In this paper, we essentially deal withKöthe-Toeplitz duals of fuzzy level sets defined using a partial metric. Since the utilization of Zadeh'sextension principle is quite difficult in practice, we prefer the idea of level sets in order to constructsome classical notions. In this paper, we present the sets of bounded, convergent, and null series and theset of sequences of bounded variation of fuzzy level sets, based on the partial metric. We examine therelationships between these sets and their classical forms and give some properties including definitions,propositions, and various kinds of partial metric spaces of fuzzy level sets. Furthermore, we study some oftheir properties like completeness and duality. Finally, we obtain the Köthe-Toeplitz duals of fuzzy levelsets with respect to the partial metric based on a partial ordering. We define the classical sets bs(H), cs(H), and cs0(H) consisting of the sets of all bounded, convergent, and null series, respectively; that isbs(H), cs(H), and cs0(H) are complete metric spaces with the partial metric Hs defined byu = (uk) and v = (vk) are the elements of the sets bs(H), cs(H), or cs0(H).By bv(H), bvq(H), and bv∞(H) consisting of sequences of q-bounded variation by using the partial metric Hs with respect to the partial ordering ⊑H, as follows:Hs denotes the partial metric of fuzzy level sets defined byu, v ∈ E1 with the partial ordering ⊑H. One can conclude that the sets bv(H), bvq(H), and bv∞(H) are complete metric spaces with the following partial metrics:u = (uk) and v = (vk) are the elements of the sets bv(H), bvq(H), or bv∞(H) and (Δu)k = uk − uk+1 for all k ∈ N.Secondly, we introduce the sets A some of which reduced to the Maddox's spaces l∞, c, c0, and l of sequences of fuzzy numbers for the special cases of that matrix A. Quite recently, Talo and Başar ⊑p if and only if  is a subset of .For the partial metric max{X, p) be a partial metric space and (xn) a sequence in . Then, we say the following:xn) converges to a point x ∈ X if and only if p = limn→∞ p.A sequence (xn) is a Cauchy sequence if there exists (and is finite) limm,n→∞ p.A sequence  is said to be complete if every Cauchy sequence (xn) in X converges, with respect to the topology τp, to a point x ∈ X such that p = limm,n→∞ p. It is easy to see that every closed subset of a complete partial metric space is complete.A partial metric space ) ⊂ Bp(f(x0), ɛ).A mapping xn) in a partial metric space  converges to a point x ∈ X, for any ϵ > 0 such that x ∈ Bp, there exists n0 ≥ 1 so that for any n ≥ n0, xn ∈ Bp.A sequence  be a partial metric space. Then, (i)xn) is a Cauchy sequence in  if and only if it is a Cauchy sequence in the metric space ,( (ii)X, p) is complete if and only if the metric space  is complete. Furthermore, limn→∞ ps = 0 if and only if p = limn→∞ p = limm,n→∞ p.a partial metric space , the limit of the sequence (−1/n) is 0 since one has limn→∞ ps, where ps is the usual metric induced by p on R−.In the partial metric space u is normal; that is, there exists an x0 ∈ R such that u(x0) = 1. (ii)u is fuzzy convex; that is, u[λx + (1 − λ)y] ≥ min{u(x), u(y)} for all x, y ∈ R and for all λ ∈ . (iii)u is upper semicontinuous. (iv)x ∈ R : u(x) > 0} in the usual topology of R.The set f. Zadeh ), where AR by E1 and call it as the space of fuzzy numbers. λ-level set [u]λ of u ∈ E1 is defined byu]λ is closed, bounded, and nonempty interval for each λ ∈  which is defined by [u]λ = . R can be embedded in E1, since each r ∈ R can be regarded as a fuzzy number We denote the set of all fuzzy numbers on  Let [u]λ =  for u ∈ E1 and for each λ ∈ . Then the following statements hold. (i)u− is a bounded and nondecreasing left continuous function on ]0,1]. (ii)u+ is a bounded and nonincreasing left continuous function on ]0,1]. (iii)The functions u− and u+ are right continuous at the point λ = 0. (iv)u−(1) ≤ u+(1).Representation Theorem 1 (see ). Let λ =  for each λ ∈ . The fuzzy number u corresponding to the pair of functions u− and u+ is defined by u : R → , u(x) = sup{λ : u−(λ) ≤ x ≤ u+(λ)}.λ-level set.Now we give the definitions of triangular fuzzy numbers with the μu) is interpreted, as follows:u]λ : =  =  holds for each λ ∈ .The membership function u, v, w ∈ E1 and α ∈ R. Then the operations addition, scalar multiplication, and product defined on E1 by u + v = w⇔[w]λ = [u]λ + [v]λ for all  λ ∈  then w−(λ) = u−(λ) + v−(λ)  and  w+(λ) = u+(λ) + v+(λ)  for all  λ ∈ .Let W be the set of all closed bounded intervals A of real numbers with endpoints d on W by d is a metric on W  is a complete metric space, (cf. Nanda .A sequence (The following statements hold.un) ∈ ω(F) is equivalent to the fact thatuk) ∈ ω(F) is bounded then the sequences of functions {uk−(λ)} and {uk+(λ)} are uniformly bounded in .The boundedness of the sequence (l∞(H), c(H), c0(H), and lp(H) consisting of the bounded, convergent, null, and p-summable sequences of fuzzy level sets with the partial metric Hs, as follows:l∞(H), c(H), and c0(H) are complete metric spaces with the partial metric H∞ defined byu = (uk) and v = (vk) are the elements of the sets c(H), c0(H), or l∞(H). Also, the space lp(H) is complete metric space with the partial metric Hp defined byu = (uk) and v = (vk) are the points of lp(H).Following Kadak and Ozluk , we giveμ(H) denote any of the spaces bs(H), cs(H), and cs0(H), and u = (uk), v = (vk) ∈ μ(H). Define the partial distance function H∞p on μ(H) byμ(H), H∞p) is a complete metric space.Let cs(H) and cs0(H), we prove the theorem only for the space bs(H). Let u = (uk), v = (vk), and w = (wk) ∈ bs(H). Then, (i)by using the axiom (P1) in  (ii)By using the axiom (P2) in  (iii)By using the axiom (P3) in  (iv)H ≤ H + H − H and Hs = 0, we haveBy using the axiom (P4) in Therefore, one can conclude that (bs(H), H∞p) is a partial metric space on bs(H). It remains to prove the completeness of the space bs(H).  Let (um) be any Cauchy sequence on bs(H), where um = {u1m) in Theorem 3.1 [ukm) = {uk(1), uk(2),…} is a Cauchy sequence and is uniformly convergent. Now, we suppose that limm→∞ ukm). We must show thatN ∈ N for all m > N, taking the limit for r → ∞ in (k ∈ N. Since (ukm) ∈ bs(H), there exists a number k ∈ N. Thus,  is bounded sequence of fuzzy numbers hence u ∈ bs(H). Also, from  = 0. Since (um) is an arbitrary Cauchy sequence, bs(H) is complete.Since the proof is similar for the spaces orem 3.1 , we say m, r > NHs∑k=0nuk → ∞ in , HqΔ, and H∞Δ byu = (uk), v = (vk) are the element of the spaces bv(H), bvq(H), or bv∞(H), respectively. Then, (bv(H), HΔ), (bvq(H), HqΔ), and (bv∞(H), H∞Δ) are complete metric spaces.Define the distance functions bv(H) and bv∞(H), we prove the theorem only for the space bvq(H). One can easily establish that HqΔ defines a metric on bvq(H). Let xi = {x0i). Then for every ϵ > 0, there exists a positive integer n0(ϵ) ∈ N for all i, j > n0, such thatx)n = xn − xn−1 and n ∈ N from , which leads us to the fact that the sequence {(Δx)ni} is a Cauchy sequence and is convergent. Now, we suppose that (Δx)ni → (Δx)n as n → ∞. We have from , thati > n0(ϵ). Let firstly j → ∞ and nextly m → ∞ in  ≤ ϵ. Finally, by using Minkowski's inequality for each m ∈ Nx ∈ bvq(H). Since HqΔ ≤ ϵ for all i > n0(ϵ), it follows that xi → x as i → ∞. Since (xi) is an arbitrary Cauchy sequence, the space bvq(H) is complete. This step concludes the proof.Since the proof is similar for the spaces uch thatHqΔxi,xj: N from , that∑k=0mHsΔxα-duals. An account of the duals of sequence spaces can be found in Köthe [The idea of dual sequence space, which plays an important role in the representation of linear functionals and the characterization of matrix transformations between sequence spaces, was introduced by Köthe and Toeplitz , whose min Köthe . One canin Köthe .λ(H), μ(H), and S(λ(H), μ(H)) of sequences defined byλ(H) and μ(H) for all k ∈ N. One can easily observe for a sequence set ν(H) of fuzzy level sets that the inclusionsλ(H)}α, {λ(H)}β, and {λ(H)}γ of a set λ(H) ⊂ ω(F) are, respectively, defined bywkzk) the coordinatewise product of the sequences w and z of level sets for all k ∈ N. Then {λ(H)}β is called β-dual of λ(H) or the set of all factor sequences of λ(H) are in cs(H). Firstly, we give a remark concerning with the convergence factor sequences of fuzzy level sets with partial metric.In this section, we focus on the alpha-, beta- and gamma-duals of the classical sets of sequences of fuzzy numbers with partial metric. For the sets ∅ ≠ λ(H) ⊂ ω(F). Then the following statements are valid.(a)λ(H)}β is a set of sequence and φ(F)<{λ(H)}β < ω(F) (“<” stands for “is a linear subset of ”) where{(b)λ(H) ⊂ μ(H) ⊂ ω(F) then {μ(H)}β < {λ(H)}β.If (c)λ(H)⊂{λ(H)}ββ : = ({λ(H)}β)β.(d)φ(F)}β = ω(F) and {ω(F)}β = φ(F).{Let m = (mk) and n = (nk)∈{λ(H)}β.(a)l ∈ λ(H). Then we get (mklk) ∈ cs(H); (nklk) ∈ cs(H) and (mk + nk)lk = (mklk)+(nklk) ∈ cs(H). Since l is arbitrary, m + n ∈ {λ(H)}β. For any α ∈ R and w = (wk)∈{λ(H)}β we haveαw ∈ {λ(H)}β. Therefore, {λ(H)}β is a linear subset of ω(F).Let (d)ω(F)}β ⊂ φ(F). Suppose that w = (wn)∈{ω(F)}β and z = (zn) be given with geometric division by zn : = (1/wn) if φ(F) from the case (a), then there exists an integer N ∈ N for all n ≥ N such that wnzn) ∈ cs(H) implies that w ∈ φ(F). The rest is an immediate consequence of this part; we omit the detail.Using (a) we need only show {Since the proof is trivial for conditions (b) and (c), we prove only (a) and (d). Let c0(H)}β = {c(H)}β = {l∞(H)}β = l1(H).{l1(H)}β = l∞(H).{The following statements hold. l∞(H)}β ⊂ {c(H)}β ⊂ {c0(H)}β by l1(H)⊂{l∞(H)}β and {c0(H)}β ⊂ l1(H). Now, consider w = (wk) ∈ l1(H) and z = (zk) ∈ l∞(H) are given. Thenwz ∈ cs(H). So the condition l1(H)⊂{l∞(H)}β holds.(a) Obviously {y = (yk) ∈ ω(F)∖l1(H) we prove the existence of an x ∈ c0(H) with yx ∉ cs(H). According to y ∉ l1(H) we may take an index sequence (np) which is a strictly increasing real valued sequence with n0 = 0 and x = (xk) ∈ c0(H) by xk : = ((sgnyk)/p), where the real signum function defined byu = (uk) ∈ E1, thus, we getnp−1 ≤ k < np. Therefore yx ∉ cs(H) and thus y ∉ {c0(H)}β. Hence {c0(H)}β ⊂ l1(H).Conversely, for a given l∞(H)⊂({l∞(H)}β)β = {l1(H)}β since {l∞(H)}β = l1(H). Now we assume the existence of a w = (wn)∈{l1(H)}β∖l∞(H). Since w is an unbounded sequence there exists a subsequence (wnk) of (wn) such that k ∈ N1. The sequence (xn) is defined by xn : = (sgn(wnk)/(k + 1)2) if n = nk and x ∈ l1(H). Howeverw ∉ {l1(H)}β, which contradicts our assumption and {l1(H)}β ⊂ l∞(H). This step completes the proof.(b) From the condition (c) of ζ-duals .Further to the statements in ∅ ≠ λ(H) ⊂ ω(F). Then the following statements are valid.