Input. A matrix with maximum copy numbers assigned to all columns and weighted rows;

Output. A subset of rows of maximum cumulative weight such that the obtained submatrix is component-mCi1P.

Note that it is equivalent if some sequences are not required to be circular, so it handles well the case where both circular and linear chromosomes are allowed. It is a relaxation of the previous problems, so the NP-hardness does not follow from them. And in fact, the problem for degree 2 matrices (adjacencies) happens to be polynomial, as we now show in the next subsection.

For a degree 2 matrix M, let G_M be the corresponding graph with a node for each column and a weighted edge for each (weighted) row. Let m : V(G_M ) → ℕ be the function specifying the maximum copy number of each column, i.e., the multiplicity limit for each vertex of G_M . We say that matrix M (resp., the corresponding graph G_M ) is component-mCi1P for m if there exists a collection of cyclic walks (resp., corresponding cyclic sequences) that satisfies the following two conditions: (i) G_M is a subgraph of the union of cyclic walks; and (ii) the total number of occurrences of each vertex v in all cyclic walks is at most m(v).

A 2m-matching of a graph G is a spanning subgraph of G such that the degree of each vertex v ∈ V(G) is at most 2m(v). The following lemma shows the correspondence between spanning subgraphs of G that are component-mCi1P for m and 2m-matchings of G.

Lemma 1 A spanning subgraph of a graph G is component-mCi1P for m if and only if it is a 2m-matching of G.

We give a sketch of the proof. For more details, we refer the reader to 21 , where a similar proof was given.

Proof. First, assume a spanning subgraph G' of G is component-mCi1P for m. Then there is a collection of cyclic walks satisfying conditions (i) and (ii). Since each vertex v appears at most m(v) times in these cyclic walks and each occurrence has only two neighbors, the degree of v in G' is at most 2m(v). Hence, G' is a 2m-matching of G.

Conversely, assume G' is a 2m-matching of G. If deg _G' (v) < 2m(v) for some v ∈ V(G'), then we add a new vertex v ₀ and for each v such that deg _G' (v) < 2m(v), we add a new edge {v ₀, v} with multiplicity 2m(v) − deg _G' (v) to G'. Since now every vertex of G' has even degree, each component C of G' is Eulerian, i.e., there is a cyclic walk which contains all edges of C, and each v ∈ V(C) appears exactly m(v) times in the walk. If C does not contain v ₀ then this cyclic walk satisfies conditions (i) and (ii) for vertices in V(C). If C contains v ₀, then after omitting all occurrences of v ₀ we obtain a cyclic walk satisfying conditions (i) and (ii) for vertices in V(C). Hence, G' is component-mCi1P for m. QED

It follows that solutions to the MAX-ROW-component-mCi1P for matrix M and m correspond to maximum weight 2m-matchings of G_M . Next, we give an algorithm for finding a maximum weight f-matching of G with running time (O((|V(G)| + |E(G)|)^3/2)), where f : V(G) → ℕ. We will use a more general form of Tutte's reduction for reducing the maximum weight f-matching problem to the maximum weight matching problem similar to the ones presented in 24 25 .

Given an edge weighted graph G and function f, construct G' in the following way: For all x in V(G), let x ₁, x ₂, ..., x _f(x) be in V(G'); and for all e = {x, y}in E(G), let e_x and e_y be in V(G'). Now, for all e = {x, y} in E(G), let {x ₁, e_x }, ..., {x _f(x), e_x }, {e_x , e_y }, {y ₁, e_y }, ...,{y _f(y), e_y } be edges of G', and all these edges have weight w(e). This reduction is illustrated in Figure 1.

Property 1 There is an f-matching in G with weight w if and only if there is a matching in G' with weight w + W, where W = ∑ e ∈ E ( G ) w ( e ) .

An unweighted version of this property was shown in 25 . The weighted version can be shown in the same way, and hence, we omit the proof.

