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An alternative algorithm for the n–Queens puzzle n皇后谜题的另一种算法
Pub Date : 2021-10-01 DOI: 10.2478/rmm-2021-0003
David Luque Sacaluga
Abstract In this paper a new method for solving the problem of placing n queens on a n×n chessboard such that no two queens directly threaten one another and considering that several immovable queens are already occupying established positions on the board is presented. At first, it is applied to the 8–Queens puzzle on a classical chessboard and finally to the n Queens completion puzzle. Furthermore, this method allows finding repetitive patterns of solutions for any n.
摘要本文提出了一种解决在n×n棋盘上放置n个皇后的问题的新方法,在没有两个皇后直接相互威胁的情况下,考虑到几个不可移动的皇后已经占据了棋盘上的既定位置。首先将其应用于经典棋盘上的8皇后谜题,最后将其应用于n皇后完成谜题。此外,这种方法允许找到任意n的解的重复模式。
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引用次数: 0
The Five-Button Door Lock – Experiment and Discovery in Mathematics 五钮门锁——数学中的实验与发现
Pub Date : 2021-10-01 DOI: 10.2478/rmm-2021-0006
S. Simonson, T. Woodcock
Experimentation, gathering data, and computation are an integral part of the process whereby mathematicians discover theorems and their proofs. Mathematics may be a science of exact proof, but the data and the process used in the discovery of certain proofs adds an experimental component. And, computers can help by creating data that reveal patterns. This productive methodology is one worth emphasizing at every level of mathematical instruction, but especially in primary, middle and high school curricula, where the emphasis on algorithms, methods, technique, and vocabulary leaves experiment and discovery as an afterthought. Inspired by a question posted by Marc Dostie on the Rediscovering Mathematics Facebook page 1, we consider a number of problems related to the following door lock. We offer this exploration as an example of how to incorporate discovery, experiment, and calculation into mathematics and pedagogy. An early version of this work was presented in an invited lecture at the Mathematics and Computer Science Colloquium at Providence College in 2015.2 Programmable door locks such as the one in the figure, commonly found in schools, hospitals, and office buildings, provide a flexible way to maintain selective security and entry to different rooms and areas of buildings. To enter a room, a person presses certain buttons, then enter, and turns the handle. In this particular model of the lock, once a button is pressed, it cannot be pressed again, however, buttons can be pressed simultaneously, and the order in which the presses occur is significant.
实验、收集数据和计算是数学家发现定理及其证明过程的组成部分。数学可能是一门精确证明的科学,但在发现某些证明时所使用的数据和过程增加了实验成分。而且,计算机可以通过创建揭示模式的数据来提供帮助。这种富有成效的方法值得在每一级数学教学中强调,尤其是在小学、初中和高中课程中,因为在这些课程中,对算法、方法、技术和词汇的强调使实验和发现成为事后的想法。受Marc Dostie在“重新发现数学”Facebook页面上发布的一个问题的启发,我们考虑了一些与以下门锁相关的问题。我们提供这个探索作为如何将发现,实验和计算纳入数学和教育学的一个例子。这项工作的早期版本于2015年在普罗维登斯学院的数学和计算机科学研讨会上被邀请演讲。图中的可编程门锁,通常在学校,医院和办公楼中发现,提供了一种灵活的方式来保持选择性安全,并进入建筑物的不同房间和区域。要进入房间,一个人按下某些按钮,然后进入,然后转动把手。在这种特殊型号的锁中,一旦按下一个按钮,就不能再按了,然而,按钮可以同时按下,而且按下的顺序是重要的。
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引用次数: 0
Fun with Latin Squares 拉丁方块的乐趣
Pub Date : 2021-09-02 DOI: 10.2478/rmm-2023-0003
Mi-Hee Han, Ella Kim, Evin Liang, Miriam Mira Lubashev, Oleg Polin, Vaibhav Rastogi, Benjamin Taycher, Ada Tsui, Cindy Wei, T. Khovanova
Abstract Do you want to know what an anti-chiece Latin square is? Or what a non-consecutive toroidal modular Latin square is? We invented a ton of new types of Latin squares, some inspired by existing Sudoku variations. We can’t wait to introduce them to you and answer important questions, such as: do they even exist? If so, under what conditions? What are some of their interesting properties? And how do we generate them?
你想知道什么是反切拉丁方块吗?或者非连续环面模拉丁方是什么?我们发明了大量新型的拉丁方块,其中一些受到了现有数独游戏的启发。我们迫不及待地想把它们介绍给你,并回答一些重要的问题,比如:它们真的存在吗?如果是,在什么条件下?它们有哪些有趣的特性?我们如何生成它们呢?
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引用次数: 0
Dots-and-Polygons Dots-and-Polygons
Pub Date : 2020-12-17 DOI: 10.2478/rmm-2022-0002
Jessica L. Dickson, R. Perrier
Abstract Dots-and-Boxes is a popular children’s game whose winning strategies have been studied by Berlekamp, Conway, Guy, and others. In this article we consider two variations, Dots-and-Triangles and Dots-and-Polygons, both of which utilize the same lattice game board structure as Dots-and-Boxes. The nature of these variations along with this lattice structure lends itself to applying Pick’s theorem to calculate claimed area. Several strategies similar to those studied in Dots-and-Boxes are used to analyze these new variations.
