Adventure in Prolog Dennis Merritt | Springer Published in 1990, 186 pages An Introduction to Logic Programming through Prolog Michael Spivey | Prentice Hall Published in 2008, 258 pages Applications of Prolog Attila Csenki | BookBoon Published in 2009, 203 pages Artificial Intelligence through Prolog Neil C. Rowe | Prentice-Hall Published in 1988, 481 pages … Read more
TL;DR: Don’t. The cut prunes Prolog’s search tree. That is, given a pure Prolog program without cut and the same program with cuts the only difference is that the program with cuts might spend less time in fruitless branches, and thus is more efficient ; might have fewer answers ; it might also terminate whereas … Read more
There are two issues involved. How to get the variable name X into the system, and how to get a term with such a variable into the atom. The X you type in is read by the top level which converts it to a regular variable which does not have a name associated. Let’s see … Read more
For elegance and didactic reasons alone, dif/2 is clearly preferable here and also in the vast majority of other cases, since as you already note “a lot of unnecessary unifications might take place” otherwise, and also because dif/2 is a pure and nicely declarative predicate that can be used in all directions and at any … Read more
What is the idiomatic approach to this class of problems? Is there a way to simplify the problem? Many of the following remarks could be added to many programs here on SO. Imperative names Every time, you write an imperative name for something that is a relation you will reduce your understanding of relations. Not … Read more
Use CLP(FD) constraints, for example: :- use_module(library(clpfd)). binary_number(Bs0, N) :- reverse(Bs0, Bs), foldl(binary_number_, Bs, 0-0, _-N). binary_number_(B, I0-N0, I-N) :- B in 0..1, N #= N0 + B*2^I0, I #= I0 + 1. Example queries: ?- binary_number([1,0,1], N). N = 5. ?- binary_number(Bs, 5). Bs = [1, 0, 1] . ?- binary_number(Bs, N). Bs = … Read more
shuffle([], B, B). shuffle([H|A], B, [H|S]) :- shuffle(B, A, S). In this kind of problems, usually the difficult part is not Prolog but identifying the simplest recursive relation that solves it.
Those prefix operators, in this context, represent instantiation modes, i.e. they tell you which arguments should be variables or instantiated when calling the predicate. They also tell you if an argument will be (possibly further) instantiated by the call. They can also be used to tell you that an argument is going to be meta-interpreted … Read more
Let’s use a DCG – a Definite Clause Grammar. We start with your original definition: lea(T, L) :- phrase(values(T), L). values(nil) –> []. values(t(X,L,R)) –> [X], values(L), values(R). Now, we need to restrict ourselves to those t/3 that are leaves. One possibility is to enumerate all cases: lea2(T, L) :- phrase(leaves(T), L). leaves(nil) –> []. … Read more
First, let’s translate the clauses into something more understandable: append([], List, List) :- !. can be written append([], List2, Result) :- Result = List2, !. and append([H|L1], List2, [H|L3]) :- append(L1, List2, L3). can be written append(List1, List2, Result) :- List1 = [Head1 | Tail1], Result = [HeadR | TailR], Head1 = HeadR, append(Tail1, List2, … Read more