Since a maximum weight matching can be found in time O ( | V ( G ′ ) | ⋅ | E ( G ′ ) | ) 26 , we have polynomial O((|V(G)| + |E(G)|)^3/2) algorithms for the maximum weight f-matching problem and for the MAX-ROW-component-mCi1P problem with multiplicities on matrices of degree 2.

The tractability does not generalize to matrices, that is, the MAX-ROW-component-Ci1P is already NP-complete for unweighted matrices with rows of degrees 2 and 3. Note that the result for unweighted matrices implies NP-completeness also for the cases when rows are weighted and/or columns have multiplicities.

We will first show that the following hypergraph covering problem is NP-complete. Here we say that a hypergraph H = (V, E) is 2, 3-uniform when all of its hyperedges are either 2-edges or 3-edges, that is, hyperedges that contain exactly two or three vertices. We will also denote 2,3-uniform hypergraphs H = (V, E) as H = (V, E ₂, E ₃), where E ₂ (resp., E ₃) is its set of 2-edges (resp., 3-edges). We denote the power set of a set S with P ( S ) (also known as 2 ^S ).

Definition 1 A graph covering of a 2, 3-uniform hypergraph H = (V, E ₂, E ₃) is a graph G = (V, E') such that there exist a map c : E 2 ∪ E 3 → P ( E ′ ) , satisfying the following for every h ∈ E ₂ ∪ E ₃ :

(a) for every h ∈ E, and for every e ∈ c(h), e ⊆ h;

(b) |c(h)| = 1 if h ∈ E ₂ and |c(h)| = 2 if h ∈ E ₃; and

(c) ∪ h ∈ E 2 ∪ E 3 c(h) = E'.

Here, we say that each set of edges c(h) covers the hyperedge h.

Informally, a graph covering of a 2,3-uniform hypergraph is a graph constructed by picking an edge from each 2-edge, and a pair of edges from each 3-edge.

Problem 1 (The 2,3-Uniform Hypergraph Covering by Cycles and Paths by Edge Removal Problem (23UCR Problem)) Given a 2, 3-uniform hypergraph H = (V, E) and an integer k, is there a graph covering of H that consists of a collection of disjoint cycles and paths after removing at most k hyperedges from E?

Here we will show that Problem 1, the 23UCR Problem, is NP-complete. Later in this section we will show that this implies that the MAX-ROW-component-Ci1P Problem is NP-complete for matrices with rows of degrees 2 and 3. First, we must define the following NP-complete version of 3SAT, which we will use to show NP-completeness of Problem 1.

Problem 2 (The 3SAT(2,3) Problem) Given a CNF formula ϕ   with the following three properties, is ϕ   satisfiable?

(a) Formula ϕ   has only 2-clauses and 3-clauses.

(b) Each variable x of ϕ   has exactly two positive occurrences and one negative occurrence in the clauses.

(c) Exactly one positive occurrence of x appears in the 3-clauses, while the other two occurrences appear in the 2-clauses.

We show that this version of 3SAT is NP-complete using a very similar proof to the one in 27 , by reduction from 3SAT.

Theorem 1 The 3SAT(2,3) Problem is NP-complete.

Proof. Clearly, the problem is in NP. We now show that it is NP-hard by reduction from 3SAT by transforming a given formula ϕ   that is an instance of 3SAT to a formula ϕ ′ that is an instance of 3SAT(2,3) that is satisfiable if and only if ϕ   is satisfiable. For each variable x of ϕ   that has k occurrences, we first replace its k occurrences with x ¹, x ²,..., x^k , i.e., replace the i-th occurrence of x (as literal x or ¬x) with xⁱ . We then add the following 2-clauses: x ¯ i ⇒ x ¯ i + 1 ( i . e . , ¬ x ¯ i ∨ x ¯ i + 1 ) for i = 1, ..., k - 1 and also x ¯ k ⇒ x ¯ 1 , where for each i,

x ˜ i = { x i if   the   i - th occurrence   of   variable   x   is   positive ,   and ¬ x i otherwise .

