2015Fall - 7.4.ExpectedValue
2015Fall - 10.6-Shortest Path Problem
Shortest Path Problem Dijkstra’s Algorithm
Complexity The worst-case running time for the Dijkstra algorithm on a graph with n nodes and m edges is O(n2 ) because it allows for directed cycles. It even finds the shortest paths from a source node s to all other nodes in the graph. This is basically O(n2 ) for node selection and O(m) for distance updates.
6 / 12
Shortest Path Problem Finding the shortest path between any two vertices
Definition (Distance Matrix (Adjacent Matrix with Weights)) Let G be a directed graph with n vertices. The distance matrix of G is D = (dij )n×n . dij represent the the weights of the edge (vi , vj ). If there’s no edge between vi and vj then dij = ∞.
10 / 12
Shortest Path Problem Finding the shortest path between any two vertices
北京工业大学研究生《概率论与数理统计》2013-2014考题
北京工业大学2013-2014学年第一学期期末数理统计与随机过程(研) 课程试卷学号 姓名 成绩注意：试卷共七道大题，请写明详细解题过程。
数据结果保留3位小数。
考试方式：半开卷，考试时只允许看教材《概率论与数理统计》 浙江大学 盛骤等编第三版（或第四版）高等教育出版社，不能携带和查阅任何其他书籍、纸张、资料等。
考试时允许使用计算器。
考试时间120分钟。
一、（10分）设学生某次考试成绩服从正态分布),(2σμN ，现从中随机抽取36位的考试成绩, 算得平均分为66.5，标准差为15分。
问在显著性水平0.05下，从样本看，(1)是否接受“70=μ”的假设？ (2)是否接受“2216≤σ”的假设？解：已知 05.0,36,15,5.66====αn S X（1）70:,70:10≠=μμH H由书中结论知，检验问题的拒绝域为)1(702-≥-n t nSX α4.13615705.6670=-=-nSX ，查表得0301.2)35()1(025.02==-t n t α，所以，接受原假设。
，（2）22122016:,16:>≤σσH H检验问题的拒绝域为)1(16)1(222-≥-n S n αχ7617.301615)136(16)1(2222=-=-S n ，802.49)136()1(205.02=-=-χχαn ，所以，接受原假设。
二、（15分）在某公路上观察汽车通过情况，取15秒为一个时间单位，记下锅炉汽车分布？(显著性水平取0.05α=)解：805.020014113282681920ˆ=*+*+*+*+*==x λ并组后k=4,而此处r=1,故自由度为k-r-1=2,200.932-200=0.932<991.5)2(205.0=χ,所以是Poisson 分布 三、（15分）为考察某种维尼纶纤维的耐水性能，安排了一组试验，测得甲醇浓度x(1)建立“缩醇化度” y 对甲醇浓度x 的一元线性回归方程； (2)对建立的回归方程进行显著性检验：（取01.0=α）； (3)在0x =36时，给出相应的y 的预测区间（取01.0=α）。
2015丘成桐数学竞赛algebra-individual
S.-T.Yau College Student Mathematics Contests2015Algebra and Number TheoryIndividual(5problems)This exam of160points is designed to test how much you know rather than how much you don’t know.You are not expected tofinish all problems but do as much asyou can.Problem1.(20pt)Let G be anfinite Z-module(i.e.,afinite abelian group with additive group law)with a bilinear,(strongly)alternative,and non-degenerate pairingℓ:G×G→Q/Z.Here“(strongly)alternating”means for every a∈G,ℓ(a,a)=0;“non-degenerate”means for every nonzero a∈G there is a b∈G such thatℓ(a,b)=0.Show in steps the following statement:(S):G is isomorphic to H1⊕H2for somefinite abelian groups H1≃H2such that =0.ℓ|Hi×H i(1.1)(5pt)For every a∈G,write o(a)for the order of a andℓa:G−→Q/Z for thehomomorphismℓa(b)=ℓ(a,b).Show that the image ofℓa is o(a)−1Z/Z.(1.2)(5pt)Show that G has a pair of elements a,b with the following properties:(a)o(a)is maximal in the sense that for any x∈G,o(x)|o(a);(b)ℓ(a,b)=o(a)−1mod Z.(c)o(a)=o(b)We call the subgroup<a,b>:=Z a+Z b a maximal hyperbolic subgroup of G. (1.3)(5pt)Let<a,b>be a maximal hyperbolic subgroup of G.Let G′be the orthog-onal complement of<a,b>consisting of elements x∈G such thatℓ(x,c)=0 for all c∈<a,b>.Show that G is a direct sum as follows:G=Z a⊕Z b⊕G′.(1.4)(5pt)Finish the proof of(S)by induction.Problem2(40pt).Let O n(C)denote the group of n×n orthogonal complex matrices, and M n×k(C)the space of n×k complex matrices,where n and k are two positive integers.For i=0,1,let F i be the space of rational function f on M n×k(C)such that (∗)f(gx)=det(g)i f(x)for all g∈O n(C)and x∈M n×k(C).We want to study in steps the structures of F0and F1.1(2.1)(10pt)For each x ∈M n ×k ,let V x denote the subspace of C n generated by columnsof x ,and let Q (x )=x t ·x ∈M k ×k (C ).Show the following are equivalent:(a)the space V x has dimension k ,and the Euclidean inner product (·,·)is non-degenerate on V x in the sense that V ⊥x ∩V x =0.