(a)φ(F)<{λ(H)}α < {λ(H)}β < {λ(H)}γ < ω(F); in particular, {λ(H)}ζ is a set of sequence.(b)λ(H) < μ(H) < ω(F) then {μ(H)}ζ < {λ(H)}ζ.If (c)I is an index set, if λ(H)i are sets of sequences and if λ(H): = ⋃i∈Iλ(H)i, thenR.If (d)λ(H)<{λ(H)}ζζ : = ({λ(H)}ζ)ζ. If Let l∞(H) < cs(H) < bs(H). We only show conditions (c) and (d) taking ζ = α. Other parts can be obtained in a similar way.(c)λ(H)i ⊂ 〈λ(H)〉 that the following conditionsNow, as an immediate consequence  y ∈ ⋂i∈I{λ(H)i}α, that is y ∈ {λ(H)i}α, then xy ∈ l1(H) for all x ∈ λ(H) and therefore y ∈ {λ(H)}α ⊂ 〈λ(H)〉α.hold by (b). On the other hand, if (d)λ(H)<{λ(H)}αα. Let w ∈ λ(H); then, wz ∈ l1(H) for all z ∈ {λ(H)}α; thus, w ∈ {λ(H)}αα and λ(H)<{λ(H)}αα by (a).We prove Condition (b) is obviously true, and (a) follows from λ(H)≠{λ(H)}ζζ as we get from ζ = β and λ(H): = c0(H). We have {c0(H)}ββ = l∞(H) ≠ c0(H). This remark gives rise to the following definition.In general ζ ∈ {α, β, γ}, and let λ(H) be a set of sequence. λ(H) is called ζ-space if λ(H) = {λ(H)}ζζ. Further, an α-space is also called a Köthe space or perfect sequence space.Let From Remarks 4.3(d) and (b) we obtain immediately the following remark.λ(H) is a set of sequence over real field and , then {λ(H)}ζ is a ζ-space; that is, {λ(H)}ζ = {λ(H)}ζζζ.If λ(H)}α = {λ(H)}β = {λ(H)}γ. This gives rise to the notion of solidity.Now we look for sufficient conditions for {λ(H) be a set of sequence over the field R. Then λ(H) is called solid ifLet λ(H) < ω(F) is any set of sequence over the field R; then, the following statements hold.λ(H) is a Köthe space, then λ(H) is solid.If λ(H) is solid, then {λ(H)}α = {λ(H)}β = {λ(H)}γ.If λ(H) is a Köthe space, then λ(H) is a ζ-space.If Consider λ(H) < ω(H) be a set of sequence over the field R.(a)λ(H) is a Köthe space and u ∈ ω(F), then u ∈ {λ(H)}αα if and only if the condition uz ∈ l1(H) holds for all z ∈ {λ(H)}α. Besides this we obtain u = (uk) ∈ λ(H) and v = (vk) ∈ ω(F) and the statementz ∈ {λ(H)}α. Therefore vz ∈ l1(H). Hence v ∈ λ(H) and λ(H) is solid over the real field.If (b)λ(H) is solid. To show {λ(H)}α = {λ(H)}β = {λ(H)}γ, it suffices to verify {λ(H)}γ < {λ(H)}α as we have v = (vk)∈{λ(H)}γ; that is,λ(H), for z = (zk) ∈ λ(H), where zk = uksgn(ukvk) and the condition u = (uk) ∈ λ(H) for all k ∈ N. Therefore by combining this with the inclusion (uv ∈ l1(H). Hence v ∈ {λ(H)}α and {λ(H)}γ < {λ(H)}α.Consider that is,supn∈N∑k(c)This is an obvious consequence of Let φ(F), ω(F), lp(H), c0(H), and l∞(H) of sequences are solid.The sets c(H) and bv(H) of sequences are not solid; therefore, none of them is a Köthe space.The sets ζ ∈ {α, β, γ}, thenFor each l1(H)}ζ = l∞(H) and {l∞(H)}ζ = l1(H),{ω(F)}ζ = φ(F) and {φ(F)}ζ = ω(F).{ζ ∈ {α, β, γ} and c0(H) < μ(H) < l∞(H), then {μ(H)}ζ = l1(H) and μ(H)⊂{μ(H)}ζζ = l∞(H). In particular {c0(H)}ζ = {c(H)}ζ = l1(H), and each of c(H), c0(H) is not a ζ-space.If The following statements hold.c0(H) and l∞(H) are solid, we know that {c0(H)}ζ = {l∞(H)}ζ = l1(H). So the statements in (d) obtain from Given specified sets are solid in (a) and (b) is an immediate consequence of their definition. Additionally, the parts (i) and (ii) of (c) can be obtained ζ-duals of the spaces cs(H), bs(H), bv(H), and bv0(H). We will find that none of these sets is solid; in particular, none of them is a Köthe space.Next, we determine the cs(H)}α = {bv(H)}α = {bv0(H)}α = {bs(H)}α = l1(H).{cs(H)}β = bv(H), {bv(H)}β = cs(H), {bv0(H)}β = bs(H), {bs(H)}β = bv0(H).{cs(H)}γ = bv(H), {bv(H)}γ = bs(H), {bv0(H)}γ = bs(H), {bs(H)}γ = bv(H).{The following statements hold.cs(H), bs(H), bv(H), and bv0(H) of sequences are β-spaces, but they are not Köthe spaces. Moreover, the sets bs(H) and bv(H) of sequences are γ-spaces, whereas both cs(H) and bv0(H) are not γ-spaces. None of the spaces cs(H), bs(H), bv(H), and bv0(H) is solid.In particular the sets cs(H)}ζ, ζ ∈ {α, β, γ} and the proofs of all other cases are quite similar.(a)x = (xk) ∈ cs(H) and y = (yk) ∈ l1(H). Then,Let  y ∈ {cs(H)}α which gives that l1(H)⊂{cs(H)}α.Therefore,  y = (yk)∈{cs(H)}α∖l1(H). Then we can construct an index sequence (np) with np < np+1 and x = (xk) byConversely, suppose that  x = (xk) ∈ cs(H). According to the choice of np the inequalitiesThen  xy ∉ l1(H), which implies y ∉ {cs(H)}α. This contradicts that y ∈ {cs(H)}α. Therefore {cs(H)}α ⊂ l1(H).hold. Thus  xk) byAs well if we take the sequence (We prove the cases for the spaces { bv0(H)}α ⊂ l1(H) holds.the condition {(b)u = (uk)∈{cs(H)}β and w = (wk) ∈ c0(H). Define the sequence v = (vk) ∈ cs(H) by vk = (wk − wk+1) for all k ∈ N. Therefore, ∑kukvk converges, butLet  l1(H) ⊂ cs(H) yields that (uk)∈{cs(H)}β ⊂ {l1(H)}β = l∞(H). Then we derive by passing to the limit in (n → ∞ which implies thatand the inclusion limit in  as n → ∞ k ∈ N. Hence (uk − uk−1)∈{c0(H)}β = {c0(H)}α = l1(H); that is, u ∈ bv(H). Therefore, {cs(H)}β⊆bv(H).for every  u = (uk) ∈ bv(H). Then, (uk − uk−1) ∈ l1(H). Further, if v = (vk) ∈ cs(H), the sequence (wn) defined by wn = ∑k=0nvk for all k ∈ N, is an element of the space c(H). Since {c(H)}α = l1(H), the series ∑kwk(uk − uk+1) is convergent. Also, we haveConversely, suppose that  wn) ∈ c(H) and (uk) ∈ bv(H) ⊂ c(H), the right-hand side of inequality uk or ∑k=0∞ukvk converges and bv(H)⊆{cs(H)}β.Since bv(H)⊆{cs(H)}β and since {cs(H)}β ⊂ {cs(H)}γ, so bv(H)⊂{cs(H)}γ. We need to show that {cs(H)}γ ⊂ bv(H). Let u = (un)∈{cs(H)}γ and v = (vn) ∈ c0(H). Then, for the sequence (wn) ∈ cs(H) defined by wn = (vn − vn+1) for all n ∈ N, we can find a number K > 0 such that n ∈ N. Since (vn) ∈ c0(H) and (un)∈{cs(H)}γ ⊂ l∞(H), there exists a real number M > 0 such that n ∈ N. Therefore,By using (a), it is known that  uk − uk−1)∈{c0(H)}γ = {c0(H)}α = l1(H); that is, (un) ∈ bv(H). Therefore, since the inclusion {cs(H)}γ ⊂ bv(H) holds, we conclude that {cs(H)}γ = bv(H), as desired.Hence (Partial metrics are more flexible than metrics; they generate partial orders and their topological properties are more general than the one for metrics, argued by the fact that the self-distance of each point need not be zero. They are useful in partially defined information for the study of domains and semantics in computer science.The concept of level sets associated with a fuzzy set was originally introduced by Zadeh. With the aid of level sets we are able to provide a formulation for a fuzzy set in terms of crisp subsets via the representation theorem. The importance of having such a representation is that it can allow us to extend operations defined on crisp sets to the case of fuzzy sets. Our focus here is on using the idea of level sets to construct the sets of sequences of fuzzy numbers within partial metric spaces.This work presents the alpha-, beta-, and gamma-duals of the sets of bounded, convergent, and null series and the set of sequences of bounded variation of fuzzy level sets, based on the partial metric. The potential applications of the obtained results include the characterization of matrix transformations between these sets of sequences.


We shall explore a nonlinear discrete dynamical system that naturally occursin population systems to describe a transmission of a trait from parents to their offspring. We considera Mendelian inheritance for a single gene with three alleles and assume that to form a new generation,each gene has a possibility to mutate, that is, to change into a gene of the other kind. We investigatethe derived models and observe chaotic behaviors of such models. Recently, chaotic dynamical systems become very popular in science and engineering. Besides the original definition of the Li-Yorke chaos , there hIn this paper, we introduce and examine a family of nonlinear discrete dynamical systems that naturally occurs to describe a transmission of a trait from parents to their offspring. Here, we shall present some essential analytic and numerical results on dynamics of such models.In , it was A and a  represent a gene pool for a population; that is, x1, x2 are the percentage of the population which carries the alleles A and a, respectively. For the convenience, we express it as a linear combination of the alleles A and ax1,  x2 ≤ 1 and x1 + x2 = 1. The rules of the Mendelian inheritance indicate that the next generation has the following form:As the first example, we consider a Mendelian inheritance of a single gene with two alleles d a (see ). Let anPAA,A  is the probability that the child receives the allele A  from parents with the allele A; PAa,A  is the probability that the child receives the allele A  from parents with the alleles A and a, respectively; and Paa,A  is the probability that the child receives the allele A  from parents with allele a. It is evident thatHere, A and a, respectively, if the distribution of the current generation is known.Thus, the evolution  is a nonPAA,A = Paa,a = 1 and PAA,a = Paa,A = 0, the dynamical system ( = V(xn−1), is called a trajectory of V starting from an initial point x0. Recall that a point x is called a fixed point of V if V(x) = x. We denote a set of all fixed points by Fix(V). A dynamical system V is called regular if a trajectory {xn) were deeply studied in –16 and qV : X → X is said to be ergodic if the limitx ∈ X.A dynamical system V0 : S2 → S2 is not ergodic:Based on some numerical calculations, Ulam has conjectured  that anyIn , Zakharekth order Cesaro mean by the following formula:k ≥ 1 and Ces0n) = Vn(x). It is clear that the first order Cesaro mean Ces1n) is nothing but (1/n)∑i=0n−1Vi(x). Based on these notations, Zakharevich's result says that the first order Cesaro mean {Ces1n)}n=0∞ of the trajectory of the operator V0 given by (kn)}n=0∞, for any k ∈ N, of the trajectory of the operator V0 diverges for any initial point except fixed points. This leads to a conclusion that the operator V0 might have unpredictable behavior. In fact, in [V0 exhibits the Li-Yorke chaos. It is worth pointing out that some strange properties of Volterra QSO were studied in [We define the given by  divergesfact, in , it was udied in , 25. any non Volterra QSO acting on the finite dimensional simplex is ergodic, that is, operators having chaotic behavior can be only found among Volterra QSO. However, in this paper, we are aiming to present the continual family of nonergodic and chaotic QSO which are non Volterra QSO.In the literature, all examples of nonergodic QSO have been found in the class of Volterra QSO see , 20, 21), 2120, 2 a single gene with two alleles always exhibits an ergodic behavior . It is of independent interest to study the evolution of a mutation in population system having a single gene with three alleles. In the next section, we consider an