Intersection is associative and commutative: (intersection A B) =
(intersection B A) and (intersection A (intersection B C)) = (intersection
(intersection A B) C) = the elements common to all three sets. So it's
actually perfectly well-founded for use with reducers, at least in
principle, and intersecting A B C D can be parallelized sensibly by
parallel intersecting A B and C D and then intersecting the two resulting
sets.


On Fri, Jan 24, 2014 at 6:43 PM, Jarrod Swart <jcsw...@gmail.com> wrote:

> If I understand you correctly I am in agreement.  I don't think you could
> take this problem to clojure.core.reducers/reduce or fold because the
> problem is inherently sequential is it not?
>
> The reduction is basically (intersection (intersection (intersection A B)
> C) D).
>
> I was curious of this myself, how do I abstract out the order of the
> (reduce set/intersection ...).  I couldn't think of one.
>
> Breaking this problem out into 'parallel' units of reduction isn't
> possible because the problem is dependent on order.  Which reducers can't
> have, or so I think after what I have read today.
>
>
> On Friday, January 24, 2014 3:56:23 PM UTC-5, Cedric Greevey wrote:
>
>> An interesting question this raises is if there is any sensible way to
>> define (intersection). It would need to behave as an identity element for
>> intersection, so would need to behave as a set (so, (set? (intersection))
>> => truthy) that contained everything (so, (contains? (intersection) foo) =>
>> foo no matter what foo is; (partial contains? (intersection)) => identity).
>> The problem would be what to do with seq? Ideally an infinite seq that will
>> produce any particular value after finite time would be produced, but
>> there's no way to sensibly produce "any particular value" given the wide
>> variety of constructor semantics, builders, factory methods, things not
>> known to this particular runtime instance but that conceptually exist
>> somewhere, etc.; of course, the seq return is a dummy of sorts anyway since
>> you couldn't really use it sensibly to it might as well just return
>> (range). Printing should likely be overridden to just print
>> "(intersection)" rather than b0rk the REPL with a neverending stream of
>> integers (or whatever).
>>
>> But then it also subtly violates another property of Clojure set objects:
>> if (= a b), (not (identical? a b)), and (identical? (a-set a) a), then
>> (identical? (a-set b) a) and thus (not (identical? (a-set b) b)). The
>> latter is true under the hypothesis for every "real" set but would be false
>> for (intersection).
>>
>> Perhaps this is why (intersection) is not supported at this time, even
>> though (union) returns an empty set object, the identity element for the
>> union operation.
>>
>>
>> On Fri, Jan 24, 2014 at 3:34 PM, Jarrod Swart <jcs...@gmail.com> wrote:
>>
>>> Ah cool, thanks for posting your solution!
>>>
>>> On Friday, January 24, 2014 3:29:49 PM UTC-5, Tassilo Horn wrote:
>>>
>>>> Jarrod Swart <jcs...@gmail.com> writes:
>>>>
>>>> > The reason you can't get this to work is that r/map returns a
>>>> <reducible>
>>>> > not a <coll> for reduce to operate on.
>>>>
>>>> Ah, indeed.  I couldn't see the forest for the trees.
>>>>
>>>> > I'm not sure of a solution because I'm not familiar with
>>>> > core.reducers.
>>>>
>>>> This works:
>>>>
>>>>   (reduce set/intersection (r/foldcat (r/map set [[1 2] [3 1] [1 3]])))
>>>>
>>>> Bye,
>>>> Tassilo
>>>>
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