On Approximation Lower Bounds for TSP with Bounded Metrics Richard Schmied† arXiv:1201.5821v2 [cs.CC] 8 Aug 2012 Marek Karpinski∗ Abstract We develop a new method for proving explicit approximation lower bounds for TSP problems with bounded metrics improving on the best up to now known bounds. They almost match the best known bounds for unbounded metric TSP problems. In particular, we prove the best known lower bound for TSP with bounded metrics for the metric bound equal to 4. 1 Introduction We give first the basic definitions and an overview of the known results. Traveling Salesperson (TSP) Problem We are given a metric space (V, d) and the task consists of constructing a shortest tour visiting each vertex exactly once. The TSP problem in metric spaces is one of the most fundamental NP-hard optimization problems. The decision version of this problem was shown early to be NP-complete by Karp [K72]. Christofides [C76] gave an algorithm approximating the TSP problem within 3/2, i.e., an algorithm that produces a tour with length being at most a factor 3/2 from the optimum. As for lower bounds, a reduction due to Papadimitriou and Yannakakis [PY93] and the PCP Theorem [ALM+ 98] together imply that there exists some constant, not better than 1 + 10−6 , such that it is NP-hard to approximate the TSP problem with distances either one or two. For discussion of bounded metrics TSP, see also [T00]. This hardness result was improved by Engebretsen [E03], who proved that it is NP-hard to approximate the TSP problem restricted to distances one and two with an approximation factor better than 5381/5380 (1.00018). Böckenhauer and Seibert [BS00] studied the TSP problem with distances one, two and three, and Dept. of Computer Science and the Hausdorff Center for Mathematics, University of Bonn. Supported in part by DFG grants and the Hausdorff Center grant EXC59-1. Email: marek@cs.uni-bonn.de † Dept. of Computer Science, University of Bonn. Work supported by Hausdorff Doctoral Fellowship. Email: schmied@cs.uni-bonn.de ∗ 1 obtained an approximation lower bound of 3813/3812 (1.00026). Then, Papadimitriou and Vempala [PV06] showed that approximating the general problem with a constant approximation factor better than 220/219 (1.00456) is NP-hard. Recently, this bound was improved to 185/184 (1.00543) by Lampis [L12]. The restricted version of the TSP problem, in which the distance function takes values in {1, . . . , B}, is referred to as the (1, B)–TSP problem. The (1, 2)–TSP problem can be approximated in polynomial time with an approximation factor 8/7 due to Berman and Karpinski [BK06]. On the other hand, Engebretsen and Karpinski [EK06] proved that it is NPhard to approximate the (1, B)–TSP problem with an approximation factor less than 741/740 (1.00135) for B = 2 and 389/388 (1.00257) for B = 8. In this paper, we prove that the (1, 2)–TSP and the (1, 4)–TSP problem are NPhard to approximate with an approximation factor less than 535/534 and 337/336, respectively. Asymmetric Traveling Salesperson (ATSP) Problem We are given an asymmetric metric space (V, d), i.e., d is not necessarily symmetric, and we would like to construct a shortest tour visiting every vertex exactly once. The best known algorithm for the ATSP problem approximates the solution within O( log n/ log log n), where n is the number of vertices in the metric space [AGM+ 10]. On the other hand, Papadimitriou and Vempala [PV06] proved that the ATSP problem is NP-hard to approximate with an approximation factor less than 117/116 (1.00862). It is conceivable that the special cases with bounded metric are easier to approximate than the cases when the distance between two points grows with the size of the instance. Clearly, the (1, B)–ATSP problem, in which the distance function is taking values in the set {1, . . . , B}, can be approximated within B by just picking any tour as the solution. When we restrict the problem to distances one and two, it can be approximated within 5/4 due to Bläser [B04]. Furthermore, it is NP-hard to approximate this problem with an approximation factor better than 321/320 [BK06]. For the case B = 8, Engebretsen and Karpinski [EK06] constructed a reduction yielding the approximation lower bound 135/134 for the (1, 8)–ATSP problem. In this paper, we prove that it is NP-hard to approximate the (1, 2)–ATSP and the (1, 4)–ATSP problem with an approximation factor less than 207/206 and 141/140, respectively. Maximum Asymmetric Traveling Salesperson (MAX-ATSP) Problem We are given a complete directed graph G and a weight function w assigning each edge of G a nonnegative weight. The task is to find a tour of maximum weight visiting every vertex of G exactly once . This problem is well-known and motivated by several applications (cf. [BGS02]). A good approximation algorithm for the MAX–ATSP problem yields 2 a good approximation algorithm for many other optimization problems such as the Shortest Superstring problem, the Maximum Compression problem and the (1, 2)–ATSP problem. The MAX–(0, 1)–ATSP problem is the restricted version of the MAX-ATSP problem, in which the weight function w takes values in the set {0, 1}. Vishwanathan [V92] constructed an approximation preserving reduction proving that any (1/α)–approximation algorithm for the MAX–(0, 1)–ATSP problem problem transforms in a (2−α)– approximation algorithm for the (1, 2)–ATSP problem. Due to this reduction, all negative results concerning the approximation of the (1, 2)–ATSP problem imply hardness results for the MAX–(0, 1)–ATSP problem. Since the (1, 2)–ATSP problem is APX-hard [PY93], there is little hope for polynomial time approximation algorithms with arbitrary good precision for the MAX–(0, 1)–ATSP problem. Due to the explicit approximation lower bound for the (1, 2)–ATSP problem given in [EK06], it is NP-hard to approximate the MAX– (0, 1)–ATSP problem with an approximation factor less than 320/319. The best known approximation algorithm for the restricted version of this problem is due to Bläser [B04] and achieves an approximation ratio 5/4. For the general problem, Kaplan et al. [KLS+ 05] designed an algorithm for the MAX–ATSP problem yielding the best known approximation upper bound of 3/2. On the approximation hardness side, Karpinski and Schmied [KS11] constructed a reduction yielding the approximation lower bound 208/207 for the MAX– ATSP problem. In this paper, we prove that approximating the MAX–(0, 1)–ATSP problem with an approximation ratio less than 206/205 is NP-hard. Overview of Known Explicit Approximation Lower Bounds and Our Results (1, B)–ATSP problem Previously known results B=2 B=4 B=8 unbounded 321/320 321/320 135/134 117/116 (1.00312) (1.00312) (1.00746) (1.00862) [EK06] [EK06] [EK06] [PV06] Our results 207/206 141/140 (1.00485) (1.00714) B=2 B=4 B=8 unbounded 741/740 741/740 389/388 220/219 (1.00135) (1.00135) (1.00257) (1.00456) [EK06] [EK06] [EK06] [PV06] (1, B)–TSP problem Previously known results Our results 535/534 337/336 337/336 (1.00187) (1.00297) (1.00297) Figure 1: Known explicit approximation lower bounds and the new results. 3 MAX–(0, 1)–ATSP problem Previously known 320/319 results (1.00314) [EK06] Our results 206/205 (1.00487) MAX–ATSP problem 208/207 (1.00483) [KS11] 206/205 (1.00487) Figure 2: Known explicit approximation lower bounds and the new results. 2 Preliminaries In this section, we define the abbreviations and notations used in this paper. Given a natural number k and a finite set V , we use the abbreviation [k] for {1, . . . , k} and (V2 ) for the set { S ⊆ V ∣ ∣S∣ = 2}. Given an asymmetric metric space (V, d) with V = {v1 , . . . , vn } and d ∶ V × V → R≥0 , a Hamiltonian cycle in V or a tour in V is a cycle visiting each vertex vi in V exactly once. In the remainder, we specify a tour σ in V by σ = {(v1 , vi1 ), . . . , (vin−1 , v1 )} ⊆ V × V or explicitly by σ = v1 Ð→ vi1 Ð→ ⋯ Ð→ vin−1 Ð→ v1 . Given a tour σ in (V, d), the length of a tour ℓ(σ) is defined by ℓ(σ) = ∑ d(a). a∈σ Analogously, given a metric space (V, d) with V = {v1 , . . . , vn } and d ∶ (V2 ) → R>0 , we specify a tour σ in V by σ = { {v1 , vi1 }, . . . , {vin−1 , v1 } } ⊆ (V2 ) or explicitly by σ = v1 − vi1 − ⋯ − vin−1 − v1 . In order to specify, an instance (V, d) of the (1, 2)–ATSP problem, it suffices to identify the arcs a ∈ V × V with weight one. The same instance is specified by a directed graph D = (V, A), where a ∈ A if and only if d(a) = 1. Analogously, in the (1, 2)–TSP problem, an instance is completely specified by a graph G = (V, E). In the remainder, we refer to an arc and an edge with weight x ∈ N as a x-arc and x-edge, respectively. 