[Haskell-cafe] GHC magic optimization ?

[Matt Morrow]

Although, in Luke's example,

> x = sum [1..10^6] + product [1..10^6]
> x' = let l = [1..10^6] in sum l + product l

We can do much much better, if we're sufficiently smart.

-- Define:
bar m n = foo (enumFromTo m n)
foo xs  = sum xs + prod xs

-- We're given:
sum = foldl (+) 0
product = foldl (*) 1
foldl f z xs =
  case xs of
    [] -> []
    x:xs -> foldl f (f z x) xs
enumFromTo m n =
  case m < n of
    True -> []
    False -> m : enumFromTo (m+1) n

-- The fused loop becomes:
foo xs = go0 0 1 xs
  where go0 a b xs =
          case xs of
            [] -> a+b
            x:xs -> go0 (a+x) (b*x) xs

-- Now inline foo in bar:
bar m n = go2 0 1 m n
  where go2 = go0 a b (go1 m n)
        go0 a b xs =
          case xs of
            [] -> a+b
            x:xs -> go0 (a+x) (b*x) xs
        go1 m n =
          case m < n of
            True -> []
            False -> m : go1 (m+1) n

-- considering go2
go2 = go0 a b (go1 m n)

    ==> case (go1 m n) of
          [] -> a+b
           x:xs -> go0 (a+x) (b*x) xs

    ==> case (case m < n of
               True -> []
               False -> m : go1 (m+1) n) of
          [] -> a+b
          x:xs -> go0 (a+x) (b*x) xs

    ==> case m < n of
          True -> case [] of
                    [] -> a+b
                    x:xs -> go0 (a+x) (b*x) xs

          False -> case (m : go1 (m+1) n) of
                     [] -> a+b
                     x:xs -> go0 (a+x) (b*x) xs

    ==> case m < n of
          True -> a+b
          False -> go0 (a+m) (b*m) (go1 (m+1) n)

-- So,
go2 = case m < n of
        True -> a+b
        False -> go0 (a+m) (b*m) (go1 (m+1) n)

-- And by the original def of go2
go2 = go0 a b (go1 m n)

-- We get
go2 = case m < n of
        True -> a+b
        False -> go2 (a+m) (b*m) (m+1) n

-- go0 and go1 and now dead in bar
bar m n = go2 0 1 m n
  where go2 = case m < n of
                True -> a+b
                False -> go2 (a+m) (b*m) (m+1) n

-- (furthermore, if (+) here is for Int/Double etc,
-- we can reduce go2 further to operate on machine
-- ints/doubles and be a register-only non-allocating loop)

-- So now finally returning to our original code:
> x = sum [1..10^6] + product [1..10^6]
> x' = let l = [1..10^6] in sum l + product l

-- We get:
x' = bar 1 (10^6)

And the intermediate list never exists at all.

Matt




On 12/4/09, Luke Palmer <lrpalmer at gmail.com> wrote:
> On Fri, Dec 4, 2009 at 3:36 AM, Neil Brown <nccb2 at kent.ac.uk> wrote:
>> But let's say you have:
>>
>> g x y = f x y * f x y
>>
>> Now the compiler (i.e. at compile-time) can do some magic.  It can spot
>> the
>> common expression and know the result of f x y must be the same both
>> times,
>> so it can convert to:
>>
>> g x y = let z = f x y in z * z
>
> GHC does *not* do this by default, quite intentionally, even when
> optimizations are enabled.  The reason is because it can cause major
> changes in the space complexity of a program.  Eg.
>
> x = sum [1..10^6] + product [1..10^6]
> x' = let l = [1..10^6] in sum l + product l
>
> x runs in constant space, but x' keeps the whole list in memory.  The
> CSE here has actually wasted both time and space, since it is harder
> to save [1..10^6] than to recompute it!  (Memory vs. arithmetic ops)
>
> So GHC leaves it to the user to specify sharing.  If you want an
> expression shared, let bind it and reuse.
>
> Luke
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