The Complexity of Approximately Counting Retractions to Square-Free Graphs
Jacob Focke, Leslie Ann Goldberg, Stanislav \v{Z}ivn\'y
TL;DR
This paper classifies the computational complexity of approximately counting retractions and homomorphisms in square-free graphs, revealing a rich structure with some problems being as hard as counting independent sets.
Contribution
It provides a complete complexity classification for approximately counting retractions to square-free graphs, including new IS-easiness results for homomorphisms.
Findings
Identifies a class of graphs where counting retractions is IS-complete.
Extends easiness results from retractions to homomorphisms.
Settles the complexity of counting homomorphisms for several non-trivial graph classes.
Abstract
A retraction is a homomorphism from a graph to an induced subgraph of that is the identity on . In a long line of research, retractions have been studied under various algorithmic settings. Recently, the problem of approximately counting retractions was considered. We give a complete trichotomy for the complexity of approximately counting retractions to all square-free graphs (graphs that do not contain a cycle of length ). It turns out there is a rich and interesting class of graphs for which this problem is complete in the class . As retractions generalise homomorphisms, our easiness results extend to the important problem of approximately counting homomorphisms. By giving new -easiness results we now settle the complexity of approximately counting homomorphisms for a whole class of non-trivial graphs which were previously unresolved.
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The Complexity of Approximately Counting Retractions to Square-Free Graphs
Jacob Focke, Leslie Ann Goldberg and Stanislav Živný
The research leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP7/2007-2013) ERC grant agreement no. 334828 and under the European Union’s Horizon 2020 research and innovation programme (grant agreement No 714532). Jacob Focke has received funding from the Engineering and Physical Sciences Research Council (grant ref: EP/M508111/1). Stanislav Živný was supported by a Royal Society University Research Fellowship. The paper reflects only the authors’ views and not the views of the ERC or the European Commission. The European Union is not liable for any use that may be made of the information contained therein.
(24 February 2021)
Abstract
A retraction is a homomorphism from a graph to an induced subgraph of that is the identity on . In a long line of research, retractions have been studied under various algorithmic settings. Recently, the problem of approximately counting retractions was considered. We give a complete trichotomy for the complexity of approximately counting retractions to all square-free graphs (graphs that do not contain a cycle of length ). It turns out there is a rich and interesting class of graphs for which this problem is complete in the class . As retractions generalise homomorphisms, our easiness results extend to the important problem of approximately counting homomorphisms. By giving new -easiness results we now settle the complexity of approximately counting homomorphisms for a whole class of non-trivial graphs which were previously unresolved.
1 Introduction
A function that maps the vertices of a graph to the vertices of a graph is a homomorphism from to if preserves the edges of , i.e. if for every pair of adjacent vertices we have . It is well-known that homomorphisms represent graph-theoretic structures including proper vertex colourings and independent sets. For example, consider the graphs , and given in Figure 1.
A homomorphism from a graph to corresponds to an independent set in , whereas a homomorphism from to corresponds to a proper -colouring of . Finally, a homomorphism from to corresponds to a -colouring of that uses the colours red (), blue (), green () and yellow (), but for which - and -coloured edges are forbidden.
If for each vertex we specify a so-called “list” and set , then is a homomorphism from to if it is a homomorphism from to such that for all it holds that . This generalisation of a homomorphism is known as a list homomorphism. For example one could consider list homomorphisms from the Petersen graph (see Figure 2 on the left) to the graph . These homomorphisms then correspond to proper -colourings of the Petersen graph, where some vertices have pre-assigned colours.
If every list contains either just a single vertex or all of the vertices of , then such a list homomorphism is called a retraction. This definition of a retraction is equivalent to the definition from the abstract, where a retraction was defined as a homomorphism from a graph to an induced subgraph of that is the identity on . The equivalence (in the sense of parsimonious polynomial-time interreducibility) was shown by Feder and Hell [15, Theorem 4.1]. The intuition behind the equivalence is that one can identify all of the vertices in that have the same single-vertex list with the corresponding vertex of . Homomorphisms from this new graph to are retractions in the sense of the abstract. For the Petersen graph in our example, the modified graph is displayed in Figure 2 on the right and contains as an induced subgraph.
Retractions are also known under the names one-or-all list homomorphisms (e.g. [15, 16]) and pre-colouring extensions (e.g. [3, 5, 41, 44, 52, 46, 17]). Related work on retractions is described in Section 1.1.
In the study of approximate counting there are three important classes of problems [12]: (1) problems that have fully-polynomial-time randomised approximation schemes (FPRASes), (2) problems that are approximation-equivalent to , the problem of counting independent sets in a bipartite graph, and (3) problems that are approximation-equivalent to , the problem of counting satisfying assignments to a Boolean formula (these problems have no FPRAS, unless ). It is believed that these three classes are disjoint, so there are no FPRASes for the -equivalent and -equivalent problems. The problems that are interreducible with under approximation-preserving (AP-)reductions are complete in a complexity class which is sometimes called and is sometimes just called . For convenience we say that a graph is -easy or -hard if the problem of approximately counting retractions to is -easy or -hard, respectively. We similarly use the terms -easy, and -hard.
In this work we give a complete complexity trichotomy for approximately counting retractions to all square-free graphs (graphs that do not contain a -cycle) and we show that all of these problems fall within the three given complexity classes. An interesting feature is that the class of -equivalent graphs turns out to be surprising and rich.
First we give some illustrative examples. Afterwards we describe the class of -equivalent graphs in detail. A key idea that emerges in the proofs is the role of triangles (3-cycles). It turns out that triangles in graphs can induce hardness, but they can also “turn” -hard cases into -easy ones. For example, consider Figure 3. The graph on the left was shown to be -hard [21, Lemma 2.15]. In comparison, we will show that the graph on the right is actually -easy. Note that this is not because a vertex was added: If one deletes any of the edges of the triangle, the resulting graph is -hard again.
To give an even more striking example, it seems surprising that the graph on the left in Figure 3 is -hard, but the graph depicted in Figure 4 turns out to be -easy.
There exists an interesting underlying balancing process between looped cliques in the neighbourhood of a looped vertex and its number of unlooped neighbours, which decides whether a graph is -easy or -hard.
We now informally define the -easy class (it is defined formally in Definition 10). This class also contains graphs with squares and is more general than what we will need to classify all square-free graphs. A graph in is a path of looped vertices with some attached unlooped degree- vertices (bristles, depicted below the path in Figure 4) and some attached looped vertices forming cliques with two consecutive vertices of (depicted above the path in Figure 4). For each vertex of the path the number of attached bristles satisfies the following properties:
- •
If is an endpoint of then it does not have a bristle.
- •
If is not an endpoint of then it is part of exactly two reflexive cliques (“to the left” of ) and (“to the right” of ). Then has at most (\mathopen{}\mathclose{{}\left|K_{L}}\right|-1)\cdot(\mathopen{}\mathclose{{}\left|K_{R}}\right|-1) bristles.
Consider again the graph in Figure 4. Note that the graph is not square-free as this is not required by the definition of . The third vertex from the right on the path is part of a reflexive -clique to the left and a reflexive -clique to the right. Hence the vertex can have at most bristles (and has in fact only bristles).
Theorem 1**.**
Let be a graph in . Then approximately counting retractions to is -equivalent under approximation-preserving reductions.
is a fairly broad class of graphs but it turns out that approximately counting retractions to any member of is -easy. For the class of square-free graphs, completely captures the truth — together with the class of non-trivial irreflexive caterpillars (defined momentarily) they form precisely the class of -equivalent square-free graphs. Here (following a couple of necessary definitions) is the full complexity classification.
A graph is reflexive if every vertex of is looped and it is irreflexive if it contains only unlooped vertices. A square is a cycle of length . A connected irreflexive graph is a star if it contains at most one vertex of degree greater than , and it is a caterpillar if it contains a path such that all vertices outside of have degree . A graph is trivial if it is a reflexive clique or an irreflexive complete bipartite graph.
Theorem 2**.**
Let be a square-free graph.
- i)
If every connected component of is trivial then approximately counting retractions to is in FP. 2. ii)
Otherwise, if every connected component of is
- •
trivial,
- •
in the class , or
- •
is an irreflexive caterpillar
then approximately counting retractions to is -equivalent. 3. iii)
Otherwise, approximately counting retractions to is -equivalent.
As a second central contribution of this work we emphasise the implications of Theorem 1 on the complexity of approximately counting homomorphisms. This problem has been studied intensively in the past [12, 29, 13, 25, 24, 43, 31, 21], but despite these efforts its complexity is still mostly open. For example, there are graphs with as few as four vertices for which its complexity is unresolved. In a partial result, Galanis, Goldberg and Jerrum [24] proved that approximately counting homomorphisms to a connected graph is -hard unless is an irreflexive complete bipartite graph or a reflexive clique. However, the knowledge about -easiness to complement this result is very fragmented. Since approximately counting homomorphisms reduces to approximately counting retractions ([21, Observation 1.2]), our -easiness results from Theorem 1 carry over to the homomorphism-counting domain. Thus, we are able to resolve the complexity of approximately counting homomorphisms to graphs in .
Corollary 3**.**
Let be a graph in . Then approximately counting homomorphisms to is -equivalent under approximation-preserving reductions.
Remark 4**.**
As a side note, the -easiness result given in Corollary 3 also extends to the problems of approximately counting surjective homomorphisms and approximately counting compactions. This follows from the reductions given in [21, Theorem 1.5].
1.1 Related Work
The concept of a retraction has been studied as early as the 1930’s and originates in work on continuous functions from topological spaces into subspaces [7]. Subsequently, retractions between discrete structures, and graphs in particular, have received a lot of research attention [32, 33, 36, 47, 19]. See [35] for an overview. The algorithmic problem of deciding whether a retraction exists can be expressed naturally as a constraint satisfaction problem (CSP) and has been studied in e.g., [34, 53, 54, 55, 18]. Retractions have also been studied in the context of directed graphs. A work by Larose [45] surveys CSPs with a digraph as the fixed template. The survey has a special focus on CSPs with additional unary constraints, such as the digraph retraction problem. Applications of retractions reach from classical results such as Brouwer’s fixed-point theorem and its equivalent no-retraction theorem [38, pp. 272-273] to specific problems such as solving Sudoku puzzles [37].
The complexity of deciding whether there is a retraction to a fixed graph has been studied in different contexts, such as CSPs and list homomorphisms [15, 16, 35, 18, 34], surjective homomorphisms [53, 54, 55, 4], as well as pre-colouring extensions and scheduling [3, 39, 40, 5, 41, 44, 52, 46, 17]. A complete complexity dichotomy for the retraction decision problem is now known as a consequence of the CSP dichotomy [8, 58] (assuming ). However, a corresponding graph-theoretical characterisation is not known. A characterisation is known [18] for pseudoforests, which are graphs in which each connected component has at most one cycle.
The complexity of exactly counting retractions to a fixed graph is also classified completely [14] (assuming ). In this case, there is a characterisation – if every connected component of is trivial, then counting retractions is in . Otherwise, counting retractions is -complete.
The first result on approximately counting retractions is the following classification for graphs which are both square-free and triangle-free.
Theorem 5** ([21, Theorem 1.1]).**
Let be a graph of girth at least .
- i)
If every connected component of is an irreflexive star, a single looped vertex, or an edge with two loops, then approximately counting retractions to is in . 2. ii)
Otherwise, if every connected component of is an irreflexive caterpillar or a partially bristled reflexive path, then approximately counting retractions to is approximation-equivalent to . 3. iii)
Otherwise, approximately counting retractions to is approximation-equivalent to .
Counting homomorphisms to square-free graphs has been studied before [27, 42] though those results apply to counting modulo a prime number, and so they cannot be applied here.
In the past, graphs without squares have been studied in various other combinatorial settings: For example, [2] investigates the complexity of finding -colourings of square-free graphs. Polarity graphs are another natural class of square-free graphs that have been studied, e.g. [6, 1]. Wrochna showed that square-free graphs are multiplicative [56, 57], solving a problem related to Hedetniemi’s conjecture111This conjecture has been refuted lately [50]. [51, Problem 7.1] and improving earlier results about the multiplicativity of square-free graphs that contain triangles [11]. Furthermore, square-free graphs play an important role in extremal graph theory and the study of the Turán number, e.g. [48, 22, 23, 20, 26, 10]. Recently, it has been shown that certain bounds on the chromatic number with respect to distance-two colouring of planar graphs are unique to square-free graphs [9].
1.2 Preliminaries
For a non-negative integer we use to denote the set . For sets and we define as an unordered version of the Cartesian product. The elements of are multisets of size exactly . Using this notation the set of edges of a graph is a subset of . An edge with two identical elements is a loop. Correspondingly, a vertex is called looped if and unlooped otherwise. The girth of a graph is the length of a shortest cycle in . All cycles have length at least .
We have already defined reflexive and irreflexive graphs. A graph is a mixed graph if it contains both looped and unlooped vertices, i.e. if it is neither reflexive nor irreflexive. Given a graph and a subset of , is the subgraph of induced by .
