Programming the Quadratic Formula in Prolog

Question

First of all, I saw this same question being asked earlier today and I decided to give it a try myself (and had trouble with it). I hope it's alright that I created a new question for this!

I'm trying to make a prolog program that solves the quadratic formula X = ​(​​−b±√​b​2​​−4ac)​/2a. The implemented predicate quadratic([A,B,C], [Result]) takes a list of a, b and c as an the first argument and a list of the result(s) as the second. The lists are giving me trouble however; I wrote this code:

quadratic([A,B,C], [X]):-
  underRoot([A,B,C], UnderRootResult),
    UnderRootResult<0,
      X is 0.
quadratic([A,B,C], [X]):-
  underRoot([A,B,C], UnderRootResult),
    UnderRootResult=:=0,
      X is -B/2*A.
quadratic([A,B,C], [X]):-
  underRoot([A,B,C], UnderRootResult),
    UnderRootResult>0,
      X is -B - sqrt(UnderRootResult)/2*A,
      X is -B + sqrt(UnderRootResult)/2*A.

(This is probably full of mistakes, so excuse me in advance) Running this will give me the undefined procedure error for underRoot/2. I don't really get why this happens. I feel like I have the general idea of how to program this but that I'm making lots of newbie mistakes. I can;t seem to find out what the problem is though, so I would appreciate any help!

edit: Also, if I'm allowed to ask two questions at once, how would I get both X's in the case of >0 in a list as one result?

Answer 1

There are many things here. First of all, usually all calls in the body are put on the same column, but that is not really an error.

Furthermore I do not find an underRoot/2 predicate. We can implement one like:

underRoot([A,B,C],X) :-
    X is B*B-4*A*C.

Furthermore you always put [X] in the head of the clauses. But in case UnderRoot is less than zero, there are no solutions, so the list should be []. Furthermore in case UnderRoot > 0, there are two solutions, so the solution should be [X1,X2].

Finally if you write -B/2*A, it will be interpreted like: (-B/2)*A, so you will multiply with A. So you will need to use -B/(2*A).

So this brings us to a following proposal:

quadratic(L, []) :-
    underRoot(L, U),
    U < 0.
quadratic([A,B,C], [X]) :-
    underRoot([A,B,C],0),
    X is -B/(2*A).
quadratic([A,B,C], [X1, X2]) :-
    underRoot([A,B,C],U) :-
    SU is sqrt(U),
    X1 is (-B-SU)/(2*A),
    X2 is (-B+SU)/(2*A).