, Observe that is an equivalence relation. For a set A ? ?(H), we define reduce H (A) as the operation which returns a set containing one element of each equivalence class of A/ . The main idea of our algorithm is to call reduce H at each step of our dynamic programming algorithm in order to keep the size of a set of partial solutions manipulated small

, For every A ? ?(H), we have |reduce H (A)| ? n k · 2 k(log 2 (k)+1) and we can moreover compute reduce H (A)

To prove that reduce H (A) ? n k · 2 k(log 2 (k)+1) , it is enough to bound the number of equivalence classes of ,

, First observe that i?[k] deg aux H (P) (v i ) = 2|V (H)| when P = ?, since each isolated vertex in H |P gives a loop in aux H (P). Moreover, when P contains an edge, removing an edge from a partial solution P of H increases i?[k] deg aux H (P) (v i ) by two; indeed, this edge removal splits a maximal path of H |P into two maximal paths. Therefore, any partial solution P satisfies that i?[k] deg aux H (P) (v i ) ? 2|V (H)|; in particular each vertex of aux H (P) has degree at most 2|V (H)|. As aux H (P) contains k vertices, We claim that, for every P ? ?(H), we have i?[k] deg aux H (P) (v i ) ? 2|V (H)|

We conclude that partitions ?(H) into at most n k · 2 k(log 2 k+1) equivalences classes. It remains to prove that we can compute reduce H (A) in time O(|A| · nk log 2 (nk)). First observe that, for every P ? ?(H), we can compute aux H (P) in time O(nk). Moreover, we can also compute the degree sequence of aux H (P) and the connected components of aux H (P) in time O(nk). Thus, by using the right data structures, we can decide whether P 1 P 2 in time O(nk). Furthermore, by using a self-balancing binary search tree, we can compute reduce H (A) in time O(|A| · nk log 2 (|reduce H (A)|)). Since log 2 (|reduce H (A)|) ? k log 2 (2nk), Since the number of partitions of {v 1 ,

, The rest of this section is dedicated to prove that, for a set of partial solutions A of H, the set reduce H (A) is equivalent to A, i.e., if A contains a partial solution that forms a Hamiltonian cycle with a complement solution, then reduce H (A) also. Our results are based on a kind of equivalence between Hamiltonian cycles and red-blue Eulerian trails

, If P ? ?(H) and Q ? ?(H) form a Hamiltonian cycle, then the multigraph aux H (P) aux H (Q) admits a red-blue Eulerian trail

, From the definitions of a partial solution and of a complement solution, Suppose that P ? ?(H) and Q ? ?(H) form a Hamiltonian cycle C. Let M := aux H (P) aux H (Q)

, Q are all the H-paths in G |Q

P , Q appear in C in this order, ? for each x ? [ ], the first end-vertices of P x is the last end-vertex of Q x?1 and the last end-vertex of P x is the first end-vertex of Q x ,

, Since aux H (P) has the same set of connected components as aux H (P ), we know that aux H (P ) M is also connected. Moreover, for every i ? [k], we have deg aux H (P) (v i ) = deg aux H (P ) (v i ) = deg M

, ) M admits a red-blue Eulerian trail. Thus, for every P ? A and multigraph M with blue edges such that aux H (P) M admits a red-blue Eulerian trail, there exists P ? reduce H (A) such that aux H (P ) M admits a red-blue Eulerian trail. Hence

, One easily checks that H is a transitive relation. Now, assuming that A H B, we have reduce H (A) B because reduce H (A) H A

, Our algorithm computes recursively, for every k-labeled graph H arising in the k-expression of G, a set A H such that A H H ?(H) and |A H | ? n k · 2 k(log 2 (k)+1) . In order to prove the correctness of our algorithm

observe that H has the same set of vertices and edges as D. Thus, we have ?(H) = ?(D) and ?(H) = ?(D) ,

To prove the lemma, it is sufficient to prove that there exists P ? A D such that aux H (P ) M contains a red-blue Eulerian trail. Let f be a bijective function such that ? for every edge e of aux D (P) with endpoints v and v i , for some , f (e) is an edge of aux H (P) with endpoints v and v j , and ? for every loop e with endpoint v i , f (e) is a loop of aux H (P) with endpoint v j . By construction of aux D (P) and aux H (P), such a function exists, We construct the multigraph M from M and T by successively doing the following: ? For every edge e of the multigraph aux D (P) with endpoints v and v i , take the subwalk W = (v , f (e) ,

, ? For every loop e with endpoint v i in the multigraph aux D (P), take the subwalk W =

