% CPSC 312 2019 – Midterm 3 vwersion A
:- use_module(library(clpfd)). % defines #< for integers.. %Question 1 % "g is not a logical consequence of KB" means there is a model of KB in which g is false % KB is :- dynamic t/0. w :- x, y. x :- u, v. u :- s. s. v :- y. y :- t. % (a) x :- u,v. % (b) s. % (c) yes :- y,y. % There are no clauses for t % Question 2 :- dynamic m/0. a :- b,c,d. a :- e. b :- g. b :- m. g. c :- m. e :- g. d. % Load this file then do % ?- [trace]. % ?- a. % Question 3 elem(E, set(E,_,_)). elem(V, set(E,LT,_)) :- V #< E, elem(V,LT). elem(V, set(E,_,RT)) :- E #< V, elem(V,RT). %?- elem(X, set(7,set(2,empty,empty),set(9,empty,set(11,empty,empty)))). % ?- elem(3,S),elem(8,S). % S = set(3, _1094, set(8, _1108, _1110)) % Note that Prolog doesn't know anything about empty, so the subtrees can't be empty. The subtrees are different variables. % insert(E,S,S1) is true if S1 is a set containing E and the elements of set S insert(V,empty,set(V,empty,empty)). insert(V,set(V,L,R),set(V,L,R)). insert(V,set(E,L,R),set(E,L1,R)) :- V #< E, insert(V,L,L1). insert(V,set(E,L,R),set(E,L,R1)) :- E #< V, insert(V,R,R1). %?- insert(3, set(7,set(2,empty,empty),set(9,empty,set(11,empty,empty))), R). % Question 4 % myapply(Lst, Sub, Res) where sub is a list of (X,Y) pairs, replaces each element of X by Y myapply([], _, []). myapply([H|T], Sub, [H1|T1]) :- app(H, Sub, H1), myapply(T, Sub, T1). app(E,[],E). app(E,[(E,V)|_],V). app(E,[(E1,_)|R],V) :- dif(E,E1), app(E,R,V). %?- myapply([a,b,c,d,e,c], [(a,f), (c,3), (g,7)], R). %?- myapply([b,a,a,b], [(a,b),(b,a)], R). %?- myapply([b,a,a,b], S, [c,c,c,c]).