Minimal and Canonical Images

TL;DR

This paper introduces new algorithms that efficiently find the canonical image of point sets under permutation groups by leveraging orbit structures and subgroup chains, improving over previous methods.

Contribution

The paper presents a novel family of algorithms that use orbit and subgroup chain structures to reduce search complexity in finding canonical images.

Findings

Algorithms are correct, as proven formally.

Experimental results show improved efficiency over prior methods.

Applicable to various permutation groups.

Abstract

We describe a family of new algorithms for finding the canonical image of a set of points under the action of a permutation group. This family of algorithms makes use of the orbit structure of the group, and a chain of subgroups of the group, to efficiently reduce the amount of search which must be performed to find a canonical image. We present both a formal proof of correctness of our algorithms and experiments on different permutation groups, which compare our algorithms with the previous state of the art.

Equations

$$\{2,3,5\},\{1,3,5\},\{3,1,2\},\{3,2,6\},\{3,6,1\},\{6,3,5\}.$$

$$\{1,5,4\},\{6,5,3\},\{4,6,1\},\{5,1,2\},\{3,2,6\},\{2,3,5\}.$$

$$\{6,5,3\},\{4,6,1\},\{3,2,6\}.$$

$$[1,\dots ,n|n+1,\dots ,2n|\dots |n\cdot n,\dots ,(n+1)\cdot n].$$

$$\mathrm{Min}(G,S,\le )$$

$$=\underset{\le }{\min }\{S^{\mathrm{ext}(g)}\mid g\in G\}\cap \{1,\dots ,n\}$$

$$=\underset{\le }{\min }\{S^h\mid h\in H\}\cap \{1,\dots ,n\}$$

$$=\mathrm{Min}(H,T,\le )\cap \{1,\dots ,n\}.$$

$$X{\preccurlyeq }_{\sigma }Y\iff X^{\sigma }\preccurlyeq Y^{\sigma }$$

$$\mathrm{Min}(G,S,{\preccurlyeq }_{\sigma })=\mathrm{Min}(G^{\sigma },S^{\sigma },\preccurlyeq {)}^{{\sigma }^{-1}}.$$

$$T_i=T^{q_i}=(S^g{)}^{q_i}=S^{t_i}=S^{p_{i^{\sigma }}{h}_i}=(S^{p_{i^{\sigma }}}{)}^{h_i}=(S_{i^{\sigma }}{)}^{h_i},$$

$$l^{\gamma }:=(c_l+1{)}^{\sigma }\cdot k+r_l^{{\alpha }_{c_l+1}}.$$

$$X_{l^{\gamma }}^{\prime h}=(X_{(c_l+1{)}^{\sigma }}^p{)}^h=((X_{c_l+1}^{\sigma }{)}^p{)}^h=Y_{c_l+1}^q=Y_l^{\prime }.$$

$$\text{textsc}CanImage(\Omega ,G,X,P)=\text{textsc}CanImage(\Omega ,G,Y,P).$$

$$H=G_{\mathcal{S}(\Omega ,G,X,P)}=G_{\mathcal{S}(\Omega ,G,Y,P)}.$$

$$P^{\prime }=\mathcal{O}(\Omega ,H,X,P)=\mathcal{O}(\Omega ,H,Y,P)$$

$$\text{textsc}CanImage(\Omega ,H,X^{\prime },P)=\text{textsc}CanImage(\Omega ,G,Y^{\prime },P),$$

$$\text{textsc}CanImage(\Omega ,G,X,P)=\text{textsc}CanImage(\Omega ,G,Y,P)\text{ if and only if }Y\in X^G.$$

$${(i,j)}^{(\sigma ,\tau )}:=(i^{\sigma },j^{\tau }).$$

Full text

We present a formal proof of correctness of our algorithms and describe experiments on different permutation groups that compare our algorithms with the previous state of the art.

††journal: Journal of Algebra

url]http://caj.host.cs.st-andrews.ac.uk/

url]https://www.morphism.de/ markusp/

url]http://conway1.mathematik.uni-halle.de/ waldecker/index-english.html

1 Background

Many combinatorial and group theoretical problems [14, 4, 2] are equivalent to finding, given a group $G$ that acts on a finite set $\Omega$ and a subset $X\subseteq\Omega$, a partition of $X$ into subsets that are in the same orbit of $G$.

We can solve such problems by taking two elements of $X$ and searching for an element of $G$ that maps one to the other. However, this requires a possible $O(|X|^{2})$ checks, if all elements of $X$ are in different orbits.

Given a group $G$ acting on a set $\Omega$, a canonical labelling function maps each element of $\Omega$ to a distinguished element of its orbit under $G$. Using a canonical labelling function we can check if two members of $\Omega$ are in the same orbit by applying the canonical labelling function to both and checking if the results are equal. More importantly, we can solve the problem of partitioning $X$ into orbit-equivalent subsets by performing $O(|X|)$ canonical image calculations. Once we have the canonical image of each element, we can organize the canonical images into equivalence classes by sorting in $O(|X|log(|X|))$ comparisons, or expected $O(|X|)$ time by placing them into a hash table. This is because checking if two elements are in the same equivalence class is equivalent to checking if their canonical images are equal.

The canonical image problem has a long history. Jeffrey Leon [9] discusses three types of problems on permutation groups – subgroup-type problems (finding the intersection of several groups), coset-type problems (deciding whether or not the intersection of a series of cosets is empty, and if not, finding their intersection) and canonical-representative-type problems. He claims to have an algorithm to efficiently solve the canonical-representative problem, but does not discuss it further. His comments have inspired mathematicians and computer scientists to work on questions related to minimal images and canonical images.

One of the most well-studied canonical-image problems is the canonical graph problem. Current practical systems derive from partition refinement techniques, which were first practically used for graph automorphisms by McKay [11] in the Nauty system. There have been a series of improvements to this technique, including Saucy [1], Bliss [8] and Traces [12]. A comparison of these systems can be found in [12].

We cannot, however, directly apply the existing work for graph isomorphism to finding canonical images in arbitrary groups. The reason is that McKay’s Graph Isomorphism algorithm only considers finding the canonical image of a graph under the action of the full symmetric group on the set of vertices. Many applications require finding canonical images under the action of subgroups of the full symmetric group.

One example of a canonical labelling function is, given a total ordering on $X$, to map each value of $X$ to the smallest element in its orbit under $G$. This Minimal image problem has been treated by Linton in [10]. Pech and Reichard [13] apply techniques similar to Linton’s to enumerate orbit representatives of subsets of $\Omega$ under the action of a permutation group on $\Omega$. Linton gives a practical algorithm for finding the smallest image of a set under the action of a given permutation group. Our new algorithm, inspired by Linton’s work, is designed to find canonical images: we extend and generalize Linton’s technique using a new orbit-based counting technique. In this paper we first introduce some notation and explain the concepts that go into the algorithm, then we prove the necessary results and finish with experiments that demonstrate how this new algorithm is superior to the previously published techniques.

2 Minimal and Canonical Images

Throughout this paper, $\Omega$ will be a finite set, $G$ a subgroup of $Sym(\Omega)$, and $\Omega$ will be ordered by some (not necessarily total) order $\leq$. If $\alpha\in\Omega$, then we denote the orbit of $\alpha$ under $G$ by $\alpha^{G}$. Simlarly, if $A\subseteq\Omega$ and $g\in G$, then $A^{g}:=\{a^{g}\mid a\in A\}$ and $A^{G}:=\{A^{g}\mid g\in G\}$.

In this paper, we want to efficiently solve the problem of deciding, given two subsets $A,B\subseteq\Omega$, if $A\in B^{G}$. We do this by defining a canonical image:

Definition 2.1**.**

A canonical labelling function $C$ for the action of $G$ on a set $\Omega$ is a function $C:\mathcal{P}(\Omega)\rightarrow\mathcal{P}(\Omega)$ such that, for all $A\subseteq\Omega$, it is true that

• (i)

$C(A)\in A^{G}$, and

• (ii)

${C(A^{g})}=C(A)$* for all $g\in G$.*

In this situation we call $C(A)$ the canonical image of $A\subseteq\Omega$ (with respect to $G$ in this particular action).

Further, we say that $g_{A}\in G$ is a canonizing element for $A$ if and only if $A^{g_{A}}=C(A)$.

