范的定义。网站上提供了更多链接
  • 有关98年后的语言扩展,请参阅
  • GHC、Hugs和其他现代Haskell实现也提供了比Haskell 98中指定的更丰富的标准库。的完整文档也可在线获取
  • 是扩展Haskell标准库的专用搜索引擎。与此类似,但也涵盖了Haskell库的集合,远远超出了标准发行版
我希望你们的类提供了类似的资源,但如果没有,以上所有内容都很容易从谷歌搜索中发现

如果有适当的参考资料,任何有价值的程序员都应该能够在几个小时内学会任何新语言的语法,并在几天内对运行时有一个有效的理解。当然,掌握一个新的范式可能需要很多时间,要求学生达到同样的标准有些不公平,但这正是这门课的目的

关于堆栈溢出的更高级别问题的答案可能会更少,但它们也不会那么烦躁:)在大多数人眼里,家庭作业问题被归类为“为我做我的工作!”


扰流器 请不要作弊。然而,只是想让你尝一尝在Haskell中可以做多么棒的事情

{-# LANGUAGE FlexibleInstances, UndecidableInstances #-}
{-# LANGUAGE OverlappingInstances, PatternGuards #-}

module Expr (Ring(..), (=:>), Expr(..), vars, eval, evalAll) where

import Control.Monad.Error

infixl 5 =:>, :=>
infixl 6 +:, -:, :+, :-
infixl 7 *:, :*

class (Eq a) => Ring a where
    (+:) :: a -> a -> a; (-:) :: a -> a -> a; x -: y = x +: invert y
    (*:) :: a -> a -> a; invert :: a -> a; invert x = zero -: x
    zero :: a; one :: a
(=:>) :: (Ring a) => a -> a -> a
(=:>) = flip (-:)

instance (Num a) => Ring a where
    (+:) = (+); (-:) = (-); (*:) = (*)
    invert = negate; zero = 0; one = 1

instance Ring Bool where
    (+:) = (||); (*:) = (&&)
    invert = not; zero = False; one = True

data Expr a b
  = Expr a b :+ Expr a b | Expr a b :- Expr a b
  | Expr a b :* Expr a b | Expr a b :=> Expr a b
  | Invert (Expr a b) | Var a | Const b

paren :: ShowS -> ShowS
paren ss s = '(' : ss (')' : s)

instance (Show a, Show b) => Show (Expr a b) where
    showsPrec _ (Const c) = ('@':) . showsPrec 9 c
    showsPrec _ (Var v) = ('$':) . showsPrec 9 v
    showsPrec _ (Invert e) = ('!':) . showsPrec 9 e

    showsPrec n e@(a:=>b)
      | n > 5 = paren $ showsPrec 0 e
      | otherwise = showsPrec 7 a . ('=':) . ('>':) . showsPrec 5 b

    showsPrec n e@(a:*b)
      | n > 7 = paren $ showsPrec 0 e
      | otherwise = showsPrec 7 a . ('*':) . showsPrec 7 b

    showsPrec n e | n > 6 = paren $ showsPrec 0 e
    showsPrec _ (a:+b) = showsPrec 6 a . ('+':) . showsPrec 6 b
    showsPrec _ (a:-b) = showsPrec 6 a . ('-':) . showsPrec 6 b

vars :: (Eq a) => Expr a b -> [a]
vars (a:+b) = vars a ++ vars b
vars (a:-b) = vars a ++ vars b
vars (a:*b) = vars a ++ vars b
vars (a:=>b) = vars a ++ vars b
vars (Invert e) = vars e; vars (Var v) = [v]; vars _ = []

eval :: (Eq a, Show a, Ring b, Monad m) => [(a, b)] -> Expr a b -> m b
eval m (a:+b) = return (+:) `ap` eval m a `ap` eval m b
eval m (a:-b) = return (-:) `ap` eval m a `ap` eval m b
eval m (a:*b) = return (*:) `ap` eval m a `ap` eval m b
eval m (a:=>b) = return (=:>) `ap` eval m a `ap` eval m b
eval m (Invert e) = return invert `ap` eval m e
eval m (Var v)
  | Just c <- lookup v m = return c
  | otherwise = fail $ "Unbound variable: " ++ show v
eval _ (Const c) = return c

