Let $G$ be a connected graph with $N$ vertices. Let $k$ be the number of vertices in a longest path of $G$ such that every vertex on the path is a cut vertex of $G$, and every intermediate vertex of the path is a degree-two vertex of $G$. We conventionally set $k = 1$ when $G$ is $2$-edge-connected. Let $P=\{1,\ldots,n\}$ be a set of pebbles with $k < N-n$. A \textit{configuration} of $P$ on $G$ is defined as a function $f$ from $V(G)$ to $\{0, 1, \ldots, n \}$ with $|f^{-1}(i)| = 1$ for $1 \le i \le n$, where $f^{-1}(i)$ is a vertex occupied with the $i$th pebble for $1 \le i \le n$ and $f^{-1}(0)$ is a set of unoccupied vertices. A \textit{move} is defined as shifting a pebble from a vertex to some unoccupied neighbor. The {\it pebble motion problem on the pair $(G,P)$} is to decide whether a given configuration of pebbles is reachable from another by executing a sequence of moves. Let $\D(G)$ denote the diameter of the graph $G$, and let $\CL(G)$ denote the maximum length of a shortest cycle containing a vertex $v$, taken over all vertices $v$ in all $2$-connected components of $G$. For completeness, we define $\CL(G) := 1$ when $G$ is a tree. In this paper, we show that the length of the shortest solution sequences for the pebble motion problem on a pair $(G, P)$ is in $\Ord\left(n\D(G) + \min\left\{k n \D(G),\ n^{2} \log\big(1+\min\{n, k\}\big)\right\}\right)$ if $G$ is an $N$-vertex tree, and in $\Ord\left(n\D(G)+\frac{n^2\min\{n,\CL(G)\}}{N-n}+n^2\log(1+\min\{n, N-n\})\right)$ if $G$ is a connected general $N$-vertex graph. Furthermore, in the case where $G$ is a connected general $N$-vertex graph and the number of unoccupied spaces $N - n$ is bounded by some constant, this length admits an upper bound of $\Ord(n \CL(G) \D(G))$. Keywords: pebble motion, motion planning, multi-agent path finding, $15$-puzzle, tree

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