Counting Linear Extensions: Parameterizations by Treewidth
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Abstract
We consider the \(\#\hbox {P}\)complete problem of counting the number of linear extensions of a poset \((\textsc {\#LE})\); a fundamental problem in order theory with applications in a variety of distinct areas. In particular, we study the complexity of \(\textsc {\#LE}\) parameterized by the wellknown decompositional parameter treewidth for two natural graphical representations of the input poset, i.e., the cover and the incomparability graph. Our main result shows that \(\textsc {\#LE}\) is fixedparameter intractable parameterized by the treewidth of the cover graph. This resolves an open problem recently posed in the Dagstuhl seminar on Exact Algorithms. On the positive side we show that \({\textsc {\#LE}}\) becomes fixedparameter tractable parameterized by the treewidth of the incomparability graph.
Keywords
Partially ordered sets Linear extensions Parameterized complexity Structural parameters Treewidth1 Introduction
Counting the number of linear extensions of a poset is a fundamental problem of order theory that has applications in a variety of distinct areas such as sorting [30], sequence analysis [25], convex rank tests [27], sampling schemes of Bayesian networks [28], and preference reasoning [24]. Determining the exact number of linear extensions of a given poset is known to be #Pcomplete [6] already for posets of height at least 3. Informally, #Pcomplete problems are as hard as counting the number of accepting paths of any nondeterministic Turing machine, implying that such problems are not tractable unless P = NP. The currently fastest known method for counting linear extensions of a general nelement poset is by dynamic programming over the lattice of downsets and runs in time \(\mathcal {O}(2^n \cdot n)\) [10]. Polynomial time algorithms have been found for various special cases such as seriesparallel posets [26] and posets whose cover graph is a (poly)tree [2]. Fully polynomial time randomized approximation schemes are known for estimating the number of linear extensions [7, 13].
Due to the inherent difficulty of the problem, it is natural to study whether it can be solved efficiently by exploiting the structure of the input poset. In this respect, the parameterized complexity framework [9, 12] allows a refined view of the interactions between various forms of structure in the input and the running time of algorithms. The idea of the framework is to measure the complexity of problems not only in terms of input sizes, but also with respect to an additional numerical parameter. The goal is then to develop socalled fpt algorithms, which are algorithms that run in time \(f(k)n^{\mathcal {O}(1)}\) where n is the input size and f is a computable function depending only on the parameter k. A less favorable outcome is a socalled XP algorithm, which runs in time \(n^{f(k)}\); the existence of such algorithms then gives rise to the respective complexity classes \({\textsf {FPT}}\) (fixedparameter tractable) and XP.
The first steps in this general direction have been taken, e.g., in [19], using the decomposition diameter as a parameter, in [15] using a parameter called activity for Nfree posets, and very recently in [22], where the treewidth of the socalled cover graph was considered as a parameter. Also the exact dynamic programming algorithm [10] can be shown to run in time \(\mathcal {O}(n^w\cdot w)\) for a poset with n elements and width w (the size of the largest antichain). Interestingly, none of these efforts has so far led to an fpt algorithm.
The theory gets interesting with those counting problems that are harder than their corresponding decision versions.
1.1 Results
In this paper we study the complexity of counting linear extensions when the parameter is the treewidth—a fundamental graph parameter which has already found a plethora applications in many areas of computer science [17, 18, 29]. In particular, we settle the fixedparameter (in)tractability of the problem when parameterizing by the treewidth of two of the most prominent graphical representations of posets, the cover graph (also called the Hasse diagram) and the incomparability graph.
Our main result then provides the first evidence that the problem does not allow for an fpt algorithm parameterized by the treewidth of the cover graph unless \({\textsf {FPT}}={{{\textsf {W}}}}{ {[1]}}\). We remark that this complements the XP algorithm of [22] and resolves an open problem recently posed in the Dagstuhl seminar on Exact Algorithms [21]. The result is based on a socalled fpt turing reduction from Equitable Coloring parameterized by treewidth [14], and combines a counting argument with a finetuned construction to link the number of linear extensions with the existence of an equitable coloring. To the best of our knowledge, this is the first time this technique has been used to show fixedparameter intractability of a counting problem.
We complement this negative result by obtaining an fpt algorithm for the problem when the parameter is the treewidth of the incomparability graph of the poset. To this end, we use the socalled combined graph (also called the coverincomparability graph [5]) of the poset, which is obtained from the cover graph by adding the edges of the incomparability graph. We employ a special normalization procedure on a decomposition of the incomparability graph to show that the treewidth of the combined graph must be bounded by the treewidth of the incomparability graph. Once this is established, the result follows by giving a formulation of the problem in Monadic Second Order Logic and applying an extension of Courcelle’s Theorem for counting.
1.2 Organization of the Paper
The paper is organized as follows. Section 2 introduces the required preliminaries and notation. Section 3 is then dedicated to proving the fixedparameter intractability of the problem when parameterized by the treewidth of the cover graph, and the subsequent Sect. 4 presents our positive results for the problem. Concluding notes are then provided in Sect. 5.
2 Preliminaries
For standard terminology in graph theory, such as the notions of a graph, digraph, path, etc. we refer readers to [11]. Given a graph G, we let V(G) denote its vertex set and E(G) its edge set. The (open) neighborhood of a vertex \(x \in V(G)\) is the set \(\{y\in V(G):\{x,y\}\in E(G)\}\) and is denoted by N(x). The closed neighborhood N[v] of v is defined as \(N(v)\cup \{v\}\). A path between two disjoint vertex sets \(A,B\subseteq V(G)\) is a path with one endpoint in A, one endpoint in B, and all internal vertices disjoint from \(A\cup B\). A set \(X\subseteq V(G)\) is a separator in G if \(GX\) contains at least two connected components.
We use [i] to denote the set \(\{0,1,\dots ,i\}\). The following fact about prime numbers will also be useful later.
Fact 1
([6]) For any \(n\ge 4\), the product of primes strictly between n and \(n^2\) is at least \(n!2^n\).
2.1 Treewidth
 (T1)
\(\cup _{t\in V(T)}X_{t}=V(G)\),
 (T2)
for every \(u\in V(G)\), the set \(T_{u}=\{t\in V(T): u\in X_{t}\}\) induces a connected subtree of T (monotonicity), and
 (T3)
for each \(uv\in E(G)\) there exists \(t\in V(T)\) such that \(u,v\in X_{t}\).

\(X_r=\emptyset \) and \(X_\ell =\emptyset \) for every leaf \(\ell \) of T. In other words, all the leaves as well as the root contain empty bags.
 Every nonleaf node of T is of one of the following three types:

Introduce node A note t with exactly one child \(t'\) such that \(X_t=X_{t'}\cup \{v\}\) for some vertex \(v\not \in X_{t'}\); we say that v is introduced at t. If \(u\in X_{t'}\) and uv is an edge in G, then we also say that uv is introduced at t.

Forget node A note t with exactly one child \(t'\) such that \(X_t=X_{t'}{\setminus } \{w\}\) for some vertex \(w\in X_{t'}\); we say that w is forgotten at t.

Join node A node t with two children \(t_1\), \(t_2\) such that \(X_{t}=X_{t_1}=X_{t_2}\).

We list some useful facts about treewidth and pathwidth.
Fact 2
[3, 4] There exists an algorithm which, given a graph G and an integer k, runs in time \(\mathcal {O}(k^{\mathcal {O}(k^3)}n)\) and either outputs a treedecomposition of G of width at most k or correctly identifies that \(tw (G)>k\). Furthermore, there exists an algorithm which, given a graph G and an integer k, runs in time \(\mathcal {O}(k^{\mathcal {O}(k^3)}n)\) and either outputs a pathdecomposition of G of width at most k or correctly identifies that \(pw (G)>k\).
Fact 3
(Folklore) Let \(\mathcal {T}\) be a treedecomposition of G and \(t\in T\). Then each connected component of \(GX_t\) lies in a single subtree of \(Tt\). In particular, for each connected component C of \(GX_t\) there exists a subtree \(T'\) of \(Tt\) such that for each vertex \(a\in C\) there exists \(t_a\in T'\) such that \(a\in X_{t_a}\).
We note that if G is a directed graph, then \(tw (G)\) and a treedecomposition of G refer to the treewidth and a treedecomposition of the underlying undirected graph of G, i.e., the undirected graph obtained by replacing each directed edge with an edge (and removing duplicate edges).
2.2 Monadic Second Order Logic
We consider Monadic Second Order (MSO) logic on (edge)labeled directed graphs in terms of their incidence structure whose universe contains vertices and edges; the incidence between vertices and edges is represented by a binary relation. We assume an infinite supply of individual variables\(x,x_1,x_2,\dots \) and of set variables\(X,X_1,X_2,\dots \) The atomic formulas are Vx (“x is a vertex”), Ey (“y is an edge”), Ixy (“vertex x is incident with edge y”), Hxy (“vertex x is the head of the edge y”), Txy (“vertex x is the tail of the edge y”), \(x=y\) (equality), \(x\ne y\) (inequality), \(P_a x\) (“vertex or edge x has label a”), and Xx (“vertex or edge x is an element of set X”). MSO formulas are built up from atomic formulas using the usual Boolean connectives \((\lnot ,\wedge ,\vee ,\rightarrow ,\leftrightarrow )\), quantification over individual variables (\(\forall x\), \(\exists x\)), and quantification over set variables (\(\forall X\), \(\exists X\)).
