Dup Foo?

[Chris Kaltwasser]

Hello,

I am curious whether there is an easier way to find duplicate fact instances along with the "duplication count" than the one I have come up with.

I have a fairly simple solution, but feel there should be something less "convoluted" than my solution.

An example, I think best explains this:

foo(a,aa).
foo(a,ac).
foo(a,aa).
foo(b,aa).
foo(b,xy).
foo(b,xy).
foo(b,xy).
foo(b,ac).
foo(b,aa).

foo(X,Y,1) :- foo(X,Y).

dupfoo(X,Y,C) :-
    bagof(Z,foo(X,Y,Z),L), length(L,C), C>1.

The goal is to locate the cases where foo(X,Y) produces multiple identical solutions.

My definition for dupfoo(X,Y,C) shown above gives the correct results on backtracking.  Namely:

?- dupfoo(X,Y,C).
X = a,
Y = aa,
C = 2 ;
X = b,
Y = aa,
C = 2 ;
X = b,
Y = xy,
C = 3.

What I don't like, in particular, is that I had to create the foo/3 predicate whose only function is to assist in collecting the duplicate count (C).  It seems to me that there should be some way to make use of the bagof predicate without having to make use of the foo/3 predicate.

Any thoughts or insights would be most appreciated.

Thanks,
Chris Kaltwasser

[Jan Wielemaker]

On Tuesday 13 May 2008 20:17:53 Chris Kaltwasser wrote:

> Hello,
>
> I am curious whether there is an easier way to find duplicate fact
> instances along with the "duplication count" than the one I have come up
> with.
>
> I have a fairly simple solution, but feel there should be something less
> "convoluted" than my solution.
>
> An example, I think best explains this:
>
> foo(a,aa).
> foo(a,ac).
> foo(a,aa).
> foo(b,aa).
> foo(b,xy).
> foo(b,xy).
> foo(b,xy).
> foo(b,ac).
> foo(b,aa).
>
> foo(X,Y,1) :- foo(X,Y).
>
> dupfoo(X,Y,C) :-
>     bagof(Z,foo(X,Y,Z),L), length(L,C), C>1.
>
> The goal is to locate the cases where foo(X,Y) produces multiple identical
> solutions.
>
> My definition for dupfoo(X,Y,C) shown above gives the correct results on
> backtracking.  Namely:
>
> ?- dupfoo(X,Y,C).
> X = a,
> Y = aa,
> C = 2 ;
> X = b,
> Y = aa,
> C = 2 ;
> X = b,
> Y = xy,
> C = 3.
>
> What I don't like, in particular, is that I had to create the foo/3
> predicate whose only function is to assist in collecting the duplicate
> count (C).  It seems to me that there should be some way to make use of the
> bagof predicate without having to make use of the foo/3 predicate.

        bagof(1, foo(X,Y), L), length(L,C), C>1.

Nobody ever said the template must contain variables from the goal :-)

Intuitively I would have choosen something along these lines:

        findall(X-Y, foo(X, Y), Solutions),
        msort(Solutions, Sorted),
        <find duplicates in sorted list>

But this is a lot more typing and probably only a bit more efficient.

        Cheers --- Jan



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