The skiving stock problem and its relation to hypergraph matchings

Discrete Optimization

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Discrete Optimization www.elsevier.com/locate/disopt

The skiving stock problem and its relation to hypergraph matchings J. Martinovic*, G. Scheithauer Institute of Numerical Mathematics, Technische Universität Dresden, 01062 Dresden, Germany

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Article history: Received 13 January 2017 Received in revised form 26 February 2018 Accepted 5 March 2018 Available online xxxx Keywords: Cutting and packing Skiving stock problem (Fractional) matching Hypergraph Additive integrality gap

abstract We consider the one-dimensional skiving stock problem which is strongly related to the dual bin packing problem: find the maximum number of products with minimum length L that can be constructed by connecting a given supply of m ∈ N smaller item lengths l1 , . . . , lm with availabilities b1 , . . . , bm . For this N P-hard optimization problem, we investigate the quality of the proper relaxation by considering the proper gap, i.e., the difference between the optimal objective values of the proper relaxation and the skiving stock problem itself. In this regard, we introduce a theory to obtain the proper gap on the basis of hypergraph matchings. As a main contribution, we characterize those hypergraphs that belong to an instance of the skiving stock problem, and consider the special case of 2-uniform hypergraphs in more detail. Moreover, this particular class is shown to correspond one-to-one and onto a certain power set. Based on this result, the number of related non-isomorphic hypergraphs and their possible proper gaps can be calculated explicitly. © 2018 Elsevier B.V. All rights reserved.

1. Introduction and preliminaries In this paper, we consider the one-dimensional skiving stock problem (SSP) [1,2] which is strongly related to the dual bin packing problem (DBPP) in literature (see e.g. [3–5]). In the classical formulation, m ∈ N := {1, 2, . . .} different item lengths l1 , . . . , lm with availabilities b1 , . . . , bm are given, the so-called item supply. We aim at maximizing the number of products with minimum length L that can be constructed by connecting the items on hand, see Fig. 1. Although representing similar problem statements, both denotations (DBPP and SSP) are separated in the following sense in literature1 : the term DBPP rather refers to highly heterogeneous input lengths, meaning that the quantities bi are very small (mostly even equal to one) for all i ∈ I := {1, . . . , m}, whereas larger values of bi are considered whenever the term SSP is used, in general (see the typology presented

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Corresponding author. E-mail addresses: [email protected] (J. Martinovic), [email protected] (G. Scheithauer). 1 This is the same differentiation as in the cutting context, where the bin packing problem and the cutting stock problem are considered. https://doi.org/10.1016/j.disopt.2018.03.001 1572-5286/© 2018 Elsevier B.V. All rights reserved.

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Fig. 1. A schematic for a skiving stock problem of small problem size.

by [6]). In other words, the DBPP focuses more on the individual character of each single item, and the SSP treats identical items (that are grouped together) collectively. Depending on the specific mathematical context both descriptions can be advantageous (as explained in the following sections). The SSP can be considered as a natural counterpart of the extensively studied one-dimensional cutting stock problem (CSP) [7–11], where larger items have to be cut into smaller ones such that a given demand is satisfied. Although both problems share a certain common structure (e.g. as regards their input data), they are not dual formulations in the sense of mathematical optimization. Hence, the SSP represents an independent challenge in the field of discrete optimization. For the case bi = 1 (i ∈ I), the considered optimization problem was firstly mentioned by Assmann et al. [3,12] and denoted by dual bin packing problem. Therein, the authors mainly investigate heuristic approaches and provide results regarding the quality and the average case behavior of the presented methods. Further contributions, especially in terms of exact approaches to the DBPP, have been studied in [4] and [5] where two branching algorithms are introduced. Based on practical preliminary thoughts [1], a generalization for larger availabilities bi ∈ N (i ∈ I) has been considered in [2] also introducing the term skiving stock problem. In that article, the author formulates a pattern-oriented model of the SSP with an infinite number of variables and provides first (numerical) results regarding the additive integrality gap of this optimization problem, i.e., the difference between the optimal values of the continuous relaxation of the SSP and the SSP itself. Further modeling approaches and detailed computational experiments have recently been presented in [13]. Despite being a relatively young field of research with only a limited number of publications, such computations are of high interest in many real world applications, e.g. industrial production processes [1,2], politico-economic problems [3,4] or wireless communications [14]. In particular, the SSP plays an important role whenever an efficient and sustainable use of given resources is desired. Referring to this, some main areas of applications are given by: • recycling offcuts in industrial scenarios [2], even within a holistic framework as a combined cuttingand-skiving procedure [1,15], • stimulating economic activity (e.g. in periods of recession) [4], • efficient allocation of wireless users in a given frequency range by means of spectrum-aggregation based resource allocation [16,17]. Furthermore, also neighboring tasks, such as dual vector packing problems [18,19] or the maximum cardinality bin packing problem [5,20], are often associated with the DBPP. These formulations are of practical use as well since they are applied in multiprocessor scheduling problems [21] or surgical case plannings [22]. ⊤ ⊤ Usually, the abbreviation E := (m, l, L, b) with l = (l1 , . . . , lm ) and b = (b1 , . . . , bm ) is used for an instance of the skiving stock problem. However, note that, whenever the individual character of each single Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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∑ item shall be highlighted, we can also apply the equivalent notation2 E = (n, l, L, e) with n = i∈I bi not necessarily distinct item lengths (for simplicity, also referred to as) l1 ≥ l2 ≥ · · · ≥ ln and e = (1, . . . , 1)⊤ ∈ Zn+ . The latter interpretation rather corresponds to the dual bin packing problem. In this article, we will apply both notations of an instance depending on the specific mathematical context. For the sake of simplicity, both of them will be referred to as instances of the skiving stock problem (or simply skiving stock instances) since the SSP represents the more common name in recent literature. However, in order to avoid misunderstandings, we clearly indicate which interpretation is actually used at various positions within the following sections. Without loss of generality, we assume all input data to be positive integers with L > max{li | i ∈ I}. Since the skiving stock problem is known to be N P-hard, a common (approximate) solution approach consists in considering a relaxed problem (for instance the continuous relaxation)3 and/or the application of appropriate heuristics. Obviously, the success of such techniques strongly depends on • the quality (i.e. the tightness) of the considered relaxation, and • the performance of the considered heuristic. This work is mainly devoted to the investigation of the first point mentioned above, where an appropriate measure of tightness is given by the absolute difference ∆(E) (called additive integrality gap or simply gap of the instance E) between the optimal values of the respective relaxation and the SSP itself. Practical experience and numerical simulations [2,13,23] have shown that there is only a very small gap pointing out the good quality of the continuous relaxation. Indeed, the largest known gap results to ∆(E) = 325/276 ≈ 1.1775 and has recently been published in [23]. It is even conjectured, see [2], that the skiving stock problem satisfies ∆(E) < 2 for any instance E (the so called modified integer round down property, see also Definition 5 in Section 2), but this result could only be obtained so far for some special classes of instances such as the divisible case [24, Sect. 3]. In this manuscript we will particularly focus on the proper relaxation which usually provides better bounds4 than the continuous relaxation, see [23]. After a short introduction to the skiving stock problem and the proper relaxation in the next section, we present an alternative interpretation of the skiving stock problem. Therefore, we reformulate the skiving stock problem and its proper relaxation as a matching problem in a hypergraph, and describe how the proper gap can be obtained within this new framework. Of course, also this approach cannot suppress the N P-hardness of the underlying problem, but it introduces an alternative method to investigate the proper gap of the skiving stock problem which possibly paves the way for further research activities in the future. Note that approaches based on graph theory have been investigated and successfully applied in different fields of discrete optimization. One of the earliest references is given by [25] where theoretical foundations on the relationship between certain integer linear programs and corresponding flow formulations have been proposed. Although their numerical behavior and benefits are not discussed in detail, some properties suggesting computational advantages of such formulations are presented therein. In the context of cutting and packing, arcflow models are a well known tool for the exact solution of the CSP [7,11] and the SSP [13]. In the last years, much research has been done to investigate and improve the corresponding models and algorithms, even with respect to related problems [26]. Observe that all these contributions focus on exact and improved modeling or solution techniques. Contrary to that, we will present an approach to measure the tightness of the proper relaxation by means of particular hypergraph properties which describes a completely new application of the graph theoretical view in cutting and packing. To our best knowledge such methods have not been dealt with in literature. 2 This representation of an instance will mainly be used when referring to the hypergraph context, i.e. from Section 3 onwards. In the preliminary section we will stick to the usual representation. 3 Since the SSP represents a maximization problem, relaxations usually lead to an upper bound for the optimal objective value. 4 This implies that, in most cases, the corresponding proper gap is smaller than the gap related to the continuous relaxation.

