0

I would like to know if someone could help me with a problem in prolog. I have to define two predicates but before that, I need to find out a way how to parse a list and form a number. For example [1,2,3] => 123. I tried different ways to do this but nothing works good. My code looks like this, I know it is not good but I can't find another way.

num([H|T],I,RI,RES):-
    H2 is H * I,
    R1 is RI + H2,
    I2 is I/10,
    RES2 is RES + R1,
    num(T,I2,R1,RES2).
lurker
  • 56,987
  • 9
  • 69
  • 103
user3671805
  • 87
  • 5
  • 2
    [This](http://stackoverflow.com/a/28442760/772868) works for binary numbers. So you need to adapt it for decimal numbers. – false Oct 31 '15 at 13:53
  • @false, good link (+1). I was going to add the CLPFD solution to my initial answer, but was oversimplifying for what I thought was the scope for the OP's problem. Ultimately, CLPFD is the correct way to go. – lurker Oct 31 '15 at 22:23

1 Answers1

2

Your hunch is correct that this can be much simpler. Recursively, you want to think of the digits [A, B, C] representing number as, (((A * 10) + B) * 10) + C.

To start, you want to think about what kind of predicate you desire. That would simply be, num(Digits, Number). which yields Number given Digits. You'll need an accumulator for intermediate results using the above formula concept, so your num/2 needs to call a num/3 that will include an accumulator argument that you carry along:

num(Digits, Num) :-
    num(Digits, 0, Num).    % Accumulator initially 0

If the input list is empty, the result is the accumulator:

num([], A, A).              % Result is accumulator when there are no digits

Then your recursive case:

num([D|T], A, R) :-
    NewA is (D * 10) + A,  % New accumulator is (current digit * 10) + old accumlator
    ...                    % Need recursive call

I'll leave the ... for you to fill in as an exercise. :)

lurker
  • 56,987
  • 9
  • 69
  • 103