A monad for expressing SAT or QSAT problems using observable sharing.

A monad for expressing SAT or QSAT problems using observable sharing.

For example, we can express a full-adder with:

full_adder :: Bit -> Bit -> Bit -> (Bit, Bit)
full_adder a b cin = (s2, c1 || c2)
  where (s1,c1) = half_adder a b
        (s2,c2) = half_adder s1 cin

half_adder :: Bit -> Bit -> (Bit, Bit)
half_adder a b = (a `xor` b, a && b)

Longer Examples

Included are a couple of examples included with the distribution. Neither are as fast as a dedicated solver for their respective domains, but they showcase how you can solve real world problems involving 10s or 100s of thousands of variables and constraints with ersatz.

ersatz-sudoku

% time ersatz-sudoku
Problem:
┌───────┬───────┬───────┐
│ 5 3   │   7   │       │
│ 6     │ 1 9 5 │       │
│   9 8 │       │   6   │
├───────┼───────┼───────┤
│ 8     │   6   │     3 │
│ 4     │ 8   3 │     1 │
│ 7     │   2   │     6 │
├───────┼───────┼───────┤
│   6   │       │ 2 8   │
│       │ 4 1 9 │     5 │
│       │   8   │   7 9 │
└───────┴───────┴───────┘
Solution:
┌───────┬───────┬───────┐
│ 5 3 4 │ 6 7 8 │ 9 1 2 │
│ 6 7 2 │ 1 9 5 │ 3 4 8 │
│ 1 9 8 │ 3 4 2 │ 5 6 7 │
├───────┼───────┼───────┤
│ 8 5 9 │ 7 6 1 │ 4 2 3 │
│ 4 2 6 │ 8 5 3 │ 7 9 1 │
│ 7 1 3 │ 9 2 4 │ 8 5 6 │
├───────┼───────┼───────┤
│ 9 6 1 │ 5 3 7 │ 2 8 4 │
│ 2 8 7 │ 4 1 9 │ 6 3 5 │
│ 3 4 5 │ 2 8 6 │ 1 7 9 │
└───────┴───────┴───────┘
ersatz-sudoku  1,13s user 0,04s system 99% cpu 1,179 total

ersatz-regexp-grid

This solves the "regular crossword puzzle" (grid.pdf) from the 2013 MIT mystery hunt.

% time ersatz-regexp-grid

SPOILER

ersatz-regexp-grid  2,45s user 0,05s system 99% cpu 2,502 total

Ersatz

Ersatz is a library for generating QSAT (CNF/QBF) problems using a monad. It takes care of generating the normal form, encoding your problem, marshaling the data to an external solver, and parsing and interpreting the result into Haskell types.

What differentiates Ersatz is the use of observable sharing in the API.

For instance to define a full adder:

full_adder :: Bit -> Bit -> Bit -> (Bit, Bit)
full_adder a b cin = (s2, c1 || c2)
  where (s1,c1) = half_adder a b
        (s2,c2) = half_adder s1 cin

half_adder :: Bit -> Bit -> (Bit, Bit)
half_adder a b = (a `xor` b, a && b)

as opposed to the following code in satchmo:

full_adder :: Boolean -> Boolean -> Boolean
           -> SAT ( Boolean, Boolean )
full_adder a b c = do
  let s x y z = sum $ map fromEnum [x,y,z]
  r <- fun3 ( \ x y z -> odd $ s x y z ) a b c
  d <- fun3 ( \ x y z -> 1   < s x y z ) a b c
  return ( r, d )

half_adder :: Boolean -> Boolean
           -> SAT ( Boolean, Boolean )
half_adder a b = do
  let s x y = sum $ map fromEnum [x,y]
  r <- fun2 ( \ x y -> odd $ s x y ) a b
  d <- fun2 ( \ x y -> 1   < s x y ) a b
  return ( r, d )

This enables you to use the a much richer subset of Haskell than the purely monadic meta-language, and it becomes much easier to see that the resulting encoding is correct.

To allocate fresh existentially or universally quantified variables or to assert that a Bit is true and add the attendant circuit with sharing to the current problem you use the SAT monad.

verify_currying :: (MonadQSAT s m) => m ()
verify_currying = do
  (x::Bit, y::Bit, z::Bit) <- forall_
  assert $ ((x && y) ==> z) === (x ==> y ==> z)

We can then hand that off to a SAT solver, and get back an answer:

main = solveWith depqbf verify_currying >>= print

Support is offered for decoding various Haskell datatypes from the solution provided by the SAT solver.

Examples

Included are a couple of examples included with the distribution. Neither are as fast as a dedicated solver for their respective domains, but they showcase how you can solve real world problems involving 10s or 100s of thousands of variables and constraints with ersatz.

sudoku

% time ersatz-sudoku
Problem:
┌───────┬───────┬───────┐
│ 5 3   │   7   │       │
│ 6     │ 1 9 5 │       │
│   9 8 │       │   6   │
├───────┼───────┼───────┤
│ 8     │   6   │     3 │
│ 4     │ 8   3 │     1 │
│ 7     │   2   │     6 │
├───────┼───────┼───────┤
│   6   │       │ 2 8   │
│       │ 4 1 9 │     5 │
│       │   8   │   7 9 │
└───────┴───────┴───────┘
Solution:
┌───────┬───────┬───────┐
│ 5 3 4 │ 6 7 8 │ 9 1 2 │
│ 6 7 2 │ 1 9 5 │ 3 4 8 │
│ 1 9 8 │ 3 4 2 │ 5 6 7 │
├───────┼───────┼───────┤
│ 8 5 9 │ 7 6 1 │ 4 2 3 │
│ 4 2 6 │ 8 5 3 │ 7 9 1 │
│ 7 1 3 │ 9 2 4 │ 8 5 6 │
├───────┼───────┼───────┤
│ 9 6 1 │ 5 3 7 │ 2 8 4 │
│ 2 8 7 │ 4 1 9 │ 6 3 5 │
│ 3 4 5 │ 2 8 6 │ 1 7 9 │
└───────┴───────┴───────┘
ersatz-sudoku  1,13s user 0,04s system 99% cpu 1,179 total

regexp-grid

This solves the regular crossword puzzle from the MIT mystery hunt.

% time ersatz-regexp-grid

SPOILER

ersatz-regexp-grid 2,45s user 0,05s system 99% cpu 2,502 total

Contact Information

Contributions and bug reports are welcome!

Please feel free to contact me through github or on the #haskell IRC channel on irc.freenode.net.

-Edward Kmett.