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Properties of uniformly hard languages
Information Processing Letters 95 (2005) 329–332 www.elsevier.com/locate/ipl
Properties of uniformly hard languages Piotr Faliszewski a,∗,1 , Janusz Jarosz b a Department of Computer Science, University of Rochester, Rochester, NY 14627-0226, USA b Institute of Computer Science, AGH University of Science and Technology, al. Mickiewicza 30, Kraków, Poland
Received 2 April 2004; received in revised form 30 September 2004 Available online 25 April 2005 Communicated by P.M.B. Vitányi
Abstract Uniformly hard languages are languages that do not contain too many easy instances. We show that certain problems are uniformly hard, prove the uniformly hard analog of the Time Hierarchy Theorem, and explore the properties of the polynomial hierarchy with respect to uniform hardness. 2005 Elsevier B.V. All rights reserved. Keywords: Computational complexity; Uniform hardness; Time hierarchy; Polynomial hierarchy
1. Introduction Uniformly hard languages, defined by Downey and Fortnow [4], are languages that do not contain too many easy instances. Uniform hardness is especially interesting in the context of the languages that are neither NP-complete nor in P. By Ladner’s Theorem ([5], but see also [1,4]), we know that if P = NP then NP-intermediate languages exist. However, all known
* Corresponding author.
E-mail address: [email protected] (P. Faliszewski). 1 Supported in part by grant NSF-CCF-0426761. Work done in
part while the author was at the Institute of Computer Science, AGH University of Science and Technology, al. Mickiewicza 30, Kraków, Poland. 0020-0190/$ – see front matter 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.ipl.2004.11.014
proofs of this fact produce languages that are not uniformly hard. See the paper of Downey and Fortnow [4] for a discussion of uniform hardness with respect to the Ladner’s Theorem. In this paper we prove several properties of uniformly hard languages. Section 2 provides the formal definition of uniform hardness and some simple, yet unpublished, results regarding uniform hardness. In Section 3 we show that EXP contains at least one uniformly hard language, and building on that result, we show that all EXP-complete languages are uniformly hard. We also prove that an analog of the Time Hierarchy Theorem holds for uniformly hard languages. In Section 4 we prove that either all levels of the polynomial hierarchy (except for P) contain uniformly hard languages or none of them does.
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2. Definitions and basic properties of uniformly hard languages We use standard definitions of the complexity theory as in [6] except for the polynomial hierarchy levp p els which are denoted by Σk and Πk . We assume that all considered languages are over the alphabet Σ = {0, 1}. If A is some language then by A[k1 , k2 ] we mean the set {x: x ∈ A ∧ k1 |x| k2 }. The join of languages A and B is the language A ⊕ B = {0x: x ∈ A} ∪ {1x: x ∈ B}. Let x1 , . . . , xk be some strings over Σ. By x1 , . . . , xk we mean a string x, such that x is not longer than 2(|x1 | + · · · + |xk |) and from which each of the xi ’s can be extracted in polynomial time. Definition 1. We say that the language A is uniformly hard if it holds that: ∀C ∈ P: ∃k ∈ N: ∀m 2: A[m, mk ] = C[m, mk ]. Let UH be the set of all uniformly hard languages. By the definition, A ∈ UH may only contain polynomially long runs of easy instances (in terms of input length). We say that some class C is uniformly hard if C ∩ UH = ∅. Lemma 2. The class UH has the following properties: (1) A ∈ UH ⇒ A ∈ UH; (2) If A ∈ UH and B is some language then A ⊕ B ∈ UH. Proof. The first claim follows from the fact that P is closed under complementation. To prove the second, assume that A is uniformly hard but A ⊕ B is not. So there is a language C = C0 ⊕ C1 ∈ P such that for all k ∈ N there exists an m 2 such that (A ⊕ B)[m, mk ] = (C0 ⊕ C1 )[m, mk ], and thus A[m − 1, mk − 1] = C0 [m − 1, mk − 1]. However, since A is assumed to be uniformly hard and C0 ∈ P , there exists k such that for all m 2 it holds that A[m , mk ] = C0 [m , mk ]. Taking m = m and k sufficiently larger than k we arrive at a contradiction. 