Alex K

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# Consistently Logical

In the recent 538 Riddler, a particular logical puzzle was posted which involves determining who is telling the truth if you assume their claims must all be consistent and that they meet an extra constraint that all liars are older than all truth-tellers. I'm going to show how we can proceed to solve this in code, and specifically using functional programming with Haskell.

# Modelling the problem

The first place to start with any problem is deciding how to represent the information in the puzzle. Often the operations on these inputs flow naturally from the structures that we define.

The problem takes the form of a set of claims about the ages of other participants in the puzzle, where each claim states that the age of some person is a value, is not a value, is more than a value, or is less than a value. So lets represent the claims as a tuple of the operation and the value:

``````type Claim = (Op, Int)
``````

We need to decide what the `Op` is. The most straightforward approach is a sum-type with a constructor for each of the four operations we need, so:

``````data Op = Is | Isnt | Gt | Lt deriving (Show, Eq)
``````

However, when a claimant is lying, it is not that we gain no information, instead we can infer the inverse of the operator - for example if someone is lying that X is 17 then that is equivalent to someone telling the truth that X is not 17. The same goes for `>` and `<`, but with the caveat that these require the `>=` and `<=` operators to convey the inverses.

``````data Op = Is | Isnt | Gt | Gte | Lt | Lte deriving (Show, Eq, Ord)

invert :: Op -> Op
invert Is   = Isnt
invert Isnt = Is
invert Gt   = Lte
invert Gte  = Lt
invert Lt   = Gte
invert Lte  = Gt
``````

Then we need to model the claimants themselves. For our purposes, each of the islanders is identified by a single letter, so we will use `Char`s for our islander-names:

``````type Islander = Char
``````

And a claimant is an islander along with the set of their claims.

``````data Claimant = Claimant
{ claimantName   :: Islander
, claimantClaims :: Set (Islander, Claim)
} deriving (Show, Eq)
``````

For example, if 'X says Y is more than 10 years old', we can encode this as:

``````Claimant 'X' (S.fromList [('Y', (Gt, 10))])
``````

And this is all we need to encode the problem as specified in the 538 description, which, with a couple of helpers, we can do in such a way that it appears pretty similar to the problem statement itself.

``````-- the islanders in the problem, with their claims
islanders :: [Claimant]
islanders = ['a' `saysThat` ['b' ==> (Gt, 20)
,'d' ==> (Gt, 16)
]
,'b' `saysThat` ['c' ==> (Gt, 18)
,'e' ==> (Lt, 20)
]
,'c' `saysThat` ['d' ==> (Lt, 22)
,'a' ==> (Is, 19)
]
,'d' `saysThat` ['e' ==> (Isnt, 17)
,'b' ==> (Is, 20)
]
,'e' `saysThat` ['a' ==> (Gt, 21)
,'c' ==> (Lt, 18)
]
]
where saysThat who cs = Claimant who (S.fromList cs)
(==>) = (,)
``````

# What are you implying?

Once we know what each islander claims to be true, how can we find out whether their claims are mutually consistent. For example we can see that 'A' says that 'B is more than 20', but 'D' says that 'B is 20'. Clearly they cannot both be right, but how would we determine that algorithmically?

Well, we can look at what each claim implies. Each claim is equivalent to saying that a value lies in one or more range of values. For example `Gt 10` is the same as saying 'in the range 11 ..', and `Is 12` is the same as 'in the range 12 .. 12'. We can model `Isnt` as saying that a value must lie in one of two ranges, each of which excludes the value, so `Isnt 12` means 'must lie in the range ..11, or in the range 13..'.

This gives us a concept of the implications of each claim, and how we can represent them:

``````data InclusiveRange = Maybe Int :..: Maybe Int
deriving (Show, Eq)

newtype Implication = OneOf { ranges :: NonEmpty InclusiveRange }
-- deriving Semigroup makes 'or'-ing ranges easy
deriving (Show, Eq, Semigroup)
``````

We can then go ahead and write smart constructors for the different implications of the various claims we can express:

``````-- helper that makes these functions tidier
inRange :: InclusiveRange -> Implication
inRange = OneOf . pure

is :: Int -> Implication
is x = inRange \$ pure x :..: pure x

isnt :: Int -> Implication
isnt x = lt x <> gt x

gt :: Int -> Implication
gt x = inRange \$ pure (x + 1) :..: Nothing

gte :: Int -> Implication
gte x = inRange \$ pure x :..: Nothing

lt :: Int -> Implication
lt x = inRange \$ Nothing :..: pure (x - 1)

lte :: Int -> Implication
lte x = inRange \$ Nothing :..: pure x
``````