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As a main contribution we characterize all hypergraphs that correspond to an instance of the skiving stock problem, the so called skiving hypergraphs. Since the matching theory is particularly well-studied for ordinary graphs as a special case of hypergraphs we will deal with this case and derive a result for the proper gap of this class of instances. Within this framework we also compute the total number of non-isomorphic skiving hypergraphs by introducing a bijection to an appropriate power set. Moreover, some introductory approaches and observations to the more general case of (actual) hypergraphs are also discussed at the end of Section 4. 2. Preliminaries on the skiving stock problem Although it already has been mentioned in the introductory section of this manuscript, we start with a brief repetition of a very important definition. Definition 1. Let m ∈ N item types with (pairwise different) lengths li and availabilities bi (i ∈ I), and a length L ∈ N be given. The tuple E = (m, l, L, b) with l = (l1 , . . . , lm )⊤ ∈ Nm and b = (b1 , . . . , bm )⊤ ∈ Nm is called instance of the skiving stock problem. We may again emphasize that it is also possible to refer to an instance by means of E = (n, l, L, e), ∑ where n = i∈I bi items of possibly repeating lengths (instead of item types) are considered. Without loss of generality, we further assume that li < L holds for all i ∈ I since items larger than or equal to L already represent an object of desired length and do not have to be dealt with in the optimization. Definition 2. Any feasible arrangement of items leading to a final product of minimum length L is called (packing) pattern of E. ⊤

We always represent a pattern by a nonnegative vector a = (a1 , . . . , am ) ∈ Zm + , where ai ∈ Z+ denotes the number of items of type i ∈ I being contained in the considered pattern. For a given instance E, the set of all patterns is defined by ⏐ ⊤ { } ⏐ P (E) := a ∈ Zm + l a≥L . Definition 3. A pattern a ∈ P (E) is called minimal if there does not exist any pattern ˜ a ∈ P (E) such that ˜ a ̸= a and ˜ a ≤ a hold (componentwise). The set of all minimal patterns is denoted by P ⋆ (E). Let xj ∈ Z+ denote the number how often the ⊤ ⋆ ⋆ minimal pattern aj = (a1j , . . . , amj ) ∈ Zm + (j ∈ J ) of E is used, where J represents an index set of all minimal patterns of E. Example 1. The following patterns are chosen in the solution illustrated in Fig. 1: a1 = (1, 1, 1)⊤ , a2 = (1, 2, 0)⊤ , and a3 = (0, 1, 2)⊤ . Then the standard model of the skiving stock problem can be formulated as ⎧ ⎫ ⏐ ⎨∑ ⎬ ∑ ⏐ z ⋆ (E) = max xj ⏐⏐ aij xj ≤ bi , i ∈ I, xj ∈ Z+ , j ∈ J ⋆ . ⎩ ⎭ j∈J ⋆

(1)

j∈J ⋆

In most cases, it is not recommendable or even not possible to compute all minimal patterns prior to the optimization due to the huge cardinality of P ⋆ (E). Hence, problem (1) cannot be tackled directly by means Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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of standard ILP solvers. However, at least the continuous relaxation of (1), i.e., ⎫ ⎧ ⏐ ∑ ⎬ ⎨∑ ⏐ aij xj ≤ bi , i ∈ I, xj ≥ 0, j ∈ J ⋆ , xj ⏐⏐ zc⋆ (E) = max ⎭ ⎩ ⋆ ⋆ j∈J

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(2)

j∈J

can be solved efficiently within a column generation approach. Hence, some (approximate) solution approaches for the skiving stock problem are given by: • branch-and-bound procedures, where upper bounds can be computed based on (2), • applying appropriate heuristics to the solution of the continuous relaxation in order to obtain an approximate solution of (1), • considering alternative exact modeling approaches [13]. Note that the availability of reasonably tight bounds is a key issue when focusing on the first two ideas of the previous list. Mainly after the publication of the famous book by Nemhauser and Wolsey [27], this idea has been placed as a central concern in combinatorial optimization and IP modeling. Definition 4. Let E = (m, l, L, b) denote an instance of the skiving stock problem. Then the difference ∆(E) := zc⋆ (E) − z ⋆ (E)

(3)

is called (additive integrality) gap (of E). Note that, similar to the one-dimensional cutting stock problem [7,10,11,28], there are also other modeling approaches, e.g. the arcflow model and the onestick model, which have been introduced in [13]. Therein, the equivalence of the corresponding continuous relaxations is proved. This implies that, among these mentioned approaches, the gap is independent of the considered modeling framework. Definition 5. A set T of instances possesses the integer round-down property (IRDP), if ∆(E) < 1 holds for all E ∈ T , and it has the modified integer round-down property (MIRDP), if ∆(E) < 2 holds for all E ∈ T . Whenever T = {E} is a singleton, we briefly say that E, instead of {E}, has the (modified) integer round-down property. An instance E with ∆(E) ≥ 1 is called non-IRDP instance. Besides possessing theoretical importance these properties are of practical interest as well. For example, if the IRDP of an instance E is known, then the optimal value of the (hard) integer problem (1) can easily be obtained by solving the linear program (2) and applying z ⋆ (E) = ⌊zc⋆ (E)⌋. Similarly, the MIRDP of an instance E only allows the two possibilities z ⋆ (E) ∈ {⌊zc⋆ (E)⌋, ⌊zc⋆ (E)⌋ − 1} for the optimal value of the integer problem. However, the computation of a corresponding feasible integer solution may still be difficult in some cases. Practical experience and numerical simulations [2,13,23] have shown that there is only a very small gap pointing out the good quality of the continuous relaxation (2). However, in most cases, the set P ⋆ (E) contains patterns that cannot appear in a feasible integer solution due to the item limitations. Hence, we can restrict our investigations to the following subset of patterns: Definition 6. Let E = (m, l, L, b) be an instance of the skiving stock problem. A pattern a ∈ P (E) with a ≤ b is called proper pattern, and a pattern a ∈ P ⋆ (E) with a ≤ b is called minimal proper pattern. A pattern a ∈ P (E) with ai > bi for some i ∈ I is called non-proper. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Furthermore, we define ⏐ { } Pp (E) := a ∈ P (E) ⏐ a ≤ b

⏐ { } and Pp⋆ (E) := a ∈ P ⋆ (E) ⏐ a ≤ b

as the sets of all proper patterns and all minimal proper patterns, respectively. Moreover, let Jp⋆ := Jp⋆ (E) denote an index set of Pp⋆ (E). Since a feasible integer solution of E cannot contain any pattern from P (E) \ Pp (E), the restricted pattern set does not influence the optimal objective value of the ILP (1). But, for the continuous relaxation, this observation is not true in general. The proper relaxation is given by ⎫ ⎧ ⏐ ∑ ⎬ ⎨∑ ⏐ aij xj ≤ bi , i ∈ I, xj ≥ 0, j ∈ Jp⋆ . (4) xj ⏐⏐ zp⋆ (E) = max ⎭ ⎩ j∈Jp⋆

j∈Jp⋆

This relaxation usually provides tighter bounds for z ⋆ := z ⋆ (E) than (2), and may therefore be of special interest for the (approximate) solution strategies mentioned above. Moreover, note that the proper relaxation has also been proved beneficial in the context of one-dimensional cutting; see, for instance, [29] and [30]. Lemma 1. Let E = (m, l, L, b) be an instance of the skiving stock problem. Then zp⋆ (E) ≤ zc⋆ (E) holds. The reverse inequality is not true in general. Proof . Due to Pp⋆ (E) ⊆ P ⋆ (E) the inequality zp⋆ (E) ≤ zc⋆ (E) is clear by definition. Hence, it suffices to construct an example where zp⋆ (E) < zc⋆ (E) holds. For this purpose, consider the instance E = (3, (21, 14, 6), 42, (1, 2, 6)). Here we obtain z ⋆ (E) = 1 and zc⋆ (E) = 85/42. Note that the unique solution of the continuous relaxation is given by x1 = 1/2, x2 = 2/3, and x3 = 6/7 for the patterns a1 = (2, 0, 0)⊤ , a2 = (0, 3, 0)⊤ , and a3 = (0, 0, 7)⊤ . Obviously, none of these satisfies the condition aj ≤ b for j ∈ {1, 2, 3}. Therefore, they are not feasible for the proper relaxation, and, due to the uniqueness of the solution, the optimal objective value has to be smaller than zc⋆ (E) which proves the assertion. In fact, we have zp⋆ (E) = 78/42 which is, for instance, provided by x4 = 1, x5 = 30/42, and x6 = 6/42 for the patterns a4 = (1, 1, 2)⊤ , a5 = (0, 1, 5)⊤ , and a6 = (0, 2, 3)⊤ . □ As to be clearly seen, a main advantage of the proper relaxation is that it might provide a better approximation (compared to the continuous relaxation) of the optimal objective value of the integer problem (1). Additionally, in many cases, rounding approaches are used to obtain a sub-optimal feasible integer solution from the solution of a certain relaxation. The example in the proof shows, as a representative for the whole class of instances from [24, Theorem 14], that the continuous relaxation may provide the approximate integer solution z = 0 whereas better rounded solutions can be obtained by means of the proper relaxation. Due to these two advantages, a more detailed consideration of the proper relaxation is expected to be worthwhile. Definition 7. Let E = (m, l, L, b) be an instance of the skiving stock problem. Then, the difference ∆p (E) := zp⋆ (E) − z ⋆ (E)