3 Related Work 3.1 Hybrid Problem Berman and Karpinski [BK99], see also [BK01] and [BK03] introduced the following Hybrid problem and proved that this problem is NP-hard to approximate with some constant. Definition 1 (Hybrid problem). Given a system of linear equations mod 2 containing n variables, m2 equations with exactly two variables, and m3 equations with exactly 4 three variables, find an assignment to the variables that satisfies as many equations as possible. In [BK99], Berman and Karpinski constructed special instances of the Hybrid problem with bounded occurrences of variables, for which they proved the following hardness result. y9 y10 y11 y12 y13 y7 y14 y6 y15 y5 y16 y17 y4 y3 y18 y2 y19 y1 y20 y21 y8 x9 x10 x11 x12 x13 x7 x14 x6 x15 x5 x16 x4 x17 x3 x18 x2 x19 x1 x21 x20 x8 hyperedge e z8 z9 z10 z11 z7 z6 z5 z4 z3 z2 z1 z21 z20 z12 z13 z14 z15 z16 z17 z18 z19 Figure 3: An example of a Hybrid instance with circles C x , C y , C z , and hyperedge e = {z7 , y21 , x14 }. Theorem 1 ([BK99]). For any constant ǫ > 0, there exists instances of the Hybrid problem with 42ν variables, 60ν equations with exactly two variables, and 2ν equations with exactly three variables such that: (i) Each variable occurs exactly three times. (ii) Either there is an assignment to the variables that leaves at most ǫν equations unsatisfied, or else every assignment to the variables leaves at least (1 − ǫ)ν equations unsatisfied. (iii) It is NP-hard to decide which of the two cases in item (ii) above holds. The instances of the Hybrid problem produced in Theorem 1 have an even more special structure, which we are going to describe. The equations containing three variables are of the form x⊕y ⊕z = {0, 1}. These equations stem from the Theorem of Håstad [H01] dealing with the hardness of approximating equations with exactly three variables. We refer to it as the MAX– E3–LIN problem, which can be seen as a special instance of the Hybrid problem. 5 Theorem 2 (Håstad [H01]). For any constant δ ∈ (0, 12 ), there exists systems of linear equations mod 2 with 2m equations and exactly three unknowns in each equation such that: (i) Each variable in the instance occurs a constant number of times, half of them negated and half of them unnegated. This constant grows as Ω(21/δ ). (ii) Either there is an assignment satisfying all but at most δ ⋅ m equations, or every assignment leaves at least (1 − δ)m equations unsatisfied. (iii) It is NP-hard to distinguish between these two cases. For every variable x of the original instance E3 of the MAX-E3-LIN problem, Berman and Karpinski introduced a corresponding set of variables Vx . If the variable x occurs tx times in E3 , then, Vx contains 7tx variables x1 , . . . , x7tx . The variables contained in Con(Vx ) = {xi ∣ i ∈ {7ν ∣ ν ∈ [tx ]}} are called contact variables, whereas the variables in C(Vx ) = Vx /Con(Vx ) are called checker variables. All variables in Vx are connected by equations of the form xi ⊕ xi+1 = 0 with i ∈ [7tx − 1] and x1 ⊕ x7tx = 0. In addition to it, there exists equations of the form xi ⊕ xj = 0 with {i, j} ∈ M x , where M x defines a perfect matching on the set of checker variables. In the remainder, we refer to this construction as the circle C x containing the variables x ∈ Vx . Let E3 be an instance of the MAX–E3–LIN problem and H be its corresponding instance of the Hybrid problem. We denote by V (E3 ) the set of variables which occur in the instance E3 . Then, H can be represented graphically by ∣V (E3 )∣ circles C x with x ∈ V (E3 ) containing the variables V (C x ) = {x1 , . . . , xtx } as vertices. The edges are identified by the equations included in H. The equations with exactly three variables are represented by hyperedges e with cardinality ∣e∣ = 3. The equations xi ⊕ xi+1 = 0 induce a cycle containing the vertices {x1 , . . . , xtx } and the matching equations xi ⊕ xj = 0 with {i, j} ∈ M x induce a perfect matching on the set of checker variables. An example of an instance of the Hybrid problem is depicted in Figure 3. In summary, we notice that there are four type of equations in the Hybrid problem (i) the circle equations xi ⊕ xi+1 = 0 with i ∈ [7tx − 1], (ii) circle border equations x1 ⊕ x7tx , (iii) matching equations xi ⊕ xj = 0 with {i, j} ∈ M x , and (iv) equations with three variables of the form x ⊕ y ⊕ z = {0, 1}. In the remainder, we may assume that equations with three variables are of the form x⊕y ⊕z = 0 or x̄⊕y ⊕z = 0 due to the transformation x̄⊕y ⊕z = 0 ≡ x⊕y ⊕z = 1. 3.2 Approximation Hardness of TSP Problems Reducing the Hybrid Problem to the (1, 2)–(A)TSP Problem Engebretsen and Karpinski [EK06] constructed an approximation preserving reduction from the Hybrid problem to the (1, 2)–ATSP problem to prove explicit approximation lower bounds for the latter problem. They introduced graphs (gadgets), which simulate variables, equations with two variables and equations with three variables. In particular, the graphs corresponding to equations of the form x ⊕ y ⊕ z = 0 and to variables are displayed in Figure 4 (a) and (b), respectively. 6 vc3 sc vc1 sc+1 vc2 (a) Graph for x ⊕ y ⊕ z = 0. (b) Variable graph. Figure 4: Gadgets used in [EK06]. For the graph corresponding to an equation with three variables, they proved the following statement. Proposition 1 ([EK06]). There is a Hamiltonian path from sc to sc+1 in Figure 4 (a) if and only if an even number of ticked edges is traversed. A similar reduction was constructed in order to prove explicit approximation lower bounds for the (1, 2)–TSP problem. The corresponding graphs are depicted in Figure 5. The graph contained in the dashed box in Figure 5 (b) will play a crucial role in our reduction and we refer to it as parity graph. In particular, our variable gadget consists only of a parity graph. In the reduction of the (1, 2)–TSP problem, the following statement was proved for the graph corresponding to equations of the form x ⊕ y ⊕ z = 0. Proposition 2 ([EK06]). There is a simple path from sc to sc+1 in Figure 5 (a) containing the vertices v ∈ {vc1 , vc2 } if and only if an even number of parity graphs is traversed. The former mentioned reductions combined with Theorem 1 yield the following explicit approximation lower bounds. Theorem 3 ([EK06]). It is NP-hard to approximate the (1, 2)–ATSP and the (1, 2)–TSP problem with an approximation ratio less than 321/320 (1.00312) and 741/740 (1.00135), respectively. Explicit Approximation Lower Bounds for the MAX–ATSP problem By replacing all edges with weight two of an instance of the (1, 2)–ATSP problem by edges of weight zero, we obtain an instance of the MAX–(0, 1)–ATSP problem, which relates the (1, 2)–ATSP problem to the MAX–ATSP problem in the following sense. 7 vc2 sc sc+1 vc1 (a) Graph corresponding to x ⊕ y ⊕ z = 0. (b) Variable graph. Figure 5: Gadgets used in [EK06] to prove approximation hardness of the (1, 2)– TSP problem. Theorem 4 ([V92]). An (1/α)–approximation algorithm for the MAX–(0, 1)–ATSP problem implies an (2 − α)–approximation algorithm for the (1, 2)–ATSP problem. This reduction transforms every hardness result addressing the (1, 2)–ATSP problem into a hardness result for the MAX–(0, 1)–ATSP problem. In particular, Theorem 3 implies the best known explicit approximation lower bound for the MAX– (0, 1)–ATSP problem. Corollary 1. It is NP-hard to approximate the MAX–(0, 1)–ATSP problem within any better than 320/319 (1.00314). 4 Our Contribution We now formulate our main results. Theorem 5. Suppose we are given an instance H of the Hybrid problem with n circles, m2 equations with two variables and m3 equations with exactly three variables with the properties described in Theorem 1. (i) Then, it is possible to construct in polynomial time an instance DH of the (1, 2)– ATSP problem with the following properties: (a) If there exists an assignment φ to the variables of H which leaves at most u equations unsatisfied, then, there exist a tour σφ in DH with length at most ℓ(σφ ) = 3m2 + 13m3 + n + 1 + u. 8 (b) From every tour σ in DH with length ℓ(σ) = 3m2 + 13m3 + n + 1 + u, we can construct in polynomial time an assignment ψσ to the variables of H that leaves at most u equations in H unsatisfied. (ii) Furthermore, it is possible to construct in polynomial time an instance (VH , dH ) of the (1, 4)–ATSP problem with the following properties: (a) If there exists an assignment φ to the variables of H which leaves at most u equations unsatisfied, then, there exist a tour σφ in (VH , dH ) with length at most ℓ(σφ ) = 4m2 + 20m3 + 2n + 2u + 2. (b) From every tour σ in (VH , dH ) with length ℓ(σ) = 4m2 + 20m3 + 2n + 2u + 2, we can construct in polynomial time an assignment ψσ to the variables of H that leaves at most u equations in H unsatisfied. The former theorem can be used to derive an explicit approximation lower bound for the (1, 2)–ATSP problem by reducing instances of the Hybrid problem of the form described in Theorem 1 to the (1, 2)–ATSP problem. Corollary 2. It is NP-hard to approximate the (1, 2)–ATSP problem with an approximation factor less than 207/206 (1.00485). Proof. Let E3 be an instance of the MAX–E3–LIN problem. We define k to be the minimum number of occurences of a variable in E3 . According to Theorem 2, we 207 207−δ may choose δ > 0 such that 206+δ+ 7 ≥ 206 − ǫ holds. Given an instance E3 of the k MAX–E3–LIN problem with δ ′ ∈ (0, δ), we generate the corresponding instance H of the Hybrid problem. Then, we construct the corresponding instance DH of the (1, 2)-ATSP problem with the properties described in Theorem 5. We conclude according to Theorem 1 that there exist a tour in DH with length at most 3 ⋅ 60ν + 13 ⋅ 2ν + δ ′ ν + n + 1 ≤ (206 + δ ′ + 6+1 n+1 )ν ≤ (206 + δ ′ + )ν ν k or the length of a tour in DH is bounded from below by 3 ⋅ 60ν + 13 ⋅ 2ν + (1 − δ ′ )ν + n + 1 ≥ (206 + (1 − δ ′ ))ν ≥ (207 − δ ′ )ν. From Theorem 1, we know that the two cases above are NP-hard to distinguish. Hence, for every ǫ > 0, it is NP-hard to find a solution to the (1, 2)-ATSP problem 207 207−δ′ with an approximation ratio 206+δ ′ + 7 ≥ 206 − ǫ. k Analogously, we derive the following statement. Corollary 3. It is NP-hard to approximate the (1, 4)–ATSP problem with an approximation factor less than 141/140 (1.00714). For the symmetric version of the problems, we construct reductions from the Hybrid problem with similar properties. Theorem 6. Suppose we are given an instance H of the Hybrid problem with n circles, m2 equations with two variables and m3 equations with exactly three variables with the properties described in Theorem 1. 