组合数学 第五章
组合数学_第5章
an c1an 1 c2 an 2 0 a0 d 0 , a1 d 1
的任一解。 将初值条件代入 an
Kr K r
n 1 1
n 2 2
a 0 K1 K 2 d 0 a1 K1r1 K 2 r2 d1
第五章 递 归 关 系 由r1≠r2知，
| P2x |=|{(i2, i3, …, in)|it+1≠t, t=1, 2, …, n-1}|=D(n-1) 从而dn=D(n-2)+D(n-1)，故知 D(n)=(n-1)(D(n-2)+D(n-1)), n≥3, D(0)=0, D(2)=1
第五章 递 归 关 系 例2（Hanoi塔问题） n个圆盘，从A柱经B柱移到C柱(参见 图1.1)。 要求每次只移一个圆盘； 移动过程始终保持大盘在 下，小盘在上；中途 A 、 B 、 C 柱均可作临时柱，最终移至 C 柱， 则A到C的最少移动次数为
h(n)=2h(n-1)+1
…
A
B
C
图5.1.1 Hano塔示意图
第五章 递 归 关 系 解 设 h(n) 为 问 题 的 解 ， 则 h(1)=1, h(2)=3=2h(1)+1,
h(3)=5=2h(2)+1, …, h(n)=2h(n-1)+1。 又若设利用 C 柱把 A 柱上 n-1 个圆盘移到 B 柱，移动次数为 h(n-1)，则将A柱所剩最大圆盘移到C柱只需移动一次，再将 B上的n-1个圆盘利用A柱移到C柱（已有一个最大圆盘在下面） 共需移动h(n-1)次。 故总的移动次数为h(n)=2h(n-1)+1。
d n n 1 n2 n n 1 n2 x ( x c x c x ) nx c ( n 1 ) x c ( n 2 ) x 0 1 2 1 2 dx
组合数学与图论
● 02
第2章 图论基础
什么是图论
图论是研究图结构的 数学分支，用于描述 对象之间的关系。图 由节点和边组成，节 点表示对象，边表示 对象之间的关系。
基本概念
无向图
边没有方向的图
权重图
边带有权重的图
度
节点相连的边数 称为节点的度
91%
有向图
边有方向的图
图的表示方法
01 邻接矩阵
02 邻接表
判断图中的节点是否都是连通的
02 组合数学方法
连通性定理和算法可以用于判断和求解
03
总结
组合数学和图论相互结合，能够解决图的同构、 着色、匹配和连通性等各种问题，通过组合数学 方法的运用，可以更好地探索图论中的难题。
● 04
第四章 组合数学与图论在计 算机科学中的应用
图数据库与图搜索
图数据库是一种专门用于存储和查询图结构数据 的数据库系统。在计算机科学中，图搜索算法如 Dijkstra算法、A*算法等被广泛应用于图数据库 的查询和分析，帮助用户快速准确地获取所需信 息。
03
● 05
第五章 组合数学与图论在统 计学中的应用
基于图的统计分 析
利用组合数学和图论 的方法进行统计学分 析，如图的频繁模式 挖掘、图数据的聚类 分析等。这些方法能 够帮助研究人员从大 量数据中提取出有用 的信息并进行深入分 析。
网络数据采样与推断
节点采样
通过在网络中随 机选择节点来获
取样本数据
使得相邻节点颜 色不同
图的匹配问题
图的匹配问题是指在 图中找到一些相互不 相邻的边，使得边的 数量最大化。组合数 学的匹配定理和匹配 算法可以用于解决图 的匹配问题。
图的连通性问题
Richard组合数学第5版-第5章课后习题答案（英文版）
Richard组合数学第5版-第5章课后习题答案（英⽂版）Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul TerwilligerSelected solutions for Chapter51.For an integer k and a real number n,we shown k=n?1k?1+n?1k.First assume k≤?1.Then each side equals0.Next assume k=0.Then each side equals 1.Next assume k≥1.RecallP(n,k)=n(n?1)(n?2)···(n?k+1).We haven k=P(n,k)k!=nP(n?1,k?1)k!.n?1 k?1=P(n?1,k?1)(k?1)!=kP(n?1,k?1)k!.n?1k(n?k)P(n?1,k?1)k!.The result follows.2.Pascal’s triangle begins111121133114641151010511615201561172135352171182856705628811936841261268436911104512021025221012045101···13.Let Z denote the set of integers.For nonnegative n∈Z de?ne F(n)=k∈Zn?kk.The sum is well de?ned since?nitely many summands are nonzero.We have F(0)=1and F(1)=1.We show F(n)=F(n?1)+F(n?2)for n≥2.Let n be /doc/6215673729.htmling Pascal’s formula and a change of variables k=h+1,F(n)=k∈Zn?kk=k∈Zn?k?1k?1=k∈Zn?k?1k+h∈Zn?h?2h=F(n?1)+F(n?2).Thus F(n)is the n th Fibonacci number.4.We have(x+y)5=x5+5x4y+10x3y2+10x2y3+5xy4+y5and(x+y)6=x6+6x5y+15x4y2+20x3y3+15x2y4+6xy5+y6.5.We have(2x?y)7=7k=07k27?k(?1)k x7?k y k.6.The coe?cient of x5y13is35(?2)13 185.The coe?cient of x8y9is0since8+9=18./doc/6215673729.htmling the binomial theorem,3n=(1+2)n=nk=0nSimilarly,for any real number r,(1+r)n=nk=0nkr k./doc/6215673729.htmling the binomial theorem,2n=(3?1)n=nk=0(?1)knk3n?k.29.We haven k =0(?1)k nk 10k =(?1)n n k =0(?1)n ?k n k 10k =(?1)n (10?1)n =(?1)n 9n .The sum is 9n for n even and ?9n for n odd.10.Given integers 1≤k ≤n we showk n k =n n ?1k ?1.Let S denote the set of ordered pairs (x,y )such that x is a k -subset of {1,2,...,n }and yis an element of x .We compute |S |in two ways.(i)To obtain an element (x,y )of S there are n k choices for x ,and for each x there are k choices for y .Therefore |S |=k n k .