/* arg(+N,+Term,?Arg) is true if Arg is the Nth argument of Term. It is used either to find a particular item in a compound term or to instantiate a variable argument of a term. */ ?-arg(1,color(red),X). X=red ?-arg(1,color(X),yellow). X=yellow /* functor(+Term,?F,?Arity) or functor(?Term,+F,+Arity) is true if Term is a term whose functor has name F and arity Arity. It is used either to find the functor name and its arity in a given term or to build a term with a particular functor name and arity: */ ?- functor(color(x),F,N). F = color, N = 1 ?- functor(X,color,1). X = color(_A) /* the univ predicate =.. is defined as +Term=..?List or ?Term=..+List Used to create a list from a predicate term, as in foo(a,b,c)=..Y where Y is unified with [foo,a,b,c] or to create a predicate term from a list, as in X=..[foo,a,b,c] where X is unified with foo(a,b,c) */ /* example of use of =.. This checks if all elements of a list satisfy a property. The property is expressed as a unary predicate. */ satisfy(,_). satisfy([X|L],P):- R=..[P,X], R, satisfy(L,P). color(red). color(green). color(blue). ?- satisfy([red,blue,green],color). yes ?- satisfy([red,yellow],color). no /* call(X) or simply X allows executing a predicate that can be passed as an argument to another predicate. */ ?- Goal=..[member,a,L],Goal. G = member(a, [a|_G367]) L = [a|_G367] ; G = member(a, [_G366, a|_G370]) L = [_G366, a|_G370] ; G = member(a, [_G366, _G369, a|_G373]) L = [_G366, _G369, a|_G373] Yes /* this and =.. come handy to define a map predicate, which applies its first argument to each element of a list and returns a list of results */ map(F,,):-!. map(F,[X|Y],[U|V]):- T=..[F,X,U], T, map(F,Y,V). double(X,Y):- Y is 2*X. ?- map(double,[1,2,3],L2). L2 = [2, 4, 6] /* clause(A,B) retrieves a clause whose head matches A and body matches B If the clause has no body (i.e. it is a fact) the value of Body is True. */ ?- clause(member(X,L),Body) X = _G256 L = [_G256|_G397] Body = true ; X = _G256 L = [_G399|_G400] Body = member(_G256, _G400) ; no
/* bagof(?Template, +Goal, ?Bag) is true if Bag is the bag of all the instances of Template for which Goal is true. If no solutions are found it fails */ ?- Input=[1,2,3], bagof((Left, Right), append(Left, Right, Input), Bag). Bag = [(,[1,2,3]),(,[2,3]),([1,2],),([1,2,3],)], Input = [1,2,3] ? ; no /* setof(?Template, +Goal, ?Set) Set is the set of all instances of Template such that Goal is satisfied when the set is non empty. Goal is a goal or set of goals (as in call(Goal)), Set is a set of terms represented as a list without duplicates in the standard order for terms. If no solutions are found it fails */ likes(bill,cider). likes(dick,beer). likes(harry,beer). likes(ian,cider). likes(tom,beer). likes(tom,cider). ?- setof(X, likes(X,Y), S). S=[dick,harry,tom], Y=beer ? ; S=[bill,ian,tom], Y=cider ? ; no ?- setof(X, Y^likes(X,Y), S). S=[bill,dick,harry,ian,tom] ? ; no /* The _ produces the same results as when using the variable Y */ ?- setof(X, likes(X,_), S). S = [dick,harry,tom] ? ; S = [bill,ian,tom] ? ; no ?- setof((Y,S), setof(X, likes(X,Y), S), SS). SS = [(beer,[dick,harry,tom]),(cider,[bill,ian,tom])] ? ; no ?- setof(X, (member(X,[1,2,2]), X>1),L). L=; no /* findall(?Template, +Goal, ?Bag) Bag is a list of instances of Template in all proofs of Goal found by Prolog. The list may be empty, all variables are considered as being existentially quantified. Each invocation of findall/3 succeeds once and no variables in Goal get bound. */ ?- findall(X, likes(X,Y),S). S = [bill,dick,harry,ian,tom,tom] ? ; no
% intersect/3: given two lists, finds their intersection % that is, fidn all the members of the first list which also % appear in the second list. If the lists contain no duplicates % then this computes set intersection % intersect(+L1, +L2, ?Result) intersect(,_,). intersect([H|T],L,[H|Rest]):- member(H,L),!,intersect(T,L,Rest). intersect([_|T],L,L1]):- intersect(T,L,L1).