RE: "distinct" predicate

[Robert Wolf]

Abhinav-Bhardwaj,

There is a "beutiful" prolog "native" solution to "dsitinct" problem.
You can wirite it easily yourself.
Here are the hints:
1. Empty list is basically "distinct".
2. The list is "distinct" if:
        a) the "head" element is not on the "tail" list AND
        b) the "tail" list is "distinct"
3. Your dialect of prolog may already have a predicate checking if
something is on the list.
4. If not, you can write it easily yourself.
5. An element is on the list if:
        a) it is the same as the "head" element of the list OR
        b) it is on the "tail" list

Something like: (I use another dialect, so you may have to modify
syntax, sorry)

distinct ([]) :- .
distinct ([a]) :- .
distinct ([head : tail]) :-
        NOT (is_on_the_list (head, tail)),
        distinct (tail).

is_on_the_list (head, [head :tail]) :- .
is_on_the_list (head, [something :tail]) :-
        is_on_the_list (head, tail).

Best regards and "happy prologing"...

Robert P. Wolf
address@hidden

-----Original Message-----
From: Abhinav-Bhardwaj [mailto:address@hidden
Sent: Thursday, 31 October 2002 7:55 PM
To: address@hidden
Subject: "distinct" predicate


hi,
is there any way by which i can specify a rule say
distinct..which succeeds only when all the terms in a list
are distinct.
for eg:
?- distinct([a, b, c, d]).
yes
?- distinct([a, b, a, d]).
no

thanks in advance,
regards,
Abhinav




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