2 Definition 3. We say that a language A honestly manyp one reduces to B, A mh B, if there is a function f and two strictly increasing polynomials p and q such that:
• x ∈ A ⇔ f (x) ∈ B; • p(|f (x)|) > |x| (honesty); • f (x) is computable in time q(|x|). The honesty condition says that the output of the reduction function cannot be too small. Note that if p is a polynomial from Definition 3 then we can replace it by m + p(m). Therefore, we can assume that p(m) m and, analogously, q(m) m. The proof of the next theorem, stated without a proof in [4], results in a useful corollary. p
Theorem 4. If A ∈ UH and A mh B then B ∈ UH. The idea of the proof is to note that there are areas of A where strings need to reduce to strings from the easy areas of B. Proof. Let f be an honest reduction from A to B and p, q be the appropriate polynomials from Definition 3. Assume that B is not uniformly hard. There must be a language C ∈ P such that for every k ∈ N there is an integer m 2 such that B[m, mk ] = C[m, mk ]. We will show that this leads to a contradiction. Let D be a language D = {x: f (x) ∈ C}. It is clear that D ∈ P since C ∈ P and f is computable in polynomial time. Let us fix a positive integer l. There exists a constant K such that: ∀m 2: ∀k K:
m p(m) p(m)l q p(m)l mk .
B is not uniformly hard so there is an integer m 2 such that B[m, mK ] = C[m, mK ]. Let x be a word, such that p(m) |x| p(m)l . Since q is f ’s time bound it holds that |f (x)| q(|x|). Hence, we have |f (x)| q(p(m)l ). The honesty condition of f says that p(|f (x)|) |x|. The polynomial p is strictly increasing so we have |f (x)| p −1 (|x|). Bounding |x| from below by p(m) gives m |f (x)|. All in all, we have m |f (x)| q(p(m)l ) mk . Between lengths m and mk languages B and C are the same. Hence, for words x such that p(m) |x| p(m)l the function f is a reduction from A to C. Thus, we have A p(m), p(m)l = x: f (x) ∈ B ∧ p(m) |x| p(m)l = x: f (x) ∈ C ∧ p(m) |x| p(m)l .
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As a consequence, A[p(m), p(m)l ] = D[p(m), p(m)l ] and A ∈ / UH as l was an arbitrary positive integer. 2 p
Corollary 5. If B ∈ / UH, and A mh B, then: ∃C, D ∈ P: ∀k ∈ N: ∃m 2: A[m, mk ] = C[m, mk ] ∧ B[m, mk ] = D[m, mk ]. To prove this corollary it is enough to take the proof of the above theorem and note that the assumptions we make there to reach a contradiction are actually the assumptions of the corollary. What there gives a contradiction here, without the assumption on the uniform hardness of A, leads to the conclusion of the corollary.
3. Time hierarchy of uniformly hard languages This section is devoted to proving the Time Hierarchy Theorem for uniformly hard languages. First, we need some language from UH ∩ EXP. Theorem 6. The class EXP contains uniformly hard languages. Proof. We will do a proof by diagonalization. Let Mi be an enumeration of polynomial time Turing machines over the alphabet Σ = {0, 1}. Each machine Mi has a clock that stops it after ni + i steps (where n is the length of the input string). Define A to be a language containing the following words. If x is the lth word of length n (in lexicographic order) then: x∈A
⇔
l n and Ml rejects x.
Obviously, A ∈ EXP. We want to show that A ∈ UH. Assume the opposite: ∃C ∈ P: ∀k ∈ N: ∃m 2: A[m, mk ] = C[m, mk ].