9 (i) Then, it is possible to construct in polynomial time an instance GH of the (1, 2)– TSP problem with the following properties: (a) If there exists an assignment φ to the variables of H which leaves at most u equations unsatisfied, then, there exist a tour σφ in GH with length at most ℓ(σφ ) = 8m2 + 27m3 + 3n + 1 + u. (b) From every tour σ in GH with length ℓ(σ) = 8m2 + 27m3 + 3n + 1 + u, we can construct in polynomial time an assignment ψσ to the variables of H that leaves at most u equations in H unsatisfied. (ii) Furthermore, it is possible to construct in polynomial time an instance (VH , dH ) of the (1, 4)–TSP problem with the following properties: (a) If there exists an assignment φ to the variables of H which leaves at most u equations unsatisfied, then, there exist a tour σφ in (VH , dH ) with length at most ℓ(σφ ) = 10m2 + 36m3 + 6n + 2 + 2u. (b) From every tour σ in (VH , dH ) with length ℓ(σ) = 10m2 +36m3 +6n+2+2u, we can construct in polynomial time an assignment ψσ to the variables of H that leaves at most u equations in H unsatisfied. Analogously, we combine the former theorem with the explicit approximation lower bound for the Hybrid problem of the form described in Theorem 1 yielding the following approximation hardness result. Corollary 4. It is NP-hard to approximate the (1, 2)–TSP and the (1, 4)–TSP problem with an approximation factor less than 535/534 (1.00187) and 337/336 (1.00297), respectively. From Theorem 4 and Corollary 2, we obtain the following explicit approximation lower bound. Corollary 5. It is NP-hard to approximate the MAX–(0, 1)–ATSP problem with an approximation factor less than 206/205 (1.00487). 5 Approximation Hardness of the (1, 2)–ATSP problem Before we proceed to the proof of Theorem 5 (i), we describe the reduction from a high-level view and try to build some intuition. 5.1 Main Ideas As mentioned above, we prove our hardness results by a reduction from the Hybrid problem. Let E3 be an instance of the MAX-E3-LIN problem and H the corresponding instance of the Hybrid problem. Every variable xl in the original instance E3 introduces an associated circle Cl in the instance H as illustrated in Figure 3. 10 outer loop inner loop D6 D2 D1 D23 D13 D4 D3 D5 Figure 6: An illustration of the instance DH and a tour in DH . The main idea of our reduction is to make use of the special structure of the circles in H. Every circle Cl in H corresponds to a graph Dl in the instance DH of the (1, 2)-ATSP problem. Moreover, Dl is a subgraph of DH , which builds almost a cycle. An assignment to the variable xl will have a natural interpretation in this reduction. The parity of xl corresponds to the direction of movement in Dl of the underlying tour. The circle graphs of DH are connected and build together the inner loop of DH . Every variable xli in a circle Cl possesses an associated parity graph Pil (Figure 7) in Dl as a subgraph. The two natural ways to traverse a parity graph will be called 0/1-traversals and correspond to the parity of the variable xli . Some of the parity graphs in Dl are also contained in graphs Dc3 (Figure 4(a) and Figure 9 for a more detailed view) corresponding to equations with three variables of the form gc3 ≡ x ⊕ y ⊕ z = 0. These graphs are connected and build the outer loop of DH . The whole construction is illustrated in Figure 6. The outer loop of the tour checks whether the 0/1-traversals of the parity graphs correspond to an satisfying assignment of the equations with three variables. If an underlying equation is not satisfied by the assignment defined via 0/1-traversal of the associated parity graph, it will be punished by using a costly 2-arc. 5.2 Constructing DH from a Hybrid Instance H Given a instance of the Hybrid problem H, we are going to construct the corresponding instance DH = (V (DH ), A(DH )) of the (1, 2)-ATSP problem. For every type of equation in H, we will introduce a specific graph or a specific way to connect the so far constructed subgraphs. In particular, we will distinguish between graphs corresponding to circle equations, matching equations, circle border equations and equations with three variables. First of all, we introduce graphs corresponding to the variables in H. 11 Variable Graphs Let H be an instance of the hybrid problem and Cl a circle in H. For every variable xli in the circle Cl , we introduce the parity graph Pil consisting of the vertices vil1 , vil– and vil0 depicted in Figure 7. vil0 vil– vil1 Figure 7: Parity graph Pil corresponding to the variable xli in circle Cl . Matching and Circle Equations Let H be an instance of the hybrid problem, Cl a circle in H and Ml the associated perfect matching. Furthermore, let xli ⊕ xlj = 0 with e = {i, j} ∈ Ml and i < j be a matching equation. Due to the construction of H, the circle equations xli ⊕ xli+1 = 0 and xlj ⊕ xlj+1 = 0 are both contained in Cl . Then, we introduce the associated parity vil0 vil– vil1 velj l0 vi+1 l– vi+1 l1 vi+1 vjl– vjl0 vel– l(i+1) ve l0 vj+1 l1 vj+1 l– vj+1 vjl1 Figure 8: Connecting the parity graph Pel l(i+1) graph Pel consisting of the vertices velj , vel– and ve . In addition to it, we connect l l l l l the parity graphs Pi , Pi+1 , Pj , Pj+1 and Pe as depicted in Figure 8. Graphs Corresponding to Equations with Three Variables Let gc3 ≡ xli ⊕ xsj ⊕ xkt = 0 be an equation with three variables in H. Then, we introduce the graph Dc3 (Figure 4) corresponding to the equation gc3. The graph Dc3 includes the vertices sc , vc1 , vc2 , vc3 and sc+1 . Furthermore, it contains the parity graphs Pel , Pbs and Pak as subgraphs, where e = {i, i+1}, b = {j, j+1} and a = {t, t+1}. Exemplary, we display Dc3 with its connections to the graph corresponding to the circle equation xli ⊕ xli+1 = 0 in Figure 9. In case of gc3 ≡ x̄li ⊕ xsj ⊕ xuk = 0, we connect the parity graphs with arcs (vil1 , vel0 ), l0 l0 ). (vi+1 , vil1 ) and (vel1 , vi+1 12 vc3 vil1 vil0 vak1 vil– vak– vak0 vel0 l1 vi+1 vc1 sc vel– vbs1 vel1 l– vi+1 sc+1 vbs– l0 vi+1 vbs0 vc2 Figure 9: The graph Dc3 corresponding to gc3 ≡ xli ⊕ xsj ⊕ xuk = 0 connected to graphs corresponding to xli ⊕ xli+1 = 0. Graphs Corresponding to Circle Border Equations vc1 bl+1 l– l1 v{1,n} v{1,n} vc2 l0 v{1,n} v1l1 v1l0 v2l0 l1 vn−1 v1l– vnl0 vblj vnl– vnl1 valn bl Figure 10: The graph corresponding to xl1 ⊕ xln = 0 Let Cl and Cl+1 be circles in H. In addition, let xl1 ⊕ xln = 0 be the circle border equation of Cl . Then, we introduce the vertex bl and connect it to v1l0 and vnl1 . Let bl+1 be the vertex corresponding to the circle Cl+1 . We draw an arc from v1l0 to l0 bl+1 . Finally, we connect the vertex v{n,1} to bl+1 . Recall that xn also occurs in the 3 equation gc , which is an equation with three variables in H. This construction is illustrated in Figure 10, where only a part of the corresponding graph Dc3 is depicted. Let Cn be the last circle in H. Then, we set bn+1 = s1 as s1 is the starting vertex of the graph D13 corresponding to the equation g13 . This is the whole description of the graph DH . Next, we are going to describe the associated tour σφ in DH given an assignment to the variables in H. 13 5.3 Constructing the Tour σφ from an Assignment φ Let H be an instance of the Hybrid problem and DH = (VH , AH ) the corresponding instance of the (1, 2)-ATSP problem as defined in Section 5.2. Given an assignment φ ∶ V (H) → {1, 0} to the variables of H, we are going to construct the associated Hamiltonian tour σφ in DH . In addition to it, we analyze the relation between the length of the tour σφ and the number of satisfied equations by φ. Let H be an instance of the Hybrid problem consisting of circles C1 , C2 , . . . , Cm and equations with three variables gj3 with j ∈ [m3 ]. The associated Hamiltonian tour σφ in DH starts at the vertex b1 . From a high level view, σφ traverses all graphs corresponding to the equations associated with the circle C1 ending with the vertex b2 . Successively, it passes all graphs for each circle in H until it reaches the vertex bm = s1 as s1 is the starting vertex of the graph D13 . At this point, the tour begins to traverse the remaining graphs Dc3 which are simulating the equations with three variables in H. By now, some of the parity graphs appearing in graphs Dc3 already have been traversed in the inner loop of σφ . The outer loop checks whether for each graph Dc3 , an even number