Zadaci Prolog 15/11/2016 ( Resenja )

/* Prvi */
max(A, B, C, A):- A >= B, A >= C.
max(A, B, C, B):- B >= A, B >= C.
max(A, B, C, C):- C >= A, C >= B.

/* Drugi */
sump(N, 0):- N < 2.
sump(N, R):- N >= 2, N mod 2 =:= 0, N1 is N - 2, sump(N1, R1), R is R1 + N.
sump(N, R):- N >= 2, N mod 2 =\= 0, N1 is N - 1, sump(N1, R1), R is R1.

/* Treci */
bin0(N, N):- N =< 1.
bin0(N, R):- N > 1, N1 is N div 2, bin0(N1, R1), R is R1 + N mod 2.

/* Cetvrti */
nsqr(N, 0, 1):- N >= 0.
nsqr(N, X, R):- X > 0, X1 is X - 1, nsqr(N, X1, R1), R is R1 * N.
st10(0, 0).
st10(N, R):- N > 0, N1 is N div 10, st10(N1, R1), R is R1 + 1.
fourth(N, K, R):- st10(N, X), sets(N, X, K, R).
sets(N, X, K, R):- X >= K, O is X - K, nsqr(10, O, ST10),
    R is N div ST10.
sets(N, X, K, R):- X > 0, X1 is X - 1, nsqr(10, X1, ST10),
    N1 is N mod ST10, sets(N1, X1, K, R).

/* Peti */
fifth(N, R):- cinioci(N, 2, R).
cinioci(N, X, R):- N > 1, (N mod X) =:= 0, R is X.
cinioci(N, X, R):- N > 1, (N mod X) =\= 0, X1 is X + 1, cinioci(N, X1, R).
cinioci(N, X, R):- N > 1, (N mod X) =:= 0, N1 is N div X, cinioci(N1, X, R).