“…Generating the set of linear extensions of a given poset P is equivalent to generating all topological sorts of its Hasse diagram [9]. For the poset P whose Hasse diagram is Much attention has been given to the combinatorial problems of counting [5,6] and generating the linear extensions of a given poset [8,19,25,28,32]. Brightwell and Winkler [6] prove that the problem of determining the number of linear extensions of a given poset is #P-complete.…”
Section: Preliminariesmentioning
confidence: 99%
“…For 1 ≤ i ≤ n e + 3, define the i-swap pair to be (3i 1,2,3,4,6,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37). Let x = n e + 1, y = n e + 2, and z = n e + 3.…”
Section: Cubic Vertex Covermentioning
confidence: 99%
“…The first is also known as the poset dimension problem, which has been shown to be NP-Complete [35] and hard to approximate [17]. The second is a well-studied combinatorial problem for which many polynomial-time algorithms exist [25,28,29,32]. Since a poset can be viewed as a generator of linear extensions (orders), the underlying data mining problem is the converse of the well-studied latter problem.…”
There has been much research on the combinatorial problem of generating the linear extensions of a given poset. This paper focuses on the reverse of that problem, where the input is a set of linear orders, and the goal is to construct a poset or set of posets that generates the input. Such a problem finds applications in computational neuroscience, systems biology, paleontology, and physical plant engineering. In this paper, two algorithms are presented for efficiently finding a single poset, if such a poset exists, whose linear extensions are exactly the same as the input set of linear orders. The variation of the problem where a minimum set of posets that cover the input is also explored. This variation is shown to be polynomially solvable for one class of simple posets (kite(2) posets) but NP-complete for a related class (hammock(2,2,2) posets).General Terms: Algorithms.
“…Generating the set of linear extensions of a given poset P is equivalent to generating all topological sorts of its Hasse diagram [9]. For the poset P whose Hasse diagram is Much attention has been given to the combinatorial problems of counting [5,6] and generating the linear extensions of a given poset [8,19,25,28,32]. Brightwell and Winkler [6] prove that the problem of determining the number of linear extensions of a given poset is #P-complete.…”
Section: Preliminariesmentioning
confidence: 99%
“…For 1 ≤ i ≤ n e + 3, define the i-swap pair to be (3i 1,2,3,4,6,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37). Let x = n e + 1, y = n e + 2, and z = n e + 3.…”
Section: Cubic Vertex Covermentioning
confidence: 99%
“…The first is also known as the poset dimension problem, which has been shown to be NP-Complete [35] and hard to approximate [17]. The second is a well-studied combinatorial problem for which many polynomial-time algorithms exist [25,28,29,32]. Since a poset can be viewed as a generator of linear extensions (orders), the underlying data mining problem is the converse of the well-studied latter problem.…”
There has been much research on the combinatorial problem of generating the linear extensions of a given poset. This paper focuses on the reverse of that problem, where the input is a set of linear orders, and the goal is to construct a poset or set of posets that generates the input. Such a problem finds applications in computational neuroscience, systems biology, paleontology, and physical plant engineering. In this paper, two algorithms are presented for efficiently finding a single poset, if such a poset exists, whose linear extensions are exactly the same as the input set of linear orders. The variation of the problem where a minimum set of posets that cover the input is also explored. This variation is shown to be polynomially solvable for one class of simple posets (kite(2) posets) but NP-complete for a related class (hammock(2,2,2) posets).General Terms: Algorithms.
“…Generation of D k (t) was first discussed by Zaks [10]. A general result by Pruesse and Ruskey implies that D k (t) has a 2-adjacent-transposition Gray code [4] and a result by Canfield and Williamson [1] proves that D k (t) can be generated by a loopless algorithm 3 . More recently, Vajnovszki and Walsh [9] found a two-close transposition Gray code and created a loopless algorithm that requires twelve if-statements and O(n) additional variables stored in three additional arrays e, s, and p. Results on k-ary trees date back to Ruskey [5] and Trojanowski [8].…”
Abstract. A binary string B of length n = kt is a k-ary Dyck word if it contains t copies of 1, and the number of 0s in every prefix of B is at most k−1 times the number of 1s. We provide two loopless algorithms for generating k-ary Dyck words in cool-lex order: (1) The first requires two index variables and assumes k is a constant; (2) The second requires t index variables and works for any k. We also efficiently rank k-ary Dyck words in cool-lex order. Our results generalize the "coolCat" algorithm by Ruskey and Williams (Generating balanced parentheses and binary trees by prefix shifts in CATS 2008) and provide the first loopless and ranking applications of the general cool-lex Gray code by Ruskey, Sawada, and Williams (Binary bubble languages and cool-lex order under review).
Background
k-ary Dyck WordsLet B(n, t) be the set of binary strings of length n containing t copies of 1. A string B ∈ B(kt, t) is a k-ary Dyck word if the number of 0s in each prefix is at most k−1 times the number of 1s. Let D k (t) be the set of k-ary Dyck words of length kt. For example, the k-ary Dyck words with k = t = 3 are given below D 3 (3) = {111000000, 110100000, 101100000, 110010000, 101010000, 100110000, 110001000, 101001000, 100101000, 110000100, 101000100, 100100100}.The k-ary Dyck words of length kt have simple bijections with a number of combinatorial objects including k-ary trees with t internal nodes [2] [3]. The 2-ary Dyck words are known as balanced parentheses when 1 and 0 are replaced by '(' and ')' respectively, and the cardinality of D 2 (t) is the tth Catalan number. A simple property of k-ary Dyck words is that they can be "separated" according to the following remark. We let αβ denote the concatenation of the binary strings α and β, and we say that α and β have the same content if they have equal length and an equal number of 1s. Remark 1. If αβ, γδ ∈ D k (t) and α and γ have the same content, then αδ, βγ ∈ D k (t). In other words, prefixes (or suffixes) of k-ary Dyck words with the same content can be separated and recombined.
“…Note that an LE-graph comes with a natural edge colouring, the swap colouring, where each edge is coloured with the incomparable pair of elements that is swapped along it. LE-graphs were originally defined by Pruesse and Ruskey in [8]. That article is mainly concerned with the existence of a Hamilton path in the LE-graph; see also [11] and [13].…”
Given a finite poset P, we consider pairs of linear extensions of P with maximal distance, where the distance between two linear extensions L 1 , L 2 is the number of pairs of elements of P appearing in different orders in L 1 and L 2 . A diametral pair maximizes the distance among all pairs of linear extensions of P. Felsner and Reuter defined the linear extension diameter of P as the distance between a diametral pair of linear extensions.We show that computing the linear extension diameter is NPcomplete in general, but can be solved in polynomial time for posets of width 3.Felsner and Reuter conjectured that, in every diametral pair, at least one of the linear extensions reverses a critical pair. We construct a counterexample to this conjecture. On the other hand, we show that a slightly stronger property holds for many classes of posets: We call a poset diametrally reversing if, in every diametral pair, both linear extensions reverse a critical pair. Among other results we show that interval orders and 3-layer posets are diametrally reversing. From the latter it follows that almost all posets are diametrally reversing.
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