For a given directed graph $G = (V, E)$ (possibly with loops), and some $S\subseteq V$ define the operation $G(S) = \{ v\mid (u,v)\in E\text{ for some } u\in S \}$.

Now consider the infinite sequence $\rho(G, S) = S, G(S), G(G(S)), G(G(G(S))), ...$ and call $|\{\rho(G, S)\}|$ the number of unique elements in the sequence.

I'm interested in the asymptotical behavior of the function $f(n)$ defined as the maximum size of $|\{\rho(G, S)\}|$ among all graphs $G$ with $n$ nodes.

There is an obvious upper bound of $2^n$ and there is a lower bound of $\Omega( e^{\sqrt{n \log n}})$ by using a $G$ that is made of disjoint cycles of different sizes that are coprime to one another. This bound is given by Landau's function. My question is:

Is it true that $f(n)\in \Theta(g(n))$, where $g(n)$ is Landau's function?

Obs. 1: This is closely related to the problem of determinizing an NFA over a unary alphabet. The sequence A157656 on the OEIS describes the exact value for each $n$ when considering a single initial state.

More relevantly, there is a classic result by Marek Chrobak (link) that says for any $n$-state unary NFA, there exists an equivalent unary DFA with $O(g(n))$ states. There should be a straightforward way of using this to solve the question in the positive that I'm not seeing.


1 Answer 1


The argument in Chrobak’s paper can be applied to this problem as well, with the same bounds.

Let $\{D_i:i<k\}$ be the set of strongly connected components of $G$ that contain a cycle (i.e., other than a lone loopless vertex), let $y_i$ be the gcd of the lengths of all cycles in $D_i$, and let $y=\operatorname{lcm}(\{y_i:i<k\})$. Observe that any closed walk in $G$ is included in some $D_i$, and has length divisible by $y_i$ (as it can be decomposed into cycles).

Since $\sum_{i<k}y_i\le\sum_{i<k}|D_i|\le n$, we have $y\le g(n)$. The proof of Chrobak’s Lemma 4.3 actually shows the following:

Lemma. For all $s,t\in V$ and $x\ge n$, $x'\ge2n^2$ such that $x\equiv x'\pmod y$, if there exists a walk from $s$ to $t$ of length $x$, then there exists a walk from $s$ to $t$ of length $x'$.

(More precisely: for every walk of length $x\ge n$, there is $i<k$ such that for every $x'\ge2n^2$, $x'\equiv x\pmod{y_i}$, there exists a walk of length $x'$.)

Proof: A walk $w$ of length $x\ge n$ contains a cycle, hence it intersects some $D_i$. Let $D_{i_0},\dots,D_{i_r}$ be the list of all $D_i$’s that $w$ intersects, and let $y'=\gcd(\{y_{i_j}:j\le r\})\mid y$. By successively removing cycles from $w$, we obtain a walk $w_0$ from $s$ to $t$ that still intersects each $D_{i_j}$, and that is simple, hence of length $x_0<n$, $x_0\equiv x\pmod{y'}$. If $x'\ge x_0+(n-1)^2$ and $x'\equiv x\pmod{y'}$, then standard bounds on the Frobenius problem show that $x'=x_0+\sum_ia_i|c_i|$, where $a_i\ge0$, and each $c_i$ is a cycle in some $D_{i_j}$. Thus, we may construct a walk from $s$ to $t$ of length $x'$ by attaching to $w_0$ an appropriate number of copies of each $c_i$. The only issue is that $c_i$ does not necessarily share a vertex with $w_0$ [Chrobak seems to ignore this point]. To fix this, we first attach to $w_0$ for each $j\le r$ a closed walk that connects all points of $D_{i_j}$; such a walk exists of length at most $|D_{i_j}|^2$ (and necessarily divisible by $y_{i_j}$, thus by $y'$), hence the total added length is at most $\sum_j|D_{i_j}|^2\le n^2$. It follows that we can construct $w'$ of length $x'$ whenever $x'\equiv x\pmod{y'}$ satisfies $x'\ge n^2+(n-1)^2+x_0$, which certainly holds if $x\ge2n^2$.

Corollary. For any $S\subseteq V$, $G^{2n^2}(S)=G^{y+2n^2}(S)$, thus $|\{\rho(G,S)\}|\le y+2n^2\le g(n)+2n^2$. That is, $g(n)\le f(n)\le g(n)+2n^2$.

Proof: $t\in G^x(S)$ iff there is $s\in S$ and a walk of length $x$ from $s$ to $t$; thus, the Lemma implies that $G^x(S)=G^{x'}(S)$ whenever $x,x'\ge2n^2$ and $x\equiv x'\pmod y$.

NB: The length of a closed walk that connects all vertices of $D_{i_j}$, $|D_{i_j}|=n_{i_j}$, can be improved to $\lfloor(n_{i_j}+1)^2/4\rfloor$ (this is tight). This reduces the overall bound to $f(n)\le g(n)+\frac54n^2$.

On second thought, we don’t need a closed walk connecting all cycles in $D_{i_j}$, but only a set of cycles such that the gcd of their lengths is $y_{i_j}$. This reduces the length of the walk to $\sim n_{i_j}\frac{\log n_{i_j}}{\log\log n_{i_j}}$ (this bound is tight), as $\sim\frac{\log n_{i_j}}{\log\log n_{i_j}}$ cycles suffice (we take an arbitrary cycle $c$, and then for each prime $p$ dividing $|c|/y_{i_j}$, a cycle whose length is not divisible by $py_{i_j}$).

Thus, $f(n)\le g(n)+n^2+(1+o(1))n\frac{\log n}{\log\log n}$.

  • $\begingroup$ Thank you for writing this down! Do you have a reference for the $\lfloor(n_{i_j}+1)^2/4\rfloor$ bound? $\endgroup$
    – alsips-cl
    Dec 25, 2023 at 17:10
  • $\begingroup$ I don’t have a reference, but it’s a simple argument. To simplify the notation, let $G$ be a strongly connected digraph of size $n$; I want to show that there is a closed walk connecting all vertices of $G$ of length at most $\lfloor(n+1)^2/4\rfloor$. Let $p=(x_0,\dots,x_l)$ be a simple path in $G$ of maximal length, and let $x_{l+1},\dots,x_{n-1}$ be the remaining vertices of $G$. We augment $p$ with simple paths from $x_l$ to $x_{l+1}$, from $x_{l+1}$ to $x_{l+2}$, etc., and from $x_{n-1}$ to $x_0$; these are $n-l$ paths, each with at most $l$ edges (by the maximality of $p$), thus ... $\endgroup$ Dec 25, 2023 at 18:54
  • $\begingroup$ ... in total, the closed walk has length at most $(n-l+1)l$. This expression is maximized for $l=(n+1)/2$, in which case we get $(n+1)^2/4$ (or rather, $\lfloor(n+1)^2/4\rfloor$ as it must be an integer). $\endgroup$ Dec 25, 2023 at 18:58
  • $\begingroup$ To see that the bound is tight, consider a graph consisting of a path with $\lfloor n/2\rfloor$ vertices, whose endpoint connects to $\lceil n/2\rceil$ other vertices (call these “special”), each of which connects back to the start of the path. Any path connecting one special vertex to another has at least $\lfloor n/2\rfloor+1$ edges, hence a closed walk that (wlog) starts in one special vertex and traverses all the other has length at least $\lceil n/2\rceil(\lfloor n/2\rfloor +1)=\lfloor(n+1)^2/4\rfloor$. $\endgroup$ Dec 25, 2023 at 19:03

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