Moreover, a comparison of ∨-rules, ∨-∧ mixed rules, and ∧-rules ismade from the perspectives of inclusion and inference relationships. Finally,some real examples and numerical experiments are conducted to comparethe proposed rule acquisition algorithms with the existing one in terms ofthe running efficiency.Rule acquisition is one of the main purposes in the analysis of formaldecision contexts. Up to now, there have been several types of rules informal decision contexts such as decision rules, decision implications, andgranular rules, which can be viewed as  Formal concept analysis (FCA) is a field of applied mathematics based on the mathematization of formal concepts and conceptual hierarchy [ formal context  consisting of an object set G, an attribute set M, and an incidence relation I between G and M [ierarchy . This th G and M . Its key G and M , machine G and M , knowled G and M –15, and  G and M , 17. Wha implications [ association rules [In FCA, a basic way of describing dependencies between the attributes of a formal context is viaications , 19 or aon rules . Consideon rules –23 discuon rules . formal decision context [ training context [ decision rules, in formal decision contexts through combining conditional formal concepts with decision formal concepts. In order to eliminate superfluous decision rules, Li et al. [ granular computing into decision rules for decreasing computation time. Moreover, the notion of a decision rule was extended into the cases of incomplete formal decision contexts [ real formal decision contexts [ decision implications, whose premises and conclusions are taken from conditional attributes and decision attributes, respectively. It should be pointed out that decision rules are special decision implications. Following the above discussion, Zhai et al. [ granular rules, which are special decision rules, with their premises and conclusions being the intents of conditional formal concepts and decision formal concepts, respectively. A detailed investigation on relation between granular rules and decision rules can also be found in [In the real world, a formal context often contains target attributes for the purpose of making decision analysis. A formal context equipped with additional target attributes is called a context   including an object set G, an attribute set M, and an incidence relation I⊆G × M, in which  ∈ I indicates that the object x has the attribute a and  ∉ I means the opposite.A formal context is a triple  is said to be regular if, for any  ∈ G × M, the following conditions hold: a1, a2 ∈ M such that  ∈ I and  ∉ I;there exist x1, x2 ∈ G such that  ∈ I and  ∉ I.there exist A formal context  forms an antitone Galois connection, while the pairs of operators  and  form isotone Galois connections . More prG, M, I) be a formal context. For X, X1, X2⊆G and B, B1, B2⊆M, the following properties hold:X1⊆X2⇒X2↑⊆X1↑, X1□⊆X2□, X1⋄⊆X2⋄;B1⊆B2⇒B2↓⊆B1↓, B1□⊆B2□, B1⋄⊆B2⋄;X⊆X↑↓, X□⋄⊆X⊆X⋄□;B⊆B↓↑, B□⋄⊆B⊆B⋄□;X↑↓↑ = X↑, X□⋄□ = X□, X⋄□⋄ = X⋄;B↓↑↓ = B↓, B□⋄□ = B□, B⋄□⋄ = B⋄.Let  be a formal context, X⊆G, and B⊆M. If X↑ = B and B↓ = X, then  is called a formal concept; if X□ = B and B⋄ = X, then  is called an object-oriented concept; if X⋄ = B and B□ = X, then  is called a property-oriented concept. For each of the cases, X and B are called the extent and intent of , respectively.Let  are, respectively, ordered byG, M, I), respectively. Hereinafter, we denote formal concept lattice by When the formal, object-oriented, and property-oriented concepts of a formal context (d by(2)X,B1≤F and  are, respectively, defined byIn the concept lattices It should be pointed out that the relation among formal, object-oriented, and property-oriented concept lattices was discussed in , 39, andG, M, I, N, J), where  and  with M∩N = ∅ are two formal contexts. Here, M and N are called the conditional attribute set and the decision attribute set of , respectively.A formal decision context is a quintuple  is also said to be regular [G, M, I) and  are regular. Hereinafter, the concerned formal decision contexts are all assumed to be regular. Moreover, for convenience, Like the formal context, a formal decision context Π =  formal concept lattices only, and the derived rules ironment .In order to widen the domain of application of rule acquisition, now we continue to put forward two new types of rules, called ∨-rules and ∨-∧ mixed rules, based on formal, object-oriented, and property-oriented concept lattices.In this subsection, we propose the notion of a ∨-rule in formal decision contexts based on formal and object-oriented concept lattices.G, M, I, N, J) be a formal decision context, let G, M, I), and let G, N, J). For any X ≠ ∅, Y ≠ G, and X⊆Y, then the expression B→∨  C is called a ∨-rule generated between the object-oriented concept  and formal concept . Here, B and C are called the premise and conclusion of the ∨-rule B → C, respectively. The set of all the ∨-rules generated between the object-oriented concepts in RO(Π).Let Π = (B→∨  C ∈ RO(Π), we conclude that each x ∈ G having at least one conditional attribute in B has all the decision attributes in C. More specifically, if B = {b1, b2,…, bs} and C = {c1, c2,…, ct}, then B→∨  C means that “if b1∨b2∨⋯∨bs, then c1∧c2∧⋯∧ct,” where ∨ and ∧ denote logical disjunction and conjunction operators.Thus, for any B→∨  C with B = {b1, b2,…, bs} and C = {c1, c2,…, ct} can be integrated by the following attribute implication rules (or association rules with their confidences being one):It should be pointed out that the ∨-rules have something to do with both the attribute implication rules and the association rules  be a formal decision context. For B1→∨  C1, B2→∨  C2 ∈ RO(Π), if B2⊆B1 and C2⊆C1, one says that B2→∨  C2 can be implied by B1→∨  C1. One denotes this implication relationship by B1→∨  C1⇒B2→∨  C2. For any B→∨  C ∈ RO(Π), if there exists B0→∨  C0 ∈ RO(Π)∖{B→∨  C} such that B0→∨  C0⇒B→∨  C, then B→∨  C is said to be redundant in RO(Π); otherwise, B→∨  C is said to be nonredundant in RO(Π). We denote by RO*(Π) the set of all of the nonredundant ∨-rules in RO(Π).Let Π =  be a formal decision context. DenoteLet Π = , B→∨  C ∈ RO(Π). Then, B→∨  C is redundant in RO(Π) if and only if αO = 0 or βO = 0, or, equivalently, B→∨  C is nonredundant in RO(Π) if and only if αO = 1 and βO = 1.For a formal decision context Π = (B→∨  C is redundant in RO(Π), there exists B0→∨  C0 ∈ RO(Π)∖{B→∨  C} such that B0→∨  C0⇒B→∨  C. By B⊆B0 and C⊆C0, which implies that X⊆X0 and Y0⊆Y. Noting that X0⊆Y0 and B0→∨  C0 is different from B→∨  C, we conclude that X ⊂ X0⊆Y0⊆Y or X⊆X0⊆Y0 ⊂ Y. Consequently, αO = 0 or βO = 0.Necessity. If αO = 0 or βO = 0, we can prove that B→∨  C is redundant in RO(Π). In fact, when αO = 0, there exists X ⊂ X0⊆Y. Suppose B0→∨  C⇒B→∨  C. As a result, B→∨  C is redundant in RO(Π). When βO = 0, there exists X⊆Y0 ⊂ Y. Suppose B→∨  C0⇒B→∨  C. Consequently, B→∨  C is redundant in RO(Π).Sufficiency. If Now, we are ready to put forward a method to derive the nonredundant ∨-rules from a formal decision context. The method can briefly be described as in G, M, I) can be derived from the formal concept lattice of the complementary formal context of  [LO| denotes the cardinality of the object-oriented concept lattice of  and |LF| denotes that of the formal concept lattice of .Note that the object-oriented concept lattice of  . Then, iG, M, I, N, J), where G = {1,2, 3,4, 5}, M = {a, b, c, d, e, f}, and N = {d1, d2, d3}. The object-oriented concept lattice of  is shown in G, N, J) is shown in  r1: if c, then d1 and d2; r2: if e, then d2 and d3; r3: if b, c, d, e, or f, then d2.According to r3 can be divided into the following attribute implication rules: r3(1): if b, then d2; r3(2): if c, then d2; r3(3): if d, then d2; r3(4): if e, then d2; r3(5): if f, then d2.It should be pointed out that In this subsection, we continue to put forward the notion of a ∨-∧ mixed rule in formal decision contexts based on formal and property-oriented concept lattices.G, M, I, N, J) be a formal decision context, let G, M, I), and let G, N, J). For any X ≠ ∅, Y ≠ G, and X⊆Y, then the expression B→∨∧  C is called a ∨-∧ mixed rule generated between the property-oriented concept  and the formal concept . Here, B and C are called the premise and conclusion of the ∨-∧ mixed rule B → C, respectively. The set of all of the ∨-∧ mixed rules generated between the property-oriented concepts in RP(Π).Let Π = (B→∨∧  C ∈ RP(Π), we conclude that each object having at least one conditional attribute in B and no conditional attribute in M∖B has all the decision attributes in C. More specifically, if B = {b1, b2,…, bs}, M∖B = {bs+1, bs+2,…, bn}, and C = {c1, c2,…, ct}, then B→∨∧  C means the following: “if b1  ∨b2  ∨⋯∨  bs and ¬bs+1  ∧  ¬bs+2∧  ⋯  ∧  ¬bn, then c1  ∧  c2  ∧  ⋯  ∧  ct,” where ∨, ∧, and ¬ denote logical disjunction, conjunction, and negation operators, respectively.Thus, for any B→∨∧C with B = {b1, b2,…, bs}, M∖B = {bs+1, bs+2,…, bn}, and C = {c1, c2,…, ct} can be integrated by the following attribute implication rules (or association rules with their confidences being one):It should be pointed out that the ∨-∧ mixed rules have something to do with both the attribute implication rules and the association rules. Concretely, a ∨-∧ mixed rule G, M, I, N, J) be a formal decision context. For any B1→∨∧  C1, B2→∨∧  C2 ∈ RP(Π), if B2⊆B1 and C2⊆C1, one says that B2→∨∧  C2 can be implied by B1→∨∧  C1. One denotes this implication relationship by B1→∨∧  C1⇒B2→∨∧  C2. For any B→∨∧  C ∈ RP(Π), if there exists B0→∨∧  C0 ∈ RP(Π)∖{B→∨∧  C} such that B0→∨∧  C0⇒B→∨∧  C, then B→∨∧  C is said to be redundant in RP(Π); otherwise, B→∨∧  C is said to be nonredundant in RP(Π). One denotes by RP*(Π) the set of all the nonredundant ∨-∧ mixed rules in RP(Π).Let Π =  be a formal decision context. DenoteLet Π = , B→∨∧  C ∈ RP(Π). Then, B→∨∧C is redundant in RP(Π) if and only if αP = 0 or βP = 0, or, equivalently, B→∨∧C is nonredundant in RP(Π) if and only if αP = 1 and βP = 1.For a formal decision context Π =  can be derived from the formal concept lattice of the complementary formal context of  [LP| denotes the cardinality of the property-oriented concept lattice of  and |LF| denotes that of the formal concept lattice of .Note that the property-oriented concept lattice of  . Then, iG, M, I, N, J) be the formal decision context in G = {1,2, 3,4, 5}, M = {a, b, c, d, e, f}, and N = {d1, d2, d3}. The property-oriented concept lattice of  is shown in G, N, J) can be found in Let Π =  be a formal decision context, let G, M, J), and let G, N, J). For any X, B, Y, and C are all nonempty and X⊆Y, then the expression B→∧  C is called a decision rule generated between the formal concepts  and . Here, B and C are called the premise and conclusion of the decision rule B→∧  C, respectively. The set of all the decision rules generated between the formal concepts in RF(Π).Let Π =  be a formal decision context. For B1→∧  C1, B2→∧  C2 ∈ RF(Π), if B1⊆B2 and C2⊆C1, one says that B2→∧  C2 can be implied by B1→∧  C1. One denotes this implication relationship by B1→∧  C1⇒B2→∧  C2. For any B→∧  C ∈ RF(Π), if there exists B0→∧  C0 ∈ RF(Π)∖{B→∧  C} such that B0→∧  C0⇒B→∧  C, then B→∧  C is said to be redundant in RF(Π); otherwise, B→∧  C is said to be nonredundant in RF(Π). One denotes by RF*(Π) the set of all the nonredundant decision rules in RF(Π).Let Π = (B→∧  C ∈ RF(Π), one concludes that each object having all the conditional attributes in B also has all the decision attributes in C. More specifically, if B = {b1, b2,…, bs} and C = {c1, c2,…, ct}, then B→∧  C means the following: “if b1∧b2∧⋯∧bs, then c1∧c2∧⋯∧ct,” where ∧ denotes logical conjunction operator. Moreover, it is easy to observe that decision rules, ∨-rules, and ∨-∧ mixed rules are different from each other in terms of their logical reasoning methodologies.Thus, for any The following example is used to show that there does not exist inclusion relationship among decision rules, ∨-rules, and ∨-∧ mixed rules. That is, we need to confirm three statements: (1) a decision rule may not be a ∨-rule or ∨-∧ mixed rule; (2) a ∨-rule may not be a decision rule or ∨-∧ mixed rule; (3) a ∨-∧ mixed rule may not be a decision rule or ∨-rule.G, M, I, N, J) be the formal decision context in G = {1,2, 3,4, 5}, M = {a, b, c, d, e, f}, and N = {d1, d2, d3}. The formal concept lattice of  is shown in G, N, J) can be found in Let Π = Ω: G, M, I, N, J) be the formal decision context in G = {1,2, 3,4, 5}, M = {a, b, c, d, e, f}, and N = {d1, d2, d3}. Then, according to the discussion in Examples r1′′ cannot be implied by the ∨-rules r1, r2, r3(1), r3(2), r3(3), r3(4), and r3(5) based on the inference rule Ω, neither can the ∨-∧ mixed rules r1′, r2′, r3′, and r4′;the decision rule r1, r2, r3(1), r3(2), r3(3), r3(4), and r3(5) cannot be implied by the decision rules r1′′, r2′′, and r3′′ based on the inference rule Ω, neither can the ∨-∧ mixed rules r1′, r2′, r3′, and r4′;each of the ∨-rules r1′, r2′, and r4′ cannot be implied by the decision rules r1′′, r2′′, and r3′′ based on the inference rule Ω, neither can the ∨-rules r1, r2, r3(1), r3(2), r3(3), r3(4), and r3(5).each of the ∨-∧ mixed rules Let Π =  as the decision attribute(s) and other attributes as the conditional attributes. Then, the scaling approach  was usedRO*(Π), RP*(Π), and RF*(Π) are the nonredundant ∨-rules, ∨-∧ mixed rules, and decision rules, respectively. It can be seen from In the experiments, we still denote the proposed rule acquisition algorithms by Algorithms m hold, then decisions hold.” In order to enrich the existing rule acquisition theory in formal decision contexts, we have proposed two new types of rules, called ∨-rules and ∨-∧ mixed rules, based on formal, object-oriented, and property-oriented concept lattices. Moreover, a comparison of ∨-rules, ∨-∧ mixed rules, and ∧-rules has been made from the perspectives of inclusion and inference relationships. Finally, some numerical experiments have been conducted to compare the proposed rule acquisition algorithms with the existing one in terms of the running efficiency.Rule acquisition is one of the main purposes in the analysis of formal decision contexts. Although there have been several types of rules in formal decision contexts such as decision rules, decision implications, and granular rules, these rules are all ∧-ones since they have the following form: “if conditions 1,2,…, and From the point of view of real applications, the results obtained in this paper need to be further extended to the cases of fuzzy formal decision contexts , incompl


If the limit A ∈ R exists in eitherr→∞(logp][Mh−1(Mf(r))/logp][Mh−1(r)) = A orlimsupr→∞(logp][Mh−1(Mf(r))/logp][Mh−1(r)) = A,liminf thenLet Mh−1(r) is an increasing function of r, it follows from Lemmas δ > 0, for all sufficiently large values of r, that(i) Since r thatr thatr thatTherefore from  we get frly from  it follohereforelogpMh−1ows thatlimsuprr.Henceliminfr (r thatlimsupr(ii) follows with a similar argument.f, g, and h be any three transcendental entire functions with g(0) = 0. If p, q, and m are any three positive integers with m > q, then ρhq) = ∞ under any of the following conditions: ρhp, q)((g) = ∞;ρhp, q)((f), λg) > 0;min, λhp, q)((f)) > 0.min((g) = ∞, since Mh−1(r) is an increasing function of r, it follows from r, that(i) If ρhp, q)((f) > 0 and λg > 0.(ii) Suppose Mh−1(r) is an increasing function of r, we get from δ > 0 and any ɛ > 0, for all sufficiently large values of r,As In the line of f, g, and h be any three transcendental entire functions with g(0) = 0. If p, q, and m are any three positive integers with m > q, then λhp, q)((f∘g) = ∞ if any of the following facts happens:λhp, q)((g) = ∞;λhp, q)((f), λg) > 0.min = 0. If p, q, and m are any three positive integers with m > q and any of the following two facts happensρhp, q)((f), λg) > 0 ormin((f), λg) > 0,min( thenLet (i) Since(ii) The proof can be carried out in the line of (i) and p, q)th order and relative th lower order. In this process, After modifying the notion of relative order of higher dimensions in case of entire functions in , where a


We show that 1-quasiconformal mappings on Goursat groups are CR or anti-CR mappings. This can reduce the determination of 1-quasiconformal mappings to the determination of CR automorphisms of CR manifolds, which is a fundamental problem in the theory of several complex variables. Rn are elements of group SO. Quasiconformal mappings on the Heisenberg group were introduced by Mostow  ∈ C∞, if Z1, Z2 ∈ C∞. The number n is called CR dimension of M and k = dim M − 2n is called the codimension of M. A smooth mapping f : M1 → M2 is called CR ifH is then defined as H = TM∩iTM and J is the restriction of the complex structure in the ambient complex manifold to H. We recall that, for a manifold M = {ρ1 = 0, ρ2 = 0,…, ρk = 0} with M is given by T1,0M = T1,0Cn∩ CTM =  {Z ∈ T1,0Cn | Zρ1 = 0, Zρ2 = 0,…, Zρk = 0}, and dimT1,0M = n − k.AGn can be realized as the real submanifold of Cn:z = x + iy, wj = uj + ivj, j = 2,…, n, by the diffeomorphism T : Sn → Gn defined byz = (1/2)((∂/∂x) − i(∂/∂y)), ∂/∂wj = (1/2)((∂/∂uj) − i(∂/∂vj)), j = 2,…, n, is the complex tangent vector of Sn, and we can define the standard CR structure on Sn by T1,0Sn = CZ and dimT1,0Sn = 1. We can also define quasiconformal mappings on Sn by  : Ω → f(Ω) is a 1-quasiconformal mapping with P; then f is CR or anti-CR; that is, eitherp is some P-differential point.Let Cn. According to the extension theorem , locallP in In the cases of Heisenberg group and Engel group, the condition of Gn, given byd, given byp, q ∈ Gn. The metric is left-invariant and is related to the dilations by the formulaGn.There is a natural homogeneous norm ||·|| on  horizontal tangent space at p ∈ Gn is a subspace HpGn of the tangent space TpGn, and it is spanned by the vector fields X0(p) and X1(p). An absolutely continuous curve γ :  → Gn is horizontal if its tangent vectors t ∈ , lie in the horizontal tangent space Hγ(t)Gn. By Chow ,,Df(p) isB on g byX0, X1] = X2, the corresponding matrix isDefine the bilinear form f be a 1-quasiconformal mapping between domains in Gn; since Df∗(p) is grading-preserving, then Df∗(p) maps Vj   onto itself, respectively. In particular V1 = span{X0, X1}, so for any X ∈ V1,S. Since V2 = RX2, thenλ.Let S above can be written as the following form:θ ∈ [0,2π), where c = |Df∗(p)X|G for any X ∈ V1 with |X|G = 1, λ2/c4 = 1; λ, θ, c depend on p ∈ Gn.The matrix Df∗(p)X = SX, X ∈ V1, now supposeSince Df∗(p) is Lie algebra homomorphism, that is,Df∗(p)X2 = λX2, we find thatBecause f, we haveG is defined by S is an orthogonal matrix. Together with (λ2/c4 = 1. Note that ((1/c)S)T((1/c)S) = I, where I is the 2 × 2 unit matrix; hences11 = ccosθ, θ ∈ [0,2π), we have .Thuss21=−λc2sx; hences112+s122 we have . The resP-differential and the usual derivatives. For the corresponding result on Heisenberg group see p.31 of [C will be used to denote various constants, and the various uses of the letter do not, however, denote the same constant.We get the following relation between f is a quasiconformal mapping between domains in Gn with f(0) = 0, and if f is P-differentiable at 0 and Df(0) preserves the plan P in (0)∂xk and Df∗(0) is the P-derivative of f at 0.Suppose lan P in , then f f is P-differentiable at 0, by definition, there exists 0 < δ < 1 such that, for c < δ and |||| ≤ 1,Df(0) commutes with δc; then for points  in particular, we haveSince Df(0) preserves the plane  P, that is,x0′, x1′, x2′,…, xn′) = f; we getNow compare this with Euclidean distances. Since Df(0) is Lipschitz; then by the equivalence of dc and d, we haveC > 0, namely,From Lemma 3.3 in , Df(0) if is continuous, then when |z| < δ < 1, we haveM2 is a positive constant. Now substituting f is conttituting  and thenThen by , for anyz = , thenIf we let P-differential and the usual tangential mapping. See Proposition 3.2 in [We need the following relation between n 3.2 in  for the f : Ω1 → Ω2 be a quasiconformal mapping between domains in the Goursat group Gn. Suppose f is P-differentiable at p and Df(p) preserves the plan P in X(p) = ∑k=0nXfk(p)∂xk and τq∗ are the tangential mappings of τq for q ∈ Gn.Let lan P in ; then−1∘f∘τp; then F is also a quasiconformal mapping and F(0) = 0. Moreover, by Chain rule  preserves the plan P.Suppose a 3.7 in ), F is PX0(0) = ∂x0 and X1(0) = ∂x1. Therefore,By ,1.From  we can sWe need the following proposition due to Capogna f be a 1-has form , and λ =CV1, the corresponding complex basis is ZG, ZG is as Df~∗(p)ZGf in the formk = 2,3,…, n. ThenFrom φ ∈ Ck(Sn),In fact, for any q = T−1(p) ∈ Sn; by ,  is unknown now. It is also very interesting to determine it, since if this can be solved, we can get the 1-quasiconformal mappings on these groups.The CR automorphism of