A canonical image can be seen as a well-defined representative of a $G$-orbit on $\Omega$ with respect to the defined action. While in this paper we will only consider the action of $G$ on a set of subsets of $\Omega$, canonical images are defined similarly for any group and action. In practice we want to be able to find canonical images effectively and efficiently. In some situations we are interested in computing the canonizing element, which might not be uniquely determined. Our algorithms will always produce a canonizing element as a byproduct of search. We choose to make this explicit here to make the exposition clearer.

Minimal images are a special type of canonical image.

Remark 2.2**.**

Suppose that $\preccurlyeq$ is a partial order on $\Omega$ such that any two elements in the same orbit can be compared by $\preccurlyeq$.

Let $\operatorname{Min}_{\preccurlyeq}$ denote the function that, for all $\omega\in\Omega$, maps $\omega$ to the smallest element in its orbit. Then $\operatorname{Min}_{\preccurlyeq}$ is a canonical labelling function.

In practical applications we are interested in more structure, namely in structures that $G$ can act on naturally via the action on a given set $\Omega$. These structures include subsets of $\Omega$, graphs with vertex set $\Omega$, sets of maps with domain or range $\Omega$, and so on.

In this paper, our main application will be finding canonical images when acting on a set of subsets of $\Omega$.

Definition 2.3**.**

Suppose that $\leq$ is a total order of $\Omega$. Then we introduce a total order $\preccurlyeq$ on $\mathcal{P}(\Omega)$ as follows:

We say that $A$ is less than $B$ and write $A\preccurlyeq B$ if and only if $A$ contains an element $a$ such that $a\notin B$ and $a\leq b$ for all $b\in B\setminus A$.

Example 2.4**.**

Let $\Omega:=\{1,2,3,4,5,6,7\}$ with the natural order and let $A:=\{1,3,4\}$, $B:=\{3,5,7\}$, $C:=\{3,6,7\}$, $D:=\{1,3\}$ and $E:=\{2\}$.

Now $A\preccurlyeq B$, because $1\in A$, $1\notin B$ and $1$ is smaller than all the elements in $B$, in particular those not in $A$. Moreover $A\preccurlyeq C$ for the same reason. Furthermore, $B\preccurlyeq C$, because $5\in B$, $5\notin C$, and if we look at $C\setminus B$, then this only contains the element $6$ and $5$ is smaller.

Next we consider $A$ and $D$. As $4\in A\setminus D$ and $D\setminus A=\varnothing$, we see that $A\preccurlyeq D$. Also $A\preccurlyeq E$ because $1\in A\setminus E$ and $1$ is smaller than all elements in $E\setminus A=E$. Finally $E\preccurlyeq B$ because $2\in E\setminus B=E$ and $2$ is smaller than all elements in $B\setminus E=B$.

Remark 2.5**.**

The example illustrates that this new order introduced above reduces to lexicographical order for sets of the same size. But for sets of different sizes, it might seem counter-intuitive. Our reason for choosing this different ordering is that it satisfies the following property:

If $n\in\mathbb{N}$ and if $A$ and $B$ are sets of integers, then $A\cap\{1,\dots,n\}<B\cap\{1,\dots,n\}$ implies $A<B$. This means that, when building $A$ and $B$ incrementally, we know the order of $A$ and $B$ as soon as we find the first integer that is contained in one of the sets but not in the other. This is not true for lexicographic ordering of sets, as $\{1\}<\{1,2\}$ but $\{1,1000\}>\{1,2,1000\}$.

If $G$ is a subgroup of $Sym(\Omega)$ and $\omega\in\Omega$, then we denote by $G_{\omega}$ the point stabilizer of $\omega$ in $G$. For distinct elements $x,y\in\Omega$, we denote by $G_{x\mapsto y}$ the set of all elements of $G$ that map $x$ to $y$. This set may be empty.

We remark that the above information is readily available from a stabilizer chain for the group $G$, which can be calculated efficiently. For further details we refer the reader to [5]. We now introduce some notation and then prove a basic result about cosets.

Definition 2.6**.**

Let $G$ be a permutation group acting on a totally ordered set $(\Omega,\leq)$, and let $\preccurlyeq$ denote the induced ordering as explained in Definition 2.3. Let $H$ be a subgroup of $G$ and $S\subseteq\Omega$. Then we define the minimal image of $S$ under $H$ to be the smallest element in the set $\{S^{h}\mid h\in H\}$ with respect to $\preccurlyeq$.

In order to simplify notation, we will from now on write $\leq$ for the induced order and then we write $\operatorname{Min}(H,S,\leq)$ for the minimal image of $S$ under $H$.

Lemma 2.7**.**

Let $G$ be a permutation group acting on a totally ordered set $(\Omega,\leq)$, and let $H$ be a subgroup of $G$ and $S\subseteq\Omega$. Then the following hold for all $x,y\in\Omega$:

1. (i)

For all $\sigma\in H_{x\mapsto y}$ it is true that $\sigma\cdot H_{y}=H_{x\mapsto y}=H_{x}\cdot\sigma$. 2. (ii)

If $\sigma\in H_{x\mapsto y}$, then $\operatorname{Min}(\sigma\cdot H,S,\leq)=\operatorname{Min}(H,S^{\sigma},\leq)$.

Proof.

If $\sigma\in H_{x\mapsto y}$, then multiplication by $\sigma$ from the right or left is a bijection on $H$, respectively. For all $\alpha\in H_{x}$ we have that $\alpha\cdot\sigma$ maps $x$ to $y$ and for all $\beta\in H_{y}$ we see that $\sigma\cdot\beta$ also maps $x$ to $y$. This implies the first statement.

For (ii) we just look at the definition: $\operatorname{Min}(\sigma\cdot H,S,\leq)$ denotes the smallest element in the set $\{S^{\sigma\cdot h}\mid h\in H\}$ and $\operatorname{Min}(H,S^{\sigma},\leq)$ denotes the smallest element in the set $\{(S^{\sigma})^{h}\mid h\in H\}$, which is the same set. ∎

2.1 Worked Example

We will find minimal, and later canonical, images using similar techniques to Linton in [10]. This algorithm splits the problem into small sub-problems, by splitting a group into the cosets of a point stabilizer. We will begin by demonstrating this general technique with a worked example.

Example 2.8**.**

In the following example we will look at $\Omega=\{1,2,3,4,5,6\}$, the subgroup $G=\langle(14)(23)(56),(126)\rangle\leq S_{6}$, and $S=\{2,3,5\}$. We intend to find the minimal image $\operatorname{Min}(G,S,\leq)$, where the ordering on subsets of $\Omega$ is the induced ordering from $\leq$ on $\Omega$ as explained in Definition 2.3.

We split our problem into pieces by looking at cosets of $G_{1}=\langle(3,4,5)\rangle$. The minimal image of $S$ under $G$ will be realized by an element contained in (at least) one of the cosets of $G_{1}$, so if we find the minimal image of $S$ under elements in each coset, and then take the minimum of these, we will find the global minimum.

Lemma 2.7 gives that, for all $g\in G$, it holds that $\operatorname{Min}(g\cdot G_{1},S,\leq)=\operatorname{Min}(G_{1},S^{g},\leq)$, and so we can change our problem from looking for the minimal image of $S$ with respect to cosets of $G_{1}$ to looking at images of $S^{g}$ under elements of $G_{1}$ where $g$ runs over a set of coset representatives of $G_{1}$ in $G$.

For each $i\in\{1,\dots,6\}$ we need an element $g_{i}\in G_{i\mapsto 1}$ (where any exist), so that we can then consider $S^{g_{i}}$.

We choose the elements id, $(162)$, $(146523)$, $(14)(23)(56)$, $(142365)$ and $(126)$ and obtain six images of $S$:

[TABLE]

As we are looking at the images of these sets under $G_{1}$, we know that all images of a set containing $1$ will contain $1$, and all images of a set not containing $1$ will not contain $1$. From Definition 2.3, all subsets of $\{1,\dots,6\}$ containing $1$ are smaller than all subsets not containing $1$. This means that we can filter our list down to $\{1,3,5\},\{3,1,2\}$ and $\{3,6,1\}$.

Furthermore, $G_{1}$ fixes $2$, so by the same argument we can filter our list of sets not containing $2$, leaving only $\{3,1,2\}$. The minimal image of this under $G_{1}$ is clearly $\{3,1,2\}$ (in this particular case we could of course also have stopped as soon as we saw $\{3,1,2\}$, as this is the smallest possible set of size 3).

