[erlang-questions] Intersection of two ordered sets

Richard Carlsson richardc@REDACTED
Sun May 10 21:02:16 CEST 2009


Nicholas Dronen wrote:
> Does this look like a sane implementation for finding the intersection
> of two ordered sets?  There are implementations in the standard erlang
> libraries (e.g. ordsets.erl), but they're a bit different and I wanted
> to make sure I'm not doing anything wildly wrong, either in terms of
> correctness or performance.  (I'm not reinventing the wheel here --
> just trying to learn erlang by implementing whatever comes to mind.)
> 
>     intersection(L1, L2) -> intersection(L1, L2, []).
> 
>     intersection([H1|T1], [H2|T2], Result) when H1 == H2 ->
>         intersection(T1, T2, [H1|Result]);
>     intersection([H1|T1], [H2|T2], Result) when H1 > H2 ->
>         intersection([H1|T1], T2, Result);
>     intersection([H1|T1], [H2|T2], Result) when H1 < H2 ->
>         intersection(T1, [H2|T2], Result);
>     intersection(_, _, Result) -> lists:reverse(Result).

Sane, yes. However:

First, there's probably no point in using an accumulator and reversing
at the end; a plain recursive version has about the same efficiency,
generates less garbage, and is easier to read.

Second, by always testing for equality (which is the least likely case)
in the first clause, you're generally wasting a test per element.

Third, if you put the equality test after the > and < tests instead,
you don't really need to actually do the test at all. If neither of
those tests are true, the elements have to be equal.

And fourth, in clause 2 you are deconstructing [H1|T1] and then
reconstructing it again (the compiler won't catch this reuse),
and similarly for [H2|T2] in clause 3.

     /Richard



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