Are there complexity classes with provably no complete problems? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:11:01Z http://mathoverflow.net/feeds/question/27572 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems Are there complexity classes with provably no complete problems? Akhil Mathew 2010-06-09T10:53:11Z 2010-09-26T02:56:29Z <p>A problem is said to be complete for a complexity class $\mathcal{C}$ if a) it is in $\mathcal{C}$ and b) every problem in $\mathcal{C}$ is log-space reducible to it. There are natural examples of NP-complete problems (SAT), P-complete problems (circuit-value), NL-complete problems (reachability), and so on.</p> <p>Papadimitrou states that "semantic" complexity classes like (I think) BPP are less likely to admit complete problems. Then again, we do not actually know whether P = BPP or not, and if so, there would be BPP-complete problems. (There exist IP-complete problems, since IP=PSPACE, and determining whether a quantified boolean formula is satisfiable is PSPACE-complete.)</p> <blockquote> <p><strong>Question:</strong> Are there any natural complexity classes that can be shown <em>not</em> to have complete problems?</p> </blockquote> <p>I think this question has to be modified slightly, because I would imagine $Time(n)$ has no complete problems because log-space reductions can introduce a polynomial factor. So, in the word "natural," I include the assumption that the complexity class should be invariant under polynomial transformations (I don't know how to make this precise, but hopefully it is clear).</p> <p>(Also, time or space bounds should be at least $\log n$, of course.)</p> <p><strong>Edit:</strong> A commenter has pointed me to an interesting result of Sipser that $\mathrm{BPP}^M$ for $M$ a suitable oracle does not have complete problems. Is the same true for a (less fancy) class of the form $\bigcup_f TIME(f)$, where the union is over a class of recursive functions $f$ that are all polynomially related to each other? (Same for $\bigcup_f NTIME(f)$, and ditto for space.</p> http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems/27575#27575 Answer by Joel David Hamkins for Are there complexity classes with provably no complete problems? Joel David Hamkins 2010-06-09T12:27:44Z 2010-06-09T12:27:44Z <p>The <a href="http://qwiki.stanford.edu/wiki/Complexity_Zoo" rel="nofollow">zoo of complexity classes</a> extends naturally into the realm of <a href="http://en.wikipedia.org/wiki/Computability_theory" rel="nofollow">computability theory</a> and beyond, to <a href="http://en.wikipedia.org/wiki/Descriptive_set_theory" rel="nofollow">descriptive set theory</a>, and in these higher realms there are numerous natural classes which have no members that are complete, even with respect to far more generous notions of reduction than the one you mention.</p> <ul> <li><p>For example, consider the class Dec of all decidable sets of natural numbers. There can be no decidable set $U$ such that every other decidable set $A$ reduces to it in time uniformly bounded by any computable function $f$ (even iterated exponentials, or the Ackerman function, etc.). In particular, it has no member that is complete in your sense. If there were such a member, then we would be able to consruct a computable enumeration $A_0$, $A_1$, $\ldots$ containing all decidable sets, which is impossible since then we could diagonalize against it: the set of $n$ such that $n\notin A_n$ would be computable, but it can't be on the list.</p></li> <li><p>The class of all arithmetically definable sets is obtained by closing the decidable sets (or much less) under projection from $\mathbb{N}^{n+1}\to \mathbb{N}^n$ and under Boolean combinations. The members of this class are exactly the sets that are defined by a first order formula over the structure $\langle\mathbb{N},+,\cdot,\lt\rangle$, and the hierarchy is stratified by the complexity of these definitions. This class has no member that is complete with respect to any computable reduction, and even with respect to any arithmetically definable reduction of bounded complexity, for in this case the hierarchy would collapse to some level $\Sigma_n$, which is known not to occur.</p></li> <li><p>A similar argument works for the hyperarithmetic hiearchy, which can have no universal member hyperarithmetic reductions of any fixed complexity. </p></li> <li><p>And similarly for the projective hiearchy on sets of reals. </p></li> </ul> <p>The general phenomenon is that there are numerous hierarchies growing from computability theory into descriptive set theory which are all known to exhibit strictly proper growth in such a way that prevents them from having universal members.