(b)det Q (x )=0.(2.2)(10pt)Show that F 0is a field generated by entries of Q (x ).(2.3)(10pt)Assuem k <n and let f ∈F 1.Show that f =0by the following two steps:(a)for any x ∈M n ×k (C )with det Q (x )=0,construct a g ∈O n (C )such thatg |V x =1and det g =−1.(b)Show that f vanishes on a general point x ∈M n ×k (C )with det Q (x )=0,thus f ≡0.(2.4)(10pt)Assume k ≥n .Show that F 1is a free vector space of rank 1over F 0.Problem 3.(40pt)Consider the equation f (x ):=x 3+x +1=0.We want to show in steps thatfor any prime p ,if (31p)=−1,then x 3+x +1is solvable mod p .Let x 1,x 2,x 3be three roots of f (x ):=x 3+x +1=0.Let F =Q (x 1),and L =Q (x 1,x 2,x 3),and K =Q (√∆)where ∆is the discriminant of f (x ):∆=[(x 1−x 2)(x 2−x 3)(x 3−x 1)]2.(3.1)(10pt)Show that f is irreducible,that ∆=−31,and that F is not Galois overQ ;(3.2)(10pt)Show that Gal(L/Q )≃S 3,the permutation group of three letters,thatGal(L/K )≃Z /3Z ,and that Gal(L/F )≃Z /2Z ;(3.3)(20pt)Let O F ,O K ,O L =be rings of integers of F,K,L respectively.Let p be aprime such that x 3+x +1=0is not soluble in Z /p Z .Show the following:(a)(5pt)pO F is still a prime ideal in O F ,(b)(5pt)pO L is product of two prime ideals in O L ,and(c)(5pt)pO K is product of two primes ideals in O K ,and(d)(5pt)x 2+31=0is soluble in F p .2Problem 4.(40pt)Let p be a prime and Z p the ring of p -adic integers with a p -adic norm normalized by |p |=p −1.Let ϕ:Z p −→Z p be a map defined by a power series of the form ϕ(x )=x p +p ∑a n x n ,a n ∈Z p ,|a n |−→0.Let E be a field,and F the E -vector space of locally constant E -valued functions on Z p with an operator ϕ∗defined by ϕ∗f =f ◦ϕ.We want to show in steps the following statement:The set of eigenvalues of ϕ∗on F is {0,1}.(4.1)(10pt)Show that ϕis a contraction map on each residue class R ∈Z p /p Z p :|ϕ(x )−ϕ(y )|≤p −1|x −y |,∀x,y ∈R.(4.2)(10pt)Show that there is a ϵR ∈R for each residue class R such thatlim n ϕn (x )=ϵR ,∀x ∈R.Here ϕn is defined inductively by ϕ1=ϕ,ϕn =ϕn −1◦ϕ.(4.3)(10pt)Let F 0(resp.F 1)be the subspace of functions f vanishing on each ϵR(resp.constant on R )for all residue class R .Show that ϕ∗=1on F 1,and that for each f ∈F 0ϕ∗n f =0for some n ∈N .(4.4)(10pt)Show that for any a ∈E ,a =0,1,the operator ϕ∗−a is invertible on F .Problem 5(20pt).Check if the following rings are UFD (unique factorization domain).(5.1)(5pt)R 1=Z [√6];(5.2)(5pt)R 2=Z [(1+√−11)/2];(5.3)(5pt)R 3=C [x,y ]/(x 2+y 2−1);(5.4)(5pt)R 4=C [x,y ]/(x 3+y 3−1).3。
英国UKMT数学竞赛2015年Junior Kangaroo详解
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15. C
Let the length and width of the garden be a metres and b metres respectively and let the width of the path be x metres. The perimeter of the garden is 2(a + b) metres and the perimeter of the larger rectangle formed by the garden and the path is 2(a + 2x + b + 2x) metres. Hence
11. B
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To obtain the smallest number of gold coins, the least possible number of boxes must be opened. Therefore, Blind Pew must open the trunk and all five chests, leaving only three boxes to be opened. Hence the smallest number of gold coins he could take is 3 × 10 = 30.