Let \(\varPhi (X)\) be an MSO formula with a free set variable X. For a labeled graph \(G=(V,E)\) and a set \(S\subseteq E\) we write \(G \models \varPhi (S)\) if the formula \(\varPhi \) holds true on G whenever X is instantiated with S.
The following result (an extension of the wellknown Courcelle’s Theorem [8]) shows that if G has bounded treewidth then we can count the number of sets S with \(G \models \varPhi (S)\).
Fact 4
[1] Let \(\varPhi (X)\) be an MSO formula with a free set variable X and w a constant. Then there is a lineartime algorithm that, given a labeled directed graph \(G=(V,E)\) of treewidth at most w, outputs the number of sets \(S \subseteq E\) such that \(G \models \varPhi (S)\).
We note that the above result requires a treedecomposition of width at most w to be provided with the input. However, as seen in Fact 2, for a graph of treewidth at most w such a tree decomposition can be found in linear time, hence we can drop this requirement from the statement of the theorem.
2.3 Posets
A partially ordered set (poset) \(\mathcal {P}\) is a pair \((P,\le ^P)\) where P is a set and \(\le ^P\) is a reflexive, antisymmetric, and transitive binary relation over P. The size of a poset \(\mathcal {P}=(P,\le ^P)\) is \({\mathcal P}:=P\). We say that pcovers\(p'\) for \(p,p' \in P\), denoted by \(p' \lhd ^P p\), if \(p' \le ^P\! p\), \(p \ne p'\), and for every \(p''\) with \(p' \le ^P\! p''\le ^P\! p\) it holds that \(p''\in \{p,p'\}\). We say that p and \(p'\) are incomparable (in \(\mathcal {P}\)), denoted \(p \parallel ^P p'\), if neither \(p \le ^P\! p'\) nor \(p' \le ^P\! p\).
A chainC of \(\mathcal {P}\) is a subset of P such that \(x \le ^P y\) or \(y \le ^P x\) for every \(x,y \in C\). An antichainA of \(\mathcal {P}\) is a subset of P such that for all \(x,y \in A\) it is true that \(x \parallel ^P y\). A family \(C_1,\dots ,C_\ell \) of pairwise disjoint subsets of P forms a total order if for each \(i,j\in [\ell ]\) and each \(a\in C_i\), \(b\in C_j\), it holds that \(a\le b\) iff \(i<j\). Furthermore, for each \(i\in [\ell 1]\) we say that \(C_i\) and \(C_{i+1}\) are consecutive. We call a poset \(\mathcal {P}\) such that every two elements of \(\mathcal {P}\) are comparable a linear order. A linear extension of a poset \(\mathcal {P}= (P,\le ^P)\) is a reflexive, antisymmetric, and transitive binary relation \(\preceq \) over P such that \(x\preceq y\) whenever \(x\le ^P y\) and the poset \(\mathcal {P}^*=(P, \preceq )\) is a linear order.
We consider the following graph representations of a poset \(\mathcal {P}=(P,\le ^\mathcal {P})\). The cover graph of \(\mathcal {P}\), denoted \(C(\mathcal {P})\), is the undirected graph with vertex set P and edge set \(\{\{a,b\}~~a\lhd b\}\). The incomparability graph of \(\mathcal {P}\), denoted \(I(\mathcal {P})\), is the undirected graph with vertex set P and edge set \(\{\{a,b\}~~a\parallel b\}\). The combined graph of \(\mathcal {P}\), denoted \(I_C(\mathcal {P})\), is the directed graph with vertex set P and edge set \(\{(a,b)~~(a\lhd b)\vee (a\parallel b)\}\). Finally, the poset graph of \(\mathcal {P}\), denoted \(P_G(\mathcal {P})\), is the directed graph with vertex set P and edge set \(\{(a,b)~~a\le b\}\). We will use the following known fact about treedecompositions and pathdecompositions of incomparability graphs.
Fact 5
[20, Theorem 2.1] Let \(\mathcal {P}\) be a poset. Then \(tw (I(\mathcal {P})) = pw (I(\mathcal {P}))\).
2.4 Parameterized Complexity
We refer the reader to [9, 12, 16] for an indepth introduction to parameterized complexity; here we only briefly summarize the most important notions required by our results.
A parameterized counting problem\(\mathcal {A}\) is a function \(\varSigma ^* \times \mathbb {N}\rightarrow \mathbb {N}\) for some finite alphabet \(\varSigma \). We call a parameterized counting problem \(\mathcal {A}\)fixedparameter tractable (\({\textsf {FPT}}\)) if \(\mathcal {A}\) can be computed in time \(f(k) \cdot x^{\mathcal {O}(1)}\) where f is an arbitrary computable function and (x, k) is the instance. The complexity class W[1] can be seen an the analog of NP for parameterized decision problems. To prove that a parameterized problem \(\mathcal A\) is W[1]hard, one can give an fpt turing reduction from a known W[1]hard problem \(\mathcal B\) to \(\mathcal A\); such an fpt turing reduction is a deterministic algorithm solving \(\mathcal A\) with an oracle to \(\mathcal B\) with the following properties: (a) the algorithm is FPT, and (b) the parameter for \(\mathcal B\) in each oracle query is bounded by a function of the parameter for \(\mathcal A\). To avoid confusion, we remark that there also exists the complexity class \(\#\)W[1] which is an analog of \(\#\)P for parameterized counting problems.
We denote by \(\#EC (G,r)\) the number of equitable colorings of graph G with r colors. We remark that the problem remains W[1]hard even if we restrict to the instances, where V(G) is divisible by r. This can be seen, for example, by padding of the instance by a single isolated clique.
3 FixedParameter Intractability of Counting Linear Extensions
The goal of this section is to prove Theorem 1, stated below.
Theorem 1
#LE parameterized by the treewidth of the cover graph of the input poset does not admit an fpt algorithm unless W[1]=\({\textsf {FPT}}\).
We begin by giving a brief overview of the proof, whose general outline follows the #Phardness proof of the problem [6]. However, since our parameter is treewidth, we needed to reduce from a problem that is not fixedparameter tractable parameterized by treewidth. Consequently, instead of reducing from SAT, we will use Equitable Coloring. This made the reduction considerably more complicated and required the introduction of novel gadgets, which allow us to encode the problem without increasing the treewidth too much.
The proof is based on solving an instance (G, r) of Equitable Coloring[tw] in FPT time using an oracle that solves #LE in FPT time parameterized by the treewidth of the cover graph (i.e., an fpt turing reduction). The first step is the construction of an auxiliary poset \(\mathcal {P}(G,r)\) of size \(2(r1)V(G)+(r^21)E(G)\). Then, for a given sufficiently large (polynomially larger than V(G)) prime number p, we show how to construct a poset \(\mathcal {P}(G,r,p)\) such that \(e(\mathcal {P}(G,r,p))\equiv e(\mathcal {P}(G,r))\cdot \#EC (G,r)\cdot A_p \mod p\), where \(A_p\) is a constant that depends on p and is not divisible by p. Therefore, if we choose a prime p that does not divide \(e(\mathcal {P}(G,r))\cdot \#EC (G,r)\), then \(e(\mathcal {P}(G,r,p))\) will not be divisible by p. Using Fact 1 we show that if \(\#EC (G,r)\ne 0\), then there always exists a prime p within a specified polynomial range of V such that p does not divide \(e(\mathcal {P}(G,r))\cdot \#EC (G,r)\).
From the above, it follows that there exists an equitable coloring of G with r colors if and only if, for at least one prime p within a specified (polynomial) number range, the number of linear extensions of \(\mathcal {P}(G,r,p)\) is not divisible by p. Moreover, we show that all inputs for the oracle will have size polynomial in the size of G and treewidth bounded by a polynomial in \(tw (G)+r\). Before proceeding to a formal proof of Theorem 1, we state two auxiliary lemmas which will be useful for counting linear extensions later in the proof.
Lemma 1
Proof
In the following we say that a set C of elements of a poset \(\mathcal {P}\) is a connected component of \(\mathcal {P}\) if C is a connected component of \(C(\mathcal {P})\).
Lemma 2
Let p be a prime number and \(\mathcal {Q}\) be a connected component of a poset \(\mathcal {P}\) such that \(\mathcal {Q}=p1\). If the number of linear extensions of \(\mathcal {P}\) is not divisible by p, then the number of elements in each connected component of \(\mathcal {P}\) other than \(\mathcal {Q}\) is divisible by p.
Proof
We now proceed to the proof of our main theorem.
Proof
(of Theorem 1) The proof is structured as follows. We begin by giving the construction of \(\mathcal {P}(G,r)\) and \(\mathcal {P}(G,r,p)\), after which we establish the desired properties of \(\mathcal {P}(G,r,p)\) and \(\mathcal {P}(G,r)\), and summarize in the conclusion.