This "cycle" of implications (2-clauses) on x ¹,..., x^k , ensures that for any truth assignment to the variables of ϕ ′ , the values of x ¯ 1 , … , x ¯ k are either all set to true or all set to false. In the first case, the xⁱ 's corresponding to the positive occurrences of x, are set to true and the xⁱ 's corresponding to the negated occurrence of x, are set to false. In the second case, the situation is reversed. Hence, any satisfying truth assignment to the variables of ϕ ′ can be translated into a satisfying truth assignment to the variables of ϕ   , and vice versa, i.e., ϕ ′ is satisfiable if and only if ϕ   is satisfiable. Since it is easy to verify that this transformation can be done in polynomial time, and that ϕ ′ is indeed an instance of 3SAT(2,3), it follows that the 3SAT(2,3) Problem is NP-complete. QED

We now show that the 23UCR Problem is NP-complete by reduction from 3SAT(2,3).

Theorem 2 The 23UCR Problem is NP-complete.

Proof. Clearly, the problem is in NP. We will show that it is also NP-hard by reduction from 3SAT(2,3).

Given a 3SAT(2,3) formula ϕ   with variables X = {x ₁, ..., x_n } and set C 2 = c 1 , … , c m 2 of 2-clauses (resp., set C 3 = c 1 , … , c m 3 of 3-clauses), we construct a 2,3-uniform hypergraph H_ϕ = (V, E). Hypergraph H_ϕ is composed of variable gadgets and clause gadgets which contains, among other vertices, a vertex for each literal of ϕ   (what we will refer to as literal vertices: there are 3n = 2m ₂ + 3m ₃ such vertices). The design of H ϕ is such that there is a graph covering G of H ϕ that consists of a collection of disjoint cycles and paths after removing at most m ₂ + n edges from E if and only if ϕ   is satisfiable. For this proof, we call such a G a valid covering of H. Note that a valid covering does not contain any vertex of degree 3 or more.

Figure 2a depicts the variable gadget for variable x ∈ X with its two positive occurrences, labeled as x ¹ and x ², and its one negative occurrence ¬x in the clauses. We call the 3-edge {x'', x''', x'''' } the auxiliary hyperedge, while the other two, {x ¹, x ², x'} and {¬x, x', x''}, are called the main hyperedges of the variable gadget.

Figure 2b (resp., 2c) depicts the clause gadget for the 2-clause containing literals p, q (resp., and also r for a 3-clause). For the 2-clause gadget, we call the 2-edge {c, c'} the auxiliary hyperedge. We will refer to literals of ϕ and the literal vertices of the gadgets of H ϕ interchangeably when the context is clear. We have the following claim.

Lemma 2 Formula ϕ   has a satisfying assignment if and only if H ϕ has a valid covering.

Proof. "⇒" We first show that a satisfying assignment of ϕ   can be used to construct a valid covering of H ϕ .

For the variable gadget corresponding to x ∈ X, we first remove the main hyperedge that contains the literal(s) that is satisfied in the assignment, and then cover the remaining two edges as depicted in Figure 3: Figure 3a (resp., 3b) depicts how to cover the clause gadget when x is false (resp., true) in the assignment. In this figure (as in all remaining figures of this paper), hyperedges drawn with dashed lines are removed, while the straight lines are edges picked in the covering.

For a 2-clause (resp., 3-clause) c containing literals p, q (resp., and also r for a 3-clause), without loss of generality let p be a literal that is satisfied in c (there has to be such a literal since it is a satisfying assignment). If c is a 2-clause (resp., 3-clause), we cover the corresponding gadget as depicted in Figure 4a (resp., 4b).

In the above covering, since exactly m ₂ + n hyperedges were removed, and since it is easy to verify that each vertex has degree at most 2, it follows that it is a valid covering of H ϕ .

