scholarly journals An O(nlogn/logw) Time Algorithm for Ridesharing

2021 ◽  
Vol 14 (1) ◽  
pp. 8
Author(s):  
Yijie Han ◽  
Chen Sun
Keyword(s):  
Word Length ◽  
Time Algorithm ◽  
Np Complete ◽  
A Chain ◽  
Special Case

In the ridesharing problem different people share private vehicles because they have similar itineraries. The objective of solving the ridesharing problem is to minimize the number of drivers needed to carry all load to the destination. The general case of ridesharing problem is NP-complete. For the special case where the network is a chain and the destination is the leftmost vertex of the chain, we present an O(nlogn/logw) time algorithm for the ridesharing problem, where w is the word length used in the algorithm and is at least logn. Previous achieved algorithm for this case requires O(nlogn) time.

scholarly journals COMPUTING SIGNED PERMUTATIONS OF POLYGONS

2011 ◽  
Vol 21 (01) ◽  
pp. 87-100
Author(s):  
GREG ALOUPIS ◽  
PROSENJIT BOSE ◽  
ERIK D. DEMAINE ◽  
STEFAN LANGERMAN ◽  
HENK MEIJER ◽  
...  
Keyword(s):  
Mirror Symmetry ◽  
Time Algorithm ◽  
Worst Case ◽  
Signed Permutations ◽  
Np Complete ◽  
Edge List ◽  
The Given ◽  
Special Case ◽  
Planar Polygon

Given a planar polygon (or chain) with a list of edges {e1, e2, e3, …, en-1, en}, we examine the effect of several operations that permute this edge list, resulting in the formation of a new polygon. The main operations that we consider are: reversals which involve inverting the order of a sublist, transpositions which involve interchanging subchains (sublists), and edge-swaps which are a special case and involve interchanging two consecutive edges. When each edge of the given polygon has also been assigned a direction we say that the polygon is signed. In this case any edge involved in a reversal changes direction. We show that a star-shaped polygon can be convexified using O(n2) edge-swaps, while maintaining simplicity, and that this is tight in the worst case. We show that determining whether a signed polygon P can be transformed to one that has rotational or mirror symmetry with P, using transpositions, takes Θ(n log n) time. We prove that the problem of deciding whether transpositions can modify a polygon to fit inside a rectangle is weakly NP-complete. Finally we give an O(n log n) time algorithm to compute the maximum endpoint distance for an oriented chain.

NEAREST NEIGHBOR PROBLEMS

1992 ◽  
Vol 02 (04) ◽  
pp. 383-416 ◽  
Author(s):  
GORDON WILFONG
Keyword(s):  
Nearest Neighbor ◽  
Polynomial Time Algorithm ◽  
Time Algorithm ◽  
Minimum Size ◽  
Minimum Cardinality ◽  
Nearest Neighbor Rule ◽  
Np Complete ◽  
Cover Problem ◽  
Consistent Subset ◽  
Special Case

Suppose E is a set of labeled points (examples) in some metric space. A subset C of E is said to be a consistent subset ofE if it has the property that for any example e∈E, the label of the closest example in C to e is the same as the label of e. We consider the problem of computing a minimum cardinality consistent subset. Consistent subsets have applications in pattern classification schemes that are based on the nearest neighbor rule. The idea is to replace the training set of examples with as small a consistent subset as possible so as to improve the efficiency of the system while not significantly affecting its accuracy. The problem of finding a minimum size consistent subset of a set of examples is shown to be NP-complete. A special case is described and is shown to be equivalent to an optimal disc cover problem. A polynomial time algorithm for this optimal disc cover problem is then given.