Let G be a graph, r be an integer and p be a prime number as above. Recall that V(G) is divisible by r and let \(s={{V(G)}\over {r}}\) (note that this implies that each color in an equitable coloring of G must occur precisely s times in G). We proceed with a description of the poset \(\mathcal {P}(G,r,p)\). The poset \(\mathcal {P}(G,r,p)\) is split into \(r+3\) “levels” \(L_1,\dots ,L_{r+3}\) by linearly ordered elements \(a_0\le a_1\le \dots \le a_{r+2} \le a_{r+3}\), called the anchors. Each of these levels, besides \(L_{r+3}\), will consist of some flowers and a chain of \(p1\) elements which we call a stick; each of these flowers and the stick will always be pairwise incomparable. The anchors \(a_0\) and \(a_{r+3}\) are the unique minimum and maximum elements, respectively. The stick and all the stalks of flowers in level \(L_{i}\) will always lie between two consecutive elements \(a_{i1}\) and \(a_{i}\), and the petals of these flowers will be incomparable with \(a_i\) as well as some anchors above that (as defined later). Observe that while the relative position of any stalk and any anchor is fixed in every linear extension, petals can be placed above \(a_i\).
The first r levels are socalled color class levels, each representing one color class. We use these levels to make sure that every color class contains exactly s vertices. Aside from the stick, each such level contains a single (V(G), s)flower. Recall that the stalk and the stick on level \(1\le i\le r\) both lie between anchors \(a_{i1}\) and \(a_i\), and that the stalk and the flower are incomparable. We associate each petal of the flower at level \(L_i\) with a unique vertex \(v\in V(G)\) and denote the petal \(v_i\). Each petal \(v_i\) will be incomparable with all anchors above \(a_{i1}\) up to \(a_{r+3}\), i.e., \(v_i\parallel a_j\) for \(i\le j\le r+2\) and \(v_i\le a_{r+3}\). Intuitively, the flower in each color class level will later force a choice of s vertices to be assigned the given color.
Level \(L_{r+1}\) is called the vertex level and consists of one stick and V(G)many (r, 1)flowers; the purpose of this level is to ensure that every vertex is assigned exactly one color. Each flower is associated with one vertex \(v\in V(G)\) and we denote the petals of the flower associated with vertex v as \(v^i\) for \(1\le i\le r\). We set \(v_i \le v^i\) for all \(v\in V(G)\) and \(1\le i\le r\).
Level \(L_{r+2}\) is called the edge level, and its purpose is to ensure that the endpoints of every edge have a different color. It consists of a stick and E(G)many \((r^2,1)\)flowers. Each flower is associated with one edge \(e=uv\in V(G)\) and we denote the petals of the flower associated with e as \(e_{i,j}\) for \(1\le i\le r\) and \(1\le j\le r\). Moreover, for edge \(e=uv\) we set \(u^i \le e_{i,j}\), \(v^j\le e_{i,j}\), and we set \(a_{r+2} \le e_{i,j}\) whenever \(i=j\). Observe that this forces any petal \(e_{i,i}\) to lie between \(a_{r+2}\) and \(a_{r+3}\) in every linear extension (i.e., prevents \(e_{i,i}\) from being “selected”).
Level \(L_{r+3}\) is called the trash level. It does not contain any new elements in the poset, but it plays an important role in the reduction: we will later show that any petals which are interpreted as “not selected” must be located between \(a_{r+2}\) and \(a_{r+3}\) in any linear extension that is not automatically “canceled out” due to counting modulo p.
Establishing the desired properties of\(\mathcal {P}(G,r,p)\)and\(\mathcal {P}(G,r)\) We begin by formalizing the notion of selection. Let a configuration be a partition \(\phi \) of petals of all flowers into \(r+3\) sets \(L_1^\phi , \dots , L_{r+3}^\phi \). Let \(\varPhi \) denote a set of all configurations. We say that a linear extension \(\preceq \) of \(\mathcal {P}(G,r,p)\)respects the configuration \(\phi \) if \(L_1^\phi \preceq a_1\preceq L_2^\phi \preceq a_2\preceq \dots \preceq a_{r+2}\preceq L_{r+3}^\phi \) and we denote the set of all linear extensions of \(\mathcal {P}(G,r,p)\) that respects \(\phi \) by \(\mathcal {L}^\phi \). We say that a configuration \(\phi \) is consistent if \(\mathcal {L}^\phi \) is nonempty; this merely means that \(L_1^\phi \le a_1\le L_2^\phi \le a_2\le \dots \le a_{r+2}\le L_{r+3}^\phi \) does not violate any inequalities in \(\mathcal {P}(G,r,p)\). Observe that if \(\phi \) is consistent, then \(\mathcal {L}^\phi \) is exactly the set of linear extension of the partial order \(\mathcal {P}^\phi (G,r,p)\), where \(\mathcal {P}^\phi (G,r,p)\) is obtained by enriching \(\mathcal {P}(G,r,p)\) with the relations \(L_1^\phi \le a_1\le L_2^\phi \le a_2\le \dots \le a_{r+2}\le L_{r+3}^\phi \) and performing transitive closure (in other words, \(\mathcal {P}^\phi (G,r,p)\) is obtained by enforcing \(\phi \) onto \(\mathcal {P}(G,r,p)\)).
Since every linear extension of \(\mathcal {P}(G,r,p)\) respects exactly one configuration, it is easy to see that \(e(\mathcal {P}(G,r,p)) = \sum _{\phi \in \varPhi } \mathcal {L}^\phi  = \sum _{\phi \in \varPhi } e(\mathcal {P}^\phi (G,r,p))\). Intuitively, a configuration \(\phi \)contributes to the above sum modulo p if \(e(\mathcal {P}^\phi (G,r,p))\) is not divisible by p. We shall prove that the only configurations which contribute to this sum modulo p are those where from every (a, b)flower there are exactly b petals in the same level as the stalk, and the remaining \(ab\) petals are in the trash. Furthermore, in each configuration \(\phi \) which contributes to the above sum modulo p, the petals in \(L_{r+1}^\phi \) represent a proper equitable coloring of G with r colors, and each such configuration is respected by the same number of linear extensions.
Let us first remark that for any configuration \(\phi \), the anchors \(a_0,a_1,\dots ,a_{r+3}\) are comparable to all elements of \(\mathcal {P}^\phi (G,r,p)\). Now, let \(\mathcal {P}_{L_i}^\phi \) be the poset induced by all elements \(e\in \mathcal {P}^\phi (G,r,p)\) such that \(a_{i1}\le e\le a_i\) and \(e\ne a_{i1}\), \(e\ne a_i\). It is readily seen that \(e(\mathcal {P}^\phi (G,r,p)) = \prod _{i=1}^{r+3}e(\mathcal {P}_{L_i}^\phi )\). We proceed by stating a series of claims about our construction.
Claim 1
For each \(i\in \{1,\dots , r\}\), it holds that either \(e(\mathcal {P}_{L_i}^\phi ) \equiv 0 \mod p\), or \(e(\mathcal {P}_{L_i}^\phi ) = s!{{2p1} \atopwithdelims (){p}}\) and \(L_i^\phi \) contains exactly s petals of \(A_i\) and no other petals.
Proof
(of the Claim) Assume that \(e(\mathcal {P}_{L_i}^\phi ) \not \equiv 0 \mod p\) and recall that level \(L_i\) contains a stick, which is a chain of \(p1\) elements that is incomparable with all elements of \(\mathcal {P}_{L_i}^\phi \) in every configuration \(\phi \). By Lemma 2 this implies that every connected component of \(\mathcal {P}_{L_i}^\phi \) distinct from the stick has size divisible by p. Clearly, \(L_i^\phi \) contains only those stalks that are associated with the level \(L_i\), and it contains all such stalks. It is readily seen from the construction that any petal in \(\cup _{j<i}A_j\) would necessarily form a component of size one in \(\mathcal {P}_{L_i}^\phi \). Hence, \(\mathcal {P}_{L_i}^\phi \) contains only elements associated with level \(L_i\), namely elements of the stick and elements of a (V(G), s)flower. Moreover, by Lemma 2 and the fact that \(V(G)+ps<2p\), each such flower has exactly p elements in level \(\mathcal {P}_{L_i}^\phi \). Since the \(ps\) elements of the stalk must be in \(\mathcal {P}_{L_i}^\phi \), the poset \(\mathcal {P}_{L_i}^\phi \) contains exactly s elements of \(A_i\). Clearly, the number of linear extensions of the petals of the (V(G), s)flower in \(\mathcal {P}_{L_i}^\phi \) is s! and hence by Lemma 1\(e(\mathcal {P}_{L_i}^\phi ) = s!{{2p1} \atopwithdelims (){p}}\), which concludes the proof. \(\square \)
Claim 2
Either \(e(\mathcal {P}_{L_{r+1}}^\phi ) \equiv 0 \mod p\), or \(e(\mathcal {P}_{L_{r+1}}^\phi ) = {{(V(G)p+p1)!} \over {(p1)}!(p!)^{V(G)}}\) and \(L_{r+1}^\phi \) contains exactly V(G) elements of \(A_{r+1}\), specifically one petal for each (r, 1)flower on level \(L_{r+1}\).