(1)
Since C ∈ P, there is an Mi such that C = L(Mi ). Let K be a positive integer such that 2K i. Eq. (1) guarantees the existence of m 2 such that A[m, mK ] = C[m, mK ]. For every m 2 it holds that mK i. Let ω be the ith word of length mK . Since i mK we have that ω ∈ A ⇔ Mi rejects ω and ω ∈ C ⇔ Mi accepts ω—a contradiction. Therefore, A ∈ UH. 2
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The language A defined above can be decided in time O(nn ); however, this time bound can be reduced to any superpolynomial function f . The only difference in the proof is that one would need a larger K in order to let A compute all the steps of Mi . This means that UH languages lay as close to P as possible. In particular, there is a uniformly hard language U computable in time O(2n ). Define EXP-H ALTING to be the problem asking if a string x is accepted by a Turing machine M in time 2|t| , where the input string is M, x, t. Every problem from EXP can be honp estly reduced to EXP-H ALTING; in particular, U mh EXP-H ALTING and by Theorem 4 EXP-H ALTING ∈ UH. All EXP-complete languages are honestly reducible to each other [2], therefore all languages complete for EXP are in UH. The classical version of the Time Hierarchy Theorem (as in [6]) says that if f (n) > n is an increasing, time constructible function, then there is a language A such that A ∈ / TIME(f (n)), yet A ∈ TIME(f (2n + 1)3 ). Using language U and Lemma 2 we can prove a similar result for UH. Theorem 7. If f (n) > 2n is an increasing, time constructible function, then there is a uniformly hard language A such that A ∈ TIME(f (2n + 1)3 ) and A ∈ / TIME(f (n)). To see the theorem holds, it is enough to take the language A from the Time Hierarchy Theorem and take its join with U. A similar argument can be devised for any increasing, time constructible, superpolynomial function f .
4. Polynomial hierarchy and uniform hardness The Time Hierarchy Theorem from the preceding section deals with languages computable in superpolynomial time. We now turn to the investigation of UH with respect to the polynomial hierarchy. We need a complete language for each of its levels. Define p Σk -SAT to be the language containing words of the form φ, x1 , . . . , xk such that: • φ is a boolean formula over the variables v1,1 , . . . , v1,|x1 | , v2,1 , . . . , vk,|xk | ;
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• It holds that ∃v1,1 . . . v1,|x1 | : ∀v2,1 . . . v2,|x2 | : . . . Qvk,1 . . . vk,|xk | : φ, where Q is an appropriate quantifier in the sequence of alternating ∃ and ∀.
∃C ∈ P, D ∈ NP, E ∈ P: ∀k ∈ N: ∃m 2: A [m, mk ] = C[m, mk ] ∧ A[m, mk ] = D[m, mk ] ∧ D[m, mk ]
p
Σk -SAT is complete for the kth level of the polynop mial hierarchy, Σk [7]. Theorem 8. If NP is not uniformly hard, then for any p p k 1 neither Σk nor Πk is uniformly hard. Proof. We will do a proof by induction on k. Our assumption implies that, by Lemma 2, coNP is not uniformly hard; the theorem holds for k = 1. We need to prove the induction step. Assume that the p p theorem holds for Σk and Πk (k = 1, 2, . . . , n). We p will prove that it holds for Σn+1 . Let A be the lanp guage Σn+1 -SAT and let A be the same language for the formulas without the first quantifier. That is, A asks about the value of the formula ∀v2,1 . . . v2,|x2 | : ∃v3,1 . . . v3,|x3 | : . . . Qvk,1 . . . vk,|xk | : φ (the values for variables v1,1 , . . . , v1,|x1 | come from consecutive symp bols of the string x1 ). Clearly, A is Πn -complete and by the inductive assumption not in UH. If C is a language in P then there is a language D in NP such that whenever it holds that A [m, mk ] = C[m, mk ] for natural numbers m, k, then it also holds that A[m, mk ] = D[m, mk ]. The proof of this fact is based on the following nondeterministic algorithm for A. For each input string y = φ, x1 , . . . , xn+1 guess a string x1 (over the alphabet Σ = {0, 1} of length |x1 |). Accept if and only if y = φ, x1 , x2 , . . . , xn+1 ∈ A . Note that |y | = |y|. Consider the same algorithm with a slight modification—accept if and only if y ∈ C. This algorithm decides some language D ∈ NP. If m |y| mk then both algorithms give the same answer since we assumed A [m, mk ] = C[m, mk ]. p Since D ∈ NP and A is complete for Πn (n 1) p p we have that A is complete for Σn and D m A . We can make this reduction honest using the technique from the Cook’s Theorem [3]. An application of Corollary 5 tells us that there is a language E ∈ P such that for every natural k there is an m 2 such that A [m, mk ] = C[m, mk ] ∧ D[m, mk ] = E[m, mk ]. This means that:
= E[m, mk ].