We show with examples that our new class of mappings is a real generalization of several known classes of mappings. We also establish fixed point results for such mappings in metric spaces. Applying our new results, we obtain fixed point results on ordinary metric spaces, metric spaces endowed with an arbitrary binary relation, and metric spaces endowed with graph.We extend the notion of generalized weakly contraction mappings due to Choudhury et al. (2011) to generalized  The well-known Banach's contraction principle has been generalized in many ways over the years –6. One oOn the other hand, the concept of the altering distance function was introduced by Khan et al. . In 2011α-ψ-contraction mappings and α-admissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Afterwards, many fixed point results via the concepts of α-admissible mappings occupied a prominent place in many aspects →∪{2,3, 4,…}. From , we get that T satisfies condition , d}) = 2x − 1 and soy < x, we have ϕ, d}) = 2x − 1, and hencey > x, we have ϕ, d}) = 2y − 1, and henceNow we obtain thatT satisfies condition ) ≥ 1 and soIf T satisfies condition . A self-mapping T : X → X is said to be β0-subadmissible if the following condition holds:Let X be a nonempty set and α : X × X → [0, ∞) a mapping.α is said to be forward transitive  if for each x, y, z ∈ X for which α ≥ 1 and α ≥ 1 one has α ≥ 1;α is said to be 0-backward transitive  if for each x, y, z ∈ X for which 0 < α ≤ 1 and 0 < α ≤ 1 one has 0 < α ≤ 1.Let α-β-weakly contraction mappings type A.In this subsection, we give the fixed point results for generalized X, d) be a complete metric space, α, β : X × X → [0, ∞) two given mappings, and T : X → X a generalized α-β-weakly contraction type A; then the following conditions hold: (a)T is continuous;(b)T is α-admissible and β0-subadmissible;(c)α is forward transitive and β is 0-backward transitive;(d)x0 ∈ X such that 0 < β ≤ 1 ≤ α.there exists Then T has a fixed point in X.Let (x0 ∈ X in assumption (d) and letting xn+1 = Txn for all n ∈ N ∪ {0}, if there exists n0 ∈ N ∪ {0} such that xn0 = xn0+1, then xn0 is a fixed point of T. This finishes the proof. Therefore, we may assume that xn ≠ xn+1 for all n ∈ N ∪ {0}. Since 0 < β ≤ 1 ≤ α, we getT is α-admissible and β0-subadmissible thatxn} is a sequence in X such that xn+1 = Txn andn ∈ N ∪ {0}. By using the generalized α-β-weakly contractive condition type A of T, we haven ∈ N ∪ {0}. Now we obtain thatn ∈ N ∪ {0}. From  ≤ d for some n ∈ N ∪ {0}. Then we haved < d for all n ∈ N ∪ {0}. This means that {d} is a monotone decreasing sequence. Since {d} is bounded below, there exists r ≥ 0 such thatn ∈ N ∪ {0}. Taking n → ∞ in the above inequality, we haver = 0; that is,Suppose that r.Using , we getxn} is a Cauchy sequence. Suppose that {xn} is not a Cauchy sequence. Then there exists ε > 0 such thatnk > mk ≥ k, where k ∈ N. Further, corresponding to mk, we can choose nk in such a way that it is the smallest integer with nk > mk ≥ k satisfying .From  and 44)xn} is a).From .From (limk→∞d(.Using (limk→∞d(.Using (limk→∞d(an provelimk→∞d(α-β-weakly contraction mapping type A in C) ≤ 1 ≤ α for all n ∈ N and xn → x as n → ∞, then 0 < β ≤ 1 ≤ α for all n ∈ N.if {In the next theorem, the continuity of a generalized X, d) be a complete metric space, α, β : X × X → [0, ∞) two given mappings, and T : X → X a generalized α-β-weakly contraction type A; then the following conditions hold:(a)C) holds;condition ((b)T is α-admissible and β0-subadmissible;(c)α is forward transitive and β is 0-backward transitive;(d)x0 ∈ X such that 0 < β ≤ 1 ≤ α.there exists Then T has a fixed point in X.Let , we obtain thatn ∈ N. Now, let us claim that Tx* = x*. Supposing the contrary, from the fact that T is a generalized α-β-weakly contractive type A and =C′) and defined byTn+1x0 = T(Tnx0) and T0x0 = x0.Let C′) to the hypotheses of O exists, where z is an element in X satisfying (T has a unique fixed point.By adding condition (tisfying . Then T x and x* are two fixed points of T. By condition (C′) there exists z ∈ X such thatT is α-admissible and β0-subadmissible thatn ∈ N. Since the limit of O exists, we get that {Tnz} converges to some element in X. Let us claim that Tnz → x as n → ∞. Suppose the contrary; that isα-β-weakly contractive condition type A of T, we haven ∈ N. On the other hand, we haven ∈ N. From =α-β-weakly contraction mappings type B.In this subsection, we obtain the existence and uniqueness of fixed point theorems for generalized X, d) be a complete metric space, α, β : X × X → [0, ∞) two given mappings, and T : X → X a generalized α-β-weakly contraction type B; then the following conditions hold:(a)T is continuous;(b)T is α-admissible and β0-subadmissible;(c)α is forward transitive and β is 0-backward transitive;(d)x0 ∈ X such that 0 < β ≤ 1 ≤ α.there exists Then T has a fixed point in X.Let  ≤ d for some n ∈ N ∪ {0}. From  < d for all n ∈ N ∪ {0}. This means that {d} is a monotone decreasing sequence. It follows from a sequence {d} bounded below that there exists r ≥ 0 such thatn ∈ N ∪ {0}. Taking n → ∞ in the above inequality, we haver = 0; that is,As in the proof of , we getψ be a complete metric space, α, β : X × X → [0, ∞) two given mappings, and T : X → X a generalized α-β-weakly contraction type B; then the following conditions hold:(a)C) holds;condition ((b)T is α-admissible and β0-subadmissible;(c)α is forward transitive and β is 0-backward transitive;(d)x0 ∈ X such that 0 < β ≤ 1 ≤ α.there exists Then T has a fixed point in X.Let  ≠ 0. By using  to the hypotheses of O exists, where z is an element in A satisfying  = β = 1 for all x, y ∈ X in Setting X, d) be a complete metric space and T : X → X a continuous generalized weakly contraction mapping. Then T has a fixed point in X.Let  be a complete metric space and T : X → X a continuous mapping andx, y ∈ X, where k ∈ [0,1) andT has a fixed point in X.Let  be a complete metric space and T : X → X a continuous mapping andx, y ∈ X, whereT has a fixed point in X.Let  be a complete metric space and let T : X → X be a continuous mapping andx, y ∈ X, whereT has a fixed point in X.Let  be a metric space and R a binary relation over X. A mapping T : X → X is said to be a generalized weakly contraction with respect to R if for each x, y ∈ X for which xRy one hasψ : [0, ∞)→[0, ∞) is altering distance function, and ϕ : [0, ∞)→[0, ∞) is a continuous function with ϕ(t) = 0 if and only if t = 0.Let  be a metric space and R a binary relation over X and T : X → X a generalized weakly contraction with respect to R; then the following conditions hold: (A)T is continuous;(B)X has a transitive property with respect to R;(C)T is comparative mapping with respect to R;(D)x0 ∈ X such that x0R(Tx0).there exists Then T has a fixed point in X.Let  defined byα = β = 1. Since T is comparative mapping with respect to R, we get T is α-admissible and β0-subadmissible. Also, α is forward transitive and β is 0-backward transitive since X has a transitive property with respect to R. Since T is a generalized weakly contraction with respect to R, we have, for all x, y ∈ X,T is generalized α-β-weakly contraction mapping types A and B. Now all the hypotheses of T follows from Consider two mappings T, we need the following condition:CR) be a metric space and R a binary relation over X and T : X → X a generalized weakly contraction with respect to R; then the following conditions hold: (A)CR) holds on X;the condition ((B)X has a transitive property with respect to R;(C)T is comparative mapping with respect to R;(D)x0 ∈ X such that x0R(Tx0).there exists Then T has a fixed point in X.Let (α and β given by (CR) implies property (C).The result follows from given by  and by oCR′) to the hypotheses of O exists, where z is an element in X satisfying (T has a unique fixed point.By adding condition (tisfying . Then T α and β given by (CR′) implies property (C′).The result follows from given by  and by oX, d) be a metric space. A set { : x ∈ X} is called a diagonal of the Cartesian product X × X and is denoted by Δ. Consider a graph G such that the set V(G) of its vertices coincides with X and the set E(G) of its edges contains all loops; that is, Δ⊆E(G). We assume G has no parallel edges, so we can identify G with the pair (V(G), E(G)). Moreover, we may treat G as a weighted graph by assigning to each edge the distance between its vertices. A graph G is connected if there is a path between any two vertices.Throughout this section, let  be a metric