(完整版)新加坡SM2 2015年真题
Additional math test 2015Paper 1 [80']1、The position vectors of points A, B and C ，relative to an origin O ，are j i j i35,9 and)3(j i krespectively ，where k is a constant ，given that C lies on the line AB ，find the value ofk. [4]2、A youth club has facilities for members to play pool, darts and table-tennis. Every member plays at least one of the three games. P, D and T represent the sets of members who play pool, darts and table-tennis respectively. Express each of the following in set language and illustrate each by means of a venn diagram. (i) the set of members who only play pool. [2] (ii) the set of members who play exactly 2 games, neither of which is darts. [2]3、Without using a calculator, solve, for x and y, the simultaneous equations 6428 yx, 81)91(314 y x . [5]4、The diagram shows a sectors COD of a circle, centre O, in which angle 34COD radians. The points A and B lie on OD and OC respectively, and AB is an arc of a circle, centre O, of radius 7 cm. Given that the area of the shaded region ABCD is 482cm , find the perimeter of this shaded region. [6]5、Given taht the expansion of nx x a )21)(( in asending powers of x is (432)bx xfind the values of the constants a, n and b. [6]6、The function f is defined,for x 0, by x x f 4cos 35)( . Find(i) the amplitude and the period of f. [2](ii) the coordinates of the maximum and minimum points of the curve y = f (x). [4]OA DCBrad 347 cm7、(a) Find the number of different arrangements of the 9 letters of the word SINGAPORE in which S does not occure as the first letter. [2](b) 3 students are selected to form a chess team from a group of 5 girls and 3 boys. Find the number of possible teams that can be selected in which there are more girls than bos. [4]8、The function f is defined, for R x , by 3,3113: x x x x f. (i) find 1 f in terms of x and explain what this implies about the symmetry of the graph of y = f (x) . [3] The function g is defined, for R x , by 23: x x f. (ii) find the values of x for which )()(1x g x f . [3] (iii) state the value of x for which gf(x) = 2 . [1]。
2015年7月gmat 数学机经(3)
2015年7月gmat 数学机经(3)前几天小编给大家整理了部分2015年7月gmat 数学机经，现在继续给大家放送最新收集的gmat 数学机经，以下是机经详细内容：1.PS 一个principal interest 问题，公式 A-b*(1.1)……n说一个人投钱，投了2000，第一年1900了，第二年多少。
记不太清了V2：答案: 1790解题思路：如果该公式代表n年后获得钱数，那么：n=0，A-B*1.10=2000n=1，A-B*1.1=1900求得：A=3000，B=1000，n=2，A-B*1.12=3000-1000*1.21=17902.PSV1：x=1, Ask: 答案: 1/4(原构筑的3/7怎么出来的好奇怪...)解题思路：So easy~算错的去面壁!3.DSV1：有个停车场，200辆车，有black和non black，sedan和non sedan。
问neither black andsedan是多少。
(1)sedan and black的车有多少。
(2)sedan占200的几分之几，black占几分之几答案: C解题思路：画表格很好理解：A+B+C+D=200，所求为：D唯(1)：即给出A但是只知道A是无法求得红色区域哒~唯(2): 可求G,H，E,F，不知道ABCD分布(1)+(2)：知道A，G，可求B。
知道B,F，可求D，bingo~选C4.DSV1：一道很简单的数轴题。
┑(￣Д￣)┍V2：数轴上的4个数按顺序排列，问四个数的乘积是否为负?(1)RX>0(2)YW<0答案: B解题思路：唯(1)：RX同号：若RX均大于0，00;若RX均小于0，当R唯(2): YW异号，且Y5.PSV1：6张不同的CD，但只有买3张的钱，问买到多少种不同的3张CD答案:解题思路：是我脑残还是题目略残...6.PSV1：这里面En代表本应该出现次数，On代表实际出现次数。
2015年 AMC 10B真题含答案
2015AMC10BProblem1What is the value of?的值是多少？Problem2Marie does three equally time-consuming tasks in a row without taking breaks.She begins the first task at1:00PM and finishes the second task at2:40PM.When does she finish the third task?Marie连续做了3项耗时相同的任务，中间没有休息。
她从下午1:00开始做第一个任务，在下午2:40完成第二个任务。
她何时完成第三个任务?Problem3Isaac has written down one integer two times and another integer three times.The sum of the five numbers is,and one of the numbers is.What is the other number?Isaac把一个整数写了2遍，另一个整数写了3遍，这5个数之和为100.其中一个数是28，另一个数是多少？Four siblings ordered an extra large pizza.Alex ate,Beth,and Cyril of the pizza.Dan got the leftovers.What is the sequence of the siblings in decreasing order of the part of pizza they consumed?4个兄弟姐妹订购了一块大的比萨饼.Alex吃了整块的，Beth吃了，Cyril吃了，Dan吃了剩下的。