Now, let us consider what would happen if the ordering of the integers was reversed, so we are looking for $\operatorname{Min}(G,S,\geq)$, again with the induced ordering.

For the same reasons as above, we begin by calculating $G_{6}=\langle(3,5,4)\rangle$ and by finding images of $S$ for some element from each coset of $G_{6}$ in $G$.

An example of six images is

[TABLE]

We can ignore anything that does not contain $6$, so we are left with:

[TABLE]

As $5$ is not fixed by $G_{6}$, we can not reason about the presence or absence of $5$ in our sets. There is an image of every set that contains $5$, and there are even two distinct images of $\{6,5,3\}$ that contain $5$. Therefore we must continue our search by considering $G_{6,5}$.

Application of an element from each coset of $G_{6,5}$ to $S$ generates nine sets, of which four contain the element $5$. In fact we reach $\{6,3,5\},\{6,5,4\}$ from the set $\{6,4,3\}$, we reach $\{5,6,1\}$ from the set $\{4,6,1\}$ and we reach $\{5,2,6\}$ from the set $\{3,2,6\}$. From these we extract the minimal image $\{6,5,4\}$.

In this example, different orderings of $\{1,2,3,4,5,6\}$ produced different sized searches, with different numbers of levels of search required.

3 Minimal Images under alternative orderings of $\Omega$

As was demonstrated in Example 2.8, the choice of ordering of the set our group acts on influences the size of the search for a minimal image. In this section we will show how to create orderings of $\Omega$ that, on average, reduce the size of search for a minimal image.

We begin by showing how large a difference different orderings can make. We do this by proving that, for any choice $\preccurlyeq$ of ordering of $\Omega$, group $G$ and any input set $S$, we can construct a minimal image problem that is as hard as finding $\operatorname{Min}(G,S,\preccurlyeq)$, but where reversing the ordering on $\Omega$ makes the problem trivial.

We make this more precise: Given $n\in\mathbb{N}$, a permutation group $G$ on $\{1,\dots,n\}$ with some ordering $\leq$ and a subset $S\subseteq\{1,\dots,n\}$, we construct a group $H$ and a set $T$ such that $\operatorname{Min}(G,S,\leq)=\operatorname{Min}(H,T,\leq)\cap\{1..n\}$, which shows that finding $\operatorname{Min}(H,T,\leq)$ is at least as hard as finding $\operatorname{Min}(G,S,\leq)$. On the other hand, we will show that $\operatorname{Min}(H,T,\geq)=T$ and that this can be deduced without search. This is done in Lemma 3.5. An example along the way will illustrate the construction.

Definition 3.1**.**

We fix $n\in\mathbb{N}$ and we let $k\in\mathbb{N}$. For all $j\in\mathbb{N}$ we define $q(j)\in\mathbb{N}$ (where $q$ stands for “quotient”) and $r(j)\in\{1,\dots,n\}$ (where $r$ stands for “remainder”) such that $j=q(j)\cdot n+r(j)$.

Let $\operatorname{ext}:G\rightarrow S_{k\cdot n}$ be the following map: For all $g\in G$ and all $j\in\{1,\dots,k\cdot n\}$, the element $\operatorname{ext}(g)$ maps $j$ to $q(j)\cdot n+r(j)^{g}$.

Example 3.2**.**

Let $n=4$ and $G=S_{4}$. Then we extend the action of $G$ to the set $\{1,\dots,12\}$ using the map $\operatorname{ext}$.

For example $g=(134)$ maps $4$ to $1$. We write $12=2\cdot 4+4$ and then it follows that $\operatorname{ext}(g)$ maps $12$ to $2\cdot 4+4^{g}=8+1=9$. In fact $g$ acts simultaneously on the three tuples $(1,2,3,4)$, $(5,6,7,8)$ and $(9,10,11,12)$ as it does on $(1,2,3,4)$.

Definition 3.3**.**

Fixing $n,k\in\mathbb{N}$ and a subgroup $G$ of $S_{n}$, and using the map $\operatorname{ext}$ defined above, we say that $H$* is the extension of $G$ on $\{1,\dots,k\cdot n\}$** if and only if $H=\{\operatorname{ext}(g)\mid g\in G\}$ is the image of $G$ under the map $\operatorname{ext}$.*

The extension $H$ of $G$ on a set $\{1,\dots,k\cdot n\}$ is a subset of $S_{k\cdot n}$. We show now that even more is true:

Lemma 3.4**.**

Let $n,k\in\mathbb{N}$ and $G\leq S_{n}$. Then the extension of $G$ onto $\{1,\dots,k\cdot n\}$ is a subgroup of $S_{k\cdot n}$ that is isomorphic to $G$.

Proof.

Let $H:=\operatorname{ext}(G)$ be the image of $G$ under the map $\operatorname{ext}$ and let $a,b\in G$ be distinct. Then let $j\in\{1,\dots,n\}$ be such that $j^{a}\neq j^{b}$. By definition $\operatorname{ext}(a)$ and $\operatorname{ext}(b)$ map $j$ in the same way that $a$ and $b$ do, so we see that $\operatorname{ext}(a)\neq\operatorname{ext}(b)$. Hence the map $\operatorname{ext}$ is injective. Therefore $\operatorname{ext}:G\rightarrow H$ is bijective.

Next we let $a,b\in G$ be arbitrary and we let $j\in\{1,\dots,k\cdot n\}$. Then the composition $ab$ is mapped to $\operatorname{ext}(ab)$, which maps $j$ to $q(j)\cdot n+r(j)^{ab}$. Now $r(j)^{ab}=(r(j)^{a})^{b}$ and therefore the composition $\operatorname{ext}(a)\operatorname{ext}(b)\in S_{k\cdot n}$ maps $j$ to $(q(j)\cdot n+r(j)^{a})^{\operatorname{ext}(b)}=q(j)\cdot n+(r(j)^{a})^{b}$. This is because $r(j)^{a}\in\{1,\dots,n\}$.

Hence $\operatorname{ext}(ab)=\operatorname{ext}(a)\operatorname{ext}(b)$. That implies $\operatorname{ext}$ is a group homomorphism and hence that $G$ and its image are isomorphic. ∎

Lemma 3.5**.**

Let $n\in\mathbb{N}$ and $G\leq S_{n}$. Let $H$ denote the extension of $G$ on $\{1,\dots,(n+1)\cdot n\}$ and let $S\subseteq\{1,\dots,n\}$. Let $T:=S\cup\{l\cdot n+l\mid l\in\{1..n\}\}$, let $\leq$ denote the natural ordering of the integers, and let $\geq$ denote its reverse. For simplicity we use the same symbols for the ordering induced on $\mathcal{P}(\Omega)$, respectively. Then

• (i)

$\operatorname{Min}(H,T,\leq)\cap\{1,\dots,n\}=\operatorname{Min}(G,S,\leq)$.

• (ii)

$\operatorname{Min}(H,T,\geq)=T$.

Proof.

Let $h\in H$. Then by construction $h$ stabilizes the partition

[TABLE]

Moreover, for all $i\in\{1,\dots,n\}$ and $g\in G$ we have that $i^{g}=i^{\operatorname{ext}(g)}$ and so Lemma 3.4 implies that

[TABLE]

This proves the first statement.

For the second statement we notice that $(n+1)\cdot n$ is now the smallest element of $T$, and it cannot be mapped to anything smaller, because it also is the smallest element available. So if we let $h\in H$ be such that $T^{h}=\operatorname{Min}(H,T,\geq)$, then $h$ fixes the point $n(n+1)$. By definition of the extension, it follows that $h$ also fixes $k\cdot n$ for all $k\in\{1\ldots n\}$. The next point of $T$ under the ordering is $n^{2}-1$. It cannot be mapped by $h$ to $n^{2}$, because $n^{2}$ is already fixed, hence $h$ has to fix $n^{2}-1$, too.

Arguing as above it follows that all points are fixed by $h$, thus in particular $\operatorname{Min}(H,T,\geq)=T$, as stated. Furthermore, any algorithm that stepped through the elements of $T$ in the order we describe would find this smallest element without having to perform a branching search, as at each step there is no choice on which element of $T$ is the next smallest. ∎

3.1 Comparing Minimal Images Cheaply

We describe some important aspects of Linton’s algorithm for computing the minimal image of a subset of $\Omega$.