Proof
(of the Claim) Assume \(e(\mathcal {P}_{L_{r+1}}^\phi ) \not \equiv 0 \mod p\), and let us first examine elements that are not associated with level \(L_{r+1}\). Clearly, no element associated with level \(L_{r+2}\) can appear in \(\mathcal {P}_{L_{r+1}}^\phi \) and the only elements associated with any level \(i<r+1\) that can end up in \(\mathcal {P}_{L_{r+1}}^\phi \) are petals. Each of these elements is smaller than exactly one petal at level \(L_{r+1}\) and incomparable to all other elements associated with this level. It is easy to see that largest possible size of a connected component of \(\mathcal {P}_{L_{r+1}}^\phi \) is \(p1+2r<2p\). By Lemma 2, every connected component in \(\mathcal {P}_{L_{r+1}}^\phi \) (except for the stick) will have size p, and therefore \(\mathcal {P}_{L_{r+1}}^\phi \) will contain exactly one element for every set of petals associated with \(L_{r+1}\) and no other elements. Hence, \(\mathcal {P}_{L_{r+1}}^\phi \) consists of V(G) chains of length p and one chain of length \(p1\). Then \(e(\mathcal {P}_{L_{r+1}}^\phi ) = {{(V(G)p+p1)!} \over {(p1)}!(p!)^{V(G)}}\) follows from Lemma 1. \(\square \)
Claim 3
Either \(e(\mathcal {P}_{L_{r+2}}^\phi ) \equiv 0 \mod p\), or \(e(\mathcal {P}_{L_{r+2}}^\phi ) = {{(E(G)p+p1)!} \over {(p1)}!(p!)^{E(G)}}\) and \(L_{r+2}^\phi \) contains exactly E(G) elements of \(A_{r+2}\), specifically one petal for each \((r^2,1)\)flower on level \(L_{r+2}\).
Proof
(of the Claim) The idea of the proof is similar to the proof of the previous claim, with one additional obstacle: that several flowers can be connected with petals from lower levels into one connected component on level \(L_{r+2}\) through the petals of flowers on level \(L_{r+1}\). So, assume \(e(\mathcal {P}_{L_{r+2}}^\phi )\) contains a connected component C which contains at least a single stalk. If C contains precisely the single stalk, then by Lemma 2 we have \(e(\mathcal {P}_{L_{r+2}}^\phi ) \equiv 0 \mod p\). Otherwise, for each stalk in C, there must be at least one petal in the same flower (otherwise the stalk cannot be connected to the rest of C); in other words, the intersection of each flower and C contains at least p vertices. Let a denote the number of flowers which intersect C, \(b_2\) denote \(A_{r+2}\cap C\), \(b_1\) denote \(A_{r+1}\cap C\) and \(b_0\) denote \(\sum _{i=1}^rA_r\cap C\). Then it follows that \(C=p\cdot a+(b_2a)+b_1+b_0\le p\cdot a+r^2E+rV+rV\), and recall that \(r^2E+rV+rV<p\). Furthermore, if \(b_1>0\) (and at least one petal from \(A_{r+1}\) is required unless C contains only a single flower), we have \(a\cdot p<C<(a+1)\cdot p\). Hence any such C cannot have size divisible by p and by Lemma 2 we have \(e(\mathcal {P}_{L_{r+2}}^\phi ) \equiv 0 \mod p\). Otherwise, if no two flowers are connected through a petal of a flower associated with level \(L_{r+1}\), then every connected component of \(\mathcal {P}_{L_{r+2}}^\phi \) of size p must consist of a stalk and exactly one petal and the claim follows analogously as the proof of Claim 2. \(\square \)
Claim 4
If \(\phi \) is a consistent configuration and for all \(i\in \{1,\dots , r+2\}\) it holds that \(e(\mathcal {P}_{L_{i}}^\phi ) \not \equiv 0 \mod p\), then 1) the petals in \(L_{r+1}^\phi \) encode a proper equitable coloring of V(G) where vertex v receives color i iff the petal \(v_i\) lies in \(L_i^\phi \), and 2)\(\mathcal {P}_{L_{r+3}}^\phi \) is isomorphic with \(\mathcal {P}(G,r)\).
Proof
(of the Claim) From Claims 1, 2 and 3 together with the assumption that \(e(\mathcal {P}_{L_{i}}^\phi ) \not \equiv 0 \mod p\), it follows that each of the levels \(L_1^\phi ,\dots ,L_{r}^\phi \) contains exactly s petals associated with the corresponding level, level \(L_{r+1}^\phi \) contains exactly one petal for each vertex of G and level \(L_{r+2}^\phi \) contains exactly one petal for each edge of G.
For the first part of this claim, we observe that each pair of petals in \(L_1^\phi ,\dots , L_r^\phi \) are associated with distinct vertices of G. If this were not the case, then since \(V(G)=rs\) there would exist a vertex v such that no element of \(L_1^\phi ,\dots , L_r^\phi \) is associated with v. But due to the construction at level \(r+1\) there exists some \(i\in \{1,\dots ,r\}\) such that \(v^i\in L_{r+1}^\phi \). Then, since \(v_i\le v^i\) and \(v_i\) can only occur either in level \(L_{i}^\phi \) or \(L_{r+3}^\phi \) (the latter of which lies above \(v^i\) in the linear extension due to the configuration \(\phi \)), this would lead to a contradiction. In particular, we conclude that there is a matching between the petals in level \(r+1\) (encoding the color for each vertex) and the union of petals in levels \(1,2,\dots r\) (encoding the vertices assigned to each color class), and by Claim 1 it follows that there are exactly s petals in \(L_{r+1}\) associated with each color class.
We now argue that the coloring is proper. Observe that by the same argument as above, if an edge \(e=uv\) satisfies \(e_{i,j}\in L_{r+2}^\phi \), then \(u^i\in L_{r+1}^\phi \) and \(v^j\in L_{r+1}^\phi \). From the construction of \(\mathcal {P}(G,r,p)\) it follows that if \(i=j\), then \(e_{i,j}\not \in L_{r+2}\). Combining these two facts we get that the coloring encoded in \(L_{r+1}^\phi \) is indeed proper.
Now let us take a look at level \(L_{r+3}^\phi \). To prove the claim, we will construct an isomorphism f from elements of \(\mathcal {P}_{L_{r+3}}^\phi \) to elements of \(\mathcal {P}(G,r)\). For every vertex \(v\in V(G)\), precisely one element \(v^i\in L_{r+1}^\phi \) and precisely one of the first r levels contains an element associated with v; to be precise, \(v_i\in L_{i}^\phi \) and \(v_j\in L_{r+3}^\phi \) and hence also \(v^j\in L_{r+3}^\phi \) for all \(j\not = i\). We set \(f(v^j)=v_{j,0}\) and \(f(v_j)=v_{j,1}\), whenever \(j\not = i\) and \(j<r\). For the last remaining elements, we set \(f(v^r)=v_{i,0}\) and \(f(v_r)=v_{i,1}\). Next, for every edge \(e=uv\) there is exactly one \(e_{a,b}\in L_{r+2}^\phi \). Moreover, if \(e_{a,b}\in L_{r+2}^\phi \) then \(u^a\in L_{r+1}^\phi \) and \(v^b\in L_{r+1}^\phi \), and all other petals for this edge e are in \(L_{r+3}^\phi \). Let \(g_i(r)=i\), \(g_i(i)=0\), and \(g_i(k)=k\) otherwise. Then we set \(f(e_{i,j})=e_{g_a(i),g_b(j)}\). Observe that, since \(e_{a,b}\) does not lie in \(L_{r+3}^\phi \), no edge is mapped to the nonexistent element \(e_{0,0}\) in \(\mathcal {P}(G,r)\). It is straightforward to verify that f is really bijective mapping between elements of \(\mathcal {P}_{L_{r+3}}^\phi \) and \(\mathcal {P}(G,r)\). Moreover, \(f(u)\le f(v)\) in \(\mathcal {P}(G,r)\) if and only if \(u\le v\) in \(\mathcal {P}_{L_{r+3}}^\phi \). Therefore, \(\mathcal {P}_{L_{r+3}}^\phi \) is isomorphic with \(\mathcal {P}(G,r)\) and the claim holds. \(\square \)
Claim 5
\(e(\mathcal {P}(G,r,p))\not \equiv 0 \mod p\) if and only if \(e(\mathcal {P}(G,r))\cdot \#EC (G,r)\not \equiv 0 \mod p\).
Proof
Claim 6
If \(\#EC (G,r)\ne 0\), then there is a prime number p greater than \(2rV(G)+r^2E(G)\) and smaller than \((2rV(G)+r^2E(G))^2\) such that p does not divide \(e(\mathcal {P}(G,r))\cdot \#EC (G,r)\).