, there exists P ? A D such that aux D (P ) M contains a red-blue Eulerian trail. Observe that aux H (P) (respectively M) is obtained from aux D (P ) (resp. M ) by replacing each edge associated with {v i , v k } or {v i } in aux D (P ) (resp. M ) with an edge associated with {v j , v k } or {v j } respectively, By construction, one can construct from T a red-blue Eulerian trail of aux D (P) M . Since A D D ?(D)

So, we can obtain a red-blue Eulerian trail of aux H (P ) M from a red-blue Eulerian trail of aux H (P) M by replacing (v i , e, v j ) with the sequence (v i , e 1 , v i , f, v j , e 2 , v j ) where f is the blue edge we add to M to obtain M . It implies the claim. Now, since A H B, there exists P ? A such that aux H (P ) M admits a red-blue Eulerian trail T . Let W be the subwalk of T such that W = ,

, Since every partial solution of H is obtained from the union of a partial solution of D and a subset of E H i,j of size at most n

D) + 0(i, j) ,

, disjoint union of two graphs or relabeling cannot create a Hamiltonian cycle. Thus, by minimality, we have H = ? i,j (D) such that ? D is a k-labeled graph arising in ? and i

,

By Lemma 4.69, the multigraph aux D (P) aux D (Q) contains a red-blue Eulerian trail. Since A D D ?(D), there exists P ? A D such that aux D (P ) aux D (Q) contains a red-blue Eulerian trail. As Q ? E H i,j , we deduce that, for every ? [k] \ {i, j}, we have deg aux D (P ) (v ) = 0 and deg aux H (P ) (v i ) = deg aux H (P ) (v j ). For the other direction, suppose that the latter condition holds. Let Q be the graph on the vertex set V (G) such that it contains exactly deg aux H (P) (v i ) many edges between the set of vertices labeled i and the set of vertices labeled j. Observe that aux H (Q) consists of deg aux H (P) (v i ) many edges between v i and v j . Therefore ,

, By Lemma 4.68, for every A ? ?(H), we can compute reduce H (A) in time O(|A| · nk 2 log 2 (nk)). Observe that, for every k-labeled graph D arising in ? and such that A D is computed before A H , we have |A D | ? n k · 2 k(log 2 (k)+1) . It follows that: ? If H = D ? F , then we have |A D ? A F | ? n 2k · 2 2k(log 2 (k)+1) . Thus, we can compute A H := reduce H (A D ? A F ) in time O, Running time. Let H be a k-labeled graph arising in ?. Observe that if H = i(v) or H = ? i?j (D), then we compute A H in time O(1)

,

, First observe that, for every partial solution P of H, we have |P + (i, j)| ? n 2 and we can compute the set P + (i, j) in time O(n 2 ). Moreover, by Lemma 4.68, for every ? {0, . . . , n ? 1}, we have |A D | ? n k · 2 k(log 2 (k)+1) and thus, we deduce that |A D + (i, j)| ? n k+2 ·2 k(log 2 (k)+1) and that A +1 D can be

,

,

, Since ? uses at most O(n) disjoint union operations and O(nk 2 ) unary operations

, M has a red-blue Eulerian trail

, M * has an Eulerian trail

, The underlying undirected graph of M * has at most one connected component containing an edge, and, for each vertex v of M * , deg + M * (v) = deg ? M * (v)

, Even though G is strongly connected, it does not have a red-blue Eulerian trail, and one can check that M * has two connected components containing an edge. To decide whether the underlying undirected graph of M * has at most one connected component containing an edge, multiple arcs are useless. So, it is enough to keep one partial solution P for each degree sequence in aux H (P) and for each set, the condition that "for each vertex v of M * , deg + M * (v) = deg ? M * (v)" can be translated to that, for each vertex v of M , the number of blue incoming arcs is the same as the number of red outgoing arcs, and the number of red incoming arcs is the same as the number of blue outgoing arcs

, v k }, deg + aux H (P 1 ) (v) = deg + aux H (P 2 ) (v) and deg ? aux H (P 1 ) (v) = deg ? aux

, If P 1 P 2 , then P 1 is part of a directed Hamiltonian cycle in G if and only if P 2 is part of a directed Hamiltonian cycle in G. From the definition of

Thus we can follow the lines of the proof for undirected graphs, and easily deduce that one can solve Directed Hamiltonian Cycle in time n O(k), vol.2 ,