(5)

is called proper gap (of E). As a direct consequence of Lemma 1, a first important observation is given by the following result. Lemma 2. Let E = (m, l, L, b) be an instance of the skiving stock problem. Then ∆p (E) ≤ ∆(E) holds. The reverse inequality is not true in general. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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This lemma has important implications as it tells us that every upper bound derived for the gap of an instance also holds for its proper gap. Definition 8. A set T of instances has the proper integer round-down property (proper-IRDP), if ∆p (E) < 1 holds for all E ∈ T , and it has the proper modified integer round-down property (proper-MIRDP), if ∆p (E) < 2 holds for all E ∈ T . Whenever T = {E} is a singleton, we briefly say that E, instead of {E}, has the proper (modified) integer round-down property. An instance E with ∆p (E) ≥ 1 is called non-proper-IRDP instance. In what follows, we briefly sum up the most important results that can be stated for the proper gap as a consequence of Lemma 2 and the considerations in [31]. Corollary 3. Let E = (m, l, L, b) be an instance of the skiving stock problem. Then ∆p (E) < max{2, m − 2} holds. Corollary 4. Let E = (m, l, L, b) be an instance of the skiving stock problem. Then ∆p (E) < ⌈m/2⌉ holds. Corollary 5. An instance E = (m, l, L, b) of the skiving stock problem possesses the proper-IRDP if m = 2, and it possesses the proper-MIRDP if m ∈ {2, 3, 4}. Corollary 6. Let E = (m, l, L, b) be an instance of the divisible case 5 of the skiving stock problem. Then E possesses the proper-MIRDP. In particular, we also obtain the following implications. Lemma 7. Let E = (m, l, L, b) be an instance of the skiving stock problem. If E has the IRDP, then it also has the proper-IRDP. If E has the MIRDP, then it also has the proper-MIRDP. Note that the reverse of the first implication is not true in general. Therefore, consider the instance E = (3, (21, 14, 6), 42, (1, 2, 6)) which is given in the proof of Lemma 1. Here we have ∆(E) > 1 and ∆p (E) < 1. Since it is conjectured that the skiving stock problem has the MIRDP the previous lemma would also entail the proper-MIRDP of the skiving stock problem. Currently, the largest known proper gap is given by ∆p (E) = 44/39 ≈ 1.1282, see [23]. Although the proper MIRDP of the skiving stock problem is conjectured and all current numerical computations support this supposition, it is very hard to obtain upper bounds of sufficiently good quality for the proper gap by means of studying the underlying optimization problems. Therefore, we want to look at the proper gap from a different point of view by reformulating the skiving stock problem and its proper relaxation as a matching problem in a hypergraph. Since, in general, the theory of ordinary graphs [32–34] has appeared to be an extremely useful tool to address and solve large classes of combinatorial problems, by 1970 the hypergraph theory started to generalize as well as to unify (and simplify) the classical theorems about graphs [35–37]. In this regard, a first attempt to provide a systematic account of hypergraphs was probably presented in [38], giving a very good overview of the theory as a whole. These early works also paved the way for transferring the matching concept to the new hypergraph context. Some of the most important cornerstones related to the content of this article are due to F¨ uredi who studied (fractional) matchings of hypergraphs over several years [39–41], 5

An instance E = (m, l, L, b) belongs to the divisible case if li | L holds for all i ∈ I.

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also with special interest on the subclass of so called uniform hypergraphs. We will point out the relation between his ideas and our contributions at appropriate positions within the following sections. For a more detailed introduction to the general topic of (fractional) matchings of hypergraphs the famous books [42,43] can be recommended. Altogether, the matching theory is particularly well-studied for ordinary graphs as a special case of hypergraphs, see [44] or [45]. Hence, after a short introduction, we will deal with this case and derive first results for the proper gap of a certain class of instances. 3. The proper gap and hypergraphs From now on we always use the alternative representation E = (n, l, L, e) for an instance of the skiving stock problem. For the sake of simplicity, we will also use the abbreviation I := {1, . . . , n}. Since all of our investigations in this section are based on hypergraphs, we start with the following definition. Definition 9. A pair H = (V, E) consisting of a finite (vertex) set V ̸= ∅ and a family E of non-empty and non-singleton subsets of V (referred to as hyperedges) is called hypergraph.6 A hypergraph H = (V, E) is ( ) called k-uniform (k ∈ N) if E is a subset of Vk . ( ) Note that symbols like Vk (for some k ∈ N) are standard terminology in the field of graph theory and shall be interpreted as the set of all subsets of V that contain exactly k elements. Moreover, we will assume n := |V| ≥ 2 in all of our considerations since, otherwise, the hypergraph H cannot possess any hyperedge.7 Remark 8. A 2-uniform hypergraph is an ordinary graph. For each hypergraph, there are two numbers that are of special interest for our approach. Definition 10. Let H = (V, E) be a hypergraph: 1. A function f : E → B is called matching (of H) if the inequality ∑ f (γ) ≤ 1 γ∈E: v∈γ

holds for all v ∈ V. 2. A matching f is called maximal if the value µ(f ) :=

∑

f (γ)

γ∈E

is maximal among all possible matchings of H. If f is maximal, then the value µ := µ(H) := µ(f ) is called matching number of H. 3. A function f : E → [0, 1] is called fractional matching (of H) if the inequality ∑ f (γ) ≤ 1 γ∈E: v∈γ

holds for all v ∈ V. 6 Note that, theoretically, also singletons could be allowed in the definition of the set of hyperedges. As we will see later, considering hyperedges that consist of only one vertex is not necessary for our intended purposes since this would correspond to item lengths larger than or equal to L in the skiving stock context. 7 A second reason for this assumption will be clear when the relationship of a hypergraph H to an instance E of the skiving stock problem is elaborated. Thereby, it will turn out that the number n of vertices corresponds to the total number of items. Hence, concerning this point, the consideration of instances consisting out of one single item is not interesting from a mathematical point of view.

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4. A fractional matching f is called maximal if the value ∑ µF (f ) := f (γ) γ∈E

is maximal among all possible fractional matchings of H. If f is maximal, then the value µF := µF (H) := µF (f ) is called fractional matching number of H. With this nomenclature, we can define the matching gap ∆M (H) of a hypergraph as the difference ∆M (H) := µF − µ.

(6)

Definition 11. A hypergraph H = (V, E) is called skiving hypergraph if there exists an instance E = (n, l, L, e) of the SSP with n = |V|, e = (1, . . . , 1)⊤ ∈ Zn+ and L > l1 ≥ l2 ≥ · · · ≥ ln ≥ 1, such that the mapping k ∑ { } gH : E → Pp⋆ (E), vi1 , vi2 , . . . , vik ↦→ e ij

(7)

j=1

(with ei ∈ Rn denoting the ith unit vector) is well-defined and bijective. In other words, this definition means that the hyperedges of H are somehow identical to the minimal proper patterns of an appropriately chosen instance E. This observation enables the computation of proper gaps from a graph theoretic point of view. Proposition 9. Let H = (V, E) be a skiving hypergraph with respect to the instance E = (n, l, L, e). Then we have z ⋆ (E) = µ(H), zp⋆ (E) = µF (H), and ∆p (E) = ∆M (H). Proof . We only prove the equation z ⋆ (E) = µ(H), the second equation can be shown analogously. Then, |J ⋆ (E)|

both equations immediately lead to the statement ∆p (E) = ∆M (H). To this end, let x ∈ Z+ p denote a solution of the skiving stock problem that belongs to E. Since H is a skiving hypergraph, the mapping k ∑ { } gH : E → Pp⋆ (E), vi1 , vi2 , . . . , vik ↦→ e ij j=1

is well-defined and bijective. Furthermore, there is a bijection g˜ : Pp⋆ (E) → Jp⋆ (E), aj ↦→ j. Thus, we can define f (γ) := x˜ g ◦g

H (γ)

for all γ ∈ E. Indeed, we obtain a maximal matching:

1. Due to bi = 1 for all i ∈ I, we have xj ∈ B for all j ∈ Jp⋆ (E) which leads to f (γ) ∈ B for all γ ∈ E. 2. Let v := vi ∈ V with i ∈ I be given. Then, due to the bijectivity of g˜ ◦ gH , we have ∑ ∑ ∑ f (γ) = aij = aij ≤ bi = 1 j∈Jp⋆ (E): aij =1

γ∈E: v∈γ

implying that f is a matching of H. 3. The matching f is also maximal since we have ∑ f (γ) = γ∈E

∑

j∈Jp⋆ (E)

xj = z ⋆ (E)

j∈Jp⋆ (E)

and larger values are not possible due to the optimality of x. □ Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Fig. 2. A graph that does not belong to an instance of the skiving stock problem.

