The simultaneous conjugacy problem in the symmetric group

TL;DR

This paper presents an efficient algorithm for solving the transitive simultaneous conjugacy problem in the symmetric group, improving upon previous methods and demonstrating promising experimental performance.

Contribution

The paper introduces a new algorithm that solves the problem in near-optimal time complexity, surpassing earlier approaches and providing empirical evidence of its efficiency.

Findings

Algorithm runs in near-quadratic time with respect to n

Expected running time is nearly linear in n for fixed d

New method outperforms previous algorithms on random instances

Abstract

The transitive simultaneous conjugacy problem asks whether there exists a permutation $\tau \in S_n$ such that $b_j = \tau^{-1} a_j \tau$ holds for all $j = 1,2, \ldots, d$, where $a_1, a_2, \ldots, a_d$ and $b_1, b_2, \ldots, b_d$ are given sequences of $d$ permutations in $S_n$, each of which generates a transitive subgroup of $S_n$. As from mid 70' it has been known that the problem can be solved in $O(dn^2)$ time. An algorithm with running time $O(dn \log(dn))$, proposed in late 80', does not work correctly on all input data. In this paper we solve the transitive simultaneous conjugacy problem in $O(n^2 \log d / \log n + dn\log n)$ time and $O(n^{3/ 2} + dn)$ space. Experimental evaluation on random instances shows that the expected running time of our algorithm is considerably better, perhaps even nearly linear in $n$ at given $d$.

Tables

Table 1. Table 1: Experimental results in miliseconds on random instances for the general transitive case.

		isomorphic pairs			non-isomorphic pairs
$n$		subquadratic	quadratic		subquadratic	quadratic
10 000		46	10 578		13	17 865
15 000		69	82		10	39 571
20 000		111	18 398		12	67 958
25 000		147	63 750		14	136 404
30 000		283	14 378		16	196 014
35 000		294	69 366		18	265 281
40 000		378	28 471		24	347 392
45 000		478	161 269		88	390 613
50 000		556	105 583		27	438 985

Table 2. Table 2: Experimental results in miliseconds on random instances for a special case with an n 𝑛 n -cycle.

		isomorphic pairs			non-isomorphic pairs
$n$		linear	subquadratic		linear	subquadratic
100 000		2 079	5 818		2 927	106
150 000		4 674	11 581		3 457	124
200 000		4 157	18 666		4 888	1 877
250 000		7 089	29 622		6 424	191
300 000		8 879	40 810		7 805	255
350 000		8 042	63 686		9 382	634
400 000		10 942	75 584		10 810	355
450 000		13 308	93 833		12 495	634
500 000		12 697	119 640		14 585	429

Equations

$$a_1$$

$$a_2$$

$$I(G_a,v)=I(G_b,w).$$

$$I^{\kappa}(G_{a},v)=\left\{\begin{array}[]{@{}ll@{}}{\tt C},&\text{if}\ \text{the walk $W(\kappa,v)$ in $G_{a}$ is closed}\\ {\tt O},&\text{otherwise}\end{array}\right.$$

$$O_{\kappa }(B)=\{v\in B|I^{\kappa }(G,v)=\mathtt{O}\}\text{ and }{C}_{\kappa }(B)=\{v\in B|I^{\kappa }(G,v)=\mathtt{C}\}.$$

$${\mathcal{P}}_{I^{\kappa }}(G_a)=\{O_{\kappa }(V(G_a)),C_{\kappa }(V(G_a))\},{\mathcal{P}}_{I^{\kappa }}(G_b)=\{O_{\kappa }(V(G_b)),C_{\kappa }(V(G_b))\}.$$

$$V_{i+1}(G_a)=C_{{\kappa }_i}(V_i(G_a))\in {\mathcal{P}}_{{\mathcal{I}}_{i+1}}(G_a)\text{ and }{V}_{i+1}(G_b)=C_{{\kappa }_i}(V_i(G_b))\in {\mathcal{P}}_{{\mathcal{I}}_{i+1}}(G_b).$$

$$\sum _{i=1}^{\lfloor \log n\rfloor +1}O(dn+T_{\text{sc}P}(i))$$

$$\displaystyle=O\bigg{(}dn\log n+\sum_{i=0}^{\lfloor\log n\rfloor+1}T_{\sc P}(i)\bigg{)}.$$

$$\sum _{i=0}^{\lfloor \log n\rfloor +1}{T}_{\text{sc}P}(i)$$

$$\displaystyle=O\bigg{(}\sum_{i=0}^{\lfloor\log n\rfloor+1}n^{2}/\,{2^{i-1}}\bigg{)}=O\bigg{(}n^{2}\sum_{i=0}^{\lfloor\log n\rfloor+1}\frac{1}{2^{i}}\bigg{)}.$$

$$O\bigg{(}dn\log n+n^{2}\sum_{i=0}^{\lfloor\log n\rfloor+1}\frac{1}{2^{i}}\bigg{)}=O\bigg{(}dn\log n+n^{2}\sum_{i=0}^{\infty}\frac{1}{2^{i}}\bigg{)}=O(n^{2}+dn\log n).$$

$${\kappa }_t=k_{t,1}{k}_{t,2}\cdots k_{t,\ell },$$

$$d^{2^{\nu }}\le \frac{m}{2^{\nu }}.$$

$$\frac{1}{4}{\log }_{d}m<2^{\nu }\le \frac{1}{2}{\log }_{d}m.$$

$$|{\kappa }_{\nu }|=\frac{m}{2^{\nu }}<\frac{4m}{{\log }_{d}m}=4\cdot \frac{m\log d}{\log m}=O(m\log d/\log m).$$

$$\sum_{t=1}^{\nu}O(m/2^{t-1}+nd^{\,2^{t}})=O\bigg{(}m+n\sum_{t=1}^{\nu}d^{\,2^{t}}\bigg{)}.$$

$$\sum _{t=1}^{\nu }{d}^{2^t}\le \sum _{t=0}^{\infty }\frac{d^{2^{\nu }}}{2^t}=2d^{2^{\nu }}<2d^{{\log }_d\sqrt{m}}=2\sqrt{m}.$$

$$O\bigg{(}n+n^{3/2}+n^{2}\frac{\log d}{2^{i-1}\log n}\bigg{)}$$

$$O\left(dn\log n+n^{3/2}\log n+n^{2}\frac{\log d}{\log n}\right)=O\bigg{(}dn\log n+n^{2}\frac{\log d}{\log n}\bigg{)}.$$

$$\kappa =(j^{{\alpha }_1}{)}^{p_1}{\kappa }_1(j^{{\alpha }_2}{)}^{p_2}{\kappa }_2(j^{{\alpha }_3}{)}^{p_3}\cdots {\kappa }_{m-1}(j^{{\alpha }_m}{)}^{p_m},$$

$$c_j^2,c_j^{2^2},\dots ,c_j^{2^{\lfloor \text{log}(n)\rfloor }}$$

$$O(dn\log n+\lambda n\log n)+O(n\log n)=O((d+\lambda )n\log n),$$

$$\sigma^{j}(i,a_{k})=\left\{\begin{array}[]{@{}ll@{}}(a_{k}(i)-i)\bmod n,&\text{if}\ k\neq j\\ n,&\text{otherwise.}\end{array}\right.$$

$${\Delta }_i^j(G_a)={\sigma }^j(i,a_1)\cdot {\sigma }^j(i,a_2)\cdots {\sigma }^j(i,a_d),$$

Full text

Keywords: divide-and-conquer, fast multiplication algorithm, graph isomorphism, permutation multiplication, simultaneous conjugacy problem, truncated iteration.

MCS: 05C85, 05C60.

1 Introduction

The $d$-Simultaneous conjugacy problem (or $d$-SCP for short) in a group $G$ asks the following: given two ordered $d$-tuples $(a_{1},a_{2},\ldots,a_{d})$ and $(b_{1},b_{2},\ldots,b_{d})$ of elements from $G$, does there exists an element $\tau$ in $G$ such that $b_{j}=\tau^{-1}a_{j}\tau$ holds for all indices $j=1,2,\ldots,d$?

This problem arises naturally in various fields of mathematics, computer science, and their applications in numerous other fields such as biology, chemistry etc. – most notably when deciding whether two objects from a given class of objects are isomorphic (which usually means structural equivalence). We quickly review some easy examples from the theory of maps on surfaces, the theory of covering graphs, computational group theory, representation theory, interconnection networks, and cryptography. We refer the reader to relevant literature for more information on these topics [4, 7, 12, 13, 14, 15, 16].