of parity graphs has been traversed in the inner loop. In every situation, in which φ does not satisfy the underlying equation, the tour needs to use a 2-arc. This paths, consisting of 1-arcs, will be aligned by means of 2-arcs in order to build a Hamiltonian tour in DH . For each circle Cl , we will have to use 2-arcs in order to obtain a Hamiltonian path from bl to bl+1 traversing all graphs associated with Cl in some order except in the case when all variables in the circle have the same parity. Since we specify only a part of the tour σφ we rather refer to a representative tour from a set of tours having the same length and the same specification. In order to analyze the length of the tour in relation to the number of satisfied equation, we are going to examine the part of σφ passing the graphs corresponding to the underlying equation and account the local length to the analyzed parts of the tour. Let us begin to describe σφ passing through parity graphs associated to variables in H. vil0 vil– vil1 vil0 1-traversal of Pil given φ(xli ) = 1. vil– vil1 0-traversal of Pil given φ(xli ) = 0. Figure 11: Traversal of the graph Pil given the assignment φ. Traversing Parity Graphs Let xli be a variable in H. Then, the tour σφ traverses the parity graph Pil using l[1−φ(xli )] the path vi lφ(xli ) → vil– → vi . In the remainder, we call this part of the tour a 14 φ(xli )-traversal of the parity graph. In Figure 11, we depicted the corresponding traversals of the graph Pil given the assignment φ(xli ), where the traversed arcs are illustrated by thick arrows. In both cases, we associate the local length 2 with this part of the tour. l0 vi+1 vil1 l0 vi+1 vil1 velj velj vel– vel– l(i+1) ve l0 vj+1 l(i+1) ve l0 vj+1 vjl1 vjl1 (a) (b) Figure 12: 1. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 0. vil1 l0 vi+1 2 2 l0 vi+1 velj vil1 velj vel– vel– l(i+1) ve l(i+1) l0 vj+1 l0 vj+1 2 ve vjl1 vjl1 (a) (b) Figure 13: 2. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 0. Traversing Graphs Corresponding to Matching Equations Let Cl be a circle in H and xli ⊕ xlj = 0 with e = {i, j} ∈ Ml a matching equation. Given xi ⊕ xi+1 = 0, xi ⊕ xj = 0, xj ⊕ xj+1 = 0 and the assignment φ, we are going 15 to construct a tour through the corresponding parity graphs in dependence of φ. We begin with the case φ(xi )⊕φ(xi+1 ) = 0, φ(xi )⊕φ(xj ) = 0 and φ(xj )⊕φ(xj+1 ) = 0. 1. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 0: In this case, we traverse the corresponding parity graphs as depicted in Figure 12. In Figure 12 (a), we have φ(xi ) = φ(xi+1 ) = φ(xj ) = φ(xj+1 ) = 1, whereas in Figure 12 (b), we have φ(xi ) = φ(xi+1 ) = φ(xj ) = φ(xj+1 ) = 0. In both cases, this part of the tour has local length 5. 2. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 0: The tour σφ is pictured in Figure 13 (a) and (b). In the case φ(xi ) = φ(xi+1 ) = 0 and φ(xj ) = φ(xj+1 ) = 1 depicted in Figure 13 (a), we are forced to enter and leave the parity graph Pel via 2-arcs. So far, we associate the local length 6 with this part of the tour. In Figure 13 (b), we have φ(xi ) = φ(xi+1 ) = 1 and φ(xj ) = φ(xj+1 ) = 0. This part of the tour σφ contains one 2-arc yielding the local length 6. 3. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 1: In dependence of φ, we traverse the corresponding parity graphs in the way as depicted in Figure 14. l0 vi+1 vil1 l0 vi+1 vil1 velj velj vel– vel– l(i+1) l0 vj+1 2 2 l(i+1) l0 vj+1 ve ve vjl1 vjl1 2 (a) 2 (b) Figure 14: 3. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 1. The situation, in which φ(xi ) = φ(xi+1 ) = 1 and φ(xj ) = 1 ≠ φ(xj+1 ) holds, is depicted in Figure 14 (a). On the other hand, if we have φ(xi ) = φ(xi+1 ) = 0 and φ(xj ) = 0 ≠ φ(xj+1 ), the tour is pictured in Figure 14 (b). In both cases, we associate the local length 6. 4. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 1: The tour σφ is displayed in Figure 15. In Figure 15 (a), we are given φ(xi ) = 16 l0 vi+1 l0 vi+1 2 vil1 velj vil1 velj vel– vel– l(i+1) l0 vj+1 l(i+1) ve l0 vj+1 vjl1 ve vjl1 2 2 2 (a) (b) Figure 15: 4. Case φ(xi ) ⊕ φ(xi+1 ) = 0, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 1. φ(xi+1 ) = 1 and φ(xj ) ≠ φ(xj+1 ) = 1, whereas in Figure 15 (b), we have φ(xi ) = φ(xi+1 ) = 0 and φ(xj ) ≠ φ(xj+1 ) = 0. In both cases, we associate the local length 6 with this part of the tour. vil1 vil1 l0 vi+1 velj l0 vi+1 2 vel– vel– 2 l(i+1) l(i+1) ve l0 vj+1 ve l0 vj+1 2 velj vjl1 vjl1 2 (a) (b) Figure 16: 5. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 1, φ(xj ) ⊕ φ(xj+1 ) = 1. 5. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 1: In this case, we traverse the corresponding parity graphs as depicted in Figure 16. In Figure 16 (a), we notice that φ(xi ) ≠ φ(xi+1 ) = 0 and φ(xj ) ≠ φ(xj+1 ) = 1, whereas in (b), we have φ(xi ) ≠ φ(xi+1 ) = 1 and φ(xj ) ≠ φ(xj+1 ) = 0. This part of the tour has local length 7. 17 2 2 2 vil1 l0 vi+1 l0 vi+1 vil1 velj velj vel– vel– l(i+1) l0 vj+1 l(i+1) l0 vj+1 2 2 ve ve vjl1 2 2 vjl1 2 (a) (b) Figure 17: 6. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 0, and φ(xj ) ⊕ φ(xj+1 ) = 1. 6. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 1: In this case, we traverse the corresponding parity graphs as depicted in Figure 17. In Figure 17 (a), we have φ(xi ) ≠ φ(xi+1 ) = 0 and φ(xj ) ≠ φ(xj+1 ) = 0, whereas in (b), we have φ(xi ) ≠ φ(xi+1 ) = 1 and φ(xj ) ≠ φ(xj+1 ) = 1. This part of the tour has local length 7. 2 vil1 2 l0 vi+1 2 l0 vi+1 vil1 velj velj vel– l0 vj+1 vel– l(i+1) l(i+1) ve l0 vj+1 ve 2 vjl1 vjl1 (a) (b) Figure 18: 7. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 0. 7. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 0 and φ(xj ) ⊕ φ(xj+1 ) = 0: In this case, we traverse the corresponding parity graphs as depicted in Figure 18. In Figure 18 (a), we have φ(xi ) ≠ φ(xi+1 ) = 1 and φ(xj ) = φ(xj+1 ) = 0, whereas in (b), we have φ(xi ) ≠ φ(xi+1 ) = 0 and φ(xj ) = φ(xj+1 ) = 1. This part of the tour has 18 2 vil1 2 2 l0 vi+1 2 l0 vi+1 vil1 velj velj vel– l0 vj+1 vel– l(i+1) l(i+1) l0 vj+1 ve ve vjl1 vjl1 (a) (b) Figure 19: 8. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 0. local length 6. 8. Case φ(xi ) ⊕ φ(xi+1 ) = 1, φ(xi ) ⊕ φ(xj ) = 1 and φ(xj ) ⊕ φ(xj+1 ) = 0: In the final case, we traverse the corresponding parity graphs as depicted in Figure 19. In Figure 19 (a), we notice that φ(xi ) ≠ φ(xi+1 ) = 0 and φ(xj ) = φ(xj+1 ) = 0, whereas in (b), we have φ(xi ) ≠ φ(xi+1 ) = 1 and φ(xj ) = φ(xj+1 ) = 1. This part of the tour has local length 6. Our analysis yields the following proposition. Proposition 3. Let xli ⊕xli+1 = 0, xli ⊕xlj = 0 and xlj ⊕xlj+1 = 0 be equations in H. Given an assignment φ to the variables in H, the associated tour σφ has local length at most 5 + u, where u denotes the number of unsatisfied equations in {xli ⊕ xli+1 = 0, xli ⊕ xlj = 0, xlj ⊕ xlj+1 = 0} by φ. Traversing Graphs Corresponding to Equations with Three Variables Let gc3 ≡ xli ⊕ xsj ⊕ xkt = 0 be an equation with three variables in H. Furthermore, let xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0 and xkt ⊕ xkt+1 = 0 be circle equations in H. For notational simplicity, we introduce e = {i, i + 1}, a = {t, t + 1} and b = {j, j + 1}. In Figure 20, we display the construction involving the graphs Dc3 , Pak , Pbs , Pel , Pil and l l . Exemplary, we depicted the connections of the graphs Dc3 , Pel , Pil and Pi+1 Pi+1 in this figure. We are going to construct the tour σφ traversing the corresponding graphs and analyze the dependency of the local length of σφ and the number of satisfied participating equations. Recall from Proposition 1 that there is a Hamiltonian path from sc to sc+1 containing the vertices vc1 , vc2 and vc3 in Dc3 if and only if an even number of parity graphs P ∈ {Pak , Pbs , Pel } is traversed. The outer loop traverses the graph Dc3 starting at sc and ending at sc+1 . Further19 vc3 vil1 vil0 vak1 vil– vak– vak0 vel0 l1 vi+1 vc1 sc vel– vbs1 vel1 l– vi+1 sc+1 vbs– l0 vi+1 vbs0 vc2 l . Figure 20: The graph Dc3 with its connections to Pil and Pi+1 more, it contains the vertices vc1 , vc2 and vc3 in some order. If σφ traverses an even number of parity graphs P ∈ {Pak , Pbs , Pel } in the inner loop, it is possible to construct a Hamiltonian path with associated local length 3 ⋅ 3 + 4. In the other case, we have to use a 2-arc yielding the local length 14. vc3 vc2 vel0 vel– vel1 l1 vi+1 vil0 vil– l0 vi+1 vil1 l– vi+1 Figure 21: A part of the graphs corresponding to xli ⊕ xli+1 = 0 and xli ⊕ xsj ⊕ xkt = 0. Let us analyze the part of σφ traversing graphs corresponding to xli ⊕ xli+1 = 0. For this reason, we will examine the situation depicted in Figure 21. Let us begin with the case φ(xli ) ⊕ φ(xli+1 ) = 0. 1. Case φ(xli ) ⊕ φ(xli+1 ) = 0 : l0 ). Afterwards, the If φ(xli ) = φ(xli+1 ) = 1 holds, the tour σφ uses the arc (vil1 , vi+1 l parity graph Pe will be traversed when the tour leads through the graph Dc3 . More precisely, it will use the path vc3 → vel0 → vel– → vel1 → vc2 . In Figure 22, we illustrated this part of the tour. l0 → vel1 → vel– → vel0 → vil1 . In the other case φ(xli ) = φ(xli+1 ) = 0, we use the path vi+1 Afterwards, the tour σφ contains the arc (vc2 , vc3 ). In both cases, we associate the local length 1 with this part of the tour. 20 vc3 vc2 vel0 vel1 vel– l1 vi+1 vil0 vil– l0 vi+1 vil1 l– vi+1 Figure 22: Case φ(xli ) = φ(xli+1 ) = 1. 