(ii)Toobtain an element (x,y )of S there are n choices for y ,and for each y there are n ?1k ?1 choices for x .Therefore |S |=n n ?1k ?1.The result follows.11.Given integers n ≥3and 1≤k ≤n .We shown k ? n ?3k = n ?1k ?1 + n ?2k ?1 + n ?3k ?1.Let S denote the set of k -subsets of {1,2,...,n }.Let S 1consist of the elements in S thatcontain 1.Let S 2consist of the elements in S that contain 2but not 1.Let S 3consist of the elements in S that contain 3but not 1or 2.Let S 4consist of the elements in S that do|S |= n k ,|S 1|= n ?1k ?1 ,|S 2|= n ?2k ?1 ,|S 3|= n ?3k ?1 ,|S 4|= n ?3k .The result follows.12.We evaluate the sumnk =0(?1)k nk 2.First assume that n =2m +1is odd.Then for 0≤k ≤m the k -summand and the (n ?k )-summand are opposite.Therefore the sum equals 0.Next assume that n =2m is even.Toevaluate the sum in this case we compute in two ways the the coe?cient of x n in (1?x 2)n .(i)By the binomial theorem this coe?cient is (?1)m 2m m .(ii)Observe (1?x 2)=(1+x )(1?x ).We have(1+x )n =n k =0n k x k,(1?x )n =n k =0nk (?1)k x k .3By these comments the coe?cient of x n in(1?x2)n isn k=0nn?k(?1)knk=nk=0(?1)knk2.2=(?1)m2mm.13.We show that the given sum is equal ton+3k .The above binomial coe?cient is in row n+3of Pascal’s /doc/6215673729.htmling Pascal’s formula, write the above binomial coe?cient as a sum of two binomial coe?ents in row n+2of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n+1of Pascal’s triangle.Write each of these as a sum of two binomial coe?ents in row n of Pascal’s triangle.The resulting sum isn k+3nk?1+3nk?2+nk?3.14.Given a real number r and integer k such that r=k.We showr k=rr?kr?1k.First assume that k≤?1.Then each side is0.Next assume that k=0.Then each side is 1.Next assume that k≥1.ObserverP(r?1,k?1)k!,andr?1k=P(r?1,k)k!=(r?k)P(r?1,k?1)k!.The result follows.15.For a variable x consider(1?x)n=nk=0nk(?1)k x k.4Take the derivative with respect to x and obtain n(1x)n1=nk=0nk(?1)k kx k?1.Now set x=1to get(?1)k k.The result follows.16.For a variable x consider(1+x)n=nk=0nkx k.Integrate with respect to x and obtain(1+x)n+1 n+1=nk=0nkx k+1k+1+Cfor a constant C.Set x=0to?nd C=1/(n+1).Thus (1+x)n+1?1n+1=nk=0nkx k+1k+1.Now set x=1to get2n+1?1 n+1=k+1.17.Routine.18.For a variable x consider(x?1)n=nk=0nk(?1)n?k x k.Integrate with respect to x and obtain(x?1)n+1 n+1=nk=0nk(?1)n?kx k+1k+1+Cfor a constant C.Set x=0to?nd C=(?1)n+1/(n+1).Thus (x?1)n+1?(?1)n+1n+1=nk=0nk(?1)n?kx k+1k+1Now set x =1to get(?1)n n +1=n k =0n k(?1)n ?k 1k +1.Therefore1n +1=n k =0 n k (?1)k 1k +1 .19.One readily checks2 m 2 + m 1=m (m ?1)+m =m 2.Therefore n k =1k 2=nk =0k 2=2nk =0 k 2 +n k =0k1=2 n +13 +n +12 =(n +1)n (2n +1)6.20.One readily checksm 3=6 m 3 +6 m 2 + m1.Thereforen k =1k3=n=6nk =0 k3+6n k =0 k2 +n k =0k1 =6 n +14 +6 n +13 +n +12 =(n +1)2n 24= n +12 2.621.Given a real number r and an integer k .We showrk=(?1)kr +k ?1k .First assume that k <0.Then each side is zero.Next assume that k ≥0.Observe r k =(r )(r 1)···(r k +1)k !=(?1)kr (r +1)···(r +k ?1)k !=(?1)kr +k ?1k.22.Given a real number r and integers k,m .We showr m m k = r k r ?km ?k.First assume that mObserver m m k =r (r ?1)···(r ?m +1)m !m !k !(m ?k )!=r (r ?1)···(r ?k +1)k !(r ?k )(r ?k ?1)···(r ?m +1)(m ?k )!= r k r ?k m ?k .23.(a) 2410.(b) 94 156.(c) 949363.(d)94156949363.24.The number of walks of length 45is equal to the number of words of length 45involving10x ’s,15y ’s,and 20z ’s.This number is45!10!×15!×20!.725.Given integers m 1,m 2,n ≥0.Shown k =0m 1k m 2n ?k = m 1+m 2n .Let A denote a set with cardinality m 1+m 2.Partition A into subsets A 1,A 2with cardinalitiesm 1and m 2respectively.Let S denote the set of n -subsets of A .We compute |S |in two ways.