A linear-time algorithm for the orbit problem over cyclic groups

Abstract

The orbit problem is at the heart of symmetry reduction methods for model checking concurrent systems. It asks whether two given configurations in a concurrent system (represented as finite strings over some finite alphabet) are in the same orbit with respect to a given finite permutation group (represented by their generators) acting on this set of configurations by permuting indices. It is known that the problem is in general as hard as the graph isomorphism problem, whose precise complexity (whether it is solvable in polynomial-time) is a long-standing open problem. In this paper, we consider the restriction of the orbit problem when the permutation group is cyclic (i.e. generated by a single permutation), an important restriction of the problem. It is known that this subproblem is solvable in polynomial-time. Our main result is a linear-time algorithm for this subproblem.

Appendix: Completing proof of Theorem 2

Let $\begin{bmatrix}n \\ k\end{bmatrix}$ denote the unsigned Stirling number of the first kind, and $\left( {\begin{array}{c}n\\ k\end{array}}\right)$ denote n choose k. The harmonic number $H_n$ is defined as

$$\begin{aligned} H_n = \sum _{k=1}^n \frac{1}{k}. \end{aligned}$$

In general, for an integer $s \ge 1$, the generalized harmonic number of order s is defined as

$$\begin{aligned} H_n^{(s)} = \sum _{k=1}^n \frac{1}{k^s}. \end{aligned}$$

It is known that

$$\begin{aligned} \frac{1}{n!} \sum _{k=1}^n k \begin{bmatrix}n \\ k\end{bmatrix} = H_n. \end{aligned}$$

Define

$$\begin{aligned} f(n) = \frac{1}{n!} \sum _{k=1}^n k^2 \begin{bmatrix}n \\ k\end{bmatrix}, \quad g(n) = \frac{1}{n!} \sum _{k=1}^n k^3 \begin{bmatrix}n \\ k\end{bmatrix}. \end{aligned}$$

It is known that

$$\begin{aligned} \sum _{k=m}^{n} \begin{bmatrix}n \\ k\end{bmatrix} \left( {\begin{array}{c}k\\ m\end{array}}\right) = \begin{bmatrix}n+1 \\ m+1\end{bmatrix}, \end{aligned}$$

(see [4]). In particular, we have

$$\begin{aligned} \begin{bmatrix}n+1 \\ 3\end{bmatrix}= & {} \sum _{k=2}^{n} \begin{bmatrix}n \\ k\end{bmatrix} \left( {\begin{array}{c}k\\ 2\end{array}}\right) = \sum _{k=1}^{n} \begin{bmatrix}n \\ k\end{bmatrix} \frac{k(k-1)}{2} = \frac{1}{2} n! (f(n) - H_n), \\ \begin{bmatrix}n+1 \\ 4\end{bmatrix}= & {} \sum _{k=3}^{n} \begin{bmatrix}n \\ k\end{bmatrix} \left( {\begin{array}{c}k\\ 3\end{array}}\right) = \sum _{k=1}^{n} \begin{bmatrix}n \\ k\end{bmatrix} \frac{k(k-1)(k-2)}{6} = \frac{1}{6} n! (g(n) - 3f(n) + 2H_n). \end{aligned}$$

That is, we have

$$\begin{aligned}&\displaystyle f(n) = \frac{2}{n!} \begin{bmatrix}n+1 \\ 3\end{bmatrix} + H_n, \text { and}\\&\displaystyle g(n) = \frac{6}{n!} \begin{bmatrix}n+1 \\ 4\end{bmatrix} + 3f(n) - 2H_n = \frac{6}{n!} \begin{bmatrix}n+1 \\ 4\end{bmatrix} + \frac{6}{n!} \begin{bmatrix}n+1 \\ 3\end{bmatrix} + H_n. \end{aligned}$$

It is known (cf. page 217 of [11]) that

$$\begin{aligned}&\displaystyle \frac{1}{n!} \begin{bmatrix}n+1 \\ 3\end{bmatrix} = \frac{1}{2} (H_n^2 - H_n^{(2)})\\&\displaystyle \frac{1}{n!} \begin{bmatrix}n+1 \\ 4\end{bmatrix} = \frac{1}{6} (H_n^3 - 3 H_n H_n^{(2)} + 2 H_n^{(3)}). \end{aligned}$$

It is also known that $H_n = \gamma + \ln n$, $\lim _{n \rightarrow \infty } H_n^{(2)} = \zeta (2) = \frac{\pi ^2}{6}$ and $\lim _{n \rightarrow \infty } H_n^{(3)} = \zeta (3) \approx 1.202$, where $\gamma \approx 0.577$ is Euler’s constant and $\zeta (s) = \sum _{k=1}^{\infty } \frac{1}{k^s}$ is the Riemann zeta function. Hence we obtain

$$\begin{aligned} f(n) = H_n^2 - H_n^{(n)} + H_n \sim \ln ^2 n. \end{aligned}$$

Putting all together, we obtain

$$\begin{aligned} g(n) = \frac{6}{n!} \begin{bmatrix}n+1 \\ 4\end{bmatrix} + 3f(n) - 2H_n \sim \ln ^3 n. \end{aligned}$$