A finite graph embedded in a closed oriented surface is conveniently given in terms of two permutations $R$ and $L$ on the set of arcs in the graph, where $R$ represents local rotations of arcs around each vertex and $L$ is the arc-reversing involution. It is generally known that two embeddings of the same graph given by $(R,L)$ and $(R^{\prime},L^{\prime})$ are combinatorially equivalent if and only if there exists a permutation $\tau$ of the arc set such that $R^{\prime}=\tau^{-1}R\tau$ and $L^{\prime}=\tau^{-1}L\tau$. Thus, the problem translates to a $2$-SCP in the symmetric group $S_{n}$.

An $n$-fold covering projection of graphs can be given by assigning to each arc of the base graph a certain permutation from the symmetric group $S_{n}$, the so-called voltage of the arc; voltage assignments to arcs naturally extend to all walks, and in particular, to all fundamental closed walks at a base vertex which generate the fundamental group. Now, two covering projections given by two voltage assignments are equivalent (which roughly means that the covering graphs are the same modulo relabeling of vertices and arcs) if and only there is a permutation in $S_{n}$ that simultaneously conjugates the two ordered tuples of voltages assigned to fundamental closed walks at a given base vertex. Thus, the problem translates to $d$-SCP in $S_{n}$, where $d$ is the Betti number of the graph.

In computational group theory, an important variant of SCP in $S_{n}$ occurs when the two tuples are equal, that is, $a_{j}=b_{j}$ for all $j=1,2,\ldots,d$; in this case we ask whether there exists a non-trivial element $\tau\in S_{n}$ that belongs to the centralizer of the permutation group generated by $a_{1},a_{2},\ldots,a_{d}$. Often we are interested in actual construction of such permutations $\tau$, if they exsist. For, this is encountered in computing the generators of automorphism groups of embedded graphs on surfaces.

In representation theory, the problem of whether two representations of a given group on a vector space are equivalent translates to the simultaneous conjugacy problem in the general linear group, where the two tuples are matrices representing the generators of the group.

As for problems encountered in computer science, the equivalence of two permutation networks under the permutation of inputs and outputs also translates to SCP in the symmetric group. Last but not least, SCP in braid groups has been used in attacking cryptographic protocols.

The paper aims to study $d$-SCP in the symmetric group $S_{n}$. For $d=1$ the problem translates to the question of whether two permutations have the same cycle type; this can be tested in $O(n)$ time. So, the problem is interesting if $d>1$. The condition that, for each $j=1,2,\ldots,d$, the permutations $a_{j}$ and $b_{j}$ have the same cycle type is clearly necessary for the two tuples to be simultaneously conjugated. However, it is not sufficient.

1.1 Related work and motivation

Starting with a purely group-theoretic context of finding centralizers, in 1977, Fontet gave the $O(dn^{2})$-time algorithm for solving the $d$-SCP in $S_{n}$ by translating it into a special case of a colored digraph isomorphism problem [5]. More precisely, let $G_{a}$ be an arc-colored digraph with the vertex set $\{1,2,\ldots,n\}$ where there is an arc $(u,v)$ colored $j$ if and only if $a_{j}(u)=v$, $j=1,2,\ldots,d$. (Note that such a digraph can have multiple parallel arcs and loops.) The arc-colored digraph $G_{b}$ is defined similarly on the vertex set $\{1,2,\ldots,n\}$ with an arc $(u,v)$ colored $j$ if and only if $b_{j}(u)=v$, $j=1,2,\ldots,d$. The permutation $\tau$ from $S_{n}$ that simultaneously conjugates the two tuples is now precisely a color and direction preserving digraph isomorphism from $G_{a}$ onto $G_{b}$ (assuming that permutations in $S_{n}$ are multiplied from left to right). Fontet’s algorithm was independently discovered by Hoffmann in 1982 [8].

In studying permutation networks, Sridhar exhibited an $O(dn\,\mathrm{log}(dn))$-time algorithm in 1989 [15] for the case in which the corresponding digraphs are strongly connected, that is, when the two tuples generate transitive subgroups in $S_{n}$. In his solution, he applied techniques used by Hopcroft and Tarjan for testing isomorphism of planar 3-connected graphs [10]. The underlying idea of this approach is to compute the automorphism partition of the disjoint union of the two digraphs, that is, the orbits of the corresponding automorphism group. An isomorphism between the two digraphs exists if and only if there is an orbit that contains a vertex from each digraph. In the context of arc-colored digraphs $G_{a}$ and $G_{b}$ associated with the simultaneous conjugacy problem, this amounts to finding the orbits of the color and direction preserving automorphism group of the disjoint union $G_{a}+G_{b}$. The construction consists of the preprocessing stage and the partition refinement stage. The main problem to be resolved is to find an initial partition such that its refinement provides the automorphism partition. The key observation is that if the automorphism partition is nontrivial, then we must start with a nontrivial initial partition, otherwise the refinement process returns the trivial partition.

Unfortunately, as we have recently discovered, Sridhar’s algorithm does not work correctly on all inputs, for example, in certain cases when all cycles of a given color are of the same length. Sridhar’s preprocessing stage for the case when all cycles of a given color are of the same length is presented in algorithm ArcLabeling in Appendix A. It labels each arc of a digraph with a pair of the form $\langle\alpha,\beta\rangle$, where $\alpha$ is one of 0, 1, and 2, and $\beta$ is a nonnegative integer. Assigning $\alpha=0$ to the arcs coloured 1 (which defines reference cycles), let other arcs receive $\alpha=1$ if they join vertices on the same reference cycle, and $\alpha=2$ otherwise; the parameter $\beta$ records “relative jumps” with respect to reference cycles, where arcs of the reference cycles receive $\beta=0$. Two arcs of the same color are then in the same cell of the initial partition if and only if they have the same label.

We now give an example on which the algorithm ArcLabeling returns the trivial initial partition, although the automorphism partition is nontrivial. Let

[TABLE]

be permutations in $S_{12}$, and consider the corresponding arc-colored digraph $G_{a}$, which is shown in Figure 1. It is obvious that by calling ArcLabeling on $G_{a}$, arcs of color 1 (solid line) receive label $\langle 0,0\rangle$ while arcs of color 2 (dashed line) receive label $\langle 2,0\rangle$. Consequently, the initial partition on the set of arcs of $G_{a}$ is trivial, which in turn results in the trivial automorphism partition of $G_{a}$. However, the automorphism partition of $G_{a}$ is actually $\{\{1,4,7,10\},\{2,5,8,11\},\{3,6,9,12\}\}$, which is nontrivial.

1.2 Our results

The question that arises naturally is, therefore, the following:

Given two ordered $d$-tuples $(a_{1},a_{2},\ldots,a_{d})$ and $(b_{1},b_{2},\ldots,b_{d})$ of elements from the symmetric group $S_{n}$, can we find an element $\tau\in S_{n}$ such that $b_{j}=\tau^{-1}a_{j}\tau$ holds for all indices $j=1,2,\ldots,d$ in time $o(dn^{2})$?

The following main result answers the question affirmatively subject to the condition that the two given tuples of permutations generate transitive permutation groups. We call this restricted problem the transitive $d$-SCP.

Theorem 1.1**.**

The transitive $d$-SCP in the symmetric group $S_{n}$ can be solved in $O(n^{2}\log d/\log n+dn\log n)$ time, using $O(n^{3/2}+dn)$ space.

Our second result implies that for a large subclass of tuples generating transitive subgroups the problem can be solved in strongly subquadratic time in $n$ at a given $d$. More precisely, we prove the following.

Theorem 1.2**.**

Let $(a_{1},a_{2},\ldots,a_{d})$ and $(b_{1},b_{2},\ldots,b_{d})$ be $d$-tuples of permutations that generate transitive subgroups of the symmetric group $S_{n}$, and let $\lambda$ be the minimal number for which there are permutations $a_{j}$ and $b_{j}$ in the two tuples that have precisely $\lambda$ cycles in their cycle decompositions. Then we can test whether one tuple is simultaneously conjugate to the other in $S_{n}$ in $O((d+\lambda)n\log n)$ time, using $O((d+\log n)n)$ space.

The last theorem implies strongly subquadratic time in $n$ as soon as $\lambda=O(n^{\epsilon})$ for some constant $0\leq\epsilon<1$. In particular, for $\lambda=1$ we give a linear time solution.