Hence, the proper gap of a skiving stock instance can also be calculated by the matching gap of the corresponding skiving hypergraph. We emphasize that the matching gap of hypergraphs is known to be unbounded for increasing values of n (see [46, Subsection 2.4.1] for a special sequence of hypergraphs). Therefore, there is some hope that large proper gaps can be obtained by means of this new approach. Note that these considerations may also have consequences for the validity of the conjecture ∆p (E) < 2 for all instances E of the SSP. However, observe that not all hypergraphs correspond to an instance of the skiving stock problem. Example 2. Consider the (hyper-)graph H = (V, E) depicted in Fig. 2. For the sake of contradiction, let us assume that there exists a corresponding instance E = (n, l, L, e). Due to the previous definition, we obtain n = 4, and the bijectivity of the mapping gH leads to ⎧⎛ ⎞ ⎛ ⎞⎫ 0 ⎪ 1 ⎪ ⎪ ⎬ ⎨⎜ ⎟ ⎜ ⎟⎪ 1 ⋆ ⎟ , ⎜0⎟ . Pp (E) = ⎜ ⎠ ⎠ ⎝ ⎝ 1 ⎪ 0 ⎪ ⎪ ⎪ ⎭ ⎩ 1 0 But, in this situation, li ≥ L/2 is satisfied for two non-adjacent indices i1 , i2 ∈ I (i1 ̸= i2 ) which requires a = ei1 + ei2 ∈ Pp⋆ (E). Since the latter does not hold, H is not a skiving hypergraph. Consequently, we need some further conditions in order to obtain a skiving hypergraph. Therefore, the following theorem presents a characterization. Theorem 10. A hypergraph H = (V, E) is a skiving hypergraph if and only if there exists a weight function c : V → N and a number C ∈ N with ⏐ { } max c(v) ⏐ v ∈ V < C

(8)

such that the equivalence

{

} vi1 , vi2 , . . . , vik ∈ E ⇐⇒

⎧ ⎪ ⎪ ⎪ ⎪ ⎨

k ( ) ∑ c vij ≥ C, j=1

(9)

∑ ⎪ ⎪ ⎪ c (vs ) < C ⎪ ⎩ ∀S ⊂ {i1 , i2 , . . . , ik } : s∈S

holds for all k ≥ 2 and all i1 , i2 , . . . , ik ∈ I with |{i1 , i2 , . . . , ik }| = k. Proof . See Appendix. □ Obviously, condition (9) states that a subset Y ⊆ V of vertices is a hyperedge if and only if it is an inclusion-minimal set with respect to the property that the corresponding c-values add up to at least C. This alternative interpretation may sometimes be used to simplify specific parts of the proofs, especially in those cases where no formal distinction between the indices of the vertices and the vertices themselves is required. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Hence, we have found a characterization for skiving hypergraphs. Let Hn denote the set of all skiving hypergraphs on n vertices. Then we introduce an equivalence relation ∼iso on Hn by means of H1 = H2 :⇐⇒ H1 ∼iso H2 , i.e., two hypergraphs are equivalent if and only if they are isomorphic. The corresponding factor set is denoted by Hn . Nevertheless, we will refer to its elements as H ∈ Hn meaning that H is a representative of the equivalence class [H]. 4. The matching gap of 2-uniform skiving hypergraphs As we have written in the introduction of this paper, and mainly from a historical point of view, the theory of fractional matchings and the matching gap is particularly well-studied for the case where 2-uniform hypergraphs (i.e. ordinary graphs) are considered, see [45] for a good overview on the most important results. In contrast to that, although there are many results on (fractional) hypergraph matchings in literature [39–41], at the moment no applicable general approach to calculate their matching gap (comparable to [45, Corollary 2.4]) is known in literature.8 However, the following theorem currently published in [44] might give hope that, also within the field of ordinary graphs, large proper gaps can be obtained on the basis of this new approach: Theorem 11. Each graph G = (V, E) with |V| = n satisfies µM (G) ≤ (n − 2)/6. Moreover this inequality is sharp, and equality holds for an infinite family of graphs.

4.1. An approach to characterize 2-uniform skiving hypergraphs We aim at investigating the matching gap of 2-uniform skiving hypergraphs (mostly simply referred to as skiving graphs) in order to calculate the proper gaps of the corresponding class of skiving stock instances. To this end, we have to characterize the 2-uniformity of skiving hypergraphs first. For readers already being very familiar with that particular field of graph theory, it may be disclosed that this consideration will eventually lead to the class of threshold graphs [33, Chapter 10]; see Remark 19 at the end of this subsection for the full details. Proposition 12. A skiving hypergraph H = (V, E) ∈ Hn is 2-uniform if and only if the implication ∑ ∑ ∀S ⊆ I, |S| ≥ 3 : c (vs ) ≥ C =⇒ ∃T ⊂ S, |T | = 2 : c (vt ) ≥ C s∈S

(10)

t∈T

holds for a corresponding weight function c : V → N and a number C ∈ N (both according to Theorem 10) of H. Proof . Obviously, condition (10) states that there is no inclusion-minimal subset Y ⊆ V of vertices of ∑ cardinality |Y | ≥ 3 with y∈Y c(y) ≥ C. As we have excluded singleton hyperedges from our consideration, the graph is clearly 2-uniform if and only if (10) holds. □ Let Hn2 denote the set of all skiving graphs. Again, we intend to only consider non-isomorphic graphs, so the factor set of Hn2 with respect to ∼iso is denoted by H2n . As above, H ∈ H2n shall be understood in such a way that H is an element of [H] ∈ H2n . 8 Most probably, as this problem is in N P in contrast to the matching problem in graphs, no “good” characterization of maximal (fractional) matchings in hypergraphs can be expected.

Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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In order to calculate the matching gap of this class of graphs, we need a characterization of skiving graphs that is, in a certain extent, more familiar than the current one. To this end, we aim at successively constructing a bijection from the power set P ({1, . . . , n − 1}) to H2n . Therefore, we define the mapping ⏐ { } j ⋆ : I → (S ∪ {n}) ∪ {−∞}, j ⋆ (τ ) := jS⋆ (τ ) := sup d ∈ S ∪ {n} ⏐ d ≤ τ .

(11)

for each subset S = {i1 < i2 < · · · < ik } ⊆ {1, . . . , n − 1}. Note that j ⋆ acts as the identity on the set S ∪ {n}, whereas it denotes the predecessor of τ ∈ I \ (S ∪ {n}) within the set S ∪ {n} otherwise (or returns the value −∞ to indicate that such a predecessor does not exist). Based on this observation, the following result is obvious: Lemma 13. The statement j ⋆ (τ ) ≤ τ is true for all τ ∈ I. Equality holds if and only if τ ∈ S ∪ {n}. Now the bijection can be described by the following algorithm: Algorithm 1 Description of the bijection Let S = {i1 < i2 < . . . < ik } ⊆ {1, . . . , n − 1} be given. We define a graph H = (V, E) on the vertex set V = {v1 , . . . , vn } by means of the following steps: { } Step 1: Construct a complete graph on the vertex set vi1 , vi2 , . . . , vik , vn . Step 2: For each remaining index τ ∈ I \ (S ∪ {n}): Compute j ⋆ (τ ) as in (11) and, if j ⋆ (τ ) is finite, connect vτ to each of the vertices vi1 , vi2 , . . . , vj ⋆ (τ ) . Otherwise, vτ represents an isolated vertex of the graph. The mapping that is described in Algorithm 1 will be denoted by ιn : P ({1, . . . , n − 1}) → H2n , S ↦→ ιn (S). At first, consider some examples to get a better overview on how ιn acts on a given subset. Example 3. Consider the extreme cases at first. If we choose S = {1, . . . , n − 1} the first step of the previous construction leads to ιn (S) = Kn , i.e., the complete graph on n vertices. Corresponding skiving stock instances are given by E = (n, l = (l1 , . . . , ln ), L, e) with li ≥ L/2 for all i ∈ I. Instead, if we take S = ∅ ¯ n , the complement of the we obtain by the second step that all vertices have to be isolated, i.e., ιn (S) = K ∑ complete graph. This graph refers to instances E = (n, l = (l1 , . . . , ln ), L, e) with i∈I li < L. Three further examples that are constructed by Algorithm 1 are depicted in Fig. 3. Possible representative instances of the skiving stock problem corresponding to these graphs are given by (from the left to the right) E1 = (n, l = (L − 1, 1, . . . , 1), L, e) , E2 = (n, l = (1, L − t, t, . . . , t), L, e) , E3 = (n, l = (L − 1, L − 1, 1, . . . , 1), L, e) , where, only for the second instance, L has to be chosen sufficiently large such that t := ⌊(L − 1)/(n − 2)⌋ ≥ 2 holds. The special role of the vertex vn will become more clear in the proof of Lemma 17. 4.2. The proof of the bijectivity Up to now, we only know that ιn maps to the set of graphs on n vertices. Hence, we start with the following important observation. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Fig. 3. For n ≥ 5 these examples show ιn ({1}) on the left side, ιn ({2}) in the middle, and ιn ({1, 2}) on the right side. The corresponding complete subgraphs, see the first step of Algorithm 1 , are painted red. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)

