CODEWITHSUNDEEP
prologmeta-predicate
annita
annita

Reputation: 41

Checking all Members Of List

Let's say that I have the following list in my prolog:

L=[10,11,2,3,5]

Is there a way that we can check all the members of a list L to make sure that each member is less than 5?

Upvotes: 1

Views: 348

Answers (2)

Éric López Clavero
Éric López Clavero

Reputation: 71

Here goes another solution without using any built-in function:

all_less_five([]).
all_less_five([X|L]):-
        X < 5,
        all_less_five(L).

This solution uses the typical recursion over lists. The predicate stands true for the empty list, and then we only call the recursion over the tail if the head is less than five.

Here are some questions over the predicate:

?- all_less_five([10,11,2]).
false.
?- all_less_five([2,3,6,5]).
false.
?- all_less_five([1,2,3,4]).
true.

Now it should be easy to implement it to any given X. Try it!

Upvotes: 0

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477881

We can make use of maplist/2 here. This is a predicate that:

maplist(:Goal, ?List)

True if Goal can successfully be applied on all elements of List. Arguments are reordered to gain performance as well as to make the predicate deterministic under normal circumstances.

So we can here check the elements with:

all_less_five(L) :-
    maplist(>(5), L).

Here for every element x ∈ L, it will thus call >(5, x), or in inline form 5 > x. So if all these elements are less than five, the the all_less_five/1 predicate will succeed.

For example:

?- all_less_five([10,11,2,3,5]).
false.

?- all_less_five([2,3,5]).
false.

?- all_less_five([2,3]).
true.

Upvotes: 1

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