inheritance of a single gene with three alleles a1, a2, and a3 and show that a nonlinear dynamical system corresponding to the mutation exhibits a nonergodic and Li-Yorke chaotic behavior.Note that if QSO is regular, then it is ergodic. However, the reverse implication is not always true. It is known that the dynamical system  is eitheIn this section, we shall derive a mathematical model of an inheritance of a single gene with three alleles.x represents a linear combination x = x1a1 + x2a2 + x3a3 of the alleles a1, a2, and a3 in which the following conditions are satisfied 0 ≤ x1, x2, x3 ≤ 1 and x1 + x2 + x3 = 1, that is, x1, x2, x3 are the percentages of the population which carry the alleles a1, a2, and a3 respectively.As it was in the previous section, an element  both parents have the same alleles. Specifically, we will consider two types of the simplest mutations; assume thata1 → a2, a2 → a3, and a3 → a1 occur with probability α;mutations a1 → a3, a3 → a2, and a2 → a1 occur with probability α.mutations We assume that prior to a formation of a new generation each gene has a possibility to mutate, that is, to change into a gene of the other kind. We assume that the mutation occurs ifIn this case, the corresponding dynamical systems are defined on the two-dimensional simplexLet us first recall the definition of the Li-Yorke chaos , 3, 4.X, d) be a metric space. A continuous map V : X → X is called Li-Yorke chaotic if there exists an uncountable subset S ⊂ X such that for every pair  ∈ S × S of distinct points, we have thatS is a scrambled set and  ∈ S × S is a Li-Yorke pair.Let Vα, Wα  systems  and (13) systems . As we aα = 1. In the first case, the operator V1 is a permutation of Zakharevich's operator V0 + αV1 and Wα = (1 − α)W0 + αW1.It is easy to check that Vα is a convex combination of two Li-Yorke chaotic operators V0, V1, meanwhile, in the second case, the evolution operator Wα is a convex combination of the Li-Yorke chaotic and regular operators W0, W1. These operators Vα, Wα were not studied in [Vα and Wα. The reason is that, in the first case, the convex combination presents a transition from one chaotic biological system to another chaotic biological system ; meanwhile, in the second case, the convex combination presents a transition from the ordered biological system to the chaotic biological system. In the next section, we are going to present some essential analytic and numerical results on dynamics of the operators Vα and Wα given by Vα is a Vα : S2 → S2:Vα(x) = x′ =  and 0 < α < 1. As we already mentioned, this operator can be written in the following form: Vα = (1 − α)V0 + αV1 for any 0 < α < 1, whereWe are aiming to present some analytic results on dynamics of LetVα : S2 → S2 be the evolution operator given by . Let Fix(Vα) and ω(x0) be sets of fixed points and omega limiting points of  Vα, respectively. Then the following statements hold true.P and Vα are commutative, that is, P∘Vα = Vα∘P.Operators x ∈ Fix(Vα) then Px ∈ Fix(Vα).If Vα) is a finite set then |Fix(Vα)| ≡ 1(mod3).If Fix(P(ω(x0)) = ω(Px0), for any x0 ∈ S2.One has that Let given by , where αVα), where α ∈ . It is worth mentioning that Fix(V0) = {e1, e2, e3, C} and Fix(V1) = {C}, where e1 = , e2 = , and e3 =  are vertices of the simplex S2 and C =  is a center of the simplex S2.We are aiming to study the fixed point set Fix(x0 ∈ Fix(Vα) is nondegenerate [x0:Recall that a fixed point generate  if and oVα : S2 → S2 be the evolution operator given by . Let C =  be a center of the simplex S2. Then the following statements hold true. All fixed points are nondegenerate.Fix(Vα) = {C} for any α ∈ .One has that Let given by , where αx ∈ Fix(Vα) be a fixed point. One can easily check that(i) Let α < 1, then the above expression is positive. Therefore, all fixed points are nondegenerate.If 1/2 ≤ α < 1/2. In this case, the above expression is equal to zero if and only if x1x2 + x1x3 + x2x3 = (1 − 2α)/(4(1 − α + α2)). Since x1 + x2 + x3 = 1, we have that x12 + x22 + x32 = (1 + 2α2)/(2(1 − α + α2)).Let 0 < x1 ≥ max{x2, x3} (See x2 ≥ x3. Since x ∈ Fix(Vα), we have that x2 = (1 − α)x22 + 2x2x3 + αx32. We then obtain thatx3 ≥ x2. This shows that, in the case 0 < α < 1/2, all fixed points are nondegenerate.Without loss of generality, we may assume that Vα) = {C}. The simple calculation shows that C ∈ Fix(Vα). It is clear that Vα(∂S2) ⊂ intS2. This means that the operator Vα does not have any fixed point on the boundary ∂S2 of the simplex S2, that is, Fix(Vα)∩∂S2 = ∅. Moreover, all fixed points are nondegenerate. Due to Theorem 8.1.4 in [Vα)| should be odd. On the other hand, due to Corollary 8.1.7 in [Vα)| ≤ 4. In Vα)| = 1. Therefore, we get that Fix(Vα) = {C}.(ii) We shall show that Fix is as follows.A local behavior of the fixed point Vα : S2 → S2 be the evolution operator given by . Then the following statements hold true.α ≠ 1/2, then the fixed point C =  is repelling.If α = 1/2, then the fixed point C =  is nonhyperbolic.If Let given by , where αx1 + x2 + x3 = 1, the spectrum of the Jacobian matrix of the operator Vα : S2 → S2 at the fixed point C =  must be calculated as follows:α ≠ 1/2, then the fixed point C =  is repelling and if α = 1/2, then the fixed point C =  is nonhyperbolic. This completes the proof.It is worth mentioning that, since α ≠ 1/2 and α = 1/2.We shall separately study two cases Vα : S2 → S2 be the evolution operator given by (α ≠ 1/2. Then ωVα(x0) ⊂ intS2 is an infinite compact subset, for any x0 ≠ C.Let given by , where αα ≠ 1/2. Since Vα is continuous and Vα(S2) ⊂ intS2, an omega limiting set ω(x0) is a nonempty compact set and ω(x0) ⊂ intS2, for any x0 ≠ C. We want to show that ω(x0) is infinite, for any x0 ≠ C. Since C is repelling, we have that C ∉ ω(x0). Let us pick up any point x* ∈ ω(x0) from the set ω(x0). Since the operator Vα does not have any periodic point, the trajectory {Vαn)((x*)}n=1∞ of the point x* is infinite. Since Vα is continuous, we have that {Vαn)((x*)}n=1∞ ⊂ ω(x0). This shows that ω(x0) is infinite for any x0 ≠ C.Let ωV0(x0) and ωV1(x0) of the operators V0 and V1 are infinite. However, unlike the operator Vα, we have inclusions ωV0(x0)⊂∂S2 and ωV1(x0)⊂∂S2. Moreover, both operators V0 and V1 are nonergodic [It is worth mentioning that the sets of omega limiting points nergodic , 27.Vα : S2 → S2 given by  is nonhyperbolic and the spectrum of the Jacobian matrix of the operator V1/2 at the fixed point C, calculated by  = |x1 − x2||x1 − x3||x2 − x3| is a Lyapunov function.C = .Every trajectory converges to the fixed point Let given by . The folV1/2 be an operator given by (ϕ(V1/2(x)) ≤ ϕ(x), for any x ∈ S2. This means that ϕ is decreasing a long the trajectory of V1/2. Consequently, ϕ is a Lyapunov function.(i) Let given by . It follgiven by  that(25ϕ(V1/2n)((x))}n=1∞ is a decreasing bounded sequence. Therefore, the limit limn→∞ϕ(V1/2n)((x)) = λ exists. We want to show that λ = 0. Suppose that λ ≠ 0. It means that λ ≠ 0, we get that(ii) We know that {ɛ0 such that for any n one has thatOn the other hand, since λ = 0.This is a contradiction. It shows that ω(x0) ⊂ l1 ∪ l2 ∪ l3. We want to show that ω(x0) = l1∩l2∩l3.Therefore, We know that ows from  that(29Vα : S2 → S2 given by (We are going to present some pictures of attractors (an omega limiting set) of the operator given by .α = 0 and α = 1, the corresponding operators V0, V1 have similar spiral behaviors which reel along the boundary of the simplex [ωV0(x0)⊂∂S2 and ωV1(x0)⊂∂S2.In the cases  simplex , 20. HowVα while α approaches to 1/2 from both left and right sides. In order to see some antisymmetry, we shall provide attractors of Vα and Vα1− at the same time.We are interested in the dynamics of the evolution operator α is an enough small number, then we can see that the omega limiting sets of operators Vα and Vα1− are separated from the boundary ∂S2 W0 + αW1, for any 0 < α < 1, whereW0 = V0 is Zakharevich's operator  which is attracting.The operator e1, e2, e3 are 3-periodic points.The vertexes of the simplex ϕ(x) = x12 + x22 + x32 − 1/3 is a Lyupanov function.W1 is regular in the set intS2.The operator Let Vα, we can prove the following results.By means of the same methods and techniques which are used for the operator Wα : S2 → S2 be the evolution operator given by ; that is, Fix(Wα) = {C}. Moreover, one has thatif if if Let given by . Then itWα : S2 → S2 be the evolution operator given by (ω(x0) ⊂ intS2 is an infinite compact set, for any x0 ≠ C.If ω(x0) = {C}, for any x0 ∈ S2.If Let given by . Then thWα : S2 → S2 given by  of the operator given by .α = 0 and α = 1, the operator W0 is chaotic and the operator W1 is regular. Since Wα = (1 − α)W0 + αW1, the evolution operator Wα gives the transition from the regular behavior to the chaotic behavior. Consequently, we are aiming to find the bifurcation point in which we can see the transition from the regular behavior to the chaotic behavior.In the cases α is a very small number then attractors of the operator Wα are separated from the boundary of the simplex  in the system can be considered as a transition between two different types of systems having Mendelian inheritances. Namely, the first mutation presents the transition between two chaotic biological systems; meanwhile the second mutation presents the transition between regular and chaotic systems.Vα : S2 → S2 given by ⊂∂S2 and ωV1(x0)⊂∂S2. If α is an enough small number then we observed that the omega limiting sets of operators Vα and Vα1− are separated from the boundary ∂S2  chaotic systems, at some point of the time, the system should become stable.In the first mutation, we have presented some pictures of attractors of the operator given by . In the ∂S2 see . If α beWα : S2 → S2 given by  and regular transformations, it is natural to expect the bifurcation scenarios in this evolution. Namely, in order to have a transition from regular to chaotic behavior we have to cross from the bifurcation point. Numerical result α⋍0.13397 also confirms the theoretical result about the exact value of bifurcation point. However, the biological plausiblity of this value is unknown for the authors.In the second mutation, we have presented some pictures of attractors of the operator given by . In the plex see . HoweverIn this paper, we have considered two types of mutations of three alleles which occurred with the same probability. It is natural to consider mutations with different probabilities among alleles. In this case, it is expected to have more complicated dynamics in the biological system. The future research is to study the dynamics of the mutated biological system having a single gene with a finite number of alleles.