2. Case φ(xli+1 ) ⊕ φ(xli+1 ) = 1 : Assuming φ(xli ) ≠ φ(xli+1 ) = 1, the tour σφ uses a 2-arc entering vel1 and the path l0 . vel1 → vel– → vel0 → vil1 . Furthermore, we need another 2-arc in order to reach vi+1 The situation is depicted in Figure 23. vc3 vc2 vel0 vel– vel1 l1 vi+1 vil0 vil– 2 vil1 l0 l– vi+1 vi+1 2 Figure 23: Case φ(xli ) ≠ φ(xli+1 ) = 1. l0 and vil1 . In the other case, namely φ(xli ) ≠ φ(xli+1 ) = 0, we use 2-arcs leaving vi+1 Afterwards, the tour uses the path vc3 → vel0 → vel– → vel1 → vc2 while traversing the graph Dc3 . In both cases, we associate the local length 2 with this part of the tour. We obtain the following proposition. Proposition 4. Let xli ⊕ xsj ⊕ xkt = 0 be an equation with three variables in H. Furthermore, let xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0 and xkt ⊕ xkt+1 = 0 be circle equations in H. Given an assignment φ to the variables in H, the associated tour σφ has local length at most 3 ⋅ 3 + 4 + 3 + u, where u denotes the number of unsatisfied equations in {xli ⊕ xsj ⊕ xkt = 0, xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0, xkt ⊕ xkt+1 = 0} by φ. Traversing Graphs Corresponding to Circle Border Equations Let Cl be a circle in H and xl1 ⊕ xln = 0 its circle border equation. Recall that the variable xln is also included in an equation with three variables. We are going to describe the part of the tour passing through the graphs depicted in Figure 24 in dependence of the assigned values to the variables xl1 and xln . Let us start with the case φ(xl1 ) ⊕ φ(xln ) = 0. 21 vc3 l1 v{1,n} bl+1 l– v{1,n} vc2 l0 v{1,n} v1l1 v1l0 v2l0 l1 vn−1 v1l– vnl0 vblj vnl– vnl1 valn bl Figure 24: Traversing Graphs Corresponding To Circle Border Equations. 1. Case φ(xl1 ) ⊕ φ(xln ) = 0 The starting point of the tour σφ passing through the graph corresponding to xl1 ⊕ xln = 0 is the vertex bl . Given the values φ(xl1 ) = φ(xln ), we use in each case the φ(xl1 )-traversal of the parity graphs P1l and Pnl ending in bll+1 . Note that in the case l φ(xl1 ) = φ(xln ) = 0, we use the 1-traversal of the parity graph P{1,n} . Exemplary, we l l display the situation φ(x1 ) = φ(xn ) = 1 in Figure 25. vc3 l1 v{1,n} bl+1 l– v{1,n} vc2 l0 v{1,n} v1l1 v1l0 v2l0 l1 vn−1 v1l– vnl0 vblj vnl– vnl1 valn bl Figure 25: Case φ(xl1 ) = φ(xln ) = 1. In both cases, we associated the local length 2 with this part of the tour. 2. Case φ(xl1 ) ⊕ φ(xln ) = 1 Given the assignment φ(xl1 ) ≠ φ(xln ) = 0, we traverse the arc (bl , v1l0 ). Due to the construction, we have to use 2-arcs to enter bl+1 and vnl1 as depicted in Figure 26. In the other case, we have to use a 2-arc in order to leave the vertex bl . In l– l1 l0 → v{1,n} → bl+1 . addition, the tour contains the path vnl1 → v{1,n} → v{1,n} Hence, in both cases, we associate the local length 3 with this part of σφ . We obtain the following proposition. 22 vc3 2 l1 v{1,n} bl+1 l– v{1,n} vc2 l0 v{1,n} v1l1 v1l0 v2l0 l1 vn−1 v1l– vnl0 vblj vnl– vnl1 2 valn bl Figure 26: Case φ(xl1 ) ≠ φ(xln ) = 0. Proposition 5. Let xl1 ⊕ xln = 0 be an circle border equations in H. Given an assignment φ to the variables in H, the associated tour σφ has local length 2 if the equation xl1 ⊕ xln = 0 is satisfied by φ and 3, otherwise. 5.4 Constructing the Assignment ψσ from a Tour σ Let H be an instance of the Hybrid problem, DH = (VH , AH ) the associated instance of the (1, 2)-ATSP problem and σ a tour in DH . We are going to define the corresponding assignment ψσ ∶ V (H) → {0, 1} to the variables in H. In addition to it, we establish a connection between the length of σ and the number of satisfied equations by ψσ . In order to define an assignment, we first introduce the notion of consistent tours in DH . Definition 2 (Consistent Tour). Let H be an instance of the Hybrid problem and DH the associated instance of the (1, 2)–ATSP problem. A tour in DH is called consistent if the tour uses only 0/1-traversals of all in DH contained parity graphs. Due to the following proposition, we may assume that the underlying tour is consistent. Proposition 6. Let H be an instance of the Hybrid problem and DH the associated instance of the (1, 2)–ATSP problem. Any tour σ in DH can be transformed in polynomial time into a consistent tour σ ′ with ℓ(σ ′ ) ≤ ℓ(σ). Proof. For every parity graph contained in DH , it can be seen by considering all possibilities exhaustively that any tour in DH that is not using the corresponding 0/1-traversals can be modified into a tour with at most the same number of 2-arcs. The less obvious cases are shown in Figure 51 (see 10. Figure Appendix). In the remainder, we assume that the underlying tour σ is consistent with all parity graphs in DH . Let us now define the corresponding assignment ψσ given σ. 23 Definition 3 ( Assignment ψσ ). Let H be an instance of the Hybrid problem, DH = (VH , AH ) the associated instance of the (1, 2)-ATSP problem. Given a tour σ in DH , in which all parity graphs are consistent with respect to σ, the corresponding assignment ψσ ∶ V (H) → {0, 1} is defined as follows. ψσ (xli ) = 1 =0 if σ uses a 1-traversal of Pil otherwise We are going to analyze the local length of σ in dependence of the number of corresponding satisfied equations by ψσ . In some cases, we will have to modify the underlying tour improving in this way on the number of satisfied equations by the corresponding assignment ψσ . Let us start with the analysis. Transforming σ in Graphs Corresponding to Matching Equations Given the equations xi ⊕ xi+1 = 0, xi ⊕ xj = 0, xj ⊕ xj+1 = 0 and a tour σ, we are going to construct an assignment in dependence of σ. In particular, we analyze the relation between the length of the tour and the number of satisfied equations by ψσ . 1. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 0, ψσ (xi ) ⊕ ψσ (xj ) = 0 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 0: x z y 2 2 2 z l0 vi+1 velj vil1 velj vel– l0 vj+1 y l0 vi+1 2 vil1 x vel– 2 v l(i+1) ve v l(i+1) ve l0 vj+1 2 vjl1 vjl1 2 (a) (b) Figure 27: 1.Case ψσ (xi )⊕ψσ (xi+1 ) = 0, ψσ (xi )⊕ψσ (xj ) = 0 & ψσ (xj )⊕ψσ (xj+1 ) = 0. Given ψσ (xi ) = ψσ (xj ) = ψσ (xj ) = ψσ (xj+1 ) = 1, it is possible to transform the l0 l0 , velj , , vj+1 underlying tour such that no 2-arcs enter or leave the vertices vil1 , vi+1 l(i+1) ve and vjl1 . Exemplary, we display in Figure 27 such an transformation, where Figure 27 (a) and Figure 27 (b) illustrate the underlying tour σ and the transformed tour σ ′ , respectively. The case ψσ (xi ) = ψσ (xi+1 ) = ψσ (xj ) = ψσ (xj+1 ) = 0 24 vil1 l0 vi+1 2 2 l0 vi+1 velj vil1 velj vel– vel– l(i+1) ve l(i+1) l0 vj+1 l0 vj+1 ve vjl1 2 vjl1 (a) (b) Figure 28: 2.Case ψσ (xi )⊕ψσ (xi+1 ) = 0, ψσ (xi )⊕ψσ (xj ) = 1 & ψσ (xj )⊕ψσ (xj+1 ) = 0. can be discussed analogously. In both cases, we obtain the local length 5 for this part of σ while ψσ satisfies all 3 equations. l0 vi+1 vil1 l0 vi+1 vil1 velj velj vel– vel– l(i+1) l0 vj+1 2 2 l(i+1) l0 vj+1 ve ve vjl1 vjl1 2 (a) 2 (b) Figure 29: 3.Case ψσ (xi )⊕ψσ (xi+1 ) = 0, ψσ (xi )⊕ψσ (xj ) = 0 & ψσ (xj )⊕ψσ (xj+1 ) = 1. 25 2 vil1 2 2 l0 vi+1 l0 vi+1 vil1 velj velj vel– vel– l(i+1) l0 vj+1 l(i+1) l0 vj+1 ve ve vjl1 2 2 2 2 vjl1 2 (a) (b) Figure 30: 4. Case ψσ (xi )⊕ψσ (xi+1 ) = 1, ψσ (xi )⊕ψσ (xj ) = 0 & ψσ (xj )⊕ψσ (xj+1 ) = 1. 2. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 0, ψσ (xi ) ⊕ ψσ (xj ) = 1 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 0: Assuming ψσ (xi ) = ψσ (xi+1 ) = 1 and ψσ (xj ) = ψσ (xj+1 ) = 0, we are able to transl0 ) and (v l0 , v l1 ). Due to the conform the tour such that it uses the arcs (vil1 , vi+1 j+1 j struction and our assumption, the tour cannot traverse the arcs (vjl1 , velj ), (velj , vil1 ), l(i+1) l(i+1) l0 l0 , ve ). Consequently, we have to use 2-arcs entering and (ve , vj+1 ) and (vi+1 leaving the parity graph Pel . The situation is displayed in Figure 28 (a). We associate only the cost of one 2-arc yielding the local length 6, which corresponds to the fact that ψσ leaves the equation xi ⊕ xj = 0 unsatisfied. Note that a similar situation holds in case of ψσ (xi ) = ψσ (xi+1 ) = 0 and ψσ (xj ) = ψσ (xj+1 ) = 1 (cf. Figure 28 (b)). 3. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 0, ψσ (xi ) ⊕ ψσ (xj ) = 0 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 1: Let us start with the case ψσ (xi ) = ψσ (xi+1 ) = 1 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0. The situation is displayed in Figure 29 (a). Due to the construction, we are able to l0 ). Note that the tour cannot traverse transform σ such that it uses the arc (vil1 , vi+1 l(i+1) l0 l1 l0 the arcs (ve , vj+1 ) and (vj+1 , vj ). Hence, we are forced to use two 2-arcs increasing the cost by 2. All in all, we obtain the local length 6. The case ψσ (xi ) = ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 1 can be analyzed analogously (cf. Figure 29 (b)). A similar argumentation holds for ψσ (xi ) ⊕ ψσ (xi+1 ) = 1, ψσ (xi ) ⊕ ψσ (xj ) = 0 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 0. 4. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 1, ψσ (xi ) ⊕ ψσ (xj ) = 0 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 1: Given ψσ (xi ) ≠ ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0, we are able to transform the l(i+1) l0 tour such that it uses the arc (vi+1 , ve ). This situation is depicted in Figure 30 (a). Notice that we are forced to use four 2-arcs in order to connect all vertices. Consequently, it yields the local length 7. 26 The case, in which ψσ (xi ) ≠ ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0 holds, vil1 l0 vi+1 velj z 2 vel– l(i+1) ve l0 vj+1 l1 vj−1 vjl1 2 x (a) l0 vi+1 vil1 velj z vel– 2 l(i+1) ve l0 vj+1 l1 vj−1 vjl1 x (b) 2 Figure 31: 5.Case with ψσ (xi ) = ψσ (xi+1 ) = 1 and ψσ (xj ) ≠ ψσ (xj+1 ) = 1. is displayed in Figure 30 (b) and can be discussed analogously. 5. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 0, ψσ (xi ) ⊕ ψσ (xj ) = 1 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 1: Let the tour σ be characterized by ψσ (xi ) = ψσ (xi+1 ) = 1 and ψσ (xj ) ≠ ψσ (xj+1 ) = 1. l0 ). Then, we transform σ in such a way that we are able to use the arc (vil1 , vi+1 The corresponding situation is illustrated in Figure 31 (a). In order to change the value of ψσ (xj ), we transform the tour by traversing the parity graph Pjl in the other direction and obtain ψσ (xj ) = 1. This transformation induces a tour with at most the same cost. On the other hand, the corresponding assignment ψσ satisfies at least 2 − 1 more equations since xlj ⊕ xlj−1 = 0 might get unsatisfied. In this case, we associate the local costs of 6 with σ. In the other case, in which ψσ (xi ) = ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0 27 vil1 l0 vi+1 velj vel– velk′ l(i+1) ve l0 vj+1 vjl1 (a) 2 2 z x vil1 l0 vi+1 velj vel– velk′ l(i+1) ve l0 vj+1 vjl1 z 2 2 x (b) Figure 32: 5.Case with ψσ (xi ) = ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0. holds, we may argue similarly. (a) − (b). The transformation is depicted in Figure 32 6. Case ψσ (xi ) ⊕ ψσ (xi+1 ) = 1, ψσ (xi ) ⊕ ψσ (xj ) = 1 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 1: Given a tour σ with ψσ (xi ) ≠ ψσ (xi+1 ) = 1 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0, we transform σ such that it traverses the parity graph Pjl in the opposite direction meaning l0 ψσ (xj ) = 0 (cf. Figure 33). This transformation enables us to use the arc (vj+1 , vjl1 ). Furthermore, it yields at least one more satisfied equation in H. In order to connect the remaining vertices, we are forced to use at least two 2-arcs. In summary, we associate the local length 7 with this situation in conformity with the at most 2 unsatisfied equations by ψσ . 28 l0 vi+1 vil1 velj 2 2 vel– l1 vj−1 l(i+1) ve l0 vj+1 vjl1 (a) l0 vi+1 vil1 2 velj vel– 2 l1 vj−1 l(i+1) l0 vj+1 ve vjl1 (b) Figure 33: 6.Case with ψσ (xi ) ≠ ψσ (xi+1 ) = 1 and ψσ (xj ) ≠ ψσ (xj+1 ) = 0. If we are given a tour σ with ψσ (xi ) ≠ ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 1, we obtain the situation displayed in Figure 34 (a). By applying local transformations without increasing the length of the underlying tour, we achieve the scenario depicted in Figure 34 (b). We argue that the associated local length of the tour is 7. The case, in which ψσ (xi ) ⊕ ψσ (xi+1 ) = 1, ψσ (xi ) ⊕ ψσ (xj ) = 1 and ψσ (xj ) ⊕ ψσ (xj+1 ) = 0 holds, can be discussed analogously. We obtain the following proposition. Proposition 7. Let xli ⊕ xli+1 = 0, xli ⊕ xlj = 0 and xlj ⊕ xlj+1 = 0 be equations in H. Then, it is possible to transform in polynomial time the given tour σ passing through the graphs corresponding to xli ⊕ xli+1 = 0, xli ⊕ xlj = 0 and xlj ⊕ xlj+1 = 0 such that it has local length 5 + u and the number of unsatisfied equations in {xli ⊕ xli+1 = 0, xli ⊕ xlj = 0, xlj ⊕ xlj+1 = 0} by ψσ is bounded from above by u. 29 l0 vi+1 vil1 velj 2 vel– 2 l0 vj+1 l(i+1) ve l1 vj−1 vjl1 (a) vil1 l0 vi+1 velj vel– 2 l(i+1) ve l0 vj+1 l1 vj−1 vjl1 2 (b) Figure 34: 6.Case with ψσ (xi ) ≠ ψσ (xi+1 ) = 0 and ψσ (xj ) ≠ ψσ (xj+1 ) = 1. Transforming σ in Graphs Corresponding to Equations With Three Variables Let gc3 ≡ xli ⊕ xsj ⊕ xrk = 0 be an equation with three variables in H. Furthermore, let Cl be a circle in H and xli ⊕ xli+1 = 0 a circle equation. For notational simplicity, we set e = {i, i + 1}. We are going to analyze the the number of satisfied equations l , Pel and Dc3 . by ψσ in relation to the local length of σ in the graphs Pil , Pi+1 l l First, we transform the tour traversing the graphs Pi , Pi+1 and Pel such that it uses the ψσ (xli )-traversal of Pel . Afterwards, due to the construction of Dc3 and Proposition 1, the tour can be transformed such that it has local length 3 ⋅ 3 + 4 if it passes an even number of parity graphs P ∈ {Pel , Pak , Pbs } by using a simple path through Dc3 , otherwise, it yields a local length of 13 + 1. Let us start with the case ψσ (xli ) = 1 and ψσ (xli+1 ) = 1. 1. Case ψσ (xli ) = 1 and ψσ (xli+1 ) = 1: l and P l with In Figure 35 (a) and (b), we display the tour passing through Pil , Pi+1 e 30 vc2 vc1 vel0 vel– vil0 vel1 l1 vi+1 2 vil1 vil– l0 vi+1 2 l– vi+1 (a) vc2 vc1 vel0 vel– vil0 vel1 l0 vi+1 vil1 vil– l– vi+1 l1 vi+1 (b) Figure 35: 1. Case ψσ (xli ) = 1 and ψσ (xli+1 ) = 1 ψσ (xli ) = 1 and ψσ (xli+1 ) = 1 before and after the transformation, respectively. vc1 vc2 vel0 vel– vel1 l1 vi+1 vil0 vil1 vil– l0 vi+1 2 l– vi+1 (a) vc2 vil0 (b) vil– vel0 vel– 2 vel1 vc1 l0 vi+1 vil1 l– vi+1 l1 vi+1 Figure 36: 2. Case ψσ (xli ) = 0 and ψσ (xli+1 ) = 0 It is possible to transform the tour σ without increasing the length such that it 31 l0 ). In the outer loop, the tour may use at least one of the traverses the arc (vil1 , vi+1 2 l0 l1 1 arcs (vc , ve ) and (ve , vc ) depending on the parity check in Dc3 . We associate the local length 1 with this part of the tour. vc2 vel0 vil0 vil– 2 (a) vc2 l1 vi+1 z 2 vil1 vc1 vel1 vel– l0 vi+1 l– vi+1 x vel0 vel– vc1 vel1 l1 vi+1 vil0 vil– vil1 2 (b) 2 z l0 vi+1 l– vi+1 x Figure 37: 3. Case ψσ (xli ) = 1, ψσ (xli+1 ) = 0 and ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 vc2 vel0 vel– vel1 vc1 l1 vi+1 l1 vi−1 vil0 vil– 2 y (a) l0 l– vi+1 vi+1 vil1 x vc2 vel0 vel– 2 vel1 vc1 l1 vi+1 l1 vi−1 vil0 vil– 2 (b) y l0 l– vi+1 vi+1 vil1 2 x Figure 38: 3. Case ψσ (xli ) = 1, ψσ (xli+1 ) = 0 and ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1. 2. Case ψσ (xli ) = 0 and ψσ (xli+1 ) = 0 : In Figure 36, we display the underlying scenario with ψσ (xli ) = 0 and ψσ (xli+1 ) = 0. 32 The transformed tour uses the 0-traversal of the parity graph Pel . The vertices vc2 and vc1 are connected via a 2-arc. We assign the local length 1 to this part of the tour. 3. Case ψσ (xli ) = 1 and ψσ (xli+1 ) = 0 : Let us assume that ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 holds. Hence, it is possible to transform the tour such that it uses the path vc2 → vel0 → vel– → vel1 → vc1 and thus, the 0-traversal of the parity graph Pel as displayed in Figure 37. vc2 vc1 vel0 vel– vel1 l1 vi+1 l1 vi−1 vil0 vil– 2 vil1 l0 l– vi+1 vi+1 2 y x (a) vc2 vc1 vel0 vel– vel1 l1 vi+1 l1 vi−1 vil0 vil– 2 vil1 x (b) l0 l– vi+1 vi+1 2 y Figure 39: 4. Case ψσ (xli ) = 0, ψσ (xli+1 ) = 1 and ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 In the other case, namely ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1, we will change the value of ψσ (xli ) achieving in this way at least 2 − 1 more satisfied equation. Let us examine the scenario in Figure 38. The tour uses the 0-traversal of the parity graph Pel , which enables σ to pass the parity check in Dc3 . In both cases, we obtain the local length 2 in conformity with the at most one unsatisfied equation by ψσ . 4. Case ψσ (xli ) = 0 and ψσ (xli+1 ) = 1 : Assuming ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 and the scenario depicted in Figure 39 (a), the tour will be modified such that the parity graphs Pil and Pel are traversed in the same direction. Since we have ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0, we are able to uncouple the parity graph Pel from the tour through Dc3 without increasing its length. We display the transformed tour in Figure 39 (b). Assuming ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1 and the scenario depicted in Figure 40 (a), we transform σ such that the parity graph Pel is traversed when σ is passing through Dc3 meaning vc2 → vel0 → vel– → vel1 → vc1 is a part of the tour. In addition, 33 vc2 vc1 vel0 vel– vel1 l1 vi+1 l1 vi−1 vil0 vil– 2 vil1 l0 l– vi+1 vi+1 2 y x (a) vc2 vc1 vel0 vel– vel1 l1 vi+1 l1 vi−1 vil0 vil– 2 (b) l0 l– vi+1 vi+1 vil1 2 x y Figure 40: 4. Case ψσ (xli ) = 0, ψσ (xli+1 ) = 1 and ψσ (xli ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1 we change the value of ψσ (xli ) yielding at least 2 − 1 more satisfied equations. The transformed tour is displayed in Figure 40 (b). In both cases, we associate the local length 2 with σ. On the other hand, ψσ leaves at most one equation unsatisfied. We obtain the following proposition. Proposition 8. Let gc3 ≡ xli ⊕ xsj ⊕ xrk = 0 be an equation with three variables in H. Furthermore, let xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0 and xrk ⊕ xrk+1 = 0 be circle equations in H. Then, it is possible to transform in polynomial time the given tour σ passing through the graph corresponding to xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0, xrk ⊕ xrk+1 = 0 and gc3 such that it has local length 4 + 3 ⋅ 3 + 3 + u and the number of unsatisfied equations in {xli ⊕ xli+1 = 0, xsj ⊕ xsj+1 = 0, xrk ⊕ xrk+1 = 0, gc3 } by ψσ is at most u. Transforming σ in Graphs Corresponding to Circle Border Equations Let Cl be a circle in H and xl1 ⊕ xln = 0 its circle border equation. Furthermore, let gc3 ≡ xln ⊕ xsj ⊕ xrk = 0 be an equation with three variables contained in H. We are going to transform a given tour σ passing through the graph corresponding to xl1 ⊕ xln = 0 such that it will have the local length 2 if xl1 ⊕ xln = 0 is satisfied by ψσ and 3, otherwise. In each case, we modify σ such that it uses a ψ(xln )-traversal of l P{1,n} . Afterwards, σ will be checked in Dc3 whether it passes the parity test. Let us begin with the analysis starting with the case ψσ (x1 ) = 0 and ψσ (xn ) = 0. 