Theorem 14. The mapping ιn is well-defined, i.e., ιn (P ({1, . . . , n − 1})) ⊆ H2n . Proof . For a given set S ⊆ {1, . . . , n − 1}, we have to show that its image ιn (S) represents a skiving graph. Since the corresponding argumentations are quite extensive we split this proof in several parts. Part 1: The case S = ∅: ¯ n . For this graph, we can set C = n + 1 and c(v) = 1 for all v ∈ V If S = ∅ is given we obtain ιn (S) = K leading directly to (8) and (9) since the right hand side of the equivalence in (9) can never be satisfied. Thus, ( ) we have ιn (S) ∈ Hn . Moreover, since ιn (S) has no (hyper-)edges at all, we obtain E ⊆ V2 which proves the 2-uniformity of ιn (S) and also ιn (S) ∈ H2n . So, henceforth, we will assume S ̸= ∅. Part 2: On the construction of c and C: Let S = {i1 < i2 < · · · < ik } be given. At first, we define C := n ·

∏ t∈S

n

ik +2−t

and

( ) C nik +2−d − 1 c (vd ) := nik +2−d

for all d ∈ S. Furthermore, we set c(vn ) := (n − 1)/n · C and ⎧ C ⎨ if j ⋆ (τ ) ∈ N, ⋆ (τ ) , i +2−j k c (vτ ) := n ⎩ 1, if j ⋆ (τ ) = −∞,

(12)

for each τ ∈ I \ (S ∪ {n}). Note that the condition j ⋆ (τ ) ∈ N is equivalent to τ > i1 . Part 3: On the feasibility of c and C: For d ∈ S we obtain c (vd ) =

( ) C nik +2−d − 1 =C− nik +2−d

C ik +2−d n   

∈N

∈[C/nn ,C/n2 ]∩N

due to the following observations: • According to d ∈ S, we have d ≤ ik leading to C/nik +2−d ≤ C/n2 . • According to d, ik ∈ S, we have d ≥ i1 ≥ 1 and ik ≤ n − 1 leading to ik − d ≤ n − 2. This immediately implies C/nik +2−d ≥ C/nn . Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Additionally, we have c (vd ) < C for all d ∈ S, c (vn ) =

∏ n−1 · C = (n − 1) · nik +2−t ∈ N n t∈S

and c(vn ) < C. For τ ∈ I \ (S ∪ {n}), we either have c(vτ ) = 1 < C or c (vτ ) =

C C ≤ 2. ik +2−j ⋆ (τ ) n n    ∈N

Thus, c : V → N is well-defined and (8) is satisfied as well. Part 4.1: On the verification of property (9): Remember that we have to prove ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

˜k ( ) ∑ c vrj ≥ C,

{ } j=1 (13) vr1 , vr2 , . . . , vr ∈ E ⇐⇒ { } ∑ ˜k ⎪ ⎪ ˜ ⎪ ∀S ⊂ r1 , r2 , . . . , r˜k : c (vs ) < C ⎪ ⎪ ⎩ ˜ s∈S ⏐{ }⏐ ⏐ ⏐ for all ˜ k ≥ 2 and all r1 , r2 , . . . , r˜k ∈ I with ⏐ r1 , r2 , . . . , r˜k ⏐ = ˜ k. At first, let us consider the case ˜ k = 2. Since (8) is satisfied, the second condition on the right hand side of the previous equivalence is always true for ˜ k = 2. Thus, it remains to prove that {vr1 , vr2 } ∈ E ⇐⇒ c (vr1 ) + c (vr2 ) ≥ C

(14)

holds for all r1 < r2 ∈ I. According to the construction in Algorithm 1, a pair {vr1 , vr2 } with r1 < r2 can represent an edge of ιn (S) only in those cases where either in step one or in step two the vertices vr1 and vr2 were connected. This is equivalent to either r1 , r2 ∈ S ∪ {n}

(15)

r1 ∈ S ∪ {n}, r2 ̸= S ∪ {n}, r1 ≤ j ⋆ (r2 )

(16)

(which represents the first step) or

which represents an edge that is added in the second step. Note that r2 > r1 leads to j ⋆ (r2 ) ̸= −∞, i.e., we do not have to demand r2 > i1 separately in order to ensure the vertex vr2 to be non-isolated. Due to the considerations in Part 3 of this proof, the first possibility (15) implies ( ) ( ) C C c (vr1 ) + c (vr2 ) ≥ C − 2 + C − 2 > C n n for r1 , r2 ∈ S, according to our initial assumption n ≥ 2. If r2 = n holds, we obtain ( ) ( ) C C > C, c (vr1 ) + c (vn ) ≥ C − 2 + C − n n also with respect to n ≥ 2. Moreover, the second case (16) leads to ( ) ( ) C nik +2−r1 − 1 C nik +2−r1 − 1 C C c (vr1 ) + c (vr2 ) = + i +2−j ⋆ (r ) ≥ + i +2−r = C, i +2−r i +2−r 1 1 1 2 nk nk nk nk Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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where r1 ≤ j ⋆ (r2 ) has been used. Hence the first implication of the equivalence (14) has been proved for ˜ k = 2. On the other hand, let r1 < r2 ∈ I with c (vr1 ) + c (vr2 ) ≥ C be given. Then, at least one of these indices has to belong to the set S ∪ {n} since, otherwise, we obtain C ≤ c (vr1 ) + c (vr2 ) ≤

C C + 2
due to the definition (12) and n ≥ 2. There remain three cases: 1. Case 1: r1 < r2 ∈ S ∪ {n} This corresponds exactly to the case (15). 2. Case 2: r1 ∈ S ∪ {n}, r2 ̸∈ S ∪ {n} At first observe that j ⋆ (r2 ) = −∞ is impossible due to r2 > r1 ≥ i1 . Furthermore, we definitely have r1 ̸= n since r1 < r2 has to hold. This means that r1 ∈ S is true. Altogether, we obtain: ( ) ] C nik +2−r1 − 1 C C [ ⋆ C ≤ c (vr1 ) + c (vr2 ) = + i +2−j ⋆ (r ) = C + i +2 nj (r2 ) − nr1 i +2−r 1 2 nk nk nk which is satisfied if and only if j ⋆ (r2 ) ≥ r1 holds. But then, this case corresponds to (16). 3. Case 3: r1 ̸∈ S ∪ {n}, r2 ∈ S ∪ {n} At first note that r1 > i1 has to hold. Otherwise, we obtain c (vr1 ) = 1 leading to the following contradiction C ≤ c (vr1 ) + c (vr2 ) = c (vr1 ) + 1 < C. Moreover, we cannot have r2 = n, since this implies C ≤ c (vr1 ) + c (vr2 ) ≤

C n−1 + C < C. n2 n

Hence, we can assume r2 ∈ S which gives C ≤ c (vr1 ) + c (vr2 ) =

C nik +2−j ⋆ (r1 )