(2)
Removing C, D, and A from the above formula we p get that A ∈ / UH. Since A is complete for Σn+1 and p the reduction of every problem from Σn+1 to A can be made honest (e.g., by techniques from the proof of p the Cook’s Theorem [3]), no language in Σn+1 can be in UH. The inductive step is proved, and since UH is closed under complementation, the whole theorem is proved as well. 2 p
On the other hand, let Σn be the first level of PH which is uniformly hard. But then the levels p p Σ1 , . . . , Σn−1 would not be uniformly hard. By the above proof of the inductive assumption this would p mean that Σn is not uniformly hard—a contradiction unless n = 1. Therefore, either all levels of the polynomial hierarchy (except for P) are uniformly hard or none of them is.
Acknowledgements Thanks to Lance Fortnow who suggested the topic.
References [1] J.L. Balcazar, J. Diaz, A note on a theorem by Ladner, Inform. Process. Lett. 15 (3) (1982) 84–86. [2] L. Berman, Polynomial reducibilities and complete sets, Ph.D. thesis, Cornell University, 1977. [3] S.A. Cook, The complexity of theorem-proving procedures, in: Proc. 3rd Annual ACM Symp. on Theory of Computing, ACM Press, New York, 1971, pp. 151–158. [4] R. Downey, L. Fortnow, Uniformly hard languages, Theoret. Comput. Sci. 298 (2) (2003) 303–315. [5] R.E. Ladner, On the structure of polynomial time reducibility, J. ACM 22 (1) (1975) 155–171. [6] C. Papadimitrou, Computational Complexity, Addison-Wesley, Reading, MA, 1994. [7] C. Wrathall, Complete sets and the polynomial-time hierarchy, Theoret. Comput. Sci. 3 (1) (1976) 23–33.
Properties of uniformly hard languages Piotr Faliszewski a,∗,1 , Janusz Jarosz b a Department of Computer Science, University of Rochester, Rochester, NY 14627-0226, USA b Institute of Computer Science, AGH University of Science and Technology, al. Mickiewicza 30, Kraków, Poland
Received 2 April 2004; received in revised form 30 September 2004 Available online 25 April 2005 Communicated by P.M.B. Vitányi
Abstract Uniformly hard languages are languages that do not contain too many easy instances. We show that certain problems are uniformly hard, prove the uniformly hard analog of the Time Hierarchy Theorem, and explore the properties of the polynomial hierarchy with respect to uniform hardness. 2005 Elsevier B.V. All rights reserved. Keywords: Computational complexity; Uniform hardness; Time hierarchy; Polynomial hierarchy
1. Introduction Uniformly hard languages, defined by Downey and Fortnow [4], are languages that do not contain too many easy instances. Uniform hardness is especially interesting in the context of the languages that are neither NP-complete nor in P. By Ladner’s Theorem ([5], but see also [1,4]), we know that if P = NP then NP-intermediate languages exist. However, all known
* Corresponding author.
