Reversible numerical calculations in Prolog

Question

While reading SICP I came across logic programming chapter 4.4. Then I started looking into the Prolog programming language and tried to understand some simple assignments i

Answer 1

There are some things which you must remember when looking at Prolog:

• There is no implicit return value when you call a predicate. If you want to get a value out of a call you need to add extra arguments which can be used to "return" values, the second argument in your f/2 predicate. While being more verbose it does have the benefit of being easy to return many values.

• This means that automatically "evaluating" arguments in a call is really quite meaningless as there is no value to return and it is not done. So there are no nested calls, in this respect Prolog is flat. So when you call f(A-1, D) the first argument to f/2 is the structure A-1, or really -(A, 1) as - is an infix operator. So if you want to get the value from a call to foo into a call to bar you have to explicitly use a variable to do it like:

  foo(..., X), bar(X, ...),

• So you need a special predicate which forces arithmetic evaluation, is/2. It's second argument is a structure representing an arithmetic expression which it interprets, evaluates and unifies the result with its first argument, which can be either a variable or numerical value.

• While in principle you can run things backwards with most things you can't. Usually it is only simple predicates working on structures for which it is possible, though there are some very useful cases where it is possible. is/2 doesn't work backwards, it would be exceptional if it did.

This is why you need the extra variables C and D and can't replace C is A-1, f(C, D) by f(A-1,D).

(Yes I know you don't make calls in Prolog, but evaluate goals, but we were starting from a functional viewpoint here)