# Proof Systems that Take Advice

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10 M on inputs x. In the sequel, we refrain from always mentioning U explicitly. Thus we simply write statements like M is a t-time bounded Turing machine with the precise meaning that U always spends at most t(n) steps to simulate M on inputs of length n. Likewise, to simulate a machine M on input x always means executing U(M, x). A nondeterministic Turing machine M is consistent with a language L (or L-consistent), if L(M) L. We can now give the definition of nondeterministic instance complexity from [1]. Definition 11 (Arvind et al. [1]). For a set L and a time bound t, the t- time-bounded nondeterministic instance complexity of x with respect to L is defined as nic t (x : L) = min{ M : M is an L-consistent t-time-bounded nondeterministic machine, and M decides correctly on x }. Similarly as in the deterministic case in [17], we collect all languages with prescribed upper bounds on the running time and nondeterministic instance complexity in a complexity class. Definition 12. Let F 1 and F 2 be two classes of functions. We define NIC[F 1, F 2 ] = {L : there exist s F 1 and t F 2 such that for all x Σ nic t (x : L) s( x )}. A particularly interesting choice for the classes F 1 and F 2 is to allow polynomial running time, but only logarithmic descriptions for the machines. This leads to the class NIC[log, poly] which plays a central role in this paper. Similarly as in the deterministic case (cf. [17]), the next proposition locates this class between the nonuniform classes NP/log and NP/poly. Proposition 13. NP/log NIC[log, poly] NP/poly. Proof. For the first inclusion, let L NP/log. Let M be a nondeterministic Turing machine with logarithmic advice that decides L and let a n be the advice given to M for inputs of length n. We define a collection of programs M n,an for L as follows. On input x the machine M n,an first checks, whether the length of the input is n. For this we need to code the number n into M n,an. If x n, then M n,an rejects. Otherwise, M n,an simulates M on input x with advice a n which is also coded into M n,an. Essentially, the machines M n,an are constructed by hardwiring n and a n into M, and thus the size of M n,an is logarithmic in n. Therefore L NIC[log, poly]. For the second inclusion, let L NIC[log, poly]. Then there exist a constant c and a polynomial p such that for all x we have nic p (x : L) c log x + c. We construct a nondeterministic Turing machine M with polynomial advice that accepts exactly L. The advice of M for length n consists of all nondeterministic Turing machines M 1,..., M m of size at most c log n + c which are consistent with L. Note that for each input length n, there are only polynomially many machines of the appropriate size c log n + c. Hence polynomial advice suffices 10

11 to encode the whole list M 1,..., M m. On input x, the machine M simulates each M i on x for at most p( x ) steps. If any of the M i accepts, then M accepts as well, otherwise it rejects. We claim, that L(M) = L. For, if x L, then there is a nondeterministic L- consistent Turing machine M i such that M i (x) accepts and M i c log x + c. Thus, also M(x) accepts. If, on the other hand, M accepts x, then so does some M i which is consistent with L. Therefore, x L because L(M i ) L. In fact, the inclusions in Proposition 13 are proper as we will show in Theorem 15 below. For the proof we need the following notion: Definition 14 (Buhrman, Fortnow, Laplante [9]). For a time bound t, the nondeterministic decision complexity of x, denoted CND t (x), is the minimal size of a t-time-bounded nondeterministic Turing machine M with L(M) = {x}. As already noted in [1], the CND measure provides an upper bound to the nic measure, i.e., for any language L and time bound t there is a constant c > 0 such that nic t (x : L) CND t (x) + c for all x Σ. By a simple counting argument, it follows that for any length n there exist strings x of length n with CND(x) n, where CND(x) is the minimal size of a nondeterministic Turing machine M with L(M) = {x} (i.e., the time-unbounded CND measure). Inspired by a similar result in [17], we now prove the following separations: Theorem For every constant c > 0, NP/n c NIC[log, poly]. 2. NIC[log, poly] P/lin. Proof. For the first item, let 0 < c < d be natural numbers. Diagonalizing against all NP machines and all advice strings, we inductively define a set A with A NIC[log, poly], but A NP/n c. Let (N i ) i N be an enumeration of all NP machines, in which every machine occurs infinitely often. In step n we diagonalize against the machine N n and every advice string of length n c which N n might use for length n. Let x 1,..., x 2 n be the lexicographic enumeration of all strings in Σ n and let S n = {x 1,..., x n d} Σ n. For each string w of length at most n c, let A w = {x S n : N n (x) accepts under advice w}. Since there are only 2 nc such sets, but 2 nd subsets of S n, there must be one which is not equal to any A w. For every n, let A n be one such set, and let A = n A n. By construction, A NP/n c. We still have to show A NIC[log, poly]. For each string s, let s be the substring of s which has all leading zeros deleted. For each n and each a A n, let M n,ea be the following machine: on input x, the machine M n,ea checks whether x = n and x = ã. If this test is positive, then M n,ea accepts, otherwise it rejects. The machine M n,ea is of size O(log n), as both n and ã are of length O(log n) (Observe that the first n d elements in the lexicographic order of Σ n have no 1 s appearing before the last log n d bits). Thus A NIC[log, poly]. For the second item, let A be a set that contains exactly one element x per length with CND(x) x. Obviously, A P/lin because A contains exactly one string per length and this element can be given as advice. On the other hand, A NIC[log, poly]. Assume on the contrary, that A NIC[log, poly]. Then there 11