Definition 3.6**.**

Suppose that $(\Omega,\leq)$ is a totally ordered set and that $G\leq Sym(\Omega)$. Then $\operatorname{Orb}(G)$ denotes the list of orbits of $G$ on $\Omega$. This list of orbits is ordered with respect to the smallest element in each orbit under $\leq$.

A $G$-orbit will be called a singleton if and only if it has size $1$.

If $S\subseteq\Omega$, then we say that a $G$-orbit is empty in S if and only if it is disjoint from $S$ as a set, and we say that it is full in S if and only if it is completely contained in $S$.

Example 3.7**.**

Let $\Omega:=\{1,\dots,8\}$, with the natural ordering on the integers, let $G:=\langle(1,4),(2,8),(5,6),(7,8)\rangle$ and let $S:=\{1,3,5,6\}$.

Then $\operatorname{Orb}(G)=[\{1,4\},\{2,7,8\},\{3\},\{5,6\}]$ because this list contains all the $G$-orbits and they are ordered by the smallest element in each orbit, namely $1$ in the first, $2$ in the second, $3$ in the third, which is a singleton, and $5$ in the last (because $4$ is already in an earlier orbit).

The orbits $\{3\}$ and $\{5,6\}$ are full in $S$, the orbit $\{2,7,8\}$ is empty in $S$ and $\{1,4\}$ is neither.

Lemma 3.8**.**

Suppose that $(\Omega,\leq)$ is a totally ordered finite set and that $G\leq Sym(\Omega)$. If $\operatorname{Min}(G,S,\leq)=\operatorname{Min}(G,T,\leq)$ and $\omega\in\Omega$, then $\omega^{G}$ is empty in $S$ if and only if it is empty in $T$, and $\omega^{G}$ is full in $S$ if and only if it is full in $T$.

Proof.

Let $\omega\in\Omega$ and suppose that $\omega^{G}$ is empty in $S$. As $\omega^{G}$ is closed under the action of $G$, and $S_{0}:=\operatorname{Min}(G,S,\leq)$ is an image of $S$ under the action of $G$, we see that $\omega^{G}$ is empty in $S_{0}$ and hence in $T_{0}:=\operatorname{Min}(G,T,\leq)$. Thus $\omega^{G}$ is empty in $T$, which is an image of $T$ under the action of $G$. The same arguments work vice versa.

Next we suppose that $\omega^{G}$ is full in $S$. Then it is full in $S_{0}=T_{0}$ and hence in $T$, and the same way we see the converse. ∎

We can now prove Theorem 3.9, which provides the main technique used to reduce search. This allows us to prove that the minimal image of some set $S$ will be smaller than or equal to the minimal image of a set $T$, without explicitly calculating the minimal image of either $S$ or $T$.

Theorem 3.9**.**

Suppose that $G$ is a permutation group on a totally ordered finite set $(\Omega,\leq)$ and that $S$ and $T$ are two subsets of $\Omega$ where $|S|=|T|$.

Suppose further that $o$ is the first orbit in the list $\operatorname{Orb}(G)$ that is neither full in both $S$ and $T$ nor empty in both $S$ and $T$. If $o$ is empty in $T$, but not in $S$, then $\operatorname{Min}(G,S,\leq)$ is strictly smaller than $\operatorname{Min}(G,T,\leq)$.

Proof.

Suppose that $o$ is empty in $T$, but not in $S$. Then $o$ is empty in $T_{0}:=\operatorname{Min}(G,T,\leq)$, but not in $S_{0}:=\operatorname{Min}(G,S,\leq)$, and in particular $T_{0}$ and $S_{0}$ are distinct, as we have seen in Lemma 3.8.

Let $\alpha$ denote the minimum of the orbit $o$ with respect to $\leq$ and let $\omega\in\Omega$. If $\omega<\alpha$, then $\omega\notin o$, so the the orbit $\omega^{G}$ appears in the list $\operatorname{Orb}(G)$ before $o$ does. Then the choice of $o$ implies that one of the following two cases holds:

1. (i)

$\omega^{G}$ is full in both $S$ and $T$. In particular, for all $g\in G$ we have that $\omega\in S^{g}\cap T^{g}$. 2. (ii)

$\omega^{G}$ is empty in both $S$ and $T$. In particular, for all $g\in G$ we have that $\omega^{G}\cap S^{g}=\varnothing$ and $\omega^{G}\cap T^{g}=\varnothing$.

If $S_{0}$ contains an element $\omega\in\Omega$ such that $\omega<\alpha$, then Case (i) above holds and $\omega\in T_{0}$. So $S_{0}\cap\{\omega^{\prime}\in\Omega\mid\omega^{\prime}<\alpha\}=T_{0}\cap\{\omega^{\prime}\in\Omega\mid\omega<\alpha\}$.

Since $S_{0}$ and $T_{0}$ are distinct, they must differ amongst the elements at least as large as $\alpha$ and, since they have the same cardinality, the smallest such element determines which of $S_{0}$ and $T_{0}$ is smaller.

We recall that $o=\alpha^{G}$ is empty in $T$ and non-empty in $S$, so there exists some $g\in G$ such that $\alpha\in S^{g}$. Then $S^{g}=S_{0}$ and $\alpha\notin T_{0}$, so $S_{0}$ is strictly smaller that $T_{0}$. ∎

Here is an example how to use Theorem 3.9.

Example 3.10**.**

Let $\Omega:=\{1,\dots,10\}$ with natural ordering, and let $G:=\langle(12),(45),(56),(89)\rangle$. We consider the sets $S:=\{3,6,7\}$ and $T:=\{3,7,9\}$ and we want to calculate the smallest of $\operatorname{Min}(G,S)$ and $\operatorname{Min}(G,T)$. Hence, we want to know which one is smaller as cheaply as possible, to avoid superfluous calculations.

We first list the orbits of $G$: $[\{1,2\},\{3\},\{4,5,6\},\{7\},\{8,9\},\{10\}]$.

Going through the orbits as listed, we see that the first one is empty in $S$ and $T$, the second one is full in $S$ and $T$, and the third one gives a difference for the first time. It is empty in $T$, but not in $S$, so Theorem 3.9 yields that the minimal image of $S$ is strictly smaller than that of $T$.

3.2 Static Orderings of $\Omega$

In this section we look at which total ordering of $\Omega$ should be used to minimize the amount of time taken to find minimal images of subsets of $\Omega$.

Given a group $G$ we will choose an ordering on $\Omega$ such that orbits with few elements appear as early as possible. In particular, singleton orbits should appear first.

This is justified by the fact that singleton orbits are always either full or empty. Also, we would expect smaller orbits to be more likely to be empty or full than larger orbits. This means that small orbits placed early in the ordering of $\Omega$ are more likely to lead to Theorem 3.9 being applicable, leading to a reduction in search.

Algorithm 1 heuristically chooses a new ordering for an ordered set $\Omega$, only depending on the group $G$, under the assumption that the algorithm that computes minimal images will pick a point from a smallest non-singleton orbit to branch on. This will not always be true – in practice Linton’s algorithm branches on the first orbit which contains some point contained in one of the current candidates for minimal image.

However, we will show that in Section 5 that Algorithm 1 produces substantially smaller, and therefore faster, searches in practice.

It is not necessary in Line 8 of Algorithm 1 to choose the smallest element of $\mathit{Points}$, choosing an arbitrary element will, on average, perform just as well. By fixing which point is chosen, we ensure that independent implementations will produce the same ordering and therefore the same canonical image.

We will also consider one simple modification of Algorithm 1, namely FixedMaxOrbit (which is the same as FixedMinOrbit) with line 6 changed to pick orbits of maximum size.

If our intuition about Theorem 3.9 is correct, then MaxOrbit should almost always produce a larger search than MinOrbit or a random ordering of $\Omega$.

3.3 Implementing alternative orderings of $\Omega$

Having calculated an alternative $Order$ using FixedMinOrbit or FixedMaxOrbit, we could create a version of MinimalImage which accepted an explicit ordering. However, rather than editing the algorithm, we can instead perform a pre-processing step, using Lemma 3.11.

Lemma 3.11**.**

Consider a group $G$ that acts on $\Omega=\{1,\dots,n\}$ and a permutation $\sigma\in Sym(\Omega)$. We define an ordering $\leq_{\sigma}$ on $\{1,\dots,n\}$, where for all $x,y\in\Omega$ we have that $x\leq_{\sigma}y$ if and only if $x^{\sigma}\leq y^{\sigma}$.