, The vertices of out-degree zero are called leaves of D. The Min Leaf Out-Branching problem asks for a given digraph D and an integer , whether there is a spanning out-tree of D with at most leaves. This problem generalizes Hamiltonian Path (and also Hamiltonian Cycle) by taking = 1. Ganian, Hlin?ný, and Obdr?álek [66] showed that there is an n 2 O(k) -time algorithm for solving Min Leaf Out-Branching problem

, In Section 5.2, we prove that one can compute in polynomial time the number of minimal transversals of ?-acyclic hypergraphs. In Subsection 5.2.3, we show that, as a corollary, we can count in polynomial time the minimal dominating sets of strongly chordal graphs and we discuss about the possible extensions of this corollary. In Subsection 5.2.4, we conclude this chapter by some open questions concerning the counting of minimal transversals and in particular, we discuss about how some existing parameters on hypergraphs could be used to extend our result. , some results are known concerning width measures. Arnborg, Lagergren, and Seese [3] extended Courcelle's theorem by showing that every counting problem expressible in MSO 2 are FPT parameterized by the tree-width of the input graph. Makowsky [105] proved that evaluating the Tutte polynomial is FPT when parameterized by tree-width. Moreover, Courcelle, Makowsky, and Rotics [33] generalized Theorem 2.52 by showing that every counting problem expressible in MSO 1 is FPT parameterized by the clique-width of the input graph. Efficient algorithms are known for the counting variants of NP-hard problems parameterized by tree-width. For example, Bodlaender et al. [9] designed a framework called determinant approach which provides 2 O(tw(G)) ·n time algorithms for the counting variants of many connectivity problems such as Hamiltonian Cycle and Steiner Tree. Recently, Golovach et al. [73, Theorem 36] proved the following theorem about counting the 1-minimal and 1-maximal (?, ?)-dominating sets with the d-neighbor-width as parameter. For a graph G, a (?, ?)-dominating set D ? V (G) of G is 1-maximal (resp. 1-minimal )

, The minimal transversals of H containing x are {a, x, c}, {a, x, d} and {b, x, d}. Observe that removing x from these sets directly yields a minimal transversal of H \ H(x) = {{a, b}, {c, d}}. However, adding x to a minimal transversal of H \ H(x) does not give necessarily a minimal transversal of H. For example

, In fact, we can show in general that T ? x ? mtr(H) if and only if T is a minimal transversal of H \ H(x) and T is not a transversal of H. Consequently, the number of minimal transversals of H containing x is #mtr(H \ H(x)) ? #(tr(H) ? mtr

we define the B-blocked transversals of H to be the transversals T of H such that each vertex x of T has a private in H \ H (B). In particular, if y is a vertex of B, then y cannot be in a B-blocked transversal of H . Observe also that if y ? V (H) \ V (H ), then y cannot be in a B-blocked transversal of H . Fact 5.5. Given a sub-hypergraph H of a hypergraph H and B ? V (H), T is a B-blocked transversal of H if and only if T ,

, As an example, let H be the hypergraph depicted in Figure 5.1, H = H and B = {x}. The only B-blocked transversal of H is {b, c}. While {b, d} is a minimal transversal of H \ H(x), it is not a B-blocked transversal as the hyperedge {x, c} does not intersect {b, d}. We call the set H (B) the blocked hyperedges. Intuitively, H (B) is the set of hyperedges that cannot be used as privates in a transversal of H . We denote by btr B (H ) the set of B-blocked transversals of H

, Observe that by definition, mtr(H) = btr ? (H). Moreover, if H(B) = H = ?, then btr H (H) = ? as mtr(?) = {?} and ? / ? tr(H). When B = {x}, we denote btr B (H) by btr x (H)

We extend this notation to mtr and btr as well. The following summarizes observations about blocked transversals, we denote by tr(H, S) := {T ? tr(H) : T ? S} ,

, Fact 5.6. Let H be a sub-hypergraph of a hypergraph H and B, S ? V (H). Then, 1. btr B (H ) = btr B?V (H ) (H )

,

, One checks easily that #mtr(H) = #mtr(H, V (H) \ {x}) + #mtr(H)(x), Therefore, we have #mtr(H) = #btr ? (H) = #btr ? (H, V (H) \ {x}) + #btr ? (H \ H(x)) ? #btr x (H)

x q such that, for all 1 ? i ? q, #btr x i (H i ) and #btr ? (H i ) can be computed in polynomial time if #btr x j (H j ) and #btr ? (H j ) are known for all j < i. As a consequence, one can compute #mtr(H) by classical dynamic programming for any ?-acyclic hypergraph. The end of this subsection is dedicated to the proof of several crucial lemmas concerning recursive formulas for computing the number of blocked transversals, and that will be useful in our algorithm ,