"⇐" Now we show that if H ϕ has a valid covering then ϕ   is satisfiable.

For hypergraph H ϕ = ( V , E ) , we say that a graph G = (V, E') selects a literal vertex for x ∈ X of H ϕ if x is adjacent to two edges of G in some clause gadget of H ϕ . Obviously, selected vertices of G correspond to a satisfying truth assignment of ϕ   if and only if

(i) in every clause gadget, at least one literal vertex is selected, and

(ii) for every x ∈ X, at most one of x and ¬x is selected.

We call a graph G = (V, E') an expected behavior covering of H ϕ = ( V , E ) when each variable (resp., clause) gadget of H ϕ is covered in a way depicted in Figure 3 (resp., 4). It is easy to verify the following observation.

Observation 1 If a valid covering G = (V, E') of H ϕ = ( V , E ) is also an expected behavior covering of H ϕ , then G corresponds to a satisfying truth assignment of ϕ.

In the remainder of this lemma, we will give a set of transformations that converts a valid covering into an expected behavior covering while preserving the validity of the covering at each step. Assume that we have a valid covering of H_ϕ .

We say that a variable gadget is undecided in a valid covering of H_ϕ if neither of its two main hyperedges is removed. We first show that we can assume that there are no undecided variable gadgets.

Claim 1 We can transform a valid covering of H_ϕ into a valid covering that contains no undecided variable gadgets.

Proof. To prove this claim we do a case analysis on the possible configurations that an undecided variable gadget can have in a valid covering of H_ϕ , and show how we can locally transform each one to a decided configuration without affecting the validity of the covering of H_ϕ .

First, assume that the auxiliary hyperedge is removed in an undecided variable gadget. The set of possible configurations that the gadget can be in is depicted on the left in Figure 5. In this figure (as in all remaining figures) double-headed arrows pointing to two vertices in a 3-edge represent the two coverings of this 3-edge as explained in Figure 6.

We can transform any configuration of Figure 5 to the decided configuration on the right. It is easy to see that in the transformed configuration, the number of hyperedges removed is the same as in any initial configuration, and that vertices x', x'', x''' and x'''' (which do not intersect any vertex outside this variable gadget) have degree at most 2. Finally, since each initial configuration of Figure 5 is part of a valid covering of H_ϕ , and the degree of any literal vertex (x ¹, x ² and ¬x) affected by the transformation has only decreased or remained the same, it follows that the covering of H_ϕ remains valid after this local transformation.

Hence, we can assume that the auxiliary hyperedge is present in any undecided variable gadget. Without loss of generality we can then assume that any configuration of the undecided variable gadget must be in one of the two forms depicted on the left in Figure 7. In each case, we can perform the corresponding transformation shown in Figure 7. Again it is easy to see that the number of hyperedges removed has not increased, that vertices x', x'', x''', x'''', c and c' (which do not intersect any vertex outside of what is shown here) have degree at most 2, and that degree of any involved literal vertex has not increased. Hence, the covering remains valid. QED

We have the following claim.

Claim 2 In any valid covering of H_ϕ , at least one hyperedge is removed from each 2-clause gadget.

Proof. If no hyperedge is removed from the 2-clause gadget (i.e., all hyperedges are covered) in a valid covering of H_ϕ , then vertex c (see Figure 2b) has degree 3, which contradicts the fact that this 2-clause gadget is part of a valid covering of H_ϕ . QED

By Claims 1 and 2, at least n + m ₂ hyperedges have been removed from the variable and 2-clause gadgets, and since in any valid covering this is the maximum number of hyperedges which can be removed, we have the following corollary.

Corollary 1 We can transform a valid covering of H_ϕ into a valid covering where:

(a) exactly one hyperedge is removed from each variable gadget and each 2-clause gadget of H_ϕ , and

(b) no hyperedge is removed from any 3-clause gadget.

We have the following claim.

Claim 3 We can transform a valid covering of H_ϕ into an expected behavior valid covering.

