Chapter 6 Exercises from Learn Prolog Now! by Blackburn and Bos

chapter6.pl

% chapter6.pl

append2([], L, L).
append2([X|T], L2, [X|L3]) :- append(T, L2, L3).

prefix(P, L) :- append2(P, _, L).

suffix(S, L) :- append2(_, S, L).

sublist(Sub, L) :- suffix(S, L), prefix(Sub, S).


reverse2([], []).
reverse2([X|T], Y) :- reverse2(T, Z), append2(Z, [X], Y).


reverse3(L, R) :- reverse3(L, [], R).
reverse3([X|TX], A, R) :- reverse3(TX, [X|A], R).
reverse3([], Acc, Acc).


% exercise 6.1
doubled(L) :- append2(X, X, L).


% exercise 6.2
palindrome(L) :- reverse3(L, L).


% exercise 6.3.1
second(X, [_, X|T]).


% exercise 6.3.2
swap12([X,Y|T], [Y,X|T]).
swap12([], []).
swap12([X], [X]).


% exercise 6.3.3
final(X, L) :- reverse3(L, [X|_]).


% exercise 6.3.4
% toptail : reverse the tail, then reverse the tail of that
toptail([X|T], Out) :- reverse3(T, [Y|OutR]), reverse3(OutR, Out).


% exercise 6.3.5
swapfl(L1, L2) :-
    L1 = [X|TX],
    toptail(L1, B),
    toptail(L2, B),
    final(X, L2).


% exercise 6.4
colors(L) :- member(red, L), member(blue, L), member(green, L).
nationalities(L) :- member(england, L), member(spain, L), member(japan, L).
pets(L) :- member(jaguar, L), member(snail, L), member(zebra, L).

% rightOf: elem A is to the right of B in list L
rightOf(A, B, L) :-
    prefix(P, L),
    final(A, P),
    member(B, P).

% leftOf: elem A is to the left of B in list L
leftOf(A, B, L) :-
    suffix(P, L),
    P=[A|_],
    member(B, P).

street(L) :-
    L=[house(C1, N1, P1), house(C2, N2, P2), house(C3, N3, P3)],
    colors([C1, C2, C3]),
    nationalities([N1, N2, N3]),
    pets([P1, P2, P3]),
    member(house(red, england, _), L),
    member(house(_, spain, jaguar), L),
    rightOf(house(_, japan,_ ), house(_, _, snail), L),
    leftOf(house(_, _, snail), house(blue, _, _), L).

zebra(N) :- street(L), member(house(_, N, zebra), L).

what about exercise 6.3.4 with append as it states in the hint

This toptail([_|Xs],Outlist) :- append(Outlist,[_],Xs). could be a possible solution.