Given graphs and , is the set of homomorphisms from to and {N\bigl{(}G\rightarrow H\bigr{)}} denotes its size. Analogously, given a corresponding set of lists , is the set of homomorphisms from to and {N\bigl{(}(G,\mathbf{S})\rightarrow H\bigr{)}} denotes its size.
We use to denote the problem of approximately counting retractions to (for a fixed graph which may have loops but does not have multi-edges). We use \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right) to denote the problem of approximately counting homomorphisms to . Formally, these problems are defined as follows.
Name:
.
Input:
An irreflexive graph and a collection of lists such that, for all , \mathopen{}\mathclose{{}\left|S_{v}}\right|\in\{1,\mathopen{}\mathclose{{}\left|V(H)}\right|\}.
Output:
{N\bigl{(}(G,\mathbf{S})\rightarrow H\bigr{)}}.
Name:
\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right).
Input:
An irreflexive graph .
Output:
{N\bigl{(}G\rightarrow H\bigr{)}}.
The list homomorphisms counting problem, defined as follows, is a generalisation of .
Name:
Input:
An irreflexive graph and a collection of lists .
Output:
{N\bigl{(}(G,\mathbf{S})\rightarrow H\bigr{)}}.
If there is an approximation-preserving reduction [12] from a problem to a problem , we write .
Observation 6** ([21, Observation 1.2]).**
Let be a graph. Then \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right)\leq_{\mathrm{AP}}\mathrm{\#\textsc{Ret}}(H)\leq_{\mathrm{AP}}\mathrm{\#\textsc{LHom}}(H).
For approximately counting list homomorphisms Galanis, Goldberg and Jerrum [25] give the following complete classification.
Theorem 7** ([25]).**
Let be a connected graph.
- i)
If is an irreflexive complete bipartite graph or a reflexive complete graph, then is in . 2. ii)
Otherwise, if is an irreflexive bipartite permutation graph or a reflexive proper interval graph, then is approximation-equivalent to . 3. iii)
Otherwise, is approximation-equivalent to .
1.3 Paper Outline
Theorem 1 is proved in Section 2. Theorem 2 is proved in Section 4. The corresponding -easiness follows mainly from Theorem 1. The corresponding -hardness results are collected in Section 3. Proving -hardness is the bulk of this work because of the combinatorial complexity of designing reductions which establish -hardness for all square-free graphs (apart from reflexive cliques, irreflexive caterpillars and those in ).
2 -Easiness Results
In this section we prove Theorem 1, which states that approximately counting retractions to any graph from the class (Definition 10) is -equivalent. The proof is built on a method for generating -easiness results from [21, Section 2.2.1] which uses the framework of constraint satisfaction problems. Intuitively, the method takes as input two CSP instances, say and , and produces a graph for which . The challenge is to find the right instances and and to identify and generate corresponding general classes of -easy graphs. For the convenience of the reader we repeat some definitions introduced in [21]. Let be a set of Boolean relations.
Name:
.
Input:
A set of variables and a set of constraints , where each constraint applies a relation from to a list of variables from .
Output:
The number of assignments that satisfy all constraints in .
is an arity-two Boolean relation. The constraint ensures that, in any satisfying assignment , we have .
Definition 8** ([21, Definition 2.6]).**
Let and be instances of . We define the undirected graph as follows. The vertices of are the satisfying assignments of . Given any assignments and in , there is an edge in if and only if the following holds: For every constraint in , we have and .
Lemma 9** ([21, Lemmas 2.5 and 2.8]).**
Let and be instances of . Then .
Definition 10**.**
A graph is in if it can be defined as follows. For some positive integer , the vertex set is of the form where induce reflexive cliques in , and are disjoint sets of unlooped degree- vertices (called bristles). There are vertices in such that each clique contains both and . The intersection of the cliques is given as follows.
- •
For , .
- •
For with \mathopen{}\mathclose{{}\left|j-i}\right|>1, .
The size of each set of bristles satisfies 0\leq\mathopen{}\mathclose{{}\left|B_{i}}\right|\leq\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left|K_{i-1}}\right|-1}\right)\cdot\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left|K_{i}}\right|-1}\right). Finally, the edge set of is given as follows.
[TABLE]
For an example graph from the class see Figure 5.
Let be as defined in Definition 10. The high-level-structure of the proof of Theorem 1 is as follows. We first define two instances and of , then we establish that is isomorphic to (Lemma 21), which then allows us to apply Lemma 9.
To give more intuition we will use a running example where is the graph depicted in Figure 5. To separate this example from the rest of the proof we use text boxes.
Let be the set of looped vertices in , i.e. and let be a set of Boolean variables. We fix an ordering “” on the vertices of with two properties: (1) In , is the smallest vertex and is the largest; (2) The order of the ’s is respected in the sense that, for any pair of distinct vertices , if there is an such that and , then . We define .
Consider and . The graph is simply a reflexive path on \mathopen{}\mathclose{{}\left|V^{*}}\right| vertices. We will construct from by choosing a subset of — this allows the creation of bristles. Similarly we will construct from by choosing a subset of — this creates reflexive cliques amongst the vertices of the reflexive path in . In order to define and , for , we define the sets of constraints (the constraints that we will delete from to define ) as follows.
[TABLE]
For (i.e. for ), we define the sets of constraints (the constraints we will delete from to define ). The definition of is a bit more involved and uses the following sets:
[TABLE]
In order to model the set of bristles we will “delete” exactly \mathopen{}\mathclose{{}\left|B_{i}}\right| constraints that belong to from . As a means to specify which constraints will be deleted we define an order on the constraints in (which uses the order on the vertices in which we fixed previously). There are several orders which would work.
Definition 11**.**
We define an order “” on . Let . Then if one of the following holds:
- •
.
- •
and .
If and the ordered pair is distinct from the ordered pair then .
Now let be the \mathopen{}\mathclose{{}\left|B_{i}}\right| smallest elements of with respect to as given in Definition 11. Note that by the definition of (Equation (2)) and the bound on \mathopen{}\mathclose{{}\left|B_{i}}\right| (Definition 10) this is well-defined since
[TABLE]
Finally, we define and as follows.
[TABLE]
In our running example, order the variables in from left to right. We have as the set of variables of and . Since \mathopen{}\mathclose{{}\left|B_{1}}\right|=4 the set contains the smallest elements of , which means . Similarly, since \mathopen{}\mathclose{{}\left|B_{2}}\right|=2, the set contains the smallest elements of , which means . Finally, since \mathopen{}\mathclose{{}\left|B_{3}}\right|=1, we have . Thus,
(4)
Regarding the edge constraints we have , , and and hence
(5)
Recall that the satisfying assignments of correspond to the vertices of . For let be the assignment with
[TABLE]
where (i.e. ). Note that is the all-zero assignment since is the minimum vertex in . The reason that we did not introduce a variable for in the definition of is that its role is captured by the all-zero assignment to . We will call assignments of the form path assignments. The path assignments inherit an order from the order on the set that we fixed, i.e. if and only if .
Lemma 12**.**
All path assignments satisfy . If are the path assignments ordered by , then they form a reflexive path in . The reflexive path is not necessarily induced by its vertices.
Proof.
This proof merely requires that and — it does not use the detailed definitions of and .
First, note that the assignments satisfy the -instance . Thus, they will still be satisfying assignments if we delete constraints from .
We now investigate the edges between the vertices of . From Definition 8 recall that, given any satisfying assignments and of , there is an edge in if and only if the following holds:
[TABLE]
Using (6) it can easily be checked that for (and therefore for all ), the vertices form a reflexive path, i.e. these vertices are looped and for and we have that is an edge in . ∎
Lemma 12 shows that even if we were to enforce all constraints in both in the modelling of vertices () and of edges (), the graph always contains a reflexive path on \mathopen{}\mathclose{{}\left|V^{*}}\right| vertices. As noted earlier, is precisely this reflexive path. Our definition of , given in (3), ensures that so it uses a subset of the constraints, which leads to the possibilitiy of more edges in . The idea behind the construction is that a vertex will be modelled by the satisfying assignment , which is a vertex in . The previous lemma shows that these assignments form a reflexive path. We now show how the clique structure is modelled. Intuitively, the deleted constraints allow shortcuts that form cliques along the underlying path of path assignments. The following observation follows immediately from the definition of in (1).
Observation 13**.**
For all , with and all , and consequently (since ).
Lemma 14**.**
Two vertices are adjacent in if and only if is adjacent to in .
Proof.
First we consider the case where are adjacent. Then there exists an index such that . Consequently and . Therefore, for , if then and , and if then and . Let be some constraint in . Then and, by the Definition of in (1), we have at least one of or . Using these facts, we can verify (6) for and which shows that is adjacent to in .
Now assume that and are not adjacent. Then there exist indices from such that , , and . Without loss of generality assume . Then so and . However, by Observation 13, and consequently and are not adjacent in . ∎
In our example, the vertices form a reflexive path (as guaranteed by Lemma 12). Furthermore, is an edge since by (5), and is an edge since by (5). The subgraph of induced by the path assignments is as depicted in Figure 6 and is isomorphic to (as is guaranteed by Lemma 14).
\sigma_{p_{0}}=(0,0,0,0,0,0)$$\sigma_{r_{1}}=(1,0,0,0,0,0)$$\sigma_{p_{1}}=(1,1,0,0,0,0)$$\sigma_{r_{2}}=(1,1,1,0,0,0)$$\sigma_{p_{2}}=(1,1,1,1,0,0)$$\sigma_{p_{3}}=(1,1,1,1,1,0)$$\sigma_{p_{4}}=(1,1,1,1,1,1)
With Lemma 14 we have established the clique structure of the looped vertices. Next we will model the bristles that are attached to the vertices . To do this, we consider satisfying assignments of that are not path assignments, i.e. that are not of the form . These assignments are called bristle assignments.
Definition 15**.**
An assignment is a bristle assignment if and only if there exist with such that and .
Definition 16**.**
For , and let be the assignment with
[TABLE]
where . Hence is of the following form.
[TABLE]
Note that is the minimum index for which takes value [math], and is the maximum index for which takes value . We say that a bristle assignment is good if it is of the form .
Observation 17**.**
Since the following are equivalent:
- (I)
* is a satisfying assignment of .* 2. (II)
For all with and it holds that .
Lemma 18**.**
Every bristle assignment that satisfies is good.
Proof.
The property of that is used in this proof is that and therefore if a constraint from is not in it has to be in one of the sets .
Let be a bristle assignment that satisfies . Let be the minimum index with and let be the maximum index with . Since is a bristle assignment, . Then, since is a satisfying assignment, . Therefore for some , and therefore (by the definition of ), and . This is consistent with Definition 16. We will show that . Let . We investigate the value of depending on to show that takes values as given in (7):
- •
If , by the minimality of we have .
- •
If , then by the choice of .
- •
If , then which implies and consequently, since is in , we have (also using the fact that cannot be in any of the other sets since ). Therefore it holds that .
- •
If , then which implies and consequently, since is in , we have (also using the fact that cannot be in any of the other sets since ). Therefore it holds that .
- •
If , then by the choice of .
- •
If , by the maximality of we have .
∎
We can now check which of the (good) bristle assignments satisfy in the running example. First, consider the set . We already showed, in the text box containing (4), that . From the definition of (Equation (3)),
(i)
.
(ii)
.
(iii)
.
(iv)
The good bristle assignments of the form have and so they are , , and , which are as follows:
\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{x_{r_{1}}} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{x_{p_{1}}} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{x_{r_{2}}} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{x_{p_{2}}}
\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} [math] [math]
\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} [math] [math]
\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} [math] [math]
\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} \color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1} [math] [math]
We now apply Observation 17. From (i) it follows that satisfies . Similarly, from (i) and (ii) it follows that satisfies . From (i) and (iii) it follows that satisfies . Finally, from (i), (ii), (iii) and (iv) it follows that satisfies . The four bristle assignments that we have checked correspond to the four bristles in in Figure 5. Similarly, one can check, from the definitions of and , that , and are the only other satisfying bristle assignments.
Lemma 19**.**
For every bristle assignment that satisfies there exists such that is the only neighbour of in .
Proof.
Here we will use the fact that and the fact that C_{\mathrm{e}}=U\setminus\mathopen{}\mathclose{{}\left(\bigcup_{k=0}^{Q}D_{e}(k)}\right).
Let be a bristle assignment that satisfies . By Lemma 18, is good, i.e. there exist , and such that . We will show that is the only neighbour of in .
Let be some satisfying assignment of which is adjacent to . Note that . This fact together with Observation 13 (which states that for all with we have ) implies that, for all , .
Let be the minimum vertex with . Then and therefore . The choice of ensures that for all vertices , iff . Therefore, if we have (since but and hence ) and consequently . Summarising, we obtain
[TABLE]
Thus, .
It remains to check that and are in fact adjacent. To this end we verify (6):
**Claim: If then . **
Proof of the claim: We check for possible violations. The only relevant and are those for which , and , i.e., all and satisfying . However, constraints of this form are in and hence are not in . (End of the proof of the claim.)
**Claim: If then . **
Proof of the claim: Again we check for violations. The only relevant and are those for which , and , i.e., all and satisfying . However, constraints of this form are in and hence are not in . (End of the proof of the claim.) ∎
Lemma 20**.**
For each there are exactly \mathopen{}\mathclose{{}\left|B_{i}}\right| good bristle assignments that satisfy and are adjacent to in .