namedProduct :: [(a, [b])] -> [[(a, b)]]
namedProduct = foldr (\(v, cs) l -> concatMap (\c -> map ((v, c):) l) cs) [[]]

evalAll :: (Eq a, Show a, Ring b) => [b] -> a -> Expr a b -> [[(a, b)]]
evalAll range name e =
    [ vs ++ [(name, either error id $ eval vs e)]
    | vs <- namedProduct $ zip (vars e) (repeat range)
    ]
{-#语言灵活实例,不可判定实例#-}
{-#语言重叠实例,PatternGuards}
模块Expr(环(..),(=:>),Expr(..),vars,eval,evalAll)其中
导入控制.Monad.Error
infixl 5=:>,:=>
infixl 6+:,-:,:+,:-
infixl 7*:,:*
类别(等式a)=>环a,其中
(+:)::a->a->a;(-)::a->a->a;x-:y=x+:反向y
(*:)::a->a->a;反转::a->a;反转x=零-:x
零::a;a::a
(=:>)::(环a)=>a->a->a
(=:>)=翻转(-:)
实例(Num a)=>环a,其中
(+:) = (+); (-:) = (-); (*:) = (*)
倒置=否定;零=0;一=1
实例环布尔在哪里
(+:) = (||); (*:) = (&&)
反转=不;零=假;一=真
数据表达式a b
=Expr a b:+Expr a b | Expr a b:-Expr a b
|Expr a b:*Expr a b | Expr a b:=>Expr a b
|反向(表达式a b)|变量a |常数b
paren::ShowS->ShowS
paren ss s='(':ss(')':s)
实例(Show a,Show b)=>Show(Expr a b)其中
showsPrec(Const c)=(@:)。showsPrec 9 c
showsPrec(Var v)=(“$”:)。showsPrec 9 v
showsPrec(倒e)=(“!”:)。showsPrec 9 e
showsPrec n e@(a:=>b)
|n>5=价格$showsPrec 0 e
|否则=显示建议7 a。('=':) . ('>':) . showsPrec 5 b
展品展览(a:*b)
|n>7=paren$showsPrec 0 e
|否则=显示建议7 a。('*':) . showsPrec 7 b
showsPrec n e | n>6=paren$showsPrec 0 e
showsPrec(a:+b)=showsPrec 6 a。('+':) . showsPrec 6 b
showsPrec(a:-b)=showsPrec 6 a。('-':) . 表演
    step :: Variable -> [[(Variable, Bool)]] -> [[(Variable, Bool)]]
    initial :: [[(Variable, Bool)]]
{-# LANGUAGE FlexibleInstances, UndecidableInstances #-}
{-# LANGUAGE OverlappingInstances, PatternGuards #-}

module Expr (Ring(..), (=:>), Expr(..), vars, eval, evalAll) where

import Control.Monad.Error

infixl 5 =:>, :=>
infixl 6 +:, -:, :+, :-
infixl 7 *:, :*

class (Eq a) => Ring a where
    (+:) :: a -> a -> a; (-:) :: a -> a -> a; x -: y = x +: invert y
    (*:) :: a -> a -> a; invert :: a -> a; invert x = zero -: x
    zero :: a; one :: a
(=:>) :: (Ring a) => a -> a -> a
(=:>) = flip (-:)

instance (Num a) => Ring a where
    (+:) = (+); (-:) = (-); (*:) = (*)
    invert = negate; zero = 0; one = 1

instance Ring Bool where
    (+:) = (||); (*:) = (&&)
    invert = not; zero = False; one = True

data Expr a b
  = Expr a b :+ Expr a b | Expr a b :- Expr a b
  | Expr a b :* Expr a b | Expr a b :=> Expr a b
  | Invert (Expr a b) | Var a | Const b

paren :: ShowS -> ShowS
paren ss s = '(' : ss (')' : s)

