Re: Counting propositions
From: x <x-false_at_yahoo.com>
Date: Fri, 18 Jun 2004 12:13:06 +0300
Message-ID: <40d2b121_at_post.usenet.com>
Date: Fri, 18 Jun 2004 12:13:06 +0300
Message-ID: <40d2b121_at_post.usenet.com>
- Post for FREE via your newsreader at post.usenet.com ****
A related Prolog question.
Guess and then check what is the meaning of "s(X)" from the following
Prolog clauses.
p(1). p(2). p(3).
s(X):-r(X,[]).
r(R,L):-q(X,L),r(R,[X|L]).
r(L,L).
q(X,L):-p(X), not(member(X,L)).
member(X,[X|_]).
member(X,[_|L]):-member(X,L).
It is possible to write a "pure" Prolog predicate that return the number of
elements that make p(X) true ?
(without using findall, bagof, setof, assert, etc.)
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
- Usenet.com - The #1 Usenet Newsgroup Service on The Planet! ***
http://www.usenet.com
Unlimited Download - 19 Seperate Servers - 90,000 groups - Uncensored
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=