Remove Symbol typeclass
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6e09b77cc1
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1 changed files with 27 additions and 46 deletions
73
lambda.hs
73
lambda.hs
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@ -1,57 +1,37 @@
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import Data.List
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data StrSymbol = StrSymbol { symBase :: String -- lowercase a to z
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, symLen :: Int
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}
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-- A symbol denoting a variable; consists of a textual name and some (or no) apostrophes to ensure uniqueness
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data Symbol = Symbol { symBase :: String -- lowercase a to z
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, symLen :: Int -- nonnegative
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}
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instance Show StrSymbol where
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show (StrSymbol s n) = s ++ (replicate n '\'')
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instance Show Symbol where
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show (Symbol s n) = s ++ (replicate n '\'')
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class Symbol a where
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-- Should we use a set instead of a list here? ~G
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-- Hardly five minutes away and you're already overengineering. ~X
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-- Overengineering is fun! ~G
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-- Also, what is the second parameter? ~X
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-- The original symbol for which a new name should be found ~G (grammer is herd)
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-- Do we need that or can we just let it return a new symbol? ~X
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-- If we just let it return a new symbol, we would lose the thing where something called a will be renamed to a'
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-- and later back to a (keep the same name) ~G
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-- How about we define a findNewName that just finds a new unique symbol that is not in the list, so we can try out both
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-- things later on. ~G
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findName :: [a] -> a -> a
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findNewName :: [a] -> a
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instance Symbol StrSymbol where
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findName other (StrSymbol base n) =
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let sameBase = filter ((base ==) . symBase) other
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lengths = map symLen sameBase
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freeLengths = [0..] \\ (nub lengths)
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in StrSymbol base (head freeLengths) -- [0..] is infinite
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-- findNewName basically finds a new unique base ~G
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-- try to find a name in the sequence
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-- a..z, aa..zz, aaa........
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{-
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findNewName other =
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let bases = nub $ map symBase other
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-}
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-- Present symbols (not including symbol to change) -> symbol to change ->
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-- unique symbol with same base and minimal amount of apostrophes
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findName :: [Symbol] -> Symbol -> Symbol
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findName other (Symbol base n) =
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let sameBase = filter ((base ==) . symBase) other
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lengths = map symLen sameBase
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freeLengths = [0..] \\ (nub lengths)
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in Symbol base (head freeLengths) -- [0..] is infinite
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-- An expression. Can be a mere symbol, a lambda application, or a lambda abstraction.
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data Expression s = ESymbol s
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| EExpr (Expression s) (Expression s)
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| ELambda s (Expression s)
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instance (Show s) => Show (Expression s) where
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show (ESymbol s) = show s
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-- show (EExpr a e@(EExpr _ _)) = "(" ++ show a ++ " " ++ (drop 1 $ show e) -- hack hack hack
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show (EExpr a@(EExpr _ _) b) = (init $ show a) ++ " " ++ show b ++ ")" -- hack hack hack
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-- ((a b) c) is equivalent to (a b c)
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show (EExpr a@(EExpr _ _) b) = (init $ show a) ++ " " ++ show b ++ ")"
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show (EExpr a b) = "(" ++ show a ++ " " ++ show b ++ ")"
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show (ELambda s e) = "\\" ++ show s ++ "." ++ show e
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_s :: String -> Expression StrSymbol
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_s s = ESymbol (StrSymbol s 0)
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_ss :: String -> StrSymbol
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_ss s = (StrSymbol s 0)
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_s :: String -> Expression Symbol
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_s s = ESymbol $ Symbol s 0
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main = do
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print (EExpr (_s "a") (EExpr (_s "b") (_s "c")))
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@ -59,11 +39,12 @@ main = do
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print (EExpr (EExpr (_s "a") (_s "b")) (_s "c"))
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print (EExpr (EExpr (EExpr (_s "a") (_s "b")) (_s "c")) (_s "d"))
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print (EExpr (EExpr (_s "a") (EExpr (_s "b") (_s "c"))) (_s "d"))
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print (ELambda (StrSymbol "a" 0) (_s "a"))
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print (ELambda (StrSymbol "a" 0) (EExpr (EExpr (_s "a") (EExpr (_s "b") (_s "c"))) (_s "d")))
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print (ELambda (Symbol "a" 0) (_s "a"))
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print (ELambda (Symbol "a" 0) (EExpr (EExpr (_s "a") (EExpr (_s "b") (_s "c"))) (_s "d")))
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-- test of findName (seems to be working) ~G
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print $ findName [(StrSymbol "a" 0), (StrSymbol "b" 0), (StrSymbol "a" 1)] (StrSymbol "a" 4)
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print $ findName [(StrSymbol "a" 0), (StrSymbol "b" 0), (StrSymbol "a" 1)] (StrSymbol "b" 3)
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print $ findName [(StrSymbol "a" 0), (StrSymbol "b" 0), (StrSymbol "a" 1)] (StrSymbol "c" 2)
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print $ findName [(StrSymbol "a" 1), (StrSymbol "a" 3), (StrSymbol "a" 0)] (StrSymbol "a" 1)
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print $ findName [(Symbol "a" 0), (Symbol "b" 0), (Symbol "a" 1)] (Symbol "a" 4)
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print $ findName [(Symbol "a" 0), (Symbol "b" 0), (Symbol "a" 1)] (Symbol "b" 3)
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print $ findName [(Symbol "a" 0), (Symbol "b" 0), (Symbol "a" 1)] (Symbol "c" 2)
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print $ findName [(Symbol "a" 1), (Symbol "a" 3), (Symbol "a" 0)] (Symbol "a" 1)
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