34 vc3 l– l1 v{1,n} v{1,n} bl+1 v1l1 v1l0 vc2 l0 v{1,n} 2 l1 vn−1 v2l0 v1l– vnl0 vblj vnl1 vnl– valn bl (a) 2 vc3 l1 v{1,n} bl+1 vc2 l– v{1,n} l0 v{1,n} v1l1 v1l0 l1 vn−1 v2l0 v1l– vnl0 vblj (b) vnl– vnl1 valn bl Figure 41: Case ψσ (x1 ) = 0 and ψσ (xn ) = 0 1. Case ψσ (x1 ) = 0 and ψσ (xn ) = 0: Let us assume that ψσ leaves gc3 unsatisfied meaning ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1. l– l0 l1 → v{1,n} → vc2 is a part In addition to it, we assume that the path vc3 → v{1,n} → v{1,n} l– l1 l0 → vc2 of σ. Notice that σ fails the parity check in Dc3 if vc3 → v{1,n} → v{1,n} → v{1,n} is not used by σ. First, we modify the tour such that it includes the arc (bl , v1l0 ). For the same reason, we may assume that vnl1 and bl+1 is connected via a 2-arc. We obtain the scenario depicted in Figure 41 (a). As for the next step, we transform l0 l1 , bl+1 ). Consequently, we use σ such that it contains the arcs (vnl1 , v{1,n} ) and (v{1,n} l the 1-traversal of the parity graph P{1,n} and connect vc3 and vc2 via a 2-arc. The modified tour is depicted in Figure 41 (b). l– l0 l1 → vc2 , If ψσ satisfies gc3 and σ contains the path vc3 → v{1,n} → v{1,n} → v{1,n} we modify σ in Dc3 such that it passes the parity test in Dc3 and contains the arc (vc2 , vc3 ). In both cases, we associate the local length 2 with this part of σ. 2. Case ψσ (x1 ) = 1 and ψσ (xn ) = 1: Let us assume that ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1 holds and σ contains the arc 35 2 vc3 y bl+1 vc2 l1 v{1,n} l0 v{1,n} x v1l1 v1l0 v2l0 2 l1 vn−1 v1l– vnl0 vblj vnl1 vnl– valn bl (a) vc3 y bl+1 2 vc2 l1 v{1,n} l0 v{1,n} x v1l1 v1l0 v2l0 l1 vn−1 v1l– vnl0 vblj (b) vnl– vnl1 valn bl Figure 42: 2. Case ψσ (x1 ) = 1 and ψσ (xn ) = 1 (vc2 , vc3 ). Given this scenario, we may assume that (bl , vnl1 ) is contained in σ due to a simple modification. Then, we are going to analyze the situation depicted in Figure 42 (a). We transform σ in the way described in Figure 42 (b). Afterwards, σ will be modified in Dc3 such that it uses a simple path in Dc3 failing the parity check. The case, in which ψσ (xli ) ⊕ ψσ (xli ) ⊕ ψσ (xli ) = 0 holds and σ contains the arc (vc2 , vc3 ), can be discussed similarly since σ passes the parity check by including the l– l0 l1 → v{1,n} → vc2 . → v{1,n} path vc3 → v{1,n} In both cases, we associate the length 2 with this part of σ. 3. Case ψσ (x1 ) = 0 and ψσ (xn ) = 1: Let us assume that ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1 holds and σ traverses the path l– l0 l1 → v{1,n} → vc2 . Then, we transform the tour σ such that it vc3 → v{1,n} → v{1,n} contains the arc (v1l0 , bl+1 ). Note that neither (bl , v1l0 ) nor (bl , vnl1 ) is included in the tour. Hence, σ contains a 2-arc to connect bl . The same holds for the vertex vnl1 . This situation is displayed in Figure 43 (a). We are going to invert the value of ψσ (xln ) such that ψσ satisfies gc3 and xl1 ⊕xln = 0. In this way, we gain at least 2 − 1 more satisfied equations. The corresponding transformation is pictured in Figure 43 (b). On the other hand, if we assume that ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 holds and σ 36 vc3 bl+1 l0 v{1,n} l1 v{1,n} v1l1 v1l0 vc2 l– v{1,n} l1 vn−1 vnl0 v2l0 v1l– vnl1 2 vnl– vblj valn x 2 bl (a) vc3 bl+1 y l0 v{1,n} l1 v{1,n} v1l1 v1l0 vc2 l– v{1,n} l1 vn−1 vnl0 v2l0 v1l– vnl1 vnl– vblj 2 valn (b) x 2 bl y vc3 bl+1 vc2 l– l1 v{1,n} v{1,n} l0 v{1,n} v1l1 v1l0 v2l0 l1 vn−1 vnl0 v1l– vnl– vblj vnl1 2 valn x (c) bl 2 y Figure 43: 3. Case ψσ (x1 ) = 0 and ψσ (xn ) = 1 l– l0 l1 → v{1,n} → vc2 , we modify the tour as depicted → v{1,n} traverses the path vc3 → v{1,n} in Figure 43 (c). Note that σ passes the parity check in Dc3 and therefore, the tour may use a simple path in Dc3 . We associate the local length 3 with σ in this case. 4. Case ψσ (x1 ) = 1 and ψσ (xn ) = 0: Let us assume that ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 1 holds and σ uses the arc (vc2 , vc3 ). Then, we transform the tour σ such that it contains the arc (bl , vnl1 ). We note l0 ) is included in the tour. For this reason, σ must that neither (v1l0 , bl+1 ) nor (vnl1 , v{1,n} 37 vc3 bl+1 v1l1 v1l0 2 v2l0 2 v1l– vnl0 vnl1 bl vc3 bl+1 v1l1 v2l0 vc2 l– l1 l0 v{1,n} v{1,n} v{1,n} x 2 l1 vn−1 vnl0 2 v1l– vnl– vnl1 valn vblj y (b) vnl– valn y v1l0 bl 3 bl+1 vc vc2 l– l1 l0 v{1,n} v{1,n} v{1,n} 2 v1l1 v1l0 2 (c) x l1 vn−1 vblj (a) vc2 l– l1 l0 v{1,n} v{1,n} v{1,n} v1l– y v2l0 x l1 vn−1 vnl0 vblj vnl– vnl1 valn bl Figure 44: 4. Case ψσ (x1 ) = 1 and ψσ (xn ) = 0 l0 use a 2-arc to connect v{1,n} . The same holds for the vertex v1l0 . The corresponding situation is displayed in Figure 44 (a). We modify the tour as displayed in Figure 44 (b) and obtain at least 2 − 1 more satisfied equations. On the other hand, if we assume that ψσ (xln ) ⊕ ψσ (xsj ) ⊕ ψσ (xrk ) = 0 holds, we l– l0 l1 → v{1,n} → vc2 as depicted in Figure 44 need to include the path vc3 → v{1,n} → v{1,n} (c). In both cases, we associate the local length 3 with this part of the tour. We obtain the following statement. 38 Proposition 9. Let Cl be a circle in H and xl1 ⊕ xln = 0 its circle border equation. Then, it is possible to transform in polynomial time the given tour σ passing through the graph corresponding to xl1 ⊕ xln = 0 such that it has local length 2 if xl1 ⊕ xln = 0 is satisfied by ψσ and 3, otherwise. Thus far, we are ready to give the proof of Theorem 5(i). 5.5 Proof of Theorem 5(i) Let H be an instance of the Hybrid problem consisting of n circles C1 , . . . , Cn , m2 equations with two variables and m3 equations with three variables gc3 with c ∈ [m3 ]. Then, we construct in polynomial time the corresponding instance DH = (V (DH ), A(DH )) of the (1, 2)-ATSP problem as described in Section 5.2. (a) Let φ be an assignment to the variables in H leaving u equations in H unsatisfied. According to Proposition 3 – 5, it is possible to construct in polynomial time the tour σφ with length ℓ(σφ ) ≤ 3 ⋅ m2 + (4 + 3 ⋅ 3) ⋅ m3 + n + 1 + u. (b) Let σ be a tour in DH with length ℓ(σ) = 3 ⋅ m2 + 13 ⋅ m3 + n + 1 + u. Due to Proposition 6 we may assume that σ uses only 0/1-traversals of every parity graph included in DH . According to Definition 3, we associate the corresponding assignment ψσ with the underlying tour σ. Recall from Proposition 7 – 9 that it is possible to convert σ in polynomial time into a tour σ ′ without increasing the length such that ψσ′ leaves at most u equations in H unsatisfied. 6 Approximation Hardness of the (1, 4)-ATSP Problem In order to prove the claimed hardness results for the (1, 4)-ATSP problem, we use the same construction described in Section 5.2 with the difference that all arcs in parity graphs have weight 1, whereas all other arcs contained in the directed graph DH obtain the weight 2. The induced asymmetric metric space (VH , dH ) is given by VH = V (DH ) and distance function defined by the shortest path metric in DH bounded by the value 4. In other words, given x, y ∈ VH , the distance between x and y in VH is dH (x, y) = min{length of a shortest path from x to y in DH , 4}. The only difficulty that remains is to prove that tours remain consistent. Thus, we have to prove that given a tour σ in VH , we are able to transform σ in polynomial time into a tour σ ′ , which uses only 0/1-traversals in parity graphs contained in DH , without increasing ℓ(σ). This statement can be proved by considering all possibilities exhaustively. Some cases are displayed in Figure 52 – Figure 54. We are ready to give the proof of Theorem 5 (ii). 39 6.1 Proof of Theorem 5 (ii) Given H an instance of the Hybrid problem consisting of n circles C1 , . . . , Cn , m2 equations with two variables and m3 equations with three variables gc3 with c ∈ [m3 ], we construct in polynomial time the associated instance (VH , d) of the (1, 4)ATSP problem. Given an assignment φ to the variables of H leaving u equations unsatisfied in H, then there exists a tour with length at most m2 ⋅(2+2)+m3 ⋅(3⋅4+2⋅4)+2⋅u+2(n+1). On the other hand, if we are given a tour σ in VH with length 4m2 +20m3 +2n+2+2⋅u, it is possible to transform σ in polynomial time into a tour σ ′ without increasing the length such that the associated assignment ψσ′ leaves at most u equations in H unsatisfied. 