I have this theorem (not sure if that's the right word), and I want to get all the solutions.
pairCube limit = do
m <- natural exists "m"
n <- natural exists "n"
a <- natural exists "a"
constrain $ m^3 .== n^2
constrain $ m .< limit
return $ m + n .== a^2
res <- allSat (pairCube 1000)
-- Run from ghci
extractModels res :: [[Integer]]
This is trying to solve the problem:
There are infinite pairs of integers (m, n) such that m^3 = n^2 and m + n is a perfect square. What is the pair with the greatest m less than 1000?
I know the actual answer, just through brute forcing, but I want to do with SBV.
However, when I run the code it gives only the following values (in the form [m, n, a]): [[9,27,6],[64,512,24],[]]
However, there are several other solutions with an m value less than 1000 that aren't included.
It's always good to give a full program:
{-# LANGUAGE ScopedTypeVariables #-}
import Data.SBV
pairCube :: SInteger -> Symbolic SBool
pairCube limit = do
(m :: SInteger) <- exists "m"
(n :: SInteger) <- exists "n"
(a :: SInteger) <- exists "a"
constrain $ m^(3::Integer) .== n^(2::Integer)
constrain $ m .< limit
return $ m + n .== a^(2::Integer)
main :: IO ()
main = print =<< allSat (pairCube 1000)
When I run it, I get:
Main> main
Solution #1:
m = 0 :: Integer
n = 0 :: Integer
a = 0 :: Integer
Solution #2:
m = 9 :: Integer
n = 27 :: Integer
a = -6 :: Integer
Solution #3:
m = 1 :: Integer
n = -1 :: Integer
a = 0 :: Integer
Solution #4:
m = 9 :: Integer
n = 27 :: Integer
a = 6 :: Integer
Solution #5:
m = 64 :: Integer
n = 512 :: Integer
a = -24 :: Integer
Solution #6:
m = 64 :: Integer
n = 512 :: Integer
a = 24 :: Integer
Unknown
Note the final Unknown.
Essentially, SBV queried Z3, and got 6 solutions; when SBV asked for the 7th, Z3 said "I don't know if there's any other solution." With non-linear arithmetic, this behavior is expected.
To answer the original question (i.e., find the max m
), I changed the constraint to read:
constrain $ m .== limit
and coupled it with the following "driver:"
main :: IO ()
main = loop 1000
where loop (-1) = putStrLn "Can't find the largest m!"
loop m = do putStrLn $ "Trying: " ++ show m
mbModel <- extractModel `fmap` sat (pairCube m)
case mbModel of
Nothing -> loop (m-1)
Just r -> print (r :: (Integer, Integer, Integer))
After running about 50 minutes on my machine, Z3 produced:
(576,13824,-120)
So, clearly the allSat
based approach is causing Z3 to give-up way earlier than what it can actually achieve if we fix m
and iterate ourself. With a non-linear problem, expecting anything faster/better would be too much to ask of a general purpose SMT solver..