( ) ] C nik +2−r2 − 1 C [ j ⋆ (r1 ) r2 + = C + n − n . nik +2−r2 nik +2

Obviously, this inequality holds if and only if j ⋆ (r1 ) ≥ r2 is true. But the latter cannot happen due to r1 < r2 . Hence, this case cannot occur. Altogether, we obtain that r1 < r2 and c (vr1 ) + c (vr2 ) ≥ C either imply r1 , r2 ∈ S ∪ {n} (which is (15)) or r1 ∈ S ∪ {n}, r2 ̸∈ S ∪ {n}, and r1 ≤ j ⋆ (r2 ) (which is (16)). Hence equivalence (14) is proved. Part 4.2: On the verification of (9) for ˜ k ≥ 3: ˜ For k ≥ 3, it suffices to show that the right hand side of the (13) cannot be satisfied, since, by Algorithm 1, the left hand side of this equivalence is always wrong, too. Let ˜ k ≥ 3 pairwise distinct indices R = {r1 , r2 , . . . , r˜k } ⊆ I be given. For the sake of contradiction, we assume that R satisfies the right hand side of (13). Then, at most one element of R belongs to S ∪ {n}. ˜ := {τ1 , τ2 } leading to (see Part 4.1) Otherwise, if there exist τ1 < τ2 ∈ R ∩ (S ∪ {n}), we can set R ∑ c (vr ) ≥ C, ˜ r∈R ∑ ˜ ⊂ R with i.e., we have found a proper subset R ˜c (vr ) ≥ C which contradicts to the assumption of this r∈R part of the proof. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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On the other hand, we also know that at least one element of R belongs to S ∪ {n}. Otherwise, we could conclude ∑ ∑ C C C C≤ c (vr ) ≤ =˜ k· 2 ≤
r∈R

by means of (12), and this is also a contradiction. Consequently, exactly one element of R, say r1 , belongs to S ∪ {n}. Now, there are two cases. 1. Case 1: r1 = n, r2 , . . . , r˜k ̸∈ S ∪ {n} In this case, we have ∑

c (vr ) = c(vn ) +

r∈R

∑

c (vr ) ≤

r∈R\{r1 }

C n−1 · C + (˜ k − 1) 2 n n

n−1 C ≤ · C + (n − 1) · 2 < C n n giving the contradiction. 2. Case 2: r1 ∈ S, r2 , . . . , r˜k ̸∈ S ∪ {n} Since R satisfies the right hand side of (13) we have c (vr ) <

C nik +2−r1

(17)

˜ = {r1 , r⋆ } ⊂ R with c(vr ) + c(vr⋆ ) ≥ C for all r ∈ R \ {r1 }. Otherwise, there would be a subset R 1 which is not possible due to the assumption of this part of the proof. Since the values of the weight function c are always multiples of n, we can even state c (vr ) ≤

C C = i +3−r 1 nk nik +2−(r1 −1)

for all r ∈ R \ {r1 } as a direct consequence of (17). Thus, altogether, we have ( ) ∑ ∑ C C C≤ c (vr ) = c (vr1 ) + c (vr ) ≤ C − i +2−r + (˜ k − 1) · i +3−r 1 1 nk nk r∈R r∈R\{r1 } ( ) C C + (n − 1) · i +3−r ≤ C − i +2−r 1 1 nk nk ( ) C C < C − i +2−r + i +2−r = C 1 1 nk nk which gives the contradiction. Hence, our assumption was wrong and, for ˜ k ≥ 3, there is no possibility to satisfy the right hand side of (13) which completes this part of the proof. Now, we know that ιn (S) ∈ Hn . But this part of the proof has also shown that there is no hyperedge that contains more than two elements. Consequently, ιn (S) is also 2-uniform and we are done. □ In what follows, we aim at proving the bijectivity of the mapping ιn . Lemma 15. The mapping ιn is injective. Proof . Consider two subsets S1 = {i1 < i2 < · · · < ik } ⊆ {1, . . . , n − 1} and S2 = {j1 < j2 < · · · < jd } ⊆ {1, . . . , n − 1} Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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with ιn (S1 ) = ιn (S2 ). Then, as a necessary condition, both of them have to possess the same number of vertices, i.e., we obtain k = d. Now define i0 := 0 and j0 := 0, and consider ⏐ { } τ := τk := inf ρ ∈ {1, . . ., k} ⏐ iρ ̸= jρ . Then, all vertices vs with s ∈ {iρ−1 + 1, . . . , iρ − 1} have degree ρ − 1 due to Step 2 of Algorithm 1. If τk = ∞ holds we have S1 = S2 . Otherwise, we can assume, without loss of generality, that iτ > jτ holds. But then, due to the second step of the construction, ιn (S1 ) possesses more vertices of degree τ − 1 than ιn (S2 ). Hence, we have ιn (S1 ) ̸= ιn (S2 ) which gives the contradiction. Thus, τ ̸= ∞ is not possible implying that S1 = S2 has to hold, i.e., ιn is injective. □ The proof of the surjectivity of ιn is more complicated since we have to find an appropriate preimage of a given element of H2n . We therefore need the following lemma. Lemma 16. Let H = (V, E) ∈ H2n be a skiving graph. Then, the corresponding weight function c : V → N can be assumed to be injective. Proof . For the sake of contradiction, we assume that there is no injective weight function ˜ c for the graph 2 H. Since H ∈ Hn holds there is a weight function c and an integer C satisfying (8) and (9). Now choose ˜ = n2 C and C { } ˜ c(v) ∈ n2 c(v), . . . , n2 c(v) + n − 1 (18) for v ∈ V such that the resulting function ˜ c is injective. Note that even in the case where the original weight { } function is constant the values of ˜ c can be chosen pairwise different due to | n2 c(v), . . . , n2 c(v) + n − 1 | = n. Now, it suffices to show that ∑ ∑ ˜ c (vs ) ≥ C ⇐⇒ ˜ c (vs ) ≥ C (19) s∈S

s∈S

˜ So, let a subset S ⊆ I be given. holds for all S ⊆ I since this implies properties (8) and (9) for ˜ c and C. (=⇒) In this case, we have ∑

˜ c (vs ) ≥ n2

s∈S

∑

˜ c (vs ) ≥ n2 C = C.

s∈S

(⇐=) In this case, we have ˜≤ n2 C = C

∑

˜ c (vs ) ≤ n2

s∈S

leading to n2 C ≤ n2

∑

∑

c (vs ) + n(n − 1)

s∈S

+ n(n − 1). Rearranging the terms results in ( ) ∑ n(n − 1) 1 c (vs ) ≥ C − = C − 1 − >C −1 n2 n

s∈S c (vs )

s∈S

which implies

∑

s∈S c (vs )

≥ C due to the integrality of the left hand side.

□

This result helps us to prove the surjectivity of ιn . Lemma 17. The mapping ιn is surjective. ¯ n we can choose S = ∅ and obtain ιn (S) = H. So, Proof . Let H = (V, E) ∈ H2n be given. If H = K ¯ henceforth, let us assume H ̸= Kn , i.e., H possesses at least one edge. Since H is 2-uniform, all edges can be Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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represented by {vi1 , vi2 } for some i1 , i2 ∈ I. Due to the previous lemma, the weight function c of H can be chosen such that all of its function values are pairwise different. For each {v, w} ∈ E mark that element whose function value (of c) is the larger one. With this procedure, we obtain a set M of |M | = k marked vertices whose weight function values are always larger than C/2. Consequently, these vertices have to build a complete subgraph. Define i1 as the index of that element of M with the largest weight function value, i2 as the index of the element of M with the second largest weight function value, and so on. We now prove that there is an unmarked vertex that is adjacent to vi1 , vi2 , . . . , vik : At first, note that the set ⏐ ( ) { } W := v ∈ V ⏐ {vik , v} ∈ E, c vik > c(v) is nonempty, since otherwise vik would not have been marked by the previous procedure. Among these vertices, let w be that one with the largest value of c, i.e. we define ⏐ } { w := argmax c(v) ⏐ v ∈ W . Note that w is the unique maximum due to the injectivity of the function c. Furthermore, we have w ̸∈ M since ( ) min c(v) = c vik > c(w) v∈M

holds by construction. Hence, w is an unmarked vertex, and, by the definition of the hyperedge set (9), w has to be connected with all vertices vi1 , vi2 , . . . , vik . Setting vn := w, we obtain that the vertices vi1 , vi2 , . . . , vik , vn form a complete subgraph of H. Note that this complete subgraph is maximal: If there were another vertex v ∈ V that is adjacent to all of the vertices vi1 , vi2 , . . . , vik , vn , especially the condition {vn , v} ∈ E has to be satisfied. Since vn = w was unmarked we have c(v) > c(w) which means that v is marked and, thus, an element of M . So the current subgraph is the largest complete subgraph of H which means that H satisfies the first construction step of Algorithm 1 for S = {i1 , i2 , . . . , ik }. For all remaining vertices τ ̸∈ S ∪ {n}, we compute, if possible, the largest j ∈ S with c (vj ) + c (vτ ) ≥ C and denote it as j ⋆ := j ⋆ (τ ). Then, due to the property (9), the vertex vτ is connected to exactly the vertices vi1 , vi2 , . . . , vij ⋆ . If such an index j ⋆ does not exist, then vτ has to be isolated due to (9). But this is exactly the second step of Algorithm 1 for S = {i1 < i2 < · · · < ik }. Consequently, we have ιn (S) = H which proves the surjectivity of ιn . □ Hence, we have proven the bijectivity of the function ιn . 4.3. On some implications of this characterization A first interesting consequence of the previous observations is given by the fact that we can state the number of non-isomorphic skiving graphs explicitly: Theorem 18. There are exactly ⏐ 2⏐ ⏐Hn ⏐ = 2n−1 non-isomorphic skiving graphs on n vertices. Proof . This result follows immediately from the bijectivity of ιn . □ Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Remark 19. Note that the number 2n−1 given in the previous theorem also counts all non-isomorphic threshold graphs on n vertices. This property appears as an unproved exercise in [33, Chapter 10], whereas from [47, Theorem 1 and Remark 5] it follows that there are at most (n − 1)! · 2n−1 threshold graphs (which would lead to the upper bound 2n−1 if only equivalence classes of isomorphic graphs are considered). Among others, these graphs G = (V, E) can be characterized by the existence of an integer labeling a of V and an integer threshold t such that {x, y} ∈ E ⇐⇒ a(x) + a(y) > t