7 Approximation Hardness of the (1, 2)-TSP Problem In order to prove Theorem 6 (i), we apply the reduction method used in Section 5 to the (1, 2)-TSP problem. As for the parity gadget, we use the graph depicted in Figure 45 with its corresponding traversals. The traversed edges are pictured by thick lines. The parity graph Pil 1-traversal of Pil . 0-traversal of Pil . Figure 45: Traversal of the graph Pil given the assignment φ. l Pi+1 Pil l P{i,j} l Pj+1 Pjl Figure 46: Graphs corresponding to equations xli ⊕xlj = 0, xli ⊕xli+1 = 0 & xlj ⊕xlj+1 = 0. 40 Let H be an instance of the hybrid problem. Given a matching equation ⊕ xlj = 0 in H and the corresponding circle equations xli ⊕ xli+1 = 0 and l l l ⊕ xlj+1 = 0, we connect the associated parity graphs Pil , Pi+1 , P{i,j} , Pjl and Pj+1 as displayed in Figure 46. xli xlj For equations with three variables gc3 ≡ x ⊕ y ⊕ z = 0 in H, we use the graph G3c depicted in Figure 47. Recall from Proposition 2 that there is a simple path from sc to sc+1 in Figure 47 containing the vertices v ∈ {vc1 , vc2 } if and only if an even number of parity graphs is traversed. vc2 sc sc+1 vc1 Figure 47: The graph G3c corresponding to x ⊕ y ⊕ z = 0. Let Cl be a circle in H with variables {xl1 , . . . , xln }. For the circle border equation of Cl , we introduce the path pl = b1l − b2l − b3l . In addition, we connect b3l and b1l+1 to the parity graphs P1l and Pnl in a similar way as in the reduction from the Hybrid problem to the (1, 2)–ATSP problem. This is the whole description of the corresponding graph GH = (V (GH ), E(GH )). We are ready to give the proof of Theorem 6 (i). 7.1 Proof of Theorem 6 (i) Given H an instance of the Hybrid problem consisting of n circles C1 , . . . , Cn , m2 equations with two variables and m3 equations with three variables gc3 with c ∈ [m3 ], we construct in polynomial time the associated instance GH of the (1, 2)– TSP problem. Given an assignment φ to the variables of H leaving u equations unsatisfied in H, then, there is a tour with length at most 8 ⋅ m2 + (3 ⋅ 8 + 3) ⋅ m3 + 3n + 1. 41 On the other hand, if we are given a tour σ in GH with length 8 ⋅ m2 + (3 ⋅ 8 + 3) ⋅ m3 + 3n + 1, it is possible to transform σ in polynomial time into a tour σ ′ such that it uses 0/1-traversals of all contained parity graphs in GH without increasing the length. Some cases are displayed in Figure 50. Moreover, we are able to construct in polynomial time an assignment to the variables of H, which leaves at most u equations in H unsatisfied. 8 Approximation Hardness of the (1, 4)-TSP Problem In order to prove the claimed approximation hardness for the (1, 4)-TSP problem, we cannot use the same parity graphs as in the construction in the previous section since tours are not necessarily consistent in this metric. For this reason, we introduce the parity graph depicted in Figure 48 with the corresponding traversals. Parity graph Pil 1-Traversal of Pil . 0-Traversal of Pil . Figure 48: 0/1-Traversals of the graph Pil . Pil l Pi+1 l P{i,j} l Pj+1 Pjl Figure 49: Graphs corresponding to xli ⊕ xlj = 0, xli ⊕ xli+1 = 0 and xlj ⊕ xlj+1 = 0. Given a matching equation xli ⊕ xlj = 0 in H and the circle equations xli ⊕ xli+1 = 0 and xlj ⊕ xlj+1 = 0, we connect the corresponding graphs as displayed in Figure 49. In order to define the new instance of the (1, 4)–TSP problem, we replace all parity graphs in GH by graphs displayed in Figure 48. In the remainder, we refer to this graph as HH . All edges contained in a parity graph have weight 1, whereas all other edges have weight 2. The remaining distances in the associated metric space VH are induced by the graphical metric in HH bounded by the value 4 meaning dH ({x, y}) = min{length of a shortest path from x to y in HH , 4}. 42 This is the whole description of the associated instance (VH , dH ) of the (1, 4)– TSP problem. We are ready to give the proof of Theorem 6 (ii). 8.1 Proof of Theorem 6 (ii) Given H an instance of the Hybrid problem consisting of n circles C1 , . . . , Cn , m2 equations with two variables and m3 equations with three variables gc3 with c ∈ [m3 ], we construct in polynomial time the associated instance (VH , dH ) of the (1, 4)–TSP problem. Given an assignment φ to the variables of H leaving u equations unsatisfied in H, there is a tour in VH with length at most m2 ⋅(2+8)+m3 ⋅(3⋅10+2⋅3)+6n+2+2⋅u. On the other hand, if we are given a tour σ in (VH , dH ) with length 10m2 + 36m3 + 6n + 2 + 2 ⋅ u, it is possible to transform σ in polynomial time into a tour σ ′ such that it uses 0/1-traversals of all contained parity graphs in GH without increasing the length. Some cases are displayed in Figure 55. Then, we are able to construct in polynomial time an assignment to the variables of H, which leaves at most u equations in H unsatisfied. 9 Further Research We have improved (modestly) the best known approximation lower bounds for TSP with bounded metrics problems. They almost match the best known bounds for the general (unbounded) metrics TSP problems. Because of a lack of the good definability properties of metric TSP, further improvements seem to be very difficult. A new less local method seems now to be necessary to achieve much stronger results. References [ALM+ 98] S. Arora, C. Lund, R. Motwani, M. Sudan and M. Szegedy, Proof Verification and the Hardness of Approximation Problems, J. ACM 45 (1998), pp. 501–555. [AGM+ 10] A. Asadpour, M. Goemans, A. Madry, S. Gharan and A. Saberi, An O(log n/ log log n)-Approximation Algorithm for the Asymmetric Traveling Salesman Problem, Proc. 21st SODA (2010), pp. 379–389. [BGS02] A. Barvinok, E. Gimadi and A. Serdyukov, The Maximum Traveling Salesman Problem, in: The Traveling Salesman Problem and Its Variations, pp. 585–607, G. Gutin and A. Punnen, eds., Kluwer, 2002. [BK99] P. Berman and M. Karpinski, On Some Tighter Inapproximability Results, Proc. 26th ICALP (1999), LNCS 1644, Springer, 1999, pp. 200 –209. [BK01] P. Berman and M. Karpinski, Efficient Amplifiers and Bounded Degree Optimization, ECCC TR01-053, 2001. 43 [BK03] P. Berman and M. Karpinski, Improved Approximation Lower Bounds on Small Occurrence Optimization, ECCC TR03-008, 2003. [BK06] P. Berman and M. Karpinski, 8/7-approximation algorithm for (1, 2)-TSP, 17th SODA (2006), pp. 641–648. [B04] M. Bläser, A 3/4-Approximation Algorithm for Maximum ATSP with Weights Zero and One, Proc. 8th APPROX-RANDOM (2004), pp. 61–71. [BS00] H.-J. Böckenhauer and S. Seibert, Improved Lower Bounds on the Approximability of the Traveling Salesman Problem, Theor. Inform. Appl. 34 (2000), pp. 213–255. [C76] N. Christofides, Worst-Case Analysis of a New Heuristic for the Traveling Salesman Problem, Technical Report CS-93-13, Carnegie Mellon University, Pittsburgh, 1976. [E03] L. Engebretsen, An Explicit Lower Bound for TSP with Distances One and Two, Algorithmica 35 (2003), pp. 301–318. [EK06] L. Engebretsen and M. Karpinski, TSP with Bounded Metrics, J. Comput. Syst. Sci. 72 (2006), pp. 509–546. [H01] J. Håstad, Some Optimal Inapproximability Results, J. ACM 48 (2001), pp. 798–859. [KLS+ 05] H. Kaplan, M. Lewenstein, N. Shafrir and M. Sviridenko, Approximation Algorithms for Asymmetric TSP by Decomposing Directed Regular Multigraphs, J. ACM 52 (2005), pp. 602–626. [K72] R. Karp, Reducibility Among Combinatorial Problems, Compl. of Computer Computations, pp. 85–103, 1972. [KS11] M. Karpinski and R. Schmied, Improved Lower Bounds for the Shortest Superstring and Related Problems, ECCC TR11-156, 2011. [L12] M. Lampis, Improved Inapproximability for TSP, CoRR abs/1206.2497, 2012. [PV06] C. Papadimitriou and S. Vempala, On the Approximability of the Traveling Salesman Problem, Combinatorica 26 (2006), pp. 101–120. [PY93] C. Papadimitriou and M. Yannakakis, The Traveling Salesman Problem with Distances One and Two, Math. Oper. Res. 18 (1993), pp. 1–11. [T00] L. Trevisan, When Hamming Meets Euclid: The Approximability of Geometric TSP and Steiner Tree, SIAM J. Comput. 30 (2000), pp. 475–485. [V92] S. Vishwanathan, An Approximation Algorithm for the Asymmetric Travelling Salesman Problem with Distances One and Two, Inf. Process. Lett. 44 (1992), pp. 297–302. 44 10 Figure Appendix 2 2 (a) (b) 2 2 (c) (d) 2 2 2 (e) (f ) 2 2 (g) (h) Figure 50: Transformations yielding a consistent tour. 45 2 2 (a) (b) 2 2 (c) (d) 2 2 2 2 (e) (f ) 2 2 2 2 (g) (h) Figure 51: Transformations yielding a consistent tour. 46 2 2 3 2 1 1 2 1 4 2 (a) (b) 2 4 4 2 4 1 4 1 2 2 4 2 (c) (d) 2 3 4 2 1 4 3 4 1 2 4 2 2 2 (e) (f ) 4 4 2 1 2 1 1 2 2 2 (g) (h) Figure 52: Transformations yielding a consistent tour. 47 2 2 4 4 2 1 4 3 1 3 4 2 2 2 2 (a) (b) 1 2 2 3 2 4 3 2 1 4 4 1 3 4 (c) (d) 2 4 2 2 1 2 1 3 1 2 (e) (f ) 2 4 2 2 1 4 1 1 2 2 (g) (h) Figure 53: Transformations yielding a consistent tour. 48 2 4 3 2 2 1 1 1 2 2 (a) (b) 1 2 2 3 2 4 3 2 1 4 4 1 3 4 (c) (d) 2 4 2 2 1 2 1 3 1 2 (e) (f ) 2 4 2 2 1 4 1 1 2 2 (g) (h) Figure 54: Transformations yielding a consistent tour. 49 2 3 2 2 2 4 2 (a) (b) 2 4 4 2 (c) (d) 4 2 2 2 2 2 4 4 2 (e) (f ) 4 4 2 2 2 4 2 (g) (h) Figure 55: Transformations yielding a consistent tour. 50