Proof
(of the Claim) Let us first upperbound \(e(\mathcal {P}(G,r))\#EC (G,r)\). Clearly, \(\mathcal {P}(G,r)\) contains \(m = 2(r1)V(G)+(r^21)E(G)\) elements, hence \(e(\mathcal {P}(G,r))\le m!\). It can easily be verified that the number of possibilities of dividing \(V(G)=rs\) vertices into r color classes with exactly s colors each is \({{(rs)!} \over {(s!)^r}}\). By Fact 1, the product of all primes between \(2rV(G)+r^2E(G)\) and \((2rV(G)+r^2E(G))^2\) is at least \((2rV(G)+r^2E(G))!2^{2rV(G)+r^2E(G)}\). However, \(2(r1)V(G)+(r^21)E(G)+ V(G) \le 2rV(G)+r^2E(G)\) and hence \(e(\mathcal {P}(G,r))\#EC (G,r)\le (2(r1)V(G)+(r^21)E(G))!+ {{V(G)!}\over {(s!)^r}}\) is clearly less than the product of all primes between \(2rV(G)+r^2E(G)\) and \((2rV(G)+r^2E(G))^2\). Note that if a natural number N is divisible by set of primes \(p_1,\dots , p_\ell \) then N is divisible by the product of these primes and in particular N is bigger than the product of these primes. Therefore, \(e(\mathcal {P}(G,r))\#EC (G,r)\) cannot be divisible by all primes between \(2rV(G)+r^2E(G)\) and \((2rV(G)+r^2E(G))^2\), from which the claim follows. \(\square \)
Claim 7
\(tw (C(\mathcal {P}(G,r,p)))\le r\cdot (tw (G)+3)+6\).
Proof
 1.
All bags of \(\mathcal {T}'\) will contain the anchors \(a_0,\dots , a_{r+3}\) as well as the topmost element of each stalk of the (V(G), s)flowers in the first r levels; let \(\delta \) denote this set of \(2r+4\) elements.
 2.
For every bag \(t\in \mathcal {T}\), the treedecomposition \(\mathcal {T}'\) will contain a node \(t'\) such that if \(v\in X_{t}\) then \(\{v^1,\dots , v^r\}\in X_{t}'\).
 3.
Afterwards, every introduce node \(t\in \mathcal {T}\) that introduces a vertex v will be replaced by a long path \(P_t\) which gradually introduces and subsequently forgets all remaining elements associated with the flower of v at level \(r+1\) (i.e., the stalk) as well as every petal from the first r levels associated with v .
 4.
For each edge e, we pick an introduce node \(t\in \mathcal {T}\) which contains both endpoint of e and extend the path \(P_t\) by a new segment which introduces and subsequently forgets all elements associated with the flower of e at level \(r+2\).
 5.
The root node is replaced by a path that takes care of all elements which are not associated with any vertex or edge in G.
It is easy to see now, that when we are forgetting vertex v in node t in \(\mathcal {T}\), we can forget elements \(v^1,\dots , v^r\) in \(\mathcal {T}'\), because we already introduced all its adjacent edges in \(C(\mathcal {P}(G,r,p))\) either in the path corresponding to the node of \(\mathcal {T}\) introducing v or the one introducing a neighbor of v.
Finally, when we get to the root node of \(\mathcal {T}\), we have already forgotten all elements associated with any specific vertex or edge of G. Therefore, the only elements besides \(\delta \) which need to be included in \(\mathcal {T}'\) are the remaining elements in the stalks in the first r levels and the sticks in every level. However, it is easy to see that at this point they all form separate chains in \(C(\mathcal {P}(G,r,p)){\setminus } \delta \). Hence there once again exists a pathdecomposition of width at most \(\delta +2\) which gradually introduces and subsequently forgets all of these elements.
One can readily see that the properties (T1), (T2), and (T3) are satisfied and we are only left with computing the width of \(\mathcal {T}'\). By construction, every join and forget node in \(\mathcal {T}\) will become a node in \(\mathcal {T}'\) whose bag has size at most \(r\cdot X_t+2r+4\). On the other hand, every introduce node in \(\mathcal {T}\) will become a path in \(\mathcal {T}'\), and the largest bag on this path has size at most \(r\cdot X_t+2r+6\), from which the claim follows. \(\square \)
Concluding the proof Let us summarize the fpt turing reduction used to prove Theorem 1. Given an instance (G, r) of Equitable Coloring[tw], we loop over all primes p such that \(2rV(G)+r^2E(G)<p<(2rV(G)+r^2E(G))^2\), and for each such prime we construct the poset \(\mathcal {P}(G,r,p)\); from Claim 6 it follows that if \(\#EC (G,r)\ne 0\), then at least one such prime will not divide \(e(\mathcal {P}(G,r))\cdot \#EC (G,r)\), and by Claim 7 the cover graph of each of the constructed posets P(G, r, p) has bounded treewidth. For each such poset \(\mathcal {P}(G,r,p)\), we compute \(e(\mathcal {P}(G,r,p))\) by the blackbox procedure provided as part of the reduction. If for any prime p we get \(e(\mathcal {P}(G,r,p))\not \equiv 0\mod p\), then we conclude that (G, r) is a yesinstance, and otherwise we reject (G, r), and this is correct by Claim 5. \(\square \)
4 FixedParameter Tractability of Counting Linear Extensions
This section is dedicated to proving our algorithmic result, stated below.
Theorem 2
#LE is fixedparameter tractable parameterized by the treewidth of the incomparability graph of the input poset.
The proof of Theorem 2 is divided into two steps. First, we apply a transformation process to a pathdecomposition \(\mathcal {Q}\) of small width (the existence of which is guaranteed by Corollary 1) of \(I(\mathcal {P})\) which results in a treedecomposition \(\mathcal {T}\) of \(I(\mathcal {P})\) satisfying certain special properties. The properties of \(\mathcal {T}\) are then used to prove that \(I_C(\mathcal {P})\) has treewidth bounded by the treewidth of \(I(\mathcal {P})\). In the second step, we construct an MSO formulation which enumerates all the linear extensions of \(\mathcal {P}\) using \(I_C(\mathcal {P})\), and apply Fact 4.
4.1 The Treewidth of Combined Graphs
We begin by arguing a useful property of separators in incomparability graphs.
Lemma 3
Let \(S\subseteq V(I(\mathcal {P}))\). Then for each pair of distinct connected components \(C_1, C_2\) in \(I(\mathcal {P}) S\), it holds that for any \(a_1,b_1\in C_1\) and any \(a_2,b_2\in C_2\) we have \(a_1\le a_2\) iff \(b_1\le b_2\). Namely, the poset contains a total order of all connected components in \(I(\mathcal {P}) S\).
Proof
We begin by proving the following claim.
Claim 8
Let a, b, c be three distinct elements of \(\mathcal {P}\) such that \(a \parallel b\) and both pairs a, c and b, c are comparable. Then \(a \le c\) iff \(b \le c\).
Proof
(of the Claim) Suppose that, w.l.o.g., \(a\le c\) and \(c\le b\). Then by the transitivity of \(\le \), we get \(a\le b\) which contradicts our assumption that \(a\parallel b\). \(\square \)
Now to prove Lemma 3, assume for a contradiction that, w.l.o.g., there exist \(a_1,b_1\in C_1\) and \(a_2, b_2\in C_2\) such that \(a_1\le b_1\) and \(b_2\le a_2\). Let \(Q_1\) be an \(a_1\)\(a_2\) path in \(I[C_1]\). By Claim 8, \(a_1\le b_1\) implies that every element q on \(Q_1\) satisfies \(q\le b_1\), and in particular \(a_2\le b_1\). Next, let \(Q_2\) be a \(b_1\)\(b_2\) path in \(I[C_2]\). Then Claim 8 also implies that each element \(q'\) on \(Q_2\) satisfies \(a_2\le q'\). Since \(b_2\) lies on \(Q_2\), this would imply that \(a_2\le b_2\), a contradiction. \(\square \)
To proceed further, we will need some additional notation. Let \(\mathcal {T}=(T,\mathcal {X})\) be a rooted treedecomposition and \(t\in T\). We denote by L(t) the set of all vertices which occur in the “branch” of \(Tt\) containing the root r; formally, \(L(t)=\{v\in X_{t'}{\setminus } X_t ~  ~ t'\) lies in the same connected component as \(r \text { in }Tt\}\). We then set \(R(t)=V(G){\setminus } (L(t)\cup X_t)\). We also let \(T^r_t\) denote the connected component of \(Tt\) which contains the root r.
Next, recall that each connected component of the graph obtained after deleting \(X_t\) must lie in a subtree of \(Tt\) (Fact 3). Let \(\varUpsilon _t\) be the set of connected components of \((I(\mathcal {P}){\setminus } X_t)\cap R(t)\). Recall that because of Lemma 3, the components of \(\varUpsilon _t\) are totally ordered by \(\mathcal {P}\). We say that two components \(B_1\), \(B_2\in \varUpsilon _t\) are consecutive if there is no element \(v\in V(I(\mathcal {P})){\setminus } (X_t\cup B_1\cup B_2)\) that is in between elements in these components; formally, for every v it holds that either \(v\le b\) for all \(b\in B_1\cup B_2\), or \(v\ge b\) for all \(b\in B_1\cup B_2\).
