CODEWITHSUNDEEP
prologmax
sonlexqt
sonlexqt

Reputation: 6469

How to return the largest found result?

I have the following facts, which describes the database of a World Cup:

% host(X) <- X is the host country of World Cup
host('South Africa').
% federation (X,Y) <- X is the federation and Y is the numbers of countries in federation X
% qualified for World Cup.
federation('AFC',4).
federation('UEFA', 10).
federation('CAF', 5).
federation('CONMEBOL', 5).
% member_of(X,Y) <- Country Y is member of federation X
member_of('AFC','Australia').
member_of('UEFA', 'England').
member_of('CAF', 'South Africa').
member_of('CONMEBOL','Brazil').
% top(X) <- Country X is top of the World.
top('Brazil').

The question is, how do I find the federation that the number of Countries qualified to World Cup is largest?
What I think I'm gonna do is something like:
solution(X):-federation(X,Y), isMax(Y).
... but I don't know how to implement it yet.

Upvotes: 0

Views: 307

Answers (3)

lurker
lurker

Reputation: 58304

I offer here as another variation and won't claim is the best of those offered so far:

solution(X) :- setof(N-F, federation(F, N), L), reverse(L, [_-X|_]).

The setof/3 will gather unique pairs of N-F where F is the federation name, and N is the number of countries in that federation qualified for World Cup. They will be in increasing order according to the natural order of N-F, which is by number N then by F. The reverse puts the largest number first in the list, and [_-X|_] just selects the number from the head of the list.

ADDENDUM

It should be noted that if the data contains more than one maximum result, @gusbro's approach will generate all of them. However, the aggregate method will not. The setof/reverse method above can provide everything in descending order of magnitude, but needs a little extra help if only the maximums are to be picked out:

pick_top([X-F|_], X, F).
pick_top([_|T], X, F) :-
    pick_top(T, X, F).

solution(X) :-
    setof(N-F, federation(F, N), L),
    reverse(L, [C-Fed|T]),
    pick_top([C-Fed|T], C, X).

This will then generate all of the top federations:

federation('AFC',4).
federation('UEFA', 10).
federation('CAF', 5).
federation('CONMEBOL', 5).
federation('PDQ', 10).

| ?- solution(X).

X = 'UEFA' ? ;

X = 'PDQ' ? ;

no

An alternative solution using bagof and avoiding reverse:

pick_top([X-F|T], Top) :-
    pick_top(T, X, [F], Top).
pick_top([X-F|T], X, A, Top) :-
    pick_top(T, X, [F|A], Top).
pick_top([Y-_|T], X, A, Top) :-
    Y < X,
    pick_top(T, X, A, Top).
pick_top([Y-F|T], X, _, Top) :-
    Y > X,
    pick_top(T, Y, [F], Top).
pick_top([], _, Top, Top).

solution(X) :-
    bagof(N-F, federation(F, N), L),
    pick_top(L, X).

Which produces a list of maximums:

| ?- solution(X).

X = ['PDQ','UEFA'] ? a

no

Upvotes: 1

CapelliC
CapelliC

Reputation: 60034

library(aggregate) offers the appropriate constructs:

solution(X) :- aggregate(max(N,C), federation(C,N), max(_,X)).

?- solution(X).
X = 'UEFA'.

Upvotes: 1

gusbro
gusbro

Reputation: 22585

Your question can be reworded as "Find the federation for which there are no other federation with more qualified countries"

That statement can be written in prolog as:

solution(X):-
  federation(X,Y),    % Find the federation for which
  \+((                % there are no
    federation(_, Z), % other federation with
    Z > Y             % more qualified countries
  )).

Upvotes: 4

Related Questions