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Correctness of short cut fusion

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1 Short cut fusion

Short cut fusion allows elimination of intermediate data structures using rewrite rules that can also be performed automatically during compilation.

The two most popular instances are the
foldr
/
build
- and the
destroy
/
unfoldr
-rule for Haskell lists.

1.1
foldr
/
build

The
foldr
/
build
-rule eliminates intermediate lists produced by
build
and consumed by
foldr
, where these functions are defined as follows:
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr c n []     = n
foldr c n (x:xs) = c x (foldr c n xs)
 
build :: (forall b. (a -> b -> b) -> b -> b) -> [a]
build g = g (:) []
Note the rank-2 polymorphic type of
build
. The
foldr
/
build
-rule now means the following replacement for appropriately typed
g
,
c
, and
n
:
foldr c n (build g) => g c n


1.2
destroy
/
unfoldr

The
destroy
/
unfoldr
-rule eliminates intermediate lists produced by
unfoldr
and consumed by
destroy
, where these functions are defined as follows:
destroy :: (forall b. (b -> Maybe (a,b)) -> b -> c) -> [a] -> c
destroy g = g listpsi 
 
listpsi :: [a] -> Maybe (a,[a])
listpsi []     = Nothing
listpsi (x:xs) = Just (x,xs)
 
unfoldr :: (b -> Maybe (a,b)) -> b -> [a]
unfoldr p e = case p e of Nothing     -> []
                          Just (x,e') -> x:unfoldr p e'
Note the rank-2 polymorphic type of
destroy
. The
destroy
/
unfoldr
-rule now means the following replacement for appropriately typed
g
,
p
, and
e
:
destroy g (unfoldr p e) => g p e

2 Correctness

If the
foldr
/
build
- and the
destroy
/
unfoldr
-rule are to be automatically performed during compilation, as is possible using GHC's RULES pragmas, we clearly want them to be equivalences.

That is, the left- and right-hand sides should be semantically the same for each instance of either rule. Unfortunately, this is not so in Haskell.

We can distinguish two situations, depending on whether
g
is defined using
seq
or not.

2.1 In absence of
seq

If
g
does not use
seq
, then the
foldr
/
build
-rule really is a semantic equivalence, that is, it holds that
foldr c n (build g) = g c n

The two sides are semantically interchangeable.

The
destroy
/
unfoldr
-rule, however, is not a semantic equivalence.

To see this, consider the following instance:

g = \x y -> case x y of Just z -> 0
p = \x -> if x==0 then Just undefined else Nothing
e = 0
These values have appropriate types for being used in the
destroy
/
unfoldr
-rule. But with them, that rule's left-hand side "evaluates" as follows:
destroy g (unfoldr p e) = g listpsi (unfoldr p e)
                        = case listpsi (unfoldr p e) of Just z -> 0
                        = case listpsi (case p e of Nothing     -> []
                                                    Just (x,e') -> x:unfoldr p e') of Just z -> 0
                        = case listpsi (case Just undefined of Nothing     -> []
                                                               Just (x,e') -> x:unfoldr p e') of Just z -> 0
                        = undefined

while its right-hand side "evaluates" as follows:

g p e = case p e of Just z -> 0
      = case Just undefined of Just z -> 0
      = 0
Thus, by applying the
destroy
/
unfoldr
-rule, a nonterminating (or otherwise failing) program can be transformed into a safely terminating one.

The obvious questions now are:

  1. Can the converse also happen, that is, can a safely terminating program be transformed into a failing one?
  2. Can a safely terminating program be transformed into another safely terminating one that gives a different value as result?

There is no formal proof yet, but strong evidence supporting the conjecture that the answer to both questions is "No!".

The conjecture goes that if
g
does not use
seq
, then the
destroy
/
unfoldr
-rule is a semantic approximation from left to right, that is, it holds that
destroy g (unfoldr p e) <math>\sqsubseteq</math> g p e