组合数学学习笔记(三)(文末附七夕彩蛋)

Combinatorial Mathematics
(TsinghuaX: 60240013x)

Inclusion-Exclusion Theorem And Pigeonhole Principle

• Calculate the number of permutations of a,b,c,d,e,f,g which don’t contain ace and df.
  

• Calculate the number of n-strings with alphabet {a,b,c,d} such that a,b,c show up at least once.
  

Euler Function Φ(n) is the number of integers which is smaller than n and relatively prime to n.

n=pa11pa22…pakk

Φ(n)=n(1−1p1)(1−1p2)…(1−1pk)

Pigeonhole Principle
If there’re n−1 pigeonholes, n pigeons, there’s at least one pigeonhole which has at least 2 pigeons.

• We know n+1 positive integers, all of them are ≤2n, prove that at least 2 of them are relatively prime.

  Hint:
  Take n boxes, put 1 and 2 in the first one, 3 and 4 in the second one, 5 and 6 in the third one, so forth, 2n−1 and 2n in the nth one.
  Now we take n+1 numbers from n boxes, so at least one box would be emptied. So there must be a pair of adjacent numbers 40 among these n+1 ones, and they are relatively prime.

• Take any n+1 integers from 1 to 2n, among them there’s at least one pair such that one is the multiple of the other.

  Proof:
  Assume the n+1 numbers are a1,a2,…,an+1. Dividing 2’s until all of them becomes odd numbers. Then it construct a sequence r1,r2,…,rn+1.
  These n+1 numbers are still in [1,2n] and they are all odd. While there are only n odd numbers in [1,2n].
  So there must be ri=rj=r, then ai=2kir, aj=2kjr
  If ai>aj, ai is a multiple of aj.

• Assume a1,a2,…,a100 is a sequence consists of 1 and 2. And any subsequence of 10 consecutive in it has a sum that is ≤16:
  ai+ai+1+⋯+ai+19≤16,1≤i≤91.
  So ∃ h and k such that k>h and ah+ah+1+⋯+ak=39

  Proof:
  Let Sj=∑ji=1ai, j=1,2,…,100
  S1≤S2≤⋯≤S100,
  S100=(a1+a2+⋯+a10)+(a11+⋯+a20)+⋯+(a91+⋯+a100)
  According to assumption, ai+ai+1+⋯+ai+19≤16,1≤i≤91.
  We have S100≤10×16=160.
  Create sequence S1,S2,…,S100,S1+39,S2+39,…,S100+39 with 200 terms.
  The largest term S100+39≤160+39=199.
  By pigeonhole principle, there must be two equal terms.
  Assume Sk=Sh+39, k>h
  So Sk−Sh=39, which is, ah+ah+1⋯+ak=39.

Pigeonhole Principle (2)
If there’re n pigeonholes, kn+1 pigeons, there is at least 1 pigeonhole which has k+1 pigeons.

Chinese Remainder Theorem
m and n are relatively prime, for any nonnegative interger a and b (a<m,b<n), there must a positive interger x which makes the equations solvable.

{x=pm+ax=qn+bp,q∈ℕ

Chinese Remainder Theorem, RT

Ramsey Problem
There must be 3 mutual friends or 3 mutual strangers among 6 people.

Ramsey Number

Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6, 6). In that case, he believes, we should attempt to destroy the aliens.
— — Joel Spencer

Group

Turnable World

Classify the arithmetical operation, learn to classify according to the degree of difficulty, but not according to their external characteristics, this is what I understand for the mathematicians’ task in the future, and this is the way I wish to go.
— — Evariste Galois

A given Set G and the binary operation of G, satisfy the following condition is called group.

• Closure
• Associativity
• Identity

• Inverse

• G={0,1,2,…,n−1}with the operation of the addition of mod n is group

• All rigid items rotation in two dimensional Euclidean space.

Concepts Related

• finite group
• infinte group
• the order of group
• subgroup
• commutative group (Abelian group)

Properties Related

• Unique Identity
  e1e2=e2=e1
• Cancellation Law stands
  ab=ac→b=c, ba=ca→b=c
• Unique Inverse element
  aa−1=a−1a=e
• (ab....c)−1=c−1…b−1a−1
• G is limited, a∈G, then exist the smallest positive integer r, that ar=e and a−1=ar−1 .
  Proof:
  Let |G|=g, then a,a2,…,ag,ag+1∈G.
  From the pigeonhole principle, there will be the identical items.
  Set am=al, 1≤m<l≤g+1, e=al−m, 1≤l−m≤g
  Let l−m=r. Then there is ar=ar−1a=e, which is a−1=ar−1.
  Since there is a positive integer r that ar=e, among which we could find the smallest, assume to be r. r is known as the order of a.
  It can be seen that H={a,a2,…,ar−1,ar=e}.
  Under its inherited operations is also a group.