instance (Show a, Show b) => Show (Expr a b) where
    showsPrec _ (Const c) = ('@':) . showsPrec 9 c
    showsPrec _ (Var v) = ('$':) . showsPrec 9 v
    showsPrec _ (Invert e) = ('!':) . showsPrec 9 e

    showsPrec n e@(a:=>b)
      | n > 5 = paren $ showsPrec 0 e
      | otherwise = showsPrec 7 a . ('=':) . ('>':) . showsPrec 5 b

    showsPrec n e@(a:*b)
      | n > 7 = paren $ showsPrec 0 e
      | otherwise = showsPrec 7 a . ('*':) . showsPrec 7 b

    showsPrec n e | n > 6 = paren $ showsPrec 0 e
    showsPrec _ (a:+b) = showsPrec 6 a . ('+':) . showsPrec 6 b
    showsPrec _ (a:-b) = showsPrec 6 a . ('-':) . showsPrec 6 b

vars :: (Eq a) => Expr a b -> [a]
vars (a:+b) = vars a ++ vars b
vars (a:-b) = vars a ++ vars b
vars (a:*b) = vars a ++ vars b
vars (a:=>b) = vars a ++ vars b
vars (Invert e) = vars e; vars (Var v) = [v]; vars _ = []

eval :: (Eq a, Show a, Ring b, Monad m) => [(a, b)] -> Expr a b -> m b
eval m (a:+b) = return (+:) `ap` eval m a `ap` eval m b
eval m (a:-b) = return (-:) `ap` eval m a `ap` eval m b
eval m (a:*b) = return (*:) `ap` eval m a `ap` eval m b
eval m (a:=>b) = return (=:>) `ap` eval m a `ap` eval m b
eval m (Invert e) = return invert `ap` eval m e
eval m (Var v)
  | Just c <- lookup v m = return c
  | otherwise = fail $ "Unbound variable: " ++ show v
eval _ (Const c) = return c

namedProduct :: [(a, [b])] -> [[(a, b)]]
namedProduct = foldr (\(v, cs) l -> concatMap (\c -> map ((v, c):) l) cs) [[]]

evalAll :: (Eq a, Show a, Ring b) => [b] -> a -> Expr a b -> [[(a, b)]]
evalAll range name e =
    [ vs ++ [(name, either error id $ eval vs e)]
    | vs <- namedProduct $ zip (vars e) (repeat range)
    ]
$ ghci GHCi, version 6.10.2: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Prelude> :l Expr.hs [1 of 1] Compiling Expr ( Expr.hs, interpreted ) Ok, modules loaded: Expr. *Expr> mapM_ print . evalAll [1..3] 'C' $ Var 'A' :* Var 'B' Loading package mtl-1.1.0.2 ... linking ... done. [('A',1),('B',1),('C',1)] [('A',1),('B',2),('C',2)] [('A',1),('B',3),('C',3)] [('A',2),('B',1),('C',2)] [('A',2),('B',2),('C',4)] [('A',2),('B',3),('C',6)] [('A',3),('B',1),('C',3)] [('A',3),('B',2),('C',6)] [('A',3),('B',3),('C',9)] *Expr> let expr = Var 'A' :=> (Var 'B' :+ Var 'C') :* Var 'D' *Expr> expr $'A'=>($'B'+$'C')*$'D' *Expr> mapM_ print $ evalAll [True, False] 'E' expr [('A',True),('B',True),('C',True),('D',True),('E',True)] [('A',True),('B',True),('C',True),('D',False),('E',False)] [('A',True),('B',True),('C',False),('D',True),('E',True)] [('A',True),('B',True),('C',False),('D',False),('E',False)] [('A',True),('B',False),('C',True),('D',True),('E',True)] [('A',True),('B',False),('C',True),('D',False),('E',False)] [('A',True),('B',False),('C',False),('D',True),('E',False)] [('A',True),('B',False),('C',False),('D',False),('E',False)] [('A',False),('B',True),('C',True),('D',True),('E',True)] [('A',False),('B',True),('C',True),('D',False),('E',True)] [('A',False),('B',True),('C',False),('D',True),('E',True)] [('A',False),('B',True),('C',False),('D',False),('E',True)] [('A',False),('B',False),('C',True),('D',True),('E',True)] [('A',False),('B',False),('C',True),('D',False),('E',True)] [('A',False),('B',False),('C',False),('D',True),('E',True)] [('A',False),('B',False),('C',False),('D',False),('E',True)]
import Data.Maybe (fromJust)
import Data.List (nub)