space endowed with a graph G and T : X → X mapping. One says that T preserves edges of G ifLet  be a metric space endowed with a graph G and T : X → X mapping. One says that X has a transitive property with respect to graph G ifLet  be a metric space endowed with a graph G. A mapping T : X → X is said to be a generalized weakly contraction with respect to graph G if for each x, y ∈ X for which  ∈ E(G) one hasψ : [0, ∞)→[0, ∞) is altering distance function, and ϕ : [0, ∞)→[0, ∞) is a continuous function with ϕ(t) = 0 if and only if t = 0.Let  be a metric space, T : X → X a given mapping, ψ : [0, ∞)→[0, ∞) an arbitrary altering distance function, and ϕ : [0, ∞)→[0, ∞) an arbitrary continuous function with ϕ(t) = 0 if and only if t = 0. If ϕ) ≤ ψ) for all x ∈ X, then T is trivially generalized weakly contraction with respect to graph G, where G = (V(G), E(G)) = .Let  be a metric space endowed with a graph G and T : X → X a generalized weakly contraction with respect to graph G; then the following conditions hold: (A)T is continuous;(B)X has a transitive property with respect to graph G;(C)T preserves edges of G;(D)x0 ∈ X such that  ∈ E(G).there exists Then T has a fixed point in X.Let  defined byα = β = 1. Since T preserves edges of G, we get T is α-admissible and β0-subadmissible. Also, α is forward transitive property and β is 0-backward transitive since X has transitive property with respect to graph G. Since T is a generalized weakly contraction with respect to graph G, we getx, y ∈ X. This implies that T is generalized α-β-weakly contraction mapping types A and B. Therefore, all the hypotheses of T follows from Consider two mappings T, we need the following condition.In order to remove the continuity of X, d) be a metric space endowed with a graph G. One says that X has G-regular property if {xn} is the sequence in X such that  ∈ E(G) for all n ∈ N and it converges to the point x ∈ X; then  ∈ E(G) for all n ∈ N.Let  be a metric space endowed with a graph G and T : X → X a generalized α-β-weakly contraction with respect to graph G; then the following conditions hold: (A)X has G-regular property;(B)X has a transitive property with respect to graph G;(C)T preserves edges of G;(D)x0 ∈ X such that  ∈ E(G).there exists Then T has a fixed point in X.Let (α and β given by (G-regular property implies property (C).The result follows from given by  and by oCG′) to the hypotheses of O exists, where z is an element in X satisfying (T has a unique fixed point.By adding condition (tisfying . Then T α and β given by (CG′) implies property (C′).The result follows from given by  and by oBy using X, d) be a metric space endowed with a graph G and T : X → X a generalized weakly contraction with respect to graph G; then the following conditions hold:(A)T is continuous;(B)G is connected graph;(C)T preserves edges of G;(D)x0 ∈ X such that  ∈ E(G).there exists Then T has a fixed point in X.Let  be a metric space endowed with a graph G and T : X → X a generalized α-β-weakly contraction with respect to graph G; then the following conditions hold:(A)X has G-regular property;(B)G is connected graph;(C)T preserves edges of G;(D)x0 ∈ X such that  ∈ E(G).there exists Then T has a fixed point in X.Let (CG′) to the hypotheses of O exists, where z is an element in X satisfying (T has a unique fixed point.By adding condition (tisfying . Then T


G-metric, metric-like, and b-metric to define new notion of generalized b-metric-like space and discuss its topological and structural properties. In addition, certain fixed point theorems for two classes of G-α-admissible contractive mappings in such spaces are obtained and some new fixed point results are derived in corresponding partially ordered space. Moreover, some examples and an application to the existence of a solution for the first-order periodic boundary value problem are provided here to illustrate the usability of the obtained results.We unify the concepts of  A function σb : X × X → R+ is a b-metric-like if, for all x, y, z ∈ X, the following conditions are satisfied: σb1)  σb = 0 implies x = y;( σb2)  σb = σb;( σb3)  σb ≤ s[σb + σb]. such that X is a nonempty set and σb is a b-metric-like on X. The number s is called the coefficient of .A b-metric-like space  if x, y ∈ X and σb = 0, then x = y, but the converse may not be true and σb may be positive for all x ∈ X. It is clear that every b-metric space is a b-metric-like space with the same coefficient s but not conversely in general.In a X = R+, let p > 1 be a constant, and let σb : X × X → R+ be defined byLet X, σb) is a b-metric-like space with coefficient s = 2p−1.Then,  be a metric-like space and σb = [σ]p, where p > 1 is a real number. Then, σb is a b-metric-like with coefficient s = 2p−1.Let  = (x + y − xy)p, where p > 1 is a real number, is a b-metric-like on X with coefficient s = 2p−1.Let X = R. Then, the mappings σbi : X × X → R+ defined byb-metric-like on X, where p > 1, a ≥ 0, and b ∈ R.Let b-metric-like σb on X generates a topology τσb on X whose base is the family of all open σb-balls {Bσb : x ∈ X, ɛ > 0}, where Bσb = {y ∈ X : |σb − σb| < ɛ} for all x ∈ X and ɛ > 0.Each b-metric-like space, or Gσb-metric space, as a proper generalization of both of the concepts of b-metric-like spaces and G-metric spaces.Now, we introduce the concept of generalized X be a nonempty set. Suppose that a mapping Gσb : X × X × X → R+ satisfies the following:Gσb1) = 0 implies x = y = z;Gσb2) = Gσb, where p is any permutation of x, y, z ;Gσb3) ≤ s[Gσb + Gσb] for all x, y, z, a ∈ X .Let Gσb is called a Gσb-metric and  is called a generalized b-metric-like space.Then, b-metric-like space.The following proposition will be useful in constructing examples of a generalized X, σb) be a b-metric-like space with coefficient s. Then,b-metric-like functions on X.Let  and  are generalized b-metric-like spaces with coefficient s = 2p−1. Note that, for x = y = z > 0, Gσbm = (2x)p > 0 and Gσbs = 3(2x)p > 0.Let X = . Then, the mappings Gσb1m, Gσb1s : X × X × X → R+ defined byp > 1 is a real number, are generalized b-metric-like spaces with coefficient s = 2p−1.Let By some straight forward calculations, we can establish the following.X be a Gσb-metric space. Then, for each x, y, z, a ∈ X, it follows that:Gσb > 0 for x ≠ y;Gσb ≤ s + Gσb);Gσb ≤ 2sGσb;Gσb ≤ sGσb + s2Gσb + s2Gσb.Let X, Gσb) be a Gσb-metric space. Then, for any x ∈ X and r > 0, the Gσb-ball with center x and radius r isLet  on X, which we call it Gσb-metric topology.The family of all X, Gσb) be a Gσb-metric space. Let {xn} be a sequence in X. Consider the following.(1)x ∈ X is said to be a limit of the sequence {xn}, denoted by xn → x, if limn,m→∞Gσb = Gσb.A point (2)xn} is said to be a Gσb-Cauchy sequence, if limn,m→∞Gσb exists (and is finite).{(3)X, Gσb) is said to be Gσb-complete if every Gσb-Cauchy sequence in X is Gσb-convergent to an x ∈ X such that be a Gσb-metric space. Then, for any sequence {xn} in X and a point x ∈ X, the following are equivalent:xn} is Gσb-convergent to x;{Gσb → Gσb, as n → ∞;Gσb → Gσb, as n → ∞;Let  and  be two generalized b-metric like spaces and let f : → be a mapping. Then, f is said to be Gσb-continuous at a point a ∈ X if, for a given ɛ > 0, there exists δ > 0 such that x ∈ X and |Gσb − Gσb|<δ imply that |Gσb′(f(a), f(a), f(x)) − Gσb′(f(a), f(a), f(a))|<ɛ. The mapping f is Gσb-continuous on X if it is Gσb-continuous at all a ∈ X. For simplicity, we say that f is continuous.Let  and  be two generalized b-metric like spaces. Then, a mapping f : X → X′ is Gσb-continuous at a point x ∈ X if and only if it is Gσb-sequentially continuous at x; that is, whenever {xn} is Gσb-convergent to x, {f(xn)} is Gσb-convergent to f(x).Let  be a Gσb-metric space and suppose that {xn}, {yn}, and {zn} are Gσb-convergent to x, y, and z, respectively. Then, we haveyn} = {zn} = a are constant, thenLet  a function. We say that T is an α-admissible mapping ifLet ψ :  with the usual ordering on R. Define the generalized b-metric-like function Gσb given by Gσb = max{(x + y)2, (y + z)2, (z + x)2} with s = 2. Consider the mapping f : X → X defined by f(x) = ln(1 + xex−/4) and the function ψ ∈ Ψb given by ψ(t) = (1/8)t,  t ≥ 0. It is easy to see that f is increasing function. Now, we show that f is a Gσb-continuous function on X.Let xn} be a sequence in X such that limn→∞Gσb = Gσb, so we have max{(limn→∞xn + x)2, 4x2} = 4x2  and, equally, limn→∞xn = α ≤ x. On the other hand, we havef is Gσb-sequentially continuous on X.Let {x, y, z ∈ X, we havef satisfies all the assumptions of u = 0).For all elements In this section, we present an application of our results to establish the existence of a solution to a periodic boundary value problem see , 31)..31]).X = C be the set of all real continuous functions on . We first endow X with the b-metric-likeu, v ∈ X where p > 1 and then we endow it with the generalized b-metric-like Gσb defined byX, Gσb) is a complete generalized b-metric-like space with parameter K = 2p−1. We equip C with a partial order given byC, ⪯) is regular; that is, whenever {xn} in X is an increasing sequence such that xn → x as n → ∞, we have xn⪯x for all n ∈ N ∪ {0}.Let r, as in , it is pt ∈ I = , with T > 0, and f : I × R → R is a continuous function.Consider the first-order periodic boundary value problemα ∈ C1 such thatt ∈ I = .A lower solution for  is a funλ > 0 such that, for all x, y ∈ C, we havea < 1. Then, the existence of a lower solution for (Assume that there exists tion for  providestion for .t ∈ I.Problem  can be rx(0) = x(T), we getx(0) in =x(0)creasing . Note thution of .x, y, z ∈ C. Then, we haveα is a lower solution for (α⪯S(α) [Let tion for , so it i (α⪯S(α) .ψ(t) = ln((a/2p−1)t + 1) where 0 ≤ a < 1. Hence, there exists a fixed point Hence, the hypotheses of  problem .