Permutation Group

Permute: 1−1 mapping on set [1,n] is called as n-order permutation.
Represented as(1a12a2……nan), a1a2…an is one of the arrangements in [1, n].

Permutation does not satisfy Commutative Law

• Rotation group of Equilateral triangle

Cycle, Odd Cycle and Even Cycle

• (a1,a2,…am)=(a1a2a2a3……am−1amama1) known as cycle representation of permutation.
• If p=(a1a2…an), then pn=(1)(2)…(n)=e.
• Any permutation can be represented into the product of numerous disjointed cycles.

Conjugate Class
Normally, it can make any permutation p in Sn decomposed into numerous disjointed cycles composition.
P=(a1a2…ak1)(b1b2…bk2)…(h1h2…hkl)
Among k1+k2+⋯+kl=n, set the number of occurrence of k order cycle as c_k, use (k)ck to represent, then the permutation format of Sn is

(1)c1(2)c2...(n)cn

∑k=1nk∗ck=n

• Any cycle can be represented by the product of swaps.
  (12…n)=(12)(13)…(1n)
  Proof:
  Set (12…n−1)=(12)(13)…(1n−1).
  (12…n−1)(1n)=(122334……n−11)(1n2233…4n−1…n1)=
  (122334……n−11nn) (2233…4n−1…1nn1)=
  (122334……n−1nn1)=(123…n)
• The decomposition of each permutation is not unique.

Odd Permutation and Even Permutation
If a permutation can be decomposed as the product of odd number of position swapping, then it is odd permutation. If it can be decomposed as the product of even number of position swapping, it is even permutation.

• Klotski of Number
  

p=(0)(115)(214)(313)(414)(511)(610)(79)(8)

is odd permutation.Start from the 0 over the right bottom corner back to bottom corner, in the horizontal direction, the vertical direction have done even-numbered times of change. An odd permutation does not equal to an even permutation.

Euler’s polyhedron formula
From Euler’s polyhedron formula: There are 5 types of regular convex polyhedron, which are: Tetrahedron, Octahedron, Icosahedron, Hexahedron (Cube) and Dodecahedron.
The sum value of any convex polyhedron vertices v and number of faces f is 2 more than the edge number e, which is v+f−e=2. This is Euler’s polyhedron formula.

Mutual Dual Polyhedron
A polyhedron and those polyhedron which used its various surface center as top, is known as Mutual Dual Polyhedron.

Burnside Lemma

The Coloring Problem of Equivalence Class

Set G as the permutation group of [1,n]. G is one of the subgroup of Sn. k∈[1,n], all the permutations which caused k element remained unchanged will compose a stabilizer of k, noted as Zk.
For Example, G=e,(12),(34),(12)(34)
Z1=e,(34)
Z2=e,(34)
Z3=Z4=e,(12)
Permutation group G’s k fixed displacement class Zk is G’s subgroup.

Orbit-stabilizer theorem
Set G as the permutation group of [1,n], Ek is [1,n] under G’s effect contains k’s equivalence class, Zk is k stablizer.
There is |Ek||Zk|=|G|.

Burnside Lemma
Set G=a1,a2,…ag is the permutation group of the target group [1,n]. Each permutation is written into the product of disjointed cycles. c1(ak) is number of the elements which are not changed by permutation ak, which is the number of cycles with the length of 1. G decomposes [1,n] into l number of equivalence classes that,

l=1|G|∑j=1gc1(aj)

Gossip of Group

鉴于今天是中国七夕节，在篇末重温一下群论创立者——伽罗瓦的悲情故事，以作为这部分学习的结束。准确地说，这三篇学习笔记其实是组合数学相关基础知识的预热，关于更多内容补充和相关习题会持续进行，一些细节也会继续修正，打call～

埃瓦里斯特·伽罗瓦（Évariste Galois，1811年10月25日-1832年5月31日），法国数学家，与尼尔斯·阿贝尔并称为现代群论的创始人。在一次几近自杀的决斗中英年早逝，引起种种揣测。

伽罗瓦的父母都是知识分子，12岁以前，伽罗瓦的教育全部由他的母亲负责，他的父亲在伽罗瓦4岁时被选为Bourg La Reine的市长。

12岁，伽罗瓦进入路易皇家中学就读，成绩都很好，却要到16岁才开始跟随 Vernier 老师学习数学，他对数学的热情剧然引爆，对于其他科目再也提不起任何兴趣。校方描述此时的伽罗瓦是“奇特、怪异、有原创力又封闭”。

