CS345 Solutions #1

  1. (25 pts.) needs(F,P) := create(F,P,G) needs(F,P) :- includes(F,G) & needs(G,P) needs(F,P) :- create(F,Q,G) & needs(G,P) It was also possible to define the IDB req (given as an example in class), and use it in needs.

  2. (25 pts.)
    int(0) :- int(s(X)) :- int(X) move(state(s(I),J,K), state(I,J,K) :- int(I) & int(J) & int(K) move(state(s(s(I)),J,K), state(I,J,K) :- int(I) & int(J) & int(K) move(state(I,s(J),K), state(I,J,K) :- int(I) & int(J) & int(K) move(state(I,s(s(J)),K), state(I,J,K) :- int(I) & int(J) & int(K) move(state(I,J,s(K)), state(I,J,K) :- int(I) & int(J) & int(K) move(state(I,J,s(s(K))), state(I,J,K) :- int(I) & int(J) & int(K) win(S) :- move(S,T) & NOT win(T) It is necessary to define int so that the rules for move will be safe.

    No, the program is not stratifed. If the dependency graph is drawn, there is a negative arc win -> win. Thus, there is a cycle involving negation, and the program cannot be stratified.

    Yes, the program is locally stratified. In any dependency graph for the program, the only negative edges would be win(state(I,J,K)) -> win(state(X,Y,Z)). However, (I+J+K) > (X+Y+Z) must hold, since the only valid moves remove one or two tokens from a row. Thus, all win -> win dependencies are acyclic. Other dependencies are win -> move, move -> int, and int -> int (where for int(X) -> int(Y), X>Y), which don't cause dependency cycles. There are no cycles involving negation in the graph, so the program is locally stratified.

  3. (25 pts.)
    Initialize: P = DELTA(P) = Q Iterate until DELTA(P) = NULL: 1. DELTA(P) := PI_(1,4) ((P JOIN P JOIN DELTA(P)) UNION (P JOIN DELTA(P) JOIN P) UNION (DELTA(P) JOIN P JOIN P)) 2. DELTA(P): = DELTA(P) - P 3. P := P UNION DELTA(P)
    p represents all paths of odd length.

  4. (25 pts.)
    There are twelve minimal models.

    1. p0(a) must be in the model. q(a) is in the EDB, so the body of the first rule (p0(X) :- q(X)) is true, thus p0(a) must true.

    2. A minimal model cannot contain p's with argument other than a. The only EDB fact is q(a), so such an IDB fact could always be removed, and we would still have a model.

    3. p_i(a) and p_(i+1)(a) can't both be false. (i = {1,2,...,8}) If p_i(a) is false, then the (i+1)th rule (p_(i+1)(X) :- q(X) & NOT p_i(X)) will have a true body, forcing p_(I+1)(a) to be true.

    4. p_i(a), p_(i+1)(a), and p_(i+2)(a) can't all be true. (i = {0,1,...7}) If all three were true, we could drop p_(i+1) and still have a model.

    5. If p8(a) is true, then p9(a) is false. If both p8(a) and p9(a) were true, we could drop p9(a) and still have a model.

    The stratified model is {q(a), p0(a), p2(a), p4(a), p6(a), p8(a)}.