Theorem 1.3**.**

Let $(a_{1},a_{2},\ldots,a_{d})$ and $(b_{1},b_{2},\ldots,b_{d})$ be $d$-tuples of permutations so that, for some $j\in[n]$, the permutations $a_{j}$ and $b_{j}$ are $n$-cycles. Then we can test whether one tuple is simultaneously conjugate in $S_{n}$ to the other one in $O(dn)$ time and $O(dn)$ space.

The structure of the paper is the following. Section 2 contains the necessary notation and basic definitions to make the paper self-contained. In Section 3 we develop an algorithm which solves the transitive $d$-SCP in the symmetric group $S_{n}$ in $O(n^{2}+dn\log n)$ time and $O(dn)$ space. In Sections 4 and 5 we then give two improvements of the basic algorithm from Section 3 to prove Theorems 1.1 and 1.2, respectively. Theorem 1.3 is proven in Section 6 by a completely different approach. Section 7 presents some empirical results. The last section concludes the paper by discussing some open problems related to our results.

2 Preliminaries

This section aims to establish some notation and terminology used in this paper. For the concepts not defined here see [3]. The notations are quite graph centered since in our solution to the transitive simultaneous conjugacy problem we use graph-based techniques in the spirit of Fontet and Hoffmann.

For a positive integer $n$, we denote the set $\{1,2,\ldots,n\}$ by $[n]$. For $i\in[n]$ and $g\in S_{n}$, we write $i^{g}$ for the image of $g$ under the permutation $g$ rather than by the more usual $g(i)$. Further, let $a=(a_{1},a_{2},\ldots,a_{d})$ be a $d$-tuple of permutations in $S_{n}$. The permutation digraph of $a$ is a pair $G_{a}=(V,A)$, where $V(G_{a})=V=[n]$ is the set of vertices, and $A(G_{a})=A$ is the set of ordered pairs $(i,a_{k})$, $i\in[n],k\in[d]$, called arcs. The degree of $G_{a}$ is $|a|$. An arc $e=(i,a_{k})$ has its initial vertex $\textrm{ini}(e)=i$, terminal vertex $\textrm{ter}(e)=i^{a_{k}}$, and color $c(e)=k$; the arc $e=(i,a_{k})$ is also referred to as outgoing from the vertex $i$. The vertices $\textrm{ini}(e)$ and $\textrm{ter}(e)$ are end-vertices of $e$.

A walk from a vertex $v_{0}$ to a vertex $v_{m}$ in a permutation digraph $G_{a}$ is an alternating sequence $W=v_{0},e_{1},v_{1},e_{2},\ldots,e_{m},v_{m}$ of vertices and arcs from $G$ such that for each $i\in[m]$, the vertices $v_{i-1}$ and $v_{i}$ are the end-vertices of the arc $e_{i}$. The vertices $v_{0}$ and $v_{m}$ are called the initial vertex and the terminal vertex of $W$, respectively. If $v_{0}=v_{m}$, then the walk $W$ is closed, and it is open otherwise. A path is a walk whose vertices (and hence arcs) are all pairwise distinct.

Walks defined so far are undirected. On the other hand we have also a directed walk in which case $v_{i-1}=\textrm{ini}(e_{i})$ and $v_{i}=\textrm{ter}(e_{i})$ for each $i\in[m]$. A digraph $G_{a}$ is weakly connected if any vertex is reachable from any other vertex by traversing a walk, and is strongly connected if any vertex is reachable from any other vertex by traversing a directed walk. Clearly, $G_{a}$ is strongly connected if and only if the tuple $a$ generates a transitive subgroup of $S_{n}$.

A subdigraph $H$ of $G_{a}$ consists of a subset $V(H)\subseteq V(G_{a})$ and a subset $A(H)\subseteq A(G_{a})$ such that every arc in $A(H)$ has both end-vertices in $V(H)$. A walk in a subdigraph $H$ of $G_{a}$ is a walk in $G_{a}$ consisting only of arcs from $A(H)$. A subdigraph $T$ of $G_{a}$ is a tree if for any two vertices $v$ and $v^{\prime}$ in $T$, there is a unique path from $v$ to $v^{\prime}$ in $T$. If $V(T)=V(G_{a})$, then $T$ is a spanning tree of $G_{a}$.

A color-isomorphism between two permutation digraphs $G_{a}$ and $G_{b}$ is a pair $(\phi_{V},\phi_{A})$ of bijections, where $\phi_{V}\colon V(G_{a})\to V(G_{b})$ and $\phi_{A}\colon A(G_{a})\to A(G_{b})$ such that $\phi_{V}(\textrm{ini}(e))=\textrm{ini}(\phi_{A}(e))$, $\phi_{V}(\textrm{ter}(e))=\textrm{ter}(\phi_{A}(e))$ and $c(e)=c(\phi_{A}(e))$ for any arc $e\in A(V_{a})$.

A partition $\mathcal{P}$ of a set $V$ is a set $\mathcal{P}=\{V_{1},V_{2},\ldots,V_{m}\}$ of nonempty pairwise disjoint subsets of $V$ whose union is $V$. Each set $V_{i}$ is called a cell of the partition. A partition is trivial if it has only one cell, $\{V\}$.

Let $\mathcal{G}$ be the set of all strongly connected permutation digraphs on $n$ vertices, each of degree $d$, and let $\mathcal{G}_{V}=\{(G_{a},v)\,:\,G_{a}\in\mathcal{G},v\in V(G_{a})\}$ be the set of all rooted digraphs from $\mathcal{G}$. Given a set $\mathcal{L}$ of labels, a vertex invariant is a function $I\colon\mathcal{G}_{V}\to\mathcal{L}$ such that whenever there exists a color-isomorphism $G_{a}\to G_{b}$ mapping $v$ onto $w$, then

[TABLE]

Since a vertex invariant $I$ assigns a label to every vertex of a permutation digraph $G_{a}$, it induces a partition $\mathcal{P}_{I}(G_{a})=\{V_{1}(G_{a}),V_{2}(G_{a}),\ldots,V_{m}(G_{a})\}$ on the set $V(G_{a})$ such that for any two vertices $u,v\in V(G_{a})$, $u$ and $v$ are in the same cell if and only if they have the same labels $I(G_{a},u)=I(G_{a},v)$. Two partitions $\mathcal{P}_{I}(G_{a})$ and $\mathcal{P}_{I}(G_{b})$ are compatible if $|\mathcal{P}_{I}(G_{a})|=|\mathcal{P}_{I}(G_{b})|=\mu$, and the cells of $\mathcal{P}_{I}(G_{b})$ can be re-enumerated such that for all $i\in[\mu]$ we have that $|V_{i}(G_{a})|=|V_{i}(G_{b})|$ and $I(G_{a},v)=I(G_{b},w)$ for all $v\in V_{i}(G_{a})$ and $w\in V_{i}(G_{b})$. The following results follow directly from definitions.

Lemma 2.1**.**

Let $I\colon\mathcal{G}_{V}\to\mathcal{L}$ be a vertex invariant, and let $G_{a},G_{b}\in\mathcal{G}$. Then:

1. (i)

If $G_{a}$ and $G_{b}$ are color-isomorphic, then the partitions $\mathcal{P}_{I}(G_{a})$ and $\mathcal{P}_{I}(G_{b})$ are compatible. 2. (ii)

Let $\mathcal{P}_{I}(G_{a})$ and $\mathcal{P}_{I}(G_{b})$ be compatible partitions, and let $v\in V_{1}(G_{a})$. Then $G_{a}$ and $G_{b}$ are not color-isomorphic if and only if there is no color-isomorphism of $G_{a}$ onto $G_{b}$ mapping $v$ to some vertex in $V_{1}(G_{b})$.

An alphabet $\Sigma$ is a finite set of characters. A word (or a string) over the alphabet $\Sigma$ is a finite sequence of characters from $\Sigma$. The length $|\kappa|$ of a word $\kappa$ is the number of characters in $\kappa$. The unique word $\epsilon$ of length 0 is the empty word. We denote the set of all words of length $m$ over the alphabet $\Sigma$ by $\Sigma^{m}$. We write $\kappa\kappa^{\prime}$ for the concatenation of words $\kappa$ and $\kappa^{\prime}$. For a word $\kappa$ and a non-negative integer $m$ we define ${\kappa}^{m}=\epsilon$ for $m=0$, and otherwise ${\kappa}^{m}={\kappa}^{m-1}\kappa$. Two words $\kappa$ and $\kappa^{\prime}$ are cyclically equivalent if there exist words $x$ and $y$ so that $\kappa=xy$ and $\kappa^{\prime}=yx$.