This paper presents new structural and algorithmic results around the scaffolding problem, which occurs prominently in next generation sequencing. The problem can be formalized as an optimization problem on a special graph, the "scaffold graph". We prove that the problem is polynomial if this graph is a tree by providing a dynamic programming algorithm for this case. This algorithm serves as a basis to deduce an exact algorithm for general graphs using a tree decomposition of the input. We explore other structural parameters, proving a linear-size problem kernel with respect to the size of a feedback-edge set on a restricted version of Scaffolding. Finally, we examine some parameters of scaffold graphs, which are based on real-world genomes, revealing that the feedback edge set is significantly smaller than the input size. During the last decade, a huge amount of new genomes have been sequenced, leading to an abundance of available DNA resources. Nevertheless, most of recent genome projects stay unfinished, in the sense that databases contain much more incompletely assembled genomes than whole stable reference genomes . One reamulation , nearly mulation , howeverscaffold graph, representing the link between already assembled contigs. The main idea is to represent each contig by two vertices linked by an edge (these "contig-edges" form a perfect matching on the scaffold graph). Other edges are constructed and weighted using complementary information on the contigs. The weight of a non-contig edge uv, with uu′, vv′ being contig-edges, corresponds to a confidence measure that the uu′ contig is succeeded by the vv′ contig (oriented as u′ − u − v − v′). The scaffold graph is a flexible tool to study the scaffolding issues. Indeed, the graph is a syntactical data-structure which may represent several semantics. For instance, the scaffold graphs used for our previous experiments have been built using Next-Generation Sequencing data, namely paired-end reads. However, we also could provide other type of information to compute the weight on the edges, like for instance ancestral support in a phylogenetic context, or comparison to other extent genomes which could be used as multiple references. The way to define the weight on the edges does not change the main goal of our method, which is to determine the optimal ordering and orientation of the contigs, given a specific criterion. It is also possible to mix two or more criteria in order to take several sources of information into account.The approach presented here relies on a combinatorial problem on a dedicated graph, called σp and σc representing the respective numbers of linear and circular chromosomes sought in the genome. Those parameters seems quite artificial at a first sight, but they are well-motivated as follows. First, in a number of species, genomes are hard to assemble with classical methods because of an heterogeneous chromosomal structure or difficulties to perform classical assembly. This is the case, for instance, with the micro-chromosomes in the chicken genome  denote the subgraph of G induced by V′ and let G − V′ := G[V \ V′]. For S ⊆ E, let G − S :=  and, for any edge or vertex x, we abbreviate G − {x} =: G − x. For a set of pairs S , we let Gr(S ) denote the graph having S as edgeset, that is, Gr(S ) := . For a function ω : E → ℕ and a set S ⊆ E, we abbreviate Σe∈S ω(e) =: ω(S ). Let G =  be a graph with a matching M and let S be a matching in G − M. The number of paths (resp. cycles) in G[S ∪ M] is denoted by G[S ∪ M] are alternating with respect to M, then, we call S a ||S||p-||S||c-cover (or simply cover) for G with respect to M (we will omit M if its clear from context). If ||S||c = 0, we may also refer to S as a ||S||p-path cover (or simply path cover). The general scaffold problem is expressed as follows see also ,17): SCAInput: A scaffold graph , M), σp ∈ ℕ, σc ∈ ℕ, k ∈ ℕQuestion: Is there a σp-σc-cover S for G with respect to M with ω(S ) ≥ k?Tree Decompositions. A tree decomposition of a graph G =  is a pair , X : T V→ 2V ) such that (1) for all uv ∈ E, there is some i ∈ T Vwith uv ⊆ X(i) and (2) for all v ∈ V, the subtree v T:= T[{X(i) | v ∈ X(i)}] is connected. We call the images of X "bags" and the size of the largest bag minus one is the width of the decomposition. A decomposition of minimum width for G is called the optimal for G and its width is called the treewidth of G. It is a folklore theorem that each graph G has an optimal tree decomposition  that is nice, that is, each bag X(i) is one of the following types:Leaf bag: i is a leaf of T and X(i) = ∅,Introduce vertex v bag: i is internal with child j and X(i) = X( j) ∪ {v} with v ∉ X(j),Forget v bag: i is internal with child j and X(i) = X(j) − v with v ∈ X(i),Introduce edge uv bag: i is internal with child j and uv ∈ E and uv ⊆ X(i) = X(j),Join bag: i is internal with children j and ℓ and X(i) = X(j) = X(ℓ).e ∈ E is introduced exactly once. Given a width-tw tree decomposition, we can obtain a nice tree decomposition of width tw in n · poly(tw) time  and u T+v := T [V(uT)∪{v}]. Let v ∈ V and 1 ≤ j ≤ |C(v)|. Then, we define vs := vs[1..0] := ⊥.In the following, for a vertex Algorithm 1: Algorithm to fill the dynamic programming table.1I ← leaves of T;2 while ∃v ∈ V \ I s.t. C(v) ⊆ I do3     foreach 1 ≤ j ≤ |C(v)| with u := vs[j] do4         foreach i ≤ σp do5             if uv ∈ M then6                 v ← maxα∈{0,1} maxi−(1−α)ℓ≤u + v;7                 v ← maxα∈{0,1} maxiℓ≤u + v;8             else9                 v ← maxi ℓ≤maxα∈{0,1}u + v;10               11 if v = r then return maxc∈{0,1}relse I ← I ∪ {v};Semantics: For  ∈ {0, 1} × [σp] × [|C(v)|] × V we let v denote the maximum weight of an i-0-cover S for v := v.I of initialized vertices and, as soon as r is initialized, the algorithm stops. Thus, we assume r ∉ I. Finally, the maximum weight of a σp-0-cover for T can be computed by maxc∈{0,1}r.We maintain a set Lemma 1 Algorithm 1 is correct, that is, for all  ∈ [σp] × [|C(v)|] × {0, 1} × V and for any maximum-weight i-0-cover S for v = ω(S).σp times per vertex. As they compute the maximum over σp values, the whole algorithm runs in Concerning the running time, we note that the body of the loop in line 3 is executed exactly once per vertex and, hence, lines 6,7,9, and 10 are executed Corollary 1 SCAFFOLDINGon trees can be solved intime.SCAFFOLDING in O(twtw) poly time. It is based on using a dynamic programming table to keep track of solutions that interact with the bags of a tree decomposition in a certain way  :=Gr(P2). For permutations P and Q, we define P □ Q as the set of pairs uv such that P(x) = ⊥ ⊕ Q(x) = ⊥ for all x ∈ uv and there is a u-v-path in Gr(P ∪ Q). Furthermore, for a function d : A → B and  ∈ A × B, we define d[x → y] as the result of setting d(x) := y (that is (d \ ({x} × B)) ∪ {}). Here, d[x → ⊥] means to remove x from the domain of d. Let T =  be a tree decomposition of G with root X(r) ∈ χ. For a bag X(i), let iGdenote the subgraph of G that contains exactly those edges of G that are introduced in a bag of the subtree of T that is rooted at X(i) and let To present our algorithm, we use special permutations (involutions) to model matchings that allow reflexive pairing . Thus, slightly abusing notation, we will consider permutations as sets of pairs. We denote the subset of reflexive pairs of a permutation X(i) will be indexed by (i) a function d : X(i) → {0, 1, 2}, (ii) a permutation P with Uuv∈P uv = d−1(1) and(iii) integers p ≤ σp and c ≤ σc. See Figure d and P. An entry will have the following semantics:A table entry for the bag Definition 2 Let i ∈ T. We call a set Si V⊆ E(iG) \ M eligible with respect to a tuple  if1 each vertex v ∈ X(i) has degree d(v) in 2 for each uv ∈ P, if u ≠ v, then there is a u-v-path in and, if u = v, then there is a path q of non-zero-length in −1(1) − u (we say that q is dangling from u).3 contains p paths and c cycles that do not intersect d−1(1),Semantics: A table entry i is the maximum weight of any set that is eligible with respect to .S for G from r.Then, we can read the maximum weight of a solution X(i) with children X(j) and X(ℓ) (possibly j = ℓ), we compute i depending on the type of the bag X(i) (entries that are not mentioned explicitly are set to −∞):Given a nice tree decomposition and a bag Leaf bag: Set i := 0.Introduce vertex v: Newly introduced vertices do not have introduced edges yet. Thus, we force the degree of v in d, P, p, c]i := [d[v → ⊥], P, p, c] jif d(v) = 0 and ∞, otherwise.Forget vertex v: A vertex v that we forget in bag X(i) can have degree 0,1, or 2 in q dangling from it. If q ends in some other vertex u ∈ X(i) \ {v} = X( j), then, the permutation for X(i) contains uu and the permutation for X(j) contains uv. Otherwise, q is dangling from v in X(j) contains vv. Formally, letIntroduce edge uv: Let d(u), d(v) ≥ 1 and, by symmetry, let d(u) ≥ d(v) ≥ 1. We define a value z (representing the assumption that uv is in S ) as follows. Let d′ := d, that is, we let d′ be the result of decreasing both d(u) and d(v) by one.Case 1 d(u) = d(v) = 2. Then, P avoids u and v. Since we assume uv ∈ S , this means that u and v have dangling paths u qand v qin d′−1(1). If u q= vq, then adding uv to S closes a cycle in X(j) contains uv ). Otherwise, uq≠ vq. Then, if u qintersects d′−1(1) \ {u, v} in a vertex x, then the permutation for X(j) contains ux  and Figure 3(d)), otherwise, it contains uu ). Likewise, if v qintersects d′−1(1) \ {u, v} in a vertex y, then the permutation for X(j) contains vy, otherwise, it contains vv. Note that, if both u qand v qintersect d′−1(1) \ {u, v} ), then we have xy ∈ P. Note further that, if neither u qnor v qintersects d′−1(1) \ {u, v} ), then u q∪ v q∪ {uv} is a path in d−1(1) and, thus, we have to decrease the number p of such paths we are looking for in Case 2: d(u) = d(v) = 1. Then, both u and v are not incident to any edges in uv incident to both. Thus, we set z only if uv ∈ P. Formally, let z := j if uv ∈ P.Case 3: d(u) = 2, d(v) = 1. Then, there is a path q containing uv and ending in v in q ends in a vertex x in d−1(1) − v, we have vx ∈ P and the permutation for X(j) contains ux. Otherwise, we have vv ∈ P and the permutation for X(j) contains uu. Note that, since v ∈ d−1(1), we know that P(v) ≠ ⊥. Formally, for vx ∈ P, let z := j, if v = x and z := j, otherwise. Finally, let i := z if uv ∈ M and i := max{z + ω(uv), j}, otherwise.Join: The join bag X(i) "glues" the (disjoint) partial solutions of its children together at the vertices of X(i) = X( j) = X(ℓ). In particular, the degrees in d. Furthermore, the permutations P1 and P2 for X(j) and X(ℓ), respectively, have to "fit" P: For example, let uv ∈ P1 and vw ∈ P2, implying that there are paths j qand ℓ qin u and v and v and w, respectively. Then, in j q∪ qℓ connecting u and w and containing v (with d(v) = 2). Finally, the numbers of paths and cycles have to "fit" p and c: For example, if the permutations for both X(j) and X(ℓ) contain uu 1), then u that is neither in Lemma 2 The described algorithm is correct, that is, the computed value i corresponds to the semantics.Theorem 1 SCAFFOLDING can be solved in O(twtw ·σp · σc · n) time, given a width-tw tree decomposition of the input instance.SCAFFOLDING, we consider a more restricted problem variant, where all paths and cycles of the solution have to be of certain, respective lengths. SCAFFOLDING. Let G∗ = G[V∗] denote the convex hull of ○G, that is, V∗ is the set of vertices on some shortest path between some vertices u, v ∈ ○V. For each v ∈ V∗, the tree rooted at v that is incident with v in G − E∗ is called the "pendant tree" vT of v.For a p. Since the correctness of this is trivial, we omit the proof.First, we remove isolated paths by cutting them into pieces of length ℓTree Rule 1  by ). Otherwise, return "NO".p that is furthest from the root of the pendant tree.The next rule cuts branching edges in pendant trees. To this end, it finds an occurrence of a path of length ℓTree Rule 2 v. Then, delete from G all edges that are incident with the least common ancestor of u and w but are not on the unique u-w-path in T.Lemma 3 Let I =  be a yes-instance of RESTRICTED SCAFFOLDING such that G is reduced with respect to Tree Rule 2 and let v ∈ V∗v is a path p that is alternating with respect to M . Then, T∩ E(vT) and |p| < ℓp and v is an endpoint of p.G is reduced with respect to Tree Rule 2 and, thus, we can reject all instances for which Lemma 3 does not hold. Hence, in the following, we assume that Lemma 3 holds for the input instance. The next reduction rule helps unify the way in which pendant trees (which are now paths) attach to G∗, simplifying the rest of the presentation.In the following, we assume that Tree Rule 3  \ M. Then, delete from G all edges incident with v that are not in M + e.G is reduced with respect to the tree reduction rules presented in the previous section. For i ∈ ℕ, let i Vdenote the set of vertices of G that have degree i in G. The goal in this subsection is to reduce the length of chains of degree-two vertices in G. Thus, we consider paths whose inner vertices are all in V2. We call these paths deg-2 path. If the solution has a cycle running through this path, then we cannot modify its length. Therefore, we focus on paths that are guaranteed to not be in a cycle in the solution:In the following, we assume that p in G is split into by the solution are recurring, that is, if the solution contains the third edge of p, then it also contains the 3 + (ℓp + 1)th edge of p. This gets slightly more complicated if there are pendants in p, but first, let us consider paths without any pendants.The path reduction rules are based on the idea that the path segments that a deg-2 path Path Rule 1  be a deg-2 path with |p| > max{ℓp, ℓc} + ℓp + 1 i and let e:= iui+1 for all i. Then, for all ei with i u≤ ℓp, add ω(ie) to ω(ei+1ℓp +) i. Finally, decrease σand contract ep by 1.The remaining two rules deal with deg-2 paths containing pendants. Note that any path in the solution that contains the pendant can only "continue" in two directions.Path Rule 2  and q =  w contain be deg-2 paths and let Tγ > 0 edges. For all i ≤ ℓpi , let e:= iui+1, let fi u:= ivi+1 and let iv+ := (i − γ) mod (ℓp + 1). Then, for all i ≤ ℓp, add ω(ie) to ω( i+f) i. Finally, decrease σand contract ep by 1.In case of two pendants, the solution is restricted by the distance between the pendants. If we can infer what the solution does by this length, we implement this right away, otherwise, we can represent the choices that a solution can take with a single pendant.Path Rule 3  be a u-v-deg-2 path such that u, v ∈ V∗. Let γu > 0 and γv > 0 u and Tv, respectively, and let w be the vertex at maximum distance to u in Tu. Let be the number of edges in Tγ := |p| mod (ℓp + 1) and let G′ be the result of replacing ux1 by wx1 of the same weight in G.1) If γ + γu≠ f ℓp + 1 = γ + γv, then delete ux1  If γ + γu = ℓp + 1 ≠ f γ + γv, then delete vxp|−1|.(3) If γ + γu = ℓp + 1 = γ + γv, then return G′  If γ = 1 and γu + γv + 1 = ℓp, then return G′  If γ = 1 and γu + γv + 1 ≠ ℓp, then delete ux1  If γ ≠ 1 and γ + γu + γv ≡ ℓp mod (ℓp + 1), then delete all edges e ∈ E(G) \ (M ∪ {ux1}).(7) In all other cases, return "NO".(p · FES(G) or 11ℓc · FES(G). To prove this, let G† =  be the result of contracting all degree-2 vertices in G∗.Finally, we can show that an input graph that is reduced with respect to these rules cannot be larger than 11ℓLemma 4 Let G be reduced with respect to all presented reduction rules. Then, |V| ≤ ℓ · (|V†| + 3|E†|) with ℓ := max{ℓc, ℓp}.Proof By Lemma 3, we know that ∀v∈V†|E(vT)| < ℓp. Therefore, if Lemma 4 is false, there is an edge uv ∈ E† such that there are > 3ℓ vertices between u and v (i.e. uv is a contraction of more than 3ℓ edges of G). Nevertheless, by irreducibility with respect to Path Rule 3, there is at most one vertex w between u and v such that wTis not empty and the distance between u and v cannot be greater than 2ℓ + 1 (by irreducibility with respect to Path Rule 1 and 2). So, |E(wT)| ≥ ℓ, contradicting Lemma 3.Theorem 2 RESTRICTED SCAFFOLDING admits a kernel containing at most 11ℓ · FES(G) vertices and (11ℓ + 1) · FES(G) edges where ℓ := max{ℓp, ℓc}.nucleotide database (http://www.ncbi.nlm.nih.gov). Then, for each of them, a set of simulated paired-end reads was generated with the tool wgsim  is the size of a smallest set of edges whose deletion leaves an acyclic graph. The degeneracy is the smallest value d for which every subgraph has a vertex of degree at most d. It is a kind of measure of sparsity of the graph. We notice that scaffold graphs look quite sparse, with few vertices of high degrees and a feedback edge set number that is usually significantly lower than the number of vertices. While degree-based graph parameters like the degeneracy d are tiny in all instances, we recall that our problem generalizes Hamiltonian Cycle, which is already N P-hard on 3-regular graphs. However, Table Table Staphylococcus Aureus genome is no longer available in the NCBI Nucleotide database  and the Escherichia Coli genome has been updated since the presented version. Errors in assemblies are quite frequent  =  for the ebola and monarch genomes and  =  for the more complicated inputs. The tests were run on an AMD Opteron(tm) Processor 6376 at 2300 MHz.We implemented the dynamic programming algorithm presented in Section 11 in C++ using boost and the treewidth optimization library TOL . We ran Figure Concerning the rice chloroplast, among the 84 contigs, only three were misplaced. All three of them are small (< 130 bp), two of them strongly overlap and the third has two occurrences in the reference genome, one complete and one partial. The seven remaining scaffolds follow exactly the right relative order and orientation of the contigs on the reference genome. Chloroplast genomes have a particularity which make them interesting as data for scaffolding. They present an inverted repeat region of approximately 20 kbp . Figure SCAFFOLDING problem which is an integral part of genome sequencing. We showed that it can be solved in polynomial time on trees or graphs that are close to being trees (constant treewidth) by a dynamic programming algorithm. We proved a linear-size problem kernel for the parameter "feedback edge set" for a restricted version in which the lengths of paths and cycles are fixed. We implemented an exact algorithm solving SCAFFOLDING in f(tw) · poly(n) time and evaluated our implementation experimentally, supporting the claim that this method produces high-quality scaffolds. Our experiments are run on data sets that are based on specific real-world genomes, which we also examined to identify a number of interesting parameters that may help design parameterized algorithms and random scaffold graph generators that produce more realistic instances. We are currently transferring the preprocessing rules to the general problem variant. We are highly interested in further graph classes that are closer to real-world instances than trees and on which the problem might be polynomial-time solvable. From an algorithmic point of view, we remark that only few bags of the used tree decompositions are small  and, in case of ties, j (ascending).Induction Base: The statement holds for all vertices v if j = 0 since Induction Step (≥): First, we show that v ≥ ω(S). To this end, let u := vs[j] and, noting that all vertices except maybe v of M, let q denote the path in G[S ∪ M] containing u. Let w be the vertex paired with v by M. Let α denote the number of edges in q ∩ E(u T+ v) \ M incident with u and let β ≤ c denote the number of edges in vs[1..(j − 1)]. Let u S:= Up∈S p ∩ E(uT) and let u S and j−1 S are path covers of u Tand Case 1: uv ∈ M and c = 0. Then, Case 2: uv ∈ M and c = 1. Then, β = 1. Furthermore, the path containing u is split over u S + uv and j-1S , implying Case 3: uv ∉ M and c = 0. Then, ||S||p = ||uS||p+ ||j−1S ||p and we haveCase 4: uv ∉ M and c = 1. If uv ∉ S , then β = 1. Furthermore, ||S||p = ||uS||p + ||j−1S||p and we haveuv ∈ S , implying α = β = 0. If w ∉ vs[1..j], then Otherwise, w ∈ vs[1..j] and the path containing u is split over u S and j−1S . Thus, Otherwise, Induction Step (≤): Next, we show that v ≤ ω(S ) by proving that a i-path cover S′ for c, i, j]v exists and, thus, v = ω(S′) ≤ ω(S ) by optimality of S. To this end, let u := vs[j] and let w := M(v).Case 1: w = u and c = 0. By line 6, there are ℓ and α such that v= u +v. By induction hypothesis, there are path covers u S and j−1 S corresponding to u and v, respectively. Then, Case 2: u = w and c = 1. By line 7, there are ℓ and α such that v= u+v. By induction hypothesis, there are path covers u S and j−1S corresponding to u and v, respectively. Then, S′ := uS ∪ j−1 S is a path cover for j−1S contains an edge incident to v. Thus, Case 3: w ≠ u and c = 0. By line 9, there are ℓ and α such that v = u + v. By induction hypothesis, there are path covers u S and j−1 S corresponding to u and v, respectively. Since c = 0 we have uv ∉ u S and, thus, Case 4: w ≠ u and c = 1.Case 4a: There are ℓ and α such that v = u + v (see line 10). By induction hypothesis, there are path covers u S and j−1 S corresponding to u and v, respectively. Since uv ∉ M, some edge incident to u in u Tis in M. Then, S′||p = ||uS ||p+ ||j−1S ||p= i. Furthermore,Case 4b: There is some ℓ such that v = ω(uv) + u + v (see line 10). Then, w ∈ vs[1..j] and, by induction hypothesis, there are path covers u S and j−1S  corresponding to u and v, respectively. Then, S′ := (uS +uv) ∪ j−1S is a path cover for w,v and u are on the same path p in u is incident to an edge of M in uT, we know that p does not end in u. Thus, ||S′||p = ||uS ||p+ ||j−1S ||p− 1 = i. Furthermore,Case 4c: There is some ℓ such that v = ω(uv) + u + v (see line 10). Then, w ∉ vs[1..j] and, by induction hypothesis, there are path covers u S and j−1 S corresponding to u and v, respectively. Thus, S′ := (u S + uv) ∪ j−1 S is a path cover for u is incident to an edge of M in uT, we have ||S′||p = ||uS ||p + ||j−1S ||p = i. Furthermore,Proof of Section 11Proof of Lemma 2 The proof is by induction on the distance of i to r (descending). In the induction base, i is a leaf of T and X(i) = ∅ and i Gis empty. Thus, the domains of d and P are empty. Thus, i = 0 and all other entries are −∞.X(i) with children X( j) and X(ℓ) (possibly j = ℓ):For the induction step, we distinguish the possible bag types of Introduce vertex v: Since i Gdoes not contain edges incident to v, only tuples with d(v) = 0 and P(v) = ⊥ are valid.Forget vertex v: Let iS be a maximum weight set that is eligible for . We show that i = ω(iS)."≤":Case 1: i = [d[v → 1], P + vv, p − 1, c] j. By induction hypothesis, there is a set j Scorresponding[d[v → 1], P + vv, p − 1, c] j. We show that j Sis eligible for  and, thus, ω(iS ) ≥ ω(jS ) = i. Since X(i) ⊂ X( j) and all paths between vertices in d−1(1) that are represented by P + vv are also represented by P, the first two conditions are satisfied by j Sfor i. Since q in v. But since v ∉ d−1(1), we know that d−1(1). Thus, p paths that do not intersect d−1(1) and j Ssatisfies the third condition.Case 2: i = [d[v → 1], (P − uu) + uv, p, c] j for some uu ∈ P. By induction hypothesis, there is a set j Scorresponding to [d[v → 1], (P−uu)+uv, p, c] j. Since uv ∈ (P − uu) + uv, there is a u-v-path q in q intersects d−1(1) in u. Thus, p paths that do not intersect d−1(1) and, thus, j Sis eligible for .Case 3: i = [d[v → x], P, p, c] j for some x ∈ {0, 2}. By induction hypothesis, there is a set j Scorresponding to [d[v → x], P, p, c] j. Then, j Sis also eligible for ."