Oracle Turing Machines and the Arithmetical Hierarchy Dr. K.W. Regan, Spring 1994 class notes, adapted for CSE596, Fall 2008 Abstract These notes supplement Section 3.9 of the Homer-Selman text with more material on reductions among problems, and on several concepts that are also useful in complexity theory. 1. Oracle Machines and Turing Reductions An oracle Turing machine M = (Q; Σ; Γ; δ; ∆; s; F; O) is defined just like a normal Turing machine, except that O Q is a special collection of oracle query states. The actions of M in a query state are controlled not⊆ by the character read on the tape, but by an \external module" which consists of a language A called the oracle set. The oracle A can be any given language of strings over some alphabet Λ Γ. (Usually we have Λ = Σ, and usually Σ = 0; 1 :) We'll assume that all OTMs discussed here⊆ are deterministic. The purpose of oracle TMsf is tog provide a formal model with which to study questions of the form, \If I could solve X, what else could I do?" To specify how this works in an intuitive manner, let us suppose that M uses the tape delimiters and $, and that Γ also contains the characters `?', `0' (for `no'), and `1' (for `yes'). (Note: ; $; ? ^ ^ are not in Λ.) Every arc into a query state q? O has the form (p; c; ?; S; q?), while there are 2 exactly two arcs out of the query state, one on `0' and one on `1'. When M enters state q?, the query string z is identified just like in the \relaxed output convention": z is the longest consecutive string of characters over Λ beginning in the cell to the right of the marker. If z belongs to the oracle set A, then the oracle module \magically" overwrites the `?'^ by a `1', and the arc on `1' is taken. If z = A, then the oracle responds `0', and the `0' arc is taken. After that, computation proceeds as2 normal, up until another query state is entered. Definition 1.1. A language A Turing-reduces to a language B, written A T B, if there is an oracle Turing machine M such that M is total with oracle B, and L(M B) = ≤A. This is like the definition of A being recursive, except for the presence of the oracle set B. It is also customary to say that A is recursive in B. In our present example, the Halting Problem is recursive in the emptiness problem for 2HDFAs; i.e., HP T 2HDFA- . Turing did indeed ≤ ; invent the idea of an oracle TM and a Turing reduction. In this particular case, M0 only needs one (1) oracle call and follows a pre-set truth-table of what to say when the oracle responds: here, 0 Accept and 1 Reject, and so we can be more specific and write HP 1tt X. 7! 7! ≤ In fact, the only examples I know of languages B; A such that B T A but not in one oracle ≤ call are too technical to give here. Consider B := \HP twice" := code(M1)code(M2)x1#x2 : f M1(x1) M2(x2) . You might think this requires two calls to an oracle X such as X = K to # ^ # g solve, but you can do it in one call by first combining M1 and M2 into a machine M3 that halts iff both of them do, and then asking a single question relevant to M3. 1 Now suppose that we replace an oracle language X by its complement X. Then we could switch the wires at the query state to change M0 to an OTM M0 whose responsese to the oracle are 1 Accept and 0 Reject. That is, M0 behaves like \Simong the oracle says: accept w iff 7! 7! zw X." Then we can be even more specificg and write HP m X. This is read, \HP many-one reduces2 e to X." (The `many' allows for the possibility that many≤ e different input strings w might lead to thee same one query string zw, but usually in practice zw uniquely depends on w, and so we could actually speak of a one-one reduction, written HP 1 X. But \many-one" carries the specifics far enough for my purposes.) In lecture you will have≤ seene the equivalent definition: Definition 1.2. A language A many-one reduces to a language B, written A m B, if there exists a total recursive function f such that for all strings w, w A f(w) B≤. 2 () 2 2. Relativized Language Classes Now we can adapt the notation used for the classes of recursive and r.e. languages to the case of oracle machines. For any oracle set A: (a) REA := L : there is an OTM M such that L(M A) = L . f g (b) RECA := L : there is an M such that L(M A) = L and M A is total . f g (c) co-REA := complements of languages in REA . f g Many theorems about the classes REC and RE go through untouched by the presence of a given oracle set A. That is to say, they relativize. Some examples: Theorem 2.1. For any oracle set A, RECA is closed under complements. A Proof. Given L REC , this means there is an OTM M0 such that M0 is total with oracle set A 2 A and L(M0 ) = L. Presuming that M0 has been put into the normal form which makes M0 \crash-free" and gives M0 distinguished accept and reject states qacc and qrej, let M1 be obtained 0 A A from M0 by interchanging qacc and qrej. Since MA is total, M1 is total. And L(M1 ) = L (the complement of L). So L RECA too. e e 2 Theorem 2.2. For any oracle set A, REA co-REA = RECA: \ Proof. First suppose L RECA. Then clearly L REA. By Theorem 2.1, L RECA, so L REA too. If L REA2, then by definition L co-RE2 A. So L REA co-REA: e 2 2 2 A A 2 2 \ e Going the othere way, let L RE co-RE . Then there are oracle TMs M1 and M2 such that A A 2 \ L(M1 ) = L and L(M2 ) = L. Thus on any input x, exactly one of M1 and M2 has an accepting computation on input x. Build an OTM M3 which on any input x makes two copies of x, simulates one step of M1 on the first copy, one step of M2 on the other copy, then back for one more step of M1 on the first copy, and so on. (The copies can be marked off with separate delimiters 1; $1; 2; $2; ^ ^ and if $1 ever runs up against 2, call the \make-more room" procedure from class.) If the step ^ of M1 or M2 enters a query state of M1 or M2, then M3 has the same query state, and the oracle A A A response works the same. M3 is total because exactly one of M1 and M2 will stop and accept; if it is M1, then M3 accepts, and if it is M2 that accepts, then M3 rejects. (If either M1 or M2 rejects A A ahead of time, M3 takes the appropriate action.) Since L = L(M ), L REC . 3 2 2 This proof too is really just like the one given in lecture for the non-oracle case. Now for two results which say when you can do away with the oracle, and when you can't: Theorem 2.3. If A REC, then REA = RE and RECA = REC: 2 Proof. Given that A is recursive, there is a total non-oracle TM MA accepting A. Let M0 be an oracle TM which accepts L with oracle A. Then instead of making q? in M0 a query state, run an arc (q?; ?; ; R; sA) to the start state sA of MA, wire the arcs ^ (fA; ; 1; S; q?); (fA; otherwise; same; L; fA) f ^ g to the accept state fA of MA (this moves left until it finds the new and changes it to a `1'), and ^ do likewise for the reject state rA of MA and `0'. This changes M0 to a non-oracle TM M1 which A still accepts L (since the \subroutine" MA always terminates), and M1 is total if M0 is total. Thus the conclusions about L follow. Theorem 2.4. It is not the case that for all B RE, REB = RE: 2 Proof. This time, let us take B to be the language of the Acceptance Problem; namely, AP := code(M)x : M accepts x . Define AP(B) := code(M)x : x L(M B) . Then AP(B) REB. Nowf the proof that AP = gREC from class is notf affected at all2 by the presenceg or absence2 of the oracle set B. To wit: If AP2 (B) RECB, then the complement AP(B) of AP(B) is in RECB, by Theorem 2.1. Hence the language2 D(B) := code(M): M is ang OTM such that with oracle B, B f B M does not accept code(M) also belongs to REC . That means there is a TM M1 such that B g L(M1 ) = D(B). What does M1 with oracle B do on input x := code(M1), pray tell? If it accepts, B B then code(M1) D(B) [by L(M1 ) = D(B)], but by definition of x D(B), M1 does not accept 2 B 2 code(M1).