Let $G_{a}$ be a permutation digraph of degree $d$ and let $\Sigma=\{k^{\alpha}\,|\,k\in[d],\alpha=\pm 1\}$ be an alphabet. A walk $W=v_{0},e_{1},v_{1},e_{2},\ldots,e_{m},v_{m}$ in $G_{a}$ defines the word $\kappa(W)=k_{1}^{\alpha_{1}}k_{2}^{\alpha_{2}}\cdots k_{m}^{\alpha_{m}}$ over $\Sigma$, where: (i) $k_{i}=c(e_{i})$; and (ii), if $\textrm{ini}(e_{i})=v_{i-1}$ and $\textrm{ter}(e_{i})=v_{i}$, then $\alpha_{i}=+1$, and $\alpha_{i}=-1$ otherwise.

Conversely, a word $\kappa^{\prime}$ over $\Sigma$ and a vertex $v_{0}\in V(G_{a})$ together define the walk $W(\kappa^{\prime},v_{0})$ in $G_{a}$ starting at $v_{0}$ such that $\kappa(W(\kappa^{\prime},v_{0}))=\kappa^{\prime}$. Finally, for a word $\kappa=k_{1}^{\alpha_{1}}k_{2}^{\alpha_{2}}\cdots k_{m}^{\alpha_{m}}$ over $\Sigma$ we define the permutation $a_{\kappa}=a_{k_{1}}^{\alpha_{1}}a_{k_{2}}^{\alpha_{2}}\cdots a_{k_{m}}^{\alpha_{m}}$.

Let $\{{\tt C},{\tt O}\}$ be a set of labels and let $\kappa$ be a word over $\Sigma$. The function $I^{\kappa}\colon\mathcal{G}\to\{{\tt C},{\tt O}\}$ defined by

[TABLE]

is vertex invariant. In group-theoretical terms, vertices labeled by C represent the set of fixed points of the permutation $a_{\kappa}$, while vertices labeled by O represent its support set. For a set $B\subseteq V(G)$ we let

[TABLE]

For $\kappa=\epsilon$, we have $I^{\epsilon}(G,v)={\tt C}$ for all $(G,v)\in\mathcal{G}$, and therefore $C_{\epsilon}(V(G))=V(G)$ and $O_{\epsilon}(V(G))=\emptyset$.

Let $G_{a}$ and $G_{b}$ be two permutation digraphs, and let $W_{1}$ and $W_{2}$ be walks in $G_{a}$ and $G_{b}$, respectively. We say that $W_{2}$ is the walk corresponding to the walk $W_{1}$ if $\kappa(W_{1})=\kappa(W_{2})$. In particular, an arc $e=(v,a_{k})\in A(G_{a})$ corresponds to an arc $f=(w,b_{k^{\prime}})\in A(G_{b})$ if $k=k^{\prime}$. Further, a word $\kappa$ over $\Sigma$ is a distinguishing word for vertices $v\in V(G_{a})$ and $w\in V(G_{b})$ if $I^{\kappa}(G_{a},v)\neq I^{\kappa}(G_{b},w)$. In this case, $v$ and $w$ are said to be distinguishable.

In this paper, we are using a $b$-bit RAM model of computation and consequently, we assume $n,m,d=O(2^{b})$.

3 Basic method

Let $G_{a}$ and $G_{b}$ be strongly connected permutation digraphs on $n$ vertices, each of degree $d$. In this section, we show that we can decide whether $G_{a}$ and $G_{b}$ are color-isomorphic by an algorithm running in $O(n^{2}+dn\log n)$ time and $O(dn)$ space. We first prove the following theorem.

Theorem 3.1**.**

Let $G_{a}$ and $G_{b}$ be strongly connected permutation digraphs on $n$ vertices and of degree $d$, and let $v_{0}\in V(G_{a})$ and $w_{0}\in V(G_{b})$. Then there exists a color-isomorphism of $G_{a}$ onto $G_{b}$ mapping $v_{0}$ onto $w_{0}$ if and only if $v_{0}$ and $w_{0}$ are indistinguishable.

Proof.

If there exists a color-isomorphism of $G_{a}$ onto $G_{b}$ mapping $v_{0}$ onto $w_{0}$, then any closed walk is mapped to a closed walk and hence $v_{0}$ and $w_{0}$ are indistinguishable.

Let $v_{0}$ and $w_{0}$ be indistinguishable vertices of $G_{a}$ and $G_{b}$, respectively. Construct an isomorphism mapping $v_{0}$ onto $w_{0}$ as follows. Take a spanning tree $T_{a}$ of $G_{a}$ rooted at $v_{0}$. For a vertex $v\neq v_{0}\in V(T_{a})$, let $v_{0},\ldots,e^{v},v$ be a unique path in $T_{a}$ from $v_{0}$ to $v$, and let $w_{0},\ldots,f^{v},w$ be the corresponding path in $G_{b}$ starting with $w_{0}$. For each vertex $v\neq v_{0}\in V(T_{a})$, map the arc $e^{v}$ to $f^{v}$ and the vertex $v$ to $w$. The subdigraph $T_{b}$ of $G_{b}$ with the set $V(G_{b})$ of vertices and with the set $\{f^{v}\in A(G_{b})\,|\,v\in V(T_{a}),v\neq v_{0}\}$ of arcs must be a spanning tree of $G_{b}$ rooted at $w_{0}$, for otherwise there exists an open walk starting with $v_{0}$ such that the corresponding walk starting with $w_{0}$ is closed. But this is not possible since $v_{0}$ and $w_{0}$ are indistinguishable. It remains to define the mapping of the cotree arcs. Let $(u,a_{k})$ be a fixed cotree arc of $T_{a}$, and suppose that $u$ is mapped to $u^{\prime}$. Then map $(u,a_{k})$ to the corresponding arc $(u^{\prime},b_{k})$ in $G_{b}$. If $a_{k}(u)$ has already been mapped to some vertex different from $b_{k}(u^{\prime})$, then there exists a closed walk starting at $v_{0}$ such that the corresponding walk starting with $w_{0}$ is open. But this is not possible. Repeating the process at the remaining cotree arcs yields the desired isomorphism. ∎

From Theorem 3.1 we derive the necessary and sufficient condition for $G_{a}$ and $G_{b}$ to be color-isomorphic. Since for an arbitrary vertex $v_{0}\in V(G_{a})$ there are $n$ possible candidates in $G_{b}$ that might be indistinguishable from $v_{0}$, we need to check, for each vertex $w_{0}\in V(G_{b})$, whether $v_{0}$ and $w_{0}$ are indistinguishable. Algorithm Indistinguishable, based on Theorem 3.1 and breadth-first search, does precisely that – with an addition of returning the empty word whenever $v_{0}$ and $w_{0}$ are indistinguishable, and a distinguishing word for $v$ and $w$ otherwise.

Proposition 3.2**.**

Let $G_{a}$ and $G_{b}$ be strongly connected permutation digraphs on $n$ vertices, each of degree $d$, and let $v_{0}\in V(G_{a})$, $w_{0}\in V(G_{b})$. The algorithm Indistinguishable$(G_{a},G_{b},v_{0},w_{0})$ correctly tests in $O(dn)$ time and $O(dn)$ space whether $v_{0}$ and $w_{0}$ are indistinguishable.

Proof.

The correctness of the algorithm follows directly from the remarks above and Theorem 3.1.

The closed walk $W$ in lines 14 or 19 is constructed at most once, and this can be done in time $O(n)$ using $O(n)$ space. Finally, since the algorithm is based on breadth-first search its total time is $O(dn)$, while the space used is also $O(dn)$. ∎

If $v_{0}$ are $w_{0}$ indistinguishable we have found a color-isomorphism. Otherwise, we get a distinguishable word $\kappa$ such that the vertex invariant $I^{\kappa}$ induces non-trivial partitions

[TABLE]

If $|O_{\kappa}(V(G_{a}))|\neq|O_{\kappa}(V(G_{b}))|$, then the partitions $\mathcal{P}_{I^{\kappa}}(G_{a})$ and $\mathcal{P}_{I^{\kappa}}(G_{b})$ are not compatible. Hence $G_{a}$ and $G_{b}$ are not color-isomorphic by (i) of Lemma 2.1. Consequently, no indistinguishable vertices exist.

Otherwise, by (ii) of Lemma 2.1 we can search for indistinguishable vertices either between $O_{\kappa}(V(G_{a}))$ and $O_{\kappa}(V(G_{b}))$ or between $C_{\kappa}(V(G_{a}))$ and $C_{\kappa}(V(G_{b}))$. We re-apply algorithm Indistinguishable on the smaller cells. This reduces the number of possible candidates from $n$ to at most $n/2$.

