-----------------------------------------------------------------------------
-- |
-- Module    : Documentation.SBV.Examples.Puzzles.Birthday
-- Copyright : (c) Levent Erkok
-- License   : BSD3
-- Maintainer: erkokl@gmail.com
-- Stability : experimental
--
-- This is a formalization of the Cheryl's birthday problem, which went viral in April 2015.
--
-- Here's the puzzle:
--
-- @
-- Albert and Bernard just met Cheryl. "When’s your birthday?" Albert asked Cheryl.
--
-- Cheryl thought a second and said, "I’m not going to tell you, but I’ll give you some clues." She wrote down a list of 10 dates:
--
--   May 15, May 16, May 19
--   June 17, June 18
--   July 14, July 16
--   August 14, August 15, August 17
--
-- "My birthday is one of these," she said.
--
-- Then Cheryl whispered in Albert’s ear the month — and only the month — of her birthday. To Bernard, she whispered the day, and only the day.
-- “Can you figure it out now?” she asked Albert.
--
-- Albert: I don’t know when your birthday is, but I know Bernard doesn’t know, either.
-- Bernard: I didn’t know originally, but now I do.
-- Albert: Well, now I know, too!
--
-- When is Cheryl’s birthday?
-- @
--
-- NB. Thanks to Amit Goel for suggesting the formalization strategy used in here.
-----------------------------------------------------------------------------

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

{-# OPTIONS_GHC -Wall -Werror #-}

module Documentation.SBV.Examples.Puzzles.Birthday where

import Data.SBV

-----------------------------------------------------------------------------------------------
-- * Types and values
-----------------------------------------------------------------------------------------------

-- | Months. We only put in the months involved in the puzzle for simplicity
data Month = May | Jun | Jul | Aug

-- | Days. Again, only the ones mentioned in the puzzle.
data Day = D14 | D15 | D16 | D17 | D18 | D19

mkSymbolicEnumeration ''Month
mkSymbolicEnumeration ''Day

-- | Represent the birthday as a record
data Birthday = BD SMonth SDay

-- | Make a valid symbolic birthday
mkBirthday :: Symbolic Birthday
mkBirthday :: Symbolic Birthday
mkBirthday = do b <- SMonth -> SDay -> Birthday
BD (SMonth -> SDay -> Birthday)
-> SymbolicT IO SMonth -> SymbolicT IO (SDay -> Birthday)
forall (f :: * -> *) a b. Functor f => (a -> b) -> f a -> f b
<$> String -> SymbolicT IO SMonth
forall a. SymVal a => String -> Symbolic (SBV a)
free String
"birthMonth" SymbolicT IO (SDay -> Birthday)
-> SymbolicT IO SDay -> Symbolic Birthday
forall a b.
SymbolicT IO (a -> b) -> SymbolicT IO a -> SymbolicT IO b
forall (f :: * -> *) a b. Applicative f => f (a -> b) -> f a -> f b
<*> String -> SymbolicT IO SDay
forall a. SymVal a => String -> Symbolic (SBV a)
free String
"birthDay"
                constrain $ valid b
                pure b

-- | Is this a valid birthday? i.e., one that was declared by Cheryl to be a possibility.
valid :: Birthday -> SBool
valid :: Birthday -> SBool
valid (BD SMonth
m SDay
d) =   (SMonth
m SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SMonth
sMay SBool -> SBool -> SBool
.=> SDay
d SDay -> [SDay] -> SBool
forall a. EqSymbolic a => a -> [a] -> SBool
`sElem` [SDay
sD15, SDay
sD16, SDay
sD19])
               SBool -> SBool -> SBool
.&& (SMonth
m SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SMonth
sJun SBool -> SBool -> SBool
.=> SDay
d SDay -> [SDay] -> SBool
forall a. EqSymbolic a => a -> [a] -> SBool
`sElem` [SDay
sD17, SDay
sD18])
               SBool -> SBool -> SBool
.&& (SMonth
m SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SMonth
sJul SBool -> SBool -> SBool
.=> SDay
d SDay -> [SDay] -> SBool
forall a. EqSymbolic a => a -> [a] -> SBool
`sElem` [SDay
sD14, SDay
sD16])
               SBool -> SBool -> SBool
.&& (SMonth
m SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SMonth
sAug SBool -> SBool -> SBool
.=> SDay
d SDay -> [SDay] -> SBool
forall a. EqSymbolic a => a -> [a] -> SBool
`sElem` [SDay
sD14, SDay
sD15, SDay
sD17])

-----------------------------------------------------------------------------------------------
-- * The puzzle
-----------------------------------------------------------------------------------------------

-- | Encode the conversation as given in the puzzle.
--
-- NB. Lee Pike pointed out that not all the constraints are actually necessary! (Private
-- communication.) The puzzle still has a unique solution if the statements @a1@ and @b1@
-- (i.e., Albert and Bernard saying they themselves do not know the answer) are removed.
-- To experiment you can simply comment out those statements and observe that there still
-- is a unique solution. Thanks to Lee for pointing this out! In fact, it is instructive to
-- assert the conversation line-by-line, and see how the search-space gets reduced in each
-- step.
puzzle :: ConstraintSet
puzzle :: ConstraintSet
puzzle = do BD birthMonth birthDay <- Symbolic Birthday
mkBirthday

            let ok    = (Birthday -> SBool) -> [Birthday] -> SBool
forall a. (a -> SBool) -> [a] -> SBool
sAll Birthday -> SBool
valid
                qe a
qb = a -> SBool
forall a. QuantifiedBool a => a -> SBool
quantifiedBool a
qb