1829年，伽罗瓦将他在代数方程解的结果呈交给法国科学院，由奥古斯丁·路易·柯西（Augustin Louis Cauchy） 负责审阅，柯西却将文章连同摘要都弄丢了（19世纪的两个短命数学天才阿贝尔与伽罗瓦不约而同地都“栽”在柯西手中）。

据说1832年3月他在狱中结识一个医生的女儿并陷入狂恋，因为这段感情，他陷入一场决斗，自知必死的伽罗瓦在决斗前夜将他的所有数学成果狂笔疾书纪录下来，并时不时在一旁写下“我没有时间”，第二天他果然在决斗中身亡，时间是1832年5月31日。这个传说富浪漫主义色彩，为后世史家所质疑。

在去世的前一天晚上，伽罗瓦仍然奋笔疾书，总结他的学术思想，整理、概述他的数学工作。他希望有朝一日自己的研究成果能大白于天下。

他的朋友 Chevalier 遵照伽罗瓦的遗愿，将他的数学论文寄给卡尔·弗里德里希·高斯与雅各比，但是都石沉大海，要一直到1843年，才由刘维尔肯定伽罗瓦结果之正确、独创与深邃，并在1846年将它发表。

我们的天才伽罗瓦是否真的死于爱情，我们不得而知，方舟子等人也提出过阴谋论等说法。我宁愿相信这是一场阴谋。

阿贝尔死于贫穷，伽罗瓦则死于愚蠢。全部科学史上，极度愚蠢战胜不可抑制的天才的例子，再没有比埃瓦里斯特·伽罗瓦过于短促的一生所提供的例子更全面了。关于他的不幸的记录，很可能作为一切自负的教书匠、无耻的政客，以及骄傲自满的院士们的一个不祥的纪念碑而竖立。伽罗瓦不是“无用的天使”，但是面对大群愚蠢的人联合反对他，就连他那非凡的力量也被粉碎了，他在同一个接着一个的不可战胜的蠢材的斗争中，耗尽了自己的生命。

（在告别人世的前夜）整个晚上，他把飞逝的时间用来焦躁地一气写出他的科学上的最后遗言，在死亡之前（他预见到死亡能够追上他）尽快地写，把他丰富的思想中那些伟大的东西尽量写一些出来。他不时中断，在纸边空白处写上“我没有时间，我没有时间”，然后又接着涂写下一个极其潦草的提纲。他在天亮之前那最后几个小时拼命写出的东西，将使世世代代的数学家们忙上几百年。他一劳永逸地给一个折磨了数学家达几个世纪之久的谜，找出了真正的解答。这个谜就是在什么条件下方程是可解的。但这只不过是许多事情中的一件。在这项伟大的工作中，伽罗瓦极其成功地用了群论。伽罗瓦的确是今天在全部数学中具有根本重要性的这一抽象论的一位伟大先驱者。

“我亲爱的朋友，”他开始写道，“我在分析方面作出了一些新的发现。”然后他在时间允许的情况下着手写出大纲。它们是划时代的。他结束说：“请雅可比或高斯公开提出他们的意见，不是对这些定理的正确性，而是对它们的重要性。我希望以后会有人发现，辨读这一堆写得很潦草的东西，对他们是有益的。满怀激情地拥抱你。E·伽罗瓦。”

1832年5月30日清晨很早的时候，伽罗瓦在“决斗场”与他的对手相遇。决斗是在25步的距离用手枪对射。伽罗瓦倒下了，肠子被射穿。没有医生在场。他被丢在他倒下的地方。9点钟的时候，一个路过那里的农民把他送到科尚医院。伽罗瓦知道他快死了。在不可避免的腹膜炎开始以前，在他的神志仍然完全清醒的时候，伽罗瓦拒绝了一个神父的祈祷。也许他记起了他的父亲。他的弟弟，他的家人中唯一得到通知的一个，流着泪赶到了。伽罗瓦努力以一种坚韧精神去安慰他的弟弟：“不要哭，”他说，“我需要我的全部勇气在20岁时死去。” 1832年5月31日上午，伽罗瓦在他生命的第21个年头去世了。他被埋葬在南公墓的普通壕沟里，所以今天伽罗瓦的坟墓已无踪迹可寻。他不朽的纪念碑是他的著作，共60页。（摘自《数学大师》）