How to find the Nth element of a list in Prolog
Solution 1
First of all, there is a builtin nth0/3
for that:
?- nth0(0,[a,b,c],X).
X = a.
?- nth0(1,[a,b,c],X).
X = b.
?- nth0(2,[a,b,c],X).
X = c.
?- nth0(3,[a,b,c],X).
false.
Get the i-th element
The problem is in the inductive case:
match([Elem|Tail],Num,Counter,MatchedNumber):-
match(Tail,Num,N,Elem),
C is N+1.
Prolog doesn't know anything about C
so the last statement doens't force Prolog to return the i-th element. It can simply return any element because N
will match with Num
in the recursive call and then set C
to Num+1
but that's not a problem because C
is not bound by anything.
A better way to solve this, is using a decrement counter:
match([H|_],0,H) :-
!.
match([_|T],N,H) :-
N > 0, %add for loop prevention
N1 is N-1,
match(T,N1,H).
Example:
?- match([a,b,c,d,e],0,X).
X = a.
?- match([a,b,c,d,e],1,X).
X = b.
?- match([a,b,c,d,e],2,X).
X = c.
?- match([a,b,c,d,e],3,X).
X = d.
?- match([a,b,c,d,e],4,X).
X = e.
?- match([a,b,c,d,e],5,X).
false.
The base case is thus that the index is 0
in which case you return the head, otherwise you query for the i-1-th element of the tail. This is also a more declarative approach.
This approach also makes use of tail recursion which in general will boost performance significantly.
Modifying the original predicate
It is rather un-Prolog to use an iterator and a bound, one in general uses a reverse iterator.
You can however modify the predicate as follows:
match([Elem|_],Num,Num,Elem) :-
!.
match([_|Tail],Num,Count,MatchedNumber) :-
Count < Num,
Count1 is Count+1,
match(Tail,Num,Count1,MatchedNumber).
So a few errors:
- Use a "cut"
!
in the first clause: since if it matches, we know Prolog should not try the second one; - Use
MatchedNumber
in the recursive call instead ofElem
; - Perform a bound check
Count < Num
, - Do the increment of the counter
Count1 is Count+1
before doing the recursive call; and - Substitute all variables you do not use by underscores
_
.
An example is then:
?- match([a,b,c,d,e],0,0,X).
X = a.
?- match([a,b,c,d,e],1,0,X).
X = b.
?- match([a,b,c,d,e],2,0,X).
X = c.
?- match([a,b,c,d,e],3,0,X).
X = d.
?- match([a,b,c,d,e],4,0,X).
X = e.
?- match([a,b,c,d,e],5,0,X).
false.
But as said before, it is inefficient to pass an additional argument, etc.
Remove the i-th element from the list
An almost equivalent approach can be used to remove the i-th element from the list:
removei([],_,[]).
removei([_|T],0,T) :-
!.
removei([H|T],N,[H|TR]) :-
N1 is N-1,
removei(T,N1,TR).
Here the base case is again that the index is 0
in which case the tail of the list is removed (thus dropping the head). The inductive case will place the head of the list in the head of the resulting list and will count on the recursive call to remove the correct item from the tail. Another base case removei([],_,[]).
is added because it is possible that i is greater than the length of the list in which case this predicate won't remove any item.
Example
?- removei([a,b,c,d,e],0,X).
X = [b, c, d, e].
?- removei([a,b,c,d,e],1,X).
X = [a, c, d, e].
?- removei([a,b,c,d,e],2,X).
X = [a, b, d, e].
?- removei([a,b,c,d,e],3,X).
X = [a, b, c, e].
?- removei([a,b,c,d,e],4,X).
X = [a, b, c, d].
?- removei([a,b,c,d,e],5,X).
X = [a, b, c, d, e].
?- removei([a,b,c,d,e],6,X).
X = [a, b, c, d, e].
Solution 2
To find the nth element of a list (where n
is relative to zero), something like this ought to suffice:
find_nth_element_of_list( 0 , X , [X|_] ) .
find_nth_element_of_list( N , X , [_|Xs] ) :-
N > 0 ,
N1 is N-1 ,
find_nth_element_of_list( N1 , X , Xs )
.
Similarly, to remove the nth element of a list, something like this ought to suffice:
remove_nth_element_of_list( 0 , [_|Xs] , Xs ) . % at n=0, toss the head and unify the tail with the result set
remove_nth_element_of_list( N , [X|Xs] , [X|Ys] ) :- % at n>0, prepend the head to the result and recurse down.