E-mail address: [email protected] (P. Faliszewski). 1 Supported in part by grant NSF-CCF-0426761. Work done in
part while the author was at the Institute of Computer Science, AGH University of Science and Technology, al. Mickiewicza 30, Kraków, Poland. 0020-0190/$ – see front matter 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.ipl.2004.11.014
proofs of this fact produce languages that are not uniformly hard. See the paper of Downey and Fortnow [4] for a discussion of uniform hardness with respect to the Ladner’s Theorem. In this paper we prove several properties of uniformly hard languages. Section 2 provides the formal definition of uniform hardness and some simple, yet unpublished, results regarding uniform hardness. In Section 3 we show that EXP contains at least one uniformly hard language, and building on that result, we show that all EXP-complete languages are uniformly hard. We also prove that an analog of the Time Hierarchy Theorem holds for uniformly hard languages. In Section 4 we prove that either all levels of the polynomial hierarchy (except for P) contain uniformly hard languages or none of them does.
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2. Definitions and basic properties of uniformly hard languages We use standard definitions of the complexity theory as in [6] except for the polynomial hierarchy levp p els which are denoted by Σk and Πk . We assume that all considered languages are over the alphabet Σ = {0, 1}. If A is some language then by A[k1 , k2 ] we mean the set {x: x ∈ A ∧ k1 |x| k2 }. The join of languages A and B is the language A ⊕ B = {0x: x ∈ A} ∪ {1x: x ∈ B}. Let x1 , . . . , xk be some strings over Σ. By x1 , . . . , xk we mean a string x, such that x is not longer than 2(|x1 | + · · · + |xk |) and from which each of the xi ’s can be extracted in polynomial time. Definition 1. We say that the language A is uniformly hard if it holds that: ∀C ∈ P: ∃k ∈ N: ∀m 2: A[m, mk ] = C[m, mk ]. Let UH be the set of all uniformly hard languages. By the definition, A ∈ UH may only contain polynomially long runs of easy instances (in terms of input length). We say that some class C is uniformly hard if C ∩ UH = ∅. Lemma 2. The class UH has the following properties: (1) A ∈ UH ⇒ A ∈ UH; (2) If A ∈ UH and B is some language then A ⊕ B ∈ UH. Proof. The first claim follows from the fact that P is closed under complementation. To prove the second, assume that A is uniformly hard but A ⊕ B is not. So there is a language C = C0 ⊕ C1 ∈ P such that for all k ∈ N there exists an m 2 such that (A ⊕ B)[m, mk ] = (C0 ⊕ C1 )[m, mk ], and thus A[m − 1, mk − 1] = C0 [m − 1, mk − 1]. However, since A is assumed to be uniformly hard and C0 ∈ P , there exists k such that for all m 2 it holds that A[m , mk ] = C0 [m , mk ]. Taking m = m and k sufficiently larger than k we arrive at a contradiction. 2 Definition 3. We say that a language A honestly manyp one reduces to B, A mh B, if there is a function f and two strictly increasing polynomials p and q such that:
• x ∈ A ⇔ f (x) ∈ B; • p(|f (x)|) > |x| (honesty); • f (x) is computable in time q(|x|). The honesty condition says that the output of the reduction function cannot be too small. Note that if p is a polynomial from Definition 3 then we can replace it by m + p(m). Therefore, we can assume that p(m) m and, analogously, q(m) m. The proof of the next theorem, stated without a proof in [4], results in a useful corollary. p
Theorem 4. If A ∈ UH and A mh B then B ∈ UH. The idea of the proof is to note that there are areas of A where strings need to reduce to strings from the easy areas of B. Proof. Let f be an honest reduction from A to B and p, q be the appropriate polynomials from Definition 3. Assume that B is not uniformly hard. There must be a language C ∈ P such that for every k ∈ N there is an integer m 2 such that B[m, mk ] = C[m, mk ]. We will show that this leads to a contradiction. Let D be a language D = {x: f (x) ∈ C}. It is clear that D ∈ P since C ∈ P and f is computable in polynomial time. Let us fix a positive integer l. There exists a constant K such that: ∀m 2: ∀k K:
m p(m) p(m)l q p(m)l mk .