, Lemma 5.7. Let H be a hypergraph

, it means that there exists a hyperedge e ? H such that e?S = ?. In this case, btr B (H, S) = ?. Moreover, there exists i ? [1, k] such that C i = {?} and e ? H i . Thus, btr B (H i , S) = ? and the equality holds in this case

We show that for all i ? k, T i = ,

Thus e ? T i = ?, that is, T i ? tr(H i , S). Moreover, let y ? T i . By definition of T , there exists e ? H \ H(B) such that e is private for y w.r.t. T . Observe that we have T i ? S ? V (H i ) = V (C i ) since T ? S. Thus, y ? V (C i ) and e ? S ? C i , because C i is a connected component ,

, It remains to show that T ? mtr(H \ H(B)). Let y ? T . By definition of T , there exists i such that y ? T i . Thus, there exists e ? H i \ H i (B) that is private for y w.r.t. T i . Moreover, since H i (B) = H i ? H(B), we know that e / ? H(B). As C 1 , . . . , C k are the connected component of H[S], we have that, for every j = i, V (C i ) ? V (C j ) = ?. Moreover, for all ? k, we have S ? V (H ) = V (C ), As H = k i=1 H i , there exists i such that e ? H i . Thus e ?T i = ? and thus e ?T = ?, that is, T ? tr(H)

, other words, e is private for y w.r.t. T and H \ H(B). That is T ? btr B (H)

, We recall that btr B (H, S)(x) is the set of B-blocked transversals T of H such that T ? S and x ? T . The following lemma shows that for any B-blocked transversal T ? S of H containing x

, By definition, x ? T and T ? S, thus we only have to show that T = T \ {x} ? btr B (H 1 ), Proof. Let H, vol.1

In other words, T is a minimal transversal of H 1 \ H 1 (B) which concludes the proof. To complete the previous lemma, we show that for each B-blocked transversal T ? S of H \ H(x), we have T ? {x} is a B-blocked transversal of H if and only if T is not a (B ? {x}, Since T is a minimal transversal of H \ H(B), there exists e ? H \ H(B) such that , otherwise e would not be private for y w.r.t. T . Thus e ? H \ (H(B) ? H(x)), that is, e is private to y w.r.t. T in H 1 \ H 1 (B) ,

, We have {{x}} btr B (H 1 , S \ {x}) \ btr B (H, S)(x) = {{x}} btr B?{x}

, We prove the lemma by proving first the left-to-right inclusion (Claim 5.9.1) and then the right-to-left inclusion (Claim 5.9.2)

, Claim 5.9.1. For every T ? {{x}} btr B (H 1 , S \ {x}) \ btr B (H, S)(x), we have T \ {x} ? btr B?{x}

, Assume towards a contradiction that T / ? tr(H 2 ), i.e., there exists e ? H 2 such that e ? T = ?. We prove that it implies T ? btr B (H, S)(x). First, observe that T ? tr(H), since T ? tr(H 1 ) = tr(H \ H(x)) and T = T ? {x}. Thus, we have e ? T = {x} and e ? H(x). As e ? H 2 = H \ (H(B) ? H(x)), we have e ? H \ H(B) and then e is a private hyperedge for x w.r.t. T and H \ H(B). Furthermore

, H 1 \ H 1 (B)). Thus, every vertex in T has a private hyperedge w.r.t. T and H \ H(B)

we can conclude that T ? mtr(H \ H(B)). Finally, we have T ? S by assumption ,

, Let y ? T . Since T ? btr B (H 1 ), there exists f ? H 1 \ H 1 (B) such that f ? T = {y}. Since H 1 \ H 1 (B) = H 2 \ H 2 (B ? {x}), every y ? T have a private hyperedge in H 2 \ H 2 (B ? {x}), that is T ? mtr, )(x) which is a contradiction. Thus, T ? tr

, Claim 5.9.2. For every T ? {x} btr B?{x} (H 2 , S \ {x}), we have T ? {x} btr B (H 1 , S \ {x}) \ btr B (H, S)(x)

, For each 1 ? i ? n and each 1 ? j ? m, let tab[i, j, 0] be #btr ? (H x i e j , [? x i ]), and for each > i, let tab[i, j, ] be #btr x (H x i e j

,

, ] from the recursive formula of #btr ? (H xi ej

, Compute tab[i, j, ] from the recursive formula of #btr x (H xi ej

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