The testing for types needs to distinguish itself from the traditional “type testing” built-ins that implicitly also test for a sufficient instantiation. So we effectively test only for sufficiently instantiated terms (si). And if they are not sufficiently instantiated, an appropriate error is issued. For a type nn, there is thus a type testing predicate … Read more
SWI-Prolog offers exclude/3 and other such meta-predicates. Your original problem can be coded like this: are_identical(X, Y) :- X == Y. filterList(A, In, Out) :- exclude(are_identical(A), In, Out). Usage example: ?- filterList(A, [A, B, A, C, D, A], Out). Out = [B, C, D].
The code as you’ve posted it is (almost) OK. The order of clauses just needs to be swapped (in order to make this predicate definition productive, when used in a generative fashion): append( [], X, X). % (* your 2nd line *) append( [X | Y], Z, [X | W]) :- append( Y, Z, W). … Read more
This site is devoted to solve such puzzles with CLP(FD). But the full power of CLP(FD) is overkill here: your assignment can be solved effectively searching the entire solution space when you have adequately described the constraints. The solution will be composed of 5 houses, where each attribute satisfy all constraints imposed by description. Be … Read more
What’s probably more useful than a slightly less nondeterministic length/2 is a proper list-length constraint. You can find an ECLiPSe implementation of it here, called len/2. With this you get the following behaviour: ?- N :: 1..3, len(Xs, N). N = N{1 .. 3} Xs = [_431|_482] % note it must contain at least one … Read more
As a general remark you will find more on SO about it under the name library(pio). Also the way to use it cleanly is rather: :- use_module(library(pio)). Your example is way too complex, so I will only consider a slightly simpler case, a newline separated list of numbers: nats([]) –> []. nats([N|Ns]) –> nat(N), newline, … Read more
The definition of flatten2/2 you’ve given is busted; it actually behaves like this: ?- flatten2([a, [b,c], [[d],[],[e]]], R). R = [a, b, c] ; false. So, given the case where you’ve already bound R to [a,b,c,d,e], the failure isn’t surprising. Your definition is throwing away the tail of lists (ListTail) in the 3rd clause – … Read more
What you actually want is something slightly different: You want to count the number of answers (so far) of a goal. The following predicate call_nth(Goal_0, Nth) succeeds like call(Goal_0) but has an additional argument which indicates that the answer found is the n-th answer. This definition is highly specific to SWI or YAP. Do not … Read more
Here’s how you can find the “powers of two” in logically-pure way! Using sicstus-prolog 4.3.5, library(reif) and library(clpz): :- use_module([library(reif), library(clpz)]). power_of_two_t(I, T) :- L #= min(I,1), M #= I /\ (I-1), call((L = 1, M = 0), T). % using (=)/3 and (‘,’)/3 of library(reif) Sample query1 using meta-predicate tfilter/3 in combination with power_of_two_t/2: … Read more
Prolog rules are read independently of each other, so you need one rule for the case where the element is unique and one where it is not. Provided the order of the elements is not relevant, you might use: ?- remUniqueVals([A,B,C], [1,1]). A = B, B = 1, dif(C, 1) ; A = C, C … Read more