            -- Albert: I do not know
            let a1 SMonth
m = (Exists (ZonkAny 2) Day -> Exists (ZonkAny 3) Day -> SBool)
-> SBool
forall a. QuantifiedBool a => a -> SBool
qe ((Exists (ZonkAny 2) Day -> Exists (ZonkAny 3) Day -> SBool)
 -> SBool)
-> (Exists (ZonkAny 2) Day -> Exists (ZonkAny 3) Day -> SBool)
-> SBool
forall a b. (a -> b) -> a -> b
$ \(Exists SDay
d1) (Exists SDay
d2) -> [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m SDay
d1, SMonth -> SDay -> Birthday
BD SMonth
m SDay
d2] SBool -> SBool -> SBool
.&& SDay
d1 SDay -> SDay -> SBool
forall a. EqSymbolic a => a -> a -> SBool
./= SDay
d2
            constrain $ a1 birthMonth

            -- Albert: I know that Bernard doesn't know
            let a2 SMonth
m = (Forall (ZonkAny 4) Day -> SBool) -> SBool
forall a. QuantifiedBool a => a -> SBool
qe ((Forall (ZonkAny 4) Day -> SBool) -> SBool)
-> (Forall (ZonkAny 4) Day -> SBool) -> SBool
forall a b. (a -> b) -> a -> b
$ \(Forall SDay
d) -> [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m SDay
d] SBool -> SBool -> SBool
.=> (Exists (ZonkAny 5) Month -> Exists (ZonkAny 6) Month -> SBool)
-> SBool
forall a. QuantifiedBool a => a -> SBool
qe (\(Exists SMonth
m1) (Exists SMonth
m2) -> [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m1 SDay
d, SMonth -> SDay -> Birthday
BD SMonth
m2 SDay
d] SBool -> SBool -> SBool
.&& SMonth
m1 SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
./= SMonth
m2)
            constrain $ a2 birthMonth

            -- Bernard: I did not know
            let b1 SDay
d = (Exists (ZonkAny 7) Month -> Exists (ZonkAny 8) Month -> SBool)
-> SBool
forall a. QuantifiedBool a => a -> SBool
qe ((Exists (ZonkAny 7) Month -> Exists (ZonkAny 8) Month -> SBool)
 -> SBool)
-> (Exists (ZonkAny 7) Month -> Exists (ZonkAny 8) Month -> SBool)
-> SBool
forall a b. (a -> b) -> a -> b
$ \(Exists SMonth
m1) (Exists SMonth
m2) -> [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m1 SDay
d, SMonth -> SDay -> Birthday
BD SMonth
m2 SDay
d] SBool -> SBool -> SBool
.&& SMonth
m1 SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
./= SMonth
m2
            constrain $ b1 birthDay

            -- Bernard: But now I know
            let b2p SMonth
m SDay
d = [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m SDay
d] SBool -> SBool -> SBool
.&& SMonth -> SBool
a1 SMonth
m SBool -> SBool -> SBool
.&& SMonth -> SBool
a2 SMonth
m
                b2  SDay
d   = (Forall (ZonkAny 9) Month -> Forall (ZonkAny 10) Month -> SBool)
-> SBool
forall a. QuantifiedBool a => a -> SBool
qe ((Forall (ZonkAny 9) Month -> Forall (ZonkAny 10) Month -> SBool)
 -> SBool)
-> (Forall (ZonkAny 9) Month -> Forall (ZonkAny 10) Month -> SBool)
-> SBool
forall a b. (a -> b) -> a -> b
$ \(Forall SMonth
m1) (Forall SMonth
m2) -> (SMonth -> SDay -> SBool
b2p SMonth
m1 SDay
d SBool -> SBool -> SBool
.&& SMonth -> SDay -> SBool
b2p SMonth
m2 SDay
d) SBool -> SBool -> SBool
.=> SMonth
m1 SMonth -> SMonth -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SMonth
m2
            constrain $ b2 birthDay

            -- Albert: Now I know too
            let a3p SMonth
m SDay
d = [Birthday] -> SBool
ok [SMonth -> SDay -> Birthday
BD SMonth
m SDay
d] SBool -> SBool -> SBool
.&& SMonth -> SBool
a1 SMonth
m SBool -> SBool -> SBool
.&& SMonth -> SBool
a2 SMonth
m SBool -> SBool -> SBool
.&& SDay -> SBool
b1 SDay
d SBool -> SBool -> SBool
.&& SDay -> SBool
b2 SDay
d
                a3  SMonth
m   = \(Forall SDay
d1) (Forall SDay
d2) -> (SMonth -> SDay -> SBool
a3p SMonth
m SDay
d1 SBool -> SBool -> SBool
.&& SMonth -> SDay -> SBool
a3p SMonth
m SDay
d2) SBool -> SBool -> SBool
.=> SDay
d1 SDay -> SDay -> SBool
forall a. EqSymbolic a => a -> a -> SBool
.== SDay
d2
            constrain $ a3 birthMonth

-- | Find all solutions to the birthday problem. We have:
--
-- >>> cheryl
-- Solution #1:
--   birthMonth = Jul :: Month
--   birthDay   = D16 :: Day
-- This is the only solution.
cheryl :: IO ()
cheryl :: IO ()
cheryl = AllSatResult -> IO ()
forall a. Show a => a -> IO ()
print (AllSatResult -> IO ()) -> IO AllSatResult -> IO ()
forall (m :: * -> *) a b. Monad m => (a -> m b) -> m a -> m b
=<< ConstraintSet -> IO AllSatResult
forall a. Satisfiable a => a -> IO AllSatResult
allSat ConstraintSet
puzzle

{- HLint ignore puzzle "Redundant lambda" -}
{- HLint ignore puzzle "Eta reduce"       -}