到这为止NP-hardness和NP-completeness就很好理解了。称问题L是NP-hard，如果任意一个NP的问题都可以多项...
北航等机构的研究者提出使用大语言模型(LLM) 来增强和加速对 P versus NP 问题的研究。
The P versus NP problem is one of the most important and unsolved problems in computer science. This consists in knowing the answer of the following question: Is P equal to NP? This incognita was first mentioned in a letter written by Kurt Gdel to John von Neumann in 1956. However, the...
The most important open question in complexity theory, the P versus NP ("P=NP") problem, asks whether polynomial time algorithms actually exist for solving NP-complete, and by corollary, all NP problems. It is widely believed that this is not the case. The complexity class NP (which have ...
The P versus NP problem is the determination of whether all NP-problems are actually P-problems. If P and NP are not equivalent, then the solution of NP-problems requires (in the worst case) an exhaustive search, while if they are, then asymptotically fa
https://www.zhihu.com/question/411543712/answer/1378895812 Gerhard J Woeginger 在去 RWTH Aachen 之前, 曾经维护了一个关于 P 和 NP 关系的各种伪证的页面 (一直更新到了 2016 年): P-versus-NP page www.win.tue.nl 通...
Among them is the central problem in theoretical computer science: the P versus NP problem, which aims to classify the possible existence of efficient solutions to combinatorial and optimization problems. The main goal is to determine whether there are questions whose answer can be quickly checked,...
Knowing whether he’s found the answer is easy. If he’s visited each house, then he’s done the job; if he’s skipped one, then he has to start over. That’s the “P” part. All P versus NP problems are tough to solve but easy to verify. ...
TheP versus NP questionconcerns the speed at which a computer can accomplish a task such as factorising a number. Some tasks can be completed reasonably quickly – in technical terms, the running time is proportional to a polynomial function of the input size – and these tasks are in class ...
NP, is of course Π2: for all Turing machines M and polynomial time bounds p, there exists a SAT instance ϕ such that M doesn't halt with the correct answer to ϕ in at most p (n) steps, where n is the size of ϕ. The NP vs. P/poly problem is similarly Π2. What ...