B is not uniformly hard so there is an integer m 2 such that B[m, mK ] = C[m, mK ]. Let x be a word, such that p(m) |x| p(m)l . Since q is f ’s time bound it holds that |f (x)| q(|x|). Hence, we have |f (x)| q(p(m)l ). The honesty condition of f says that p(|f (x)|) |x|. The polynomial p is strictly increasing so we have |f (x)| p −1 (|x|). Bounding |x| from below by p(m) gives m |f (x)|. All in all, we have m |f (x)| q(p(m)l ) mk . Between lengths m and mk languages B and C are the same. Hence, for words x such that p(m) |x| p(m)l the function f is a reduction from A to C. Thus, we have A p(m), p(m)l = x: f (x) ∈ B ∧ p(m) |x| p(m)l = x: f (x) ∈ C ∧ p(m) |x| p(m)l .
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As a consequence, A[p(m), p(m)l ] = D[p(m), p(m)l ] and A ∈ / UH as l was an arbitrary positive integer. 2 p
Corollary 5. If B ∈ / UH, and A mh B, then: ∃C, D ∈ P: ∀k ∈ N: ∃m 2: A[m, mk ] = C[m, mk ] ∧ B[m, mk ] = D[m, mk ]. To prove this corollary it is enough to take the proof of the above theorem and note that the assumptions we make there to reach a contradiction are actually the assumptions of the corollary. What there gives a contradiction here, without the assumption on the uniform hardness of A, leads to the conclusion of the corollary.
3. Time hierarchy of uniformly hard languages This section is devoted to proving the Time Hierarchy Theorem for uniformly hard languages. First, we need some language from UH ∩ EXP. Theorem 6. The class EXP contains uniformly hard languages. Proof. We will do a proof by diagonalization. Let Mi be an enumeration of polynomial time Turing machines over the alphabet Σ = {0, 1}. Each machine Mi has a clock that stops it after ni + i steps (where n is the length of the input string). Define A to be a language containing the following words. If x is the lth word of length n (in lexicographic order) then: x∈A
⇔
l n and Ml rejects x.
Obviously, A ∈ EXP. We want to show that A ∈ UH. Assume the opposite: ∃C ∈ P: ∀k ∈ N: ∃m 2: A[m, mk ] = C[m, mk ].
(1)
Since C ∈ P, there is an Mi such that C = L(Mi ). Let K be a positive integer such that 2K i. Eq. (1) guarantees the existence of m 2 such that A[m, mK ] = C[m, mK ]. For every m 2 it holds that mK i. Let ω be the ith word of length mK . Since i mK we have that ω ∈ A ⇔ Mi rejects ω and ω ∈ C ⇔ Mi accepts ω—a contradiction. Therefore, A ∈ UH. 2
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The language A defined above can be decided in time O(nn ); however, this time bound can be reduced to any superpolynomial function f . The only difference in the proof is that one would need a larger K in order to let A compute all the steps of Mi . This means that UH languages lay as close to P as possible. In particular, there is a uniformly hard language U computable in time O(2n ). Define EXP-H ALTING to be the problem asking if a string x is accepted by a Turing machine M in time 2|t| , where the input string is M, x, t. Every problem from EXP can be honp estly reduced to EXP-H ALTING; in particular, U mh EXP-H ALTING and by Theorem 4 EXP-H ALTING ∈ UH. All EXP-complete languages are honestly reducible to each other [2], therefore all languages complete for EXP are in UH. The classical version of the Time Hierarchy Theorem (as in [6]) says that if f (n) > n is an increasing, time constructible function, then there is a language A such that A ∈ / TIME(f (n)), yet A ∈ TIME(f (2n + 1)3 ). Using language U and Lemma 2 we can prove a similar result for UH. Theorem 7. If f (n) > 2n is an increasing, time constructible function, then there is a uniformly hard language A such that A ∈ TIME(f (2n + 1)3 ) and A ∈ / TIME(f (n)). To see the theorem holds, it is enough to take the language A from the Time Hierarchy Theorem and take its join with U. A similar argument can be devised for any increasing, time constructible, superpolynomial function f .