≥": Let Case 1: x ∈ {0, 2}. Then, iS is eligible for  and, by induction hypothesis, ω(iS ) ≤ [d[v → x], P, p, c] j ≤ i.Case 2: x = 1. Then, by Definition 2(1), there is a path q in v. If q has another end u in d−1(1), then i Sis eligible for  + uv, p − 1, c, j) and, thus, ω(iS ) ≤ [d[v → 1], (P − uu) + uv, p − 1, c]. Otherwise, iS is eligible for . In both cases, ω(iS ) ≤ i.Introduce edge uv: Let iS correspond to i. We show that i= ω(iS ). Let d′ and z be as described in the dynamic programming and note that d, P, p, c]i =  j, then, by induction hypothesis, there is a set jS corresponding to  j and uv ∉ M. Then, j Sis eligible for  and, thus, ω(iS ) ≥ ω(jS ) = i.d, P, p, c]i = z + ω(uv): To show i ≤ ω(iS ), we consider a set j Sthat corresponds to the entry for X( j) from which i is computed and whose existence is granted by induction hypothesis. Then, we show that jS + uv is eligible for , implying ω(iS ) ≥ ω(j S+ uv) = z + ω(uv) if uv ∉ M and ω(iS ) ≥ ω(jS ) = z if uv ∈ M. Note that, in each of the cases, the degrees of u and v in j Ssatisfies Definition 2(1) for d′.In the following, we proceed in a similar manner for the case that i if S is eligible for  and its weight ω(S ) = iis maximum among all sets that are.We say a Case 1: d(u) = d(v) = 2. Then, both u and v have degree 1 in Case 1a: z = j. Then, there is a u-v-path in d−1(1) = d′−1(1) \ {u, v}, adding uv does not touch any paths intersecting d−1(1). Thus, Definition 2(2) is satisfied. Further, adding uv closes a cycle in d−1(1) than Case 1b: z =  j. Then, both u and v have dangling paths in uv connects two paths such that the resulting path does not intersect d−1(1). Thus, there are one more such paths in Case 1c: z =  j for some x ∈ d−1(1) = d′−1(1) \ {u, v} with xx ∈ P. Then, v and a u-x-path ). Thus, adding uv connects two paths such that the resulting path intersects d−1(1) exclusively in x. Hence, there is a path dangling from x in j S+ uv satisfies Definition 2(2). Since no paths or cycles avoiding d′−1(1) are affected by adding uv, Definition 2(3) is satisfied. The case that z =  j is analogous.Case 1d: z =  j for some x, y ∈ d−1(1) with xy ∈ P. Then, u qand v qthat start in u and v, respectively, and end in x and y, respectively ). Thus, adding uv connects u qand v qto a single path q that starts in x and ends in y, thus intersecting d−1(1). Hence both Definition 2(2) and (3) are satisfied.Case 2: d(u) = d(v) = 1. Thus, u and v are not incident to any edges in uv forms a new path connecting u and v in z = jand uv ∈ P, we conclude that both Definition 2(2) and (3) are satisfied.Case 3: d(u) = 2, d(v) = 1. Then, v has no incident edges in u in u has degree 1 in q in vx ∈ P for some x ∈ d−1(1) and q′ := q + uv.Case 3a: x = v. Then, z =  j. Since q′ which is a path dangling from u, Definition 2(2) is satisfied. Since no other paths are touched, Definition 2(3) is satisfied.Case 3b: x ≠ v. Then, z = j. Since q′ which is a u-x-path, Definition 2(2) is satisfied. Since no other paths are touched, Definition 2(3) is satisfied.uv ∉ iS ∩ M, then i Sis eligible for . By induction hypothesis,  j ≥ ω(iS ) and, thus, i ≥  j} ≥ ω(iS ). Otherwise, uv ∈ i S∪ M"≥": If iS − uv is eligible for a tuple corresponding to one of the entries over which we maximize to compute z. Thus, ω(iS ) ≤ ω(i S− uv) + ω(uv) ≤ z + ω(uv) if uv tt M and ω(iS ) ≤ z if uv ∈ M. Note that, in each case, i S− uv satisfies Definition 2(1) since the degrees of u and v decrease by one when removing uv.In the following, we show that Case 1: d(u) = d(v) = 2. Then, uv is part of a path q in u nor v is an endpoint of q.Case 1a: q is closed  ). Then, q does not intersect d−1(1), implying that q − uv is a u-v-path in d′−1(1) only in u and v. Thus, i S− uv is eligible for .Case 1b: q is open and does not intersect d−1(1) ). Then, q − uv decomposes into paths u qand vqintersecting d′−1(1) only in u and v, respectively. Thus, i S− uv is eligible for .Case 1c: q is open and intersects d−1(1) in a single vertex x ). Then, q − uv decomposes into paths u qand v qin d′−1(1) in x. Since no path avoiding d−1(1) is touched, i S− uv is eligible for either  ∪ {xu, vv}, p, c, j) or  ∪ {uu, vx}, p, c, j).Case 1d: q is open and intersects d−1(1) in 2 distinct vertices x and y ). Then, q − uv decomposes into a u-x-path u qand a v-y-path v qin i S- uv is eligible for  ∪ {ux, vy}, p, c, j).Case 2: d(u) = d(v) = 1. Then, q = {uv} is a path in u and v are isolated in uv ∈ P and i S− uv is eligible for .Case 3: d(u) = 2 and d(v) = 1. Thus, q = . If q intersects d−1(1) − v in a vertex x, then q − uv intersects d′−1(1) − u and, thus, i S− uv is eligible for  + ux, p, c, j). Otherwise, q is a dangling from v in q − uv is a dangling from u in i S− uv is eligible for  + uu, p, c, j).Join: "≤": Let d1, d2, P1, P2, p1, p2, c1, c2 be such that i =  j + ℓ. By induction hypothesis, there are sets j Sand S ℓ corresponding to  j and ℓ, respectively, such thati S:= jS ∪ S t is eligible for  and, thus, ω(S ) ≥ ω(iS ) =  j + ℓ = i. First, by (2), we have that Definition 2(1) is satisfied. Second, to show that Definition 2(2) is satisfied, consider some uv ∈ P. Then, there is a path q =  in Gr(P1 ∪ P2). Note that u-x1-path, an x1-x2-path, . . . . The concatenation of these paths forms a u-v path in uu ∈ (P1)1, there is a path u in u. Analogously, a similar path uu ∈ (P2)1 in uu ∈ (P1 ∩ P2)1, there is a path u and avoiding d−1(1) and u qis neither in p1 paths avoiding p2 paths avoiding p1 + p2 + |(P1 ∩ P2)1| such paths. Similarly, c1 + c2 + |(P1 ∩ P2)2| cycles avoiding d−1(1).We show that j S:= i S∩E(jG ) and let ℓ S:= i S∩E(ℓG). We show that j Sand ℓS are eligible for tuples  and , respectively, such that (2)-(5) hold. First, for all u ∈ X(i) = X( j) = X(ℓ) let d1 (d2) be the number of edges of j G (ℓG) incident with u. Since P1 be the set of pairs uv with u, u ≠ v, then there is a u-v path in u = v, then u. Let P2 be defined analogously for d2 and uv ∈ P, since d(u) = 1 ⇐⇒ d1(u) = 1 ⊕ d2(u) = 1, we have P1(u) = ⊥ ⊕ P2(u) = ⊥ and there is a u-v-path q in P1 ∪ P2) contains a u-v-path and we conclude that (3) holds. Third, let p1 and c1 be the number of paths and cycles, respectively, in p2 and c2 in d2. Then Definition 2(3) is satisfied. Further, let p ′ denote the number of pairs uu ∈ P1 ∩ P2, that is, p ′ := |(P1 ∩ P2)1|. Then, since uu, a different path consisting of the concatenation of the two paths dangling from u in p1 + p2 + p′ paths avoiding d−1(1). Thus, (4) holds and, in complete analogy, (5) holds.    □"≥": Let Proof of Theorem 1, sketch The bottleneck in the computation are the join nodes so we focus on computing their dynamic programming table. To calculate the number of entries that have to be considered in order to compute i, assume that P1 and P2 are fixed. Then so is P and u is incident to an edge in M that has been introduced in the subtree rooted at j, then d1(u) > 0 and, thus, d1(u) = 2. However, if u is not incident with any matching edges in d1(u) < 2, since otherwise, u would be incident to two non-matching edges. Thus, u is incident to a matching edge in P1 and P2 also fixes d1 and, by extension, d2. Finally, we can choose p1 and c1 in order to compute p2 and c2. Since we need to consider only permutations that are also involutions, there are less than twtw ways to choose P1 and P2. Thus, the maximum is over at most twtw ·σp · σc elements. Since there are O(n) bags in the tree decomposition, the algorithm can be executed in the claimed running time. □Lemma 5 Tree Rule 2 is correct, that is, the instance I =  is yes if and only if the result I′ =  of applying Tree Rule 2 to I is yes.Proof Clearly, since all vertices of the graph G must be covered, then the only way to pack the vertex u is to include it into a path of length p lincluded the vertex v′ = LCA and then decrease the number of paths by one. □Proof of Lemma 3 We show that v Tdoes not contain branching vertices (vertices with at least two children) since it is straightforward that, if v Tis a path, it has to be alternating for I to be a yes-instance and, if its length exceeds ℓp, then Tree Rule 2 applies. Towards a contradiction, assume that v Thas branching vertices and let z denote such a vertex in v Tsuch that, among all branching vertices, z is furthest from v. Let u be a leaf of z Tthat has maximum distance to z and let d denote this distance. Since z is the only branching vertex in z Tand I is a yes-instance, the unique u-z path in z Tis alternating. Thus, by irreducibility with respect to Tree Rule 2, we know that d < ℓp. However, since z is branching, there is another leaf w at distance d′ ≤ d < ℓp to z in zT. Thus, if the unique u-w-path in z Tis not alternating or its length is not ℓp, then I is not a yes-instance. But otherwise, Tree Rule 2 is applicable to u and w, contradicting irreducibility. □Lemma 6 Tree Rule 3 is correct, that is, the instance I =  is a yes-instance if and only if the result of applying Tree Rule 3 to I is.Proof Let v and e be as defined in Tree Rule 3 and let e = {u, v}. To show correctness of Tree Rule 3, we prove that all optimal solutions for I contain e. To this end, let S be an optimal solution with e ∉ S . By Lemma 3, v Tis an alternating path p ending in v with |p| < ℓp. By definition, M(u) is on p, so |p| ≥ 2. But then, G − e contains an isolated path of length strictly less than ℓp, implying that S is not a solution for I. □Lemma 7 Path Rule 1 is correct, that is, the instance I =  is a yes-instance if and only if the result I′ =  of applying Path Rule 1 to I is.Proof Let G′.S be a solution for I, let G[S ∪ M] contains p since |p| > ℓc. We show that S ′ := S \ p ′ is a solution of for I′ with ω′(S ′) = ω(S ). First, note that all vertices of G′ are covered by S ′. Second, note that p \ p′ is alternating since |p| = ℓp + 1. Finally, G′[S ′ ∪ M′] contains one path less than G[S ∪ M]."⇒": Let S ′ be a solution for I′ and let u denote the vertex onto which I′ and let S for I with ω(S ) = ω′(S ′). First, since |p| > ℓc + ℓp + 1, no cycle in G[S ∪ M] contains e. If e ∉ S ′ ∪ M′, then S := S ′ ∪ (p′ − e0). If e ∈ S ′ ∪ M′, then je ∉ S ′ ∪ M′ for some ℓp <j ≤ 2ℓp. Then, "⇐": Let Lemma 8 Path Rule 2 is correct, that is, the instance I =  is a yes-instance if and only if the result I′ =  of applying Path Rule 2 to I is.Proof First, note that no solution for I or I′ covers w in a cycle since w Tis necessarily covered by a path. Also note that q exists in I′.S be a solution for I and note that there is some i ≤ ℓp such that ie∉ S ∪M . Then, however, i−γf∉ S and S ′ := S \ p is a solution for I′ and ω′(S ′) = ω(S )."⇒": Let S ′ be a solution for I′ and note that there is some i ≤ ℓp − γ such that i f∉ S ′ ∪ M′ . But then, S ′ ∪ (p − i+γe) is a solution for I and ω(S ) = ω′(S ′). □"⇐": Let Lemma 9 Path Rule 3 is correct, that is, the instance I =  is a yes-instance if and only if the result I′ =  of applying Path Rule 3 to I is.Proof "⇒": Let S be solution for I and let u qand v qdenote the paths containing u and v, respectively, in G[S ∪ M].Case 1: Neither u qnor v qcontain edges of p. Then, γ = 1 and ux1 ∉ S ∪ M. If γu +γv +1 = ℓp, then Case (4) applies and otherwise, Case (5) applies. In both cases, S is also a solution for I′.Case 2: v qcontains edges of p, but u qdoes not. Then, γv+ γ = ℓp + 1 and ux1 ∉ S ∪ M. If γu + γ = ℓp + 1, then Case (3) applies and otherwise, Case (1) applies. In both cases, S is also a solution for I′.Case 3: u qcontains edges of p, but v qdoes not. Then, γu + γ = ℓp + 1 and vxp|−1| ∉ S ∪ M and ux1 ∈ S \ M. If γu + γ = ℓp + 1, then Case (3) applies and switching ux1 for wx1 in S gives a solution of same weight for I′. Otherwise, Case (2) applies and S is also a solutionI′.for Case 4: Both u qand v qcontain edges of p. Then, γu + γv + γ ≡ ℓp mod (ℓp + 1) and ux1 ∈ S \ M. If γ ≠ 1, then Case (6) applies and S is also a solution for I′. Otherwise, Case (3) applies and switching ux1 for wx1 in S gives a solution of same weight for I′.S ′ be a solution for I′. Note that, if G′ does not contain wx1, then S is clearly also a solution for I. Thus, assume that G′ contains wx1 and, thus, either Case (3) or (4) applies to I. We show that the result S of switching wx1 for ux1 in S ′ is a solution of the same weight for I. To show this, it suffices to show that, if a path q contains wx1, then q ends at u. Assume this is false, that is, q contains wx1 and some edge e ∈ E(G′) \ M incident with u. Since w Tis not empty, no cycle in G′[S ′ ∪ M] contains u. Then, the length of the u-v-path containing p − ux1 in G′ is equivalent to γ + γu modulo ℓp + 1."⇐": Let Case 1: vxp|−1 |∈ S ∪ M. Then, since q does not end in u, we have I, implying that Case (3) applies to I. But then, γ + γv = ℓp + 1 and, hence, wx1 ∉ S ∪ M.Case 2: vxp|−1| ∉ S ∪ M. Then, since q does not end in u, we have u + γ ≠ ℓp + 1 since 0 < γ ≤ ℓp and 0 < γu < ℓp. Thus, Case (3) does not apply to I, implying that Case (4) applies to I. But then, γ = 1 and, since vxp|−1| ∉ S ∪ M, we have wx1 ∉ S ∪ M. □Observation 1 Let G be connected. Then, |V†| ≤ 2 FES(G) since |E†| ≤ |V†| + FES(G†) ≤ |V†| + FES(G) and 2|E†| ≥ 3|V†|.Proof of Theorem 2 By Lemma 4, we have The authors declare that they have no competing interests.MW, AC and RG conceived the method and the proofs. MW implemented and tested the algorithms on all datasets. MW, AC and RG wrote the paper.