For the induced orderings $\preccurlyeq$ and $\preccurlyeq_{\sigma}$ on subsets of $\Omega$ as in Definition 2.3 it holds that

[TABLE]

for all subsets $X$ and $Y$ of $\Omega$, and hence (simplifying notation)

[TABLE]

Proof.

Following Definition 2.3, $X\preccurlyeq_{\sigma}Y$ if and only if there is an $x\in X$ such that $x\not\in Y$ and for all $y\in Y\backslash X$ it holds that $x\leq_{\sigma}y$. By definition of $\leq_{\sigma}$, this is the case whenever $x^{\sigma}\leq y^{\sigma}$, and since $x^{\sigma}\in X^{\sigma}$ and for all $y^{\sigma}$ in $Y^{\sigma}\backslash X^{\sigma}$ it holds that $x^{\sigma}\leq y^{\sigma}$, it follows that $X^{\sigma}\preccurlyeq Y^{\sigma}$.

Consider the map $\varphi_{\sigma}:S^{G}\rightarrow(S^{\sigma})^{G^{\sigma}}$ that maps sets $X\in S^{G}$ to $X^{\sigma}\in(S^{\sigma})^{G^{\sigma}}$. This map is bijective, and by the above it respects the ordering, so the second claim follows.

∎

Lemma 3.11 gives an efficient method to calculate minimal images under different orderings without having to alter the underlying algorithm. The most expensive part of this algorithm is calculating $G^{\sigma}$, but this is still very efficiently implemented in systems such as GAP, and also can be cached so it only has to be calculated once for a given $G$ and $\sigma$.

4 Dynamic Ordering of $\Omega$

In Section 3.2, we looked at methods for choosing an ordering for $\Omega$ that allows a minimal image algorithm to search more quickly. There is a major limitation to this technique – it does not make use of the sets whose canonical image we wish to find.

In this section, instead of producing an ordering ahead of time, we will incrementally define the ordering of $\Omega$ as the algorithm progresses. At each stage we will consider exactly which extension of our partially constructed ordering will lead to the smallest increase in the number of sets we must consider.

We are not free to choose our ordering arbitrarily as we must still map two sets in the same orbit of $G$ to the same canonical image. However, we can use different orderings for sets that are in different orbits of $G$.

Firstly, we will explain how we build the orderings that our algorithm uses.

4.1 Orderings

When building canonical images, we build orderings as the algorithm progresses. We represent these partially built orderings as ordered partitions.

Definition 4.1**.**

Let $k\in\mathbb{N}$ and let $P=[X_{1},\dots,X_{k}]$ be an ordered partition of $\mathcal{P}(\Omega)$. Then, given two subsets $S$ and $T$ of $\Omega$, we write $S<_{P}T$ if and only if the cell that contains $S$ occurs before the cell that contains $T$ in $P$.

We say that $P$ is $G$-invariant if and only if for all $i\in\{1,\dots,k\}$ and $g\in G$ it holds that $S\in X_{i}$ if and only if $S^{g}\in X_{i}$.

A refinement of an ordered partition $[X_{1},\dots,X_{k}]$ is an ordered partition $[Y_{1,1},Y_{1,2},\dots,Y_{k,l}]$ where $l\in\mathbb{N}$ and such that, for all $i\in\{1,\dots,k\}$ and $j\in\{1,\dots,l\}$, we have that $Y_{i,j}\subseteq X_{i}$.

A completion of an ordered partition $X=[X_{1},\dots,X_{k}]$ is a refinement where every cell is of size one. Given an ordering for $\Omega$, the standard completion of an ordered partition $X$ orders the members of each cell of $X$ using the ordering on sets from Definition 2.3.

In our algorithm we need a completion of an ordered partition, but the exact completion is unimportant – it is only important that, given an ordered partition $X$, we always return the same completion. For this reason we define the standard completion of an ordered partition.

Example 4.2**.**

Let $G:=\langle(12)\rangle\leq S_{3}$ and $\Omega:=\{1,2,3\}$. Moreover let

$P:=[\{\},\{1\},\{2\},\{1,3\},\{2,3\}\mid\{3\},\{1,2\},\{1,2,3\}]$* be an ordered partition of $\mathcal{P}(\Omega)$.*

The orbits of $G$ on $\Omega$ are $\{1,2\}$ and $\{3\}$. In particular all elements of $G$ stabilize the partition $P$.

The ordered partition $Q:=[\{1,3\},\{2\}\mid\{\},\{1\},\{2,3\}\mid\{3\},\{1,2\},\{1,2,3\}]$ is a refinement of $P$ that is not $G$-invariant.

To see this, we let $g:=(12)\in G$. We have that $\{1,3\}$ is in the first cell of $Q$, but $\{1,3\}^{g}=\{2,3\}$ is not in the first cell.

In Example 4.2 we only considered a very small group, because the size of $\mathcal{P}(\Omega)$ is $2^{|\Omega|}$. In practice we will not explicitly create ordered partitions of $\mathcal{P}(\Omega)$, but instead store a compact description of them from which we can deduce the cell that any particular set is in.

In this paper, we will consider two methods of building and refining ordered partitions. We first define the orbit count of a set, which we will use when building refiners.

Definition 4.3**.**

Let $G$ be a group acting on an ordered set $\Omega$, and $S\subseteq\Omega$. Define the orbit count of $S$ in $G$, denoted $\operatorname{Orbcount}(G,S)$ as follows: Given the list $\operatorname{Orb}(G)$ of orbits of $G$ on $\Omega$ sorted by the their smallest member, the list $\operatorname{Orbcount}(G,S)$ contains the size of the intersection $|o\cap S|$ in place of $o\in\operatorname{Orb}(G)$.

We will see the practical use of $\operatorname{Orbcount}$ in Lemma 5.1.

Lemma 4.4**.**

Suppose that $(\Omega,\leq)$ is a totally ordered finite set and that $G\leq Sym(\Omega)$. Suppose further that $S,T\subseteq\Omega$ and that there is some $g\in G$ such that $S^{g}=T$. Then $\operatorname{Orbcount}(G,S)=\operatorname{Orbcount}(G,T)$.

Proof.

Let $o\in\operatorname{Orb}(G)$ and $g\in G$ with $S^{g}=T$. Then $o^{g}=o$ and $\alpha\in(o\cap S)$ if and only if $\alpha^{g}\in(o\cap S)^{g}$, if and only if $\alpha^{g}\in(o^{g}\cap S^{g})=(o\cap T)$.

∎

Definition 4.5**.**

Let $P$ be an ordered partition of $\mathcal{P}(\Omega)$.

• (i)

If $\alpha\in\Omega$, then the point refinement of $P$ by $\alpha$ is the ordered partition $Q$ defined in the following way: Each cell $X_{i}$ of $P$ is split into two cells, namely the cell $\{S\mid S\in X_{i},\alpha\in S\}$, and the cell $\{S\mid S\in X_{i},\alpha\not\in S\}$. If one of these sets is empty, then $X_{i}$ is not split.

• (ii)

If $G\leq Sym(\Omega)$ and $C=\operatorname{Orbcount}(G,T)$ for some set $T\subseteq\Omega$, then the orbit refinement of $P$ by $C$ is the ordered partition $Q$ defined as follows: Each cell $X_{i}$ of $P$ is split into two cells, namely $\{S\ \mid S\in X_{i},\operatorname{Orbcount}(G,S)=C\}$, and $\{S\ \mid S\in X_{i},\operatorname{Orbcount}(G,S)\neq C\}$. If one of these sets is empty, $X_{i}$ is not split.

4.2 Algorithm

We will now present our algorithm. First, we give a technical definition which will be used in proving the correctness of our algorithm.

Definition 4.6**.**

For all $n\in\mathbb{N}$ we define $\mathcal{L}_{n}$ to be the set of lists of length $n$ whose entries are non-empty subsets of $\Omega$. If $X\in\mathcal{L}_{n}$, then as a convention we write $X_{1},\dots,X_{n}$ for the entries of the list $X$.

If $X\in\mathcal{L}_{n}$ and $H\leq G\leq S_{n}$ is such that $|G:H|=k\in\mathbb{N}$ and $Q=\{q_{1},\dots,q_{k}\}$ is a set of coset representatives of $H$ in $G$, then we define $X^{Q}$ to be the list whose first $k$ entries are $X_{1}^{q_{1}},\dots,X_{1}^{q_{k}}$, followed by $X_{2}^{q_{1}},\dots,X_{2}^{q_{k}}$ until the last $k$ entries are $X_{n}^{q_{1}},\dots,X_{n}^{q_{k}}$. We note that $X^{Q}\in\mathcal{L}_{n\cdot k}$.