(20)

for all x, y ∈ V , see [33, Chapter 10, Theorem 10.4]. Obviously, this statement is somewhat similar to the definition of skiving graphs, for instance provided by (14) in its easiest version. Indeed, any threshold graph is a skiving graph and vice versa: given a labeling function a of V and the integer t, then we can define c(v) := a(v) for the weight function of the skiving graph. Moreover, setting ⏐ { } C := min a(x) + a(y) ⏐ {x, y} ∈ E leads to the fact that the set of edges defined by a(x) + a(y) > t is actually the same as the set of edges given by the condition c(x) + c(y) ≥ C. Conversely, given a skiving graph (with weight function c and value C), we can define a(v) := 2c(v) for all v ∈ V and t := 2C − 1. Then, again, the sets of edges defined by (14) and (20), respectively, are equivalent. Hence, Algorithm 1 also provides a new method to characterize and construct all threshold graphs on n vertices. Furthermore, now we are able to calculate the matching gap of the class H2n by means of [45] and the construction described in Algorithm 1 . Theorem 20. Let H ∈ H2n be given. Then, we have ∆M (H) ∈ {0, 1/2}. Proof . Let H ∈ H2n be given. Then, there is some set S ⊆ {1, . . . , n−1} with ιn (S) = H. If S = ∅ holds, the graph H has no edges at all. Hence, the only (fractional) matching is given by the constant function f ≡ 0 implying that ∆M (H) = 0 has to hold. So, henceforth, let us assume S ̸= ∅, say S = {i1 < i2 < · · · < ik }. Then, we partition the vertex set into I := S ∪ {n} (the vertices of the maximal complete subgraph { } vi1 , vi2 , . . . , vik , vn of H) and A := V \ (S ∪ {n}) (the other ones). Furthermore, let D(H) ⊆ V denote the set of all vertices that are missing in at least one maximal matching of H. Also this set can be partitioned into DI (H) = I ∩ D(H) and DA (H) = A ∩ D(H). Now we show that the subgraph D that is induced by the vertex set D(H) has at most one nontrivial component, i.e., at most one component with at least two vertices: For the sake of contradiction, we assume that there are two such components C1 , C2 . Each of these components has to contain a vertex from DI (H) since vertices from the set DA (H) cannot be adjacent due to step two of Algorithm 1 . But vertices from DI (H) are always adjacent since they form a complete subgraph. Thus, C1 and C2 belong to the same component. Now, by [45, Corollary 2.4] we know that ∆M (H) =

n(c) 2

where n(c) ∈ Z+ describes the number of nontrivial components of D with some extra properties. As we have seen, we can bound this value above by n(c) ≤ 1 which leads to ∆M (H) ∈ {0, 1/2}. □ In order to avoid a huge amount of further definitions, we refused to specify and use, respectively, the exact properties of the number n(c) ∈ Z+ from [45]. Nevertheless, this does not weaken our result since both Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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matching gaps are possible for all n ≥ 3. To see this, consider the complete graphs K2 (with matching gap equal to zero) and K3 (with matching gap equal to 1/2) which always can be found (with, possibly, some additional isolated vertices) in the set H2n for n ≥ 3. Note that the result of the previous theorem does not hold for arbitrary graphs on n vertices. An infinite family of counterexamples is, for instance, provided by those graphs that lead to equality in Theorem 11, see [44]. In the terminology of the skiving stock problem, i.e. with reference to the original definition of an instance, the previous theorem can be formulated as follows. Corollary 21. Each instance E = (m, l, L, b) of the skiving stock problem with ⏐ } { max e⊤ a ⏐ a ∈ Pp⋆ (E) = 2 possesses a proper gap ∆p (E) ∈ {0, 1/2}, i.e., E possesses the proper-IRDP. Note that this corollary represents a new result for the proper gap of the skiving stock problem since it does not follow from the existing theory that was presented in Section 2. Certainly, this result is not very astonishing, but our major goal was the introduction of a theory that enables the possibility to look at the skiving stock problem from an other point of view. With respect to this objective we succeeded in presenting an alternative approach to compute the proper gap of the skiving stock problem that may hopefully stimulate the scientific activity in this interdisciplinary field. Note that, vice versa, the introduced relationship also offers the possibility to compute the (fractional) matching number or the matching gap of a given hypergraph (belonging to an instance of the SSP) by means of known properties of the skiving stock problem. However, contrary to the expectations that could be connected to the initial theorem from [44], our investigations did not lead to the result that ordinary graphs (that represent a skiving stock instance) possess large matching gaps. Remark 22. The decision whether a given skiving stock instance E = (n, l, L, e) (or E = (m, l, L, b)) belongs to the 2-uniform case (as it is required for the previous corollary) can be done in polynomial time. For instance, a very simple algorithm of complexity O(n2 ) can be stated as follows: Algorithm 2 Recognition of 2-uniformity Input: Instance E = (n, l, L, e) of the skiving stock problem (as always with L > l1 ≥ . . . ≥ ln ). for all i ∈ {1, . . . , n − 1} with li + ln < L do Compute ∑ the set C(i) := {j > i : li + lj < L}. if li + j∈C(i) lj ≥ L then Return ’FALSE’ and stop. end if end for Return ’TRUE’. Obviously, any pattern a ∈ Bn destroying the 2-uniformity of an instance has to consist of a largest index i ∈ I with ai = 1 and further indices j > i with aj = 1 and li + lj < L (otherwise, it would contain a minimal subpattern with only two items). After having generated the sets C(i) ∪ {i} (for i ∈ I), we simply have to check whether their corresponding item lengths sum up to at least L. If so, C(i) ∪ {i} can be reduced to a minimal proper pattern possessing at least three items. An interesting consequence of the previous remark is given by the observation, that skiving stock instances belonging to the 2-uniform case can be solved in polynomial time. Although the matching theory is not comparably well studied yet for k-uniform hypergraphs (with k ≥ 3) or further non-uniform generalizations, their matching gap is known to be unbounded when the number n Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Fig. 4. An illustration of the Fano plane.

of vertices increases. To this end, we briefly summarize a corresponding sequence of hypergraphs that also can be found in literature, see [46, Subsection 2.4.1] for instance. Example 4. Let k ≥ 2 be a prime power. Then, there exists a (finite) projective plane Pk of order k (see [48–51] for an introduction to this topic), i.e., an incidence structure consisting of so-called points and lines satisfying the following properties: 1. 2. 3. 4.

Any two lines intersect at one point. Any two points determine one line. Every line contains k + 1 points. Every point lies on k + 1 lines.

Consequently, the projective plane of order k consists of k 2 + k + 1 points and also k 2 + k + 1 lines. For k = 2, this plane is known as the Fano plane, see Fig. 4. According to [46, Subsection 2.4.1], we obtain µ (Pk ) = 1

and

µF (Pk ) = k − 1 +

1 k

implying that ∆M (Pk ) = k − 2 + 1/k holds. These results are also (at least indirectly) contained in [52, Example 1 of Section 2] and [41, Theorem 1.3], where, in the latter, the inequality ) ( 1 µF (H) ≤ k − 1 + · µ(H) k is shown to hold for any k-uniform hypergraph H (and equality is obtained for all projective planes). However, none of these are skiving hypergraphs which follows from the properties of a projective plane. Theorem 23. The projective plane Pk is not a skiving hypergraph. Proof . For the sake of contradiction, we assume that Pk = (V, E) is a skiving hypergraph with corresponding weight function c and integer C. Consider two distinct hyperedges A = {α1 , . . . , αk+1 } ∈ E

and

B = {β1 , . . . , βk+1 } ∈ E,

where, without loss of generality, α2 ̸∈ B and β2 ̸∈ A can be assumed. Moreover, according to the first property of projective planes, there has to be a unique element, say a1 = β1 , that belongs to both of them. Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Due to (9), we have k+1 ∑

c (αj ) ≥ C

and

j=1

k+1 ∑

c (βj ) ≥ C.

