Copyright | (c) 2019-2024 Rudy Matela |
---|---|
License | 3-Clause BSD (see the file LICENSE) |
Maintainer | Rudy Matela <rudy@matela.com.br> |
Safe Haskell | Safe-Inferred |
Language | Haskell2010 |
Utilities for mapping or transforming Expr
s.
Synopsis
- mapValues :: (Expr -> Expr) -> Expr -> Expr
- mapVars :: (Expr -> Expr) -> Expr -> Expr
- mapConsts :: (Expr -> Expr) -> Expr -> Expr
- mapSubexprs :: (Expr -> Maybe Expr) -> Expr -> Expr
- (//-) :: Expr -> [(Expr, Expr)] -> Expr
- (//) :: Expr -> [(Expr, Expr)] -> Expr
- renameVarsBy :: (String -> String) -> Expr -> Expr
Documentation
mapValues :: (Expr -> Expr) -> Expr -> Expr Source #
O(n*m).
Applies a function to all terminal values in an expression.
(cf. //-
)
Given that:
> let zero = val (0 :: Int) > let one = val (1 :: Int) > let two = val (2 :: Int) > let three = val (3 :: Int) > let xx -+- yy = value "+" ((+) :: Int->Int->Int) :$ xx :$ yy > let intToZero e = if typ e == typ zero then zero else e
Then:
> one -+- (two -+- three) 1 + (2 + 3) :: Int
> mapValues intToZero $ one -+- (two -+- three) 0 + (0 + 0) :: Integer
Given that the argument function is O(m), this function is O(n*m).
mapVars :: (Expr -> Expr) -> Expr -> Expr Source #
O(n*m). Applies a function to all variables in an expression.
Given that:
> let primeify e = if isVar e | then case e of (Value n d) -> Value (n ++ "'") d | else e > let xx = var "x" (undefined :: Int) > let yy = var "y" (undefined :: Int) > let xx -+- yy = value "+" ((+) :: Int->Int->Int) :$ xx :$ yy
Then:
> xx -+- yy x + y :: Int
> primeify xx x' :: Int
> mapVars primeify $ xx -+- yy x' + y' :: Int
> mapVars (primeify . primeify) $ xx -+- yy x'' + y'' :: Int
Given that the argument function is O(m), this function is O(n*m).
mapConsts :: (Expr -> Expr) -> Expr -> Expr Source #
O(n*m). Applies a function to all terminal constants in an expression.
Given that:
> let one = val (1 :: Int) > let two = val (2 :: Int) > let xx -+- yy = value "+" ((+) :: Int->Int->Int) :$ xx :$ yy > let intToZero e = if typ e == typ zero then zero else e
Then:
> one -+- (two -+- xx) 1 + (2 + x) :: Int
> mapConsts intToZero (one -+- (two -+- xx)) 0 + (0 + x) :: Integer
Given that the argument function is O(m), this function is O(n*m).
mapSubexprs :: (Expr -> Maybe Expr) -> Expr -> Expr Source #
O(n*m).
Substitute subexpressions of an expression using the given function.
Outer expressions have more precedence than inner expressions.
(cf. //
)
With:
> let xx = var "x" (undefined :: Int) > let yy = var "y" (undefined :: Int) > let zz = var "z" (undefined :: Int) > let plus = value "+" ((+) :: Int->Int->Int) > let times = value "*" ((*) :: Int->Int->Int) > let xx -+- yy = plus :$ xx :$ yy > let xx -*- yy = times :$ xx :$ yy
> let pluswap (o :$ xx :$ yy) | o == plus = Just $ o :$ yy :$ xx | pluswap _ = Nothing
Then:
> mapSubexprs pluswap $ (xx -*- yy) -+- (yy -*- zz) y * z + x * y :: Int
> mapSubexprs pluswap $ (xx -+- yy) -*- (yy -+- zz) (y + x) * (z + y) :: Int
Substitutions do not stack, in other words a replaced expression or its subexpressions are not further replaced:
> mapSubexprs pluswap $ (xx -+- yy) -+- (yy -+- zz) (y + z) + (x + y) :: Int
Given that the argument function is O(m), this function is O(n*m).
(//-) :: Expr -> [(Expr, Expr)] -> Expr Source #
O(n*m).
Substitute occurrences of values in an expression
from the given list of substitutions.
(cf. mapValues
)
Given that:
> let xx = var "x" (undefined :: Int) > let yy = var "y" (undefined :: Int) > let zz = var "z" (undefined :: Int) > let xx -+- yy = value "+" ((+) :: Int->Int->Int) :$ xx :$ yy
Then:
> ((xx -+- yy) -+- (yy -+- zz)) //- [(xx, yy), (zz, yy)] (y + y) + (y + y) :: Int
> ((xx -+- yy) -+- (yy -+- zz)) //- [(yy, yy -+- zz)] (x + (y + z)) + ((y + z) + z) :: Int
This function does not work for substituting non-terminal subexpressions:
> (xx -+- yy) //- [(xx -+- yy, zz)] x + y :: Int
Please use the slower //
if you want the above replacement to work.
Replacement happens only once:
> xx //- [(xx,yy), (yy,zz)] y :: Int
Given that the argument list has length m, this function is O(n*m).
(//) :: Expr -> [(Expr, Expr)] -> Expr Source #
O(n*n*m).
Substitute subexpressions in an expression
from the given list of substitutions.
(cf. mapSubexprs
).
Please consider using //-
if you are replacing just terminal values
as it is faster.
Given that:
> let xx = var "x" (undefined :: Int) > let yy = var "y" (undefined :: Int) > let zz = var "z" (undefined :: Int) > let xx -+- yy = value "+" ((+) :: Int->Int->Int) :$ xx :$ yy
Then:
> ((xx -+- yy) -+- (yy -+- zz)) // [(xx -+- yy, yy), (yy -+- zz, yy)] y + y :: Int
> ((xx -+- yy) -+- zz) // [(xx -+- yy, zz), (zz, xx -+- yy)] z + (x + y) :: Int
Replacement happens only once with outer expressions having more precedence than inner expressions.
> (xx -+- yy) // [(yy,xx), (xx -+- yy,zz), (zz,xx)] z :: Int
Given that the argument list has length m, this function is O(n*n*m). Remember that since n is the size of an expression, comparing two expressions is O(n) in the worst case, and we may need to compare with n subexpressions in the worst case.
renameVarsBy :: (String -> String) -> Expr -> Expr Source #
Rename variables in an Expr
.
> renameVarsBy (++ "'") (xx -+- yy) x' + y' :: Int
> renameVarsBy (++ "'") (yy -+- (zz -+- xx)) (y' + (z' + x')) :: Int
> renameVarsBy (++ "1") (abs' xx) abs x1 :: Int
> renameVarsBy (++ "2") $ abs' (xx -+- yy) abs (x2 + y2) :: Int
NOTE: this will affect holes!