An upper bound method for the process of plane strain extrusion through a wedge-shaped die is derived. A technique for constructing a kinematically admissible velocity field satisfying the exact asymptotic singular behavior of real velocity fields in the vicinity of maximum friction surfaces  is described. Two specific upper bound solutions are found using the method derived. The solutions are compared to an accurate slip-line solution and it is shown that the accuracy of the new method is very high. The approximate methods for analysis of plane strain and axisymmetric extrusion/drawing can be conveniently divided into three groups: (i) the slab method, (ii) semianalytical solutions for flow through infinite channels, and (iii) the upper bound method. The slab method has been used, for example, in . In geneα) through which a sheet of plastic material is being pushed by force P  be taken relative to the axis of symmetry and the virtual apex O of the die. In addition, introduce Cartesian coordinates  whose origin is situated at O and whose x-axis coincides with the axis θ = 0. In the polar coordinates, the surfaces of the die are determined by the equation θ = ±α. Since θ = 0 is an axis of symmetry for the flow, it is sufficient to find the solution in the region θ ≥ 0. It is assumed that the sheet is rigid perfectly/plastic and obeys Mises yield criterion. The tensile yield stress is denoted by σ0.Consider a wedge-shaped die ξeq=ξeq=ξrr,θ = α is automatically satisfied. The equation of incompressibility in the polar coordinates reduces to ∂u/∂r + u/r = 0. Here u is the radial velocity. The general solution of the incompressibility equation isf(θ) is an arbitrary function of θ and R0 is the radial coordinate of point A  must tend to a finite limit as θ → 0. A necessary condition for that is C = 0. Then, it follows from  passes through point A and the line ρ = ρd1(θ) passes through point B (ρd0(α) = 1 and ρd1(α) = h. Substituting these conditions into (f(θ):ρ = ρd(θ) on the velocity discontinuity lines yielddρd/dθ by means of (ρd(θ) should be replaced with ρd0(θ) for the velocity discontinuity line between the plastic zone and rigid zone 1 and with ρd1(θ) for the velocity discontinuity line between the plastic zone and rigid zone 2.Assume that the circumferential velocity vanishes everywhere. Then, the velocity boundary condition at  point A . The nonation isu=−UR0fi=eri=erθ = αng form:ur=−vi.wation isu=−UR0fsinφation isρ=ρd(θ)=1ws from (ρ=ρd(θ)=t point B . Therefoed from (ρ=ρd0(θ)=ituting (ur=−v(erc zone isup=uer.Tation isu=−UR0f(θaccount ([u]=[(u+vituting (dρdθsinθ+f(θ) is independent of ρ, integrating with respect to ρ and using (F(θ) = ∫0θf(z)dz. It is evident from this equation that the value of Wd is the same for each of the velocity discontinuity lines. The radial velocity at the friction surface is found from (u|θ=α = −UR0f(α)r−1. Then, using , a, a(25)WVmeans of  as28)WmponentsWi=WV+2Wdtheorem [PuU2B=Wi,d using  isτrθ = ksin2ψ, where τrθ is the shear stress in the polar coordinates. Then, it follows from  and G1(α) are determined by numerical integration. Then, pu is readily found from (G0(α) and G1(α) with α at several m-values is illustrated in Figures G1(α) is practically a linear function of α. The variation of pu with α at several h-values and m = 1 is depicted in pu has been compared to an accurate slip-line solution derived in [h at several values of α (in the nomenclature of the present paper) is depicted in Figure 7.26 in [pu with 1 − h at α = π/12 and α = π/6 has been calculated by means of pu by meion (see ) that τ0rβc−cosψ.Here β ψ(θ) isdψdθ=c−cows from (ψα=12arcstion of (θ=∫0ψcoseans of (u=−UR0rβ( ψ(θ) isdψdθ=c−co R0sinα , 40) is ispu bythe formα=∫0ψαcosR0sinα (β=sinα[∫0eans of (u=−UR0rβ( ψ(θ) isdψdθ=c−coR0sinα (β=sinα[∫0cos2ψ),F(θ) =∫0θerefore,  is satised from (f(θ)=β(c−cos2ψ),F(θ) =∫0θund from . The varrived in  by the m 7.26 in . The varmeans of , (34), (ed from (f(θ)=β(c−cos2ψ),F(θ) =∫0θa and b are arbitrary constants. It is evident that  is  is 44) if(θ)=a+2bent that . It follf(θ)=a+2bnt that (F(θ)=aθ+b= α into  gives(4f(θ)=a+2b α into (46)f(θ)=e use of . In partA general kinematically admissible velocity field has been built up for the process of extrusion through a wedge-shaped die. This velocity field satisfies the asymptotic behavior of the real velocity field in the vicinity of maximum friction surfaces.Equations for the power dissipation within the plastic zone, at the velocity discontinuities and at the friction surface, have been developed in terms of ordinary integrals. These allow the calculation of the power required for the extrusion process.A semianalytic solution for flow through an infinite converging channel has been adopted to calculate an upper bound on the extrusion force by the method developed. The very close agreement between this upper bound solution and a slip-line solution found by the method of Riemann has been found.A very simple kinematically admissible velocity field has been proposed to calculate an upper bound on the extrusion force by the method developed. The very close agreement between the two upper bound solutions has been found. This suggests that simple kinematically admissible velocity fields satisfying  can be uThe method can be extended to axisymmetric extrusion. To this end, it is just necessary to account for  in the g


The quotient hemirings via fuzzy strong h-ideals are investigated. The relationships between fuzzy congruences and fuzzy strong h-ideals of hemirings are discussed. Pay attention to an open question on fuzzy congruences. Finally, the normal fuzzy strong h-ideals of hemirings are explored.The aim of this paper is to introduce the concepts of fuzzy strong  The concept of fuzzy sets was formulated by Zadeh , and sinSemirings, regarded as a generalization of rings, have been recently found particularly useful in solving problems in different disciplines of applied mathematics and information sciences because semirings provide an algebraic framework for modelling. A special semiring with a zero and endowed with the commutative addition is said to be a hemiring. Nowadays, semirings (hemirings) are useful in optimization theory, graph theory of discrete event dynamical systems, matrices, determinants, generalized fuzzy computation, automata, theory, formal language theory, coding theory, and analysis of computer programs.h-ideals and k-ideals of hemirings were thoroughly investigated by la Torre . Let F(S) denote the set of all fuzzy sets of S. A fuzzy set μ in S of the formy and value r and is denoted by yr. In particular, if r = 1, y1 denotes the fuzzy point with support y and value 1.A fuzzy set μ of S is called a fuzzy left ideal if for all x, y ∈ S, we have1)(Fμ(x + y) ≥ μ(x)∧μ(y),2)(Fμ(xy) ≥ μ(y).A fuzzy left ideal μ of S is called a fuzzy left h-ideal if for all a, b, x, z ∈ S, x + a + z = b + z → μ(x) ≥ μ(a)∧μ(b). A fuzzy right h-ideal is defined similarly.A fuzzy set μ and ν be fuzzy sets of S. The sum μ + ν of μ and ν is defined byLet In particular,x ∈ S and μ ∈ F(S), then y + μ = y1 + μ.Note that if μ of S is called a fuzzy strong left(right) h-ideal of S if for all x, y, z, a, b ∈ S,μ(x + y) ≥ μ(x)∧μ(y),μ(xy) ≥ μ(y)(μ(xy) ≥ μ(x)),x + a + z = y + b + z → (y1 + μ)(x) ≥ μ(a)∧μ(b). Note that if μ is a fuzzy strong h-ideal of S, then μ(0) ≥ μ(x).A fuzzy set S = {0, a, b, c} be a set with an addition operation (+) and a multiplication operation (·) as follows:Let μ in S by μ(0) = μ(a) = 0.6, μ(b) = μ(c) = 0.5. Then μ is a fuzzy strong h-ideal of S.Define a fuzzy set μ and ν be fuzzy sets of S. Then h-sum of μ and ν is defined bya′, a′′, b′, b′′, z ∈ S such that x + a′ + b′ + z = a′′ + b′′ + z; otherwise, (μ+hν)(x) = 0.Let μ be a fuzzy set of S and r ∈ . Then the sets μr = {x∣μ(x) ≥ r} and μrs = {x∣μ(x) > r} are called an r-level subset and an r-strong level of μ, respectively. We now characterize the fuzzy strong h-ideal of S by using their (strong) level subsets.Let μ of S is a fuzzy strong h-ideal of S if and only if nonempty μr is a strong h-ideal of S for all r ∈ .A fuzzy set μ be a fuzzy strong h-ideal of S, x, y ∈ μr, and a ∈ S. Then μ(x + y) ≥ μ(x)∧μ(y) ≥ r, μ(ax) ≥ μ(a)∨μ(x) ≥ r, and so x + y, ax ∈ μr. Similarly, we get xa ∈ μr, hence μr is an ideal of S.Let x, y, z ∈ S and a, b ∈ μr be such that x + a + z = y + b + z. Hence, μ(a) ≥ r, μ(b) ≥ r. Thus we haveNow, let b′ ∈ S such that x = y + b′ and μ(b′) ≥ r; that is, b′ ∈ μr, and so x ∈ y + μr. Therefore, μr is a strong h-ideal of S.This implies that there exists x′, y′ ∈ S. If possible, let μ(x′ + y′) < μ(x′) ∧ μ(y′). Choose r such that μ(x′ + y′) < r < μ(x′) ∧ μ(y′). Then x′, y′ ∈ μr, but x′ + y′ ∉ μr, a contradiction. Hence, μ(x + y) ≥ μ(x)∧μ(y) for all x, y ∈ S. Similarly, we have μ(xy) ≥ μ(x)∨μ(y) for all x, y ∈ S.Conversely, assume that the given conditions hold. Let x′, y′, z′, a′, b′ ∈ S such that x′ + a′ + z′ = y′ + b′ + z′ and (y1′ + μ)(x′) < μ(a′)∧μ(b′); choose r such that (y1′ + μ)(x′) < r < μ(a′)∧μ(b′). Then a′, b′ ∈ μr, but (y1′ + μ)(x′) = ⋁x′=y′+cμ(c) < r. That is, x′ ∉ y′ + μr, a contradiction. Hence, (y1 + μ)(x) ≥ μ(a)∧μ(b) for all x, y, z, a, b ∈ S with x + a + z = y + b + z. This implies that μ is a fuzzy strong h-ideal of S.Now assume there exist h-ideals are investigated. Finally, we give an isomorphism theorem of hemirings.In this section, the quotient hemirings via fuzzy strong θ be an equivalence relation on S. Recall that θ is called a congruence relation on S if  ∈ θ and  ∈ θ imply  ∈ θ and  ∈ θ.Let I be a strong h-ideal of S, x, y ∈ S. We call x congruent to y mod I, if and only if there exist a, b ∈ Iand z ∈ S be such that x + a + z = y + b + z I if and only if x ∈ y + I,x]I + [y]I = [x + y]I,[ab∣a ∈ [x]I, b ∈ [y]I}⊆[xy]I. {Let μ be a fuzzy strong h-ideal of S and μr an r-level subset of S. We denote by μx = {y ∈ S∣y ∈ [x]μr} the equivalence class containing x and by S/μ = {μx∣x ∈ S} the set of all equivalence classes of S.Let μ is a fuzzy strong h-ideal of S and μr2 = μr, then S/μ is a hemiring under the binary operations:x, y ∈ S.If μx = μu and μy = μv. Since μx = {y ∈ S∣y ∈ [x]μr}, we have μx = [x]μr, and so [x]μr = [u]μr, [y]μr = [v]μr. Then [x]μr + [y]μr = [u]μr + [v]μr. By x]μr + [y]μr = [x + y]μr, and so [x + y]μr = [u + v]μr; that is, μx+y = μu+v. Hence the addition is well defined.Firstly, we show that the above binary operations are well defined. In fact, if x]μr · [y]μr⊆[xy]μr. Next we show [xy]μr⊆[x]μr · [y]μr. Let z = ab ∈ [xy]μr; since μr2 = μr, then z = ab ∈ [xy]μr = xy + μr = xy + xμr + yμr + μr2 = (x + μr)(y + μr) = [x]μr · [y]μr. This implies that z = ab ∈ [x]μr · [y]μr, and so [xy]μr⊆[x]μr · [y]μr; that is, [x]μr · [y]μr = [xy]μr. This means that μx · μy = μxy. Hence the multiplication is well defined. Now it is easy to verify that S/μ is a hemiring.By f be a homomorphism from S to S′, μ a fuzzy subset of S, and μ′ a fuzzy subset of S′. Then the image f(μ) of μ and the preimage f−1(μ) of μ′ are both fuzzy sets defined, respectively, as follows:x ∈ S.Let f : S → S′ be a homomorphism of hemirings. A strong h-ideal I of S is called f-compatible, if for all x, y, z, a, b ∈ S, f(x + a + z) = f(y + b + z) implies x ∈ y + I. A fuzzy strong h-ideal μ of S is called f-compatible, if for all x, y, z, a, b ∈ S, f(x + a + z) = f(y + b + z) implies (y + μ)(x) ≥ μ(a)∧μ(b).Let f is a monomorphism, then every fuzzy strong h-ideal is f-compatible.If the above f : S → S′ be an epimorphism of hemirings. If μ is a f-compatible fuzzy strong h-ideal of S, then f(μ) is a fuzzy strong h-ideal of S′.Let x′, y′ ∈ S′; then(1) Let x′, y′ ∈ S′; then(2) Let x, y, z, a, b ∈ S and x′, y′, z′, a′, b′ ∈ S be such that x′ + a′ + z′ = y′ + b′ + z′, f(x) = x′, f(y) = y′, f(z) = z′, f(a) = a′, f(b) = b′; then we have f(x + a + z) = f(y + b + z). Since μ is f-compatible, we have (y1 + μ)(x) ≥ μ(a)∧μ(b); thus(3) Let f(μ) is a fuzzy strong h-ideal of S′.Therefore, Similarly, we can obtain the following result.f: S → S′ be a monomorphism of hemirings. If μ′ is a fuzzy strong h-ideal of S′, then f−1(μ′) is an f-compatible fuzzy strong h-ideal of S.Let The proof is similar to μ and ν be two fuzzy sets of S. If they are fuzzy strong left  h-ideals of S, then so are μ∩ν and μ+hν.Let μ∩ν is a fuzzy strong left h-ideal of S. For any x, y ∈ S,We first show that μ(xy) ≥ μ(y) and ν(xy) ≥ ν(y), it follows thatSince μ∩ν is a fuzzy left ideal of S.Therefore, a, b, x, y, z be such