ACYCLIC MATCHINGS IN SUBCLASSES OF BIPARTITE GRAPHS

2012 ◽  
Vol 04 (04) ◽  
pp. 1250050 ◽  
Author(s):  
B. S. PANDA ◽  
D. PRADHAN
Keyword(s):  
Time Algorithm ◽  
Bipartite Graphs ◽  
Maximum Cardinality ◽  
Chain Graph ◽  
Matching Problem ◽  
Permutation Graph ◽  
Complexity Result ◽  
Np Complete ◽  
A Chain ◽  
Acyclic Matching

A set M ⊆ E is called an acyclic matching of a graph G = (V, E) if no two edges in M are adjacent and the subgraph induced by the set of end vertices of the edges of M is acyclic. Given a positive integer k and a graph G = (V, E), the acyclic matching problem is to decide whether G has an acyclic matching of cardinality at least k. Goddard et al. (Discrete Math.293(1–3) (2005) 129–138) introduced the concept of the acyclic matching problem and proved that the acyclic matching problem is NP-complete for general graphs. In this paper, we propose an O(n + m) time algorithm to find a maximum cardinality acyclic matching in a chain graph having n vertices and m edges and obtain an expression for the number of maximum cardinality acyclic matchings in a chain graph. We also propose a dynamic programming-based O(n + m) time algorithm to find a maximum cardinality acyclic matching in a bipartite permutation graph having n vertices and m edges. Finally, we strengthen the complexity result of the acyclic matching problem by showing that this problem remains NP-complete for perfect elimination bipartite graphs.

scholarly journals An Efficient Algorithm to Solve the Conditional Covering Problem on Trapezoid Graphs

2011 ◽  
Vol 2011 ◽  
pp. 1-10
Author(s):  
Akul Rana ◽  
Anita Pal ◽  
Madhumangal Pal
Keyword(s):  
Time Algorithm ◽  
Chordal Graphs ◽  
Covering Problem ◽  
Shortest Distance ◽  
Trapezoid Graph ◽  
Np Complete ◽  
Conditional Covering ◽  
General Graphs ◽  
Special Case ◽  
Trapezoid Graphs

Let G=(V,E) be a simple connected undirected graph. Each vertex v∈V has a cost c(v) and provides a positive coverage radius R(v). A distance duv is associated with each edge {u,v}∈E, and d(u,v) is the shortest distance between every pair of vertices u,v∈V. A vertex v can cover all vertices that lie within the distance R(v), except the vertex itself. The conditional covering problem is to minimize the sum of the costs required to cover all the vertices in G. This problem is NP-complete for general graphs, even it remains NP-complete for chordal graphs. In this paper, an O(n2) time algorithm to solve a special case of the problem in a trapezoid graph is proposed, where n is the number of vertices of the graph. In this special case, duv=1 for every edge {u,v}∈E, c(v)=c for every v∈V(G), and R(v)=R, an integer >1, for every v∈V(G). A new data structure on trapezoid graphs is used to solve the problem.

scholarly journals Total Roman {3}-Domination: The Complexity and Linear-Time Algorithm for Trees

Mathematics ◽  
2021 ◽  
Vol 9 (3) ◽  
pp. 293
Author(s):  
Xinyue Liu ◽  
Huiqin Jiang ◽  
Pu Wu ◽  
Zehui Shao
Keyword(s):  
Linear Time ◽  
Minimum Weight ◽  
Domination Number ◽  
Time Algorithm ◽  
Simple Graph ◽  
Linear Time Algorithm ◽  
Domination Problem ◽  
Np Complete ◽  
Chordal Bipartite Graphs ◽  
Dominating Function

For a simple graph G=(V,E) with no isolated vertices, a total Roman {3}-dominating function(TR3DF) on G is a function f:V(G)→{0,1,2,3} having the property that (i) ∑w∈N(v)f(w)≥3 if f(v)=0; (ii) ∑w∈N(v)f(w)≥2 if f(v)=1; and (iii) every vertex v with f(v)≠0 has a neighbor u with f(u)≠0 for every vertex v∈V(G). The weight of a TR3DF f is the sum f(V)=∑v∈V(G)f(v) and the minimum weight of a total Roman {3}-dominating function on G is called the total Roman {3}-domination number denoted by γt{R3}(G). In this paper, we show that the total Roman {3}-domination problem is NP-complete for planar graphs and chordal bipartite graphs. Finally, we present a linear-time algorithm to compute the value of γt{R3} for trees.