Nature Communications6: Article number: 8903 10.1038/ncomms8903 (2015); Published: 07302015; Updated: 03102016y axis should be ‘cm2V−1s−1' not ‘V−1s−1cm−1'. Similarly, the second and third sentences of the second paragraph of the ‘Quantum yield calculation' section should read ‘From Fig. 2 and Table 1, we obtain ϕμ(1+c1)=11 cm2V−1s−1 and ϕk2=11 × 10−10 cm3s−1. These data compare well with the corresponding values of 8.2 cm2V−1s−1 and 9.2 × 10−10 cm3s−1 in Wehrenfennig et al.8 where the support was mesoporous alumina.'This article contains errors in the units used for carrier mobility. In Fig. 2a–d, the units on the


In recent years, with the wide application of computers, difference system has become one of the important theoretical bases for computer, information system, engineering control, ecological balance, and so forth. As typical nonlinear difference equations, rational difference equations have become a research hot spot in mathematical modelling. The behavior of solutions of the system for rational difference equation has received extensive attention.In , Ozban hIn , KurbanlThe periodicity of the positive solutions of the rational difference systemxn+1=1yn,In , Ozban seader to , 6 and tA, B ∈ [0, +∞) and x−1, x0, y−1, and y0 ∈ [0, +∞).In this paper, we investigate the behavior of positive solutions for the system of rational difference equationsBefore stating our main results, we state some definitions used in this paper.xn, yn)}n=0∞ that satisfies }n=0∞ of  = , n = 0,1,…, and T is called a period.A solution {}n=0∞ be a solution of system , all solutions of system }n=0∞ be a solution of system xA, B > 0, it can be seen easily that all solutions of }n=0∞ be an arbitrary positive solution of ((i)A < B2, then, for each integer n ≥ 0, the subsequence {xn+3l}l=0∞ → 0, and the subsequence {yn+3l}l=0∞ → ∞.If (ii)A > B2, then, for each integer n ≥ 0, the subsequence {xn+3l}l=0∞ → ∞, and the subsequence {yn+3l}l=0∞ → 0.If (iii)xn+3yn+3 = xnyn, n = 0,1,….Let {(ution of .(i)If A n, we will show thatl = 0, xn = (A/B2)0xn. Assume that  For every fixed how thatxn+3l=AB2how thatxn+3l=AB2ion thatyn+3l=B2Ahow thatxn+3l=AB2ion thatyn+3l=B2Ax−1 = 8, x0 = 1, y−1 = 3, and y0 = 7.(i)A = 1, B = 2, and n = 2, xn+3l}l=0∞ and {yn+3l}l=0∞ (shown by the black spot at the top of every peak).For (ii)A = 2, B = 1, and n = 2, xn+3l}l=0∞ and {yn+3l}l=0∞ (shown by the black spot at the top of every peak).For Let


There is an error in Table 4. The value listed for “Work + Regular Saving” under “Model 2” should be “0.382–6.623”. Please see the corrected


Reconciliation methods explain topology differences between a species tree and a gene tree by evolutionary events other than speciations. However, not all phylogenies are trees: hybridization can occur and create new species and this results into reticulate phylogenies. Here, we consider the problem of reconciling a gene tree with a species network via duplication and loss events. Two variants are proposed and solved with effcient algorithms: the first one finds the best tree in the network with which to reconcile the gene tree, and the second one finds the best reconciliation between the gene tree and the whole network. Moreover, we propose a faster algorithm solving the problem described in [Reconciliations explain topology incompatibilities between a species tree and a gene tree by evolutionary events - other than speciation - affecting genes , for a r and thisribed in  when resWe start by giving some basic definitions that will be useful in the paper.Definition 1 (Rooted phylogenetic network)A rooted phylogenetic network N on a label set (N), has indegree 0 and outdegree 2. All other internal nodes have either indegree 1 and outdegree 2 (speciation node), or indegree 2 and outdegree 1 (hybridization node).V(N), I(N), E(N), L(N) and N. The size of N, denoted by |N|, is equal to |V(N)| + |E(N)|. Given x in V(N), we denote by x Nthe subnetwork of N rooted at x, i.e. the subgraph of N consisting of all edges and nodes reachable from x. If x is a leaf of N, we denote by s(x) the label of x in x is a speciation node, we denote by p(x) the only parent of x.Denote by x and y of N, we say that x is lower or equal to y in N, denoted by x ≤N y , if and only if there exists a path (possible reduced to a single node) in N from y to x (resp. and x ≠ y). If x ≤N y, then, for every path p from y to x, denote by length(p) the number of speciation nodes in N such that x <N z ≤N y. If N is a tree, then, for every two nodes u, v of N, N LCA [N that is above or equal to both u, v.Given two nodes N   is the lx in N, two paths of N starting from x are said to be separated if each path contains a different child of x. Let x, y be two nodes of N. Denote by z of N such that there exist two separated paths in N from z to x and y. For example, in Figure N is a tree and x, y are not comparable, N M contains exactly one node, which coincides with N LCA.Given a speciation node N has at most k hybridization nodes, we say that N is of level-k [networks and trees, respectively. In this paper, we allow trees to contain artificial nodes, i.e. nodes with indegree and outdegree 1.If every biconnected component of  level-k . A rooteB be a biconnected component of a network N. Then B contains exactly one node r(B) without ancestors in B [[r(B) the root of B. If B is not trivial, i.e. B consists of more than one node, we can contract it by removing all nodes of B other than r(B), then connect r(B) to every node with indegree 0 created by this removal. Then the following definitions are well-posed.Let rs in B [, Lemma 5Definition 2 (Tree bc(N)) Given a network N, the tree bc(N) is obtained from N by contracting all its biconnected components.N and its associated tree bc(N).For example, Figure bc(N) that corresponds to a biconnected component B in N. Given two biconnected component i, Bj B, we say that i B≤ N Bj (resp. i <N Bj B) if and only if i Bis the parent (resp. a child) of Bj if bc(N). We also denote by NLCA the biconnected component corresponding to N.Let denote by Definition 3 (Elementary network) Given a network N, each biconnected component B that is not a leaf of N defines an elementary network, denoted by N(B), consisting of B and all cut-edges coming out from B.Definition 4 (Switchings of a network [) Given a network N, a switching S of N is obtained from N by choosing, for each hybridization node, an incoming edge to switch on and the other to switch off. Once this is done, we also switch off all switched-on edges with the target node having only switched-off outgoing edges  for an example). For each bi-connected component B of N, we also denote by S(B) the subgraph of S restricted to N(B). network ) Given ac) a switching of the level-2 network in (a). We denote by onV(S) the set of nodes of S that are not an endpoint of any switched-off edge. A path of S is a path of N that uses only switched-on edges.Switchings will be used in the next section to model gene histories for genes evolving in a species network. For example, Figure G will denote a tree and N a network such that there is a bijection between L(N) and G represents a gene tree such that each leaf corresponds to a contemporary gene and is labeled by the species containing this gene, while N is a species network such that each leaf represents an extant species. In the cophylogeny problem, G represents a parasite tree such that each leaf corresponds to a parasite species that is labeled by the species that hosts it, while N is a host network such that each leaf represents an extant species.Hereafter, G is associated to a node of S and an event - a speciation u of G with a leaf x of S such that s(u) = s(x). A speciation in a node u of G is constrained to the existence of two separated paths from the mapping of u to the mappings of its two children, while the only constraint given by a duplication event is that evolution of G cannot go back in time. More formally:We will now extend the definition of ation in  to netwoDefinition 5 (Reconciliation) Given a tree G and a network N such that a reconciliation between G and N is a function α that maps each node u of G to a pair (rα(u), αe(u)) where αr(u) is a node of V(N) eand α(u) is an event of type  or orsuch that: if and only if u ∈ L(G), rα(u)∈L(N) and s(u) = s(rα(u));• for every u ∈ I(G) with child nodes {u1, u2}, if then • for any two nodes u, v of V(G) G u, if such that v <r then α(v) ≤N αr(u). r Otherwise, α(v) N αr<(u).• N is a tree, this definition coincides with the one given in [Note that, if given in .α, denoted by d(α) is the number of nodes u of G such that x, y of N, the distance between x and y, denoted by dist, is defined as follows:The number of duplications of  done in . In morey ≤N x, then dist = minp length(p) over all possible paths p from x to y;• If dist = +∞.• otherwise, u ∈ I(G) with child nodes {u1, u2}, the number of losses associated with u in a reconciliation α, denoted by αl(u), is defined as follows:Then, for every αl(u) = min{dist) + dist), dist)+dist)} where x1, x2 are the two children of rα(u);• if αl(u) = dist(rα(u), rα(u1)) + dist(rα(u), rα(u2)).• if α, denoted by l(α), is the sum of αl(·) for all internal nodes of G. Thus, the cost of α, denoted by cost(α), is d(α)·δ + l(α)·λ, where δ and λ are respectively the cost of a duplication and a loss event. We use cost to denote the cost of the minimum reconciliations between G and N. A reconciliation having the minimum cost is called a most parsimonious reconciliation.The number of losses of a reconciliation S be a switching of N, then a reconciliation between G and S is defined similarly to Definition 5 except that only switched-on edges are considered when defining paths, and only nodes in onV(S) are counted for calculating the length of the shortest path in the definition of dist. This is done to model the fact that, since each gene of an hybridization species is inherited from one of its two parents, we should not count as a loss the fact that the other parent does not contribute. Moreover, note that, for every two nodes x, y of S such that S yx ≤, there is a unique path from y to x in S.Let N is a tree, there is a unique reconciliation (the LCA reconciliation) between G and N which has minimum cost and this reconciliation can be found in O(|G|) time [When G|) time -18 as fou in L(G), rα(u) is defined as the only node x of S such that s(u) = s(x), and • for each node u in I(G) with child nodes {u1, u2}, rα(u) = N LCA(rα(u1), rα(u2)); if rα(u1) N ≤rα(u2) or rα(u2) N αr≤(u1) then • for each node e αis totally defined by rα, hence it can be omitted, and we will use α to refer to rα. Note that the algorithms used on trees to find the LCA reconciliation can also be used on switchings, which - when only switched-on edges are considered - are actually trees. Hereafter, when we refer to the reconciliation between a tree and a switching, we refer to the LCA reconciliation between them. The problems in which we are interested in here are defined as follows:In the LCA reconciliation, the mapping Problem 1 BEST SWITCHING FOR THEODELInput A tree G, a network N such that and positive costs δ and λ for respectively  and  events.Output A switching S of N such that the cost is minimum over all switchings of N.Problem 2 MINIMUMECONCILIATION ON NETWORKSInput A tree G, a network N such that and positive costs δ and λ for respectively  and  events.Output A minimum reconciliation between G and N.Remark 1 For the sake of simplicity, we suppose that G does not contain any internal node u such that u are mapped to a leaf of N). If it is not the case, we can simplify the instance by replacing in G each such subtree Gu by a leaf labeled by the unique label in   For every u ∈ V(G), B(u) is defined as the lowest biconnected component B of N such that ℒ(Nr(B)) contains ℒ(uG).Then the following remark holds:Remark 2 If u ∈ L(G), then B(u) is the only leaf of N such that s(u) = s(B(u)). If u ∈ L(G), then B(u) = N LCA(B(u1), B(u2)) where u1, u2 are child nodes of u in G.N Gthe tree obtained from G by adding some artificial nodes on the edges of G and label each node of N Gby a biconnected component of N via an extension of the labeling function B(·) as follows:We define by Definition 7 (Tree N G) N is obtained from G as follows: for each internal node u in G with child nodes uThe tree G1, u2 such that there exist k biconnected components (u) and strictly above B(u1), we add k artificial nodes v1 > ... >k on the edge v, and B(j v) is fixed to 2.N G. We have the following lemma:See Figure Lemma 1 Let u be a node of I(N G), and u' be one of the children of u. If u is an artificial node, then B(u) is the parent of B(u') and a child of B(p(u)); otherwise, B(u') is either equal to B(u) or one of its children.Proof: Suppose that u is an artificial node. Then, by Definition 7, B(p(u)), B(u), B(u') are three consecutive nodes of N G, thus B(u) is the parent of B(u') and a child of B(p(u)). Suppose now that u is not an artificial node, and let u" be the child of u in G such that u" ≤Gn u' ≤Gn u. If B(u") = B(u'), then by definition, u" = u' because no artificial node is added between u and u", and thus the claim holds. If B(u') ≠ B(u"), then by Definition 7, B(u') is the highest biconnected component of N that is below B(u) and above B(u"), which is then a child of B(u).Definition 8 (B G) B is the set of all maximal connected subgraph H of GN such that BLet B be a biconnected component of N different from a leaf, then G(u) = B for every u ∈ I(H).For example, in Figure Lemma 2 Let B be a biconnected component of N different from a leaf, then we have the following:(i) for every H ∈ B, H is either a binary tree or an edge whose upper extremity is an artificial node. Moreover, for every leaf u of H, BG(u) is a child of B.