Let $X,Y\in\mathcal{L}_{n}$. We say that $X$ and $Y$ are $G$-equivalent if and only if there exist a permutation $\sigma$ of $\{1,\dots,n\}$ and group elements $g_{1},\dots,g_{n}\in G$ such that, for all $i\in\{1,\dots,n\}$, it holds that $Y_{i}=X_{i^{\sigma}}^{g_{i}}$.

We now prove a series of three lemmas about coset representatives, which form the basis for the correctness proof of our algorithm. They are used to perform the recursive step, moving from a group to a subgroup.

Lemma 4.7**.**

Suppose that $G$ is a permutation group on a set $\Omega$, that $H$ is a subgroup of $G$ of index $k\in\mathbb{N}$ and that $T$ is a set of left coset representatives of $H$ in $G$.

Then the following are true:

1. (i)

$|T|=k$. 2. (ii)

If $T=\{t_{1},\dots,t_{k}\}$ and $g\in G$ and if, for all $i\in\{1,\dots,k\}$, we define $q_{i}:=gt_{i}$, then $Q:=\{q_{1},\dots,q_{k}\}$ is also a set of coset representatives of $H$ in $G$. In particular there is a bijection from $Q$ to any set of left coset representatives of $H$ in $G$.

Proof.

By definition the index of $H$ in $G$ is the number of (left or right) cosets of $H$ in $G$.

For the second statement we let $i,j\in\{1,\dots,k\}$ be such that $q_{i}H=q_{j}H$, hence $gt_{i}H=gt_{j}H$. Then $t_{j}^{-1}t_{i}=t_{j}^{-1}g^{-1}gt_{i}\in H$ and hence $t_{i}H=t_{j}H$. Hence $i=j$ because $t_{i}$ and $t_{j}$ are from a set of coset representatives. ∎

Lemma 4.8**.**

Suppose that $G$ is a permutation group on a set $\Omega$ and that $S,T\subseteq\Omega$ are such that the lists $[S]$ and $[T]$ are $G$-equivalent.

Let $H$ be a subgroup of $G$ of index $k\in\mathbb{N}$ and let $P=\{p_{1},\dots,p_{k}\}$ and $Q:=\{q_{1},\dots,q_{k}\}$ be sets of left coset representatives of $H$ in $G$.

Then $[S]^{P}$ and $[T]^{Q}$ are $H$-equivalent.

Proof.

As $[S]$ and $[T]$ are $G$-equivalent, we know that there exists a group element $g\in G$ such that $S^{g}=T$.

We fix $g$, for all $i\in\{1,\dots,k\}$ we let $t_{i}:=gq_{i}$ and we consider the set $T:=\{t_{i}\mid i\in\{1,\dots,k\}\}$. Then $T$ is also a set of left coset representatives of $H$ in $G$, by Lemma 4.7. As $P$ is also a set of left coset representatives, we know that $T$ and $P$ have the same size, so there is a bijection from $P$ to $T$. This can be expressed in the following way:

There is a permutation $\sigma\in S_{k}$ such that, for all $i\in\{1,..,k\}$, it is true that $p_{i^{\sigma}}H=t_{i}H$. That means there is a unique $h_{i}\in H$ such that $p_{i^{\sigma}}h_{i}=t_{i}$.

Let now $S_{i}:=S^{p_{i}}$ and $T_{i}:=T^{q_{i}}$ for $i\in\{1,\dots,k\}$, then

[TABLE]

hence $[S]^{P}$ and $[T]^{Q}$ are $H$-equivalent. ∎

Lemma 4.9**.**

Suppose that $G$ is a permutation group on a set $\Omega$, that $n\in\mathbb{N}$ and that $X,Y\in\mathcal{L}_{n}$ are $G$-equivalent. Let $H$ be a subgroup of $G$ of index $k\in\mathbb{N}$ and let $P=\{p_{1},\dots,p_{k}\}$ and $Q:=\{q_{1},\dots,q_{k}\}$ be sets of left coset representatives of $H$ in $G$.

Then $X^{P}$ and $Y^{Q}$ are $H$-equivalent.

Proof.

As $X$ and $Y$ are $G$-equivalent, we know that there exist a permutation $\sigma\in S_{n}$ and $g_{1},\dots,g_{n}\in G$ such that $Y_{i}=X_{i^{\sigma}}^{g_{i}}$ for all $i\in\{1,\dots,n\}$. We fix this permutation $\sigma$.

If $i\in\{1,\dots,n\}$, then $[X_{i^{\sigma}}]$ and $[Y_{i}]$ satisfy the hypothesis of Lemma 4.8, so it follows that $[X_{i^{\sigma}}]^{P}$ and $[Y_{i}]^{Q}$ are $H$-equivalent.

So we find a permutation $\alpha_{i}\in S_{k}$ and group elements $h_{i1},\dots,h_{ik}\in H$ such that $(X_{i^{\sigma}}^{p_{j^{\alpha_{i}}}})^{h_{ij}}=Y_{i}^{q_{i}}$ for all $j\in\{1,\dots,k\}$.

Using $\sigma$ and $\alpha_{1},\dots,\alpha_{k}$ we define a permutation $\gamma$ on $1,\dots,n\cdot k$.

First we express $l\in\{1,\dots,n\cdot k\}$ uniquely as $l=c_{l}\cdot k+r_{l}$ where $c_{l}\in\{0,\dots,n-1\}$ and $r_{l}\in\{1,\dots,k\}$ and we define

[TABLE]

This is well-defined because of the ranges of $c_{l}$ and $r_{l}$ and it is a permutation because of the uniqueness of the expression and because $\sigma$ and $\alpha_{1},\dots,\alpha_{n}$ are permutations.

Then, for each $l\in\{1,\dots,n\cdot k\}$, expressed as $l=c_{l}\cdot k+r_{l}$ as we did above, we set $h:=h_{c_{l}+1,r_{l}}$, $X^{\prime}_{l}:=X_{c_{l}+1}^{p_{r_{l}}}$ and $Y^{\prime}_{l}:=Y_{c_{l}+1}^{q_{r_{l}}}$.

Then $X^{P}=[X^{\prime}_{1},\dots,X^{\prime}_{n\cdot k}]$ and $Y^{Q}=[Y^{\prime}_{1},\dots,Y^{\prime}_{n\cdot k}]$.

If we set $p:=p_{r_{l}^{a_{c_{l}+1}}}$ and $q:=q_{r_{l}}$, then we have, for all $l=c_{l}\cdot k+r_{l}$, that

[TABLE]

This is $H$-equivalence. ∎

We can now describe the algorithm we use to compute canonical images, and prove that it works correctly.

Definition 4.10**.**

Suppose that $\Omega$ is a finite set, that $G$ is a permutation group on $\Omega$, that $L\in\mathcal{L}_{k}$ and that $P$ is an ordered partition on $\mathcal{P}({\Omega})$.

• (i)

An $\Omega$-selector is a function $\mathcal{S}$ such that

• (a)

$\mathcal{S}(\Omega,G,L,P)=\omega\in\Omega$, where $|\omega^{G}|>1$;

• (b)

$\mathcal{S}(\Omega,G,L,P)=\mathcal{S}(\Omega,G,M,P)$* whenever $L$ and $M$ from $\mathcal{L}_{k}$ are $G$-equivalent.*

• (ii)

An Ordering refiner is a function $\mathcal{O}$ such that for all $G$-invariant partitions $P$ of $\mathcal{P}{({\Omega})}$

• (a)

$\mathcal{O}(\Omega,G,L,P)=P^{\prime}$, where $P^{\prime}$ is a $G$-invariant refinement of $P$;

• (b)

$\mathcal{O}(\Omega,G,L,P)=\mathcal{O}(\Omega,G,M,P)$* whenever $L$ and $M$ from $\mathcal{L}_{k}$ are $G$-equivalent.*

An ordering refiner cannot return a total ordering, unless $G$ acts trivially, because the partial ordering cannot distinguish between values that are contained in the same orbit of $G$.

Our method for finding canonical images is outlined in Algorithm 2. It recursively searches for the minimal image of a collection of lists, refining the ordering that is used as search progresses.