The British Administrator in Burma: A New View

1983 ◽  
Vol 14 (2) ◽  
pp. 372-378 ◽  
Author(s):  
Edith L. Piness
Keyword(s):  
Unique Position ◽  
Chain Of Command ◽  
A Chain ◽  
Anglo Indian ◽  
Special Case ◽  
New View

The Anglo-Burmese administrator, ostensibly another Anglo-Indian administrator and traditionally perceived as such, must be viewed as a special case. He served Company and Crown in a milieu and under circumstances that differed sharply from those known to his counterparts in the subcontinent. Burma was in a unique position in that it was a part of both the Indian and British Empires. The administrator in Burma found himself at the end of a chain of command that extended from London via Calcutta to Rangoon. London tended to view the remote province with an eye to diplomatic and mercantile considerations. Calcutta concentrated on local priorities and imposed a rigid economy that accorded with Indian priorities. Little attempt was made to understand the alien culture with its differing needs. To the Anglo-Burmese administrator fell the difficult task of governing the ancient land under the aegis of Calcutta and the eyes of London.

k-Efficient domination: Algorithmic perspective

2022 ◽  
Author(s):  
Mohsen Alambardar Meybodi
Keyword(s):  
Decision Problem ◽  
Dominating Set ◽  
Polynomial Time Algorithm ◽  
Time Algorithm ◽  
Chordal Graphs ◽  
Sparse Graphs ◽  
Fpt Algorithm ◽  
Domination Problem ◽  
Efficient Dominating Set ◽  
Np Complete

A set [Formula: see text] of a graph [Formula: see text] is called an efficient dominating set of [Formula: see text] if every vertex [Formula: see text] has exactly one neighbor in [Formula: see text], in other words, the vertex set [Formula: see text] is partitioned to some circles with radius one such that the vertices in [Formula: see text] are the centers of partitions. A generalization of this concept, introduced by Chellali et al. [k-Efficient partitions of graphs, Commun. Comb. Optim. 4 (2019) 109–122], is called [Formula: see text]-efficient dominating set that briefly partitions the vertices of graph with different radiuses. It leads to a partition set [Formula: see text] such that each [Formula: see text] consists a center vertex [Formula: see text] and all the vertices in distance [Formula: see text], where [Formula: see text]. In other words, there exist the dominators with various dominating powers. The problem of finding minimum set [Formula: see text] is called the minimum [Formula: see text]-efficient domination problem. Given a positive integer [Formula: see text] and a graph [Formula: see text], the [Formula: see text]-efficient Domination Decision problem is to decide whether [Formula: see text] has a [Formula: see text]-efficient dominating set of cardinality at most [Formula: see text]. The [Formula: see text]-efficient Domination Decision problem is known to be NP-complete even for bipartite graphs [M. Chellali, T. W. Haynes and S. Hedetniemi, k-Efficient partitions of graphs, Commun. Comb. Optim. 4 (2019) 109–122]. Clearly, every graph has a [Formula: see text]-efficient dominating set but it is not correct for efficient dominating set. In this paper, we study the following: [Formula: see text]-efficient domination problem set is NP-complete even in chordal graphs. A polynomial-time algorithm for [Formula: see text]-efficient domination in trees. [Formula: see text]-efficient domination on sparse graphs from the parametrized complexity perspective. In particular, we show that it is [Formula: see text]-hard on d-degenerate graphs while the original dominating set has Fixed Parameter Tractable (FPT) algorithm on d-degenerate graphs. [Formula: see text]-efficient domination on nowhere-dense graphs is FPT.

scholarly journals Results on a Super Strong Exponential Time Hypothesis

2020 ◽  
Vol 34 (09) ◽  
pp. 13700-13703
Author(s):  
Nikhil Vyas ◽  
Ryan Williams
Keyword(s):  
Randomized Algorithm ◽  
Time Algorithm ◽  
Critical Threshold ◽  
Polynomial Method ◽  
Variable Ratio ◽  
Sat Solving ◽  
Worst Case ◽  
Exponential Time ◽  
Exponential Time Hypothesis ◽  
Special Case