(i)By construction|S |= m 1+m 2n .(ii)For 0≤k ≤n let the set S k consist of the elements in S whose intersection with A 1has cardinality k .The sets {S k }n k =0partition S ,so |S |= nk =0|S k |.For 0≤k ≤n we now compute |S k |.To do this we construct an element x ∈S k via the following 2-stage procedure: stage to do #choices 1pick x ∩A 1 m 1k2The number |S k |is the product of the entries in the right-most column above,which comes to m 1k m 2n ?k .By these comments |S |=n k =0m 1k m 2n ?k .The result follows.26.For an integer n ≥1shown k =1 n k n k ?1 =12 2n +2n +1 ? 2n n .Using Problem 25,n k =1 n k nk ?1 =n k =0n k n k ?1 =n k =0n k nn +1?k =2n n +1 =12 2n n ?1 +12 2n n +1.8It remains to show12 2nn ?1 +12 2n n +1 =12 2n +2n +1 ? 2n n.This holds since2n n ?1 +2 2n n + 2n n +1 = 2n +1n +2n +1n +1= 2n +2n +1.27.Given an integer n ≥1.We shown (n +1)2n ?2=nk =1Let S denote the set of 3-tuples (s,x,y )such that s is a nonempty subset of {1,2,...,n }and x,y are elements (not necessarily distinct)in s .We compute |S |in two ways.(i)Call an element (s,x,y )of S degenerate whenever x =y .Partition S into subsets S +,S ?with S +(resp.S ?)consisting of the degenerate (resp.nondegenerate)elements of S .So |S |=|S +|+|S ?|.We compute |S +|.To obtain an element (s,x,x )of S +there are n choices for x ,and given x there are 2n ?1choices for s .Therefore |S +|=n 2n ?1.We compute |S ?|.To obtain an element (s,x,y )of S ?there are n choices for x,and given x there are n ?1choices for y ,and given x,y there are 2n ?2choices for s .Therefore |S ?|=n (n ?1)2n ?2.By these comments|S |=n 2n ?1+n (n ?1)2n ?2=n (n +1)2n ?2.(ii)For 1≤k ≤n let S k denote the set of elements (s,x,y )in S such that |s |=k .Thesets {S k }nk =1give a partition of S ,so |S |= n k =1|S k |.For 1≤k ≤n we compute |S k |.To obtain an element (s,x,y )of S k there are n k choices for s ,and given s there are k 2ways to choose the pair x,y .Therefore |S k |=k 2 nk .By these comments|S |=n k =1k 2 n k .The result follows.28.Given an integer n ≥1.We shown k =1k n k 2=n 2n ?1n ?1 .Let S denote the set of ordered pairs (s,x )such that s is a subset of {±1,±2,...,±n }andx is a positive element of s .We compute |S |in two ways.(i)To obtain an element (s,x )of S There are n choices for x ,and given x there are 2n ?1n ?1 choices for s .Therefore|S |=n 2n ?1n ?1.9(ii)For1≤k≤n let S k denote the set of elements(s,x)in S such that s contains exactlyk positive elements.The sets{S k}nk=1partition S,so|S|=nk=1|S k|.For1≤k≤nwe compute|S k|.To obtain an element(s,x)of S k there are nkways to pick the positiveelements of s and nn?kways to pick the negative elements of s.Given s there are kways to pick x.Therefore|S k|=k nk2.By these comments |S|=nk=1knk2.The result follows.29.The given sum is equal tom2+m2+m3n .To see this,compute the coe?cient of x n in each side of(1+x)m1(1+x)m2(1+x)m3=(1+x)m1+m2+m3.In this computation use the binomial theorem.30,31,32.We refer to the proof of Theorem5.3.3in the text.Let A denote an antichain such that|A|=nn/2.For0≤k≤n letαk denote the number of elements in A that have size k.Sonk=0αk=|A|=nn/2.As shown in the proof of Theorem5.3.3,≤1,with equality if and only if each maximal chain contains an element of A.By the above commentsnk=0αknn/2nknk≤0,with equality if and only if each maximal chain contains an element of A.The above sum