A block of a bag \(X_t\) is a maximum set of consecutive connected components in \((I(\mathcal {P})X_t)\cap R(t)\); note that each block has a natural total ordering among the contained components, given by Lemma 3. We say that a node \(t\in T\) has z blocks if there exist z distinct blocks of \(X_t\). Blocks will play an important role in the treedecomposition we wish to obtain from our initial pathdecomposition of \(I(\mathcal {P})\). The following lemma captures the operation we will use to alter our pathdecomposition.
Lemma 4
 1.
The width of \(\mathcal {T}'\) is at most the width of \(\mathcal {T}\).
 2.
The tree \(T'\) contains \(T^r_t\) as a subtree which is separated from the rest of \(T'\) by t.
 3.
The degree of t in \(T'\) is \(z+1\).
 4.
There exists a bijection \(\alpha \) between the z blocks of \(X_t\) and the z trees in \(T't\) other than \(T^r_t\) such that for each block B of \(X_t\), we have \(\bigcup _{s\in \alpha (B)}X'_s{\setminus } X_t = B\).
 5.
For each \(t'\in N[t]{\setminus } T^r_t\), we have \(X_{t'}=X_t\).
Proof
It will be useful to observe that \(T  T^r_t\) is a subtree of T and in particular it is connected. Consider the following construction of \(\mathcal {T}'\). First, we copy all nodes of \(T^r_t\cup \{t\}\) (along with their bags) into \(\mathcal {T}'\), thus ensuring that Property 2 holds. Second, for each block B of \(X_t\) we make a copy \(T^B\) of the tree \(T T^r_t\), and connect the node \(t^B\) corresponding to t in \(T{\setminus } T^r_t\) by an edge to the node t in \(T'\). Moreover, for each node \(s\in T{\setminus } T^r_t\) we set \(X'_{s^B}=X_s \cap (B\cup X_t)\). It is easy to verify that all of the required properties are now satisfied, and it remains to show that \(\mathcal {T}'\) is indeed a treedecomposition.
We argue that \(\mathcal {T}'\) satisfies all three properties of treedecompositions. Property (T1) follows directly from fact that \(\mathcal {T}\) was tree decomposition, and hence every vertex that does not occur in a bag in \(T^r_t\) must occur in some bag \(X_s\) for some node \(s\in T{\setminus } T^r_t\); then this vertex either also occurs in \(X_t\) or occurs in some block B and hence in \(X'_{s^B}\). Property (T2) is also straightforward, since each vertex either does not occur in any block or in exactly one block, and in both cases monotonicity follows from the monotonicity of \(\mathcal {T}\) and the construction. For the final Property (T3), we recall that there are no edges between the blocks of \(X_t\); in particular every edge \(e=ab\) in \(I(\mathcal {P})[X_t\cup R(t)]\) is either contained in \(X_t\), goes between a vertex of \(X_t\) and a vertex of some block B, or is contained in some block B. In all three cases, it holds that if \(a,b\in X_s\) for some \(s\in T{\setminus } T^r_t\), then \(e\in X_{s^B}\) for some block B. Therefore, \(\mathcal {T}'\) is a treedecomposition and the proof is complete. \(\square \)
We proceed by showing how Lemma 4 is applied to transform a given pathdecomposition.
Lemma 5
 1.
The degree of t in T is \(z+1\).
 2.
There exists a bijection \(\alpha \) between the z blocks of \(X_t\) and the z trees in \(T't\) other than \(T^r_t\) such that for each block B of \(X_t\), we have \(\bigcup _{s\in \alpha (B)}X_s {\setminus } X_t= B\).
 3.
For \(t'\in N(t)\cap T^r_t\) there exists a vertex v such that \(X_{t'}=X_t{\setminus } \{v\}\), and furthermore \(t'\) has degree at most 2 and 1 block.
 4.
For each pair of neighbors \(t,t'\in T\), it holds that \(X_t{\setminus } X_{t'}+X_{t'}{\setminus } X_t\le 1\).
Proof
 1.
For each pair of neighbors \(t,t'\in T\), it holds that \(X_t{\setminus } X_{t'}+X_{t'}{\setminus } X_t\le 1\).
 2.
For each vertex u that was already processed, the introduce node of u satisfies the conditions of the lemma.
 3.
Any node t of degree greater than 2 is an introduce node of an already processed vertex.
For the induction step, suppose that the conditions hold in a treedecomposition \(\mathcal {T}\) obtained by inductively applying Lemma 4 as above, and the first unprocessed vertex is v. Let t be the unique node where v is introduced, and let \(\mathcal {T}'\) be the treedecomposition we obtained by applying Lemma 4 on \(\mathcal {T}\) and t.
It is easy to verify that \(\mathcal {T}'\) then satisfies the desired Conditions 1, 2, and 4 at the node t by Lemma 4. As for Condition 3, since t is the introduce node of v and the first invariant condition holds in \(\mathcal {T}\), it is clear that for \(t'\in N(t)\cap T^r_t\) it is the case that \(X_{t'}=X_t{\setminus } \{v\}\). Moreover, \(t'\) cannot be an introduce node, since then \(t'\) would have to introduce an already processed vertex, which would imply that \(X_{t'} = X_{t}\) due to our application of Lemma 4 on introduce nodes. So, let us consider the node s on the unique \(t'\)r path that is the closest introduce node to \(t'\), and let \(s'\) be the neighbor of s on the s\(t'\) path. Since no vertex was introduced on the \(s'\)\(t'\) path, it follows that \(R(s')=R(t')\). Since \(s'\) only has 1 block by the construction, it must be the case that \(t'\) also only has one block, and so Condition 3 also holds.
We proceed by arguing that the invariant conditions remain satisfied by \(\mathcal {T}'\). Since \(\mathcal {Q}\) was nice and \(\mathcal {T}\) satisfied the first invariant condition, it is readily seen that the first invariant condition holds for all pairs of neighbors in \(T^r_t\) as well as for t with all of its neighbors. If \(s^B\) and \(s'^B\) are neighbors in a tree \(\alpha (B)\) of \(T't\), then by the construction in Lemma 4 there exists a pair of neighbors \(s,s'\in T{\setminus } T^r_t\) such that \(X'_{s^B}=X_s \cap (B\cup X_t)\) and \(X'_{s'^B}=X_{s'} \cap (B\cup X_t)\). But then \(X'_{s^B}{\setminus } X'_{s'^B}+X'_{s'^B}{\setminus } X'_{s^B} = (X_s \cap (B\cup X_t)){\setminus } (X_{s'} \cap (B\cup X_t))+(X_{s'} \cap (B\cup X_t)){\setminus } (X_s \cap (B\cup X_t)) \le X_{s}{\setminus } X_{s'}+X_{s'}{\setminus } X_{s}\le 1\) and the first invariant condition follows. As for the second invariant condition, notice that from the construction it follows that all the vertices that precede v in the order of introduction in \(\mathcal {Q}\) must have been introduced in some node of \(T^r_t\), and the application of Lemma 4 does not alter such introduce nodes for previously processed vertices. Finally, by the induction hypothesis all nodes of \(T{\setminus } T^r_t\) have degree at most 2, therefore from the construction in Lemma 4 it is clear that all nodes in \(T'{\setminus } (T^r_t\cup \{t\})\) also have degree 2. Since all introduce nodes of unprocessed vertices lie in \(T'{\setminus } (T^r_t\cup \{t\})\), we conclude that the third invariant condition also holds in \(\mathcal {T}'\).
Now, let us consider the treedecomposition \(\mathcal {T}\) obtained after processing all vertices of \(I(\mathcal {P})\) according to the procedure described above. \(\mathcal {T}\) satisfies Condition 4 due to the first invariant of our procedure, and for all other conditions it suffices to consider nodes with more than 1 block. In particular, it suffices to verify that all such nodes satisfy the conditions of Lemma 4 and additionally also condition 3 of this Lemma. So, suppose for a contradiction that there exists a node t which does not meet these conditions, but all nodes on the unique tr path do. Then there are two possibilities to consider for the unique neighbor \(t'\) of t on the tr path. If \(t'\) were to have more than 1 block, then by our assumptions \(t'\) would have to satisfy the conditions of Lemma 4, contradicting the fact that t has more than 1 block. On the other hand, if \(t'\) were to have only a single block, then by construction t must be an introduce node of some vertex v and by our invariants and the construction it follows that t in fact must satisfy all the required conditions. In particular we conclude that all nodes t satisfy Conditions 14 and the lemma follows. \(\square \)
We call a treedecomposition rooted at a leaf with \(X_r=\emptyset \) which satisfies the properties of Lemma 5 a blocked treedecomposition. The next ingredient we will need for proving that \(I_C(\mathcal {P})\) has small treewidth is the notion of coverguards.
Let \(\mathcal {T}=(T,\mathcal {X})\) be a treedecomposition of \(I(\mathcal {P})\) rooted at r and let \(t\ne r\). Then the coverguard of t, denoted \(\mathcal {A}_t\), is the set of vertices in L(t) which are incident to a cover edge whose other endpoint lies in R(t); formally, \(\mathcal {A}_t=\{v\in L(t) ~~ \exists u\in R(t): \{uv\}\in E(C(\mathcal {P})\). For a vertex \(v\in I(\mathcal {P})\), we let \(\mathcal {A}^v=\{t\in T~~v\in \mathcal {A}_t\}\) and \(X^v=\{t\in T~~v\in X_t\}\).