that x + a + z = y + b + z. ThenLet μ∩ν is a fuzzy strong left h-ideal of S.Hence μ+hν is a fuzzy strong left h-ideal. In fact,Now we show that x, y ∈ S, we have(1) for any x, y ∈ S, we have(2) For any a, b, x, y, z′ be any elements of S such that x + a + z′ = y + b + z′. If there exist c′, c′′, d′, d′′, e′, e′′, f′, f′′, z′′, z′′′ ∈ S, such thatz′′′′ = z′ + z′′ + z′′′ + a + b, and so(3) Let μ+hν)(a) = 0 or (μ+hν)(b) = 0, and so [y1 + (μ+hν)](x) ≥ 0 = (μ+hν)(a)∧(μ+hν)(b). Summing up the above statements, μ+hν is a fuzzy strong left h-ideal of S. The case for fuzzy strong right h-ideals can be similarly proved.This givesS) the set of all fuzzy strong h-ideals of S with the same tip t; that is, μ(0) = ν(0) for all μ, ν ∈ FSI(S). Then we have the following result.Now denote by FSI(S), +h, ∩) is a bounded complete lattice under the relation “⊆”.The . It follows from μ∩ν ∈ FSI(S) and μ+hν ∈ FSI(S). It is clear that μ∩ν is the greatest lower bound of μ and ν. We now show that μ+hν is the least upper bound of μ and ν, since μ(0) = ν(0); for any x ∈ S, we haveLet μ⊆μ+hν. Similarly, we have ν⊆μ+hν. Now, let ω ∈ FSI(S) be such that μ, ν⊆ω; then we have μ+hν⊆ω+hω⊆ω. Hence, μ∨ν = μ+hν. It is clear that replace the {μ, ν} with arbitrary family of FSI(S) and so (FSI(S), +h, ∩) is a complete lattice under the relation “⊆”. ∅ and χS are the minimal and the maximal elements in (FSI(S), +h, ∩), respectively. Therefore, (FSI(S), +h, ∩) is a bounded complete lattice.Hence, Finally, we give an isomorphism theorem of hemirings.f: S → S′ be an isomorphism of hemirings and ν a fuzzy strong h-ideal of S′. If νr2 = νr and f−1(ν)r2 = f−1(ν)r, thenLet S/f−1(ν) and S′/ν are both hemirings. Define ξ : S/f−1(ν) → S/ν byIt follows from Theorems ξ is well defined as follows: f−1(ν)x = f−1(v)y⇒[x]f−1(v)r = [y]f−1(ν)r⇒x + f−1(ν)r = y + f−1(ν)r⇒f(x + f−1(ν)r) = f(y + f−1(ν)r). Since f is a homomorphism, then f(x) + f(f−1(ν)r) = f(y) + f(f−1(ν)r)⇒f(x) + vr = f(y) + νr⇒[f(x)]νr = [f(y)]νr⇒νf(x) = νf(y).(1) ξ is a homomorphism:(2) ξ is an epimorphism; for any νy ∈ S′/ν, since f is epimorphism, then there exists x ∈ S, such that f(x) = y, so ξ((f−1(ν))x) = νf(x) = νy.(3) ξ is monomorphism: νf(x) = νf(y)⇒[f(x)]νr = [f(y)]νr⇒f(x) + νr = f(y) + νr⇒f(x) + f(f−1(ν)r) = f(x) + f(f−1(ν)r)⇒f(x + f−1(ν)r) = f(y + f−1(ν)r); since f is monomorphism, so x + f−1(ν)r = y + f−1(ν)r; thus [x]f−1(ν)r = [y]f−1(ν)r, so we have (f−1(ν))x = (f−1(ν))y; hence S/f−1(ν)≅S′/ν, and the proof is completed.(4) h-ideals of hemirings. The following concepts can be seen in → be an increasing function. Then a fuzzy set μf : S →  defined by μf(x) = f(μ(x)) is a fuzzy strong left h ideal of S. In particular, if f(μ(0)) = 1, then μf is normal; if f(t) ≥ t for all t ∈ , μ⊆μf.Let x, y ∈ S, then we haveF1). Similarly, μf(xy) = f(μ(xy)) ≥ f(μ(y)) = μf(y), which proves (F2) holds. Hence μf is a fuzzy left ideal of S.For all a, b, x, y, z ∈ S be such that x + a + z = y + b + z. ThenNow, let μ is a fuzzy strong left h-ideal, we haveSince soμf is a fuzzy strong left h-ideal of S; if f(μ(0)) = 1, then μf is normal; suppose that f(t) = f(μ(x)) ≥ μ(x) for all x ∈ S, which proves μ⊆μf.Therefore, N(S) denote the set of all normal fuzzy strong left h-ideal of S. Note that N(S) is a poset under the set inclusion.Let μ ∈ N(S) be nonconstant such that it is a maximal element of (N(S), ⊆). Then μ takes only two values 0 and 1.Let μ is normal, we have μ(0) = 1. For some x ∈ S, let μ(x) ≠ 1; we know that μ(x) = 0; otherwise, there exists x0 ∈ S, such that 0 < μ(x0) < 1. Now we define on S a fuzzy set ν by putting ν(x) = (μ(x) + μ(x0))/2 for some x ∈ S. Then ν is well defined.Since x, y ∈ S, we haveF1) holds.For all F2) holds. Hence v is a fuzzy left ideal of S. Now, let a, b, x, y, z ∈ S be such that x + a + z = y + b + z; since μ is a normal fuzzy strong left h-ideal, we have (y1 + μ)(x) = ⋁x=y+e μ(e) ≥ μ(a)∧μ(b), which impliesν is a fuzzy strong left h-ideal of S. By ν+ is a normal fuzzy strong left h-ideal of S. Note thatν+(x0) < 1 = ν+(0); this means that ν+ is nonconstant and μ is not a maximal of N(S). This is a contradiction.Similarly, we can obtainh-ideal μ of S is called maximal if μ+ is a maximal element of N(S).A nonconstant fuzzy strong left h-ideal μ of S is maximal, thenμ is normal,μ takes only the values 0 and 1,χμ0 = μ,μ0 is a maximal strong left h-ideal of S.If a fuzzy strong left μ be a maximal fuzzy strong left h-ideal of S. Then μ+ is a nonconstant maximal element of the poset (N(S), ⊆). It follows from μ+ takes only the values 0 and 1. Note that μ+(x) = 1 if and only if μ(x) = μ(0), and μ+(x) = 0 if and only if μ(x) = μ(0) − 1. By μ(x) = 0, and so μ(0) = 1. Hence μ is normal and μ+ = μ. This proves that (1) and (2) hold.Let (3) Obviously.μ0 = {x ∈ S∣μ(x) = 1} is a strong left h-ideal. Obviously μ0 ≠ S since μ takes two values. Now let A be a strong h-ideal that contains μ0. Then μμ0⊆μA, and in consequence, μ = μμ0⊆μA. It follows from μbeing normal that μA is also normal and takes only two values: 0 and 1. By the assumption, μ is maximal, so μ = μA or μ = ω, where ω(x) = 1 for all x ∈ S. In the last case μ0 = S, which is impossible. So, μ = μA; that is, μA = χA. Hence μ0 = A.(4) It is clear that h-ideals of hemirings. We also discuss some concept of fuzzy strong h-ideals of hemrings and then we consider quotient hemirings and their isomorphism theorem. Finally, we introduce normal fuzzy strong h-ideals of hemirings. In the future study of hemirings, we can apply fuzzy congruences of hemirings to some applied fields, such as decision making, data analysis, and forecasting.In this paper, we consider the relationships between fuzzy congruences and fuzzy strong


In this paper, we extend this result and put forward a generalized method which can completely solve the family of systems of Diophantine equations x2 − 6y2 = −5 and x = az2 − b for each pair of integral parameters a, b. The proof utilizes algebraic number theory and p-adic analysis which successfully avoid discussing the class number and factoring the ideals.Mignotte and Pethö used the Siegel-Baker method to find all the integral solutions ( Z, N, and Q be the sets of all integers, positive integers, and rational numbers, and let a, b be the integers. The system of Diophantine equationsa = 2 and b = 1; however, their method was complicated as a combination of algebraic and transcendental number theory. In 1998, Cohn [a = 2 and b = 1. In 2004, Le [D − 1 is the power of an odd prime. As an example, solutions of the equations for D = 6 and D = 8 are given in the paper so as to show the effectiveness of the method.Let quationsx2−6y2=−5quationsx2−6y2=−52004, Le  used a sp-adic method [In this paper, we use algebraic number theory and Skolem's c method  to solvec method  to compuIn order to well interpret the main result, the symbol notation used in this paper is defined as below.m ≥ 0 is an integer and A denote β2 = ±ηε0 or β2 = ±η, with β2 in Here, we assume that The main result of this paper is as follows.x, y, z) be an integral solution of (z exists only when it satisfies one of the following four equations for k ∈ Z: N1)(az + θA = ±αεk with 2N2)(az + θA = ±αεk with 2N3)(az + θA = ±αεk with 2N4), α is the nonassociated factor such that α− denotes the relative conjugate of α, and A is referred to above.Let  = , , , , , and .The system of  for a = Before the proof of the theorem, m, ε0, β, and If A, we haveRewriting A is an algebraic number. Thus, the lemma is proven.Since There are four separate cases in consideration during the proof. Since the process is very similar in each case, some details will be omitted for simplicity. Now we prove the theorem.After rewriting the first equation of , factoriAdding , we get2x=±(ηε0nCase 1. Assume that n = 2m + 1 is odd and a to obtainQ(θ), we haveCase 2. In another case, when  n = 2m is even and ¯n. Then2x=ηε02m+nserting , we getserting  denotes the standard p-adic valuation and p is an odd prime.For any integer i > 1, the p-adic valuation of the term of p-adic valuation of the term of We knowQ(θ) is computed. Furthermore, nonassociated factor of Q(θ) is also calculated. The idea of computation of fundamental unit and nonassociated factorization stems from Zhu and Chen [To complete the proof of the corollary, the fundamental unit in the totally complex quartic field and Chen  and Buchand Chen , which oCase 1. Substituting a = 2 and b = 1 into  is reduced toEquationε = 11 + 2θ − 6θ2 − 4θ3 and α = 6 + 4θ − θ3. From A is an element in the field z + θA in the basis 1, θ, θ2, and θ3 of Q(θ), we obtain 2z + θA = a + bθ + cθ2 + dθ3  ; then, from c = 0. In short, we denote this fact by (2z + θA)2 = 0.From p-adic analysis, a suitable prime p is needed. Here, we take p = 7. A straightforward computation shows thatαε0)2 ≡ 0(mod 7), (αε7)2 ≡ 0(mod 7), and (αεi)2≢0(mod 7) , we get k ≡ 0,7(mod 8).To use the k ≡ 0(mod 8); since ε8 ≡ 1(mod 7), we obtain ε8 = 1 + 7ξ; thenk = 8m. So we haveαξ)2 = (−30 + 30θ + 60θ2 + 30θ3)2≢0(mod 7), we get m = 0 and z = ±3 by working modulo 7r+2 on formula (r||m.(i) Let  formula , where 7k ≡ 7(mod 8), we havek = −1 + 8m. So we have(ii) When m = 0 and z = ±68.Similar deduction shows that Case 2. Secondly, we similarly consider the following equation:ε = −49 − 20θ + 3θ2 + (7/2)θ3 and α = 2 + θ. By the same argument we choose p = 23 and ε11 ≡ 1(mod 23). Similarly we have k ≡ 0,1(mod 11); then we can deduce z = ±1 and z = ±91, respectively.Case 3. Consider the following equation:ε = −85 + 34θ + 30θ2 + 38θ3, α = 4 + θ3, p = 5, and ε4 ≡ 1(mod 5). Direct deductions show that z = ±2 and z = ±11798.Case 4. The final equation isε = 1 − 2θ − θ2 − (1/2)θ3, α = 12 − θ, p = 13, and ε5 ≡ 1(mod 13). A similar deduction yields z = 0 and z = ±6.x, y, z) is an integral solution of  = , , , , , and .All in all, if ±z=tituting  into theThis completes the proof of the corollary.a ≤ 10 and 1 ≤ b ≤ 10 are solved and results are listed in z-value of solutions .Like before, we can solve a family of systems of Diophantine equations . As a di


Mm1 of meromorphic spirallike functions. Such results as integral representations, convolution properties, and coefficient estimates are proved. The results presented here would provide extensions of those given in earlier works. Several other results are also obtained.We introduce and investigate a new subclass  M denote the class of functions f of the formLet f, g ∈ M, where f is given by  and MK* the subclasses of f ∈ M which are defined, respectively, byθ = 0 in (f ∈ M consisting of meromorphic functions which are starlike and convex of order γ  (0 ≤ γ < 1), respectively. For some recent investigations on meromorphic spirallike functions and related topics, see, for example, the earlier works f, we introduce the following class of meromorphic functions.Using the operator θ| < π/2, 0 ≤ λ < 1/2, and η > 1, let Mml denote a subclass of M consisting of functions satisfying the condition thatHmlf is given by  becomes the class M(η).We note that, for Mml.In the present paper, we aim at proving some interesting properties such as integral representations, convolution properties, and coefficient estimates for the class The following lemma will be required in our investigation.An}n=1∞ is defined bySuppose that the sequence {n ≥ 2, we deduce from .We begin by proving the following integral representation for the class f ∈ Mml. Thenω is analytic in U with ω(0) = 0 and |ω(z)| < 1.Let f ∈ Mml andτ ∈ P, which impliesω is analytic in U with ω(0) = 0 and |ω(z)| < 1. We find from .Next, we derive a convolution property for the class ξ ∈ C and |ξ| = 1. Then f ∈ Mml if and only ifLet f ∈ Mml if and only ifFrom the definition , we knowind from  thatz.Now, we discuss the coefficient estimates for functions in the class f ∈ Mml. ThenSuppose that f ∈ Mml. Then there exists τ ∈ P such thatzn in both sides of −2λf ∈ Mws from (eiθ(1−2λ)mbining (eiθ(1−2λ)ides of (2eiθ(1−λ),β;l;m),|an|≤2(η−follows:A1=(1−2λ)follows:A1=(1−2λ) we have|an|≤An .In what follows, we present some sufficient conditions for functions belonging to the class ζ be a real number with 0 ≤ ζ < 1. If f ∈ M satisfies the conditionf ∈ Mml provided thatLet ω is analytic in U with ω(0) = 0 and |ω(z)| < 1. Thus, we haveθ ≥ (1 − ζ)/(η − 1). This completes the proof of From , it follζ = 1 − (η − 1)cosθ in If we take f ∈ M satisfies the inequalityf ∈ Mml.If f ∈ M given by .If a function given by  satisfieη − 1)cosθ, ifIn virtue of