Algorithm ColorIsomorphic formally describes this process. In the $i$-th iteration, it uses the function Partition to split the sets $V_{i}(G_{a})$ and $V_{i}(G_{b})$ with respect to the word $\kappa_{i}$. This is done by traversing the walk $W(\kappa_{i},v)$ in $G$, where either $G$ is $G_{a}$ and $v\in V_{i}(G_{a})$, or $G$ is $G_{b}$ and $v\in V_{i}(G_{b})$.

Theorem 3.3**.**

Let $G_{a}$ and $G_{b}$ be strongly connected permutation digraphs on $n$ vertices, each of degree $d$. The algorithm ColorIsomorphic$(G_{a},G_{b})$ correctly tests in $O(n^{2}+dn\log n)$ time and $O(dn)$ space whether $G_{a}$ and $G_{b}$ are color-isomorphic.

Proof.

We first show the following loop invariant.

At the start of each iteration of the repeat loop at lines 2-13 the following holds:

1. (i)

$1\leq|V_{i}(G_{a})|\leq{n}/\,{2^{i-1}}$, and 2. (ii)

$V_{i}(G_{a})$ and $V_{i}(G_{b})$ are cells of the partitions of $V(G_{a})$ and $V(G_{b})$, respectively, where both partitions are induced by a vertex invariant $\mathcal{I}_{i}$ such that the vertices in $V_{i}(G_{a})$ and the vertices in $V_{i}(G_{b})$ have the same labels.

Prior to the first iteration, we have $i=1$, $|V_{1}(G_{a})|=n$. Thus, property (i) of the invariant holds. By taking $\mathcal{I}_{1}=I^{\epsilon}$, the partitions $\mathcal{P}_{\mathcal{I}_{1}}(G_{a})$ and $\mathcal{P}_{\mathcal{I}_{1}}(G_{b})$ are trivial, and so $V_{1}(G_{a})\in\mathcal{P}_{\mathcal{I}_{1}}(G_{a}),V_{1}(G_{b})\in\mathcal{P}_{\mathcal{I}_{1}}(G_{b})$, and the vertices in $V_{i}(G_{a})$ and the vertices in $V_{i}(G_{b})$ have the same labels. Thus, property (ii) holds as well.

To see that each iteration maintains the loop invariant, we inductively assume that before the iteration $i$, properties (i) and (ii) hold and that at the end of iteration $i$ the condition in line 13 is false. Consequently, $\kappa_{i}\neq\epsilon$ and $|V_{i+1}(G_{a})|=|V_{i+1}(G_{b})|$. Without loss of generality we may assume that $V_{i+1}(G_{a})=C_{\kappa_{i}}(V_{i}(G_{a}))$ and $V_{i+1}(G_{b})=C_{\kappa_{i}}(V_{i}(G_{b}))$.

First, since $\kappa_{i}$ is a distinguishable word for some $v_{i}\in V_{i}(G_{a})$ and some $w_{i}\in V_{i}(G_{b})$, at least one of the cells$C_{\kappa_{i}}(V_{i}(G_{a}))$ and $C_{\kappa_{i}}(V_{i}(G_{b}))$ is nonempty. But then, since $|C_{\kappa_{i}}(V_{i}(G_{a}))|=|C_{\kappa_{i}}(V_{i}(G_{b}))|$, both $V_{i+1}(G_{a})$ and $V_{i+1}(G_{b})$ are nonempty. Moreover, since $|V_{i}(G_{a})|\leq{n}/\,{2^{i-1}}$, and since $|V_{i+1}(G_{a})|\leq|V_{i}(G_{a})|/\,2$, it follows that $|V_{i+1}(G_{a})|\leq{n}/\,{2^{i}}$, and the first part of the invariant holds.

To prove (ii), let $\mathcal{I}_{i+1}\colon\mathcal{G}\to\{{\tt C},{\tt O}\}^{i+1}$ be a function defined by $\mathcal{I}_{i+1}(G,v)=\mathcal{I}_{i}(G,v)\cdot I^{\kappa_{i}}(G,v).$ Note that because $\mathcal{I}_{i}$ and $I^{\kappa_{i}}$ are vertex invariants, so is $\mathcal{I}_{i+1}$. Since $V_{i}(G_{a})\in\mathcal{P}_{\mathcal{I}_{i}}(G_{a})$ and $V_{i}(G_{b})\in\mathcal{P}_{\mathcal{I}_{i}}(G_{b})$, and since $V_{i+1}(G_{a})$ and $V_{i+1}(G_{b})$ are nonempty, it follows that

[TABLE]

Furthermore, $\mathcal{I}_{i+1}(G_{a},v)=\mathcal{I}_{i}(G_{a},v)\cdot{\tt C}$ for all $v\in V_{i+1}(G_{a})$, while $\mathcal{I}_{i+1}(G_{b},w)=\mathcal{I}_{i}(G_{b},w)\cdot{\tt C}$ for all $w\in V_{i+1}(G_{b})$. But $\mathcal{I}_{i}(G_{a},v)=\mathcal{I}_{i}(G_{b},w)$ for all $v\in V_{i}(G_{a})$ and $w\in V_{i}(G_{b})$, and so $\mathcal{I}_{i+1}(G_{a},v)=\mathcal{I}_{i+1}(G_{b},w)$ for all $v\in V_{i+1}(G_{a})$ and $w\in V_{i+1}(G_{b})$. Thus, the second part of the invariant holds.

Next, by property (i), it follows that the algorithm terminates in at most $\lfloor\log n\rfloor+1$ iterations. Furthermore, we show that on termination the algorithm returns true if and only if the digraphs are color-isomorphic. We first show that, if $G_{a}$ and $G_{b}$ are color-isomorphic, the algorithm returns true. Suppose for the purpose of deriving a contradiction that at the end of some iteration, say $i$, a word $\kappa_{i}$ is nonempty and $|V_{i+1}(G_{a})|\neq|V_{i+1}(G_{b})|$. Then, by property (ii), $V_{i}(G_{a})\in\mathcal{P}_{\mathcal{I}_{i}}(G_{a})$ and $V_{i}(G_{b})\in\mathcal{P}_{\mathcal{I}_{i}}(G_{b})$, where $\mathcal{I}_{i}$ is a vertex invariant such that vertices in $V_{i}(G_{a})$ and vertices in $V_{i}(G_{b})$ have the same labels. Consider the vertex invariant $\mathcal{I}_{i+1}$ as defined above. Because $|V_{i+1}(G_{a})|\neq|V_{i+1}(G_{b})|$, it follows that the partitions $\mathcal{P}_{\mathcal{I}_{i+1}}(G_{a})$ and $\mathcal{P}_{\mathcal{I}_{i+1}}(G_{b})$ are not compatible. However, since $G_{a}$ and $G_{b}$ are assumed to be color-isomorphic, we have by (i) of Lemma 2.1 that $\mathcal{P}_{\mathcal{I}_{i+1}}(G_{a})$ and $\mathcal{P}_{\mathcal{I}_{i+1}}(G_{b})$ must be compatible, which is a contradiction. Conversely, if the algorithm returns true, then at the end of some iteration, say $i$, a word $\kappa_{i}$ must be empty, so there exists a vertex in $G_{a}$ which is indistinguishable from some vertex in $G_{b}$. By Theorem 3.1, $G_{a}$ and $G_{b}$ are color-isomorphic. This completes the proof of correctness of the algorithm.