(ii) if B = B(r(G)), B consists of one binary treethen G.Proof: (i) First, suppose that I(H) contains an artificial node u. Then this artificial node is the only internal node of H; indeed, by Lemma 1, the value of B(·) for the parent and the child of u are both different from B. Thus, H consists of only one edge whose upper extremity is u. If I(H) does not contain any artificial node, it follows that H is a binary tree. Moreover, by Lemma 1 and Definition 8, B(u) is a child of B for every leaf u of H.(ii) Suppose now that B = B(r(G)), and B Gcontains at least two components H1, H2, rooted at two different nodes r1, r2 where B(r1) = B(r2) = B. Let B(v) = B for every node v on the two paths from r to r1 and to r2, because B(r(G)) = B(r1) = B(r2) = B and H1 and H2 since they can both be extended. Hence B Gcontains only one component. Suppose that this component is an edge; thus, its upper extremity is an artificial node that has been added on the path from a node u to a node v of G where u is strictly higher than r(G). But this is not possible, because r(G) is the highest node of G. Therefore, B Gcontains one binary tree.i Bdifferent from a leaf, denote by i Sbe a switching of N(iB), and let H be a tree in u ∈ L(H), B(u) is a child of i Band thus r(B(u)) is a leaf of N(iB), which is also a leaf of iS. Hence, we can define a reconciliation between H and iS, denoted by u of H is mapped to the leaf r(B(u)) of iS.Given a biconnected component Lemma 3 Let S be a switching of N, and let α be the reconciliation between G and S. For every u ∈ I(G), there exists such that u ∈ I(H), and The proof of this lemma is deferred to the appendix. The following definition will be useful later on.Definition 9 ) i be a biconnected component of N different from a leaf, and Si a switching of NLet B(iB); then cost is defined as follows:i if B= B(r(G)), and otherwise;• where u is the only leaf of H.• H to a reconciliation between G and any switching of N that contains iS. For example, in Figure H is the edge  and, if i Sis the switching on the left, then cost = λ.As we will see later, this cost corresponds to the contribution of N :The next lemma is a central one, and it permits to solve Problem 1 independently per each biconnected component of Lemma 4 Let {B1,..., p} be the biconnected components of N that are not leaf nodes, and let S be a switching of N where each elementary network N(Bi) has Si as switching. Then BProof: Let α be the reconciliation between G and S. Denote by αd(iS) the number of nodes in I(iS) associated to a duplication by α. By Remark 1, no duplication happens at leaves of S. Additionally, u of H (Lemma 3).i Sif and only if it is also on (resp. off) in S. Let x, y be two nodes of S such that S xy ≤. Then we define z in onV(iS) \ L(iS) such that S z ≤S xy <. Then, for each internal node u of G, we define the number of losses associated with u in iS, denoted by αl, similarly to αl(u) but using the function dist. Then, u1, u2 be two children of u in G, then αl >0 only if the path from α(u) to either α(u1) or α(u2) in S contains at least a node of iB. Therefore, we have three possible cases: (1) α(u) is in is in iB, (3) α(u) is above r(iB) while either α(u1) or α(u2) is in a biconnected component below iB. In case (1), by Lemma 3, there exists a binary tree H of u∈I(H), and v∈I(iS), thus i Sthat case (2) holds for u1, then u1 must be the root of a binary tree H1 of t G, and the contribution of u1 to αl is u1 ≠ r(G). Finally, let suppose that case (3) holds for u1 and let a, ub ube the artificial nodes added on the edge  of G such that B(au) = i Band B(bu) is a child of iB. Hence, u1 to αl is dist(r(iS), r(B(bu))). Let call u in the first, second, and third case. By construction, V1 is disjoint from u has two children u1, u2 such that u1 is in iB, and u2 is below iB, then u must be in iB, i.e. cannot be above r(iB). Thus,Let us now consider the loss count. Note that a node/edge is on (resp. off) in Therefore,G are considered. Hence, the second term between squared brackets corresponds to As proved above (in case (3)), the first term between squared brackets is equal to Therefore, N for which its reconciliation with G has the smallest cost, by analyzing each biconnected component of N independently. See Figure Algorithm 1 computes the switching of Algorithm 1 Solving Problem 1Input: A species network N and a gene tree G such that δ, λ for duplication and loss events, respectively.1: Output: A switching S of N that is optimal in the sense of Problem 1.2: N Gand its labeling function B(·);3: Compute the tree i Bof N that is not a leaf;4: Compute for each biconnected component i Bof N do5: for each switching N(iB) do6:      7:         N(iB) with the lowest value of j costover all j;8:     S of N in which each elementary network N(iB) has 9: Return the switching Theorem 2 Let N be a level-k network with p biconnected components. Algorithm 1 runs in O(|N| + 2k·p·|G|) time and returns a switching S of N such that cost is minimum.Proof: Complexity: It is well-known that all biconnected components of N can be computed in linear time, i.e. O(|N|), by using depth-first-search [LCA operations on a tree can be performed in constant time [B(·) can be computed in O(|NG|). Hence, the tree N Gcan be constructed in times O(|N | G+ |N|).t-search . After aant time ,22. ThusN Gwhich takes time O(|N G|), we can compute iB. Each biconnected component i Bof N has at most k hybridization nodes, then N(iB) has at most 2k switchings. For each switching N(iB), it takes O(|B G|) to compute j cost[Bi Gis the size of N G. Therefore, the total complexity of the loop at lines 5 - 8 is O(2k ·|N G|). Each edge of G can have at most p artificial nodes added to it. Hence in the worst case O(|N G|) = O(|G|·p), i.e. the total complexity of Algorithm 1 is O(|N| + 2k·p·|G|).By a simple traversal of te costj ,24. The Correctness: Let S be a switching of N where each elementary network N(iB) has i Sas the switching. By Lemma 4, cost = cost is minimum if and only if iS. Lines 5-8 in Algorithm 1 search, for each network N(iB), the tree G and a network N, in [G on N can be solved in polynomial time, when host switchings  are also accounted for. In their model, each node of N is dated by a single date while each node of G is dated by a set of dates, and they search for a parsimonious reconciliation between G and N, i.e. one that has minimum cost, under the constraint that an event associated to a node u of G can only happen at a node/edge of N whose date is contained in the set of possible dates of u. Although the algorithm complexity stays polynomial, it is very high: O(τ3·|G|·|N|5) for a binary tree and a binary network, where τ is the number of possible dates of the nodes of G and N, which is at most O(|G| + |N|). A drawback of this model is that it requires information on the dates that is often not available. Moreover, transfers are not always possible in all parts of Tree of Life. Here, we take into account only speciation, duplication and loss events, and consider G and N as undated (see Problem 2 and Definition 5 for more details). Using a similar dynamic algorithm on this simpler model, and by some further analyses, we provide an algorithm that is a generalization of the LCA algorithm to networks that has a much smaller complexity than the algorithm in [O(h2·|G|·|N|), where h is the number of hybridization nodes of N.Given a tree rk N, in  the authrithm in , namely x, y be two nodes of N. Denote by z to x and to y passes through z'. For example, in Figure m1, m2, m3 are in m1 and m2 are in m3 to x, y must pass m2.Let For the sake of simplicity, we consider only reconciliations without duplications on hybridization nodes: indeed, since losses are not counted at hybridization nodes, a duplication on such nodes can be moved to its only child without changing the total cost of the reconciliation.Lemma 5 Let α be a reconciliation of minimum cost between G and N, then for every node u of G that has two children u1, u2, we have:(i) if  then (ii) if r then either α(u1) ≤N rα(u2) rand α(u) = rα(u2), ror α(u2) ≤N rα(u1) rand α(u) = rα(u1).Proof: (i) Suppose that rα(u) must be a node of x1, x2 be two children of rα(u) such that αl(u) = dist)+dist). Suppose that y in rα(u) to rα(u1) (resp. to rα(u2)) passes for y. Let y1, y2 be the two children of y.x1 to rα(u1) passes through y, y1, while the one from x2 to rα(u2) passes y, y2. Consider the reconciliation α' such that v ≠ u, while α' respects Definition 5, and that d(α) = d(α'). Denote by f = dist(rα(u), y), then f >0. The numbers of losses in α and α' are different on those associated with u and p(u) (if u is not the root of G). We thus have αl(p(u)) ≥ α'l(p(u)) - f if u ≠ r(G). Moreover, αl(u) = dist) + dist) = dist + 1 + dist) + dist + 1 +dist) ≥ dist(rα(u), y) + dist + dist(rα(u), y)+dist)) ≥ 2·f +α' l(u). Hence, whether u coincides with r(G) or not, l(α) > l(α'), i.e. cost(α) > cost(α'), a contradiction.First, we suppose that the shortest path from x1 to rα(u1) and to rα(u2) pass through y, and then pass through y1 (or y2). This means that y1 is above both rα(u1) and rα(u2). Let y' be one of the lowest nodes below or equal to y1 that is above both rα(u1) and rα(u2). Let y', and sup-1 2 pose, without loss of generality, that rα(u1) and rα(u2). Hence, the shortest path from x1 to rα(u1) must pass, in the order, through y, y1, y', and y', while the shortest path from x2 to rα(u2) must pass through y, y1, y', and α'as done above apart for α'(u), which is fixed to y', we arrive at a contradiction by a similar argument.Suppose now that both the shortest paths from rα(u1), rα(u2) are not comparable. Hence, rα(u) to rα(u1) and to rα(u2) must pass through y. Similarly to case (i), we have that the reconciliation α' that coincides with α apart for the fact that α, a contradiction. Hence, rα(u) N 2 ℳin(rα(u1), rα(u2)). Considering now the reconciliation α' that coincides with α but for α' (u), which is fixed to d(α) = d(α')+1, and u ≠ r(G). Let x1, x2 be the two children of rα(u). If the shortest path from rα(u) to rα(u1) passes through x1 while the shortest path from rα(u) to rα(u2) passes through x2, then cost(α) > cost(α'), a contradiction. Thus, both the two shortest paths from rα(u) to rα(u1) and rα(u2) must pass through x1 (or x2). Let y be one of the lowest nodes located on both of these paths. Then, y ∈ N ℳin(rα(u1), rα(u2)). By the same argument as in the previous case, the reconciliation α" coinciding with α but for α, a contradiction.(ii) Suppose that rα(u1) N αr ≤(u2) or rα(u2) N αr ≤(u1). Suppose that the first case holds (the second case is similar), but rα(u) ≠ rα(u2), i.e. rα(u2) N αr <(u). Let x1, x2 be two children of rα(u). If the shortest path from rα(u) to rα(u1) passes through x1 while the shortest path from rα(u) to rα(u2) passes through x2, then by replacing eα(u) by an rα(u) to rα(u1) and rα(u2) must pass through x1 (or x2). Let y be a node that is located on both paths such that there is no other node below y on these two paths. Then, y ≠ N ℳin(rα(u1), rα(u2)). By the same argument as in the case (i), the reconciliation α' such that v ≠ u, α, a contradiction.Hence, either G and N. Let α be a reconciliation between G and N. For every u ∈ V(G), denote by αcost(u) the cost of the reconciliation of α restricted to uG. Hence, if α is a most parsimonious reconciliation, then αcost(u) is the minimum cost among all reconciliations between u Gand N that maps u to rα(u). Algorithm 2 aims at computing, for each u, the set u) containing all pairs  such that c is the minimum cost among all reconciliations between u Gand N mapping u to x. It is straightforward to see that the cost of a most parsimonious reconciliation between G and N is the minimum cost involved in a pair in r(G)).Now, we are ready to describe a dynamic algorithm to compute a reconciliation of minimum cost between merge used in Algorithm 2 takes as input two lists of pairs  - where x is a node of N and c is a real number - and merges them keeping, for each x, the pair  with the smallest value of c. The method computeMin used in Algorithm 2 is detailed in Algorithm 3. This method computes, for two nodes y, z of N, the set y and z up to the root of N . For this, it labels each node v in such a way that, if v is not strictly above y and z, then label (v) = ∅. Otherwise, label(v) is the lowest node such that every path from v to y and z passes through it. This method also computes the value of the function dist between y (resp. z) and each node visited in the corresponding breath-first-search.The function Algorithm 2 Solving Problem 2Input: A network N and a tree G such that δ, λ for duplication and loss events, respectively.1: Output: The set u) of pairs  for every 2: for each node u of G in post-order do3: u) ← ∅;4:   if u is a leaf then5:   x be the leaf of S such that s(x) = s(u);6:      Let u) ← {};7:      else8:   u1, u2 be the two children of u;9:      Let for each  ∈ u1) and each  ∈ u2) do10:      computeMin;11:         C ← ∅;12:         for each do13:         x1, x2 be the two children of x in N ;14:            Let c = c1 + c2 + λ·min{dist + dist, dist + dist};15:            C ← C ∪ {};16:            if N z y ≤then17:         c = δ + λ·dist+ c1 + c2;18:            C ← C ∪ {};19:            else if N y z ≤then20:         c = δ + λ·dist + c1 + c2;21:            C ← C ∪ {};22:            u) = merge23:        24: Return The following theorem proves the correctness of Algorithm 2:Theorem 3 Algorithm 2 returns a matrix  such that, for every u ∈ V(G),  is contained in u) u and N mapping u to x with cost cif and only if there exists a most parsimonious reconciliation between G.Algorithm 