-----------------------------------------------------------------------------
-- |
-- 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
--
-- @
-- 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 ,
-- with the exception that we generate the trees filled with symbolic operators
-- and ask the SMT solver to find the appropriate fillings.
-----------------------------------------------------------------------------
{-# LANGUAGE DeriveAnyClass #-}
{-# LANGUAGE DeriveDataTypeable #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE TemplateHaskell #-}
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 :
-- 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 . (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 :: (Eq a, SymVal 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]