Playing with regular expressions: Intersection

We implement symbolic intersection among regular expression
by adapting Antimirov's regular expression rewriting algorithm.
By symbolic we mean that regular expression are not transformed to
automata. We rewrite regular expressions by using partial derivatives.

The algorithm (highlights):

 r1 `intersect` r2 =

    let [l1,...,ln] = lettersOf (r1 | r2)
    in

          (l1, partDeriv r1 l1 `intersect` partDeriv r2 l1)
       | ...
       |  (l1, partDeriv r1 ln `intersect` partDeriv r2 ln)

The idea is that each regular expression is turned into a
'disjunction' of clauses where each clause starts with
a letter, followed by the partial derivative of the regular
expression wrt that letter.

You probably will know Brzozowski's derivative operation
which says that

 derivative r l = [ w | (l,w) in L(r) ]

If we would use derivatives the above algorithm
is non-terminating. For example,

     deriv (Star (L 'A')) 'A'
-->  <<>,'A'*>

     deriv (deriv (Star (L 'A')) 'A') 'A'
-->  (<{},'A'*>|<<>,'A'*>)

and so on. The nice property of partial derivatives is that
they are 'finite'. Each regular expression has a finite
number of partial derivatives. For example,

    partDeriv (Star (L 'A')) 'A'

--> <<>,'A'*>

    partDeriv (partDeriv (Star (L 'A')) 'A') 'A'
--> <<>,'A'*>

Please check the paper

Valentin M. Antimirov: Partial Derivatives of Regular Expressions and Finite Automaton Constructions.
Theor. Comput. Sci. 155(2): 291-319 (1996)

for more details on derivatives, partial derivatives and their connection
(roughly, derivatives correspond to DFA states and partial derivatives correspond to NFA states).


There's another source of non-termination which we explain via
the following example.

    A* `intersect` A*

--> (A, A* `intersect` A*)

where we simplified partDeriv A* A --> <<>,A*> to A*

The point to note is that the proof tree is infinite. The standard approach
is to use co-induction which then yields a finite (recursive) proof.

OK, here's finally the Haskell implementation:

> {-# LANGUAGE GADTs #-} 

We're just using the GADT syntax to write 'ordinary' algebraic data types

> module RegExpIntersection where

> import List


> data RE a where
>   Phi :: RE a                     -- empty language
>   Empty :: RE a                   -- empty word
>   L :: a -> RE a                  -- single letter taken from alphabet a
>   Choice :: RE a -> RE a -> RE a  -- r1 + r2
>   Seq :: RE a -> RE a -> RE a     -- (r1,r2)
>   Star :: RE a -> RE a            -- r*
>   Var :: Int -> RE a              -- first-order variables to represent regular equations
>  deriving Eq

A word is a list of alphabet letters

> type Word a = [a]

Pretty printing of regular expressions

> instance Show a => Show (RE a) where
>     show Phi = "{}"
>     show Empty = "<>"
>     show (L c) = show c
>     show (Choice r1 r2) = ("(" ++ show r1 ++ "|" ++ show r2 ++ ")")
>     show (Seq r1 r2) = ("<" ++ show r1 ++ "," ++ show r2 ++ ">")
>     show (Star r) = (show r ++ "*")

Turning a list of regular expressions into a regular expression by using | (choice)

> resToRE :: [RE a] -> RE a
> resToRE (r:res) = foldl Choice r res
> resToRE [] = Phi


Computing the set of letters of a regular expression

> sigmaRE :: Eq a => RE a -> [a]
> sigmaRE (L l) = [l]
> sigmaRE (Seq r1 r2) = nub ((sigmaRE r1) ++ (sigmaRE r2))
> sigmaRE (Choice r1 r2) = nub ((sigmaRE r1) ++ (sigmaRE r2))
> sigmaRE (Star r) = sigmaRE r
> sigmaRE Phi = []
> sigmaRE Empty = []


Testing if a regular expression is empty (empty word)

> isEmpty :: RE a -> Bool
> isEmpty Phi = False
> isEmpty Empty = True
> isEmpty (Choice r1 r2) = (isEmpty r1) || (isEmpty r2)
> isEmpty (Seq r1 r2) = (isEmpty r1) && (isEmpty r2)
> isEmpty (Star r) = True
> isEmpty (L _) = False