4. Polynomial hierarchy and uniform hardness The Time Hierarchy Theorem from the preceding section deals with languages computable in superpolynomial time. We now turn to the investigation of UH with respect to the polynomial hierarchy. We need a complete language for each of its levels. Define p Σk -SAT to be the language containing words of the form φ, x1 , . . . , xk such that: • φ is a boolean formula over the variables v1,1 , . . . , v1,|x1 | , v2,1 , . . . , vk,|xk | ;
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• It holds that ∃v1,1 . . . v1,|x1 | : ∀v2,1 . . . v2,|x2 | : . . . Qvk,1 . . . vk,|xk | : φ, where Q is an appropriate quantifier in the sequence of alternating ∃ and ∀.
∃C ∈ P, D ∈ NP, E ∈ P: ∀k ∈ N: ∃m 2: A [m, mk ] = C[m, mk ] ∧ A[m, mk ] = D[m, mk ] ∧ D[m, mk ]
p
Σk -SAT is complete for the kth level of the polynop mial hierarchy, Σk [7]. Theorem 8. If NP is not uniformly hard, then for any p p k 1 neither Σk nor Πk is uniformly hard. Proof. We will do a proof by induction on k. Our assumption implies that, by Lemma 2, coNP is not uniformly hard; the theorem holds for k = 1. We need to prove the induction step. Assume that the p p theorem holds for Σk and Πk (k = 1, 2, . . . , n). We p will prove that it holds for Σn+1 . Let A be the lanp guage Σn+1 -SAT and let A be the same language for the formulas without the first quantifier. That is, A asks about the value of the formula ∀v2,1 . . . v2,|x2 | : ∃v3,1 . . . v3,|x3 | : . . . Qvk,1 . . . vk,|xk | : φ (the values for variables v1,1 , . . . , v1,|x1 | come from consecutive symp bols of the string x1 ). Clearly, A is Πn -complete and by the inductive assumption not in UH. If C is a language in P then there is a language D in NP such that whenever it holds that A [m, mk ] = C[m, mk ] for natural numbers m, k, then it also holds that A[m, mk ] = D[m, mk ]. The proof of this fact is based on the following nondeterministic algorithm for A. For each input string y = φ, x1 , . . . , xn+1 guess a string x1 (over the alphabet Σ = {0, 1} of length |x1 |). Accept if and only if y = φ, x1 , x2 , . . . , xn+1 ∈ A . Note that |y | = |y|. Consider the same algorithm with a slight modification—accept if and only if y ∈ C. This algorithm decides some language D ∈ NP. If m |y| mk then both algorithms give the same answer since we assumed A [m, mk ] = C[m, mk ]. p Since D ∈ NP and A is complete for Πn (n 1) p p we have that A is complete for Σn and D m A . We can make this reduction honest using the technique from the Cook’s Theorem [3]. An application of Corollary 5 tells us that there is a language E ∈ P such that for every natural k there is an m 2 such that A [m, mk ] = C[m, mk ] ∧ D[m, mk ] = E[m, mk ]. This means that:
= E[m, mk ].
(2)
Removing C, D, and A from the above formula we p get that A ∈ / UH. Since A is complete for Σn+1 and p the reduction of every problem from Σn+1 to A can be made honest (e.g., by techniques from the proof of p the Cook’s Theorem [3]), no language in Σn+1 can be in UH. The inductive step is proved, and since UH is closed under complementation, the whole theorem is proved as well. 2 p
On the other hand, let Σn be the first level of PH which is uniformly hard. But then the levels p p Σ1 , . . . , Σn−1 would not be uniformly hard. By the above proof of the inductive assumption this would p mean that Σn is not uniformly hard—a contradiction unless n = 1. Therefore, either all levels of the polynomial hierarchy (except for P) are uniformly hard or none of them is.
Acknowledgements Thanks to Lance Fortnow who suggested the topic.
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