Our next aim is to add all the coverguards into each bag. Below, we show that this does not increase the size of bags too much.
Lemma 6
Let \(\mathcal {T}=(T,\mathcal {X})\) be a blocked treedecomposition of \(I(\mathcal {P})\) of width k. Then for each \(t\in T\) it holds that \(\mathcal {A}_t\le 2 k+2\).
Proof
First, observe that if a node \(t\in T\) has 0 blocks, then \(R(t)=\mathcal {A}_t=\emptyset \). So, consider a node t which has exactly 1 block consisting of connected components \((D_1,\dots ,D_j)\) in \((I(\mathcal {P})X_t)\cap R(t)\).
Claim 9
\(\mathcal {A}_t\le 2k+2\).
Proof
(of the Claim) Assume for a contradiction that \(\mathcal {A}_t>2k+2\). By Lemma 3 we have that \((D_1,\dots ,D_j)\) are consecutive connected components in a total order of connected components in \(I(\mathcal {P})X_t\). Hence any edge in \(C(\mathcal {P})X_t\) between R(t) and L(t) must necessarily have one endpoint in \(D_1 \cup D_j\). Furthermore, an element in \(\mathcal {A}_t\) cannot be adjacent to both \(D_1\) and \(D_j\) in \(C(\mathcal {P})X_t\) due to transitivity and acyclicity. So, we may partition \(\mathcal {A}_t\) into \(\mathcal {A}_t^1=\{v\in \mathcal {A}_t~~\exists u\in D_1: v\lhd ^\mathcal {P}u\}\) and \(\mathcal {A}_t^2=\{v\in \mathcal {A}_t~~\exists u\in D_j: u\lhd ^\mathcal {P}v\}\).
By Lemma 3, it also follows that \(\mathcal {A}_t^1\) and \(\mathcal {A}_t^2\) must each lie in separate connected components of \(I(\mathcal {P})X_t\), say \(C^1\) and \(C^2\) respectively. Furthermore, each element in \(\mathcal {A}_t^1\) is maximal in \(C^1\) and each element in \(\mathcal {A}_t^2\) is minimal in \(C^2\). In particular, each of \(\mathcal {A}_t^1\), \(\mathcal {A}_t^2\) forms a clique in \(I(\mathcal {P})\). But by our assumption on the size of \(\mathcal {A}_t\), at least one of \(\mathcal {A}_t^2\) and \(\mathcal {A}_t^1\) must have size greater than \(k+1\), which implies that \(I(\mathcal {P})\) contains a clique of size at least \(k+2\). It is wellknown that each clique must be completely contained in at least one bag of a treedecomposition, and so we arrive at a contradiction with \(tw (I(\mathcal {P}))\le k\). Hence we conclude that \(\mathcal {A}_t\le 2k+2\) and the claim holds. \(\square \)
Finally, consider a node t which has at least 2 blocks. By Property 3 of Lemma 5, it holds that t has a neighbor \(t'\) in \(T^r_t\) such that \(X_{t'}=X_t{\setminus } \{v\}\) and \(t'\) has 1 block. By Claim 9 we know that \(\mathcal {A}_{t'}\le 2k+2\). Since \(L(t)=L(t')\) and \(R(t)\subseteq R(t')\), it follows that \(\mathcal {A}_t\subseteq \mathcal {A}_{t'}\), and in particular \(\mathcal {A}_t\le \mathcal {A}_{t'}\). We have now proved the desired bound for all nodes in \(\mathcal {T}\), and so the lemma holds. \(\square \)
The following lemma allows us to argue that adding coverguards into each bag still results in a treedecomposition; it is worth noting that the assumption that the decomposition is blocked is essential for the lemma to hold.
Lemma 7
Let \(\mathcal {T}=(T,\mathcal {X})\) be a blocked treedecomposition of \(I(\mathcal {P})\) rooted at r and let \(v\in I(\mathcal {P})\). Then \(T[\mathcal {A}^v\cup X^v]\) is a tree.
Proof
Since \(\mathcal {T}\) is a treedecomposition, we have that \(T[X^v]\) must be a tree. Consider a connected component A of \(T[\mathcal {A}^v]\) and its unique A\(X^v\) path Q, with endpoints \(x\in X^v\) and \(a\in A\). Since r is located at a leaf of T it must hold that \(r\not \in Q\). We consider two cases: either r lies in the same connected component as \(X^v\) in \(TQ\), or it lies in a different connected component.
In the former case, it follows that each internal vertex q of Q satisfies \(R(a)\subseteq R(q)\) and \(v\in L(q)\). But then by the definition of \(\mathcal {A}^v\) and the fact that \(a\in \mathcal {A}^v\), this would imply \(q\in \mathcal {A}^v\), contradicting our construction of Q. Hence if r lies in the same connected component as \(X^v\) in \(TQ\), then A is adjacent to \(X^v\).
In the latter case, there must exist a node \(q\in Q\) of degree at least 3 such that each of A, \(X^v\) and r occur in different components of \(Tq\). By the definition of \(\mathcal {A}^v\), there exists a vertex \(u\in R(a)\) such that \(v\lhd ^\mathcal {P}u\) or \(u\lhd ^\mathcal {P}v\). Since \(u,v\in R(q)\) due to the location of the root and there is a cover edge between them, it follows that either u, v occur in the same connected component of \(X_q\) or in two consecutive ones, but in either case u, v must lie in the same block of q, say block B. But since \(u,v\not \in X_q\), this contradicts Property 2 in Lemma 5; indeed, each tree in \(Tq\) contains at most one of v, u in its bags, and hence there exists no tree \(T'\) in \(Tq\) satisfying \(\bigcup _{t'\in T'}X_{t'} {\setminus } X_q= B\). Hence r cannot occur in a different connected component than \(X^v\) in \(TQ\).
We conclude that Q contains no internal vertices. In particular, every connected component of \(T[\mathcal {A}^v]\) is adjacent to \(X^v\). \(\square \)
With Lemmas 6 and 7, we have the tools necessary for arguing that there exists a treedecomposition of the combined graph of small width.
Lemma 8
Let \(\mathcal {T}=(T,\mathcal {X})\) be a blocked treedecomposition of \(I(\mathcal {P})\) such that \(tw (\mathcal {T})\le k\). Then there exists a treedecomposition \(\mathcal {T}'\) of \(I_C(\mathcal {P})\) of width at most \(3k+2\).
Proof
Consider the treedecomposition \(\mathcal {T}'=(T,\mathcal {X}')\) where \(\mathcal {X}'=\{X'_t~~t\in T\}\) is defined as follows. For each \(t\in T\) such that its unique neighbor s in \(T^r_t\) satisfies \(X_{t}{\setminus } X_{s}=1\), we set \(X'_t=X_t\cup \mathcal {A}_s\); it will be useful to observe that \(\mathcal {A}_s \supseteq \mathcal {A}_t\). For all other nodes \(t\in T\), we then set \(X'_t=X_t\cup \mathcal {A}_t\). We call nodes of the first type nonstandard and nodes of the second type standard.
First, we note that the size of each bag in \(\mathcal {T}'\) is at most \(3k+2\), since every node \(t\in T\) satisfies \(\mathcal {A}_t\le 2k+2\) by Lemma 6. Furthermore, \(\mathcal {T}'\) satisfies condition (T1) because \(\mathcal {T}\) was a treedecomposition of \(I(\mathcal {P})\). \(\mathcal {T}'\) also satisfies condition (T2); indeed, for each \(v\in \mathcal {P}\) it holds that \(X'^v\) restricted to standard nodes is a connected tree by Lemma 7, and by construction every nonstandard node t such that \(v\in X'_t{\setminus } X_t\) is adjacent to a standard node containing v. So, it only remains to argue condition (T3).
Obviously, condition (T3) holds for any edge of \(I(\mathcal {P})\). So, consider two elements u, v of \(\mathcal {P}\) such that \(u\lhd ^\mathcal {P}v\) or \(v\lhd ^\mathcal {P}u\). If there exists a node \(t\in T\) such that \(u,v\in X_t\), then \(u,v\in X'_{t}\) and the condition also holds for this edge in \(I_C(\mathcal {P})\). So, assume that \(X^v\) and \(X^u\) are disjoint and let Q be the unique \(X^v\)\(X^u\) path in T. By Property 4, the \(X^v\)\(X^u\) path Q in T must contain at least one internal node.
Consider the case where one of these subtrees, say w.l.o.g. \(X^v\), lies in the connected component \(T^r_t\) of \(TQ\). Then for each internal node \(q\in Q\), it holds that \(v\in L(q)\) and \(u\in R(q)\), which in turn implies that \(v\in \mathcal {A}_q\). Let \(q_u\) be the endpoint of Q in \(X^u\) and let \(q_0\) be the neighbor of \(q_u\) in Q. By Property 4 we have \(X_{q_u}{\setminus } X_{q_0}=\{u\}\), which implies that \(q_u\) is a nonstandard node and in particular \(\mathcal {A}_{q_0}\subseteq X'_{q_u}\). Since \(q_0\) is an internal node of Q, it follows that \(v\in X'_{q_u}\) which means that condition (T3) also holds for any edge uv in this case.