Finally, we compute the complexity of the algorithm. The initialization in line 1 requires $O(n)$ time. From the above analysis it follows that the repeat loop is executed at most $\lfloor\log n\rfloor+1$ times. Lines 3 and 12 require $O(1)$ time, while line 4 requires $O(dn)$ time by Proposition 3.2. Let $T_{\sc P}(i)$ denote the time of the Partition to split up $V_{i}(G_{a})$ and $V_{i}(G_{b})$ in lines 6-7 with respect to $\kappa_{i}$. Then, we can express the total time of the algorithm as being bounded from above by

[TABLE]

Since Partition requires time proportional to the product of the length of a word times the number of vertices in a cell we are splitting, and since, at each iteration $i$, we have $|\kappa_{i}|=O(n)$ and $|V_{i}(G_{a})|=|V_{i}(G_{b})|\leq n/\,{2^{i-1}}$, it holds that $T_{\sc P}(i)=O(n^{2}/\,{2^{i-1}})$. It follows that

[TABLE]

Since $\sum_{i=0}^{\infty}1/\,{2^{i}}=2$, we can bound the running time of the algorithm as

[TABLE]

The space complexity of the algorithm is clearly $O(dn)$. This completes the proof. ∎

4 Towards a subquadratic time algorithm

The running time of the algorithm ColorIsomorphic is dominated by the running time of the function Partition; in turn, this depends on the length of a distinguishing word $\kappa$. Since the output sets $O_{\kappa}(V(G_{a}))$ and $C_{\kappa}(V(G_{a}))$ of Partition$(V(G_{a}),\kappa)$ are the support set and the fixed points set of the permutation $a_{\kappa}$, respectively, we consider the following a bit more general problem:

Given a finite sequence $a_{1},a_{2},\ldots,a_{d}$ of $d$ permutations in $S_{n}$ and a word $\kappa=k_{1}k_{2}\cdots k_{m}$ over $[d]$, can we evaluate the product $a_{\kappa}=a_{k_{1}}a_{k_{2}}\cdots a_{k_{m}}$ on $s$ points in time $o(sm)$?

An especially interesting case for us is when $m,s=\Theta(n)$. In this case, we show that, at a given $d$, the product $a_{\kappa}$ can be evaluated in time $o(n^{2})$. This reduces the running time of the algorithm ColorIsomorphic to subquadratic in $n$.

Our approach is similar to the fast computation of large positive integer powers $\beta^{m}$ by repeated squaring. For technical reasons, we describe a procedure of how to make $\kappa$ always of even length. Namely, if $|\kappa|$ is odd we expand it by a single character $*\notin\Sigma$ making its length even. Moreover, defining $a_{*}$ to be the identity permutation of $[n]$, the new word $\kappa$ defines the same product $a_{k_{1}}a_{k_{2}}\cdots a_{k_{m}}$.

Let $\mathcal{S}_{0}=\{a_{1},a_{2},\ldots,a_{d}\}$, $\kappa_{0}=\kappa$, $k_{0,j}=k_{j}$, $j\in[m]$, and $\Sigma_{0}=[d]$. In case $|\kappa_{0}|$ is odd, we use the above technical procedure making its length even. In the next step we scan through the word $\kappa_{0}$, replacing a pair $k_{0,2j-1}k_{0,2j}$ by $k_{1,j}$, $j\in[\left\lceil m/2\right\rceil]$. The obtained word $\kappa_{1}=k_{1,1}k_{1,2}\cdots k_{1,\left\lceil m/2\right\rceil}$ is actually built over $\Sigma_{1}\cup\{k_{1,\left\lceil m/2\right\rceil}\}$, where $\Sigma_{1}=\{ij\,|\,i,j\in\Sigma_{0}\}$. Clearly, $|\kappa_{1}|=\frac{1}{2}|\kappa_{0}|$ and $a_{\kappa_{0}}=a_{\kappa_{1}}$. So assuming that the set $\mathcal{S}_{1}=\{a_{ij}=a_{i}a_{j}\,|\,a_{i},a_{j}\in\mathcal{S}_{0}\}$ of all products of pairs of permutations in $\mathcal{S}_{0}$ is precomputed and the permutation $a_{k_{1,\left\lceil m/2\right\rceil}}$ is known, the time of straightforward evaluation of $a_{\kappa}$ is reduced to half.

The above reduction step can be repeated. If before the $t$-th iteration the length $|\kappa_{t-1}|$ is odd, we apply the above technical procedure making $|\kappa_{t-1}|$ even. Then, after the $t$-th iteration we have the set $\Sigma_{t}=\{ij\,|\,i,j\in\Sigma_{t-1}\}$ and the word

[TABLE]

where $k_{t,j}=k_{t-1,2j-1}k_{t-1,2j}$ for $j\in[\ell]$, and $\kappa_{t}$ is over the alphabet $\Sigma_{t}\cup\{k_{t,\ell}\}$. We leave to the reader to check that $\ell=\left\lceil m/2^{t}\right\rceil$. Again, with $\mathcal{S}_{t}=\{a_{ij}=a_{i}a_{j}\,|\,a_{i},a_{j}\in\mathcal{S}_{t-1}\}$ and the permutation $a_{k_{t,\ell}}$ being precomputed, straightforward evaluation of the product $a_{\kappa_{t}}$ takes $O(n\cdot m/2^{t})$.

At this point, we truncate the iteration similarly as it is the recursion in [2]. Namely, since $|\mathcal{S}_{t}|=d^{2^{t}}$, the construction of $\mathcal{S}_{t}$ from $\mathcal{S}_{t-1}$ takes $O(n\cdot d^{2^{t}})$. Consequently, we iterate the described process for $\nu$ times until

[TABLE]

Observe that inequality (2) has no closed form solution for ${\nu}$. However, by rewriting it as $2^{\nu}\leq\log_{d}m-{\nu}\log_{d}2$, we see that $2^{\nu}\leq\log_{d}m$. Finally, we define $\nu$ to be the integer satisfying

[TABLE]

Therefore, after ${\nu}$ iterations, the length of the word $\kappa_{\nu}$ becomes bounded from above by

[TABLE]

Based on this analysis, we give a formal description of this reduction in the algorithm WordReduction. This algorithm guarantees that the size of the word is $O(m\log d/\log m)$, and simultaneously increases the set of permutations.

Lemma 4.1**.**

Let $\mathcal{S}_{0}=\{a_{1},a_{2},\ldots,a_{d}\}$ be a set of permutations of $[n]$, and let $\kappa_{0}=k_{1}k_{2}\cdots k_{m}$ be a word over $[d]$. The algorithm WordReduction$(\mathcal{S}_{0}$, $\kappa_{0})$ takes $O(m+n\sqrt{m})$ time and $O(m+n\sqrt{m})$ space.

Proof.

We may assume that $m\geq d^{\,4}$ as otherwise we are done. Lines 6-8 take $O(1)$ time. Since $|\kappa_{t-1}|=\left\lceil m/\,2^{t-1}\right\rceil$, lines 8 and 9 take $O(m/\,2^{t-1})$ time. Obviously, the construction of $\mathcal{S}_{t}$ from $\mathcal{S}_{t-1}$ in line 10 takes $O(nd^{\,2^{t}})$ time, while line 11 takes $O(n)$ time. So, we can express the total running time of the algorithm as being bounded from above by

[TABLE]

The last summation can be estimated as

[TABLE]

Thus, the algorithm takes $O(m+n\sqrt{m})$ time. Moreover, the space complexity is proportional to the sum of the length of the input word and the size of the set $\mathcal{S}_{\nu}$. However, $\mathcal{S}_{\nu}$ has $d^{\,2^{\nu}}=O(\sqrt{m})$ permutations of length $n$, and the result follows. ∎

Once the length of the word is guaranteed to be $O(m\log d/\log m)$, we evaluate its corresponding product on $s$ points in a straightforward manner.

Lemma 4.2**.**

Let $a_{1},a_{2},\ldots,a_{d}$ be permutations of $[n]$ and let $k_{1}k_{2}\cdots k_{m}$ be a word over $[d]$. Then we can evaluate the product $a_{\kappa_{1}}a_{\kappa_{2}}\ldots a_{\kappa_{m}}$ on $s$ points in $O(m+n\sqrt{m}+s\min\{m,m\log d/\log m\})$ time and $O(m+n\sqrt{m})$ space.

Proof.

The evaluation is done in two phases. First, by Lemma 4.1 we reduce the length of the word to $O(m\log d/\log m)$ in $O(m+n\sqrt{m})$ time and $O(m+n\sqrt{m})$ space. And second, we evaluate the obtained word in $O(s\min\{m,m\log d/\log m\})$ time. ∎

Remark 4.3**.**

Lemma 4.2 partially answers the question we posed at the beginning of this section. The evaluation can be done in time $o(sm)$ whenever $s=\Theta(n)$ and $\log d=o(\log m)$.

We are now ready to prove Theorem 1.1.

Proof of Theorem 1.1.

Given $d$-tuples $a$ and $b$ of permutations that generate transitive subgroups of the symmetric group $S_{n}$, the tuple $a$ is simultaneously conjugate to tuple $b$ if and only if the permutation digraph $G_{a}$ is color-isomorphic to the permutation digraph $G_{b}$. To this end, we modify the basic method in the algorithm ColorIsomorphic by speeding-up the bottleneck which is Partition.

