prolog-dif

I have a prolog assignment. I need to look at the first item in a list, see if its following items are the same until they are not and separate the lists by the first item and its duplicates. e.g if my list was a,a,a,b,c it would separate it into first: a,a,a. second: b,c. My current solution works except that the final matching item comes into the second list, not the first. I can't seem to think of a way to get it to appear in the first list instead. grab([],[],[]). grab([A,A|L],[A|L2],Rest) :- grab([A|L],L2,Rest). grab([A|L],Init,[A|L2]) :- grab(L,Init,L2). When the first two elements are

OK I am new to Prolog, so excuse me if this is something trivial, but I can't seem to find a proper elegant answer to this. I am trying to work out the exercise here on learnprolognow.org , exercise 2.4 (the crossword). The exercise provides these facts: word(astante, a,s,t,a,n,t,e). word(astoria, a,s,t,o,r,i,a). word(baratto, b,a,r,a,t,t,o). word(cobalto, c,o,b,a,l,t,o). word(pistola, p,i,s,t,o,l,a). word(statale, s,t,a,t,a,l,e). And the solution I came up with to solve the crossword placement of each word is this: crossword(V1, V2, V3, H1, H2, H3) :- word(V1, V1a, V1bH1b, V1c, V1dH2b, V1e,

(This is NOT a coursework question. Just my own personal learning.) I'm trying to do an exercise in Prolog to delete elements from a list. Here's my code : deleteall([],X,[]). deleteall([H|T],X,Result) :- H==X, deleteall(T,X,Result). deleteall([H|T],X,[H|Result]) :- deleteall(T,X,Result). When I test it, I first get a good answer (ie. with all the Xs removed.) But then the backtracking offers me all the other variants of the list with some or none of the instances of X removed. Why should this be? Why do cases where H==X ever fall through to the last clause? Aasmund Eldhuset The last clause

I come up against this all the time, and I'm never sure which way to attack it. Below are two methods for processing some season facts. What I'm trying to work out is whether to use method 1 or 2, and what are the pros and cons of each, especially large amounts of facts. methodone seems wasteful since the facts are available, why bother building a list of them (especially a large list). This must have memory implications too if the list is large enough ? And it doesn't take advantage of Prolog's natural backtracking feature. methodtwo takes advantage of backtracking to do the recursion for me,

different(Xs, Ys) :- member(X, Xs), non_member(X, Ys). different(Xs, Ys) :- member(Y, Ys), non_member(Y, Xs). While this definition using member/2 and non_member/2 is almost 1 perfect from a declarative viewpoint, it produces redundant solutions for certain queries and leaves choice points all around. What is a definition that improves upon this (in a pure manner probably using if_/3 and (=)/3 ) such that exactly the same set of solutions is described by different/2 but is determinate at least for ground queries (thus does not leave any useless choice points open) and omits (if possible) any