is nonpositive but each summand is nonnegative.Therefore each summand is zero and the sum is zero.Consequently(a)each maximal chain contains an element of A;(b)for0≤k≤n eitherαk is zero or its coe?cient is zero.We now consider two cases.10Case:n is even.We show that for0≤k≤n,αk=0if k=n/2.Observe that for0≤k≤n, if k=n/2then the coe?cient ofαk isnonzero,soαk=0.Case:n is odd.We show that for0≤k≤n,eitherαk=0if k=(n?1)/2orαk=0 if k=(n+1)/2.Observe that for0≤k≤n,if k=(n±1)/2then the coe?cient ofαk is nonzero,soαk=0.We now show thatαk=0for k=(n?1)/2or k=(n+1)/2. To do this,we assume thatαk=0for both k=(n±1)/2and get a contradiction.By assumption A contains an element x of size(n+1)/2and an element y of size(n?1)/2. De? ne s=|x∩y|.Choose x,y such that s is maximal.By construction0≤s≤(n?1)/2. Suppose s=(n?1)/2.Then y=x∩y?x,contradicting the fact that x,y are incomparable. So s≤(n?3)/2.Let y denote a subset of x that contains x∩y and has size(n?1)/2. Let x denote a subset of y ∪y that contains y and has size(n+1)/2.By construction |x ∩y|=s+1.Observe y is not in A since x,y are comparable.Also x is not in A by the maximality of s.By construction x covers y so they are together contained in a maximal chain.This chain does not contain an element of A,for a contradiction.33.De?ne a poset(X,≤)as follows.The set X consists of the subsets of{1,2,...,n}. For x,y∈X de?ne x≤y whenever x?y.Forn=3,4,5we display a symmetric chain decomposition of this poset.We use the inductive procedure from the text.For n=3,,1,12,1232,233,13.For n=4,,1,12,123,12344,14,1242,23,23424,For n=5,,1,12,123,1234,123455,15,125,12354,14,124,124545,1452,23,234,234525,23524,2453,13,134,134535,13534,345.1134.For 0≤k ≤ n/2 there are exactlyn kn k ?1symmetric chains of length n ?2k +1.35.Let S denote the set of 10jokes.Each night the talk show host picks a subset of S for his repertoire.It is required that these subsets form an antichain.By Corollary 5.3.2each antichain has size at most 105 ,which is equal to 252.Therefore the talk show host can continue for 252nights./doc/6215673729.htmlpute the coe?cient of x n in either side of(1+x )m 1(1+x )m 2=(1+x )m 1+m 2,In this computation use the binomial theorem.37.In the multinomial theorem (Theorem 5.4.1)set x i =1for 1≤i ≤t .38.(x 1+x 2+x 3)4is equal tox 41+x 42+x 43+4(x 31x 2+x 31x 3+x 1x 32+x 32x 3+x 1x 33+x 2x 33)+6(x 21x 22+x 21x 23+x 22x 23)+12(x 21x 2x 3+x 1x 22x 3+x 1x 2x 23).39.The coe?cient is10!3!×1!×4!×0!×2!which comes to 12600.40.The coe?cient is9!3!×3!×1!×2!41.One routinely obtains the multinomial theorem (Theorem 5.4.1)with t =3.42.Given an integer t ≥2and positive integers n 1,n 2,...,n t .De?ne n = ti =1n i .We shownn 1n 2···n t=t k =1n ?1n 1···n k ?1n k ?1n k +1···n t.Consider the multiset{n 1·x 1,n 2·x 2,...,n t ·x t }.Let P denote the set of permutations of this multiset.We compute |P |in two ways.(i)We saw earlier that |P |=n !n 1!×n 2!×···×n t != n n 1n 2···n t.12(ii)For1≤k≤t let P k denote the set of elements in P that have?rst coordinate x k.Thesets{P k}tk=1partition P,so|P|=tk=1|P k|.For1≤k≤t we compute|P k|.Observe that|P k|is the number of permutations of the multiset{n1·x1,...,n k?1·x k?1,(n k?1)·x k,n k+1·x k+1,...,n t·x t}. Therefore|P k|=n?1n1···n k?1n k?1n k+1···n t.By these comments|P|=tn1···n k?1n k?1n k+1···n t.The result follows.43.Given an integer n≥1.Show by induction on n that1 (1?z)n =∞k=0n+k?1kz k,|z|<1.The base case n=1is assumed to hold.We show that the above identity holds with n replaced by n+1,provided that it holds for n.Thus we show1(1?z)n+1=∞=0n+z ,|z|<1.Observe1(1?z)n+1=1(1?z)n11?z=∞k=0n+k?1kz k∞h=0z h=0c zwherec =n?1+n1+n+12+···+n+ ?1=n+.The result follows.1344.