Instead of using Partition to partition the sets $V_{i}(G_{a})$ and $V_{i}(G_{b})$ of sizes $s=O(n/2^{i-1})$ with respect to a distinguishing word $\kappa_{i+1}$ of length $m=O(n)$, we can compute, by Lemma 4.2, the cells $C_{\kappa_{i+1}}(V_{i}(G_{a}))$, $O_{\kappa_{i+1}}(V_{i}(G_{a}))$, $C_{\kappa_{i+1}}(V_{i}(G_{b}))$, and $O_{\kappa_{i+1}}(V_{i}(G_{b}))$ in

[TABLE]

time and $O(n^{3/2})$ space. Then, by replacing $T_{\sc P}(i)$ in equation (1) with (4) we can bound the running time of the algorithm to

[TABLE]

In addition to $O(dn)$ space required by the basic method, we need an extra $O(n^{3/2})$ space to compute the cells and the result follows. ∎

5 Strongly subquadratic time algorithm

If we could guarantee that a spanning tree $T_{a}$ ($a=(a_{1},a_{2},\ldots,a_{d})$) in line 1 of Indistinguishable has many arcs of the same color, then any distinguishing word arising from such a tree would contain many occurrences of some character. Consequently, the evaluation of the corresponding permutation can be done even more efficiently than in the previous section.

To achieve this, let $\lambda_{i}$ be the number of cycles in the cycle decomposition of $a_{i}$, and let us define $\lambda=\min_{i\in[d]}\lambda_{i}$. We present a construction of $T_{a}$ which for $\lambda=O(n^{\epsilon})$, where $0\leq\epsilon<1$, results in a strongly subquadratic running time in $n$ of the algorithm ColorIsomorphic.

Let $a_{j}$ be a permutation with $\lambda_{j}=\lambda$ cycles, and let $C_{1},C_{2},\ldots,C_{\lambda}$ be their corresponding subdigraphs in $G_{a}$. We construct $T_{a}$ in two steps: we first construct a weakly connected subdigraph $S_{a}$ of $G_{a}$, and then we find the required spanning tree $T_{a}$ as a spanning tree in $S_{a}$. More precisely, $V(S_{a})=V(G_{a})$ while $A(S_{a})$ consists of all arcs $A(C_{i})$, where $i\in[\lambda]$, and $\lambda-1$ arcs $e_{1},e_{2},\ldots,e_{\lambda-1}$ (of colors different from $j$) such that $S_{a}$ is weakly connected. This can be achieved in $O(dn)$ time by slightly modifying a breadth first search on $G_{a}$ such that, as soon as a vertex in $C_{i}$ is visited, all other vertices in $C_{i}$ are marked as visited as well. With $S_{a}$ in hand, we construct a subdigraph $T_{a}$ by deleting an arbitrary arc from each $A(C_{i})$. The above construction is summarized in the following lemma.

Lemma 5.1**.**

With the notation and assumptions of this section, we can compute a spanning tree with $\lambda-1$ arcs of colors different from $j$ in $O(dn)$ time and $O(dn)$ space.

Any distinguishing word $\kappa$ over $\Sigma=\{k^{\alpha}\,|\,k\in[d],\alpha=\pm 1\}$ arising from a spanning tree in Lemma 5.1 contains at most $\lambda-1$ characters different from $j^{+1}$ and $j^{-1}$, and can, therefore, be written without loss of generality as

[TABLE]

where each $0\leq p_{i}<n$, $\alpha_{i}\in\{-1,+1\}$, $\kappa_{i}\in\Sigma\setminus\{j^{+1},j^{-1}\}$, and $m\leq\lambda$. Efficient evaluation of the powers $a_{j}^{p_{i}}$ and $(a_{j}^{-1})^{p_{i}}$ using the idea of fast computation of large positive integer powers mentioned in the previous section, results in efficient evaluation of the product corresponding to equality $(\ref{eq:word})$.

Lemma 5.2**.**

Let $a_{1},a_{2},\ldots,a_{d}$ be permutations of $[n]$, and let $j^{p_{1}}\kappa_{1}j^{p_{2}}\kappa_{2}\cdots j^{p_{m}}$ be a word over $[d]$, where each $0\leq p_{i}<n$ and $\kappa_{i}\in[d]\setminus\{j\}$. Given the powers $a_{j}^{2},a_{j}^{2^{2}},\ldots,a_{j}^{2^{\lfloor\textrm{log}(n)\rfloor}}$, we can evaluate the product $a_{j}^{p_{1}}a_{\kappa_{1}}a_{j}^{p_{2}}a_{\kappa_{2}}\cdots a_{j}^{p_{m}}$ on $s$ points in $O(sm\log n)$ time.

Proof.

Note that each $p_{i}$ can be written as $p_{i}=\sum_{k=0}^{\lfloor\log n\rfloor}c_{k}{2^{k}}$, where each $c_{k}\in\{0,1\}$. For the given powers $a_{j}^{2},a_{j}^{2^{2}},\ldots,a_{j}^{2^{\lfloor\log n\rfloor}}$ we can obviously evaluate each $a_{j}^{p_{i}}$ at any point in $O(\log n)$ time. The result of lemma trivially follows. ∎

We are now ready to prove Theorem 1.2, which implies strongly subquadratic time in $n$ at a given $d$ as soon as $\lambda=O(n^{\epsilon})$ for some constant $0\leq\epsilon<1$. In particular, for $\lambda=O(1)$ we have the following obvious corollary to Theorem 1.2.

Corollary 5.3**.**

If $\lambda=O(1)$, then the transitive $d$-SCP in the symmetric group $S_{n}$ can be solved in time $O(dn\log n)$.

Proof of Theorem 1.2.

By Lemma 5.1 we can compute a spanning tree $T_{a}$ with $\lambda-1$ arcs of colors different from $j$ in $O(dn)$ time and $O(dn)$ space. Taking such a $T_{a}$ in line 1 of Indistinguishable, any distinguishing word arising from $T_{a}$ can be written as in equality (5), which is of length $O(\lambda)$. We then compute the inverse permutations $a_{1}^{-1},a_{2}^{-1},\ldots,a_{d}^{-1}$ and $b_{1}^{-1},b_{2}^{-1},\ldots,b_{d}^{-1}$. This takes $O(dn)$ time and $O(dn)$ space. Next, for each $c\in\{a_{j},a_{j}^{-1},b_{j},b_{j}^{-1}\}$ we compute the powers

[TABLE]

using the standard approach, which requires $O(n\log n)$ time and $O(n\log n)$ space. Then, to partition the sets $V_{i}(G_{a})$ and $V_{i}(G_{b})$ of sizes $s=O(n/2^{i-1})$ with respect to a distinguishing word $\kappa_{i+1}$ (of length $O(\lambda)$) we compute the cells $C_{\kappa_{i+1}}(V_{i}(G_{a}))$ and $O_{\kappa_{i+1}}(V_{i}(G_{a}))$, as well as $C_{\kappa_{i+1}}(V_{i}(G_{b}))$ and $O_{\kappa_{i+1}}(V_{i}(G_{b}))$, by the procedure described in Lemma 5.2 – instead of by using Partition in ColorIsomorphic. This can be done in $O(\lambda n\log n/\,2^{i-1})$ time, which is now, in fact, $T_{\sc P}(i)$ in equality (1). Consequently, taking into account the $O(n\log n)$ time for precomputing the powers $c_{j}^{2^{k}}$ we can bound the running time of the algorithm by

[TABLE]

as claimed.

In addition to $O(dn)$ space required by the basic algorithm, we need an extra $O(n(\log n+d))$ space for storing the inverse permutations and powers, which proves the space bound and completes the proof.

∎

6 Linear time algorithm

In this section, we consider a special case when there is a permutation which is an $n$-cycle, that is, $\lambda=1$. Here we take a completely different approach not based on a distinguishing word. Rather, to each permutation digraph, we assign a special string in such a way that the permutation digraphs are color isomorphic if and only if the corresponding strings are cyclically equivalent. This equivalence is then checked by the linear time Knuth-Morris-Pratt string-matching algorithm.

