Now we have solved an NP-hard problem in polynomial time, so we have a contradiction if we assume $P \neq NP$. Some manufacturers name commercially the measurement unit of their own interpolation. If $M$ terminates, test if the output of $M$ is an unit disk configuration of $G$. If $M$ does not terminate in $p(n)$ steps, output NO. If the disks do not overlap, it is also a unit coin graph or penny graph. Now, we can solve unit disk recognition problem in polynomial time as follows: Given a graph $G$ with $n$ vertices, run $M$ with $G$ as input for $p(n)$ steps. top A unit disk graph is the intersection graph of a family of unit disks in the plane. Is there a Turing machine $M$ and a polynomial $p(n)$, such that if the input of $M$ is a unit disk graph $G$ with $n$ vertices encoded as an adjacency matrix, then $M$ terminates in at most $p(n)$ steps, and outputs an unit disk configuration of $G$? In particular, the machine $M$ is allowed to have undefined behavior if the input is not an unit disk graph encoded as adjacency matrix.Īnswer: Suppose that there is such Turing machine $M$ and polynomial $p(n)$. Maybe there could be some other encodings in which only unit disk graphs could be represented, but formulating them would be another topic.)Įdit: I'll try to formalize the question and the answer more: (Here we assume that the input is some well-known encoding of a graph, and therefore the restriction that the input must be an unit disk graph doesn't really make the problem at all easier. We connect two points if their disks’ interiors 124F. At each point, we centre a disk of diameter d. We are given an arbitrary set of n points xed in the plane and a xed positive quantity d. Therefore the problem you pose is NP-hard in the sense that if it admits polynomial time algorithm, then P=NP. Recall the denition of a unit disk graph. If there would be a polynomial time algorithm for your problem, it could be used to solve the NP-hard recognition problem in polynomial time by just giving the input to it and checking if its output is correct.
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