Possible behaviors of `predsort/3`

问题

This is a followup to an answer to a question about sorting on a particular argument of a term, without creating a new list for a keysort (if I understood the original question correctly).

Say we wanted predsort/3 to behave exactly as sort/2: if I understand correctly, this would mean calling it as:

?- predsort(compare, List, Sorted).

Now say that we wanted to use predsort/3 to sort as implemented by msort/2 (see also this question). One way to do it would be to define a comparison predicate Pred(-Delta, +A, +B) that does not unify Delta with = when the elements are actually equal:

mcompare(Delta, A, B) :-
    compare(Delta0, A, B),
    (   Delta0 == (=)
    ->  Delta = (<)
    ;   Delta = Delta0
    ).

?- predsort(mcompare, List, Sorted).

Question: does that really simply sort without removing duplicates, as msort/2 does? It seems like it should.

Moving on: say we wanted to sort terms with arity > n, on the standard order of the nth argument in the term. The clean way to do it would be:

sort_argn(N, List, Sorted) :-
    map_list_to_pairs(arg(N), List, Pairs),
    keysort(Pairs, Sorted_pairs),
    pairs_values(Sorted_pairs, Sorted).

If we wanted to use predsort/3 to achieve the same effect, we could try using a comparison predicate as follows:

compare_argn(N, Delta, A, B) :-
    arg(N, A, AN),
    arg(N, B, BN),
    compare(Delta, AN-A, BN-B).

And to sort on the second argument:

?- predsort(compare_argn(2), List, Sorted).

However, this is not the same as sort_argn/3 above that uses keysort/2. It will remove duplicates, and it will order compound terms according to the standard order of the original full term if the second arguments of two terms happen to be equal:

?- predsort(compare_argn(2), [f(b,2), f(a,1), f(a,1), f(a,2)], Sorted).
Sorted = [f(a, 1), f(a, 2), f(b, 2)].

?- sort_argn(2, [f(b,2), f(a,1), f(a,1), f(a,2)], Sorted).
Sorted = [f(a, 1), f(a, 1), f(b, 2), f(a, 2)].

Making the assumption that for every pair of A and B passed to the comparison predicate Pred(Delta, A, B), A comes before B in the original list. Can we define a comparison:

compare_argn_stable(N, Delta, A, B) :-
    arg(N, A, AN),
    arg(N, B, BN),
    compare(Delta0, AN, BN),
    (   Delta0 == (=)
    ->  Delta = (<)
    ;   Delta = Delta0
    ).

At this point, if and only if any two elements A and B are always passed to the comparison predicate in the same order as they were in the original list, this should behave identically to sort_argn/3 above:

?- predsort(compare_argn_stable(N), List, Sorted).

Now of course it is important that compare_argn_stable/4 unifies Delta with < when the two "keys" are equal. Furthermore, the behavior is implementation dependent, and only identical to the keysort example iff predsort/3 keeps the original order of elements when passing them to the comparison predicate.

Question Is that correct?

Question Is there any standard that covers this aspect of predsort/3?

回答1:

Since no one has answered, and since I am quite certain about it now:

Yes, you could use predsort/3 to emulate any of the other sorts. The question describes in some detail how.

However: this is a bad idea for several reasons.

• The "stability" depends on the implementation of predsort/3 (see the question)
• The predsort/3 itself is not part of any standard (as far as I can tell)
• The chances are, your Prolog implementation provides an msort/2 or keysort/2 that is far more efficient than predsort/3

There might be rare cases where the size of the elements of the list is much bigger than the length of the list we are sorting, and this little dance:

list_to_keyval_pairs(List, Pairs), % defined by the user as necessary
keysort(Pairs, Sorted_pairs),
pairs_values(Sorted_pairs, Sorted)

(see here) is actually more expensive (slower) than using predsort(keycmp, List, Sorted), with keycmp/3 defined by the user. Even then, the order of results with equivalent keys depends not only on the (user) definition of keycmp/3, but also on the implementation of predsort/3.

In other words, a "stable" sort with predsort/3 is a bad idea.

来源：https://stackoverflow.com/questions/28076715/possible-behaviors-of-predsort-3