Abstract
We introduce two new metrics of “simplicity” for knight's tours: the number of turns and the number of crossings. We give a novel algorithm that produces tours with 9.25n+O(1) turns and 12n+O(1) crossings on an n×n board, and we show lower bounds of (6−ϵ)n and 4n−O(1) on the respective problems of minimizing these metrics. Hence, our algorithm achieves approximation ratios of 9.25/6+o(1) and 3+o(1). Our algorithm takes linear time and is fully parallelizable, i.e., the tour can be computed in O(n2/p) time using p processors in the CREW PRAM model. We generalize our techniques to rectangular boards, high-dimensional boards, symmetric tours, odd boards with a missing corner, and tours for (1,4)-leapers. In doing so, we show that these extensions also admit a constant approximation ratio on the minimum number of turns, and on the number of crossings in most cases.
Highlights
The game of chess is a fruitful source of mathematical puzzles
We propose two new metrics that capture simplicity and structure in a knight’s tour
Computational Complexity Consider the following decision versions of the problems: is there a knight’s tour on an n × n board with at most k turns? We do not know if these problems are in P
Summary
The game of chess is a fruitful source of mathematical puzzles. The puzzles often blend an appealing aesthetic with interesting and deep combinatorial properties [1]. A knight’s tour is closed if the last cell in the path is one knight move away from the first one The knight’s tour problem is a special case of the Hamiltonian cycle problem, in which we find a simple cycle that visits all the nodes for a specific class of graphs These graphs are formed by representing each cell on the board as a node and connecting cells a knight move apart. E.g. if {c1, c2} and {c3, c4} are two distinct pairs of consecutive cells visited along the tour, a crossing happens if the open line segments (c1, c2) and (c3, c4) intersect. Knight’s tours are typically visualized by connecting consecutive cells by a line segment. Problem 1 asks for the (self-intersecting) polygon with the smallest number of vertices that represents a valid knight’s tour
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