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Complements of Nondeterministic Classes • from P Complements of Nondeterministic Classes ² From p. 133, we know R, RE, and coRE are distinct. { coRE contains the complements of languages in RE, not the languages not in RE. ² Recall that the complement of L, denoted by L¹, is the language §¤ ¡ L. { sat complement is the set of unsatis¯able boolean expressions. { hamiltonian path complement is the set of graphs without a Hamiltonian path. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 181 The Co-Classes ² For any complexity class C, coC denotes the class fL : L¹ 2 Cg: ² Clearly, if C is a deterministic time or space complexity class, then C = coC. { They are said to be closed under complement. { A deterministic TM deciding L can be converted to one that decides L¹ within the same time or space bound by reversing the \yes" and \no" states. ² Whether nondeterministic classes for time are closed under complement is not known (p. 79). °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 182 Comments ² As coC = fL : L¹ 2 Cg; L 2 C if and only if L¹ 2 coC. ² But it is not true that L 2 C if and only if L 62 coC. { coC is not de¯ned as C¹. ² For example, suppose C = ff2; 4; 6; 8; 10;:::gg. ² Then coC = ff1; 3; 5; 7; 9;:::gg. ¤ ² But C¹ = 2f1;2;3;:::g ¡ ff2; 4; 6; 8; 10;:::gg. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 183 The Quanti¯ed Halting Problem ² Let f(n) ¸ n be proper. ² De¯ne Hf = fM; x : M accepts input x after at most f(j x j) stepsg; where M is deterministic. ² Assume the input is binary. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 184 3 Hf 2 TIME(f(n) ) ² For each input M; x, we simulate M on x with an alarm clock of length f(j x j). { Use the single-string simulator (p. 57), the universal TM (p. 118), and the linear speedup theorem (p. 64). { Our simulator accepts M; x if and only if M accepts x before the alarm clock runs out. 2 2 ² From p. 63, the total running time is O(`M kM f(n) ), where `M is the length to encode each symbol or state of M and kM is M's number of strings. 2 3 ² As `M kM = O(n), the running time is O(f(n) ), where the constant is independent of M. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 185 Hf 62 TIME(f(bn=2c)) ² Suppose TM MHf decides Hf in time f(bn=2c). ² Consider machine Df (M): if MHf (M; M) = \yes" then \no" else \yes" ² Df on input M runs in the same time as MHf on input 2n+1 a M; M, i.e., in time f(b 2 c) = f(n), where n = j M j. aA student pointed out on October 6, 2004, that this estimation omits the time to write down M; M. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 186 The Proof (concluded) ² First, Df (Df ) = \yes" ) Df ; Df 62 Hf ) Df does not accept Df within time f(j Df j) ) Df (Df ) 6= \yes" ) Df (Df ) = \no" a contradiction ² Similarly, Df (Df ) = \no" ) Df (Df ) = \yes." °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 187 The Time Hierarchy Theorem Theorem 16 If f(n) ¸ n is proper, then TIME(f(n)) ( TIME(f(2n + 1)3): ² The quanti¯ed halting problem makes it so. Corollary 17 P ( EXP. ² P ⊆ TIME(2n) because poly(n) · 2n for n large enough. ² But by Theorem 16, 2 TIME(2n) ( TIME((22n+1)3) ⊆ TIME(2n ) ⊆ EXP: ² So P ( EXP. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 188 The Space Hierarchy Theorem Theorem 18 (Hennie and Stearns (1966)) If f(n) is proper, then SPACE(f(n)) ( SPACE(f(n) log f(n)): Corollary 19 L ( PSPACE. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 189 Nondeterministic Time Hierarchy Theorems Theorem 20 (Cook (1973)) If f(n) is proper, then NTIME(nr) ( NTIME(ns) whenever 1 · r < s. Theorem 21 (Seiferas, Fischer, and Meyer (1978)) If T1(n);T2(n) are proper, then NTIME(T1(n)) ( NTIME(T2(n)) whenever T1(n + 1) = o(T2(n)). °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 190 The Reachability Method ² The computation of a time-bounded TM can be represented by a directed graph. ² The TM's con¯gurations constitute the nodes. ² Two nodes are connected by a directed edge if one yields the other. ² The start node representing the initial con¯guration has zero in degree. ² When the TM is nondeterministic, a node may have an out degree greater than one. