Non-determinism, backtracking and Monads

Graham Klyne
Wed, 11 Jun 2003 08:37:48 +0100

At 11:06 05/06/03 +0200, Jerzy Karczmarczuk wrote:
>I permit myself to observe that your powerset problem (and the restricted
>length problem, i.e. the combinations) is usually solved in Prolog, through
>backtracking, using reasoning/style which adopts this "individualistic"
>powerset(<source>,<possible result>)   ---   is the pattern. And the
>solution is
>powerset([],[]).   Since nothing else can be done. Otherwise you pick the item
>                    or not.
>powerset([X|Rest],L) :- powerset(Rest,L).
>powerset([X|Rest],[X|L) :- powerset(Rest,L).
>The xxx ++ map (x :) xxx  solution in Haskell is a particular formulation
>(and optimization) of the straightforward transformation from a version
>using the non-deterministic Monad. This one is really almost a carbon copy
>of the Prolog solution, with appropriate "lifting" of operations from
>individuals to lazy lists.

I was thinking some more about this comment of yours, and my own experience 
with the ease of using lists to implement prolog-style generators, and 
think I come to some better understanding.  If I'm right, I assume the 
following is common knowledge to experienced Haskell programmers.  So, in 
the spirit of testing my understanding...

The common thread here is a non-deterministic calculation in which there 
are several possible solutions for some problem.  The goal is to find (a) 
if there are any solutions, and (b) one, more or all of the solutions.

Prolog does this, absent ! (cut), by backtracking through the possible 

My first epiphany is that the Haskell idea of using a lazily evaluated list 
for result of a non-deterministic computation is pretty much the same thing 
(which is pretty much what you said?  Is this what you mean by "the 
non-deterministic monad"?).  The mechanisms for accessing a list mean that 
the solutions must be accessed in the order they are generated, just like 
Prolog backtracking.

So there seems to be a very close relationship between the lazy list and 
non-deterministic computations, but what about other data structures?  I 
speculate that other structures, lazily evaluated, may also be used to 
represent the results of non-deterministic computations, yet allow the 
results to be accessed in a different order.  And these, too, may be 
(should be?) monads.  If so, the Haskell approach might be viewed as a 
generalization of Prolog's essentially sequential backtracking.

In a private message concerning the powerset thread on this list, a 
correspondent offered a program to evaluate the subsets in size order, 
which I found particularly elegant:

 >ranked_powerset :: [a] -> [[[a]]]
 >ranked_powerset = takeWhile (not . null) . foldr next_powerset ([[]] : 
repeat [])
 >next_powerset :: a -> [[[a]]] -> [[[a]]]
 >next_powerset x r = zipWith (++) ([] : map (map (x:)) r) r
 >powerset :: [a] -> [[a]]
 >powerset = tail . concat . ranked_powerset

They also pointed out that "ranked_powerset is handy since you can use it 
to define combinatorial choice etc.":

 > choose :: Int -> [a] -> [[a]]
 > choose k = (!! k) . ranked_powerset

So here is an example of a different structure (a list of lists) also used 
to represent a non-deterministic computation, and furthermore providing 
means to access the results in some order other than a single linear 
sequences (e.g. could be used to enumerate all the powersets containing the 
nth member of the base set, *or* all the powersets of a given size, without 
evaluating all of the other powersets).

To test this idea, I think it should be possible to define a monad based on 
a simple tree structure, which also can be used to represent the results of 
a non-deterministic computation.  An example of this is below, at the end 
of this message, which seems to exhibit the expected properties.

So if the idea of representing a non-deterministic computation can be 
generalized from a list to a tree, why not to other data structures?  My 
tree monad is defined wholly in terms of reduce and fmap.  Without going 
through the exercise, I think the reduce function might be definable in 
terms of fmap for any data type of the form "Type (Maybe a)", hence 
applicable to a range of functors?  In particular, I'm wondering if it can 
be applied to any gmap-able structure over a Maybe type.

I'm not sure if this is of any practical use;  rather it's part of my 
attempts to understand the relationship between functors and monads and 
other things functional.


-- spike-treemonad.hs

   data Tree a = L (Maybe a) | T { l,r :: Tree a }
     deriving Eq

   instance (Show a) => Show (Tree a) where
     show t = (showTree "" t) ++ "\n"

   showTree :: (Show a) => String -> Tree a -> String
   showTree _ (L Nothing ) = "()"
   showTree _ (L (Just a)) = show a
   showTree i (T l r) = "( " ++ (showTree i' l) ++ "\n" ++
                        i'   ++ (showTree i' r) ++ " )"
                        where i' = ' ':' ':i

   instance Functor Tree where
     fmap f (L Nothing)  = L Nothing
     fmap f (L (Just a)) = L (Just (f a))
     fmap f (T l r)      = T (fmap f l) (fmap f r)

   reduce :: Tree (Tree a) -> Tree a
   reduce (L Nothing)              = L Nothing
   reduce (L (Just (L Nothing ) )) = L Nothing
   reduce (L (Just (L (Just a)) )) = L (Just a)
   reduce (L (Just (T l r     ) )) = T l r
   reduce (T l r)                  = T (reduce l) (reduce r)

   instance Monad Tree where
     -- L Nothing >>= k = L Nothing
     t         >>= k = reduce $ fmap k t
     return x        = L (Just x)
     fail s          = L Nothing

   -- tests

   t1 :: Tree String
   t1 = T (L $ Just "1")
          (T (T (T (L $ Just "211")
                   (L $ Just "311"))
                (L Nothing))
             (L $ Just "22"))

   k1 :: a -> Tree a
   k1 n = T (L $ Just n) (L $ Just n)

   r1 = t1 >>= k1

   k2 :: String -> Tree String
   k2 s@('1':_) = L $ Just s
   k2 s@('2':_) = T (L $ Just s) (L $ Just s)
   k2 _         = L Nothing

   r2 = t1 >>= k2

   t3 = L Nothing :: Tree String
   r3 = t3 >>= k1

   r4 = t1 >>= k1 >>= k2
   r5 = (return "11") :: Tree String
   r6 = r5 >>= k1

   -- Check out monad laws
   -- return a >>= k = k a
   m1a = (return t1) >>= k1
   m1b = k1 t1
   m1c = m1a == m1b

   -- m >>= return = m
   m2a = t1 >>= return
   m2b = t1
   m2c = m2a == m2b

   -- m >>= (\x -> k x >>= h) = (m >>= k) >>= h
   m3a =  t1 >>= (\x -> k1 x >>= k2)
   m3b =  (t1 >>= k1) >>= k2
   m3c = m3a == m3b

Graham Klyne
PGP: 0FAA 69FF C083 000B A2E9  A131 01B9 1C7A DBCA CB5E