3 computeMiny, store the ordered list of visited nodes in BFS(y) and compute, for each u in BFS(y), dist;1: Proceed a BFS from z;2: Do the same from for each node v ∈ BFS(y) do3: if v = y or v = z or v is not in BFS(z) then label(v) = ∅4:   else if v has only one child v1 that is in BFS(z) or BFS(y) then5:   label(v) = label(v1);6:      else7:   v1, v2 be the two children of v;8:      Let if label(v1) = label(v2) ≠ ∅ then9:      label(v) = label(v1);10:         else11:      label(v) = {v}; Add v into 12:         Proof: For each u ∈ V(G), we need to prove that, if α' is a reconciliation of minimum cost between u Gon N, then (α'(u), α'cost) is contained in u). This is obviously true for every leaf of G (by lines 5-7). Let now u be an internal node having two children u1, u2. Then, following Lemma 5, u) can be computed from u1) and u2) by using the information contained in dist. Lines 10-23 in Algorithm 2 computes u) following this lemma.v that is above y, z, denote by low(v) the lowest node such that every path from v to y, z must pass through this node. There is only one such node. Indeed, suppose that there are two such nodes m, m', then every path from v to y must pass through both m, m', i.e. either m <N m' or m' <N m, contradicting the lowest property of m, m'. To prove the claim, we need to show that if v is a node of BFS(y), then:It remains now to prove that Algorithm 3 correctly computes v is not strictly above y and z, then label(v) = ∅;(i) if label(v) = low(v) and line 12 of Algorithm 3 is performed if and only if v is actually in (ii) otherwise, v be a node of BFS(y), then (i) holds by line 4. Now consider the case where v is strictly above y and z. We will prove (ii) by recursion on v following the order of the nodes in BFS(y). The recursion begins from the set of lowest nodes that are strictly above y and z, i.e. the set of nodes of v in v. Let v1, v2 be two children of v, then by hypothesis v1, v2 are not strictly above y, z, i.e. label(v1) = label(v2) ≠ ∅; and low(v) = v. Thus, due to line 12, label(v) = v = low(v). Now let v be a node strictly above y, z such that (ii) is correct for each node below v which is strictly above or equal to y, z. If v is a hybridization node, then it is evident that low(v) = low(v1) where v1 is the only child of v. Moreover, since label(v1) = low(v1) (by the hypothesis of recurrence), then label(v) = label(v1) = low(v1) = low(v). If v is a speciation node having two children v1, v2 such that v2 is not in either BFS(y) or BFS(z), then we also have low(v) = low(v1). Hence, due to lines 5 - 6, we have label(v) = label(v1) = low(v1) = low(v). Now consider the last case, i.e. v is a speciation node having two children v1, v2 that are both in BFS(y) and BFS(z). If there exists a node q = low(v1) = low(v2), then low(v) = q because every from v to y, z must pass either through v1 or v2, i.e. always pass through q. Following line 10, we fix label(v) = label(v1) = q = low(v). Moreover, in this case v can not be in v to y, z passes through a node q below v. In the last case, we have v1, v2 are above y, z, then there exists a node q1 = label(v1), and a different node q2 = label(v2). We will prove that q such that every path from v to y, z must pass through q. Hence, every path from v1 (resp. v2) to y, z must also pass through q, so q1 N q ≤(resp. q2 N q≤). It means that there is a path from v1 to q to q2 and then to y, z that does not pass through q1, a contradiction. Hence v is in v to y, z must pass through is v itself (line 12).Let We now present some intermediate results that will be useful to prove the complexity of Algorithm 2.L be a subset of L(N ). If |L| = 1, then m of N such that m is above all leaves in L and there exist at least two separated paths from m to two distinct leaves of L.We extend the definition of u of G, N L(u) is defined as the set of leaves of N to which α maps the leaf set of uG, i.e. s(u) = s(x)}.Given a node Lemma 6 Let α be a most parsimonious reconciliation between G and N, then, for every node u of Proof: It is true for every leaf u of G. Let u now be an internal node having two children u1, u2.u1, u2 ∈ L(G), then following Remark 1, N L(u) consists of two distinct nodes rα(u1), rα(u2). By Lemma 5, rα(u) to rα(u1) and rα(u2). It means that If u1 ∉ L(G), then following Remark 1, |N L(u1)| ≥ 2, then there always exist two distinct leaves x, y of N L(u) such that x ∈ NL(u1), y ∈ NL(u2), i.e. x is below rα(u1) and y is below rα(u2). If rα(u) to rα(u1), rα(u2). By extending these two paths from rα(u1) to x, and from rα(u2) to y, we have to two separated paths from rα(u) to x, y. In other words, rα(u) ∈ N ℳ(N L(u)). If eα(u) = rα(u2) N αr ≤(u1), then rα(u) = rα(u1) following Lemma 5, and rα(u) is not a leaf of N following Remark 1. Let u' be the highest node such that u' ≤G u1 and rα(u'), i.e. from rα(u), to two distinct leaves of NL(u'), i.e. two distinct leaves of N L(u). We can prove the claim similarly for the case when rα(u1) N αr ≤(u2).Let Lemma 7 If N is a network that contains h hybridization nodes, then for every subset L of .The proof of this lemma is deferred to the appendix. We are now ready to state the complexity of Algorithm 2.Theorem 4 The time complexity of Algorithm 2 is O(h2·|G|·|N|) where h is the number of hybridization nodes of N.Proof: For every u ∈ V(G), N that u can be mapped to, which is bounded by O(h) following Lemma 7.for loop at lines 3 - 23 is performed |G| times, and, at each iteration, the for loop at lines 10-23 is performed O(h2) times. In each iteration of the second loop, the operation computeMin, as detailed in Algorithm 3, requires two breath-first-search traversals, which can be performed in time O(|N|). Moreover, for every node x of x to y, z, which can be extended to be two separated paths from x to two distinct leaves l1, l2 of N L(u) where l1 is a leaf below y of and l2 is a leaf below z. This is always possible because, by Remark 1, |N L(u)| > 1. Hence, x must be in O(h). The operation merge at lines 23 for two lists of size O(h) can be implemented in times O(h), if we know that the resulting list is also of size O(h). Hence, it takes O(|N| + h) = O(|N|) times for each iteration of the loop 10 - 23. Therefore, the total complexity is O(h2· |G|·|N|).The G and N can be then obtained by a standard back-tracking of the matrix x,c) of r(G)) such that c is the minimum value over all pairs in r(G)).Finally, a reconciliation of minimum cost between In this paper, we have studied two variants of the reconciliation problem between a gene tree and a species network. In particular, for the problem of finding the "most parsimonious" switching of the network, even though the number of switchings can be exponential with respect to the number of hybridization nodes, we proposed an algorithm that is exponential only with respect to the level of the network, which is often low for biological data. Moreover, the problem of finding a reconciliation between a gene tree and a network, which was solved in  for a mou ∈ I(G), there must exist H ∈ GB(u) such that u ∈ I(H). If H is an edge, then u is the only internal node of H, which must be an artificial node by Lemma 2. But this is not possible because nodes of G cannot be artificial. Hence, H must be a binary tree. Let denote i B= B(u), and i S= S(B(u)). We will prove now that u.By Definition 8, for every u be an internal node of G that has two children u1, u2 in G, and let H be the binary tree of u ∈ I(H). Denote j = 1, 2, if j uis a leaf, let j Hbe equal to j u, otherwise j His the binary tree of j u∈ I(jH). For the sake of convenience, if j uis a leaf, we also denote by j uto the only leaf x of N such that s(x) = s(j u). Note that s(x) = α(j u).Let j = 1, 2 (which is evidently true if j uis a leaf), and we will show that this implies that the claim is true for u.We now suppose that u in N Gsuch that Let H = H1 = H2. This implies that H will contain an artificial node. The same holds for u2. Thus, (i) If H = H2. As in point (i), this implies H that is mapped to ri Bare above all nodes of (ii) If ii)). Then, similarly as in point (ii), H mapped respectively to (iii) Suppose now that u ∈ I(G) by recursion.Therefore, in all cases we always have We will prove this lemma by recursion on the number of hybridization nodes. See the figure N is a tree, and it is evident that If there is no hybridization node, then h hybridization nodes. Let N now be a network that has h + 1 hybridization nodes. Let  be an edge of N having as target a hybridization node  such that it does not exist any hybridization node above a . Let N' be the network obtained by removing  from N , then N' has exactly h hybridization nodes. Let q in Q must be above a. Indeed, if q is not above a, then every path from q to every node of L does not contains , thus q is in a, hence all nodes of Q must be contained in a path leading a, and this path does not contain any hybridization to node. Let enumerate the nodes in Q as q1,... m qfrom the lowest to the highest one.Suppose that the claim is correct for every network having If Q| = m >1. In the following, we will define m - 1 edges of N having as target a hybridization node such that if N" is the network obtained from N' by removing these edges, then Suppose now that |* Lthe set of nodes in L that are below m qin N'. Hence * L \ Lis not empty since otherwise m qwould be in iq, i = 1,..., m, let i qsuch that a. Hence, a since there is no hybridization node above a, thus every path from i qto * L \ Lmust pass through a, b. By definition, there must exist at least two separated paths from i qto two leaves of L. Hence, for every i, there exists always a path from i qto a node of * Lthat passes through l(iq).Denote by m Lthe set of nodes in * Lsuch that, for every l ∈ mL, there is a path from q to l that passes through m Lis not empty. Recursively, for every i< m, let i Lbe the set of nodes in L* \ Lm \... Li+1 such that, for every l ∈ iL, there is a path from qi to l that passes i< m, and for every i ≠ j, i L∩ jL = ∅.Denote by i < m, an index c(i) that is strictly greater than i such that Lc(i) ≠ ∅, together with two paths i p(resp. i q(resp. qc(i)) to a node of Lc(i) as follows: if l(qi) is not in iL, then by the definition, there exists a unique j such that j Lcontains l(qi), and j > i. We fix c(i) = j. Next, we define i p(resp. i q(resp. qc(i)) to l(iq) that passes l(iq) is in iL, then let c(i) be the smallest number that is greater than i and Lc(i) ≠ ∅ . Let l' be a node of Lc(i). Since i qis in N M(L), then there must exist a path from i qto l', and we define i pas this path. The path qc(i) to l' that passes through ip, L. Denote by i hthe highest common hybridization node of i pand ihs are distinct since each i pstarts at a different node iq, and each qc(i) that is strictly greater than iq. We define  recursively in increasing order of i from 1 to m - 1 as follows. If i = 1, then i bis the highest hybridizatition node on ip. If i >1, then i bis the highest hybridization node on i pand different from all k bfor every k < i. There exists always such a node ib, for example ih. Therefore, all ibs are distinct. Denote by i athe parent of i bon the path ip.We will define, for each N" the network obtained from N' by removing all edges . For every node x in x is also in x', x" the two children of x. By definition, for every l ∈ L, there exists a path, denoted by f'(l), in N' from x to l such that at least one path among them passes through x' and one other passes through x". To prove that x is in N" ) from x to each leaf l of L, denoted by f"(l), such that at least one path among them passes through x'and one other passes through x".Denote by x is above m q contains mq, then l ∈ L* and f'(l) must pass through x' because there is no hybridization node above mq. Suppose that k be the index such that l ∈ kL, then we can choose a path f"(l) in N' from x to l that does not contain any  as follows. This path starts from x, passes through x', goes down to k q, then takes the path from k qto l that contains q". Note that this path does not include any i psince, by construction, every path i pstarts from i qand goes to a node in Lc(i) that is different from iL, while this path passes through k qand goes to a node in k L. Moreover, this path and i pcan not have any common hybridization node above i abecause i bis the highest hybridization node on ip. Hence, it can not pass through  for any i. If l is not in m qto l must pass through x, going down to l. It is evident that this path does not include any ip, and it cannot have with i pany common hybridization node above i abecause i bis the highest hybridization node on ip. Hence it does not pass through any . If f'(l) does not contain mq, then we fix f"(l) = f'(l). It is easy to see that f'(l) does not contain any edge  because otherwise f'(l) and i pmust have at least a common hybridization node above i a(since there is no hybridization node above iq). But this is not possible because i bis the highest hybridization node on ip. Remark that at least one of the paths f'(l) in this case must pass through x" since all paths in the first case must pass through x'. Hence at least two of the paths f"(l) are separated, thus x is in Consider first the case that x is not above mq, then similarly as in the previous case where f'(l) does not contain mq, we deduce that f'(l) does not contain any  for every l. Hence, by choosing f"(l) = f'(l) for every l, we are done.Consider now the case that Therefore, we have N" contains h - |Q| + 1 hybridization nodes, then following the hypothesis of recurrence, The network The authors declare that they have no competing interests.Both authors contributed to design the models, algorithms and to write the paper.