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 191 Illustration of the Reachability Method Initial configuration yes yes °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 192 Relations between Complexity Classes Theorem 22 Suppose f(n) is proper. Then 1. SPACE(f(n)) ⊆ NSPACE(f(n)), TIME(f(n)) ⊆ NTIME(f(n)). 2. NTIME(f(n)) ⊆ SPACE(f(n)). 3. NSPACE(f(n)) ⊆ TIME(klog n+f(n)). ² Proof of 2: { Explore the computation tree of the NTM for \yes." { Speci¯cally, generate a f(n)-bit sequence denoting the nondeterministic choices over f(n) steps. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 193 Proof of Theorem 22(2) ² (continued) { Simulate the NTM based on the choices. { Recycle the space and then repeat the above steps until a \yes" is encountered or the tree is exhausted. { Each path simulation consumes at most O(f(n)) space because it takes O(f(n)) time. { The total space is O(f(n)) because space is recycled. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 194 Proof of Theorem 22(3) ² Let k-string NTM M = (K; §; ¢; s) with input and output decide L 2 NSPACE(f(n)). ² Use the reachability method on the con¯guration graph of M on input x of length n. ² A con¯guration is a (2k + 1)-tuple (q; w1; u1; w2; u2; : : : ; wk; uk): °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 195 Proof of Theorem 22(3) (continued) ² We only care about (q; i; w2; u2; : : : ; wk¡1; uk¡1); where i is an integer between 0 and n for the position of the ¯rst cursor. ² The number of con¯gurations is therefore at most (2k¡4)f(n) log n+f(n) jKj £ (n + 1) £ j§j = O(c1 ) (1) for some c1, which depends on M. ² Add edges to the con¯guration graph based on M's transition function. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 196 Proof of Theorem 22(3) (concluded) ² x 2 L , there is a path in the con¯guration graph from the initial con¯guration to a con¯guration of the form (\yes"; i; : : :) [there may be many of them]. log n+f(n) ² This is reachability on a graph with O(c1 ) nodes. ² It is in TIME(clog n+f(n)) for some c because reachability 2 TIME(nj) for some j and h ij log n+f(n) j log n+f(n) c1 = (c1) : °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 197 Space-Bounded Computation and Proper Functions ² In the de¯nition of space-bounded computations earlier, the TMs are not required to halt at all. ² When the space is bounded by a proper function f, computations can be assumed to halt: { Run the TM associated with f to produce an output of length f(n) ¯rst. { The space-bounded computation must repeat a con¯guration if it runs for more than clog n+f(n) steps for some c (p. 196). { So we can prevent in¯nite loops during simulation by pruning any path longer than clog n+f(n). °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 198 The Grand Chain of Inclusions L ⊆ NL ⊆ P ⊆ NP ⊆ PSPACE ⊆ EXP: ² By Corollary 19 (p. 189), we know L ( PSPACE. ² The chain must break somewhere between L and PSPACE.a ² It is suspected that all four inclusions are proper. ² But there are no proofs yet.b aBill Gates (1996), \I keep bumping into that silly quotation at- tributed to me that says 640K of memory is enough." bCarl Friedrich Gauss (1777{1855), \I could easily lay down a mul- titude of such propositions, which one could neither prove nor dispose of." °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 199 Nondeterministic Space and Deterministic Space ² By Theorem 4 (p. 84), NTIME(f(n)) ⊆ TIME(cf(n)); an exponential gap. ² There is no proof yet that the exponential gap is inherent. ² How about NSPACE vs. SPACE? ² Surprisingly, the relation is only quadratic|a polynomial|by Savitch's theorem. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 200 Savitch's Theorem Theorem 23 (Savitch (1970)) reachability 2 SPACE(log2 n): ² Let G(V; E) be a graph with n nodes. ² For i ¸ 0, let PATH(x; y; i) mean there is a path from node x to node y of length at most 2i. ² There is a path from x to y if and only if PATH(x; y; dlog ne) holds. °c 2011 Prof. Yuh-Dauh Lyuu, National Taiwan University Page 201 The Proof (continued) ² For i > 0, PATH(x; y; i) if and only if there exists a z such that PATH(x; z; i ¡ 1) and PATH(z; y; i ¡ 1).
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