下面那个选项是将这4个兄弟姐妹按照他们所吃的比萨饼的量降序排列？Problem5David,Hikmet,Jack,Marta,Rand,and Todd were in a-person race with other people.Randfinished places ahead of Hikmet.Marta finished place behind Jack.David finished placesbehind Hikmet.Jack finished places behind Todd.Todd finished place behind Rand.Martafinished in th place.Who finished in th place?David，Hikmet，Jack，Marta，Rand，Todd和其他6人参加了一场12人的比赛。
概率论与数理统计(英文) 第七章
7. Estimation Problems7.1 Point Estimate 点估计1. conceptEstimation is one of the most important problems in statistics. Why ----the distribution depends on the parameter.Poisson distribution ),(λk PNormal distribution ),(2σμNProblems of estimation include(1) determining parameters of a population from the random samples, estimating the mean or variance of a population by sampling,(2) how to judge if such estimation is a “good” one.Point Estimation ----- use the value of a statistic to estimate a parameterFor example, if we use X to estimate the parameter λ of a Poisson population, X is then the estimator of λ and x is a point estimate of the parameter λ.Point estimation is not unique.For example, we often use X orM----- mean.2. Judge(1) Unbiased 无偏(2) asymptotically unbiased 一致(3) minimum variance unbiased estimator. 最有效无偏估计Proof.Since for a Poisson we know thatλ(PE=)and by Theorem 6.3.1E(PE=,X)()we conclude thatλE=(X)which shows that X is an unbiased estimator of λ. □Proof. Suppose μ is the mean of the population, by Theorem 6.4.1（P100）, we have thatωωμωωωωωω≠+=+-=⋅==⎰⎰∞--∞--∞--1)()()()(dx e xe dx e x X E x x x . Therefore, X is a biased estimator of ω.均值的点估计 First of all, let us discuss the situation of using point estimations. X is usually the most common choice of the point estimator of the mean. By Theorem 6.4.1, we can easily obtain the following theorem.Example 7.1.3 Let 2ˆSbe the variance of a random sample of size n from a random variable X which has a finite variance 2σ. Show that 2ˆSis an asymptotically unbiased estimator of 2σ.Proof. By Theorem 6.3.1,)()()()(11)ˆ(222122122X E X E X E X E n X X n E S E n i i n i i -=-=⎪⎭⎫ ⎝⎛-=∑∑==. Since []22)()()(X E X E X D -= and []22)()()(X E X E X D -=, we have []{}[]{}).()()()()()()ˆ(222X D X D X E X D X E X D S E -=+-+= Here,n X D n X D n X D n X n D X D n i n i i n i i 212121)(1)(1)(11)(σ====⎪⎭⎫ ⎝⎛=∑∑∑===. Thus, we get that22221)ˆ(σσσnn n S E -=-=. We can see that 2ˆSis a biased estimator of 2σ. But when n is large, 2221lim )ˆ(lim σσ=-=∞→∞→nn S E n n which leads to the conclusion that 2ˆSis an asymptotically unbiased estimator of 2σ.堂上练习Show that 22)(σ=S E ， E ∑=--=n i i X X n S 122)(11=?As mentioned before, there exists more than one unbiased estimators for the mean μ. Therefore, it is meaningful to study, under those popular distributions, whether X is also the minimum variance unbiased estimator. Here, an important tool is Cramer-Rao inequality .(最小方差无偏估计)Cramer-Rao inequality⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛∂∂≥2)(ln 1)ˆ(θθX f nE Dwhere )(x fis the population density function and n is the sample size.minimum variance unbiased estimator of μ.Proof . In this example, the population density function is22121)(⎪⎭⎫ ⎝⎛--=σμπσx e x fwhich leads to()⎪⎭⎫ ⎝⎛-=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛---∂∂=∂∂σμσσμπσμμx x x f 1212ln )(ln 2. Therefore,222222111)(ln σσμσσμσμ=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛∂∂X E X E X f E since )1,0(~N X σμ-. Thus, )(11)(ln 1222X D n n X f nE ==⋅=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛∂∂σσθ. According to Theorem 7.1.1, we conclude that X is a minimum variance unbiased estimator of μ.3. Method of Moments 矩估计的方法Example 7.1.7求总体的数学期望和方差的矩估计。