Theorem 4.11**.**

Suppose that $\Omega$ is a finite set, that $G$ is a permutation group on $\Omega$, and that $X\subseteq\Omega$. Then $\textsc{CanImageBase}(\Omega,G,X)\in X^{G}$.

Proof.

In every step of Algorithm 2 the list of considered sets is a list of elements of $X^{G}$. ∎

Theorem 4.12**.**

Let $\Omega$ be a finite set and $G$ a permutation group on $\Omega$. Let $X,Y\in\mathcal{L}_{k}$ be $G$-equivalent, and let $P$ be a $G$-invariant ordered partition of $\mathcal{P}(\Omega)$. Then

[TABLE]

Proof.

We proceed by induction on the size of $G$.

The base case is $|G|=1$. As $X$ and $Y$ are $G$-equivalent, $X$ and $Y$ contain the same sets, possibly in a different order. For a given $P$ and $\Omega$, there is only one standard completion of $P$ that gives a complete ordering on $\mathcal{P}(\Omega)$, and so $X$ and $Y$ have the same smallest element under the standard completion of $P$ and so the claim follows.

Consider now any non-trivial group $G$, and suppose for our induction hypothesis that the claim holds for all groups $H$ where $|H|<|G|$.

Definition 4.10 and the fact that $\mathcal{S}$ is an $\Omega$-selector imply that

[TABLE]

Moreover, $H$ is a proper subgroup of $G$. We take two sets $Q_{1}$ and $Q_{2}$ of coset representatives of $H$ in $G$, which are not necessarily equal.

Since $\mathcal{O}$ is an ordering refiner, it holds that

[TABLE]

By Lemma 4.9, $X^{Q_{1}}$ and $Y^{Q_{2}}$ are $H$-equivalent, and by definition $P^{\prime}$ is $G$-invariant. If we identify the cell $P^{\prime}_{i}$ of $P^{\prime}$ that contains the smallest element of $X^{Q_{1}}$, then $L^{\prime}_{X}$ contains those elements of $X^{Q_{1}}$ that are in $P^{\prime}_{i}$. Each of these elements is $H$-equivalent to an element of $Y^{Q_{2}}$, and therefore $L^{\prime}_{X}$ is $H$-equivalent to $L^{\prime}_{Y}$. Then the induction hypothesis yields that

[TABLE]

so the claim follows by induction. ∎

We see from Theorem 4.11 and 4.12 that

[TABLE]

We note that Algorithm 2 can easily be adapted to return an element $g$ of $G$ such that $X^{g}=\textsc{CanImage}(\Omega,G,X,P)$. This happens by attaching to each set, when it is created, the permutation that maps it to the original input $S$. We omit this addition for readability.

5 Experiments

In this section, we will compare how well our new algorithms perform in comparison with the MinImage function of Linton’s.

All of our experiments are performed in GAP [3] using code available in the Images package [7]. Where our algorithm requires it, we use the implementation of partition backtracking provided in the Ferret package [6].

We consider a selection of different canonical image algorithms and we analyze how they perform compared to each other, and compared to the traditional minimal image algorithm of Linton’s, which we will refer to as MinImage.

The first three algorithms that we consider come from Section 3. They produce, given a group $G$ on a set $\Omega$, an ordering of $\Omega$. This ordering is then used in MinImage.

1. (i)

FixedMinOrbit uses results from Section 3 to calculate an alternative ordering of $\Omega$, choosing small orbits first. 2. (ii)

FixedMaxOrbit works similarly to FixedMinOrbit, choosing large orbits first.

We also consider algorithms that dynamically choose which value to branch on as search progresses. We will use the following lemma for the proof of correctness for all our orderings.

Lemma 5.1**.**

Let $\Omega$ be a finite set, $G$ a permutation group on $\Omega$, and $P$ an ordered partition of $\mathcal{P}(\Omega)$.

Let $L\in\mathcal{L}_{n}$, and $\operatorname{Count}=[\operatorname{Orbcount}(G,S)\mid S\in L]$.

• (i)

Any function that accepts $(\Omega,G,L,P)$ and returns some $\omega\in\Omega$ with $|\omega^{G}|>1$ that is invariant under reordering the elements of $\operatorname{Count}$, is an $\Omega$-selector.

• (ii)

Any function that accepts $(\Omega,G,L,P)$ and returns either $P$ or the point refinement of $P$ by some $\omega^{G}$ for $\omega\in\Omega$, and is invariant under permutation of the elements of $\operatorname{Count}$, is an ordering refiner.

• (iii)

Any function that accepts $(\Omega,G,L,P)$ and returns either $P$ or the orbit refiner of $P$ by some member of $\operatorname{Count}$ and is invariant under permutation of the elements of $\operatorname{Count}$, is an ordering refiner.

Proof.

The only thing we have to show is that, if $L$ and $M$ are $G$-equivalent, then any of the functions above yield the same result for inputs $(\Omega,G,L,P)$ and $(\Omega,G,M,P)$. By Lemma 4.4, any $G$-equivalent lists $L$ and $M$ will produce the same list $\operatorname{Count}$, up to reordering of elements, and hence the claim follows. ∎

Firstly we define a list of orderings. Each of these orderings chooses an orbit, or list of orbits, to branch on – we will then make an $\Omega$-selector by choosing the smallest element in any of the orbits selected, to break ties (we could choose any point, as long as we picked it consistently). Each of these algorithms operates on a list $L\in\mathcal{L}_{k}$. In each case we look for an orbit, ignoring orbits of size one (as fixing a point that was already fixed leads to the same group).

Firstly we will consider two algorithms that only consider the group, and not $L$:

1. (i)

MinOrbit Choose a point from a shortest non-trivial orbit that has a non-empty intersection with at least one element of $L$. 2. (ii)

MaxOrbit Choose a point from a longest non-trivial orbit that has a non-empty intersection with at least one element of $L$.

We also consider four algorithms that consider both the group, and $L$.

In the following, for an orbit $o$

1. (i)

RareOrbit minimises $\sum\limits_{s\in L}\mid s\cap o|$, 2. (ii)

CommonOrbit maximises $\sum\limits_{s\in L}\mid s\cap o|$, 3. (iii)

RareRatioOrbit minimises ${\log(\sum\limits_{s\in L}\mid s\cap o|)}/{|o|}$, 4. (iv)

CommonRatioOrbit maximises ${\log(\sum\limits_{s\in L}\mid s\cap o|)}/{|o|}$.

The motivation for RareOrbit is that this is the branch which will lead to the smallest size of the next level of search – this exactly estimates the size of the next level if our ordering refiner only fixed a point in orbit $o$. We therefore expect, conversely, CommonOrbit to perform badly and to produce very large searches.

One limitation of RareOrbit is that it will favour smaller orbits – in general we want to minimize the size of the whole search. The idea here is that if we have two levels of search where we split on an orbit of size two, and each time create 10 times more sets is equivalent to splitting once on an orbit of size four, and creating 100 times more sets. RareRatioOrbit compensates for this. We expect that CommonRatioOrbit is the inverse of this, so we also expect it to perform badly.

For each of these orderings, we use the ordering refiner that takes each fixed point of $G$ in their order in $\Omega$, and performs a point refinement by each recursively in turn. By repeated application of Lemma 4.4, this is a $G$-invariant ordering refiner.

We also have a set of orderings which make use of orbit counting. To keep the number of experiments under control, we used the RareOrbit strategy in each case to choose which point to branch on next, and we also build an orbit refiner.

Given an unordered list of orbit counts,

1. (i)

RareOrbitPlusMin chooses the lexicographically smallest one. 2. (ii)

RareOrbitPlusRare chooses the least frequently occurring orbit count list (using the lexicographically smallest to break ties). 3. (iii)

RareOrbitPlusCommon chooses the most frequently occurring orbit count list (using the lexicographically smallest to break ties).

5.1 Experiments

In this section we perform a practical comparison of our algorithms, and the MinImage algorithm of Linton, for three different families of problems: grid groups, $m$-sets, and primitive groups.

5.2 Experimental Design

We will consider three sets of benchmarks in our testing. In each experiment, given a permutation group that acts on $\{1,\dots,n\}$, we will run an experiment with each of our orderings to find the canonical image of a set of size $\left\lfloor\frac{n}{2}\right\rfloor,\left\lfloor\frac{n}{4}\right\rfloor$ and $\left\lfloor\frac{n}{8}\right\rfloor$.