type Variable = Char
data LogicExpr
   = Var Variable
   | Neg LogicExpr
   | Conj LogicExpr LogicExpr
   | Disj LogicExpr LogicExpr
   | Impl LogicExpr LogicExpr
   deriving (Eq, Ord)

-- evaluates an expression
evaluate :: LogicExpr -> [(Variable, Bool)] -> Bool
evaluate (Var v) bs      = fromJust (lookup v bs)
evaluate (Neg e) bs      = not (evaluate e bs)
evaluate (Conj e1 e2) bs = evaluate e1 bs && evaluate e2 bs
evaluate (Disj e1 e2) bs = evaluate e1 bs || evaluate e2 bs
evaluate (Impl e1 e2) bs = not (evaluate e1 bs) || evaluate e2 bs
-- get variables in an expression
varsp :: LogicExpr -> [Variable]
varsp (Var v)      = [v]
varsp (Neg e)      = varsp e
varsp (Conj e1 e2) = varsp e1 ++ varsp e2
varsp (Disj e1 e2) = varsp e1 ++ varsp e2
varsp (Impl e1 e2) = varsp e1 ++ varsp e2

-- get variables in an expression without duplicates
vars :: LogicExpr -> [Variable]
vars = nub . varsp

-- possible boolean values
bools = [True, False]

-- all possible combinations of variable assignments
booltable :: [Variable] -> [[(Variable, Bool)]]
booltable [] = [[]]
booltable (a:as) = [(a,b) : r | b <- bools, r <- booltable as]

-- variable assignments and corresponding evaluation of an expression
truthtable :: LogicExpr -> [([(Variable, Bool)], Bool)]
truthtable e = [(bs, evaluate e bs) | bs <- booltable (vars e)]
-- read a right-associative infix operator
readInfix opprec constr repr prec r
   = readParen (prec > opprec)
     (\r -> [(constr e1 e2, u) |
             (e1,s) <- readsPrec (opprec+1) r,
             (op,t) <- lex s,
             op == repr,
             (e2,u) <- readsPrec (opprec) t]) r

instance Read LogicExpr where
   readsPrec prec r
      =  readInfix 1 Impl "->" prec r
      ++ readInfix 2 Disj "|" prec r
      ++ readInfix 3 Conj "&" prec r
      ++ readParen (prec > 4)
         (\r -> [(Neg e, t) |
                 ("!",s) <- lex r,
                 (e,t)   <- readsPrec 4 s]) r
      ++ readParen (prec > 5)
         (\r -> [(Var v, s) |
                 ([v], s) <- lex r]) r
showcell :: (Variable, Bool) -> String
showcell (v,b) = v : "=" ++ show b

showrow :: [(Variable, Bool)] -> Bool -> String
showrow []     b = show b
showrow [a]    b = showcell a ++ " => " ++ show b
showrow (a:as) b = showcell a ++ " && " ++ showrow as b

printrow :: ([(Variable, Bool)], Bool) -> IO ()
printrow = putStrLn . uncurry showrow

printtbl :: [([(Variable, Bool)], Bool)] -> IO ()
printtbl = mapM_ printrow
Prelude Main> printtbl $ truthtable $ read "(a -> b) & (b -> a)"
a=True && b=True => True
a=True && b=False => False
a=False && b=True => False
a=False && b=False => True

Prelude Main> printtbl $ truthtable $ read "(a | b) | (!a & !b)"
a=True && b=True => True
a=True && b=False => True
a=False && b=True => True
a=False && b=False => True