Finally, consider the case where there exists a node \(q\in Q\) of degree at least 3 such that each of \(X^u\), \(X^v\) and r occur in different components of \(Tq\). Then we reach a contradiction similarly as in the proof of Lemma 7. In particular, since \(u,v\in R(q)\) due to the location of the root and there is a cover edge between them, it follows that either u, v occur in the same connected component of \(X_q\) or in two consecutive ones, but in either case u, v must lie in the same block of q, say block B. But since \(u,v\not \in X_q\), this contradicts Property 2 in Lemma 5; indeed, each tree in \(Tq\) contains at most one of v, u in its bags, and hence there exists no tree \(T'\) in \(Tq\) satisfying \(\bigcup _{t'\in T'}X_{t'} {\setminus } X_q= B\). Hence this case in fact violates our assumptions and cannot occur.
Summarizing the above arguments, we conclude that each bag in \(\mathcal {T}'\) has size at most \(3k+2\) and that \(\mathcal {T}'\) satisfies all of the conditions of a treedecomposition. \(\square \)
Corollary 2
Let \(\mathcal {P}\) be a poset such that \(tw (I(\mathcal {P}))\le k\). Then \(tw (I_C(\mathcal {P}))\le 3k+2\).
4.2 MSO Formulation
In this subsection, we use Fact 4 to prove the following result, which forms the second ingredient required for our proof of Theorem 2.
Lemma 9
#LE is fixedparameter tractable parameterized by the treewidth of the combined graph of the input poset.
Proof
Let \(\mathcal {P}:=(P,\le ^P)\) be a poset. Let G be an (edge)labeled directed graph obtained from \(I_C(\mathcal {P})\) by directing every bidirectional edge of \(I_C(\mathcal {P})\), i.e., every edge of \(I(\mathcal {P})\), in an arbitrary way and labeling it with the label \(\parallel \).
For a set of edges \(E \subseteq E(G)\) with label \(\parallel \), let G[E] be the graph obtained from G after reversing every edge in E. Moreover, for a linear extension \(\preceq \) of \(\mathcal {P}\) let \(E_G(\preceq )\) be the set of edges (u, v) of G such that \(v \preceq u\). Note that because every linear extension of \(\mathcal {P}\) has to respect the direction of the edges in G given by \(C\), it holds that every edge in \(E_G(\preceq )\) has label \(\parallel \).
Claim 10
\(E_G(\preceq )\) defines a bijection between the set of linear extensions of \(\mathcal {P}\) and the set of subsets E of edges of G with label \(\parallel \) such that G[E] is acyclic.
Proof
(of the Claim) Let \(\preceq \) be a linear extension of \(\mathcal {P}\). Then, as observed above, \(E_G(\preceq )\) is a set of edges of G with label \(\parallel \). Moreover, because \(G[E_G(\preceq )]\) is a subgraph of \(P_G(\preceq )\) and \(P_G(\preceq )\) is acyclic so is \(G[E_G(\preceq )]\). Hence, \(E_G(\preceq )\) is a function from the set of linear extensions of \(\mathcal {P}\) to the set of subsets E of edges of G with label \(\parallel \) such that G[E] is acyclic. Towards showing that \(E_G(\preceq )\) is injective, assume for a contradiction that this is not the case, i.e., there are two distinct linear extensions \(\preceq _1\) and \(\preceq _2\) of \(\mathcal {P}\) such that \(E_G(\preceq _1)=E_G(\preceq _2)\) and let u and v be two elements of \(\mathcal {P}\) ordered differently by \(\preceq _1\) and \(\preceq _2\). Then \(\{u,v\} \in I(\mathcal {P})\) and hence either \((u,v) \in G\) or \((v,u) \in G\) the label of (u, v) or (v, u) respectively is \(\parallel \). W.l.o.g. assume that \((u,v) \in G\) with label \(\parallel \). But then, because \(\preceq _1\) and \(\preceq _2\) differ on u and v, either \((u,v) \in E_G(\preceq _1)\) but not \((u,v) \in E_G(\preceq _2)\) or \((u,v) \in E_G(\preceq _2)\) but not \((u,v) \in E_G(\preceq _1)\). In both cases we get a contradiction to our assumption that \(E_G(\preceq _1)=E_G(\preceq _2)\).
It remains to show that \(E_G(\preceq )\) is surjective. To see this let E be a subsets of the edges of G with label \(\parallel \) such that G[E] is acyclic. Because G[E] is acyclic it has a topological ordering, say \(\preceq \), of its vertices. Because G[E] contains \(C(\mathcal {P})\) as a subgraph and any topological ordering of \(C(\mathcal {P})\) is a linear extension of \(\mathcal {P}\), we obtain that \(\preceq \) is a linear extension and also \(E=E_G(\preceq )\). \(\square \)
It follows from the above that instead of counting the number of linear extensions of \(\mathcal {P}\) directly, we can count the number of subsets E of the edges of G with label \(\parallel \) such that G[E] is acyclic. We will show next that there is an MSO formula \(\varPhi (X)\), whose length is independent of G and can hence be considered constant, such that \(G \models \varPhi (X)\) if and only if X is a subset of the edges of G with label \(\parallel \) such that G[E] is acyclic. Because of Fact 4, this implies that #LE is fixedparameter tractable when parameterized by \(tw (G)\) and hence also when parameterized by \(tw (I_C(\mathcal {P}))\), concluding the proof of the lemma.

The formula \( edgesin (X):=\forall x X x \rightarrow P_\parallel x\), which holds if and only if X is a set of edges of G with label \(\parallel \).

The formula \( edgesne (C):=\exists c C c \wedge (\forall c C c \rightarrow E c)\), which holds if and only if C is a nonempty set of edges of G.

The formula \( cyclic (C,X):=\forall v V v \rightarrow degree (C,X,v)\), which holds if and only if the set C of edges of G[X] is a disjoint union of directed cycles. Note that this implies that either C is empty or C contains at least one (directed) cycle.

The formula \( degree (C,X,v):= degree _0(C,X,v) \vee degree _2(C,X,v)\), which holds if and only if either no edge in C is incident to v or there are exactly two edges in C that are incident to v such that one of them corresponds to an edge in G[X] with tail v and the other corresponds to an edge in G[X] with head v.

The formula \( degree _0(C,X,v):=\lnot \exists c C c \wedge I vc\), which holds if and only if no edge in C is incident to v.
 The formulawhich holds if and only if there are exactly two edges in C that are incident to v such that one of them corresponds to an edge in G[E] with tail v and the other corresponds to an edge in G[E] with head v.$$\begin{aligned} degree _2(C,X,v) :=&\,\, \exists c_i \exists c_o C c_i \wedge C c_o \wedge in (X,c_i,v) \wedge out (X,c_o,v) \wedge \\&\left( \forall c (C c \wedge \lnot c=c_i \wedge \lnot c=c_o) \rightarrow \lnot (H vc \vee T vc)\right) \\ \end{aligned}$$
 The formulawhich holds if and only if v is the head of the edge of G[X] represented by c.$$\begin{aligned} in (X,c,v)&:= (\lnot P_\parallel c \wedge H vc) \vee (P_\parallel c \wedge \lnot X c \wedge H vc) \vee (P_\parallel c \wedge X c \wedge T vc) \end{aligned}$$
 The formulawhich holds if and only if v is the tail of the edge of G[X] represented by c.$$\begin{aligned} out (X,c,v)&:= (\lnot P_\parallel c \wedge T vc) \vee (P_\parallel c \wedge \lnot X c \wedge T vc) \vee (P_\parallel c \wedge X c \wedge H vc) \end{aligned}$$
For completeness, we conclude the section by stating the proof of Theorem 2.
5 Conclusions and Future Work
We have given the first parameterized intractability result for counting linear extensions. We hope that the employed techniques will inspire similar results and expand our knowledge about the parameterized complexity of counting problems. In particular, even for #LE there remain many open questions concerning other very natural parameterizations such as the width of the poset (which is, in fact, upperbounded by the treewidth of the incomparability graph) or the treewidth of the poset graph. Moreover, our intractability result for the treewidth of the cover graph poses the question whether there are stronger parameterizations under which #LE becomes tractable, e.g., the treewidth of the comparability graph (i.e., the undirected graph underlying the poset graph), the treedepth or even vertex cover number of the comparability or cover graph, as well as combinations of these parameters with parameters such as the width, the dimension, or the height of the poset. These numerous examples illustrate that the parameterized complexity of #LE is still largely unexplored. As a side note it would also be interesting to establish whether our hardness result for #LE can be sharpened to \(\#\)W[1]hardness and to obtain matching membership results.
Notes
Acknowledgements
Open access funding provided by Austrian Science Fund (FWF). Eduard Eiben was supported by ParetoOptimal Parameterized Algorithms (ERC Starting Grant 715744) and by the Austrian Science Fund (FWF, projects P26696 and W1255N23). Robert Ganian is also affiliated with FI MU, Brno, Czech Republic.
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