And we can combine these in a single function from a claim to its implications:

``````implication :: Claim -> Implication
implication (op, age) = (\$ age) \$ case op of
Is   -> is
Isnt -> isnt
Gt   -> gt
Gte  -> gte
Lt   -> lt
Lte  -> lte
``````

# Making Inferences

The next step is to look at how we can combine two implications to get the logical consequent. For instance, from `Gt 10` and `Lt 15` we should be able to infer that the value lies in the range `11..14`. This implies some kind of monoidal structure, where implications can be combined to create more detailed ones. In which case we need the `mempty` base case:

``````anything :: Implication
anything = inRange \$ Nothing :..: Nothing
``````

And as a first pass we will look at what we can infer from two implications:

``````-- apply a binary operation, taking lhs if rhs is Nothing, or rhs if
-- lhs is Nothing
safeBinOp :: (a -> a -> a) -> Maybe a -> Maybe a -> Maybe a
safeBinOp f ma mb = liftA2 f ma mb <|> ma <|> mb

infer :: Implication -> Implication -> Maybe Implication
infer (OneOf lhs) (OneOf rhs) = fmap OneOf . NE.nonEmpty . catMaybes
\$ liftA2 infer' (NE.toList lhs) (NE.toList rhs)
where
infer' (a :..: b) (a' :..: b') =
let lb = safeBinOp max a a'
ub = safeBinOp min b b'
in if fromMaybe False (liftA2 (>) lb ub)
then Nothing -- invalid, lb > ub
else pure (lb :..: ub)
``````

The inference from two implications is what we get by seeing what constraints each range applies to each range on the other side. Any ranges that are invalid as a result are removed by `catMaybes`, and if we don't get any ranges at all we can't return any implication, which means that the two statements are in contradiction.

Lets define a way to pretty print an implication, and then take this inference mechanism for a spin!

``````showImpl :: Implication -> String
showImpl = L.intercalate ";" . fmap showRange . NE.toList . ranges
where
showRange (Nothing :..: Just x) = ".." <> show x
showRange (Just x :..: Nothing) = show x <> ".."
showRange (Just x :..: Just y)  = if x == y
then show x
else show x <> ".." <> show y
showRange _   = "anything"

-- Give a two-parameter kleisli arrow, apply it to two arguments
-- (hard to believe this isn't in Control.Monad tbh)
liftKleisli :: Monad m => (a -> b -> m c) -> m a -> m b -> m c
liftKleisli f a b = liftA2 (,) a b >>= uncurry f
``````

Now we can play around with `infer`:

``````λ> let p = putStrLn . maybe "Contradiction" showImpl
λ> let inferAll = foldr (liftKleisli infer) (pure anything) . fmap pure
λ> p \$ inferAll [anything]
anything
λ> p \$ inferAll [gt 5, lt 10]
6..9
λ> p \$ inferAll [gt 5, lt 10, isnt 7]
6;8..9
λ> p \$ inferAll [gt 5, lt 10, isnt 7, isnt 8]
6;9
λ> p \$ inferAll [gt 5, lt 10, isnt 7, isnt 8, isnt 6]
9
λ> p \$ inferAll [gt 5, lt 12, isnt 8, isnt 10]
6..7;9;11
λ> p \$ inferAll [gt 20, is 20]
``````

Great, we seem to have a working inference engine. We can now wrap that up in a Monoid:

``````newtype Consequence = Consequence { getConsequence :: Maybe Implication }
deriving (Show, Eq)

instance Semigroup Consequence where
(Consequence ma) <> (Consequence mb) = Consequence (liftKleisli infer ma mb)

instance Monoid Consequence where
mempty = Consequence (pure anything)
mappend = (<>)

consequence :: Implication -> Consequence
consequence = Consequence . pure
``````

# Finding a solution

We now have a lot of the pieces we need to find a solution, and one way we can go about that is to enumerate all the possible arrangements of liars and truth-tellers on the island, checking to see which ones are internally consistent, and then checking that all the liars are older than the truth-tellers. In our case that is perfectly tractable, as we only have to consider `2^5`, or 32, arrangements.

First of all we need to tag an islander as either a liar or a truth-teller, and while we can use a `Bool` for this, it is much better to use a specific data-type for our purposes:

``````data Honesty = Honest | Liar deriving (Eq, Show)
``````

Then we need all 32 arrangements of these two values, eg:

• Honest, Honest, Honest, Honest, Honest
• Honest, Honest, Honest, Honest, Liar
• Honest, Honest, Honest, Liar, Honest
• and so on...