(Problem statement contains typo)The given sum is equal to (?3)n .Observe (?3)n =(?1?1?1)n=n 1+n 2+n 3=nnn 1n 2n 3(?1)n 1+n 2+n 3=n 1+n 2+n 3=nnn 1+n 2+n 3=nnn 1n 2n 3(?1)n 2.45.(Problem statement contains typo)The given sum is equal to (?4)n .Observe (?4)n =(?1?1?1?1)n=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1+n 2+n 3+n 4=n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 1?n 2+n 3?n 4.Also0=(1?1+1?1)n= n 1+n 2+n 3+n 4=nnn 1n 2n 3n 4(?1)n 2+n 4.46.Observe√30=5=5∞ k =01/2k z k.For n =0,1,2,...the n th approximation to √30isa n =5n k =0 1/2k 5?k.We have14n a n051 5.52 5.4753 5.47754 5.47718755 5.477231256 5.4772246887 5.4772257198 5.4772255519 5.477225579 47.Observe101/3=21081/3=2(1+z)1/3z=1/4,=2∞k=01/3kz k.For n=0,1,2,...the n th approximation to101/3isnk=01/3k4?k.We haven a n021 2.1666666672 2.1527777783 2.1547067904 2.1543852885 2.1544442306 2.1544327697 2.1544350898 2.1544346059 2.15443470848.We show that a poset with mn+1elements has a chain of size m+1or an antichain of size n+1.Our strategy is to assume the result is false,and get a contradiction.By assumption each chain has size at most m and each antichain has size at most n.Let r denote the size of the longest chain.So r≤m.By Theorem5.6.1the elements of the posetcan be partitioned into r antichains{A i}ri=1.We have|A i|≤n for1≤i≤r.Thereforemn+1=ri=1|A i|≤rn≤mn, 15for a contradiction.Therefore,the poset has a chain of size m+1or an antichain of size n+1.49.We are given a sequence of mn+1real numbers,denoted{a i}mni=0.Let X denote the setof ordered pairs{(i,a i)|0≤i≤mn}.Observe|X|=mn+1.De?ne a partial order≤on X as follows:for distinct x=(i,a i)and y=(j,a j)in X,declare xof{a i}mni=0,and the antichains correspond to the(strictly)decreasing subsequences of{a i}mni=0sequence{a i}mni=0has a(weakly)increasing subsequence of size m+1or a(strictly)decreasingsubsequence of size n+1.50.(i)Here is a chain of size four:1,2,4,8.Here is a partition of X into four antichains:8,124,6,9,102,3,5,7,111Therefore four is both the largest size of a chain,and the smallest number of antichains that partition X. (ii)Here is an antichain of size six:7,8,9,10,11,12.Here is a partition of X into six chains:1,2,4,83,6,1295,10711Therefore six is both the largest size of an antichain,and the smallest number of chains that partition X.51.There exists a chain x116。
组合数学(第5章5.1)
证明1（组合证明）
(x+y)n =(x+y) (x+y)…(x+y) 用分配律展开： 每一项具有形式xnkyk, k=0,1,…,n. 共有2n项。它的系数相当于在n项因子选k个y，余 下n-k因子是x. 因此，等于组合数：
n k
这就证明了结论。
n nk k (x + y) = ∑ x y k =0 k
r r 1 r 1 注意： = k k + k 1
r r 1 r 2 r 2 = k k + k 1 + k 2
…
r r 1 r 2 r 2 r k r k 1 = k k + k 1 + k 2 + K + 1 + 0
S的一个元x, 令A是不含x的k-组合的集 合，而B是包含x的k-组合的集合，那么，X ＝A∪B，且A∩B＝。由加法原理 |X|＝|A|+|B|。
计算|A|和|B|。 A是集合S\{x}的所有k-组合集，因此
n 1 |A|＝ k
而B的元素是将x添加到S\{x}的(k1)-组合 因此，
n p(n, k ) = k
n＝3
n＝0
(0, 0)
n＝1 n＝2
(1, 0)
(3, 2)
这是二项式系数 的另 一种组合学解释。
n k
n＝4
5.2 二项式定理
定理5.2.1 令n是一个正整数, 那么对于所有的 定理 x, y有：
n nk k (x + y) = ∑ x y k k =0
n≥1
n n n （4） 1 + 2 + K + n = n2 n 1 1 2 n
Richard组合数学第5版-第5章课后习题答案(英文版)
evaluate the sum in this case we compute in two ways the the coefficient of xn in (1−x2)n. (i)
By the binomial theorem this coefficient is (−1)m
2m m
.