Proof.
Every good bristle assignment is of the form for some , and . In the proof of Lemma 19 we have shown that the only neighbour of is . Then the following claim completes the proof of the lemma:
**Claim: satisfies if and only if is among the \mathopen{}\mathclose{{}\left|B_{i}}\right| smallest elements of . **
Proof of the claim: Since and we have .
First consider the case where is one of the \mathopen{}\mathclose{{}\left|B_{i}}\right| smallest elements of . From Observation 17 we know that satisfies if and only if for all with and it holds that . Note that each such is in and by Definition 11, . Thus, and we obtain as required.
Now consider the remaining case where is not among the the \mathopen{}\mathclose{{}\left|B_{i}}\right| smallest elements of . Then, by our choice of , . Moreover, for all , since , and . Hence and consequently does not satisfy by Observation 17. (End of the proof of the claim.) ∎
Lemma 21**.**
Let and . Then is isomorphic to .
Proof.
Here we collect the previous results. By Lemma 12 the path assignments satisfy (and hence are vertices of ). Lemma 14 shows that the subgraph of induced by the path assignments is isomorphic to . By Lemma 18 all other satisfying assignments are good bristle assignments and by Lemma 19 each of these good bristle assignments has a unique neighbour, which is among . Then Lemma 20 shows that there are exactly \mathopen{}\mathclose{{}\left|B_{i}}\right| good bristle assignments adjacent to . ∎
We can now prove Theorem 1, which we re-state at this point for the convenience of the reader.
Theorem 0**.**
Let be a graph in . Then approximately counting retractions to is -equivalent under approximation-preserving reductions.
Proof.
Let . The -easiness part of Theorem 1 follows directly from Lemmas 21 and 9. The -hardness part follows from \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right)\leq_{\mathrm{AP}}\mathrm{\#\textsc{Ret}}(H) (Observation 6) together with the fact that \#\mathrm{\textsc{BIS}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right) for all connected graphs other than reflexive cliques and irreflexive stars [24, Theorem 1]. ∎
In our running example we conclude by showing in Figure 7 how is encoded as . We have already demonstrated which assignments are satisfying and thus are vertices of . Using (5) and (6) it is straightforward to verify (with some work) that each satisfying bristle assignment of the form is in fact adjacent only to .
\sigma_{p_{0}}=(0,0,0,0,0,0)$$\sigma_{r_{1}}=(1,0,0,0,0,0)$$\sigma_{p_{1}}=(1,1,0,0,0,0)$$\sigma_{r_{2}}=(1,1,1,0,0,0)$$\sigma_{p_{2}}=(1,1,1,1,0,0)$$\sigma_{p_{3}}=(1,1,1,1,1,0)$$\sigma_{p_{4}}=(1,1,1,1,1,1)$$\beta_{1}[p_{1},r_{2}]=(\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1,0,1,0},0,0)$$\beta_{1}[r_{1},r_{2}]=(\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0,0,1,0},0,0)$$\beta_{1}[p_{1},p_{2}]=(\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1,0,1,1},0,0)$$\beta_{1}[r_{1},p_{2}]=(\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0,0,1,1},0,0)$$\beta_{2}[p_{2},p_{3}]=(1,1,\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1,0,1,0})$$\beta_{2}[r_{2},p_{3}]=(1,1,\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{0,0,1,0})$$\beta_{3}[p_{3},p_{4}]=(1,1,1,\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\bm{1,0,1})
3 -Hardness Results
From previous results we already know the following:
- •
Theorem 5 classifies the complexity of approximately counting retractions to for all graphs that are both square-free and triangle-free (i.e. have girth at least ).
- •
For irreflexive , the proof of Theorem 5 does not use triangle-freeness — [21, Theorem 2.3] gives a trichotomy for approximately counting retractions to the class of irreflexive square-free graphs.
Therefore, we investigate square-free graphs that contain at least one triangle and at least one looped vertex. It turns out that we have to work through a number of technical cases to cover all -hard graphs with these properties (and hence all -hard square-free graphs).
For a positive integer the graph is a looped star on vertices (the underlying star has degree- vertices). The (reflexive) net is a looped triangle where each vertex of the triangle has an additional looped neighbour, as shown in the following illustration:
Here is an overview of the cases that we consider:
- •
In Section 3.2 we show that mixed triangles induce -hardness.
- •
In Section 3.3 we show in which cases the neighbourhood of a looped vertex induces -hardness.
- •
In Sections 3.4, 3.5 and 3.6 we show -hardness for square-free graphs with an induced , with an induced net and with an induced reflexive cycle of length at least , respectively. Essentially, these three sections deal with the graphs from the excluded subgraph characterisation of reflexive proper interval graphs (see [25, Section 1 and Appendix A] for the details about this characterisation).
3.1 Retractions and Neighbourhoods
Definition 22**.**
For a graph and a vertex we define the (distance-) neighbourhood of as . (In particular, this might include itself.) Then \deg_{H}(v)=\mathopen{}\mathclose{{}\left|\Gamma_{H}(v)}\right| is the degree of . More generally, the distance- neighbourhood of is defined as \Gamma_{H}^{k}(v)=\{u\in V(H)\mid\text{There is a walk }W=u,w_{1},\dots,w_{k-1},v\text{ (on k edges) in }H\}. Let be a subset of . Then is the set of common neighbours of the vertices in .
The following well-known and simple observation shows that, for approximately counting retractions, hardness carries over from subgraphs that are induced by the neighbourhood of a vertex.
Observation 23**.**
Let be a graph and let be a vertex of . Then .
Proof.
Let be an input to , let be the vertices of and . Let be a vertex distinct from the vertices in . Then we construct the graph with vertices and edges . We set , where
[TABLE]
Then {N\bigl{(}(G,\mathbf{S})\rightarrow H[\Gamma_{H}(u)]\bigr{)}}={N\bigl{(}(G^{\prime},\mathbf{S}^{\prime})\rightarrow H\bigr{)}}. ∎
3.2 Square-Free Graphs with Mixed Triangles
Lemma 24**.**
Let be a square-free graph which contains a triangle with exactly two looped and one unlooped vertex. Then .
Proof.
Let be a triangle in , where and are looped and is unlooped. Consider the neighbourhood . Since is square-free, is precisely the triangle . The problem \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H[\{b_{1},b_{2},r\}]}\right) corresponds to counting independent sets where vertices not in the independent set have a weight of and vertices in the independent set have weight . It is well-known that approximately counting weighted independent sets is -hard, see for instance [30, Lemma 2]. This gives \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H[\Gamma_{H}(b_{1})\cap\Gamma_{H}(b_{2})]}\right). From Observation 6 it follows immediately that .
Finally, one can easily observe that : Let be an input to and let . Let and be vertices distinct from the vertices in . Then we construct the graph with vertices and edges E(G^{\prime})=E(G)\cup\bigl{(}V(G)\times\{w_{1},w_{2}\}\bigr{)}. We set , where
[TABLE]
Then {N\bigl{(}(G,\mathbf{S})\rightarrow H[\Gamma_{H}(b_{1})\cap\Gamma_{H}(b_{2})]\bigr{)}}={N\bigl{(}(G^{\prime},\mathbf{S}^{\prime})\rightarrow H\bigr{)}}. ∎
Lemma 25**.**
Let be a square-free graph which contains a triangle with exactly two unlooped and one looped vertex. Then .
Proof.
Let be a triangle in with vertices , and , where is looped and both and are unlooped. Let . By Observation 23 it holds that . Suppose we can show that . Then is a single edge with one looped () and one unlooped vertex () and it is well-known that counting homomorphisms to this graph corresponds to counting independent sets, which in turn is known to be -hard ([12, Theorem 3]). Summarising we have
[TABLE]
where the second reduction is from Observation 6. This proves the lemma. It remains to prove the following claim.
Claim: .
Proof of the claim: For let be the number of common neighbours of and in . Then since and have two common neighbours: and , and these are their only common neighbours as is square-free. Similarly, as the “common” neighbours in this case are simply the neighbours of , which are only and (since is square-free). Now let . The vertex is a common neighbour of and since every vertex in is a neighbour of . It turns out that is the only common neighbour of and : Suppose there exists a vertex in which is a common neighbour of and . If (see Figure 8 on the left) then is adjacent to .
Additionally, is then looped and hence . Then is a square. If otherwise then is a square (see Figure 8 on the right), both cases give a contradiction. So we have shown that, for ,
[TABLE]
Intuitively, we will use this fact to “boost” the vertices and and make them exponentially more likely to be used by a homomorphism to .
Let be the number of vertices of . Now let be an -vertex input to and let be the desired precision. As usual, from we define an input to . We introduce a vertex distinct from the vertices of that will serve as a pin to the vertex in . Then, for each we introduce an independent set on vertices all of which are connected only to and . The parameter will depend on the input size, specifically we set . Intuitively it is clear that this gadget introduces a weight equal to for each vertex . For sufficiently large , the image of is likely to be or . This implies the statement of the lemma. The reader that trusts this intuition can skip reading the following calculations.
We give the full details for the sake of completeness: For each , let be an independent set of size with vertices distinct from the remaining vertices of . Then is the graph with vertices and edges E(G^{\prime})=E(G)\cup\bigcup_{v\in V(G)}\bigl{(}\{v,p\}\times I_{v}\bigr{)}. We set , where
[TABLE]
We say that a homomorphism is full if . Let be the number of full homomorphisms from to . Let be the number of non-full homomorphisms from to . Then
[TABLE]
For , let be the number of homomorphisms for which . By the construction of , every vertex with contributes a factor of to , whereas a vertex with contributes a factor of to . It follows that
[TABLE]
and
[TABLE]
Therefore,
[TABLE]
where the last inequality holds for sufficiently large by the choice . Summarising, by (9) and (10), we have
[TABLE]
and, using (9), (10) as well as (11), we obtain
[TABLE]
Hence {N\bigl{(}(G^{\prime},\mathbf{S}^{\prime})\rightarrow H^{\prime}\bigr{)}}/2^{sn}\in\bigl{[}{N\bigl{(}(G,\mathbf{S})\rightarrow H^{\prime}[\{b,r_{1}\}]\bigr{)}},{N\bigl{(}(G,\mathbf{S})\rightarrow H^{\prime}[\{b,r_{1}\}]\bigr{)}}+1/4\bigr{]}. Let be the solution returned by an oracle call to with input , i.e. an approximation of {N\bigl{(}(G^{\prime},\mathbf{S}^{\prime})\rightarrow H^{\prime}\bigr{)}}. Then the output \mathopen{}\mathclose{{}\left\lfloor Q/2^{sn}}\right\rfloor approximates {N\bigl{(}(G,\mathbf{S})\rightarrow H^{\prime}[\{b,r_{1}\}]\bigr{)}} with the desired precision as was shown in [12, Proof of Theorem 3]. (End of the proof of the claim.) ∎
3.3 Square-Free Neighbourhoods of a Looped Vertex
Now we consider graphs of the form (see Figure 9). Why are we interested in these graphs? Let be a square-free graph with a looped vertex and let not contain any mixed triangle as a subgraph. Then consider , the graph induced by the neighbourhood of . Since does not contain mixed triangles, the unlooped neighbours of do not have any neighbours in apart from . Since is square-free, is square-free as well and therefore a looped neighbour of can have at most one additional neighbour apart from and itself (within ). It follows that is of the form . Note that does not contain any cycles. Therefore the hardness results for graphs of this form come from the classification for graphs of girth at least (Theorem 5). The remaining cases () are covered in this work. As an overview in advance, we will obtain in the following cases (where in most cases we actually show the stronger result \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(X(k_{1},k_{2},k_{3})}\right) — -hardness for then follows from Observation 6.):
- •
and
- –
and (Theorem 5)
- –
and (Theorem 5)
- –
and (Theorem 5)
- •
and
- –
and (Lemma 36)
- –
and (Lemma 37)
- •
, and (Lemma 38)
- •
(Lemma 40)
Following the classification for graphs of the form we give a hardness result (Lemma 39) which uses properties of the distance- neighbourhood of a looped vertex in .
A useful and well-known tool for proving hardness results for approximate counting problems are gadgets based on complete bipartite graphs where two states dominate (see, e.g., [12, Lemma 25], [31, Section 5], [43, Lemma 5.1] and [21, Lemma 2.30]). Let . One can use the described tool to show that, under certain conditions, a homomorphism from a complete bipartite graph to will typically map one side to and the other to . In this case it is then easy to reduce from counting independent sets to obtain -hardness. Formally, we use the version stated by Kelk [43]:
Lemma 26** ([43, Lemma 5.1]).**
Let be a graph with . Suppose that, for every pair with satisfying and , at least one of the following holds:
- (1)
. 2. (2)
. 3. (3)
\mathopen{}\mathclose{{}\left|S}\right|\cdot\mathopen{}\mathclose{{}\left|T}\right|<\mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|.
Then \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right).