Denote ψt(x) = tn−ψ(x/t) with t > 0 and x ∈ Rn. Given a function f ∈ Lloc1(Rn), the Lusin area integral of f is defined bya(x) denote the usual cone of aperture onea = 1, we denote Sψ,a(f) as Sψ(f).Let g-functions and the Littlewood-Paley gμ*-functions besides the Lusin area integrals. The Littlewood-Paley g-functions, which can be viewed as a “zero-aperture” version of Sψ, and gμ*-functions, which can be viewed as an “infinite-aperture” version of Sψ, are, respectively, defined byψ to be the Poisson kernel, then the functions defined above are the classical Littlewood-Paley operators.Now let us turn to the introduction of the other two Littlewood-Paley operators. It is well known that the Littlewood-Paley operators include also the Littlewood-Paley b ∈ Lloc1(Rn),  m ≥ 1, the corresponding m-order commutators of Littlewood-Paley operators above generated by a function b are defined byμ > 0.Letting p  (Rn) in  is bounded on Lp(ω) in )(Rn) in . In 2005p(ω) in ). There p(ω) in –9 and soLp(·)(Rn) become one class of important research subject in analysis filed due to the fundamental paper ,  , and , we have the following results.For commutators  is defined by (b ∈ BMO(Rn), p(·) ∈ B(Rn), m ≥ 1,  1 < r < ∞. If, for any x ∈ Rn and r0 > 0, function u satisfiesC > 0 independent of f such that, for any function sequences {fh}h=1∞ with ||{∑h|fh|r}r1/||Mp(·),u(Rn) < ∞, the following inequality holds:Suppose that function fined by . Let b ∈bm, gψ] is defined by ,u(Rn) < ∞, the following inequality holds:Suppose that  is defined by (μ > 3 + 2(ɛ + γ)/n, and 0 < γ < ɛ. Then under the same condition as the one in C > 0 independent of f such that, for any function sequences {fh}h=1∞ with ||{∑h|fh|r}r1/||Mp(·),u(Rn) < ∞, the following inequality holds:Suppose that (fj)|q)q1/||Lp(·)(Rn) ≤ C||(∑j|fj|q)q1/||Lp(·)(Rn),|||q)q1/||Lp(·)(Rn) ≤ C||(∑j|fj|q)q1/||Lp(·)(Rn),|||q)q1/||Lp(·)(Rn) ≤ C||(∑j|fj|q)q1/||Lp(·)(Rn),||n/2, ɛ}.Let Next, let us show the proofs of Theorems fh}h||lr ∈ Mp(·),u(Rn); for any x0 ∈ Rn, r0 > 0, denotefh0 = fhχB,  fhj = fhχB∖B, j ∈ N∖{0}.Let ||{D1, notice that supp fh0 ⊂ B; using D2. To do this, we need to consider Sψ,a first. Without loss of generality, we may assume that a ≥ 1. Letx ∈ B,  z ∈ B∖B,  j ≥ 1, theny > 2j+1r0, then t > |x − y|/a ≥ (|x| − |y|)/a > (2j+1r0 − r0)/a ≥ 2jr0/a. Thus, by condition (iii), we getI, we obtainD1, D2, we obtainNoting that, in order to prove Now let us prove Theorems gψ, similar to the estimate of Sψ,a, via a simple calculation, we get that , supp fhj ⊂ B∖B,  j ≥ 1, thenFor hat (see ) if x ∈ gμ*, by the definitions of Sψ,a and gμ*, we haveSψ,a in the proof of x ∈ B, supp fhj ⊂ B∖B, j ≥ 1, thenμ > 3 + 2(ɛ + γ)/n, we obtainFor b ∈ BMO, ||{fh}h||lr ∈ Mp(·),u(Rn). For any x0 ∈ Rn, r0 > 0, denotefh0 = fhχB and fhj = fhχB∖B, j ∈ N∖{0}.Let E1, notice that supp fh0 ⊂ B; by E2. According to the estimate of Sψ,a in the proof of x ∈ B, z ∈ B∖B, j ≥ 1, thenTo finish the proof of E1, E2, we haveThus,bm, gψ], according to the estimate of gψ in the proof of x ∈ B, supp fhj ⊂ B∖B, j ≥ 1, thenFor , according to the estimate of  in the proof of x ∈ B, supp fhj ⊂ B∖B, j ≥ 1, thenFor [


After publication of  we becamIn the original publication, the images of Figures 1, 3, 5–9 are shuffled. In addition, images of Figures 1, 3, 4, 6–9 contain incorrectly encoded symbols. They should be replaced with Figures , 8, 9 prn of 11” in the main text should be removed.All four occurrences of “Yasuda and Miyano Page O(|E|2 log |V | log |E|)” should read “O(|E|2 log |V | log |E|)”All three occurrences of “E+)” should read “CC(E+)”In the Results, both of two “CC.” in subsection Circular chromosome graphThe line with the sentence “Therefore, In the Methods, a Q.E.D. symbol “☐(white box)” should be inserted at the end of the following lines:In the Background, “BCRABL” should read “BCR-ABL”Lemma 1, the expression “c = p1e1p2e2 . . . etcptc+1” should read “ In the Methods, in the proof of Lemma 2, the expression “2|V |(nN+nT )+(4|V |+1)Pe ∗ ∗ ∗ES n(e) ≤U(4|V|+1)(|E|+1)” should read “ In the same proof, just above Lemma 3, the following expressionsIn the Methods, in the proof of should readLemma 4, in the paragraph that begins with “All of these steps”, the expression “m = f(e) + f(ē)(∈ES).” should read “m = f(e) + f(ē) (e∈ES).” (Insert a white space before “(e∈ES)”.) In the Methods, in the proof of C satisfies (6).”, the expression “w)=0(e∈EL∪ER).” should read “w)=0 (e∈EL∪ER).” (Insert a white space before “(e∈EL∪ER)”.) In the same proof, just before “Therefore, Incorrectly formatted descriptions in  should bFormulation of the problem, the phrase “its computational complexity was not analyzed” should read “its computational complexity was not intensively analyzed”In the Results, in subsection Polynomial-time solvable variation, both of two “EL∪ER” should read “E”In the Results, in the first paragraph of subsection Definition 2, the phrase “if all g∈CC are good” should read “if all g∈CC are good and n(e) = 0 for e∈E − EW ”In the Results, in Definition 2, the expression “EW = {e∈ES |n(e) ≥ 1}∪{e∈EL∪ER |e is required}” should read “EW = {e∈E |e is required}”In the Results, in the paragraph just after Definition 3 should read as follows:In the Results, the last sentence that begins with “Finally, if some” just before g∈CC that are not good still remain, edges e in g are forcibly removed from EW by changing e not required and setting n(e) to 0. Finally, if n(e) > 0 for some e∈E− EW, e is changed to be required and added to EW by confirming its existence, or n(e) is forcibly set to 0.”“In addition, if some Definition 3 should read “Let G =  be a chromosome graph that satisfies WCC with respect to given VW ⊂ V and EW ⊂ E. Then, find a set C of chromosomes on G that minimizes W(C) when (3) is satisfied, each v ∈ VW is at an end of some c∈C, and each e∈EW appears in C.”In the Results, Lemma 4, the sentences “For e∈ES ∪ {eN, eT }, we set l(e) = n(e), l(ē) = 0, and u(ē) = n(e). For e∈EL∪ER, we set l(e) = 1.” should read “For e∈ES∪{eN, eT}, we set l(e) = n(e), l(ē) = 0, and u(ē) = n(e) if e is not required, whereas l(e) = max{n(e), 1}, l(ē) = 0, and u(ē) = max{n(e) − 1, 0} if e is required. We assume that eN is required because nN ≥1. We also assume that eT is required if |VW| ≥ 1. For e∈EL∪ER, we set l(e) = 1 if e is required, or l(e) = 0 otherwise.”In the Methods, in the paragraph just above Lemma 4, the description “or n(e) ≥ 1” should be removed.In the Methods, in the paragraph just after Proof of Theorem 3, the phrase “by making all edges in EL∪ER required” should read “by making all edges required”In the Methods, in subsection The following items correct inaccurate descriptions in the original manuscript. We regret any inconvenience that they might have caused.