Proof. Firstly, in the valid covering of H_ϕ we can assume, by Claim 1 and Corollary 1, that exactly one main hyperedge is removed and the auxiliary hyperedge is not removed from each variable gadget. However, this does not imply expected behavior. All possible configurations of a decided variable gadget without expected behavior and their corresponding transformations to the expected behavior covering are shown in Figure 8. Analogous to the proof of Claim 1, these local transformations do not affect validity of the covering.

Secondly, in the valid covering of H_ϕ we can assume, by Corollary 1, that exactly one hyperedge is removed from each 2-clause gadget. Assume now that a 2-clause gadget is not expected behavior covered. The only possible such configuration can be transformed to the expected behavior covering as shown in Figure 9. Again, these local transformations do not affect validity of the covering.

Thirdly, in the valid covering of H_ϕ we can assume, again by Corollary 1, that the 3-clause gadget is covered, and hence it is also expected behavior covered (see Figure 4b). Since all gadgets are expected behavior covered in this valid covering of H_ϕ , the claim holds. QED

It follows by Observation 1 and Claim 3, that if H_ϕ has a valid covering, then ϕ is satisfiable. This completes the proof of the lemma. QED

Finally, since the construction of H_ϕ is polynomial, then by Lemma 2 it follows that the 23UCR Problem is NP-complete. QED

Let the component-Ci1P by Row Removal Problem be the corresponding decision version of the MAX-ROW-component-Ci1P Problem as follows.

Problem 3 (The component-Ci1P by Row Removal Problem) Given a binary matrix M and an integer k, can we obtain a submatrix that is component-mCi1P by removing at most k rows from M?

We now show that the component-Ci1P by Row Removal Problem is NP-complete for matrices with rows of degrees 2 and 3.

The following lemma shows the correspondence between the component-Ci1P by Row Removal Problem for matrices with rows of degrees 2 and 3 and the 23UCR Problem. A 2,3-uniform hypergraph H = (V, E) can be represented by a binary matrix B_H with |V| columns and |E| rows, where for each hyperedge h ∈ E, we add a row with 1's in the columns corresponding to the vertices in h and 0's everywhere else. Obviously, there is a one-to-one correspondence between 2,3-uniform hypergraphs and such matrices.

Lemma 3 A 2,3-uniform hypergraph H = (V, E) can be covered by a collection of disjoint cycles and paths after removing at most k hyperedges from E if and only if matrix B_H has the component-Ci1P after removing at most k rows.

Proof. Assume first that H has a covering G that consists of a collection of disjoint cycles and paths after removing at most k hyperedges from E. Remove the (at most k) rows from B_H that correspond to the hyperedges removed from E. Each path (resp., cycle) O of G defines a cyclic order on its set of vertices. Consider the cyclic ordering of the columns of each component of B_H corresponding to O. It is easy to see that each such cyclic ordering is a Ci1P ordering of its corresponding component, and hence B_H has the component-Ci1P after removing at most k rows.

Conversely, assume that each component C = {v ₁,...,v _|C|} of the submatrix of B_H obtained by removing at most k rows is Ci1P with respect to cyclic order π = v i 1 , … , v i | C | of its columns. Consider the following covering G of H, after removing the (at most k) hyperedges from E that correspond to the rows removed from B_H : for every hyperedge, pick the edge between two adjacent columns/vertices in π. Note that every picked edge is { v i j , v i j + 1 } for some j, or { v i | C | , v i 1 } . Hence, G consists of a collection of disjoint cycles and paths. QED

By Theorem 2 and Lemma 3 it follows that the component-Ci1P by Row Removal Problem is NP-complete for matrices with rows of degrees 2 and 3. Since this decision problem is NP-complete, it follows that the MAX-ROW-component-Ci1P Problem is also NP-complete for matrices with rows of degrees 2 and 3.

Theorem 3 The MAX-ROW-component-Ci1P Problem is NP-complete.