Reputation: 25
Very new to prolog. I am trying to create a simple recursive rule to find nth element on the list. For example if I have a list of strings or numbers I want to be able to use the query
?- findme([dog, cat , bird], 1, R).
R = dog
?- findme([9,8,7,6,5], 3,R).
R= 7
?- findme([mouse,cheese,cat,milk], 5, R).
R = false
I wish not to use the builtin Nth0, nor have R such that R is n -1
Upvotes: 0
Views: 1091
Reputation: 37771
The following is giving expected output.
find([],N) :- write("There is no such element in the list"), nl.
find([Element|List],1) :- write("The element is ", Element), nl.
find([Element|List],N) :- N1 = N-1, find(List,N1).
Output :
find([1,2,3,4],3)
The element is 3
Yes
find([1,2,3,4],0)
There is no such element in the list
Yes
find([1,2,3,4],5)
There is no such element in the list
Yes
find([1,2,4,3],4)
The element is 3
Yes
Update
find([],N,false) :- N>0.
find([Element|List],1,Element).
find([Element|List],N,R) :- N1 = N-1, find(List,N1,R).
Upvotes: 0
Reputation: 12992
Here is my implementation:
find([],N,false):-N>0.
find([H|_],1,H).
find([_|T],N,R) :- N1 is N-1, find(T,N1,R).
Some examples:
?- find([dog, cat , bird], 1, R).
R = dog ;
false.
?- find([mouse,cheese,cat,milk], 5, R).
R = false.
?- find([9,8,7,6,5], 3,R).
R = 7 ;
false.
Upvotes: 1