In order to prove Lemmas 36 and 37 we will use Lemma 26 and, in addition, a reduction from the problem of counting large cuts, which is formally defined as follows: A cut of a graph is a partition of into two subsets (the order of this pair is ignored) and the size of a cut is the number of edges that have exactly one endpoint in each of these two subsets.
Name:
.
Input:
An integer and a connected graph in which every cut has size at most .
Output:
The number of size- cuts in .
The full details of the proof involve analysing different types of homomorphisms. The most important part of the results leading up to the proof of Lemmas 36 and 37 are the Tables 1 and 2, respectively. These tables show which types of homomorphisms represent a significant share of the overall number of homomorphisms that we are interested in. The crucial question is whether we can ensure that the right types of homomorphisms dominate this number. We desire two properties. First, the number of homomorphisms should be dominated by homomorphisms of two distinct types. Second, these two types should interact in an “anti-ferromagnetic” way.
The following definitions and preliminary technical results resemble the ones introduced in [21, Section 2.2.2]. We are going to use the graph (see Figure 10) as a vertex gadget. This gadget was originally introduced in [12]. In general it is a good candidate when looking for gadgets to prove reductions from . Here is the formal definition.
Let , and be positive integers. Let and be independent sets of size and let and be independent sets of size . The set of edges between and forms a perfect matching. Then is the graph for which the vertex set is the union of , , and . The edges are .
Let be a graph and let be a homomorphism from to . Let . We say that has type \mathopen{}\mathclose{{}\left(h(A),h(B,B^{\prime}),h(A^{\prime})}\right). In general, a tuple is an -type if and . Let , , and .
An -type is non-empty (with respect to ) if there exists a homomorphism from to that has type . Otherwise, is called an empty -type. From the definition of we observe the following.
Observation 27**.**
Let be a graph. An -type is non-empty if and only if
- (1)
, and are non-empty, 2. (2)
, 3. (3)
.
Let and be -types. We write if, for , we have . An -type is maximal if it is non-empty and every -type with , is empty. The following analysis (Lemma 28) is contained in the proof of [21, Lemma 2.25].
Lemma 28**.**
Let be a graph and let be a maximal -type. Then is completely defined by and since
- (1)
, and . 2. (2)
* and .*
Given an -type we define to be the number of homomorphisms in that have type . We also set \displaystyle\widehat{N}(T)=\mathopen{}\mathclose{{}\left|T_{1}}\right|^{pt}\mathopen{}\mathclose{{}\left|T_{2}}\right|^{qt}\mathopen{}\mathclose{{}\left|T_{3}}\right|^{pt}. For non-empty , is a close approximation to :
Lemma 29** ([21, Lemma 2.20]).**
Let be a graph. Let and be positive integers. There exists a positive integer such that for all and all -types that are non-empty with respect to , it holds that
[TABLE]
Lemma 30** ([21, Lemma 2.23]).**
Let be a connected graph with at least vertices. Let be a non-empty -type that is not maximal. Then there exists a non-empty -type such that \widehat{N}(T)\leq\mathopen{}\mathclose{{}\left(\frac{2\mathopen{}\mathclose{{}\left|E(H)}\right|-1}{2\mathopen{}\mathclose{{}\left|E(H)}\right|}}\right)^{t}\widehat{N}(T^{*}).
Let T=\bigl{(}A(T),E(B(T),B^{\prime}(T)),A^{\prime}(T)\bigr{)} be an -type. Then we call symmetric to the -type T^{\prime}=\bigl{(}A^{\prime}(T),E(B^{\prime}(T),B(T)),A(T)\bigr{)}. Clearly, .
Lemma 31**.**
Let . Then all maximal -types are listed in Table 1 (apart from those that are symmetric to a listed -type). For each listed -type the last column of the table gives the corresponding value .
Proof.
Let and let be a maximal -type. We claim that
[TABLE]
for the following reasons (we give the arguments for , they are identical for ):
- •
Since is a neighbour of every vertex in , from item (2) of Lemma 28 we obtain .
- •
If, for some , we have then as is the only neighbour of . By item (2) of Lemma 28 it follows that .
- •
If , then . By item (2) of Lemma 28 this gives , a contradiction.
- •
gives a contradiction in the same way.
Table 1 then lists all possible combinations of sets and . From Lemma 28 it follows that these sets determine completely (and are given accordingly). By definition, and . From item (1) of Lemma 28 it has to hold that . Then \displaystyle\widehat{N}(T)=\mathopen{}\mathclose{{}\left|T_{1}}\right|^{pt}\mathopen{}\mathclose{{}\left|T_{2}}\right|^{qt}\mathopen{}\mathclose{{}\left|T_{3}}\right|^{pt} can be computed from the given sets in each row. ∎
Lemma 32**.**
Let . Then all maximal -types are listed in Table 2 (apart from those that are symmetric to a listed -type). For each listed -type the last column of the table gives the corresponding value .
Proof.
Let and let be a maximal -type. We claim that
[TABLE]
The remainder of the proof is analogous to that of Lemma 31 with only one additional argument:
- •
If and , then . By item (2) of Lemma 28 it follows that . The same is true if both and .
∎
Lemma 33**.**
Let . Consider the types , given by Table 1. Then there is a and positive integers and such that, for all and all positive integers , we have .
Proof.
Let
[TABLE]
and
[TABLE]
For each of the seven possible values of we can check (for example by computer) that every member of is less than . Thus, we can choose and so that
[TABLE]
We check the sought-for bound for each , :
- :
is fulfilled for some sufficiently large if and only if which is equivalent to \displaystyle\log\mathopen{}\mathclose{{}\left((3+k_{1})^{2}/3}\right)/\log\mathopen{}\mathclose{{}\left(9+k_{1}}\right)<q/p. This is true by (12).
- :
is fulfilled for some sufficiently large if and only if . This is true by (12).
- :
is fulfilled for some sufficiently large if and only if . This is true by (12).
- :
is fulfilled for some sufficiently large if and only if . This is true by (12).
- :
is fulfilled for some sufficiently large if and only if . This is true by (12).
∎
Lemma 34**.**
Let . Consider the types , given by Table 2. Then there is a and positive integers and such that, for all and all positive integers , we have .
Proof.
Let
[TABLE]
and
[TABLE]
For each of the four possible values of we can check (for example by computer) that every member of is less than . Thus, we can choose and so that
[TABLE]
Suppose that and are types listed in Table 2 which are distinct from and have the property that for all . Then the sought-for bound automatically holds for if it holds for .
We check the sought-for bound for each , :
- :
is fulfilled for some sufficiently large if and only if . This is true by (13).
- :
.
- :
is fulfilled for some sufficiently large if and only if . This is true by (13).
- :
is fulfilled for some sufficiently large if and only if . This is true by (13).
- :
.
- :
, for all .
- :
is fulfilled for .
- :
is fulfilled for some sufficiently large if and only if . This is true by (13).
- :
is fulfilled for some sufficiently large if and only if . This is true by (13).
∎
Remark 35**.**
We point out that the proofs of Lemmas 33 and 34 break for larger . (For larger there exists some lower bound on which exceeds some upper bound on that ratio.) Lemma 34 also breaks for and . This matches the results from Section 2 which show that approximately counting retractions to is actually -easy for these values of .
Lemma 36**.**
If , then \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(X(k_{1},0,1)}\right).
Proof.
We make a case distinction depending on . The first case is the main work of the proof and we use the dominance of the type from Table 1 for as shown in Lemma 33. The second case () then follows from Lemma 26.
Case 1: . We use a reduction from , which is known to be -hard (see [12]). Let and be an input to , be the number of vertices of and be the parameter of the desired precision. To shorten notation let . From we construct an input to \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right) by introducing vertex and edge gadgets. We assume that the vertices of are labelled as in Figure 11 (on the left).
Let , be positive integers that fulfil (12). Note that and only depend on which is a parameter of and therefore does not depend on the input . We define the parameter of the gadget graph to be . We also define a new parameter .
For each vertex we introduce a vertex gadget which is a graph as given in Figure 10. We denote the corresponding sets by and , respectively. We say that two gadgets and are adjacent if and are adjacent in .
For every edge we introduce an edge gadget as follows. We introduce two size- independent sets, denoted by and . As shown in Figure 12 we construct the set of edges
[TABLE]
Putting the pieces together, is the graph with
[TABLE]
Let be a homomorphism from to , be some vertex of and be the corresponding vertex gadget. Then corresponds to a homomorphism from to and therefore has an -type.
We say that a homomorphism from to is full if its restriction to each vertex gadget is either of type (from Table 1) or of its symmetric type (let us call it ). The cut corresponding to a full homomorphism partitions into those vertices for which has type and those for which has type . We say that a full homomorphism is -large if the size of the corresponding cut is equal to , otherwise we say that the homomorphism is -small. Consider a full homomorphism from to .
- •
For an edge of suppose has type and has type . Note that by the definition of the edge gadget, we have . Then the vertices in can be mapped to any of , whereas all vertices in have to be mapped to (the sole common neighbour of , , and the vertices in ).
- •
Suppose instead that and have the same type or . Then the homomorphism has to map the vertices in both and to .
Thus, every pair of adjacent gadgets of different types contributes a factor of to the number of full homomorphisms, whereas every pair of adjacent gadgets of the same type only contributes a factor of . Recall the definition of as the number of homomorphisms from to that have type . Then for every size- cut of arises in ways as a full homomorphism from to .
Let be the number of solutions to with input and (our goal is to approximate this number). We partition the homomorphisms from to into three different sets. is the number of -large (full) homomorphisms, is the number of homomorphisms that are full but -small and is the number of non-full homomorphisms. Then we have and {N\bigl{(}G^{\prime}\rightarrow H\bigr{)}}=Z^{*}+Z_{1}+Z_{2}. Thus it remains to show that for our choice of , , and . Under this assumption we then have \displaystyle{N\bigl{(}G^{\prime}\rightarrow H\bigr{)}}/(2N(T_{5})^{n}3^{sK})\in[L,L+1/4] and a single oracle call to determine {N\bigl{(}G^{\prime}\rightarrow H\bigr{)}} with precision suffices to determine with the sought-for precision as demonstrated in [12].
Now we prove . As there are at most ways to assign a type or to the vertex gadgets in we have . Then we obtain the following bound since :
[TABLE]
We obtain a similar bound for : From Lemmas 30, 31 and 33 we know that for our choice of and there exists such that for every -type that is not or we have . Using Lemma 29 this gives for sufficiently large with respect to , and (which do not depend on the input ). Since we can assume that is sufficiently large with respect to and as otherwise the input size is bounded by a constant (in which case we can solve in constant time).
For each -type , the cardinality of each set is bounded above by \max\{\mathopen{}\mathclose{{}\left|V(H)}\right|,2\mathopen{}\mathclose{{}\left|E(H)}\right|\}=12+2k_{1} and hence there are at most \mathopen{}\mathclose{{}\left(2^{12+2k_{1}}}\right)^{3} different types. Furthermore, as has vertices, there are at most possible functions from the at most vertices in to vertices in . Since and we obtain
[TABLE]
The last inequality holds for sufficiently large as
[TABLE]
for some constant that only depends on , but not on the input , whereas . (End of Case 1)
Case 2: . We will show that in this case we can apply Lemma 26 to obtain \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right). We have . Therefore, \mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|=\mathopen{}\mathclose{{}\left|V(H)}\right|=3+k_{1}\geq 10. Let be a pair with satisfying , and both and to meet the requirements of Lemma 26. We have to show that \mathopen{}\mathclose{{}\left|S}\right|\cdot\mathopen{}\mathclose{{}\left|T}\right|<\mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|=3+k_{1}. Note that for every vertex of it holds that \mathopen{}\mathclose{{}\left|\Gamma_{H}(u)}\right|\leq 3. Therefore, \mathopen{}\mathclose{{}\left|S}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H}(T)}\right|\leq 3 and analogously \mathopen{}\mathclose{{}\left|T}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H}(S)}\right|\leq 3. Hence
[TABLE]
(End of Case 2) ∎
Lemma 37**.**
If , then \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(X(k_{1},1,1)}\right).
Proof.
This proof is very similar to the proof of Lemma 36. We give the details for the sake of completeness. As before, we make a case distinction depending on . The first case is about the dominance of the type from Table 2 for as shown in Lemma 34. Otherwise, we use Lemma 26.
Case 1: . We use a reduction from (-hard by [12]). Let and be an input to , be the number of vertices of and be the parameter of the desired precision. To shorten notation let . We assume that the vertices of are labelled as in Figure 11 (on the right).
Let , be positive integers that fulfil (13). Note that and only depend on which is a parameter of and therefore does not depend on the input . We will define the parameter of the gadget graph to be . We also define a new parameter .
For each vertex we introduce a vertex gadget which is a graph as given in Figure 10. We denote the corresponding sets by and , respectively. We say that two gadgets and are adjacent if and are adjacent in . For every edge we use exactly the same edge gadget as in the proof of Lemma 36 (see Figure 12). is the graph with
[TABLE]
Let be a homomorphism from to , be some vertex of and be the corresponding vertex gadget. Then corresponds to a homomorphism from to and therefore has an -type.
