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我们提供的组合学Combinatorics及其相关学科的代写,服务范围广, 其中包括但不限于:

  • Statistical Inference 统计推断
  • Statistical Computing 统计计算
  • Advanced Probability Theory 高等概率论
  • Advanced Mathematical Statistics 高等数理统计学
  • (Generalized) Linear Models 广义线性模型
  • Statistical Machine Learning 统计机器学习
  • Longitudinal Data Analysis 纵向数据分析
  • Foundations of Data Science 数据科学基础

数学代写|组合学代写Combinatorics代考|Bijection. Combinatorial Bijection Principle

Suppose that 59 teams are participating in a soccer cup. How many matches will be played? Even after additional explanations regarding the rules of the tournament, a large number of respondents hesitated to provide the answer, attempting the construction of various schemes and the related calculations. There were mathematicians among those who got confused about this issue, not to mention those who participate in the competition schedule. This is a kind of question to which the student can give an instant and reasonable answer, and at the same time it can make the specialist lose his balance and dig deep in search of the truth that is right on the surface. A foreword regarding the rules of cup competitions is needed. The classic system is that each match should end effectively (that is, by the victory of one of the teams), and the team that loses is no longer taking part in the tournament. This is the fundamental rule of the winner’s detection system, which is called a single-elimination, knockout, or sudden death tournament. The rest of the rules are not significant. Therefore, they are the responsibility of organizers of the competition (for example, football association). The organizers compile a schedule of the tournament, providing the rules for the creation of pairs at different stages of the competition, decide on which stage one or another team enter the tournament etc. They can also make a decision that the teams should play two matches on each stage instead of one. This does not change the essence of the knockout system, provided that after these two matches one of the two teams necessarily leaves the tournament. This alternative rule does not change our task either: the answer is simply doubled.

Therefore, assume we have a “classic competition”, when two teams play one match to determine which one of them is eliminated. How many matches will have to be played by all the teams?

The one, who focuses from the beginning on the various options of the schedule of competition, will waste a good deal of time searching for the answer. And this is the most popular route to a solution. Alternatively, the one, who realizes that the schedule of the competition is irrelevant to the task, no matter how simple or tricky it is, will get the answer almost immediately. The only important rule is the following: the losing team is eliminated from the competition. Imagine that the tournament is over. Which teams have not been knocked out? Only the cup winner. All the rest were eventually defeated and left the competition. There are 58 of them. And there were the same amount of matches, because each team lost in a single match, and each match resulted in a defeat of one of the teams. The teams, which lost in the tournament, are in such connection with the matches played, that there is no doubt that the number of matches and the number of losing teams are the same. This connection is called bijective correspondence (or one-to-one correspondence). We will have to deal with many more similar situations and use the term “bijective correspondence” or simply “bijection”, and therefore, it is time to stop and explain in detail its exact meaning.

数学代写|组合学代写Combinatorics代考|Bijection between paths

  1. Here we will deal with the summation of numbers and vectors. If one needs to calculate the sum of several (many) summands, then by the appropriate positioning of parentheses, this task can be reduced to the repeated summation of two summands. Moreover, the parentheses can be positioned in many different ways. The result does not depend on this. This is one of the fundamental arithmetical laws. It can be deduced from the associativity of addition, which refers to any three summands. The reader is well familiar with this property from the elementary school. Symbolically, it is presented as follows:
    (a+b)+c=a+(b+c) .
    Considering the sums of many summands we will adhere to the following rule: each “+” sign must correspond to a certain pair of parentheses (opening and closing parentheses). Hence, there should be the same amount of pairs of parentheses as the amount of “+” signs in the expression. In particular, under such agreement, the associativity property is expressed as follows:
    ((a+b)+c)=(a+(b+c)) .
    Actually, we are not interested in associativity law and its consequences. We are dealing with a purely combinatorial problem: how many ways are there to place parentheses correctly in the sum of $n$ summands? The word “correctly” here means that there should be
  2. equal amounts of opening and closing parentheses, and every pair of parentheses (opening parenthesis; closing parenthesis) corresponds to a certain “+” sign. In other words, pairs of parentheses (opening and closing) must be in bijective correspondence with the “+” signs.
  3. In the case of three summands, there are two ways to place parentheses: $((a+b)+c)$ and $(a+(b+c))$.


数学代写|组合学代写Combinatorics代考|Bijection. Combinatorial Bijection Principle

假设有 59 支球队参加足球杯。将进行多少场比赛?即使在对比赛规则进行了额外解释后,仍有大量受访者犹豫不决,试图构建各种方案和相关计算。被这个问题搞糊涂的不乏数学家,更何况是参加比赛日程的人。这是一种学生可以立即给出合理答案的问题,同时也可以让专家失去平衡,深入挖掘表面上的真相。关于杯赛规则的前言是必要的。经典的系统是每场比赛都应该有效结束(即由其中一支球队的胜利),输的队伍不再参加比赛。这是获胜者检测系统的基本规则,称为单淘汰赛、淘汰赛或猝死锦标赛。其余的规则并不重要。因此,他们是比赛组织者(例如足协)的责任。组织者编制比赛时间表,提供在比赛不同阶段创建配对的规则,决定一个或另一个球队进入比赛的哪个阶段等。他们还可以决定球队应该进行两场比赛在每个阶段而不是一个阶段。这并没有改变淘汰赛制度的本质,前提是在这两场比赛之后,两支球队中的一支必须退出锦标赛。



数学代写|组合学代写Combinatorics代考|Bijection between paths

  1. 这里我们将处理数字和向量的求和。如果需要计算几个 (许多) 被加数的和,那么通过 括号的适当定位,这个任务可以简化为两个被擞的重复求和。此外,括号可以以许多 不同的方式放置。结果不取决于此。这是基本的算术定律之一。它可以从加法的结合性 推导出来,它指的是任何三个被吅数。读者从小学就熟悉这个属性了。象征性地,它呈 现如下:
    (a+b)+c=a+(b+c) .
    考虑到许多加数的总和,我们将遵循以下规则: 每个” ${ }^{\prime \prime}$ 号必须对应于特定的一对括号 (左括号和右括号) 。因此,括号对的数量应该与表达式中““”号的数量相同。特别地, 在这种约定下,结合性表示如下:
    ((a+b)+c)=(a+(b+c)) .
    实际上,我们对结合律及其后果不感兴趣。我们正在处理一个纯粹的组合问题: 有多少 种方法可以正确地在总和中放置括号 $n$ 求和? 这里的“正确”一词意味着应该有
  2. 等量的左右括号,每对括号 (左括号; 右括号) 对应某个“+”号。换句话说,括号对(左 括号和右括号)必须与“”“符号双射对应。
  3. 在三个被加数的情况下,有两种放置括号的方法: $((a+b)+c)$ 和 $(a+(b+c))$.
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