    -----------------------------------------------------------------------------
    -- |
    -- Module      :  Documentation.SBV.Examples.Queries.FourFours
    -- Copyright   :  (c) Levent Erkok
    -- License     :  BSD3
    -- Maintainer  :  erkokl@gmail.com
    -- Stability   :  experimental
    --
    -- A query based solution to the four-fours puzzle.
    -- Inspired by <http://www.gigamonkeys.com/trees/>
    --
    -- @
    -- Try to make every number between 0 and 20 using only four 4s and any
    -- mathematical operation, with all four 4s being used each time.
    -- @
    --
    -- We pretty much follow the structure of <http://www.gigamonkeys.com/trees/>,
    -- with the exception that we generate the trees filled with symbolic operators
    -- and ask the SMT solver to find the appropriate fillings.
    -----------------------------------------------------------------------------

    {-# LANGUAGE FlexibleInstances  #-}
    {-# LANGUAGE TemplateHaskell     #-}
    {-# LANGUAGE StandaloneDeriving  #-}
    {-# LANGUAGE DeriveDataTypeable  #-}
    {-# LANGUAGE DeriveAnyClass      #-}
    {-# LANGUAGE ScopedTypeVariables #-}

    module Documentation.SBV.Examples.Queries.FourFours where

    import Data.SBV
    import Data.SBV.Control

    import Data.List (inits, tails)
    import Data.Maybe

    -- | Supported binary operators. To keep the search-space small, we will only allow division by @2@ or @4@,
    -- and exponentiation will only be to the power @0@. This does restrict the search space, but is sufficient to
    -- solve all the instances.
    data BinOp = Plus | Minus | Times | Divide | Expt

    -- | Make 'BinOp' a symbolic value.
    mkSymbolicEnumeration ''BinOp

    -- | Supported unary operators. Similar to 'BinOp' case, we will restrict square-root and factorial to
    -- be only applied to the value @4.
    data UnOp  = Negate | Sqrt | Factorial

    -- | Make 'UnOp' a symbolic value.
    mkSymbolicEnumeration ''UnOp

    -- | Symbolic variant of 'BinOp'.
    type SBinOp = SBV BinOp

    -- | Symbolic variant of 'UnOp'.
    type SUnOp  = SBV UnOp

    -- | The shape of a tree, either a binary node, or a unary node, or the number @4@, represented hear by
    -- the constructor @F@. We parameterize by the operator type: When doing symbolic computations, we'll fill
    -- those with 'SBinOp' and 'SUnOp'. When finding the shapes, we will simply put unit values, i.e., holes.
    data T b u = B b (T b u) (T b u)
               | U u (T b u)
               | F

    -- | A rudimentary 'Show' instance for trees, nothing fancy.
    instance Show (T BinOp UnOp) where
       show F       = "4"
       show (U u t) = case u of
                        Negate    -> "-" ++ show t
                        Sqrt      -> "sqrt(" ++ show t ++ ")"
                        Factorial -> show t ++ "!"
       show (B o l r) = "(" ++ show l ++ " " ++ so ++ " " ++ show r ++ ")"
         where so = fromMaybe (error $ "Unexpected operator: " ++ show o)
                            $ o `lookup` [(Plus, "+"), (Minus, "-"), (Times, "*"), (Divide, "/"), (Expt, "^")]

    -- | Construct all possible tree shapes. The argument here follows the logic in <http://www.gigamonkeys.com/trees/>:
    -- We simply construct all possible shapes and extend with the operators. The number of such trees is:
    --
    -- >>> length allPossibleTrees
    -- 640
    --
    -- Note that this is a /lot/ smaller than what is generated by <http://www.gigamonkeys.com/trees/>. (There, the
    -- number of trees is 10240000: 16000 times more than what we have to consider!)
    allPossibleTrees :: [T () ()]
    allPossibleTrees = trees $ replicate 4 F
      where trees :: [T () ()] -> [T () ()]
            trees [x] = [x, U () x]
            trees xs  = do (left, right) <- splits
                           t1            <- trees left
                           t2            <- trees right
                           trees [B () t1 t2]
              where splits = init $ tail $ zip (inits xs) (tails xs)

    -- | Given a tree with hols, fill it with symbolic operators. This is the /trick/ that allows
    -- us to consider only 640 trees as opposed to over 10 million.
    fill :: T () () -> Symbolic (T SBinOp SUnOp)
    fill (B _ l r) = B <$> free_ <*> fill l <*> fill r
    fill (U _ t)   = U <$> free_ <*> fill t
    fill F         = return F

    -- | Minor helper for writing "symbolic" case statements. Simply walks down a list
    -- of values to match against a symbolic version of the key.
    sCase :: (SymWord a, Mergeable v) => SBV a -> [(a, v)] -> v
    sCase k = walk
      where walk []              = error "sCase: Expected a non-empty list of cases!"
            walk [(_, v)]        = v
            walk ((k1, v1):rest) = ite (k .== literal k1) v1 (walk rest)

