SLIDE 9 Inference/Privacy Problem: Complexity
9
(u,0) (u,1) (u,2) (u,v) (u,v+1) (u,v+2) T(e,u,v) (0,0) (0,1) (0,2) (0,u+v) (0,u+v+1) (0,u+v+2) (u,0) (u,1) (u,2) (u,v) (u,v+1) (u,v+2) T(e,u,v) (0,0) (0,1) (0,2) (0,u+v) (0,u+v+1) (0,u+v+2)
We first show that given a recursively enumerable set W, there is a situation S such that W m PP[Public]. Note that m is the many-one equivalence relationship. The result is then immediate from the following reasoning. * It has been shown that there is set K which is creative. K is the set {x: the xth partial recursive function halts on input x} * The situation S that is constructed from the recursively enumerable set K will guarantee that PP[Public] is creative. This is because if the two sets A and B are many one equivalent and A is creative, then so is B. Therefore, if PP[Public] is creative then it cannot be recursive. Given a recursively enumerable set W, we create a situation S by defining a set of privacy constraints and a privacy function. Let the set of privacy constraints be {(0,0)}. That is, the only element that is assigned the private level is the pair (0,0). We consider pairs of natural numbers. This does not cause any problem due to the existence of the pairing function from NxN onto N where N is the set of all natural numbers. The set of privacy constraints is recursive (note that in this case it is also finite) and does not depend on W. We define a privacy function, which depends on W as follows. We assume that e is the index of W. The privacy function f for a pair (u,v) is defined as follows: {(u, v+1)} if u 0 AND NOT T(e, u, v) f(u,v) = {(0, u+v) if u 0 AND T(e, u, v) {(u, v-1) if u = 0 AND v 0 (the empty set) if u = 0 AND v = 0 Note that T is the Kleene’s T Predicate
Graphical Representation of the Privacy Function f
PP[Public] = {(u,v): there is a path via f from (u,v) to (0,0)}
We have shown that We m PP[Public] where We is the eth recursively enumerable set. Note that m is many-one equivalence.