Question

I’m trying to make a prolog function. The function reads in a sentence, and then tries to extract a key word. If a key word is found, it prints a message. I want it to also print a message if no keywords are found. Here is my example :

contains([word1|_]) :- write('word1 contained').
contains([Head|Tail]) :- Head \= word1, contains(Tail).
contains([word2|_]) :- write('word2 contained').
contains([Head|Tail]) :- Head \= word2, contains(Tail).
contains([word3|_]) :- write('word3 contained').
contains([Head|Tail]) :- Head \= word3, contains(Tail).

The above code will check and see if the extracted word is present. But it does not give an answer if the words ‘word1,word2 or word3’ are not contained. Does anybody know how I should go about implementing this?

I tried adding :

contains([_|_]) :- write('nothing contained'),nl.
contains([Head|Tail]) :- Head \= _, contains(Tail).

But clearly this is the wrong thing to do.

Answer 1

The standard way to write the main part of your contains predicate is:

contains([word1|_]) :- !, write('word1 contained').
contains([word2|_]) :- !, write('word2 contained').
contains([word3|_]) :- !, write('word3 contained').
contains([Head|Tail]) :- contains(Tail).

Which means:

• when you find a word, don’t search any further (this is what the cut (!) operator is for).
• when nothing else worked, recurse on tail.

To add an answer in case nothing is found, just add another cut on the recursive call, so that the later case is only called when nothing else (including recursion) worked:

contains([word1|_]) :- !, write('word1 contained').
contains([word2|_]) :- !, write('word2 contained').
contains([word3|_]) :- !, write('word3 contained').
contains([Head|Tail]) :- contains(Tail), !.
contains(_) :- write('Nothing found').