j=1

Without loss of generality, we assume c (α2 ) ≥ c (β2 ). Then, we consider the set D given by D := {β1 , α2 , β3 , . . . , βk+1 } . If we show D ∈ E, then the set D would correspond to one hyperedge of Pk . But then, the points β1 , β3 would describe the lines B and D implying that B = D has to hold (see property (ii)), which is impossible due to α2 ̸= β2 . At first note that ∑

c(d) ≥

k+1 ∑

c (βj ) ≥ C

j=1

d∈D

˜ ⊂ D with holds. If there exists a proper subset D ∑

c (d) ≥ C

˜ d∈D choose one of them with minimal cardinality and denote it as D⋆ . Then, due to the minimality, this subset D⋆ satisfies ∑ ∑ c(d) ≥ C and ∀D′ ⊂ D⋆ : c (d) < C d∈D ⋆

d∈D ′

implying that D⋆ ∈ E. But then, due to property (iii) of the projective plane, we have |D⋆ | = k + 1 which means that D = D⋆ ∈ E. In this case, our argumentation from above leads to a contradiction. Hence, Pk does not represent a skiving hypergraph. □ Observe that the previous theorem also leads to a necessary condition for skiving hypergraphs. Corollary 24. Let H = (V, E) be a skiving hypergraph. Then, H does not contain any projective plane as an induced subhypergraph. Hence, skiving hypergraphs are a special subclass of those hypergraphs that do not contain any projective plane. For such hypergraphs, the (fractional) matching theory has been studied in a few works, but only for the k-uniform case (with some k ≥ 3), see [40] for a good overview. Hence, their results can also be used for skiving hypergraphs leading to the following exemplary statement that is motivated by [40, Section 6]. Corollary 25. Let H = (V, E) be a k-uniform skiving hypergraph with k ≥ 3. Then µF (H) ≤ (k − 1) · µ(H) holds. A first (preliminary) consequence of this result can be formulated as follows: Corollary 26. Let H = (V, E) be a k-uniform skiving hypergraph (for some k ≥ 3). Then its matching gap can be bounded above by ) ( 1 , ∆M (H) < m⋆ · 1 − k−1 where m⋆ is given by ⏐ { } m⋆ := min m ∈ N ⏐ There exists an instance E = (m, l, L, b) corresponding to H . Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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Proof . Let E = (m, l, L, b) denote an instance that corresponds to H in the sense defined in Section 3. Due to µ(H) = z ⋆ (E), µF (H) = zp⋆ (E) and ∆M (H) = ∆p (E) (as also presented in Section 3), we then have zp⋆ (E) ≤ (k − 1) · z ⋆ (E). Similar to the statement of [53, Lemma 4], we can prove that in this case ( ∆p (E) < m · 1 − has to hold.9 Because of ∆p (E) = µM (H) we are done.

1 k−1

)

□

Note that the computation of the exact minimum m⋆ may be difficult, and hence appropriately chosen upper bounds will likely have to be used. In order to obtain more powerful theoretical results, more efficient characterizations of skiving hypergraphs, especially for the k-uniform case (with k ≥ 3), are of great interest (and thus a big challenge) in future research. 5. Conclusions In this manuscript we introduced a possibility to obtain the proper gap of a skiving stock instance based on hypergraph matchings. Thereby, we firstly stated the particular relationship between the matching number of hypergraphs and the optimal objective value of the skiving stock problem. Afterwards, we characterized those hypergraphs that can be interpreted as an instance of the skiving stock problem. Since the matching theory is particularly well-established for ordinary graphs, we investigated the case of 2-uniform skiving hypergraphs in more detail, and were able to calculate the explicit number of corresponding non-isomorphic hypergraphs and their possible gaps. These proofs made strong use of the remarkable result that this particular class of hypergraphs corresponds bijectively to a certain power set. Even though we did not succeed in finding (new) large gaps within that particular family of instances, the matching gap of arbitrary hypergraphs is known to be unbounded which offers an alternative approach to possibly disprove the conjecture ∆p (E) < 2 in future. Appendix. Proof of Theorem 10 We prove both implications separately. (=⇒) Let H = (V, E) with V = {v1 , . . . , vn } be a skiving hypergraph with respect to the instance E = (n, l, L, e). We set C := L and c(vi ) := li for all i ∈ I. This implies condition (8) of our characterization due to the general assumptions (in particular: li < L for all i ∈ I) for an instance of the skiving stock problem. Furthermore, the mapping k ∑ { } gH : E → Pp⋆ (E), vi1 , vi2 , . . . , vik ↦→ eij j=1

9 In the terminology of [53] we have used the parameters p = k − 1 and q = 0. Note that the statement in [53, Lemma 4] remains true (and can be proved in the same way as indicated in its supplemental material, i.e., in Reference 8 from [53]), if the continuous relaxation is replaced by the proper relaxation, and if ≤ (instead of <) appears on the left hand side of the applied implication.

Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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is well-defined and bijective by definition. Let k ≥ 2 and pairwise distinct i1 , i2 , . . . , ik ∈ I be given. Under these conditions we obtain the following equivalences { } vi1 , vi2 , . . . , vik ∈ E (7)

({

⇐⇒ gH

vi1 , vi2 , . . . , vik

})

=

k ∑

eij ∈ Pp⋆ (E)

j=1

⇐⇒ l⊤

k ∑

eij ≥ L, ∀S ⊂ {i1 , i2 , . . . , ik } : l⊤

j=1

⇐⇒

k ∑

l⊤ eij ≥ L, ∀S ⊂ {i1 , i2 , . . . , ik } :

∑

l ⊤ es < L

s∈S

k ∑

lij ≥ L, ∀S ⊂ {i1 , i2 , . . . , ik } :

j=1

⇐⇒

es < L

s∈S

j=1

⇐⇒

∑

∑

ls < L

s∈S

k ( ) ∑ ∑ c vij ≥ C, ∀S ⊂ {i1 , i2 , . . . , ik } : c (vs ) < C. j=1

s∈S

(⇐=) Let c : V → N and C ∈ N with the properties (8) and (9) be given. Without loss of generality, we can choose the labels of the vertices in such a way that c (v1 ) ≥ c (v2 ) ≥ · · · ≥ c (vn ) is satisfied. We claim that the instance E = (n, l, L, e) with n := |V|, L := C and li := c(vi ) corresponds to the hypergraph H = (V, E) where E is defined according to (9). Considering the mapping k ∑ { } gH : E → Pp⋆ (E), vi1 , vi2 , . . . , vik ↦→ e ij , j=1

we have to prove the conditions within Definition 11. Part 1: gH is well-defined: Let k ≥ 2 and pairwise distinct i1 , i2 , . . . , ik ∈ I be given, i.e., we particularly have |{i1 , i2 , . . . , ik }| = k. Under these conditions, the following equivalences hold: { } vi1 , vi2 , . . . , vik ∈ E k ( ) ∑ (9) ∑ ⇐⇒ c vij ≥ C, ∀S ⊂ {i1 , i2 , . . . , ik } : c (vs ) < C j=1

⇐⇒

k ∑

s∈S

lij ≥ L, ∀S ⊂ {i1 , i2 , . . . , ik } :

j=1

⇐⇒ l⊤

eij ≥ L, ∀S ⊂ {i1 , i2 , . . . , ik } : l⊤

j=1

({

ls < L

s∈S

k ∑

⇐⇒ gH

∑

vi1 , vi2 , . . . , vik

∑

es < L

s∈S

})

=

k ∑

eij ∈ Pp⋆ (E).

j=1

Part 2: gH is bijective: Let two index sets {i1 , i2 , . . . , ik } and {i⋆1 , i⋆2 , . . . , i⋆k⋆ } be given. Then we have ⋆

gH

k k ({ }) ∑ ∑ ({ }) ⋆ vi1 , vi2 , . . . , vik = gH vi⋆1 , vi⋆2 , . . . , vi⋆⋆ ⇐⇒ e ij = e ij k

j=1

j=1

Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.

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25

which is satisfied if and only if k = k ⋆ and {i1 , i2 , . . . , ik } = {i⋆1 , i⋆2 , . . . , i⋆k⋆ } hold. Hence, gH is injective. On the other hand, if a ∈ Pp⋆ (E) is given, we can express it as a linear combination a=

k ∑

eij

j=1

for some k ≥ 2 and uniquely determined pairwise distinct i1 , i2 , . . . , ik ∈ I. But then, due to the equivalences in the first part of the proof of this implication, the preimage of the pattern a is given { } by the hyperedge vi1 , vi2 , . . . , vik which proves that gH is surjective. References

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Please cite this article in press as: J. Martinovic, G. Scheithauer, The skiving stock problem and its relation to hypergraph matchings, Discrete Optimization (2018), https://doi.org/10.1016/j.disopt.2018.03.001.