Good points. But the "identity" thing is still what gets me. What is the
identity of an intersection?
Like you said it can't be #{}. If you seed an intersection with #{} you
get #{}, so you can't intersect from the empty set. The identity for an
intersection is whatever the common element is, but how would you know that?
On Friday, January 24, 2014 7:03:40 PM UTC-5, Cedric Greevey wrote:
>
> 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 <[email protected]<javascript:>
> > 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 <[email protected]> wrote:
>>>
>>>> Ah cool, thanks for posting your solution!
>>>>
>>>> On Friday, January 24, 2014 3:29:49 PM UTC-5, Tassilo Horn wrote:
>>>>
>>>>> Jarrod Swart <[email protected]> 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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