    -- | Evaluate a symbolic tree, obtaining a symbolic value. Note how we structure
    -- this evaluation so we impose extra constraints on what values square-root, divide
    -- etc. can take. This is the power of the symbolic approach: We can put arbitrary
    -- symbolic constraints as we evaluate the tree.
    eval :: T SBinOp SUnOp -> Symbolic SInteger
    eval tree = case tree of
                  B b l r -> eval l >>= \l' -> eval r >>= \r' -> binOp b l' r'
                  U u t   -> eval t >>= uOp u
                  F       -> return 4

      where binOp :: SBinOp -> SInteger -> SInteger -> Symbolic SInteger
            binOp o l r = do constrain $ o .== literal Divide ==> r .== 4 ||| r .== 2
                             constrain $ o .== literal Expt   ==> r .== 0
                             return $ sCase o
                                        [ (Plus,    l+r)
                                        , (Minus,   l-r)
                                        , (Times,   l*r)
                                        , (Divide,  l `sDiv` r)
                                        , (Expt,    1)   -- exponent is restricted to 0, so the value is 1
                                        ]

            uOp :: SUnOp -> SInteger -> Symbolic SInteger
            uOp o v = do constrain $ o .== literal Sqrt      ==> v .== 4
                         constrain $ o .== literal Factorial ==> v .== 4
                         return $ sCase o
                                    [ (Negate,    -v)
                                    , (Sqrt,       2)  -- argument is restricted to 4, so the value is 2
                                    , (Factorial, 24)  -- argument is restricted to 4, so the value is 24
                                    ]

    -- | In the query mode, find a filling of a given tree shape /t/, such that it evalutes to the
    -- requested number /i/. Note that we return back a concrete tree.
    generate :: Integer -> T () () -> IO (Maybe (T BinOp UnOp))
    generate i t = runSMT $ do symT <- fill t
                               val  <- eval symT
                               constrain $ val .== literal i
                               query $ do cs <- checkSat
                                          case cs of
                                            Sat -> Just <$> construct symT
                                            _   -> return Nothing
        where -- Walk through the tree, ask the solver for
              -- the assignment to symbolic operators and fill back.
              construct F           = return F
              construct (U o s')    = do uo <- getValue o
                                         U uo <$> construct s'
              construct (B b l' r') = do bo <- getValue b
                                         B bo <$> construct l' <*> construct r'

    -- | Given an integer, walk through all possible tree shapes (at most 640 of them), and find a
    -- filling that solves the puzzle.
    find :: Integer -> IO ()
    find target = go allPossibleTrees
      where go []     = putStrLn $ show target ++ ": No solution found."
            go (t:ts) = do chk <- generate target t
                           case chk of
                             Nothing -> go ts
                             Just r  -> do let ok  = concEval r == target
                                               tag = if ok then " [OK]: " else " [BAD]: "
                                               sh i | i < 10 = ' ' : show i
                                                    | True   =       show i

                                           putStrLn $ sh target ++ tag ++ show r

            -- Make sure the result is correct!
            concEval :: T BinOp UnOp -> Integer
            concEval F         = 4
            concEval (U u t)   = uEval u (concEval t)
            concEval (B b l r) = bEval b (concEval l) (concEval r)

            uEval :: UnOp -> Integer -> Integer
            uEval Negate    i = -i
            uEval Sqrt      i = if i == 4 then  2 else error $ "uEval: Found sqrt applied to value: " ++ show i
            uEval Factorial i = if i == 4 then 24 else error $ "uEval: Found factorial applied to value: " ++ show i

            bEval :: BinOp -> Integer -> Integer -> Integer
            bEval Plus   i j = i + j
            bEval Minus  i j = i - j
            bEval Times  i j = i * j
            bEval Divide i j = i `div` j
            bEval Expt   i j = i ^ j

    -- | Solution to the puzzle. When you run this puzzle, the solver can produce different results
    -- than what's shown here, but the expressions should still be all valid!
    --
    -- @
    -- ghci> puzzle
    --  0 [OK]: (4 - (4 + (4 - 4)))
    --  1 [OK]: (4 / (4 + (4 - 4)))
    --  2 [OK]: sqrt((4 + (4 * (4 - 4))))
    --  3 [OK]: (4 - (4 ^ (4 - 4)))
    --  4 [OK]: (4 + (4 * (4 - 4)))
    --  5 [OK]: (4 + (4 ^ (4 - 4)))
    --  6 [OK]: (4 + sqrt((4 * (4 / 4))))
    --  7 [OK]: (4 + (4 - (4 / 4)))
    --  8 [OK]: (4 - (4 - (4 + 4)))
    --  9 [OK]: (4 + (4 + (4 / 4)))
    -- 10 [OK]: (4 + (4 + (4 - sqrt(4))))
    -- 11 [OK]: (4 + ((4 + 4!) / 4))
    -- 12 [OK]: (4 * (4 - (4 / 4)))
    -- 13 [OK]: (4! + ((sqrt(4) - 4!) / sqrt(4)))
    -- 14 [OK]: (4 + (4 + (4 + sqrt(4))))
    -- 15 [OK]: (4 + ((4! - sqrt(4)) / sqrt(4)))
    -- 16 [OK]: (4 * (4 * (4 / 4)))
    -- 17 [OK]: (4 + ((sqrt(4) + 4!) / sqrt(4)))
    -- 18 [OK]: -(4 + (4 - (sqrt(4) + 4!)))
    -- 19 [OK]: -(4 - (4! - (4 / 4)))
    -- 20 [OK]: (4 * (4 + (4 / 4)))
    -- @
    puzzle :: IO ()
    puzzle = mapM_ find [0 .. 20]