(ii) Observe (1 − x2) = (1 + x)(1 − x).
5
Now set x = 1 to get Therefore
(−1)n n =
n (−1)n−k 1 .
n+1
k
k+1
k=0
1
n
=
n (−1)k 1 .
n+1
k
k+1
k=0
19. One readily checks
m 2
+
m
= m(m − 1) + m = m2.
2
1
Therefore
n
n
k2 =
not contain 1 or 2 or 3. Note that {Si}4i=1 partition S so |S| =
4 i=1
|Si|.
We
have
n
n−1
n−2
n−3
n−3
|S| =
, k
|S1| =
, k−1
|S2| =
, k−1
|S3| =
, k−1
|S4| =
. k
The result follows.
We have
n P (n, k) nP (n − 1, k − 1)
=
组合数学(哈工大 第五章)
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任世军 (哈尔滨工业大学)
组合数学 递推关系
December 22, 2014
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递推关系
. Definition . 设{an } 为一序列， 把该序列中an 和它前面几个ai (0 ≤ i ≤ n) 关联起来的 方程称做一个递推关系（递归关系）。 .
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Example
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任世军 (哈尔滨ember 22, 2014
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递推关系
. . 在一个平面上有一个圆和n 条直线， 这些直线中的每一条在圆内都同其他 的直线相交。如果没有多于三条的直线相交于一点， 试问这些直线将圆分 成多少个不同区域？ . 解: 设这n 条直线将圆分成的区域数为an ， 如果有n − 1 条直线将圆分 成an−1 个区域， 那么再加入第n 条直线与在圆内的其他n − 1 条直线相 交。显然， 这条直线在圆内被分成n 条线段， 而每条线段又将第n 条直线 在圆内经过的区域分成两个区域。这样， 加入第n 条直线后， 圆内就增加 了n 个区域。 而对于n = 0， 显然有a0 = 1， 于是对于每个整数 n， 可以建立 如下带初值的递推关系 a0 = 1, a1 = 2, an = an−1 + n
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组合数学(第5章5.2)
如果A是一个链而C是反链，则|AC|1
第十八页，共22页。
定理5.7.1: 令(X, )是一个有限偏序集, 并令r是最大的 链的大小, 则X可以被划分成r个但不能再少的反链.
第十二页，共22页。
（2）若k=1, 生成{1,2,…,n}的1条对称链：
AkAk{n} 注意|Ak|=(n1)/2，因此|Ak|+|Ak{n}|=n.
注意到：{1,2,…,n}的任一个组合或者 是A或者是A{n}的形式，其中A是 {1,2,…,n1}的一个组合。
那么，可以验证：{1,2,…,n}的每一个 组合恰好出现在上面构造的某个对称链 中，这些链构成了{1,2,…,n}所有组合的 一个划分。
{3}{1,3}
{2}{2,3}
第十页，共22页。
* n=4时，有6条对称链： {1}{1,2}{1,2,3}
{1}{1,2}{1,2,3}{1,2,3,4} {4}{1,4}{1,2,4} {3}{1,3}{1,3,4}
{3}{1,3}
{3,4}
{2}{2,3}{2,3,4}
{2}{2,3}
学习内容
5.4 二项式系数的单峰性 5.5 牛顿二项式定理 5.6 偏序集
第一页，共22页。
回顾
二项式系数Pascal公式
knnk1kn 11
Pascal三角形的对称性
第二页，共22页。
5.4 二项式系数的单峰性
定义1: 使得:
设序列s0,s1,s2,…,sn, 若存在一个整数t, 0tn,
第5章二项式系数2北京航空航天大学学习内容?54二项式系数的单峰性?55牛顿二项式定理56偏序集?56偏序集回顾?二项式系数pascal公式?????????k????????????????????11nnn?pascal三角形的对称性?????????????1kk54二项式系数的单峰性?定义1
组合数学第五章
{}1 ,,n a n b n c n a ⋅⋅⋅、从中取出个字母，要求的个数为偶数，问有多少种取法？
2 a b c d e n a b 、由字母，，，，组成的长为的字中，要求与的个数之和为偶数，问这样的字有多少个？
{}123412123,,,1,2,3,41,2,3,41,2,451,2,3,410n i i i S e e e e a S n e i e i e e e e e i ∞⋅∞⋅∞⋅∞⋅===、设多重集合=，表示集合满足下列条件的组合数，分别求数列的生成函数:
(1)每个()出现奇数次；
(2)每个()出现3的倍数次；
（3）不出现，至多出现次；
（4）出现13或11次，出现或次；
（5）每个()至少出现次。
{}124,,
,1,2,
,1,2,
,k n i i S e e e a S n S S e i k i e i k i ∞⋅∞⋅∞⋅==、设多重集合=，表示集合满足下列条件的组合数，分别求数列的指数型生成函数:
(1)的每个元素出现奇数次；
(2)的每个元素至少出现4次；
（3）()至少出现次；（4）()至多出现次.
