AI Cologne

Again, we are going to tackle some problems mentioned in the list of 99-Prolog problems. Today's post builds upon the gcd predicate defined last time to attack problems: P31 = is_prime P35 = prime_factors P36 = prime_factors_mult P39 = prime_list Now, building on the previously defined gcd predicate, on may define a prime inductively: a prime is a number which is only (divisible by one and itself) the smallest prime is 2 any number greater than 2 is either composite or prime Building on this definition we may translate the smallest prime into a Prolog predicate quite easily. To determine the primeness/composibility of any number greater than 2, one may use the gcd predicate. Recalling that a prime p is a number which cannot be divided by any number in ]1, p[, the following predicate may be given: is_prime(2) :- !. is_prime(3) :- !. is_prime(X) :- X >= 2, UpperBound is round(sqrt(X)) + 1, is_prime_t(X, 2, UpperBound), !. is_prime_t(_, UpperBound, UpperBound). is_prime_t(X, ...

Today's post follows the tradition of the my last post, i.e. it continues with another problem stated in the list of 99 Prolog problems. Today, we are going to have a look at problem 32 and 33. Problem 32 concerns the implementation of a greatest common divisor predicate (gcd). In an earlier post, I implemented a gcd routine in Python . We are going to reuse this definition of gcd here, using a ternary predicate/relation. The Prolog definition of gcd makes use of gcd's recursive definition, which are: the gcd of two numbers x and y is equal to the gcd of x - y and y for y the gcd of some number and 0 is the number the gcd of some number and 1 is 1 Now, given the above properties (and some more) a possible gcd predicate might look like this: gcd(A, 0, A) :- !. %includes gcd(0,0) gcd(1, _, 1) :- !. gcd(A, A, A) :- !. gcd(A, B, GCD) :- A X is B - A, gcd(X, A, GCD). gcd(A, B, GCD) :- X is A - B, gcd(X, B, GCD). Provided the above definition of gcd, the solution to proble...

Today's post is about problem 14 from the list of 99 Prolog problems (see here ): (*) Duplicate the elements of a list. Example: ?- dupli([a,b,c,c,d],X). X = [a,a,b,b,c,c,c,c,d,d] Since the problem is marked with one star, a solution should be within reach. Now, to duplicate each element of the list, we do the following: We traverse the list element by element until we reach the last element, i.e. the empty list. For each element (excluding the empty list), the element is added twice to a newly built output list. Given the above solution statement, a possible translation to Prolog code might look like this: dupli([], []). dupli(L, O) :- dupli_t(L, [], O). dupli_t([], Current, Current). dupli_t([H|T], CList, OutList) :- append(CList, [H, H], IntermediateList), dupli_t(T, IntermediateList, OutList).