14 function this allows the advice to refer to the machine M (but not to the input x which is given in binary notation). Now, since L NIC[log, poly], for every x L there is an L-consistent Turing machine M x with running time p which accepts x and M x c log x + c. Thus every element x L has a polynomial-size f-proof x, w, 1 M x where w is an accepting path of M x (x). In fact, we can prove a more general version of the preceding theorem, where we replace polynomial upper bounds for the proof length by arbitrary upper bounds. In this way we obtain: Theorem 20. For any language L and any function t : N N, t n Ω(1), the following conditions are equivalent: 1. L has a t O(1) -bounded ps/1 with input advice. 2. L has a t O(1) -bounded ps/log with input advice. 3. L NIC[O(log t), t O(1) ]. For a language L we now consider the following three assertions: A1: L has a polynomially bounded ps/log with output advice. A2: L has a polynomially bounded ps/log with input advice. A3: L has a polynomially bounded ps/poly with output advice. By our results so far, assertions A1, A2, and A3 are equivalent to the statement that L is contained in the classes NP/log, NIC[log, poly], and NP/poly, respectively. As these classes form a chain of inclusions by Proposition 13, we get the implications A1 A2 A3 for every L. Moreover, by Corollary 16, the inclusions NP/log NIC[log, poly] NP/poly are proper. Hence we obtain: Corollary 21. There exist languages L for which A2 is fulfilled, but A1 fails. Likewise, there exist languages L for which A3 is fulfilled, but A2 fails. Table 1 provides an overview of our results on question Q1 obtained so far, showing which languages possess polynomially bounded proof systems with advice. It is interesting to note that all language classes appearing in this table form a chain of strict inclusions (cf. Corollary 16). Table 1. Languages with polynomially bounded proof systems input advice output advice ps/poly NP/poly NP/poly ps/log NIC[log, poly] NP/log ps/1 NIC[log, poly] NP/1 ps/0 NP 14

18 Thus, setting δ = 2 c d, we get n δ TAUT =n. Therefore, by definition of n there exists a number m n,0 t(d n 2 ) such that n m n,0 δ TAUT =n, i.e., a δ-fraction of all tautologies of length n has a conjunctive f-proof of size m n,0. Consider now the set TAUT =n 0 of all tautologies of length n which do not have conjunctive f-proofs of size m n,0. Repeating the above argument for TAUT =n 0 yields a proof length m n,1 t(d n 2 ) such that n m n,1 δ TAUT =n 0. Iterating this argument we obtain a sequence m n,0, m n,1,..., m n,l(n), where l(n) = log TAUT =n log(1 δ) 1 n log(1 δ) 1 such that every ϕ TAUT =n has a conjunctive f-proof of size m n,i for some i {0,..., l(n)}. We will now define a proof system g which uses polynomial output advice and obeys the same proof lengths as f. Assume that f is computed by the polynomial-time Turing transducer M f with advice function h f. The system g will be computed by a polynomial-time Turing transducer M g using the advice function h g (n) = m n,0, h f (m n,0 ),..., m n,l(n), h f (m n,l(n) ). The machine M g works as follows: On input π, M g first checks whether the proof π has the form ϕ, ψ 1,..., ψ n, π, i, where ϕ, ψ 1,..., ψ n are formulas of length n such that ϕ {ψ 1,..., ψ n }, π is a string (an f-proof), and i is a number l(n). If this test fails, then g(π ) is undefined. Then M g uses its advice to check whether π = m n,i. If so, then M g simulates M f on input π using advice h f (m n,i ) (which is available through the advice function h g ). If the output of this simulation is ψ 1 ψ n, then M g outputs ϕ, otherwise g(π ) is undefined. By our analysis above, g is a propositional proof system (it is correct and complete). The advice only depends on the length n of the proven formula, so g uses output advice. To estimate the advice length, let h f (m) log a m for some constant a. Then l(n) ( ) h g (n) ( m n,i + h(m n,i ) ) (l(n) + 1) n/δ + log a (2 n/δ ) = n O(1). i=0 The size of a g-proof ϕ, ψ 1,..., ψ n, π, i for ϕ TAUT =n is dominated by π t(d n 2 ), and therefore g is s(n)-bounded for some s(n) O(t(d n 2 )). In some sense, Theorem 26 transfers the results of Theorem 18 and Corollary 23 to super-polynomial proof lengths. However, while Theorem 18 has an easy proof and holds for all languages, the last construction is rather non-trivial and uses some assumption on L. Here we stated the result for the most interesting case L = TAUT, but the same proof also works for all languages L with a polynomial-time computable AND-function., 18

20 11. J.-Y. Cai, V. T. Chakaravarthy, L. A. Hemaspaandra, and M. Ogihara. Competing provers yield improved Karp-Lipton collapse results. Information and Computation, 198(1):1 23, S. A. Cook and J. Krajíček. Consequences of the provability of NP P/poly. The Journal of Symbolic Logic, 72(4): , S. A. Cook and R. A. Reckhow. The relative efficiency of propositional proof systems. The Journal of Symbolic Logic, 44(1):36 50, R. M. Karp and R. J. Lipton. Some connections between nonuniform and uniform complexity classes. In Proc. 12th ACM Symposium on Theory of Computing, pages ACM Press, J. Köbler, J. Messner, and J. Torán. Optimal proof systems imply complete sets for promise classes. Information and Computation, 184(1):71 92, J. Krajíček and P. Pudlák. Propositional proof systems, the consistency of first order theories and the complexity of computations. The Journal of Symbolic Logic, 54(3): , P. Orponen, K. Ko, U. Schöning, and O. Watanabe. Instance complexity. Journal of the ACM, 41(1):96 121, P. Pudlák. The lengths of proofs. In S. R. Buss, editor, Handbook of Proof Theory, pages Elsevier, Amsterdam,

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