5、设有砝码重为1g 的3个，重为2g 的4个，重为4g 的2个，问能称出多少重量？各有几种方案？
6⨯、如果要把棋盘上偶数个方格涂成红色，试确定用红色、白色和蓝色对1n 棋盘的方格涂色的方法数。
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第五章 区组设计与编码1.拉丁方与正交拉丁方拉丁方：由1，2，…，n 构成n ×n 方阵n n ij a ⨯)(，如果每行每列中1，2，…，n 各出现一次，则称此方阵为拉丁方。
如： ⎪⎪⎭⎫ ⎝⎛=12212n ⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛=1233122312131323213n 正交换丁方：设()()nn ijnn ij a A a A ⨯⨯==)2(2)1(1是两个n ×n 的拉丁方，如果矩阵()()nn ijija a⨯)2()1(,中2n 个数偶())2()1(,ijij a a 互不相同，n j i ,,2,1, =，则称1A 和2A 正交，或称1A 和2A 是互相正交的拉丁方。
如：⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛=123312231,21313232121A A 由于⎪⎪⎪⎭⎫⎝⎛)1,2()2,1()3,3()3,1()1,3()2,2()2,3()3,2()1,1(中元素两两不同，故21A A ⊥ 36名军官问题：ij a 表示军官的军衔为i ，军官所在的团为j ，()66),(⨯j i 要求第一个数字构成拉丁方，第二个数字也构成拉丁方，并且正交。
定理：两两相互正交的n 阶拉丁方的个数不超过n-1个pf ：设k A A A ,,,21 是两两相互正交的n 阶拉丁方，往证1-≤n k 。
设n a a a 21是1，2，…，n 的一个全排列，(),,,2,1,)(k l a A nn l ijl ==⨯对1A 作如下置换⎪⎪⎭⎫ ⎝⎛n a a a n2121，得一方阵()nn ij a A ⨯=)1(1，则1A 与k A A ,2正交。
事实上，如果1A 不与2A 正交，则()()nn ijija a⨯)2()1(,中至少有一对数偶出现次数超过1次，设该对数偶为),(m l a a ，则),(m a l 在())2()1(,ijij a a 中至少出现2次，与21,AA 正交矛盾，由此不妨设k A A A ,,,21 的第一行均为),,2,1(n ，任取,1k m l ≤<≤则方阵()()nn m ijija a⨯)()1(,的第一行均是),,(),2,1(),1,1((r n 从而1不能出现在k A A A ,,,21 的第2行第1列元素，故这些元素只能取2，3，…, n 中一个，而且取法不能相同，故n k ≤。
2.域与有限域 域：),(,:)1(,+→⨯+≠V V V V V φ Abel 群b a b a +→),( （2） ),(:***∙→⨯∙V VVV Abel 群ab b a →),( }0/{*V V=（3）分配律成立 ac ab c b a +=+)( ca ba a c b +=+)( 则称),,(⋅+V 为域例： Q ，R ，C ， {}Q b a bi a i Q ∈+=,|)( 二元域}1,0{2=F ， p 元域}1,,2,1,0{-=p F p 有限域：元素个数有限的域叫做有限域比如： },,2,1,0{p F p =模p 的剩余类域， 取)(x f 是][x F p 中n 次不可约多项式，则{}p i n n p q F a xa x a a x f x F F ∈+++==-110))((][}{110p i n n F a a a a ∈+++=-αα 为n p q =元域如：,1)(},1,0{22++==x x x f F 则)(x f 在][2x F 中不可约， },|{))((][2101024F a a x a a x f x F F ∈+==}1,,1,0{x x +=4F 中元素的加法与乘法取模)(x f 的运算11)1(2=++=+=+x x xx x x可以证明：有限域中元素的个数必为素数的方幂。
定理：p p n ,α=为素数，3,1≥≥n α，则存在n-1个互相正交的n 阶拉丁方。
Pf ：取n 阶有限域},,,{110-==n p n F F αααα ，其中 1,010==-n αα令 ()1,,2,1)(-==⨯n k a A nn k ij k其中.1,,2,1,0,)(-=+⋅=n j i a ji k k ijααα则121,,,-k A A A 是正交拉丁方。