</p> http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems/27593#27593 Answer by Rune for Are there complexity classes with provably no complete problems? Rune 2010-06-09T15:51:41Z 2010-06-09T15:51:41Z <p>Here's a really simple class that is very natural and has no complete problems: ALL, the class of all languages. The reason is that there are uncountably many problems in ALL, but only countably many Turing machines to go around (for reductions), so every problem in ALL cannot be reduced to a single problem in ALL. </p> <p>Similarly, any class with advice, like P/poly, L/poly, BQP/qpoly, or even P/1 does not have complete problems (using the same argument).</p> http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems/27612#27612 Answer by Antonio E. Porreca for Are there complexity classes with provably no complete problems? Antonio E. Porreca 2010-06-09T19:39:12Z 2010-06-09T19:39:12Z <p>Another class without complete problems w.r.t. logspace or polytime reductions (not an union of classes <b>TIME</b>(<i>f</i>) for some family of polynomially related functions <i>f</i>, but still relatively natural in my opinion) is</p> <blockquote> <p><b>ELEMENTARY</b> = <b>TIME</b>(2<sup><i>n</i></sup>) &cup; <b>TIME</b>(2<sup>2<sup><i>n</i></sup></sup>) &cup; <b>TIME</b>(2<sup>2<sup>2<sup><i>n</i></sup></sup></sup>) &cup; <b>TIME</b>(2<sup>2<sup>2<sup>2<sup><i>n</i></sup></sup></sup></sup>) &cup; ⋯</p> </blockquote> <p>If <i>L</i> were <b>ELEMENTARY</b>-complete, then it would belong to some level of this hierarchy, and all problems above could be reduced to it. But this hierarchy is known to be proper (time hierarchy theorem), contradiction.</p> http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems/28032#28032 Answer by Akhil Mathew for Are there complexity classes with provably no complete problems? Akhil Mathew 2010-06-13T15:16:45Z 2010-06-13T18:52:36Z <p>I'm not sure that it's entirely correct, but here goes. Let $f(n)$ be a proper complexity function (a.k.a. space and time constructible, etc.) and consider the class $\mathcal{C} = \bigcup_{P,Q} DTIME Q(f(P(n)))$, where $P,Q$ rangs over polynomials with natural number coefficients. Suppose $f(n) \geq n$. Then the language $L=(M,x,P,Q)$: $M$ is a deterministic Turing machine that accepts string $x$ in at most $Q(f(P(n))$ steps should be $\mathcal{C}$-complete. Indeed, verification is just a mechanical process of simulating the Turing machine (which can be done in polynomial time on the length of $M$ and $x$), and every language decided by a Turing machine $M$ in $DTIME(Q(f(P(n)))$ should reduce to $L$ based on the machine $M$. The same should hold for nondeterministic complexity classes.</p> <p>I've seen something like this for $NP$ (this is how the Cook-Levin theorem is proved, if I understand correctly), and I think it should generalize, and that natural complexity classes based solely on a time constraint (which is sufficiently large) should admit complete problems.</p> http://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems/28086#28086 Answer by Rune for Are there complexity classes with provably no complete problems? Rune 2010-06-14T03:25:37Z 2010-06-14T03:25:37Z <p>This is more of a comment than an answer, but the comment go too long. From this thread, there seem to be two different themes to coming up with classes without complete problems. </p> <p>Completeness is defined using two properties. L is X-complete if<br> (1) L is in X<br> (2) L is X-hard (under some suitable notion of efficient reductions)</p> <p>The first theme involves classes which have hard problems, but if the hard problem were a member of the class itself, it would cause problems. The examples of POLYLOGSPACE and ELEMENTARY fall in this category. Both have hard problem, of course, but if the hard problem were a member of the class, some hierarchy theorem would be violated. (Space hierarchy and time hierarchy theorems, respectively.) Similarly one could come up with more examples of this kind.</p> <p>The second theme involves classes which have no hard problems, such as ALL or P/poly. These classes don't have a complete problem for a fundamentally different reason than the previous case.</p> <p>It would be interesting to see if there are other classes which fail to have complete problems for completely different reasons.</p>