I am looking for solution of the following problem.
Let $G$ be a weighted graph with (positive) weights. The length of a path in a weighted graph is the sum of the weights of the selected edges. The distance between two nodes is a minimal length of path between these nodes. If $N$ is a node in $G$ and $R > O$, we can define a circle with center at $N$ and radius $R$, or $N,R$-circle, as a set of all nodes in $G$, whose distance to node $N$ is $\leq R$; we say that all these nodes are covered by this $N,R$-circle.
Question: what is a minimum number of circles of radius $R$ that cover all nodes of the graph $G$.
In my case $G$ is a tree with a finite number of nodes and edges, may be it simplifies the problem?
Thank you.
I have two news for you: a good one and a bad one. The good one is that your problem is equivalent to a well-known problem, so there is no need to “re-invent a wheel”. To see that, consider new graph $G’$ with the same set of vertices as $G$, but vertices $u$ and $v$ of $G’$ are adjoined by an edge iff the distance between $u$ and $v$ in $G$ is at most $R$. Then your problem is equivalent to find a smallest dominating set of vertices in the graph $G’$. But the bad news is that this problem is NP-hard, so it should admit (known) efficient algorithms only in special cases.