Kyle Jones
• Member for 9 years, 10 months
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• Seattle, WA

In general, if NP turns out to be the subset of any tractable class then that will be a black swan event for those who depend on cryptography as it exists today. $NP \subseteq BQP$ is one such ...

As far as we know right now, no. Quantum computation makes some problems tractable that are currently intractable for classical computers, but problems that are theoretically impossible for classical ...

Plainly, evaluating $10^{19}$ positions is not going to be feasible in 7200 seconds. But if you use alpha-beta to prune the search tree and have perfect move ordering you can get that down to around $... View answer Accepted answer 1 votes Given a SAT instance A, another SAT instance B can be constructed such that if it is found satisfiable the satisfying assignment proves the unsatisfiability of A. But the proof is one-sided; if B is ... View answer 1 votes So far as we know, NP and co-NP are separate complexity classes under Karp reduction, aka many-one polynomial-time reduction. Your procedure is a Turing reduction, not a Karp reduction. Turing ... View answer 1 votes The practical limit to the number of hosts that can use a single NAT'd address is much less than 61,440. This is because even a modest Internet-capable computer will use many simultaneous TCP ... View answer 1 votes For a netmasked IP address xxx.xxx.xxx.xxx/N, convert each xxx octet in the address to binary. The masked address is the leftmost N bits with all other bits replaced with zeroes. So 10.0.0.2/2 ... View answer 1 votes Filesystems typically don't handle internal edits to files as you describe them. When you edit a program file, a text editor will pull the whole file into memory, allow you to make modifications to a ... View answer 1 votes Variable$a_i$set means class$a$is in timeslot$i$, with a similar encoding for class$b$. Higher values of$i$means later time slots. If class$a$must occur before class b then you need ... View answer 1 votes Your problem can't be any easier than DNF-TAUTOLOGIES since that problem is produced by setting$k$equal to$2^n$. Therefore your problem is NP-hard, just as DNF-TAUTOLOGIES is. View answer Accepted answer 1 votes There's no real reason not to do your checks given that they are computationally tractable, but there's very little point in doing them either. Your checks make determining the satisfiability of easy ... View answer Accepted answer 1 votes Your problem is NP-complete by the modern presentation of Schaefer's dichotomy theorem. You can also prove its NP-completeness by direct reduction of SAT to your problem. If a CNF formula has$n$... View answer 1 votes The basic algorithm is to first put all the members of S into a hash table$\mathcal{S}$. Then walk down the list T, recording its elements in a hash table$\mathcal{T}$as you go. If an element has ... View answer 1 votes We don't know if any families of SAT formulas are hard since we don't yet know if P and NP are distinct complexity classes. Schaefer's dichotomy theorem gives six families of formulas that are known ... View answer 1 votes Your code ignores quantification levels. Formulas prefixed with$\exists{x_1}\forall{x_2}$are not supposed to be evaluated the same way as those prefixed with$\forall{x_1}\exists{x_2}$. Your code ... View answer Accepted answer 1 votes Your reasoning is correct. Your "sum-of-products" is more commonly known as disjunctive normal form (DNF). It is easy to show that conversion from conjunctive normal form (CNF) to DNF is NP-hard, so ... View answer 1 votes Cached data about a domain for which a DNS server is not authoritative could only have come from prior recursive queries. If you send a query without the RD (recursion desired) bit set, the server is ... View answer Accepted answer 1 votes Since each place in a binary number is a power of two, the binary expansion of$2^k$for any$k$will be a 1 followed by$k$zeroes. Examples: $$1 = 2^0 = 1$$$$16 = 2^4 = 1\underbrace{0000}_{4\ ... View answer 1 votes The intersection of two non-NP-hard languages can be NP-hard. Example: The solutions of any 3SAT instance are the set intersection of the solutions of a HORN-3SAT instance and an ANTIHORN-3SAT ... View answer Accepted answer 1 votes I think the key is that NSSets aren't required to "maintain order," meaning that you cannot access elements with an index, nor can you extract elements in any particular order. Because NSSets don't ... View answer 1 votes The term I've seen in use is superlinear, as in "Are there super-linear time complexity lower bounds for any natural problem in NP?" View answer 1 votes The problem as described still seems amenable to doing a bipartite matching with weighted edges if you adapt the graph setup as follows. If an employee must do four jobs, represent the employee with ... View answer Accepted answer 1 votes If you could add clauses to a CNF formula to cause it to be unsatisfiable iff certain sets of variables don't have non-unique assignments among the formula's satisfying assignments, you could use this ... View answer 1 votes Multiple memory locations hash to the same location in the memory cache. The tag array serves to disambiguate which memory location's contents are stored in each cache location. If the address in ... View answer Accepted answer 1 votes Yes, each input value is weighted before being fed into the nodes in the next layer. Each input/output pair of "neurons" has its own weight and it is these weights that are adjusted by the neural ... View answer 1 votes This$\qquad \forall x \exists z \space (x = z)$is a true QBF. Whatever Boolean value$x$is set to,$z$can be set to match. Translated to quantified CNF:$\qquad \forall x \exists z \space (x \...

If, as I suspect, you are still trying to convert quantified 3-CNF formulas into quantified mixed Horn formulas, the answer is that quantifiers for the newly added bridge variables can be placed ...

Use a SAT solver that also allows you to express pseudo-Boolean constraints. Encoding the verification of the existence of the tiling of an NxN grid as a CNF formula is straightforward. Each grid ...