We run our algorithms on a randomly chosen conjugate of each primitive group, to randomize the initial ordering of the integers the group is defined over. The same conjugate is used of each group in all experiments, and when choosing a random subset of size $x$ from a set $S$, we always choose the same random subset. We use a timeout of five minutes for each experiment. We force GAP to build a stabilizer chain for each of our groups before we begin our algorithm, because this can in some cases take a long time.

For each size of set and each ordering, we measure three things. The total number of problems solved, the total time taken to solve all problems, counting timeouts as 5 minutes, and the number of moved points of the largest group solved. Our experiments were all performed on an Intel(R) Xeon(R) CPU E5-2640 v4 running at 2.40GHz, with twenty cores. Each copy of GAP was allowed a maximum of 6GB of RAM.

5.2.1 Grid Groups

In this experiment, we look for canonical images of sets in grid groups.

Definition 5.2**.**

Let $n\in\mathbb{N}$. The direct product $S_{n}\times S_{n}$ acts on the set $\{1,\ldots,n\}\times\{1,\ldots,n\}$ of pairs in the following way:

For all $(i,j)\in\{1,\ldots,n\}\times\{1,\ldots,n\}$ and all $(\sigma,\tau)\in S_{n}\times S_{n}$ we define

[TABLE]

The subgroup $G\leq Sym(\{1,\ldots,n\}\times\{1,\ldots,n\})$ defined by this action is called the $n\times n$* grid group**.*

We note that, while the construction of the grid group is done by starting with an $n$ by $n$ grid of points and permuting rows or columns independently of each other, we actually represent this group as a subgroup of $S_{n\cdot n}$, and we do not assume prior knowledge of the grid structure of the action.

We ran experiments on the grid groups for grids of size $5\times 5$ to $100\times 100$. The results of this experiment are given in Table 1.

The basic algorithm, MinImage, is only able to solve $22$ problems within the timeout. FixedMinOrbit solves $43$ problems, while being implemented as a simple pre-processing step to MinImage. The dynamic MinOrbit is able to solve $55$ problems, and the best orbit-based strategy, SingleMaxOrbit, solves $70$ problems. However the advanced techniques, which filter by orbit lists, perform much better. Even ordering by the most common orbit list leads to solving over $185$ problems, and the best strategy, RareOrbitPlusRare, solves $235$ out of the total $288$ problems.

Furthermore, for these large groups, the algorithms are still performing very small searches: For example FixedMinOrbit, on its largest solved problem with size $\left\lfloor\frac{n}{2}\right\rfloor$ sets and a grid of size $12\times 12$, generates $793,124$ search nodes, while RareOrbitPlusMin produces only $183,579$ search nodes on its largest solved problem with $\left\lfloor\frac{n}{2}\right\rfloor$ sets ($65\times 65$).

5.2.2 M-Sets

Linton [10] considers, given integers $n$ and $m$, defining a permutation group on the set $T$ of all subsets of size $m$ of $\{1,\dots,n\}$ under the action on $T$ of $S_{n}$ acting on the members of the $m$-sets. He then looks for minimal images of randomly chosen subsets of $T$ of size $k$, under the standard lexicographic ordering on sets.

We ran experiments for $m=2$ and $n\in\{10,15,\dots,100\}$, for $m=4$ and $n\in\{10,15,\dots,35\}$, for $m=6$ and $n\in\{10,15,20\}$ and finally $m=8$, $n=10$ as described in Section 5.2. We choose these 30 experiments as these were the problems that any of our techniques were able to solve in under 5 minutes. The results of our experiments are shown in Table 2.

Similarly to our experiments on grid groups, we find that the standard MinImage algorithm is only able to solve a very small set of benchmarks. Some of the better algorithms, including FixedMinOrbit, are able to solve $9$ problems. In particular, once again MinOrbit is not significantly better than FixedMinOrbit, although it is slightly faster on average over all problems.

However, the orbit-based strategies do much better, solving all the problems which we set. In the case of sets that contain an eighth of all $m$-sets, the best technique is able to solve all problems any technique can solve, in under 5 minutes. The largest solved problem, which was instance $n=35,m=4$ for a set on an eighth of all $m$-sets, is solved in only $6,594$ search nodes by RareOrbitPlusMin, while the largest solved problem of MinImage, $n=15,m=4$ takes $631,144$ search nodes.

5.2.3 Comparison to Graph Canonical Image

A set of $2$-sets can be viewed as an undirected graph, where the two sets represent the edges. The problem of finding the canonical image of this set of $2$-sets is equivalent to the traditional problem of finding a canonical image of this graph. We can therefore perform a comparison between our technique and Nauty, for these problems. Nauty is able to a find canonical image for all our $2$-set problems almost instantly. We investigated why Nauty was able to outperform us by such a large margin, and found three problems. We list the most important one first.

• (i)

The central algorithm of Nauty makes use of properties of the form “vertices with $i$ neighbors can only map to other vertices with $i$ neighbors”. Our algorithm does not make use of this property, as it represents a much more complex condition when considered on $m$-sets. Further, while we could add a special case specifically for when the group we are considering is the symmetric group operating on $m$-sets, we would prefer to find a more general technique.

• (ii)

Our algorithm spends a large proportion of its time calculating stabilizer chains, and mapping sets through elements of the group. This is not required for the graphs.

• (iii)

Our algorithm is written in GAP rather than highly optimized C.

The most important results to draw from this comparison is that our algorithms should not be viewed as a replacement for graph isomorphism algorithms. We are investigating how to close this performance gap, without special casing.

5.2.4 Primitive Groups

In this experiment we look for canonical images of sets under the action of primitive groups which move between $2$ and $1,000$ points. We remove the natural alternating and symmetric groups, as finding minimal and canonical images in these groups is trivial and can be easily special-cased. So we look at a total number of $5,845$ groups, each of which was successfully treated by at least one algorithm.

We perform the experiment as described in Section 5.2. The results are given in Table 3.

All algorithms are able to solve a large number of problems. This is unsurprising as many primitive groups are quite small (for example the cyclic groups), so any technique is able to brute-force search many problems. However, we can still see that, for the hardest problems $\left\lfloor\frac{n}{2}\right\rfloor$, many algorithms outperform MinImage, and the techniques that use extra orbit-counting filtering solve 300 more problems, and they run much faster.

For the easiest set of problems, $\left\lfloor\frac{n}{8}\right\rfloor$, we see that the algorithm RareOrbitPlusMin, which usually performs best, solves slightly fewer problems. This is because there are a small number of groups where the extra filtering provides no search reduction, but still requires a small overhead in time. However, the total time taken is still much smaller, and the algorithm only fails to solve five problems. These five problems involve groups that are isomorphic to the affine general linear groups $\operatorname{AGL}(8,2)$, $\operatorname{AGL}(6,3)$ and $\operatorname{AGL}(9,2)$, and the projective linear group $\operatorname{PSL}(9,2)$. This suggests that the linear groups may be a source of hard problems for canonical image algorithms in the future.

5.2.5 Experimental Conclusions

Our experiments show that using FixedMinOrbit is almost always superior to MinImage. As implementing FixedMinOrbit requires a fairly small amount of code and time over MinImage, this suggests that any implementations of Linton’s algorithm should have FixedMinOrbit added, because this provides a substantial performance boost, for relatively little extra coding.

Algorithms that dynamically order the underlying set, such as MinOrbit and RareOrbit provide only a small benefit over FixedMinOrbit. Algorithms which add orbit counting provide a much bigger gain, often allowing solving problems on groups many orders of magnitude larger than before, thereby greatly advancing the state of the art.

6 Conclusions

We present a general framework and a new set of algorithms for finding the canonical image of a set under the action of a permutation group. Our experiments show that our new algorithms outperform the previous state of the art, often by orders of magnitude.

Our basic framework runs on the concept of refiners and selectors and is not limited to finding only canonical images of subsets of $\Omega$. In future work we will investigate families of refiners and selectors that allow finding canonical images for many other combinatorial objects.

Acknowledgements.

All authors thank the DFG (Wa 3089/6-1) and the EPSRC CCP CoDiMa (EP/M022641/1) for supporting this work. The first author would like to thank the Royal Society, and the EPSRC (EP/M003728/1). The third author would like to acknowledge support from the OpenDreamKit Horizon 2020 European Research Infrastructures Project (#676541). The first and third author thank the Algebra group at the Martin-Luther Universität Halle-Wittenberg for the hospitality and the inspiring environment. The fourth author wishes to thank the Computer Science Department of the University of St Andrews for its hospitality during numerous visits.