We say that a homomorphism from to is full if its restriction to each vertex gadget is either of type (from Table 2) or of its symmetric type (let us call it ). The cut corresponding to a full homomorphism partitions into those vertices for which has type and those for which has type . We say that a full homomorphism is -large if the size of the corresponding cut is equal to , otherwise we say that the homomorphism is -small. Consider a full homomorphism from to .
- •
For an edge of suppose has type and has type . Note that by the definition of the edge gadget, we have . Then the vertices in can be mapped to any of , whereas all vertices in have to be mapped to (the sole common neighbour of , , , and the vertices in ).
- •
Suppose instead that and have the same type or . Then the homomorphism has to map the vertices in both and to .
Thus, every pair of adjacent gadgets of different types contributes a factor of to the number of full homomorphisms, whereas every pair of adjacent gadgets of the same type only contributes a factor of . Recall the definition of as the number of homomorphisms from to that have type . Then for every size- cut of arises in ways as a full homomorphism from to .
Let be the number of solutions to with input and (our goal is to approximate this number). We partition the homomorphisms from to into three different sets. is the number of -large (full) homomorphisms, is the number of homomorphisms that are full but -small and is the number of non-full homomorphisms. Then we have and {N\bigl{(}G^{\prime}\rightarrow H\bigr{)}}=Z^{*}+Z_{1}+Z_{2}. Thus it remains to show that for our choice of , , and . Under this assumption we then have \displaystyle{N\bigl{(}G^{\prime}\rightarrow H\bigr{)}}/(2N(T_{9})^{n}3^{sK})\in[L,L+1/4] and a single oracle call to determine {N\bigl{(}G^{\prime}\rightarrow H\bigr{)}} with precision suffices to determine with the sought-for precision as demonstrated in [12].
Now we prove . As there are at most ways to assign a type or to the vertex gadgets in we have . Then we obtain the following bound since :
[TABLE]
We obtain a similar bound for : From Lemmas 30, 32 and 34 we know that for our choice of and there exists such that for every -type that is not or we have . Using Lemma 29 this gives for sufficiently large with respect to , and . Since we can assume that is sufficiently large with respect to and as otherwise the input size is bounded by a constant (in which case we can solve in constant time).
For each -type , the cardinality of each set is bounded above by \max\{\mathopen{}\mathclose{{}\left|V(H)}\right|,2\mathopen{}\mathclose{{}\left|E(H)}\right|\}=16+2k_{1} and hence there are at most \mathopen{}\mathclose{{}\left(2^{16+2k_{1}}}\right)^{3} different types. Furthermore, as has vertices, there are at most possible functions from the at most vertices in to vertices in . Since and we obtain
[TABLE]
The last inequality holds for sufficiently large as
[TABLE]
for some constant that only depends on , but not on the input , whereas . (End of Case 1)
Case 2: . We will show that in this case we can apply Lemma 26 to obtain \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right). We have . Therefore, \mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|=\mathopen{}\mathclose{{}\left|V(H)}\right|=4+k_{1}\geq 10. Let be a pair with satisfying , and both and to meet the requirements of Lemma 26. We have to show that \mathopen{}\mathclose{{}\left|S}\right|\cdot\mathopen{}\mathclose{{}\left|T}\right|<\mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|=4+k_{1}. Note that for every vertex of it holds that \mathopen{}\mathclose{{}\left|\Gamma_{H}(u)}\right|\leq 3. Therefore, \mathopen{}\mathclose{{}\left|S}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H}(T)}\right|\leq 3 and analogously \mathopen{}\mathclose{{}\left|T}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H}(S)}\right|\leq 3. Hence
[TABLE]
(End of Case 2) ∎
Lemma 38**.**
If , then \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(X(k_{1},0,2)}\right).
Proof.
Let . To shorten notation let . Again we use Lemma 26 to obtain \#\mathrm{\textsc{SAT}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H}\right). We have and \mathopen{}\mathclose{{}\left|F(H)}\right|\cdot\mathopen{}\mathclose{{}\left|V(H)}\right|=\mathopen{}\mathclose{{}\left|V(H)}\right|=5+k_{1}\geq 10. The remainder of the proof is identical to Case 2 in the proof of Lemma 37. ∎
Lemma 39**.**
Let be a graph and be a looped vertex with an unlooped neighbour . If \mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|\geq 2 ( has at least neighbours in ) and for all we have \mathopen{}\mathclose{{}\left|\Gamma_{H}(u)\cap\Gamma_{H}(g)}\right|=1 ( is the only common neighbour of and ), then .
Proof.
Let be the graph obtained by replacing the vertex in by an independent set of size \mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|^{s}, where s=2\mathopen{}\mathclose{{}\left\lceil\log_{\mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|}(\mathopen{}\mathclose{{}\left|\Gamma_{H}(b)}\right|)}\right\rceil. This is well-defined as \mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|>1. The choice of will become clear in a moment. Within the graph , is adjacent only to by the assumption that \mathopen{}\mathclose{{}\left|\Gamma_{H}(b)\cap\Gamma_{H}(g)}\right|=1. Therefore, each vertex in shares an edge only with . The transformation is depicted in Figure 13.
First, we will show -hardness for and we will apply Lemma 26 to achieve this. Let us check that the requirements are met. First note that . Now we have to show that, for every pair with satisfying , and both and it holds that \mathopen{}\mathclose{{}\left|S}\right|\cdot\mathopen{}\mathclose{{}\left|T}\right|<\mathopen{}\mathclose{{}\left|F(H^{\prime})}\right|\cdot\mathopen{}\mathclose{{}\left|V(H^{\prime})}\right|=\mathopen{}\mathclose{{}\left|V(H^{\prime})}\right|. First note that if there exists a vertex then as is the only neighbour of in . Hence . By the same reasoning it holds that . Then \mathopen{}\mathclose{{}\left|S}\right|,\mathopen{}\mathclose{{}\left|T}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H^{\prime}}(b)\setminus I}\right|\leq\mathopen{}\mathclose{{}\left|\Gamma_{H}(b)}\right| and, by our choice s=2\mathopen{}\mathclose{{}\left\lceil\log_{\mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|}(\mathopen{}\mathclose{{}\left|\Gamma_{H}(b)}\right|)}\right\rceil, we can conclude that
[TABLE]
This proves .
To complete the proof of the lemma we show the following claim.
Claim: .
Proof of the claim:
Let be an input to . Let be a weight function on the vertices in with w(g)=\mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|^{s} and for all . By the construction of it is standard that
[TABLE]
Now consider the vertices in . For , let be the number of common neighbours of and in . (It is essential that we regard all neighbours in , not just the neighbours in .) By definition w^{\prime}(u)=\mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right| if and, by the assumptions of this lemma, if . Hence, for all we have
[TABLE]
Intuitively, we will use the fact that has larger “weight” compared to the other vertices in to “boost” the vertex and make it more likely (by a factor of \mathopen{}\mathclose{{}\left|\Gamma_{H}(g)}\right|^{s}) to be used in a homomorphism to .
Here are the details: Let be an input to . We construct a graph from in the following way. Let and be vertices that are distinct from the vertices in . Intuitively, and will serve as “pins” to and , respectively. In addition, for each , we introduce a independent set of size , see Figure 14. Then is the graph with vertices and edges E(G^{\prime})=E(G)\cup\bigl{(}V(G)\times\{\beta\}\bigr{)}\cup\bigcup_{v\in V(G)}\bigl{(}I_{v}\times\{v,\gamma\}\bigr{)}.
Consider a homomorphism from to . Then, since every vertex in is a neighbour of and , is a homomorphism from to . Furthermore, by the construction of , for each and each , it holds that is a common neighbour of and in . Recall that there are such common neighbours. Thus, using (14) and (15), we conclude
[TABLE]
∎
3.4 Square-Free Graphs with an Induced
This section can be seen as an extension of Section 3.3 as it essentially shows -hardness for graphs of the form where . Consider a square-free graph with an induced . Suppose that there is an induced such that the neighbourhood of its center does not contain any triangles. Then is subject to Theorem 5 which shows . Then, by Observation 23.
However, when considering square-free graphs as opposed to graphs of girth at least , might contain triangles. The smallest open case is displayed in Figure 15.
The goal of this section is to prove Lemma 40.
Lemma 40**.**
Let be a square-free graph. If contains a as an induced subgraph then .
First we will introduce some preliminary results.
Definition 41**.**
Let be the Stirling number of the second kind, i.e. the number of surjective functions from a set of elements to a set of elements.
Lemma 42** ([12, Lemma 18]).**
If and are positive integers and , then
[TABLE]
Corollary 43**.**
If and are positive integers with and , then
[TABLE]
The proof of Lemma 40 uses the same general idea as the proof of [28, Theorem 2] — namely that approximating the partition function of the -state ferromagnetic Potts model is -equivalent if and, in addition, we are allowed to specify that certain vertices have to have a specific spin. Crucially, -hardness is known only if this single-vertex “pinning” is allowed. In general, the complexity of approximating the partition function of the Potts model is still unresolved and an important open problem. The approach of simulating ferromagnetic Potts with “pinning” to obtain hardness results has been used before, for instance in the proofs of [29, Lemma 3.6] and [21, Lemma 2.2]. The gadgets we use here to accomplish the reduction are tailored to the specific problem and different from the gadgets used in similar reductions.
As in the proof of [28, Theorem 2] we use a reduction from the problem of counting so-called multiterminal cuts. We introduce the corresponding definitions from [28]. A multiterminal cut of a graph with distinguished vertices (called terminals) is a set of edges that disconnects the terminals (i.e. ensures that there is no path in that connects any two distinct terminals). The size of a multiterminal cut is its cardinality. We consider the following computational problem.
Name:
.
Input:
A connected irreflexive graph with distinct terminals and a positive integer . The input has the property that every multiterminal cut has size at least .
Output:
The number of size- multiterminal cuts of with terminals .
Lemma 44** ([28, Section 4]).**
Let . Then .
Definition 45**.**
Let be an instance of . is the set of separating functions from to . For let and, for , let . Finally, let \Phi^{*}(I)=\{\phi\in\Phi(I)\mid\mathopen{}\mathclose{{}\left|\mathrm{Cut}(\phi)}\right|=K\}.
Observation 46**.**
Let be an instance of . For each , is a multiterminal cut of . On the other hand, each size- multiterminal cut splits the graph into exactly connected components (as otherwise there would exist a multiterminal cut of size less than ). Hence each size- multiterminal cut corresponds exactly to the function for which if is in the same connected component as . Thus, is the subset of functions in that correspond to size- multiterminal cuts. Let be the number of size- multiterminal cuts of the instance . Then T(I)=\mathopen{}\mathclose{{}\left|\Phi^{*}(I)}\right|.
Now we have all the tools at hand to prove the main lemma of this section.
Lemma 40.
Let be a square-free graph. If contains a as an induced subgraph then .
Proof.
Suppose that contains a mixed triangle as an induced subgraph, then the statement of this lemma follows from Lemmas 24 and 25. Hence, for the remainder of this proof let be a square-free graph with an induced without any induced mixed triangle subgraphs. We choose a vertex such that is the center of an induced . We consider the graph which is the subgraph of that is induced by the neighbourhood of . For ease of notation we set . Let be the set of unlooped neighbours of . Since does not contain any mixed triangles, for each , is the only neighbour of in . Since is square-free, so is . Therefore every looped neighbour of has degree . By the choice of , has at least neighbours including itself, i.e. we have . Let be the looped neighbours of degree and be the looped neighbours of degree , where for each we have (where we use that looped vertices can only have looped neighbours since is the sole neighbour of vertices in ). The graph is depicted in Figure 16.
We will give a reduction from to . By the choice of we have . This gives the desired reduction since
[TABLE]
where the first reduction is from Lemma 44 and the last reduction is from Observation 23.
Let be an instance of and let be the desired precision bound. Let n=\mathopen{}\mathclose{{}\left|V(G)}\right| and m=\mathopen{}\mathclose{{}\left|E(G)}\right|. From the instance we construct an instance of . We will need some parameters whose relevance will become clear later in the proof. Let and .
The intuition behind the gadgets that will be used in this proof is the following. For every vertex in we introduce a huge clique . The image of such a clique under a homomorphism to tends to be a reflexive clique, i.e. tends to be of the form . There are such neighbourhoods. These will correspond to the different states that a vertex can be in. We will have to add some attachments to the clique to balance out the fact that is a clique on vertices if , whereas it is a clique on vertices if . For each edge we introduce a gadget that favours the case where and have identical states (i.e. the corresponding cliques have the same image under homomorphisms to ).
Here are the details. First we define the graph . We introduce distinct vertices which will serve as “pins” to the vertices . In the first part of the construction we will only use the vertices . For every vertex we introduce a graph (the “vertex gadget”) as follows. Let be a clique on vertices. For each vertex in this clique we introduce distinct vertices . Then is the graph with vertices
[TABLE]
and edges
[TABLE]
The graph is depicted in Figure 17. Note that the vertices are identical over all vertex gadgets whereas the remaining vertices are distinct for each .
For every edge we introduce a graph together with a set of edges (the “edge gadget”). The graph is defined in precisely the same way as but uses the parameter instead of . We denote the corresponding clique by . Further, we set E_{e}=\mathopen{}\mathclose{{}\left(V(C_{u})\cup V(C_{v})}\right)\times V(C_{e}). The edge gadget is depicted in Figure 18.
