sbv-8.7: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.

Copyright | (c) Levent Erkok
License | BSD3
Maintainer | erkokl@gmail.com
Stability | experimental
Safe Haskell | None
Language | Haskell2010

Documentation.SBV.Examples.Puzzles.Garden

Description

The origin of this puzzle is Raymond Smullyan's "The Flower Garden" riddle:

In a certain flower garden, each flower was either red, yellow, or blue, and all three colors were represented. A statistician once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound to be red. A second statistician visited the garden and made the observation that whatever three flowers you picked, at least one was bound to be yellow.

Two logic students heard about this and got into an argument. The first student said: “It therefore follows that whatever three flowers you pick, at least one is bound to be blue, doesn’t it?” The second student said: “Of course not!”

Which student was right, and why?

We slightly modify the puzzle. Assuming the first student is right, we use SBV to show that the garden must contain exactly 3 flowers. In any other case, the second student would be right.

Synopsis

- data Color
  - = Red
  - | Yellow
  - | Blue
- type Flower = SInteger
- col :: Flower -> SBV Color
- validPick :: SInteger -> Flower -> Flower -> Flower -> SBool
- count :: Color -> [Flower] -> SInteger
- puzzle :: Goal
- flowerCount :: IO ()

# Documentation

data Color Source #

Colors of the flowers

Constructors

Red |
Yellow |
Blue |

#### Instances

Instances details

Eq Color Source # |
Instance details Defined in Documentation.SBV.Examples.Puzzles.Garden Methods (==) :: Color -> Color -> Bool (/=) :: Color -> Color -> Bool
Data Color Source # |

Instance details

Defined in Documentation.SBV.Examples.Puzzles.Garden

Methods

gfoldl :: (forall d b. Data d => c (d -> b) -> d -> c b) -> (forall g. g -> c g) -> Color -> c Color

gunfold :: (forall b r. Data b => c (b -> r) -> c r) -> (forall r. r -> c r) -> Constr -> c Color

toConstr :: Color -> Constr

dataTypeOf :: Color -> DataType

dataCast1 :: Typeable t => (forall d. Data d => c (t d)) -> Maybe (c Color)

dataCast2 :: Typeable t => (forall d e. (Data d, Data e) => c (t d e)) -> Maybe (c Color)

gmapT :: (forall b. Data b => b -> b) -> Color -> Color

gmapQl :: (r -> r' -> r) -> r -> (forall d. Data d => d -> r') -> Color -> r

gmapQr :: forall r r'. (r' -> r -> r) -> r -> (forall d. Data d => d -> r') -> Color -> r

gmapQ :: (forall d. Data d => d -> u) -> Color -> [u]

gmapQi :: Int -> (forall d. Data d => d -> u) -> Color -> u

gmapM :: Monad m => (forall d. Data d => d -> m d) -> Color -> m Color

gmapMp :: MonadPlus m => (forall d. Data d => d -> m d) -> Color -> m Color

gmapMo :: MonadPlus m => (forall d. Data d => d -> m d) -> Color -> m Color

Ord Color Source # |
Instance details Defined in Documentation.SBV.Examples.Puzzles.Garden Methods compare :: Color -> Color -> Ordering (<) :: Color -> Color -> Bool (<=) :: Color -> Color -> Bool (>) :: Color -> Color -> Bool (>=) :: Color -> Color -> Bool max :: Color -> Color -> Color min :: Color -> Color -> Color
Read Color Source # |
Instance details Defined in Documentation.SBV.Examples.Puzzles.Garden Methods readsPrec :: Int -> ReadS Color readList :: ReadS [Color] readPrec :: ReadPrec Color readListPrec :: ReadPrec [Color]
Show Color Source # |
Instance details Defined in Documentation.SBV.Examples.Puzzles.Garden Methods showsPrec :: Int -> Color -> ShowS show :: Color -> String showList :: [Color] -> ShowS
HasKind Color Source # |

Instance details

Defined in Documentation.SBV.Examples.Puzzles.Garden

Methods

kindOf :: Color -> Kind Source #

hasSign :: Color -> Bool Source #

intSizeOf :: Color -> Int Source #

isBoolean :: Color -> Bool Source #

isBounded :: Color -> Bool Source #

isReal :: Color -> Bool Source #

isFloat :: Color -> Bool Source #

isDouble :: Color -> Bool Source #

isUnbounded :: Color -> Bool Source #

isUninterpreted :: Color -> Bool Source #

isChar :: Color -> Bool Source #

isString :: Color -> Bool Source #

isList :: Color -> Bool Source #

isSet :: Color -> Bool Source #

isTuple :: Color -> Bool Source #

isMaybe :: Color -> Bool Source #

isEither :: Color -> Bool Source #

showType :: Color -> String Source #

SymVal Color Source # |

Instance details

Defined in Documentation.SBV.Examples.Puzzles.Garden

Methods

mkSymVal :: MonadSymbolic m => Maybe Quantifier -> Maybe String -> m (SBV Color) Source #

literal :: Color -> SBV Color Source #

fromCV :: CV -> Color Source #

isConcretely :: SBV Color -> (Color -> Bool) -> Bool Source #

forall :: MonadSymbolic m => String -> m (SBV Color) Source #

forall_ :: MonadSymbolic m => m (SBV Color) Source #

mkForallVars :: MonadSymbolic m => Int -> m [SBV Color] Source #

exists :: MonadSymbolic m => String -> m (SBV Color) Source #

exists_ :: MonadSymbolic m => m (SBV Color) Source #

mkExistVars :: MonadSymbolic m => Int -> m [SBV Color] Source #

free :: MonadSymbolic m => String -> m (SBV Color) Source #

free_ :: MonadSymbolic m => m (SBV Color) Source #

mkFreeVars :: MonadSymbolic m => Int -> m [SBV Color] Source #

symbolic :: MonadSymbolic m => String -> m (SBV Color) Source #

symbolics :: MonadSymbolic m => [String] -> m [SBV Color] Source #

unliteral :: SBV Color -> Maybe Color Source #

isConcrete :: SBV Color -> Bool Source #

isSymbolic :: SBV Color -> Bool Source #

SatModel Color Source # | Make `Color` a symbolic value.
Instance details Defined in Documentation.SBV.Examples.Puzzles.Garden Methods parseCVs :: [CV] -> Maybe (Color, [CV]) Source # cvtModel :: (Color -> Maybe b) -> Maybe (Color, [CV]) -> Maybe (b, [CV]) Source #

type Flower = SInteger Source #

Represent flowers by symbolic integers

col :: Flower -> SBV Color Source #

The uninterpreted function `col` assigns a color to each flower.

validPick :: SInteger -> Flower -> Flower -> Flower -> SBool Source #

Describe a valid pick of three flowers `i`, `j`, `k`, assuming we have `n` flowers to start with. Essentially the numbers should be within bounds and distinct.

count :: Color -> [Flower] -> SInteger Source #

Count the number of flowers that occur in a given set of flowers.

puzzle :: Goal Source #

Smullyan's puzzle.

flowerCount :: IO () Source #

Solve the puzzle. We have:

    >>> flowerCount
    Solution #1:
      N = 3 :: Integer
    This is the only solution. (Unique up to prefix existentials.)

So, a garden with 3 flowers is the only solution. (Note that we simply skip over the prefix existentials and the assignments to uninterpreted function `col` for model purposes here, as they don't represent a different solution.)