Smart scrambling of digits onto pyramid to match eee

三角锥上的数字

The Problem A tetrahedron (as shown below) has four sides and six edges. Numbers from 1 to 11, except number 10, are each assigned either to a vertex or to an edge. The number assigned to an edge must equal to the sum of the numbers assigned to the two endpoints. Each of the numbers must be used exactly once.

问题  一个四面体有 4 个顶点和 6 条边。从 1 – 11 的数 (没有10),一一地被分配到边和顶点。每条边上的数字是该边两个端点的数字之和。上述每个数字必须刚好使用一次。

Note: Another name for the tetrahedron (四面体) is triangular pyramid (三角锥).

Tetrahedron

Can you find a solution that satisfies ALL of the above conditions?Let’s work on this problem until we find a solution!

第一步总是最难的。要大胆地试!

First step is always the most challenging, as we are usually not sure which path to take when facing a difficult problem. Don’t be afraid to try; when it does not lead to where we expect, reflect on why it does not work and have another try where we have a better chance.

(第一步)试了几次后,发现最小的两个数1 和 2应该被分配到顶点而不是边上。在连接1,2的边上的数字应是3。

(Stage 1) After a few trial-and-errors (or guess-and-checks), we come to focus on the two smallest numbers 1 and 2, and decide that they must be assigned to the vertices, not the edges.
Why? Because if one of them (say 1) is assigned to an edge, they must be the sum of two numbers assigned to its two endpoints, therefore the two numbers at the endpoints must be even smaller. However, this is impossible, as the number 1 and 2 are already the smallest.
Tetra -1-s1

Which vertices to assigned numbers ‘1’ & ‘2’? In a tetrahedron, any two vertices are adjacent, and the whole solid is highly symmetric – by rotating /reflecting, any two adjacent vertices can be transformed to the two designated ones. This is settled! We can assign numbers ‘1’ & ‘2’ to any two vertices, and the edge connecting those two vertices shall be assigned number ‘3’.

Well begun, half done! But there is other half, for which we start by some frantic trying.

(第二步) 最大的数字11 应该分配到一边上。 这条边的端点不是 1 或者2。

(Stage 2) This time we look at the largest number 11. Suppose we assign it to an edge that’s adjacent to ‘2’. For the edge, one endpoint is ‘2’, the other endpoint – it must be ‘9’. We are now in a situation that we cannot go on any further. ConTetra -1-s15sider any edge with endpoint ‘9’ (other than the edge we already assigned ’11’): which number shall we assign? It can neither be ‘1’ nor ‘2’ nor ‘3’ (since all these numbers have been used, we are not supposing to repeat numbers). Smallest available number is ‘4’, but 9 + 4 = 13, which is not in the given set of numbers.

So the edge to which number ’11’ is assigned cannot have any endpoint ‘1’ or ‘2’; and take another look at the figure, we decided that it must be the edge that is the furthest from the edge ‘3’. And we mark 11 there.

(第三步) 分解数字 11 成两数之和。11 = 4 + 7 = 5 + 6. (注意,数字 1,2,3 已被用过了。) 只有这几种可能性,试试就得了。

(Stage 3) Tetra -1-s3

Decompose 11 into the sum of two positive numbers (but without using number ‘1’ ‘2’ ‘3’), we have the following two: 11 = 4 + 7 = 5 + 6. A couple of trials will finally lead to the solution! We leave it to be worked out by the reader.

What knowledge we have used in this whole process?

(解决本题时用到了什么?)用到了数和运算,用到了形状(点对和连接的线段),加上一些逻辑推演。所以这个问题是综合性的呀!

还有,从图形的对称性,大大减少要考虑的情况。

  • Numbers and its operation – you have to understand addition and subtractions of integers.
  • Shapes – you are not asked to do any calculation like perimeters and areas, but you have to understand an edge is constituted with two endpoints, and which edges meet and which do not (we call them the furthest edges).
  • Logic deduction – see how we’ve used the fact that 1 and 2 are the SMALLEST to deduct that they MUST be assigned to vertices, not edges; and we’ve used the fact that 11 is the LARGEST to deduct that they must be assigned to an edge.
  • Also see how we have used some symmetry to reduce the situation we need to consider.

What is the lesson we learned? Try – Think – Try again where we have a better chance. Use thinking / logic reasoning as a guidance so the trial can efficiently leads to result.

