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

License | BSD3 |

Maintainer | erkokl@gmail.com |

Stability | experimental |

Safe Haskell | None |

Language | Haskell2010 |

Solves the following puzzle:

You and a friend pass by a standard coin operated vending machine and you decide to get a candy bar. The price is US $0.95, but after checking your pockets you only have a dollar (US $1) and the machine only takes coins. You turn to your friend and have this conversation: you: Hey, do you have change for a dollar? friend: Let's see. I have 6 US coins but, although they add up to a US $1.15, I can't break a dollar. you: Huh? Can you make change for half a dollar? friend: No. you: How about a quarter? friend: Nope, and before you ask I cant make change for a dime or nickel either. you: Really? and these six coins are all US government coins currently in production? friend: Yes. you: Well can you just put your coins into the vending machine and buy me a candy bar, and I'll pay you back? friend: Sorry, I would like to but I cant with the coins I have. What coins are your friend holding?

To be fair, the problem has no solution *mathematically*. But there is a solution when one takes into account that
vending machines typically do not take the 50 cent coins!

# Documentation

We will represent coins with 16-bit words (more than enough precision for coins).

mkCoin :: Int -> Symbolic Coin Source

Create a coin. The argument Int argument just used for naming the coin. Note that we constrain the value to be one of the valid U.S. coin values as we create it.

combinations :: [a] -> [[a]] Source

Return all combinations of a sequence of values.

Constraint 6: Cannot buy the candy either. Here's where we need to have the extra knowledge that the vending machines do not take 50 cent coins.