“点盒游戏”是一种流行的儿童游戏,Berlekamp、Conway、Guy等人对其制胜策略进行了研究。在本文中,我们将考虑两种变体,即“点与三角形”和“点与多边形”,它们都使用与“点与盒”相同的点阵游戏棋盘结构。这些变化的性质以及这种晶格结构使其可以应用匹克定理来计算声称的面积。一些类似于《点与盒》研究的策略被用来分析这些新的变化。
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引用次数: 0
Next Level Odd-One-Out Puzzles 下一关奇一出谜题
Pub Date : 2020-09-18 DOI: 10.2478/rmm-2020-0003
Benjamin Berger
A commonly occurring task in intelligence tests or recreational riddles is to “find the odd one out”, that is, to determine a unique element of a set of objects that is somehow special. It is somewhat arbitrary what exactly the relevant feature is that makes one object different. But once that is settled, the answer becomes obvious. Not so with a puzzle popularized by Tanya Khovanova to express her dislike for this type of puzzle. Here, it is a more complicated relation between the objects and the features that determines the odd object, because there is only one object that does not have a unique feature expression. This puzzle inspired me to look for even more complicated relations between objects, features and feature expressions that appear to be even more symmetric, but actually still single out a “special object”. This paper provides useful definitions, a theoretical basis, solution algorithms, and several examples for this kind of puzzle.
在智力测试或娱乐谜语中,一个常见的任务是“找出奇怪的东西”,也就是说,在一组物体中确定一个独特的元素,这个元素在某种程度上是特别的。让一个对象与众不同的相关特征是什么,这多少有些武断。但一旦解决了这个问题,答案就显而易见了。但Tanya Khovanova却不喜欢这种类型的谜题。在这里,决定奇数对象的是对象与特征之间更为复杂的关系,因为只有一个对象没有唯一的特征表达。这个谜题激发了我去寻找物体、特征和特征表达之间更复杂的关系,这些关系看起来更加对称,但实际上仍然挑出了一个“特殊物体”。本文为这类难题提供了有用的定义、理论基础、求解算法和几个例子。
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引用次数: 0
Stay in Command: Optimal Play for Two Person Generala 保持指挥:最佳发挥两人一般
Pub Date : 2020-09-18 DOI: 10.2478/rmm-2020-0004
Joseph Heled
Generala is a social multi-player dice game popular in Spanish and Portuguese speaking countries. Generala is similar to Yacht, Yatzy, Kniffel, and other close varieties. The most well known variant, a commercial game marketed as Yahtzee, still sells 50 million games each year. All five variants above use five dice and the player scores by matching the numbers on the dice with certain combinations. Generala has 10 such combinations: Ones, Twos, Threes, Fours, Fives, Sixes, Escalera (1-2-3-4-5 or 2-3-4-5-6), Full-House (XXXYY), Fourof-a-Kind (XXXXY), and Generala (XXXXX). In a turn, the player rolls all five dice and then may either stand or re-roll some of the dice. She may then re-roll some of the dice a second time, but after the third roll the player must score the five dice by matching them with one unused combination. Generala scores 50 points, Four-of-a-Kind 40, Full House 30 and Escalera 20. The Ones to Sixes score according to the number of dice with that value. For example, 1-1-3-4-4 scores 2 points as Ones, 3 points as Threes and 8 points as Fours. In some versions a player scores 5 or 10 bonus points when hitting a combination on the first roll, but here I will consider only the “Vanilla” version. The player may waive a combination, i.e. pick a combination which does not match the dice, for a score of zero. For example, a player might waive Generala with 1-11-1-5 if Ones is no longer available to score. Each combination can be scored exactly once. After completing one turn the dice pass to the player on the left, and this process continues until all players have completed 10 turns and either scored or waived all their combinations. The player with most points wins. As is true for most games, the origin and early history of Generala is mostly unknown, and it is simply considered a “traditional game”. Dice of various kinds pre-date recorded history and feature in 5, 000-year-old games such as Senet and the Royal Game of Ur