另辟蹊径–数学教育中的游戏方法

爱玩耍爱游戏是儿童的天性。玩游戏也可以学数学吗?当然,在一定条件下是可以的。关键是玩法得当,在高手引导下或好老师指导下收获最大,进步也快。

玩游戏学数学,非自今日提出,但是多数家长对之有戒心,可说是没成气候。究其原因呢,许多人头脑里有个框,认为学数理化就得正襟危坐,一本正经,学和玩扯不到一块。这种观念的效果不一定是促进学习,反而让孩子提不起劲学习,也给家长拿来“我的孩子不是学数学的料”,最终放松乃至完全放弃数学学习。

游戏呢不光是计算机上的数学游戏(大家对于计算机游戏已了解较多),更主要的是含有“数学元素”。

什么是数学元素呢?不只是传统的代数几何三角分析,也有逻辑,组合问题和博弈策略。这类游戏中,有的是用数学方法分析传统的游戏,让学习者看到游戏背后的数学道理(一些教师,科普期刊,网上学习站做了努力)。还有的是把数学观念或者事实融合在游戏中;在玩游戏的过程中同时应用数学概念发展游戏策略。这后一种游戏引起了教育者越来越多的认真的兴趣。这种数学游戏也可以看作是头脑实验的雏形,或者融入了数学原理的益智活动。游戏可以在计算机上玩,但是在纸上作最好。我们借这个机会分享一下。

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为不致空泛,我们来看两个简单游戏:1. 数独游戏; 2. 余数井跳跃游戏。涉及到的知识主要是逻辑和简单算术,尤其适合在小学高年级到刚进初中阶段(4-7 年级)的学习。

第一个是数独游戏 (Sudoku)。这个游戏的发明人就叫 Sudoku。常见的格式是一个正方形分成 3 × 3 9 个小正方形,每个小正方形又有 3 × 3 9个小格子;总共81 个小格子。在每一行,每一列,每个小正方形中数字 1 – 9 都要出现一次。(中文把这个游戏译成数独,很妙!)游戏开始时在若干格子里给了数字,参加游戏的人接下去把它填好。如果给的数字少,最终填数的答案是唯一的(或有两三个也可以),那末游戏的难度就大。这个游戏吸引了很多人,很益智力,也帮助理解数学方法。

余数井游戏(Jump Between Wells of Remainders) – 示例

Wells-of-Remainder第二是余数井游戏。在一张图上玩:有出发点(Start)和终点(Finish)。图的每个节点往下走常会有几条不同路径。这图比线性链条稍复杂一点,乍一看像项目管理的节点图。每节点有一个整数,出发点的上方指定了一个除数(整数)。玩法很简单,规则定步数,你选路径。每次移的步数怎末定?每到一个节点用这个节点的数除以指定的整数,商是整数(余数是0),就不能移动;不能整除就用带余除法求余,余数是几就移几步(可选不同路径),超过一步就是跳跃。从出发点开始,反复用规则。还有,只许往终点方向走,不许往回。在找到达终点的路径后游戏结束。这帮助练习带余除法,探究整除规律;也可以想出 Smart Strategy,有 Strategy 助力,更快地做完。数学知识已融入了规则,玩中也学到了。

有孩子(愿意动脑筋,理解能力较强的孩子)在规律或者Smart Strategy被揭示出来后,一下子觉得简单了。这让他们体会到数学的威力!下一步呢?我们可以让他们出游戏题给别的孩子玩。啊哈!这是新挑战。比方说余数井游戏,到达终点的路径怎样掩藏起来不易被发现?若孩子真是出了难度较大的题目,那么(如条件允许)老师可以陪孩子玩。孩子化了心思的杰作一定要多与鼓励。让孩子在兴趣盎然中动了脑筋,开始用数学工具思考,表达和发展解决问题的能力。(还有一些图形简单变换的游戏,对于孩子理解形状,提高空间想象力特别有帮助,为省篇幅从略。)

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知道数独游戏的人比较多;知道余数井的人少多了。您想了解更多吗?或想了解一下游戏的难点,挑战之处在哪里?我们会在以后通过具体实际例子作更详细的介绍。

请您多多关注。

谈到游戏,想作个声明。没有说学数学不要下功夫,只是玩玩就可以了。但是,在今天,社会日益多元化,日渐丰富的计算机游戏抓住了孩子们。让不同禀赋的孩子都有兴趣学数学,愿意配合学,主动学,多走一步去探究深想一下问题,如何能做到?就成为很重要的课题。兴趣与热爱是学习中不衰的动力;教育中有重新认识游戏方法的必要。想让孩子参加数学竞赛的家长:在小学阶段,不能纯逼孩子做题,要让孩子乐学;大家有兴趣研究一下竞赛题的方向,会发现很多游戏化的数学问题。真兴趣培养起来了,能力和乐于探索挑战的精神也是可以加上的。当然,游戏方法需要和其他配合,如讲清概念,专题训练,给孩子提示重点等。

只要引导得当,孩子在高中以后数学方面的兴趣充分发展了,而由于学习任务加重,系统化学习会更好。从以游戏和活动为主(小学到刚进初中)到更加系统化的学习,作为家长和老师,就是帮孩子保持了好奇心和探索的热情,并且把学习推进到了一个更高阶段。