k A 是拉丁方否则k A 中必存在某行（列）有两个元素相同，不妨设 情形1： )()(k ih k ija a =则 k i k ji k αααααα+=+，,h jαα=h j =情形2： )()(k ij k hj a a = j i k jh k αααααα+=+，i k h k αααα=，i h αα=h i = i A 与)(j i A j ≠相互正交否则存在g A 与h A 不相互正交，即存在 ()())()()()(h fkg fkh ijg ij a aa a =则k j f i ==,部分非素数幂的正文拉丁方的构造设k A A A ,,,21 是一组1n 阶正文拉丁方，k B B B ,,,21 是另一组2n 阶的正文拉丁方，构造一组k 个21n n 阶矩阵如下：()()()()()()()()()⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡=Br a BraBra Br a Br aBr a Br a B a B a C r mn r m r m r n r r r n r r r r r )()()()(2)(22)21)(1)(12)(11112111k r ,,2,1 =则 k C C C ,,,21 是正文拉丁方 如：⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫⎝⎛=32113221323131212321A A ⎪⎪⎪⎪⎪⎭⎫⎝⎛=⎪⎪⎪⎪⎪⎭⎫⎝⎛=2431134242133124432134122143123421B B 则 ⎪⎪⎪⎪⎪⎭⎫⎝⎛=)43()33()23()13()33()43()13()23()23()13()43()33()13()23()33()43(),3(1B ⎪⎪⎪⎪⎪⎭⎫⎝⎛=)23()43()33()13()13()33()43()23()43()23()13()33()33()13()23()43(),3(2B3. 均衡不完全的区组设计（BIBD ）定义：设b v B B B x x x X ,,,},,,,{2121 =是X 的非空子集 如果满足：（1）v k kB B B b <==== 21（2）X 中每一个元素在b B B B ,,,21 中恰出现r 次， （3）X 中任一对元素在b B B B ,,,21 中正好同时出现λ次， 则称{b B B B ,,,21 }为均衡不完全区组设律（BIBD ） ),,,,(λk r v b例 },,,,,,,,,{987654321x x x x x x x x x X =一个（12，9，4，31）设计如下：},,{3211x x x B =，},,,{5442x x x B =},,,{9873x x x B =},,{7414x x x B = },,{8525x x x B =，},,{9636x x x B =，},,{9517x x x B =，},,{7628x x x B = },,{8439x x x B =,},,{86110x x x B =，},,{94211x x x B =，},,{75312x x x B =定理：),,,,(λk r v b －设计必满足（1）vr bk = （2）)1()1(-=-v k r λ 对于一个),,,,(λk r v b －设计，构造如下矩阵 1B 2B … b B→⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=⨯bv vb v v bb v b b b b b b b b b x x x A21222211121121区组矩阵 其中⎩⎨⎧∉∈=ji j i ij B x B x b 01则（1）J I r AATλλ+-=)(其中I 为单位阵，J 为全1矩阵（2）v b ≥，如果v b =，则称BIBD 为对称的，这时r k = 例：},,,,,,{7654321x x x x x x x X =},,{4211x x x B =， },,{5322x x x B =，},,{6433x x x B =， },,{7544x x x B = ，},,{1655x x x B =， },,{2766x x x B =， },,{3177x x x B =。
则（7，7，3，3，1）－对称设计 对称BIBD 的性质对称BIBD 中任意两组都正好有λ个共同元素 只须证 ⎪⎪⎪⎪⎪⎭⎫⎝⎛=λλλλλλλλλλk k A A T 事实上 A AA A A A A A A A T T T )()(11--==A J I k A))(1λλ+-=-JA AI k 1)(-+-=λλ⎪⎪⎪⎭⎫⎝⎛=+-=k kkJ I kλλλλλλλλλλλ)( 如果v B B B ,,,21 是关于},,,{21v x x x X =的对称BIBD ，任取,i B 则 i v i i i i i i B B B B B B B B B B \,\,\,\,\111 +-是关于i B X \的BIBD 。
i v i i i i i i B B B B B B B B B B ,,,,,1121+-是关于i B 的BIBD4. Hadamard 矩阵定义：设,)(n n ij n h H ⨯=其中,11-=or h ij 如果n Tn n nI H H =则称n H 为n 阶Hadamard矩阵。
例：⎪⎪⎪⎪⎪⎭⎫ ⎝⎛------=⎪⎪⎭⎫ ⎝⎛-==1111111111111111,1111),1(421H H H性质如果2>n ，并且n H 为Hadamard 矩阵，则4mod 0≡n如果()()n n n ij n m m m ij m a H a H ⨯⨯==)()(,是Hadamard 矩阵，则()n n ij mn H a H )(=是m 阶的Hadamad 矩阵如果n H 是Hadamard 矩阵，则⎪⎪⎭⎫⎝⎛-n nn n H HH H 也是Hadamad 矩阵 规范Hadamard 矩阵↔n H 对称BIBD ⎪⎭⎫⎝⎛---14,12,1n n n。