Finally, is the graph with vertices
[TABLE]
and edges
[TABLE]
Note that the first set in the union is a set of edges for each terminal . The purpose of these edges will be to ensure that the corresponding graph is in the right “state” (the one corresponding to ). Here we now use all of the , not just the first as we did in the construction of and .
Next we define the lists . We set
[TABLE]
For a homomorphism and a vertex we say that the image is the state of (under ).
A pinned configuration is a tuple of vertices of such that, for each , and are edges of . Note that the vertices of (see Figure 17) have to map to a pinned configuration under a homomorphism from to . For let be the number of pinned configurations . We have
[TABLE]
We say that a vertex is full (under ) if the following conditions are met:
- •
There exists such that .
- •
For every element and every pinned configuration there exists a vertex in the clique such that (elementwise).
We call an edge monochromatic (under ) if and have the same state. Otherwise, we say that is dichromatic. We say that the homomorphism is full if every vertex is full under . We say that a full homomorphism is -small if there are at most dichromatic edges under , otherwise we say that it is -large.
Let be the number of full homomorphisms that are -small. Further, let be the number of full homomorphisms that are -large and let be the number of non-full homomorphisms. Then
[TABLE]
Let be the sought-for number of size- multiterminal cuts of the instance . We will now investigate the way in which the number relates to . Recall the definitions about separating functions from Definition 45. In particular, we will use the sets , and . To shorten notation within the scope of this proof, we write when we mean and when we mean . From Observation 46 we know that T=\mathopen{}\mathclose{{}\left|\Phi^{*}}\right|.
For a function we say that a homomorphism agrees with if, for each vertex of , the state of under is . Note that, by the construction of , under a full homomorphism a terminal has state . Therefore, each full homomorphism agrees with exactly one .
If agrees with , then it follows that is exactly the set of dichromatic edges under . Hence, each -small full homomorphism agrees with exactly one function and each -large full homomorphism agrees with exactly one function . Let be the number of full homomorphisms that agree with . Then
[TABLE]
Let . We are interested in the number . What are the possible full homomorphisms that agree with ?
Observation A
Let . We consider possible images of the vertices of . For to agree with , the state of is fixed to be , where this set can be either of the form or of the form . From Equations (16) and (17) it follows that there are a total of pinned configurations with . Similarly, from Equations (16) and (18) it follows that there are a total of pinned configurations with . As is full, each possible pinned configuration has to be used at least once by the vertices in . As a consequence each vertex contributes a factor of to .
Observation B
Let . What are the possible images of the vertices of ? We make a case distinction depending on .
- •
Let . Then, as is full, and are different states from the set . By the definition of we have that . It follows that . There are pinned configurations with . Each of the vertices of can have any of these pinned configurations. Therefore, contributes a factor of to .
- •
Let . Then, as is full, for some . Then , where is of the form or of the form . As before there are corresponding pinned configurations. Therefore, contributes a factor of to .
Summarising, we obtain
[TABLE]
For each we have \mathopen{}\mathclose{{}\left|\mathrm{Cut}(\phi)}\right|=K. Then, using the fact that \mathopen{}\mathclose{{}\left|\Phi^{*}}\right|=T, we obtain
[TABLE]
To shorten the notation let
[TABLE]
We want to approximately compute the value , where we have shown that . Assume for now that we have
[TABLE]
Then consider the algorithm which makes a oracle call with input to obtain a value and returns \mathopen{}\mathclose{{}\left\lfloor Q/L}\right\rfloor. This algorithm approximates with the desired error bound as is shown in [12, Proof of Theorem 3]. It remains to show the following claim.
Claim: {N\bigl{(}(J,\mathbf{S})\rightarrow H_{b}\bigr{)}}/L\in[Z^{*}/L,Z^{*}/L+1/4].
Proof of the claim: Recall that {N\bigl{(}(J,\mathbf{S})\rightarrow H_{b}\bigr{)}}=Z^{*}+Z_{1}+Z_{2}. Clearly, we have {N\bigl{(}(J,\mathbf{S})\rightarrow H_{b}\bigr{)}}/L\geq Z^{*}/L. We will show and to prove {N\bigl{(}(J,\mathbf{S})\rightarrow H_{b}\bigr{)}}/L\leq Z^{*}/L+1/4.
Recall that is the number of -large full homomorphisms. Using (20) and (21) and the fact that for each we have \mathopen{}\mathclose{{}\left|\mathrm{Cut}(\phi)}\right|\geq K+1 we obtain
[TABLE]
where the last inequality follows from the fact that there are functions in . Then
[TABLE]
where the last inequality holds for sufficiently large by our choice of and the fact that .
Recall that is the number of homomorphisms that are not full. How many non-full homomorphisms are there? In general, there are at most 2^{\mathopen{}\mathclose{{}\left|V(H_{b})}\right|} possible states for any vertex . By the same arguments as given in Observation A, each full vertex under contributes a factor of to . Since the requirements of Corollary 43 are met for sufficiently large and we obtain
[TABLE]
We will use this bound shortly. If is not full, there has to exist at least one vertex which is not full under , and consequently there are at most full vertices under .
Now assume that is not full under . Then either has state for some but not all of the possible pinned configurations are used, or (which means that either or for some , by the fact that is a large clique). In both cases the vertices in can each use at most different pinned configurations. Hence, each non-full vertex contributes a factor of at most to . In particular this factor is smaller (for sufficiently large ) than the factor contributed by full vertices (see (22)). Finally, for each edge there are at most \mathopen{}\mathclose{{}\left|V(H_{b})}\right|^{(k+1)t} mappings from the vertices in to . Therefore,
[TABLE]
Recall that . Then
[TABLE]
where the second inequality follows from (22) and the last inequality holds for sufficiently large by our choice of . This proves the claim and completes the proof. (End of the proof of the claim.) ∎
3.5 Square-Free Graphs with an Induced Net
The goal of this section is to prove -hardness for square-free graphs with an induced net (see Figure 19). Note that the subgraphs of the net that are induced by a distance- neighbourhood of some vertex of the net are of two forms. Either the corresponding subgraph is a looped edge (if ) or it is isomorphic to a looped triangle where one vertex in the triangle has a single additional looped neighbour (if ). Approximately counting retractions to either of these two graphs is -easy (see Theorem 1). Therefore we cannot use these subgraphs in our hardness proof, so we need to work harder.
In our proof (Lemma 48) we use the same general approach that we used to prove Lemma 40. To make the approach work we have to find new gadgets tailored to the net. In one part of the reduction we will need to approximate real values by integers. To achieve this we use Dirichlet’s approximation lemma, which has been used frequently in this line of research (see for instance [24]).
Lemma 47** (see e.g. [49, p. 34]).**
Let be real numbers and be a natural number. Then there exist positive integers with such that \mathopen{}\mathclose{{}\left|r\lambda_{i}-t_{i}}\right|\leq 1/N^{1/d} for every
Lemma 48**.**
Let be a square-free graph that has the net (as displayed in Figure 19) as an induced subgraph. Then .
Proof.
Let be a square-free graph with an induced net that is labelled as in Figure 19. Note that each of the vertices , and might have additional neighbours in . However, they cannot have further common neighbours as is square-free. We use a reduction from which is -hard under AP-reductions by Lemma 44. Let be an instance of and the desired precision bound. Let n=\mathopen{}\mathclose{{}\left|V(G)}\right| and m=\mathopen{}\mathclose{{}\left|E(G)}\right|. We will construct an instance of .
To construct this instance we will use a number of parameters which we introduce at this point. Let . For let s_{i}=s\cdot\log_{\mathopen{}\mathclose{{}\left|\Gamma_{H}(w_{i})}\right|/3}2. For , the value is chosen such that
[TABLE]
It will become clear later in the proof why this is useful. (The important part is that the values \mathopen{}\mathclose{{}\left(\frac{\mathopen{}\mathclose{{}\left|\Gamma_{H}(w_{i})}\right|}{3}}\right)^{s_{i}}, for , are identical and that the base of the exponent on the right-hand side is greater than .) We will now determine integers that approximate , and using Dirichlet’s approximation lemma. Let and \delta^{\prime}=\log_{\mathopen{}\mathclose{{}\left|V(H)}\right|}e^{\delta} (the choices will become clear later in the proof). By Lemma 47 we obtain integers , , and of value at most such that \mathopen{}\mathclose{{}\left|rs_{i}-t_{i}}\right|\leq\delta^{\prime}/m for .
We go on to define the graph . Intuitively, for , the terminal will serve as a “pin” to the vertex . For each vertex we introduce a graph which is simply a star with center and leaves . Formally, the vertices of are . and the edges are . For each edge we introduce a graph which is defined as follows. For , let be disjoint independent sets of size . Let and let ( is an independent set of size ). Then is depicted in Figure 20 and formally defined as the graph with vertices
[TABLE]
and edges
[TABLE]
Then is the graph with vertices
[TABLE]
and edges
[TABLE]
The corresponding set of lists is defined by
[TABLE]
Let be a homomorphism from to . By the definition of every vertex is also a vertex of . An edge is called monochromatic under if . Otherwise, it is called dichromatic under . We say that is -small if the number of dichromatic edges under is at most . Otherwise, is called -large. Let be the number of -small homomorphisms from to and let be the number of -large homomorphisms. Clearly,
[TABLE]
Recall the definitions of separating functions from Definition 45 and, to shorten notation, define the sets and . is the number of size- multiterminal cuts of the instance . Our goal is to approximate . From Observation 46 we know that T=\mathopen{}\mathclose{{}\left|\Phi^{*}}\right|.
We say that a homomorphism agrees with if, for each , we have . By definition of the lists , for each , a homomorphism from to has to map to . Furthermore, as is adjacent to all three terminals and is square-free we have .
At this point we have introduced the gadget and the graph and have defined what it means for a homomorphism from to to agree with a function in (which in turn corresponds to a multiterminal cut). All these definitions are heavily tailored to the graph . The following steps, however, are very similar to those in the proof of Lemma 40. What complicates this proof in comparison to that of Lemma 40 is the fact that we need to use Dirichlet’s approximation lemma to balance out the edge interactions. Here are the details.
Every homomorphism agrees with exactly one function . In particular, if agrees with , then the dichromatic edges of are exactly the multiterminal cut . It follows that every -small homomorphism agrees with exactly one function and every -large homomorphism agrees with exactly one function . For let be the number of homomorphisms from to that agree with . Then
[TABLE]
Let . We are interested in the number and investigate which homomorphisms agree with . For each the image of under is fixed by the lists in and the fact that . Therefore, we only need to consider possible images of the graphs . We make a case distinction depending on .
- •
Let . (This means that is dichromatic under .) By the definition of it follows that the image is a subset of . The vertices and are distinct and are from . As is square-free it follows that . In addition, each vertex of is adjacent to one of the terminals. Since the images of these vertices are also in this does not bring any additional constraints. Summarising, since has size the edge contributes a factor of to .
- •
Let for some . (This means that is a monochromatic edge under with .) Then, for with , by the same arguments as before we have . However the vertices in can be mapped to any neighbour of . Therefore, each edge in contributes a factor of \mathopen{}\mathclose{{}\left|\Gamma_{H}(w_{i})}\right|^{t_{i}}\cdot 3^{t-t_{i}} to .
Using this knowledge, for , we have
[TABLE]
Since m=\mathopen{}\mathclose{{}\left|\mathrm{Cut}(\phi)}\right|+\sum_{i\in[3]}\mathopen{}\mathclose{{}\left|\mathrm{Mon}_{i}(\phi)}\right| we can simplify this expression to
[TABLE]
For , recall the fact that \mathopen{}\mathclose{{}\left|rs_{i}-t_{i}}\right|\leq\delta^{\prime}/m, where s_{i}=s\cdot\log_{\mathopen{}\mathclose{{}\left|\Gamma_{H}(w_{i})}\right|/3}2. For an upper bound on we use the fact that . Also note that \mathopen{}\mathclose{{}\left|\Gamma_{H}(w_{i})}\right| is bounded above by \mathopen{}\mathclose{{}\left|V(H)}\right| (which we use in the second inequality of the following expression). Furthermore, \sum_{i\in[3]}\mathopen{}\mathclose{{}\left|\mathrm{Mon}_{i}(\phi)}\right|\leq m and \delta^{\prime}=\log_{\mathopen{}\mathclose{{}\left|V(H)}\right|}e^{\delta} (which we use in the third inequality of the following expression). Then
[TABLE]
Analogously, for a lower bound on we use the fact that . We obtain 3^{tm}\cdot 2^{(\sum_{i\in[3]}\mathopen{}\mathclose{{}\left|\mathrm{Mon}_{i}(\phi)}\right|)\cdot rs}\cdot e^{-\delta}\leq Z_{\phi}. Summarising, we have
[TABLE]
Putting these bounds on into the expression for in (24) gives
[TABLE]
Since \sum_{i\in[3]}\mathopen{}\mathclose{{}\left|\mathrm{Mon}_{i}(\phi)}\right|=m-K for each and \mathopen{}\mathclose{{}\left|\Phi^{*}}\right|=T we obtain
[TABLE]
We set to obtain
[TABLE]
Assume for now that {N\bigl{(}(J,\mathbf{S})\rightarrow H\bigr{)}}/L\in[Z^{*}/L,Z^{*}/L+1/4]. Then the algorithm that makes a oracle call with input returns a solution such that Z^{*}/L\cdot e^{-\varepsilon/2}\leq\mathopen{}\mathclose{{}\left\lfloor Q/L}\right\rfloor\leq Z^{*}/L\cdot e^{\varepsilon/2} as was shown in [12, Proof of Theorem 3]. Using (26) and our choice of this gives
[TABLE]
Therefore the output \mathopen{}\mathclose{{}\left\lfloor Q/L}\right\rfloor approximates with the desired precision. It remains to show the following claim.