Testing if a regular expression contains nothing

> isPhi :: RE a -> Bool
> isPhi Phi = True
> isPhi Empty = False
> isPhi (Choice r1 r2) = (isPhi r1) && (isPhi r2)
> isPhi (Seq r1 r2) = (isPhi r1) || (isPhi r2)
> isPhi (Star r) = False
> isPhi (L _) = False


Brzozowski's derivative operation 
deriv r l denotes the regular expression where the 
"leading l has been removed"
(not necessary for the intersection algorithm,
only included to illustrate the difference
to the upcoming partial derivative algorithm)

> deriv :: Eq a => RE a -> a -> RE a
> deriv Phi _ = Phi
> deriv Empty _ = Phi
> deriv (L l1) l2
>    | l1 == l2  = Empty
>    | otherwise = Phi
> deriv (Choice r1 r2) l = 
>   Choice (deriv r1 l) (deriv r2 l)
> deriv (Seq r1 r2) l =
>   if isEmpty r1 
>   then Choice (Seq (deriv r1 l) r2) (deriv r2 l)
>   else Seq (deriv r1 l) r2
> deriv (this@(Star r)) l =
>   Seq (deriv r l) this



(a variant) of Antimirov's partial derivative operation

Antimirov demands that partDeriv (Star (L 'A')) 'A' yields [A*]
whereas our version yields [<<>,'A'*>].
The difference is not essential here.

> partDeriv :: Eq a => RE a -> a -> [RE a]
> partDeriv Phi l = []
> partDeriv Empty l = []
> partDeriv (L l') l 
>     | l == l'   = [Empty]
>     | otherwise = []
> partDeriv (Choice r1 r2) l = nub ((partDeriv r1 l) ++ (partDeriv r2 l))
> partDeriv (Seq r1 r2) l 
>     | isEmpty r1 = 
>           let s1 = [ (Seq r1' r2) | r1' <- partDeriv r1 l ]
>               s2 = partDeriv r2 l
>           in nub (s1 ++ s2)
>     | otherwise = [ (Seq r1' r2) | r1' <- partDeriv r1 l ]
> partDeriv (Star r) l = [ (Seq r' (Star r)) | r' <- partDeriv r l ]


Here's finally the partial derivative based intersection algorithm.


> type Env a = [((RE a, RE a), RE a)]


> intersectRE :: Eq a => RE a -> RE a -> RE a
> intersectRE r1 r2 = intersectC [] r1 r2

> intersectC :: Eq a => Env a -> RE a -> RE a -> RE a
> intersectC env r1 r2 
>  | r1 == Phi || r2 == Phi = Phi
>  | r1 == Empty || r2 == Empty = Empty
>  | otherwise =
>      case lookup (r1,r2) env of
>       Just r -> r
>       Nothing ->
>         let letters = sigmaRE (r1 `Choice` r2)
>             env' = ((r1,r2),r):env
>             r1l l = resToRE $ partDeriv r1 l
>             r2l l = resToRE $ partDeriv r2 l
>             r' = resToRE $ map (\l -> Seq (L l) (intersectC env' (r1l l) (r2l l))) letters
>             r =  if (isEmpty r1) && (isEmpty r2)
>                  then Choice r' Empty
>                  else r'
>         in r

The problem with the algorithm is that we use recursion in Haskell to model
recursive proofs. For example, try

intersect rAstar rAstar




Converting a regular equation into a regular expression

We assume that variables x appears (if at all) at position (r,Var x)
convert2 traveres the regexp and yields (r1,r2)
where the invariant is that r1 is part of the loop and r2 is the base case

> convert :: Int -> RE a -> RE a
> convert x r = let (r1,r2) = convert2 x r
>               in Seq (Star r1) r2

> convert2 :: Int -> RE a -> (RE a, RE a)
> convert2 x Empty = (Empty, Empty)
> convert2 x (Var y)
>        | x == y    = (Empty,Empty)
>        | otherwise = (Empty, Var y) -- can this happen?
> convert2 x (r@(Seq l r1))
>        | contains x r1 = let (r2,r3) = convert2 x r1
>                          in (Seq l r2, r3)
>        | otherwise = (Empty, r)
> convert2 x (Choice r1 r2) = let (r1', r1'') = convert2 x r1
>                                 (r2', r2'') = convert2 x r2
>                             in (Choice r1' r2', Choice r1'' r2'')