Lets create a small helper for that:

``````-- produce an arrangement of xs of size n
arrangement :: [a] -> Int -> [[a]]
arrangement _  0 = [[]]
arrangement xs n = xs >>= \x -> (x:) <\$> arrangement xs (n - 1)
``````

And we can quickly check that this does what we want:

``````λ> length \$ arrangement ['a', 'b'] 5
32
λ> take 3 \$ arrangement ['a', 'b'] 5
["aaaaa","aaaab","aaaba"]
``````

Now we can generate all 32 potential solutions that we need to consider:

``````fmap (`zip` islanders) \$ arrangement [Honest, Liar] (length islanders)
``````

In order to find a solution, we need to combine all the claims about the islanders together, potentially inverting them if they come from liars, and then check to see if there were any contradictions, and whether it meets the age rule. First, lets define the types of solutions and their conclusions:

``````type Conclusions = Map Char Consequence
type Solution = [(Honesty, Claimant)]
``````

We need a way to turn a solution into a conclusion about all its claims:

``````conclusions :: [(Honesty, S.Set (Char, Claim))] -> Conclusions
conclusions solution =
M.fromListWith (<>)
. fmap (fmap (consequence . implication))
\$ [(who, (trust h op, val)) | (h, cs) <- solution
, (who, (op, val)) <- S.toList cs
]
where trust Liar = invert
trust Honest = id
``````

and a way to know if a set of conclusions contains any contradictions:

``````-- A set of conclusions is valid if there are no contradictions
viable :: Conclusions -> Bool
viable = all (isJust . getConsequence)
``````

and a way to determine if it meets the age-limit rule, where we check that the oldest honest person is younger than the youngest liar - or in the language of ranges, that the highest lower bound on the age of the honest claimants is lower than the lowest upper bound on the ages of the liars.

``````meetsAgeLimitRule :: Solution -> Conclusions -> Bool
meetsAgeLimitRule claimants conc =
let oldestHonest = getAge max lowerBound isHonest
youngestLiar = getAge min upperBound (not . isHonest)
in fromMaybe True (liftA2 (<) oldestHonest youngestLiar)
where
honest = S.fromList [who | (Honest, Claimant who _) <- claimants]
isHonest = (`S.member` honest)

lowerBound (OneOf cs) = let f (lb :..: _) = lb in safely min (f <\$> cs)
upperBound (OneOf cs) = let f (_ :..: ub) = ub in safely max (f <\$> cs)

getAge overall perPerson isRelevant
= safely overall
. fmap (getConsequence . snd >=> perPerson)
. filter (isRelevant . fst)
\$ M.toList conc

-- safely apply a bin-op over a foldable of Maybes
safely :: (Foldable f, Ord a) => (a -> a -> a) -> f (Maybe a) -> Maybe a
safely f = foldr (safeBinOp f) Nothing
``````

With that all in place, we can then proceed to search all the possible arrangements and print out the viable ones we find:

``````viables :: [(Solution, Conclusions)]
viables = filter (uncurry meetsAgeLimitRule)
. filter (viable . snd)
. fmap   (withConclusion . (`zip` islanders))
\$ arrangement [Honest, Liar] (length islanders)

withConclusion :: Solution -> (Solution, Conclusions)
withConclusion cs = (cs, conclusions (fmap claimantClaims <\$> cs))

-- print out all the viable solutions
main :: IO ()
main = mapM_ printSolution (zip [1 ..] viables)
where
printSolution (n, (sol, conc)) = do
putStrLn \$ "Solution: " <> show n
putStrLn \$ replicate 15 '-'
mapM_ (go conc) sol
putStrLn ""
go conc (h, c) = putStrLn
\$ unwords [[claimantName c]
,show h
,maybe "" showImpl
(M.lookup (claimantName c) conc >>= getConsequence)
]
``````

There is just one solution:

``````Solution: 1
---------------
a Liar 19
b Liar 20
c Honest 18
d Honest ..16
e Liar 20..
``````

And we can clearly see from the conclusions that this is internally consistent. Walking through the puzzle manually with this solution should be enough to convince you that the solution is sound, but to my mind the nicest part is how naturally it flows from observing the way the parts are stuctured, from seeing the idea of ranges just fall out of the claims, to the fact that we can just fold the claims together using their monoidal structure and then being able to read the conclusions back from the consequences we had calculated. Being able to define the solution clearly made proceeding through the solution that much easier.