Let $G_{a}$ be a permutation digraph of degree $d$ with $n$ vertices so that some permutation, say $a_{j}$, is an $n$-cycle. Without loss of generality we may assume that the vertices are labeled in such a way that $a_{j}(i)=i\bmod n+1$, $i\in[n]$, and we consider $a_{j}$ as a reference cycle. To $G_{a}$, we assign a string over $\{0,1,\ldots,n\}$ in three steps as follows. First, we label the arcs of $G_{a}$: each arc $e$ with $c(e)\neq j$ is labeled by $\sigma^{j}(e)$ that indicates the number of color-$j$ arcs along the reference cycle $a_{j}$ from its initial vertex $\textrm{ini}(e)$ to its terminal vertex $\textrm{ter}(e)$; while arcs $e$ with $c(e)=j$ are labeled by $\sigma^{j}(e)=n$. More formally, for each arc $e=(i,a_{k})$,

[TABLE]

Second, using these arc-labels we label each vertex $i$ of $G_{a}$ by the string

[TABLE]

which is a concatenation of labels of all arcs out-going from $i$, ordered by color. Finally, we encode $G_{a}$ as a string $\mathcal{S}^{j}(G_{a})=\Delta^{j}_{1}(G_{a})\cdot\Delta^{j}_{2}(G_{a})\cdots\Delta^{j}_{n}(G_{a})$.

Proposition 6.1**.**

Let $G_{a}$ and $G_{b}$ be permutation digraphs on $n$ vertices each of degree $d$ where, for some $j\in[n]$, the permutations $a_{j}$ and $b_{j}$ are $n$-cycles. Then there exists a color-isomorphism of $G_{a}$ onto $G_{b}$ if and only if the corresponding strings $\mathcal{S}^{j}(G_{a})$ and $\mathcal{S}^{j}(G_{b})$ are cyclically equivalent.

Proof.

Suppose first that there exists a color-isomorphism $f\colon G_{a}\to G_{b}$. Let $x=\Delta^{j}_{1}(G_{a})\cdot\Delta^{j}_{2}(G_{a})\cdots\Delta^{j}_{f(1)-1}(G_{a})$ and $y=\Delta^{j}_{f(1)}(G_{a})\cdot\Delta^{j}_{f(1)+1}(G_{a})\cdots\Delta^{j}_{n}(G_{a})$. It is easy to see that $\mathcal{S}^{j}(G_{b})=yx$, and so $\mathcal{S}^{j}(G_{a})$ and $\mathcal{S}^{j}(G_{b})$ are cyclically equivalent.

Conversely, suppose that the strings $\mathcal{S}^{j}(G_{a})$ and $\mathcal{S}^{j}(G_{b})$ are cyclically equivalent. By definition, in both strings the character $n$ occurs precisely $n$ times at positions $j,j+n,\ldots,j+(d-1)n.$ So, there must exists $k\in[n]$ such that $x=\Delta^{j}_{1}(G_{a})\cdot\Delta^{j}_{2}(G_{a})\cdots\Delta^{j}_{k-1}(G_{a})$, $y=\Delta^{j}_{k}(G_{a})\cdot\Delta^{j}_{k+1}(G_{a})\cdots\Delta^{j}_{n}(G_{a})$ and $\mathcal{S}^{j}(G_{b})=yx$. It is straightforward to check that the mapping $f\colon V(G_{a})\to V(G_{b})$, defined by $f(i)=(i+k-2)\bmod n+1$ extends to a color-isomorphism of $G_{a}$ onto $G_{b}$. This completes the proof. ∎

We are now ready to prove Theorem 1.3.

Proof of Theorem 1.3.

First, we can construct the corresponding strings $\mathcal{S}^{j}(G_{a})$ and $\mathcal{S}^{j}(G_{b})$ of length $dn$ in $O(dn)$ time and $O(dn)$ space. Next, by Proposition 6.1, testing $G_{a}$ and $G_{b}$ for color isomorphism is equivalent to testing whether $\mathcal{S}^{j}(G_{a})$ and $\mathcal{S}^{j}(G_{b})$ are cyclically equivalent. In turn, this is equivalent to asking whether $\mathcal{S}^{j}(G_{b})$ is a substring of the string $\mathcal{S}^{j}(G_{a})\cdot\mathcal{S}^{j}(G_{a})$. Finally, the last problem can be solved in $O(dn)$ time and $O(dn)$ space using the Knuth-Morris-Pratt string-matching algorithm [11], which concludes the proof. ∎

7 Empirical results

In empirical evaluation we compared three algorithms. The quadratic one is the algorithm described in [5, 8], the subqaudratic one is the algorithm ColorIsomorphic with WordReduction improvement, and the linear one is the algorithm described in Section 6. All algorithms have been implemented in the system GAP [6]. A breadth-first search has been used to construct a spanning tree in line 1 of Indistinguishable.

We performed two experiments: in the first experiment we compared the subquadratic algorithm with the quadratic algorithm in the general transitive case, while in the second experiment we compared the linear algorithm with the subquadratic one in a special case when the first permutation in each tuple was an $n$-cycle.

The tests were conducted on isomorphic and non-isomorphic pairs of randomly generated tuples. For isomorphic pairs in the first experiment, the first tuple was obtained by repeatedly selecting random elements of $S_{n}$ until a transitive subgroup was generated. The second tuple, isomorphic to the first one, was obtained by conjugation with a random permutation. For isomorphic pairs in the second experiment, the first tuple was obtained by repeatedly selecting $\log n$ random elements of $S_{n}$ in such a way that the first element was an $n$-cycle. The isomorphic tuple was then generated in the same way as in the first experiment.

For non-isomorphic pairs in the first experiment, we initially generated a random transitive tuple $(a_{1},a_{2},\ldots,a_{d})$ in such a way that $a_{1}^{2}\neq 1$. Next, we randomly chose a permutation $\tau$ such that $\tau a_{1}^{2}\neq a_{1}^{2}\tau$, and constructed the tuples $(a_{1},a_{2},\ldots,a_{d},a_{1}^{2})$ and $(\tau^{-1}a_{1}\tau,\tau^{-1}a_{2}\tau,\ldots,\tau^{-1}a_{d}\tau,a_{1}^{2})$. It is not hard to check that such pair of tuples is non-isomorphic. Non-isomorphic pairs in the second experiment were generated in the same way as in the first experiment with $a_{1}$ being an $n$-cycle and $d=\log n$.

The results of the first experiment for sizes of $n$ varying between 10 000 and 50 000 are presented in Table 1, while the results of the second experiment for sizes of $n$ varying between 100 000 and 500 000 are shown in Table 2. Runtimes, given in miliseconds, were measured by GAP’s NanosecondsSinceEpoch function on a Macbook Pro with 2,9 GHz Intel Core i5 processor under macOS Sierra version 10.12.6.

Table 1 shows that the subquadratic algorithm is a clear winner of the comparison. Moreover, its running time in practice is much faster than our worst-case estimate. The main reason for this is that the running time is dominated by the length of distinguishing words, and this in turn depends on the depth of the breadth-first search tree constructed in line 1 of Indistinguishable, which, in practice, is usually $O(\log n)$ with a modest constant [14].

The results in Table 2 show that the linear algorithm outperforms the subquadratic one on isomorphic pairs while on non-isomorphic pairs the opposite happens. The main reason for this is that, in contrast with the linear algorithm the subquadratic algorithm does not necessary scan the entire input to find that given tuples are non-isomorphic.

8 Concluding remarks

We have shown that the $d$-SCP in $S_{n}$ can be solved in $O(n^{2}\log d/\log n+dn\log n)$ worst-case time for the case when the two given tuples of permutations generate transitive groups. On the other hand, our experimental results on random instances suggest the following conjecture.

Conjecture 8.1**.**

If the permutations of transitive tuples are chosen uniformly at random, the expected running time of the algorithm ColorIsomorphic with WordReduction improvement is nearly-linear in $n$ at given $d$.

Next, to to the best of our knowledge, there are no known lower bounds for the SCP, except the trivial linear bound $\Omega(dn)$, which leaves open also the gap between the upper and the lower bound.

The key approach used in our solution is that, given a sequence $a_{1},a_{2},\ldots,a_{d}$ of permutations in $S_{n}$ and a word $k_{1}k_{2}\cdots k_{m}$ over $[d]$, we evaluate the product $a_{k_{1}}a_{k_{2}}\cdots a_{k_{m}}$ on $n$ points in time $o(nm)$ whenever $\log d=o(\log m)$. If $d\geq m$, the straightforward $O(nm)$-time evaluation is optimal, however, it remains open how big $d$ can be that there still exists an $o(nm)$-time solution.

Finally, we remark that we have recently developed a worst-case subquadratic algorithm in $n$ at given $d$ also for the general case, i.e., when each tuple of permutations generates an intransitive group [1].

Acknowledgment

The authors would like to thank Ilia Ponomarenko for enlightening discussions, and to the referee for valuable suggestions.

Appendix A