> contains :: Int -> RE a -> Bool
> contains x (Var y) = x == y
> contains x (Seq r1 r2) = contains x r1 || contains x r2
> contains x (Star r) = contains x r
> contains x (Choice r1 r2) = contains x r1 || contains x r2
> contains x _ = False


We use first-order syntax to represent recursive proofs/regular expressions.
We then use convert to actually build the non-recursive regular expression.


> intersectRE2 :: Eq a => RE a -> RE a -> RE a
> intersectRE2 r1 r2 = intersectC2 1 [] r1 r2

> intersectC2 :: Eq a => Int -> Env a -> RE a -> RE a -> RE a
> intersectC2 cnt env r1 r2 
>  | r1 == Phi || r2 == Phi = Phi
>  | r1 == Empty || r2 == Empty = Empty
>  | otherwise =
>      case lookup (r1,r2) env of
>       Just r -> r
>       Nothing ->
>         let letters = sigmaRE (r1 `Choice` r2)
>             env' = ((r1,r2),Var cnt):env
>             r1l l = resToRE $ partDeriv r1 l
>             r2l l = resToRE $ partDeriv r2 l
>             r' = resToRE $ map (\l -> Seq (L l) (intersectC2 (cnt+1) env' (r1l l) (r2l l))) letters
>             r =  if (isEmpty r1) && (isEmpty r2)
>                  then Choice r' Empty
>                  else r'
>         in convert cnt r


For testing purposes, it's handy to have a function which tests for (semantic)
equality among regular expressions (again written using partial derivatives).


> type EnvEq a = [(RE a, RE a)]

> eqRE :: Eq a => RE a -> RE a -> Bool
> eqRE r1 r2 = eqREC [] r1 r2

> eqREC :: Eq a => EnvEq a -> RE a -> RE a -> Bool
> eqREC env r1 r2 
>   | isEmpty r1 && (not (isEmpty r2)) = False
>   | isPhi r1 && (not (isPhi r2)) = False
>   | otherwise =
>      if elem (r1,r2) env
>      then True
>      else let letters = sigmaRE (r1 `Choice` r2)
>               env' = (r1,r2):env
>               r1l l = resToRE $ partDeriv r1 l
>               r2l l = resToRE $ partDeriv r2 l
>           in and $ map (\l -> eqREC env' (r1l l) (r2l l)) letters

examples

> rA = L 'A'
> rAstar = Star rA
> rB = L 'B'
> rBstar = Star rB
> rAorBstar = Star (Choice rA rB)

Some sample runs (using intersectRE2, intersectRE won't terminate for interesting cases)

*RegExpIntersection> intersectRE2 rAstar rBstar
<((<>|<>)|<>)*,((<'A',{}>|<'B',{}>)|<>)>
*RegExpIntersection> eqRE Empty (intersectRE2 rAstar rBstar)
True
*RegExpIntersection> intersectRE2 rAorBstar rBstar
<((<>|<>)|<>)*,((<'A',{}>|<'B',<((<>|<'B',<>>)|<>)*,((<'A',{}>|<>)|<>)>>)|<>)>
*RegExpIntersection> eqRE rBstar (intersectRE2 rAorBstar rBstar)
True
*RegExpIntersection> intersectRE2 rAorBstar rAorBstar
<((<>|<>)|<>)*,((<'A',<((<'A',<>>|<'B',<>>)|<>)*,((<>|<>)|<>)>>|<'B',<((<'A',<>>|<'B',<>>)|<>)*,((<>|<>)|<>)>>)|<>)>
*RegExpIntersection> eqRE rAorBstar (intersectRE2 rAorBstar rAorBstar)
True
*RegExpIntersection> eqRE rAstar (intersectRE2 rAorBstar rAorBstar)
False
*RegExpIntersection> eqRE rBstar (intersectRE2 rAorBstar rAorBstar)
False


The regular expressions generated by the convert function are 'very verbose'.
There's space for lots of improvement.