prolog_notes5.txt

% nthElement(X, Index, List)
nthElement(X, 0, [X|_]).
nthElement(X, N, [_|T]) :-
	nthElement(X, M, T),
	N is M + 1.

inUnion(X, A, B) :-
	member(X, A);
	member(X, B).

inIntersection(X, A, B) :-
	member(X, A),
	member(X, B).

inDifference(X, A, B) :-
	member(X, A),
	not(member(X, B)).

isSubset(A, B) :-
	not((
	member(X, A),
	not(member(X, B))
	)).

areEqual(A, B) :-
	subset(A, B),
	subset(B, A).

removeDuplicates([], []).
removeDuplicates([H|T], [H|R]) :-
	not(member(H, R)),
	removeDuplicates(T, R).
removeDuplicates([H|T], R) :-
	member(H, R),
	removeDuplicates(T, R).

1) p(L), по даден списък от различни списъци L проверява дали
всеки два различни елемента на L имат общ елемент, който не
принадлежи на някой елемент на L.

Ev. X of L, Ev. Y of L 
( 
	X \= Y => 
	Ex. Z of X 
		(
		Z of Y and Ex. T of L 
			(
			Z not of T
			)
		 )
)

not (Ex. X of L)(Ex. Y of L)
(
	X \= Y and
	not 
	(
		Ex. Z of X
		(
			Z of Y and Ex. T of L
			(
				Z not of T
			)
		)
	)
)

p(L) :- 
	not
	((
	member(X, L), 
	member(Y, L), 
	X \= Y,
		not
		((
		inIntersection(Z, X, Y),
		member(T, L),
			not
			(
			member(Z, T)
			)
		))
	)).

2) q(L) : Дали съществуват два различни елемента от L,
които имат общ елемент, който не принадлежи на никой друг елемент
на L.

(Ex. X of L)(Ex. Y of L)
(
	X \= Y and 
	(Ex. T of X)
	(
		T of Y and 
		not
		(Ex. Z of L)
		(
			T of Z and
			Z \= X and
			Z \= Y
		)
	)
)

q(L) :- 
	member(X, L), member(Y, L), 
	X \= Y,
	inIntersection(T, X, Y),
	not
	((
		member(Z, L),
			member(T, Z),
			Z \= X,
			Z \= Y
	)).

3) p(X, Y): По даден списък Х генерира в Y всички списъци,
чийто елементи са елементи на Х и броят на срещанията
на най-честосрещания елемент в Y е число, което не е елемент
на X.

Нужно: member, subset, permutation, count, countMax.

subsequence([], []).
subsequence([H|T], [H|R]) :-
	subsequence(T, R).
subsequence([_|T], R) :-
	subsequence(T, R).

subset(S, L) :-
	subsequence(Sub, L),
	permutation(Sub, S).

%count(X, N, L)
count(_, 0, []).
count(X, N, [X|T]) :- 
	count(X, M, T),
	N is M + 1.
count(X, N, [H|T]) :-
	X \= H,
	count(X, N, T).

countMax(X, N, L) :-
	member(X, L),
	count(X, N, L),
	not
	((
		member(Y, L),
		count(Y, M, L),
		M > N
	)).

p(X, Y) :-
	subset(Y, X),
	countMax(_, N, Y),
	not
	(
		member(N, X)
	).

isPalindrome([]).
isPalindrome([_]).
isPalindrome(L) :-
	append(A, B, L),
	len(A, N),
	len(B, N),
	reverse(A, B).
isPalindrome(L) :-
	append(A, [_|B], L),
	len(A, N),
	len(B, M),
	M =:= N + 1,
	reverse(B, RB),
	prefix(A, RB).