All known SAT-solving paradigms (backtracking, local search, and the polynomial method) only yield a 2n(1−1/O(k)) time algorithm for solving k-SAT in the worst case, where the big-O constant is independent of k. For this reason, it has been hypothesized that k-SAT cannot be solved in worst-case 2n(1−f(k)/k) time, for any unbounded ƒ : ℕ → ℕ. This hypothesis has been called the “Super-Strong Exponential Time Hypothesis” (Super Strong ETH), modeled after the ETH and the Strong ETH. We prove two results concerning the Super-Strong ETH:1. It has also been hypothesized that k-SAT is hard to solve for randomly chosen instances near the “critical threshold”, where the clause-to-variable ratio is 2k ln 2 −Θ(1). We give a randomized algorithm which refutes the Super-Strong ETH for the case of random k-SAT and planted k-SAT for any clause-to-variable ratio. In particular, given any random k-SAT instance F with n variables and m clauses, our algorithm decides satisfiability for F in 2n(1−Ω( log k)/k) time, with high probability (over the choice of the formula and the randomness of the algorithm). It turns out that a well-known algorithm from the literature on SAT algorithms does the job: the PPZ algorithm of Paturi, Pudlak, and Zane (1998).2. The Unique k-SAT problem is the special case where there is at most one satisfying assignment. It is natural to hypothesize that the worst-case (exponential-time) complexity of Unique k-SAT is substantially less than that of k-SAT. Improving prior reductions, we show the time complexities of Unique k-SAT and k-SAT are very tightly related: if Unique k-SAT is in 2n(1−f(k)/k) time for an unbounded f, then k-SAT is in 2n(1−f(k)(1−ɛ)/k) time for every ɛ > 0. Thus, refuting Super Strong ETH in the unique solution case would refute Super Strong ETH in general.

scholarly journals On the rectangular knapsack problem: approximation of a specific quadratic knapsack problem

2020 ◽  
Vol 92 (1) ◽  
pp. 107-132 ◽  
Author(s):  
Britta Schulze ◽  
Michael Stiglmayr ◽  
Luís Paquete ◽  
Carlos M. Fonseca ◽  
David Willems ◽  
...  
Keyword(s):  
Knapsack Problem ◽  
Polynomial Time Algorithm ◽  
Time Algorithm ◽  
Approximation Ratio ◽  
Experimental Results ◽  
Cardinality Constraint ◽  
Quadratic Knapsack Problem ◽  
Problem Instances ◽  
Quadratic Knapsack ◽  
Special Case

Abstract In this article, we introduce the rectangular knapsack problem as a special case of the quadratic knapsack problem consisting in the maximization of the product of two separate knapsack profits subject to a cardinality constraint. We propose a polynomial time algorithm for this problem that provides a constant approximation ratio of 4.5. Our experimental results on a large number of artificially generated problem instances show that the average ratio is far from theoretical guarantee. In addition, we suggest refined versions of this approximation algorithm with the same time complexity and approximation ratio that lead to even better experimental results.

scholarly journals The Complexity of Constructing Gerechte Designs

10.37236/104 ◽  
2009 ◽  
Vol 16 (1) ◽  
Author(s):  
E. R. Vaughan
Keyword(s):  
Polynomial Time ◽  
Polynomial Time Algorithm ◽  
Time Algorithm ◽  
Latin Squares ◽  
Np Complete

Gerechte designs are a specialisation of latin squares. A gerechte design is an $n\times n$ array containing the symbols $\{1,\dots,n\}$, together with a partition of the cells of the array into $n$ regions of $n$ cells each. The entries in the cells are required to be such that each row, column and region contains each symbol exactly once. We show that the problem of deciding if a gerechte design exists for a given partition of the cells is NP-complete. It follows that there is no polynomial time algorithm for finding gerechte designs with specified partitions unless P=NP.

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