On Sun, Jul 24, 2011 at 6:26 PM, Costello, Roger L. wrote:

Hi Folks,

I am creating an Int range by starting with the full range of Int:

let r1  =  [minBound::Int .. maxBound::Int]

Then I restrict the upper bound of the range, e.g.,

let r2  =  r1 `intersect` [minBound::Int .. 10]

Then I restrict the lower bound of the range, e.g.,

let r3  =  r2 `intersect` [0 .. maxBound::Int]

No error thus far.

However, when I try to display r3 I get an "Out of Memory" error.

Why is that?

While it is not obvious from the documentation alone, 'intersect' is not lazy enough. It is defined as intersect xs ys = [x | x <- xs, any (== x) ys] So, supose you have x:xs and ys such that x ∉ ys. Then 'intersect (x:xs) ys' needs to traverse all of ys before deciding that x really isn't inside ys. Now, you're doing 'r2 `intersect` [0 .. maxBound::Int]'. In this case, 'x:xs = r2' and 'ys = [0 .. maxBound::Int]'. The first element of r2 is minBound, which of course is not in ys. But to find that out, 'intersect' would need to traverse the whole 'ys' list. This should require at least 32 GiB on GHC with a 32 bit platform, and much much much more on a 64 bit platform.

I created a function that restricts a range using the above approach:

restrict                                  ::  [Int] -> Restriction-> [Int] restrict r (MinInclusive v)    =   r `intersect` [v .. maxBound::Int] restrict r (MaxInclusive v)   =   r `intersect` [minBound::Int .. v]

The arguments to "restrict" is a range, r, and an indication of whether the upper bound is to be restricted (MaxInclusive v) or the lower bound is to be restricted (MinInclusive v).

Is there a better way to do this?

This will give you the same problems.

What can I do to my restrict function to avoid getting an "Out of Memory" error?

By explicitly representing intervals, instead of using the list of elements they contain. Something like: data Interval a = Interval {mini :: a, maxi :: a} deriving (Eq, Ord, Show) type IntervalUnion a = [Interval a] Instead of '[x..y]', use '[Interval x y]'. Instead of '[a..b] ++ [c..d]', use '[Interval a b, Interval c d]'. HTH, -- Felipe.