-----------------------------------------------------------------------------
-- |
-- Module : Documentation.SBV.Examples.Existentials.Diophantine
-- Copyright : (c) Levent Erkok
-- License : BSD3
-- Maintainer: erkokl@gmail.com
-- Stability : experimental
--
-- Finding minimal natural number solutions to linear Diophantine equations,
-- using explicit quantification.
-----------------------------------------------------------------------------
module Documentation.SBV.Examples.Existentials.Diophantine where
import Data.SBV
--------------------------------------------------------------------------------------------------
-- * Representing solutions
--------------------------------------------------------------------------------------------------
-- | For a homogeneous problem, the solution is any linear combination of the resulting vectors.
-- For a non-homogeneous problem, the solution is any linear combination of the vectors in the
-- second component plus one of the vectors in the first component.
data Solution = Homogeneous [[Integer]]
| NonHomogeneous [[Integer]] [[Integer]]
deriving Show
--------------------------------------------------------------------------------------------------
-- * Solving diophantine equations
--------------------------------------------------------------------------------------------------
-- | ldn: Solve a (L)inear (D)iophantine equation, returning minimal solutions over (N)aturals.
-- The input is given as a rows of equations, with rhs values separated into a tuple. The first
-- parameter limits the search to bound: In case there are too many solutions, you might want
-- to limit your search space.
ldn :: Maybe Int -> [([Integer], Integer)] -> IO Solution
ldn mbLim problem = do solution <- basis mbLim (map (map literal) m)
if homogeneous
then return $ Homogeneous solution
else do let ones = [xs | (1:xs) <- solution]
zeros = [xs | (0:xs) <- solution]
return $ NonHomogeneous ones zeros
where rhs = map snd problem
lhs = map fst problem
homogeneous = all (== 0) rhs
m | homogeneous = lhs
| True = zipWith (\x y -> -x : y) rhs lhs
-- | Find the basis solution. By definition, the basis has all non-trivial (i.e., non-0) solutions
-- that cannot be written as the sum of two other solutions. We use the mathematically equivalent
-- statement that a solution is in the basis if it's least according to the lexicographic
-- order using the ordinary less-than relation. (NB. We explicitly tell z3 to use the logic
-- AUFLIA for this problem, as the BV solver that is chosen automatically has a performance
-- issue. See: .)
basis :: Maybe Int -> [[SInteger]] -> IO [[Integer]]
basis mbLim m = extractModels `fmap` allSatWith z3{allSatMaxModelCount = mbLim} cond
where cond = do as <- mkExistVars n
bs <- mkForallVars n
-- Tell the solver to use this logic!
setLogic AUFLIA
return $ ok as .&& (ok bs .=> as .== bs .|| sNot (bs `less` as))
n = if null m then 0 else length (head m)
ok xs = sAny (.> 0) xs .&& sAll (.>= 0) xs .&& sAnd [sum (zipWith (*) r xs) .== 0 | r <- m]
as `less` bs = sAnd (zipWith (.<=) as bs) .&& sOr (zipWith (.<) as bs)
--------------------------------------------------------------------------------------------------
-- * Examples
--------------------------------------------------------------------------------------------------
-- | Solve the equation:
--
-- @2x + y - z = 2@
--
-- We have:
--
-- >>> test
-- NonHomogeneous [[0,2,0],[1,0,0]] [[0,1,1],[1,0,2]]
--
-- which means that the solutions are of the form:
--
-- @(1, 0, 0) + k (0, 1, 1) + k' (1, 0, 2) = (1+k', k, k+2k')@
--
-- OR
--
-- @(0, 2, 0) + k (0, 1, 1) + k' (1, 0, 2) = (k', 2+k, k+2k')@
--
-- for arbitrary @k@, @k'@. It's easy to see that these are really solutions
-- to the equation given. It's harder to see that they cover all possibilities,
-- but a moments thought reveals that is indeed the case.
test :: IO Solution
test = ldn Nothing [([2,1,-1], 2)]
-- | A puzzle: Five sailors and a monkey escape from a naufrage and reach an island with
-- coconuts. Before dawn, they gather a few of them and decide to sleep first and share
-- the next day. At night, however, one of them awakes, counts the nuts, makes five parts,
-- gives the remaining nut to the monkey, saves his share away, and sleeps. All other
-- sailors do the same, one by one. When they all wake up in the morning, they again make 5 shares,
-- and give the last remaining nut to the monkey. How many nuts were there at the beginning?
--
-- We can model this as a series of diophantine equations:
--
-- @
-- x_0 = 5 x_1 + 1
-- 4 x_1 = 5 x_2 + 1
-- 4 x_2 = 5 x_3 + 1
-- 4 x_3 = 5 x_4 + 1
-- 4 x_4 = 5 x_5 + 1
-- 4 x_5 = 5 x_6 + 1
-- @
--
-- We need to solve for x_0, over the naturals. We have:
--
-- >>> sailors
-- [15621,3124,2499,1999,1599,1279,1023]
--
-- That is:
--
-- @
-- * There was a total of 15621 coconuts
-- * 1st sailor: 15621 = 3124*5+1, leaving 15621-3124-1 = 12496
-- * 2nd sailor: 12496 = 2499*5+1, leaving 12496-2499-1 = 9996
-- * 3rd sailor: 9996 = 1999*5+1, leaving 9996-1999-1 = 7996
-- * 4th sailor: 7996 = 1599*5+1, leaving 7996-1599-1 = 6396
-- * 5th sailor: 6396 = 1279*5+1, leaving 6396-1279-1 = 5116
-- * In the morning, they had: 5116 = 1023*5+1.
-- @
--
-- Note that this is the minimum solution, that is, we are guaranteed that there's
-- no solution with less number of coconuts. In fact, any member of @[15625*k-4 | k <- [1..]]@
-- is a solution, i.e., so are @31246@, @46871@, @62496@, @78121@, etc.
--
-- Note that we iteratively deepen our search by requesting increasing number of
-- solutions to avoid the all-sat pitfall.
sailors :: IO [Integer]
sailors = search 1
where search i = do soln <- ldn (Just i) [ ([1, -5, 0, 0, 0, 0, 0], 1)
, ([0, 4, -5 , 0, 0, 0, 0], 1)
, ([0, 0, 4, -5 , 0, 0, 0], 1)
, ([0, 0, 0, 4, -5, 0, 0], 1)
, ([0, 0, 0, 0, 4, -5, 0], 1)
, ([0, 0, 0, 0, 0, 4, -5], 1)
]
case soln of
NonHomogeneous (xs:_) _ -> return xs
_ -> search (i+1)