Copyright | (c) Levent Erkok |
---|---|

License | BSD3 |

Maintainer | erkokl@gmail.com |

Stability | experimental |

Safe Haskell | None |

Language | Haskell2010 |

This is a formalization of the Cheryl's birtday problem, which went viral in April 2015. (See http://www.nytimes.com/2015/04/15/science/a-math-problem-from-singapore-goes-viral-when-is-cheryls-birthday.html.)

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.

- type Month = SWord8
- type Day = SWord8
- may :: SWord8
- june :: SWord8
- july :: SWord8
- august :: SWord8
- valid :: Month -> Day -> SBool
- existsDay :: (Day -> SBool) -> SBool
- forallDay :: (Day -> SBool) -> SBool
- existsMonth :: (Month -> SBool) -> SBool
- forallMonth :: (Month -> SBool) -> SBool
- puzzle :: Predicate
- cheryl :: IO ()

# Types and values

Represent month by 8-bit words; We can also use an uninterpreted type, but numbers work well here.

Represent day by 8-bit words; Again, an uninterpreted type would work as well.

# Helper predicates

existsDay :: (Day -> SBool) -> SBool Source #

Assert that the given function holds for one of the possible days.

forallDay :: (Day -> SBool) -> SBool Source #

Assert that the given function holds for all of the possible days.

existsMonth :: (Month -> SBool) -> SBool Source #

Assert that the given function holds for one of the possible months.

forallMonth :: (Month -> SBool) -> SBool Source #

Assert that the given function holds for all of the possible months.

# 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.