Prolog - Return the n-th row of a matrix

Question

I am trying to write the predicate rowN/3 which returns the n-th element (in this case row) of a matrix. Example:

?- rowN([[1,2],[3,4],[5,6]], 2, R). 
R = [3,4];
No

I am struggling with the counter. I have tried unsuccessfully to find some pretty similar examples. So far I've managed to write this:

Code:

rowN(L,[],[]).
rowN([],X,[]).
rowN([],[],[].
rowN([H|T],X,R) :-
    A==X,
    A is A + 1,
    rowI(T,A,H).

Answer 1

This line does not make much sense:

rowN(L,[],[]).

because the second argument is an integer (if I understand correctly), and you use a list. This is the case with nearly all your arguments. Furthermore you use RowI in your recursive call?

Solution

A solution is to first specify that the first row (I = 1), is equal to the head of the matrix:

rowN([H|_],1,H).

next you need to find an iterative way to enumerate through your matrix. So the header is definitely something of the form:

rowN([H|T],I,X) :-
#   ...

Now we will assume that I is not equal to 1 (we will discuss this topic later). In that case we need to traverse the matrix further, so we will take the tail and set the counter I one back. This can be done using:

rowN([_|T],I,X) :-
    I1 is I-1,
    rowN(T,I1,X).

So our predicate reads:

rowN([H|_],1,H).
rowN([_|T],I,X) :-
    I1 is I-1,
    rowN(T,I1,X).

Now if you use this predicate, it will give the correct result:

?- rowN([[1,2],[3,4],[5,6]], 2, R).
R = [3, 4] ;
false.

The question is why does the predicate does not generate other results: after showing the first result, for rowN([[1,2],[3,4],[5,6]], 2, R) :- rowN([[3,4],[5,6]],1,[3,4])., it could try to find alternatives. It does so by using the second clause, but then it will eventually run out of rows and call for the rowN([],_,_) predicate, since not of the clauses match, it will fail.

This solution is not perfect: it does not work in all directions correctly, which is in general hard in Prolog. That's why good Prolog programmers have written libraries.

Using swi-prolog's builtin nth1/3

Instead of reinventing the wheel, you can make use of the nth1/3 predicate in swi-prolog. Although the arguments are swapped - you need to call it like nth1(2,[[1,2],[3,4],[5,6]],R). - it has the advantage that it works in more directions that what most people can come up in an fast solution, it is with near certainty bugfree (because it has been tested billions of times by all Prolog programs that use the predicate) and some of these builtins are implemented in C++ making them sometimes faster. For instance:

?- nth1(2, [[1,2],[3,4],[5,6]], R).
R = [3, 4].

?- nth1(I, [[1,2],[3,4],[5,6]], [5,6]).
I = 3.

?- nth1(I, [[1,2],[3,4],[5,6]], R).
I = 1,
R = [1, 2] ;
I = 2,
R = [3, 4] ;
I = 3,
R = [5, 6].

?- nth1(I,M,[2,3]).
I = 1,
M = [[2, 3]|_G23] ;
I = 2,
M = [_G22, [2, 3]|_G26] ;
I = 3,
M = [_G22, _G25, [2, 3]|_G29] ;
I = 4,
M = [_G22, _G25, _G28, [2, 3]|_G32] .

You can thus ask what the second row is, ask where the row [5,6] is located, make the query more generic by answering with tuples of the index I and the row R and generate a matrix with a row [2,3] somewhere.