Claim: {N\bigl{(}(J,\mathbf{S})\rightarrow H\bigr{)}}/L\in[Z^{*}/L,Z^{*}/L+1/4].
Proof of the claim: Recall that {N\bigl{(}(J,\mathbf{S})\rightarrow H\bigr{)}}=Z^{*}+Z_{1}. It is immediate that {N\bigl{(}(J,\mathbf{S})\rightarrow H\bigr{)}}/L\geq Z^{*}/L. It remains to show that .
To obtain the following expression we first use (24) and (25). The third inequality then uses the fact that, for every , we have \mathopen{}\mathclose{{}\left|\mathrm{Cut}(\phi)}\right|\geq K+1 and hence \sum_{i\in[3]}\mathopen{}\mathclose{{}\left|\mathrm{Mon}_{i}(\phi)}\right|\leq m-(K+1). Finally, in the fourth inequality we use the fact that \mathopen{}\mathclose{{}\left|\Phi\setminus\Phi^{*}}\right|\leq\mathopen{}\mathclose{{}\left|V(H)}\right|^{n}
[TABLE]
Recall the definition . It follows that
[TABLE]
where the last inequality holds for sufficiently large by our choice of and the fact that . This proves the claim and completes the proof. (End of the proof of the claim.) ∎
3.6 Square-Free Graphs with an Induced Reflexive Cycle of Length at least
The goal of this section is to show the following lemma. We build up to its proof, which is given at the end of this section.
Lemma 49**.**
Let be a square-free graph. If contains a reflexive cycle of length at least as an induced subgraph then .
Let be a connected square-free graph with an induced reflexive cycle of length at least . If all cycles in have length at least , then has girth at least and the complexity of is classified by Theorem 5. In the special case where is reflexive this classification is straightforward to see: Either there is just a single cycle in , then is a pseudotree and -hardness follows from -hardness for the decision problem [18, Corollary 4.2, Theorem 5.1] together with [12, Theorem 1] — or there are multiple cycles (all of which have length at least ), then there exists an induced (as is connected) and hardness follows from Lemma 40.
Thus, it remains to show hardness if contains both a cycle of length at least as well as a cycle of length at most , i.e. (since is square-free) it contains a triangle. The hardness proof we give in this section will handle the case where includes triangles but will not rely on this fact (i.e. it will also cover the before-mentioned case where all cycles have length at least without relying on hardness results for the decision problem).
As mentioned before, it is known that approximately counting list homomorphisms to reflexive graphs with an induced cycle of length at least is -hard [25, Lemma 3.4]. That proof makes use of a certain set of two-vertex lists. In the proof of Lemma 49 we will use single-vertex lists to simulate these two-vertex lists.
As we have already shown -hardness results for square-free graphs with an induced or an induced net in Sections 3.4 and 3.5 we now focus on graphs that do not contain such subgraphs. When considering reflexive graphs it turns out that this leaves a class of graphs which we call reflexive triangle-extended cycles. We will also make use of these reflexive triangle-extended cycles when considering square-free graphs that are not necessarily reflexive. When we do this we will restrict to the (reflexive) subgraph induced by the looped vertices of .
Definition 50**.**
A reflexive triangle-extended cycle of length consists of a reflexive cycle together with a set , and a reflexive triangle for each . An example of a reflexive triangle-extended cycle is depicted in Figure 21.
Definition 51**.**
Analogously to Definition 50, a reflexive triangle-extended path is a reflexive path together with a set , and a reflexive triangle for each .
Lemma 52**.**
Let be a connected reflexive square-free graph that does not contain an induced and also does not contain an induced net. If contains an induced cycle of length at least then is a reflexive triangle-extended cycle of length at least . Otherwise it is a reflexive triangle-extended path.
Proof.
Case 1: contains an induced cycle with . If is just the cycle , then the statement of the lemma is true (). Otherwise, consider any with a neighbour (has to exist since is connected).
Since does not contain an induced the vertex is adjacent to a neighbour of . Let and be the other neighbours of and in , respectively, i.e. and . The vertices are all distinct as has length at least . As is square-free we observe
[TABLE]
The proof of the following claim directly proves that is a reflexive triangle-extended cycle.
Claim: .
Proof of the claim: Assume there exists a neighbour of in . By (27) we have . Furthermore, since is square-free, we obtain the following.
[TABLE]
Let be the subgraph of induced by the vertices , see Figure 22. Then is a net (cf. Figure 19 where , and ).
Because of (28) we have . (The dashed edges incident to in Figure 22 cannot exist.) Because of (27) we have . (The dashed edges incident to in Figure 22 cannot exist.) Finally, since is an induced cycle, there are no edges between the vertices outside of . Therefore, is an induced net in , which gives a contradiction. This proves the claim in Case 1. (End of Case 1)
Case 2: All induced cycles in are triangles. This case is handled very similarly to the previous one: Let be a maximal induced path in . If is just the path , then the statement of the lemma is true (). Otherwise, let be a neighbour of . We show that is adjacent to a neighbour of :
- •
If is an inner vertex of then, since does not contain an induced , is also adjacent to a neighbour of in .
- •
If is an endpoint of , then has to be adjacent to a vertex with as is maximal induced. Without loss of generality assume that is the neighbour of which is closest to in . Then has to be a triangle ( has to be a neighbour of ) as is induced and all induced cycles in are triangles.
Then the proof of the following claim shows that is a reflexive triangle-extended path.
Claim: .
Proof of the claim: Assume there exists a neighbour of in . We observe the following properties:
- •
The vertex does not have a neighbour in : Assume the opposite and let be a neighbour of . Without loss of generality let be closer to than in . Furthermore, let be the neighbour of in which is closest to . Then the edge and the path close an induced cycle with . If this cycle has length greater than , a contradiction. If we obtain a contradiction to the fact that is square-free (see Figure 22).
- •
Both and are inner points of : Suppose, for contradiction, that is an end point of . Since does not have a neighbour in , replacing by in gives an induced path . Moreover, is longer than which is a contradiction to the maximality of . This shows that cannot be an end point of . Analogously cannot be an end point of .
Then let and be the other neighbours of and in (they have to exist since and are inner points of ). The remainder of the argument is analogous to the proof of the claim in Case 1. (End of Case 2) ∎
The goal of the remainder of this section is to prove Lemma 49. In order to prove Lemma 49 we work with the following parameterised version of the list homomorphism counting problem. Let be a graph and be a set of subsets of .
Name:
\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}}\right).
Input:
An irreflexive graph and a collection of lists .
Output:
{N\bigl{(}(G,\mathbf{S})\rightarrow H\bigr{)}}.
We also use the following lemma.
Lemma 53** ([25, Proof of Lemma 3.4]).**
Let be a graph that contains an induced reflexive cycle on vertices. Let . Then \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}}\right)\equiv_{\mathrm{AP}}\#\mathrm{\textsc{SAT}}.
The key to proving Lemma 49 is the following result. It states that for certain graphs that contain a reflexive triangle-extended cycle we can simulate each size- list of vertices in the corresponding cycle by gadgets using only single-vertex lists.
Lemma 54**.**
Let be a square-free graph that does not contain any mixed triangle as an induced subgraph and let be the graph induced by the looped vertices of . Suppose that contains a connected component that is a reflexive triangle-extended cycle, where is the corresponding reflexive cycle as given by Definition 50 and the length of is . Let and be sets with
[TABLE]
Let \mathcal{L}^{\prime\prime}=\mathcal{L}^{\prime}\cup\bigl{\{}S\subseteq V(H)\mid\mathopen{}\mathclose{{}\left|S}\right|\in\{1,\mathopen{}\mathclose{{}\left|V(H)}\right|\}\bigr{\}}. Then
[TABLE]
Proof.
For the reflexive triangle-extended cycle we use the notation (, and for ) as given by Definition 50. Let , and be as given in the statement of the lemma. We have for some . Let be an input to \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}}\right). Let . Since is not part of the goal is to simulate using gadgetry and lists from .
From we define an instance of \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}^{\prime\prime}}\right). To this end we will define, for each , a vertex gadget and a corresponding set of lists . There are two distinct paths in that connect and : and . The graph has two parts: a graph and a graph , which depend on the paths and . We first define and (and the corresponding sets of lists and ) and then we describe the way in which they are connected to form . The definition of depends on , the number of edges of :
- •
If is even, think of a path on edges. Let be one of the end points of this path. We pin to (the vertex in the “middle” of ). This graph is depicted in Figure 23 on the left. The graph is then a modification of this graph where each vertex of the path is replaced by a clique of size (apart from which will be pinned to anyway). This modification will ensure that only looped vertices can be in the image of . The graph is depicted in Figure 23 (on the right) and is formally defined as follows: , where all these vertices are distinct from the vertices of (and distinct from the vertices used in other gadgets). E(J_{P_{1}})=\bigl{\{}\{v_{i},v_{i}^{\prime}\}\mid i\in\{0,\dots,\ell/2-1\}\bigr{\}}\cup\bigl{\{}\{v_{i-1},v_{i-1}^{\prime}\}\times\{v_{i},v_{i}^{\prime}\}\mid i\in[\ell/2-1]\bigr{\}}\cup\bigl{\{}\{v_{\ell/2-1},v^{*}\},\{v_{\ell/2-1}^{\prime},v^{*}\}\bigr{\}}. We set and for all
- •
If is odd, then is defined very similarly to the previous case, see Figure 24. Formally, V(J_{P_{1}})=\{v_{i},v_{i}^{\prime}\mid i\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor\}\}\cup\{v_{1}^{*},v_{2}^{*}\}, where all these vertices are distinct from the vertices of (and distinct from the vertices used in other gadgets). E(J_{P_{1}})=\bigl{\{}\{v_{i},v_{i}^{\prime}\}\mid i\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor\}\bigr{\}}\cup\bigl{\{}\{v_{i-1},v_{i-1}^{\prime}\}\times\{v_{i},v_{i}^{\prime}\}\mid i\in[\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor]\bigr{\}}\cup\bigl{\{}\{v_{\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor},v_{\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor}^{\prime}\}\times\{v^{*}_{1},v^{*}_{2}\}\bigr{\}}. We set S^{1}_{v_{1}^{*}}=\{c_{\mathopen{}\mathclose{{}\left\lceil\ell/2}\right\rceil}\}, S^{1}_{v_{2}^{*}}=\{c_{\mathopen{}\mathclose{{}\left\lfloor\ell/2}\right\rfloor}\} and for all .
This completes the definition of . is defined analogously. However, the length of is instead of . Furthermore, if is even, note that the vertex in the “middle” of is rather than . (Accordingly, if is odd, we use c_{\mathopen{}\mathclose{{}\left\lceil(q+\ell)/2}\right\rceil} and c_{\mathopen{}\mathclose{{}\left\lfloor(q+\ell)/2}\right\rfloor} to “pin” to.) Formally, is defined as follows:
- •
If is even, we have and E(J_{P_{2}})=\bigl{\{}\{w_{i},w_{i}^{\prime}\}\mid i\in\{0,\dots,(q-\ell)/2-1\}\bigr{\}}\cup\bigl{\{}\{w_{i-1},w_{i-1}^{\prime}\}\times\{w_{i},w_{i}^{\prime}\}\mid i\in[(q-\ell)/2-1]\bigr{\}}\cup\bigl{\{}\{w_{(q-\ell)/2-1},w^{*}\},\{w_{(q-\ell)/2-1}^{\prime},w^{*}\}\bigr{\}}. We set and for all
- •
If is odd, we have V(J_{P_{2}})=\{w_{i},w_{i}^{\prime}\mid i\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor\}\}\cup\{w_{1}^{*},w_{2}^{*}\} and E(J_{P_{2}})=\bigl{\{}\{w_{i},w_{i}^{\prime}\}\mid i\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor\}\bigr{\}}\cup\bigl{\{}\{w_{i-1},w_{i-1}^{\prime}\}\times\{w_{i},w_{i}^{\prime}\}\mid i\in[\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor]\bigr{\}}\cup\bigl{\{}\{w_{\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor},w_{\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor}^{\prime}\}\times\{w^{*}_{1},w^{*}_{2}\}\bigr{\}}. We set S^{2}_{w_{1}^{*}}=\{c_{\mathopen{}\mathclose{{}\left\lceil(q+\ell)/2}\right\rceil}\}, S^{2}_{w_{2}^{*}}=\{c_{\mathopen{}\mathclose{{}\left\lfloor(q+\ell)/2}\right\rfloor}\} and for all .