Generala是一款在西班牙语和葡萄牙语国家流行的社交多人骰子游戏。Generala类似于Yacht, Yatzy, Kniffel和其他接近的品种。最著名的变体是一款名为Yahtzee的商业游戏,每年仍售出5000万份游戏。上述所有五种变体都使用五个骰子,玩家通过将骰子上的数字与特定组合相匹配来得分。Generala有10个这样的组合:one, two, Threes, four, five, six, Escalera(1-2-3- 5- 5或2-3-3 -5-6),fullhouse (xxxx), four -a- kind (xxxx)和Generala (XXXXX)。在一个回合中,玩家掷完所有五个骰子,然后可以站起来或重新掷一些骰子。然后她可以再次掷骰子,但在第三次掷骰子后,玩家必须通过将5个骰子与一个未使用的组合相匹配来获得5个骰子的点数。Generala 50分,Four-of-a-Kind 40分,Full House 30分,Escalera 20分。从1到6的玩家根据该值的骰子数量得分。例如,1-1-3-3 -4为1得2分,为3分,为4得8分。在某些版本中,玩家在第一次掷骰子时命中组合可以获得5或10个奖励点,但在这里我只考虑“香草”版本。玩家可以放弃一个组合,即选择一个与骰子不匹配的组合,得分为零。例如,如果one不再可以得分,玩家可能会放弃Generala的1-11-5。每个组合只能得分一次。在完成一个回合后,骰子将传递给左边的玩家,这一过程将持续到所有玩家完成10个回合并得分或放弃所有组合。得分最多的玩家获胜。与大多数游戏一样,《Generala》的起源和早期历史大多不为人知,它只是被认为是一款“传统游戏”。各种各样的骰子在有记载的历史之前就有了,并且在有5000年历史的游戏中有特色,比如塞内特和乌尔的皇家游戏
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引用次数: 0
Making the Unfair Fair 让不公平变得公平
Pub Date : 2020-09-18 DOI: 10.2478/rmm-2020-0002
R. Vallin
Although in a sequence of coin flips, any given consecutive set of, say, three flips is equally likely to be one of the eight possible, i.e., HHH, HHT, HTH, HTT, THH, THT, TTH, or TTT, it is rather peculiar that one sequence of three is not necessarily equally likely to appear first as another set of three. This fact can be illustrated by the following game: you and your opponent each ante a penny. Each selects a pattern of three, and the umpire tosses a coin until one of the two patterns appears, awarding the antes to the player who chose that pattern. Your opponent picks HHH; you pick HTH. The odds, you will find, are in your favor. By how much?
虽然在抛硬币的序列中,任何给定的连续的一组,比如说,抛三次硬币,都有同样的可能是八种可能中的一种,即HHH, HHT, HTH, HTT, THH, THT, TTH,或TTT,但相当奇怪的是,一个三次的序列不一定同样有可能首先出现在另一个三次的序列中。这个事实可以用下面的游戏来说明:你和你的对手每人下注一便士。每个人选择三种模式,裁判员抛硬币,直到出现两种模式中的一种,并将赌注授予选择该模式的玩家。你的对手选择HHH;你选HTH。你会发现,机会对你有利。减少多少?
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引用次数: 0
Counting Clues in Crosswords 计算填字游戏的线索
Pub Date : 2020-09-18 DOI: 10.2478/rmm-2020-0001
Kevin Ferland
We consider different ways to count the number of clues in American-style crossword puzzle grids. One yields a basic parity result for symmetric square grids. Another works efficiently even for non-symmetric grids that are already numbered. We further discuss the upper limit on the number of clues in a crossword puzzle with no 2-letter answers, and open questions are given. As a bonus, a mathematically-themed crossword puzzle is included!
我们考虑了不同的方法来计算美式纵横字谜网格中的线索数量。一种是对称方形网格的基本奇偶性结果。另一种方法即使对已经编号的非对称网格也有效。我们进一步讨论了没有两个字母答案的填字游戏的线索数量上限,并给出了开放问题。作为奖励,一个数学主题的填字游戏包括!
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引用次数: 1
Adjustable Coins 可调节的硬币
Pub Date : 2020-08-09 DOI: 10.2478/rmm-2021-0009
S. Moran, I. Yavneh
Abstract In this paper we consider a scenario where there are several algorithms for solving a given problem. Each algorithm is associated with a probability of success and a cost, and there is also a penalty for failing to solve the problem. The user may run one algorithm at a time for the specified cost, or give up and pay the penalty. The probability of success may be implied by randomization in the algorithm, or by assuming a probability distribution on the input space, which lead to different variants of the problem. The goal is to minimize the expected cost of the process under the assumption that the algorithms are independent. We study several variants of this problem, and present possible solution strategies and a hardness result.
在本文中,我们考虑一个场景,其中有几种算法来解决给定的问题。每个算法都与成功的概率和成本相关联,如果解决不了问题也会受到惩罚。用户可以在给定的代价下一次运行一种算法,也可以放弃并支付罚金。成功的概率可以通过算法中的随机化来暗示,或者通过假设输入空间上的概率分布来暗示,从而导致问题的不同变体。目标是在假设算法是独立的情况下最小化过程的预期成本。我们研究了这个问题的几种变体,并提出了可能的解决策略和一个硬度结果。
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引用次数: 0
Bishop Independence on the Surface of a Square Prism 正方形棱镜表面上的主教独立性
Pub Date : 2020-06-29 DOI: 10.2478/rmm-2021-0007
L. H. Harris, S. Perkins, P. Roach
Abstract Bishop Independence concerns determining the maximum number of bishops that can be placed on a board such that no bishop can attack any other bishop. This paper presents the solution to the bishop independence problem, determining the bishop independence number, for all sizes of boards on the surface of a square prism.
主教独立性是指确定棋盘上可以放置的主教的最大数量,以使任何主教都不能攻击其他主教。本文给出了确定正方形棱柱表面各种尺寸板的象独立数的象独立问题的解。
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引用次数: 0
期刊
Recreational Mathematics Magazine
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