Results 81 to 90 of about 35,538 (168)

Candy Crush is NP-hard

open access: yes, 2014
We prove that playing Candy Crush to achieve a given score in a fixed number of swaps is NP-hard.
openaire   +2 more sources

Polynomial algorithms that prove an NP-hard hypothesis implies an NP-hard conclusion

open access: yes, 1999
A number of results in Hamiltonian graph theory are of the form $\mathcal{P}$$_{1}$ implies $\mathcal{P}$$_{2}$, where $\mathcal{P}$$_{1}$ is a property of graphs that is NP-hard and $\mathcal{P}$$_{2}$ is a cycle structure property of graphs that is also NP-hard.
Bauer, D.   +3 more
openaire   +1 more source

On Percolation and NP-Hardness

open access: yes, 2016
The edge-percolation and vertex-percolation random graph models start with an arbitrary graph G, and randomly delete edges or vertices of G with some fixed probability. We study the computational hardness of problems whose inputs are obtained by applying percolation to worst-case instances.
Bennett, Huck   +2 more
openaire   +2 more sources

Misere Hackenbush is NP-Hard

open access: yes, 2012
Hackenbush is a two player game, played on a graph with coloured edges where players take it in turns to remove edges of their own colour. It has been shown that under normal play rules Red-Blue Hackenbush (all edges are coloured either red or blue) is NP-hard. We will show that Red-Blue Hackenbush is in P, but that Red-Blue-Green Hackenbush is NP-Hard,
openaire   +2 more sources

Analogy is NP-Hard

open access: yesAnalogy is NP-Hard
Analogy is described in predicate logic. This paper deals with the analogy without any function symbols except constants. We show that the problem of deciding whether a given atomic formula can be inferred by analogy is NP-hard even in such a simple case. Proc.
openaire  

Communication Complexity is NP-hard

open access: yes
In the paper where he first defined Communication Complexity, Yao asks: \emph{Is computing $CC(f)$ (the 2-way communication complexity of a given function $f$) NP-complete?} The problem of deciding whether $CC(f) \le k$, when given the communication matrix for $f$ and a number $k$, is easily seen to be in NP.
Hirahara, Shuichi   +2 more
openaire   +2 more sources

Home - About - Disclaimer - Privacy