We can now define the graph (for ): is the graph obtained from and by identifying with and with . As an example, if has even length and has odd length, the graph is depicted in Figure 25. Let denote the set that contains the two vertices and . The lists of the vertices in are the union of and . (Note that this is well-defined as .) This completes the definition of and .
We can finally define the instance : is the graph with vertices and edges
[TABLE]
Finally, with
[TABLE]
Note that for each we have and therefore is a valid input to \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}^{\prime\prime}}\right).
To show how homomorphisms from to relate to homomorphisms from to we determine some properties of the gadget . Consider the case where , the number of edges of , is even, and , the number of edges of , is odd. (The other cases of and even or odd will be analogous.) Then is the gadget depicted in Figure 25. Let the vertices of be labelled accordingly. Now let be a homomorphism from to .
Claim 1: For all , the image contains at least one looped vertex. For all j\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor\}, the image contains at least one looped vertex.
Proof of Claim 1: First consider . Assume that both and are unlooped vertices of . Since is an edge in , and have to be connected by an edge in and therefore have to be different unlooped vertices. However, the vertices and are also neighbours of which is pinned to the looped vertex . It follows that form a mixed triangle in , a contradiction. This argument can be repeated iteratively for each (using the fact that contains a looped vertex). The argument for (j\in\{0,\dots,\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor\}) is analogous.
Claim 2: There exists a vertex with .
Proof of Claim 2: Let where is looped (this can be assumed without loss of generality by Claim 1). We will show that : By Claim 1 and the construction of there exists a walk on edges in which uses looped vertices only and goes from to , which by assumption is looped itself. As this walk is looped it is in , and as it contains it is in . Hence,
[TABLE]
Similarly, by the construction of we obtain
[TABLE]
Since is a reflexive triangle-extended cycle we have
[TABLE]
and
[TABLE]
Therefore, from Equations (29) and (30) it follows that .
Claim 3: Let and . Then the image of the remaining vertices of under is determined completely. In particular, .
Proof of Claim 3: Consider the case where (the case can be treated analogously). Since is a reflexive triangle-extended cycle and a connected component of , the walk is the only -edge walk on looped vertices in that leads from to (since there are no reflexive shortcuts in ). Thus, by Claim 1 and the construction of , we have . We assume without loss of generality (by renaming) that for each we have .
Now consider the image for some . The vertex is a neighbour of , and (or alternatively if ) in . Therefore, by the fact that for each we have and by the pinning that ensures , we know that has to be a neighbour of , and in . Since there is no edge between and we have . Then as otherwise would form a square in which is a contradiction to the fact that is square-free. So we have established that for each it holds that .
Similarly one establishes that for each i\in[\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor] it holds that : Note that by the construction of the homomorphism has to map w_{\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor} and w_{\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor}^{\prime} to common neighbours of c_{\mathopen{}\mathclose{{}\left\lfloor(q+\ell)/2}\right\rfloor} and c_{\mathopen{}\mathclose{{}\left\lceil(q+\ell)/2}\right\rceil}. Since is a reflexive triangle-extended cycle and a connected component of , the walk c_{\mathopen{}\mathclose{{}\left\lceil(q+\ell)/2}\right\rceil}\dots,c_{q-1},c_{0} is the only \mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor-edge walk on looped vertices in that leads from a common neighbour of c_{\mathopen{}\mathclose{{}\left\lfloor(q+\ell)/2}\right\rfloor} and c_{\mathopen{}\mathclose{{}\left\lceil(q+\ell)/2}\right\rceil} to . Therefore, we have and \forall i\in[\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor],\ c_{q-i}\in h(\{w_{i},w_{i}^{\prime}\}). Then by the same arguments as before we establish that for each i\in[\mathopen{}\mathclose{{}\left\lfloor(q-\ell)/2}\right\rfloor] it holds that .
Finally, is a neighbour of , and . Hence, is a neighbour of , and . Since there is no edge between and we have . Then as otherwise would form a square in which is a contradiction to the fact that is square-free. We obtain . This proves Claim 3.
From the construction of together with Claim 2 and Claim 3 we directly obtain that {N\bigl{(}(G,\mathbf{S}^{G})\rightarrow H\bigr{)}}={N\bigl{(}(J,\mathbf{S}^{J})\rightarrow H\bigr{)}}, which gives the sought-for reduction in the case where is even and is odd. All other cases of even or odd and even or odd can be treated analogously. ∎
Now we can prove Lemma 49.
Lemma 49.
Let be a square-free graph. If contains a reflexive cycle of length at least as an induced subgraph then .
Proof.
Suppose that contains a mixed triangle as an induced subgraph, then the statement of this lemma follows from Lemmas 24 and 25. We can now assume that does not contain any mixed triangle as an induced subgraph.
Let be the reflexive cycle of length in . Let be the graph induced by the looped vertices in . If (and hence ) contains an induced or an induced net then by Lemmas 40 and 48, respectively. Otherwise, the connected component of that contains the cycle has to be a reflexive triangle-extended cycle by Lemma 52 and therefore fulfills the requirements of Lemma 54.
Let . Then \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}}\right)\equiv_{\mathrm{AP}}\#\mathrm{\textsc{SAT}} by Lemma 53. We can use Lemma 54 iteratively to obtain \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\mathcal{L}}\right)\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\bigl{\{}S\subseteq V(H)\mid\mathopen{}\mathclose{{}\left|S}\right|\in\{1,\mathopen{}\mathclose{{}\left|V(H)}\right|\}\bigr{\}}}\right). Note that by the problem definitions we have \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H,\bigl{\{}S\subseteq V(H)\mid\mathopen{}\mathclose{{}\left|S}\right|\in\{1,\mathopen{}\mathclose{{}\left|V(H)}\right|\}\bigr{\}}}\right)=\mathrm{\#\textsc{Ret}}(H). Summarising,
[TABLE]
∎
4 Putting the Pieces together
This section contains the proof of Theorem 2 (we restate it here for convenience). We will use the following theorem, which is a consequence of the classification in [21] for approximately counting retractions to graphs of girth at least , since that proof does not use the fact that is triangle-free for irreflexive .
Theorem 55** ([21, Theorem 2.3]).**
Let be an irreflexive square-free graph.
- i)
If every connected component of is a star, then is in . 2. ii)
Otherwise, if every connected component of is a caterpillar, then is approximation-equivalent to . 3. iii)
Otherwise, is approximation-equivalent to .
In the following lemma we collect the -hardness results which we use to prove Theorem 2.
Lemma 56**.**
Let be a connected square-free graph other than a reflexive clique, a member of , or an irreflexive caterpillar. Then is approximation-equivalent to .
Proof.
If is irreflexive then by assumption it is not a caterpillar. Thus is approximation-equivalent to by Theorem 55.
If is not irreflexive, i.e. if has at least one loop, then we collect different -hardness results proved throughout this work to show hardness. If contains a mixed triangle as induced subgraph, then by Lemmas 24 and 25. If does not contain a mixed triangle as an induced subgraph but contains a , a net or a reflexive cycle of length at least as an induced subgraph, then by Lemmas 40, 48 and 49, respectively. It remains to show if is a graph with the following properties:
- •
is connected and square-free.
- •
has at least one looped vertex.
- •
is not a reflexive clique.
- •
.
- •
does not contain any of the following as an induced subgraph: a mixed triangle, a , a net, a reflexive cycle of length at least .
Let be a connected component in the graph induced by the looped vertices in . (It will turn out that is actually the only connected component in this graph.) Then by the properties of and Lemma 52 we know that is a reflexive triangle-extended path. We recall the definition of a reflexive triangle-extended path from Definition 51: is a reflexive path together with a set , and a reflexive triangle for each . Since is not a reflexive clique it holds that .
Note that (where the set of bristles is empty and corresponds to ). For all the clique has size , for it has size . Since and is connected, there exists a vertex outside of with a neighbour in . The vertex has to be unlooped as otherwise it would be part of the reflexive connected component . We consider four disjoint cases.
- •
If there exists a vertex ( is unlooped) which is adjacent to a vertex and , then consider two different cases:
- –
If is adjacent to a vertex with , then since is unlooped and is a mixed triangle, a contradiction.
- –
If is the only neighbour of in , then the requirements of Lemma 39 are met (with and ) and hence .
- •
If there exists a vertex which is adjacent to a vertex , and , then is a graph of the form where (cf. Figure 9) and therefore
[TABLE]
by Lemma 36, Observation 6 and Observation 23 (in the order of the reductions used).
- •
Suppose there exists a vertex with that is adjacent to a vertex . Without loss of generality (by renaming the vertices of ) let . If then is part of a reflexive triangle in and is a graph of the form where . Then we have by the same arguments used in the previous case. If otherwise then is the only neighbour of in and is a graph of the form where . Then we use Theorem 5 to infer that (since has girth at least and therefore is subject to Theorem 5). It follows that by the same arguments as in the previous case.
- •
If for every pair of adjacent vertices with and we have ( is a so-called bristle) and , then is the triangle-extended path together with a number of bristles (all of which are attached to a vertex in ). To match the notation of in Definition 10 we set and, for all , we set . Further, if then and otherwise . Note that \mathopen{}\mathclose{{}\left|K_{i}}\right|\in\{2,3\} which we will use in a moment. For each let be the set of unlooped neighbours (bristles) of . By the fact that all unlooped vertices of have degree and a neighbour in we have and E(H)=\bigcup_{i=0}^{Q}\mathopen{}\mathclose{{}\left(K_{i}\times K_{i}}\right)\cup\bigcup_{i=1}^{Q}\mathopen{}\mathclose{{}\left(\{p_{i}\}\times B_{i}}\right). Since is a triangle-extended path we can also verify the properties (for ) and (for with \mathopen{}\mathclose{{}\left|j-i}\right|>1).
Therefore, since , there exists such that at least one of the following holds:
- (i)
\mathopen{}\mathclose{{}\left|K_{i-1}}\right|=\mathopen{}\mathclose{{}\left|K_{i}}\right|=2 and \mathopen{}\mathclose{{}\left|B_{i}}\right|\geq 2. 2. (ii)
\mathopen{}\mathclose{{}\left|K_{i-1}}\right|=2 and \mathopen{}\mathclose{{}\left|K_{i}}\right|=3 (or \mathopen{}\mathclose{{}\left|K_{i-1}}\right|=3 and \mathopen{}\mathclose{{}\left|K_{i}}\right|=2) and \mathopen{}\mathclose{{}\left|B_{i}}\right|\geq 3. 3. (iii)
\mathopen{}\mathclose{{}\left|K_{i-1}}\right|=\mathopen{}\mathclose{{}\left|K_{i}}\right|=3 and \mathopen{}\mathclose{{}\left|B_{i}}\right|\geq 5.
In all three cases we will show that the neighbourhood of induces a -hard subgraph, i.e. that . Then, by Observation 23, we obtain which completes the proof of this case and with it the proof of the theorem.
- –
If item (i) holds, then is of the form where . The graph has girth at least and therefore is subject to Theorem 5. Since with is a mixed graph but not a partially bristled reflexive path we obtain by Theorem 5.
- –
If item (ii) holds, then is of the form where . Then by Lemma 37.
- –
If item (iii) holds, then is of the form where . Then by Lemma 38.
∎
We will use the following remark to deal with graphs that have multiple connected components.
Remark 57** ([21, Remark 1.15]).**
Let be a graph with connected components . On the one hand it holds that . On the other hand, given an oracle for each , we can construct a polynomial-time algorithm for .
Theorem 0**.**
Let be a square-free graph.
- i)
If every connected component of is trivial then approximately counting retractions to is in FP. 2. ii)
Otherwise, if every connected component of is
- •
trivial,
- •
in the class , or
- •
is an irreflexive caterpillar
then approximately counting retractions to is -equivalent. 3. iii)
Otherwise, approximately counting retractions to is -equivalent.
Proof.
If is a trivial graph then by the result of Dyer and Greenhill [14] (see Theorem 7). From (Observation 6) it also follows that . Then item i) follows from Remark 57.
If is a graph for which item i) does not hold, then has a connected component that is not a trivial graph. By Remark 57 we have . Then -hardness in item ii) follows from the reduction \mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H^{\prime}}\right)\leq_{\mathrm{AP}}\mathrm{\#\textsc{Ret}}(H^{\prime}) (Observation 6) together with the fact that \#\mathrm{\textsc{BIS}}\leq_{\mathrm{AP}}\mathrm{\#\textsc{Hom}}\mathopen{}\mathclose{{}\left(H^{\prime}}\right) since is a non-trivial connected graph [24, Theorem 1].
We will now prove -easiness in item ii). If is a trivial graph we have already pointed out that and hence is trivially -easy. If then by Theorem 1. If is an irreflexive caterpillar then by Theorem 55. Hence, -easiness in item ii) follows from Remark 57.
If is a graph for which item ii) does not hold, then has a connected component that is not trivial, not a member of and not an irreflexive caterpillar. Then is approximation-equivalent to by Lemma 56 and by Remark 57. This proves item iii). ∎
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