N > 0 ,
N1 is N-1 ,
remove_nth_element_of_list( N1 , Xs , Ys )
.
Solution 3
If for some reason you need to achieve this using no built-in predicates besides append
here is my implementation:
my_length([], 0).
my_length([_|T], N1) :- my_length(T, N), N1 is N + 1.
% Nth
my_nth(N, List, H) :-
append(L1, [H|_], List),
my_length(L1, N),
!.
test_my_nth(X, Y, Z) :-
my_nth(0, [123, 456, 789], X),
my_nth(1, [123, 456, 789], Y),
my_nth(2, [123, 456, 789], Z).
% X = 123
% Y = 456
% Z = 789
and without append
:
my_length([], 0).
my_length([_|T], N1) :- my_length(T, N), N1 is N + 1.
% Nth
my_nth(0, [H|_], H) :- !.
my_nth(N, [_|T], Result) :-
N1 is N - 1,
my_nth(N1, T, Result),
!.
test_my_nth(X, Y, Z) :-
my_nth(0, [123, 456, 789], X),
my_nth(1, [123, 456, 789], Y),
my_nth(2, [123, 456, 789], Z).
% X = 123
% Y = 456
% Z = 789
And without !
:
my_length([], 0).
my_length([_|T], N1) :- my_length(T, N), N1 is N + 1.
% Nth
my_nth(N, [_|T], Result) :-
N > 0,
N1 is N - 1,
my_nth(N1, T, Result).
my_nth(0, [H|_], H).
test_my_nth(X, Y, Z) :-
my_nth(0, [123, 456, 789], X),
my_nth(1, [123, 456, 789], Y),
my_nth(2, [123, 456, 789], Z).
% X = 123
% Y = 456
% Z = 789
Why would anyone need it? There is a specific professor at Poznan University of Technology that requires students to write predicates like this.
Amir Jalilifard
I'v started software development since 2006 and it seems like now it consumes most every part of my life.My life is my computer and solving problems is my passion.I mostly enjoy areas like Computer Vision, Robotics and AI. Also, I am excited about Content based image recognition(CBIR).My major goal in my life is joining one of the Google software development teams because the people who are crazy enough, they think, they can change the world and i believe,Google is a legitimate proxy for putting my dent in the universe.
Updated on July 09, 2022Comments
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Amir Jalilifard almost 2 years
I am trying to write a Prolog code finding the n-th element of a list. I wrote the below code but it doesn't return the element right.
match([Elem|Tail],Num,Num,Elem). match([Elem|Tail],Num,C,MatchedNumber):- match(Tail,Num,N,Elem), C is N+1.
In the first line I say, if the requested element number is equal to counter, then give the first element of the current list to the variable called
MatchedNumber
. This code returns theNum
andCounter
right but I don't know why when I want to set theMatchedNumber
asElem
, it always returns the first element of the list.1: what is wrong with this code? 2: How can I say instead of showing the matched number, remove it from list?
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Amir Jalilifard almost 9 yearsThanks. Yes. I made a mistake. It is not "C". It is "Counter" variable. Now I changed it. So, how can I solve this problem using my code?
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Amir Jalilifard almost 9 yearsMy question is just that, how can I edit my code to solve this problem?
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false almost 9 yearsThese versions are not steadfast.
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Willem Van Onsem almost 9 years@AmirJalilifard: well thats explained in the remainder of this answer.
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Willem Van Onsem almost 9 years@false: indeed, one can use a negative index or use unbounded variables.
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Amir Jalilifard almost 9 years@ CommuSoft He just said what was the problem of just a part of this code. When I changed it, the code didn't work! This is why I am asking him to give me another hint!
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Willem Van Onsem almost 9 yearsWell the second hint is: which value do you want to compute with
is
?N
orC
... -
Amir Jalilifard almost 9 yearsI have a counter called C. I have a requested element number called N. now I want to check when N=C. When they are equal it means we are in the right place in the list. Now I just wanna show this value!
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false almost 9 years
match([a|_], 0, b)
loops whereasmatch([a|_],0,X), X=b
fails. -
Willem Van Onsem almost 9 years@false: (1) modified with
N > 0
, (2) shouldn't the second fail? -
Amir Jalilifard almost 9 yearsYes. Better. Thank you. But I've not get my answer yet! How can I modify my code to do this job for me?
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Willem Van Onsem almost 9 years@AmirJalilifard: I reformulated it in a list of items how you can modify your original predicate.