算法导论第一章答案1
USC算法导论作业4
CSCI303Homework4 Problem1(6.1-1): What are the minimum and maximum numbers of elements in a heap of height h? Solution1: A heap is a semi-complete binary tree,so the minimum number of elements in a heap of height h is2h,and the maximum number of elements in a heap of height h is2h+1?1. Problem2(6.1-6): Is the sequence 23,17,14,6,13,10,1,5,7,12 a max-heap? Solution2: The sequence is not a max-heap,because7is a child of6,but7>6. Problem3(6.2-1): Using Figure6.2as a model,illustrate the operation of Max-Heapify(A,3)on the array A= 27,17,3,16,13,10,1,5,7,12,4,8,9,10 . Solution3:
Problem4(6.2-6): Show that the worst-case running time of Max-Heapify on a heap of size n is?(lg n).(Hint:For a heap with n nodes,give node values that cause Max-Heapify to be called recursively at every node on a path from the root down to a leaf.) Solution4: Consider a heap of n nodes where the root node has been changed to contain the smallest value of all the nodes.Now when we call Max-Heapify on the root,the value will have to be exchanged down with its child at every level,until it reaches the lowest level.This is because,after every swapping,the value will still be smaller than both its children(since it is the minimum),until it reaches the lowest level where it has no more children.In such a heap,the number of exchanges to max-heapify the root will be equal to the height of the tree,which is lg n.So the worst case running time is?(lg n).
算法导论 第三版 第24章 答案 英
Chapter24 Michelle Bodnar,Andrew Lohr April12,2016 Exercise24.1-1 If we change our source to z and use the same ordering of edges to decide what to relax,the d values after successive iterations of relaxation are: s t x y z ∞∞∞∞0 2∞7∞0 25790 25690 24690 Theπvalues are: s t x y z NIL NIL NIL NIL NIL z NIL z NIL NIL z x z s NIL z x y s NIL z x y s NIL Now,if we change the weight of edge(z,x)to4and rerun with s as the source,we have that the d values after successive iterations of relaxation are: s t x y z 0∞∞∞∞ 06∞7∞ 06472 02472 0247?2 Theπvalues are: s t x y z NIL NIL NIL NIL NIL NIL s NIL s NIL NIL s y s t NIL x y s t NIL x y s t 1
Note that these values are exactly the same as in the worked example.The di?erence that changing this edge will cause is that there is now a negative weight cycle,which will be detected when it considers the edge(z,x)in the for loop on line5.Since x.d=4>?2+4=z.d+w(z,x),it will return false on line7. Exercise24.1-2 Suppose there is a path from s to v.Then there must be a shortest such path of lengthδ(s,v).It must have?nite length since it contains at most|V|?1 edges and each edge has?nite length.By Lemma24.2,v.d=δ(s,v)<∞upon termination.On the other hand,suppose v.d<∞when BELLMAN-FORD ter-minates.Recall that v.d is monotonically decreasing throughout the algorithm, and RELAX will update v.d only if u.d+w(u,v) \documentclass{ctexart} \usepackage{amsmath} \usepackage{amssymb} \usepackage{fancyhdr} \begin{document} \pagestyle{fancy} \title{算法分析与设计第一次作业} \author{XXXX XXX\XXXXXX} \date{2013/9/11} \maketitle \noindent 3.1-2\ 证明： 证明$(n+a)^b=\Theta(n^b)$等价于证明存在$c_{1},c_{2},n_{0}>0$使得对于任意的 $n>n_{0}$,都有$0\leq c_{1}n^b \leq (n+a)^b \leq c_{2}n^b$成立。 $\because$ $n+a\leq n+|a|$，$\therefore$当$n\geq |a|$时，$n+a\leq 2n$。 又$\because$ $n+a\geq n-|a|$，$\therefore$当$|a|\leq \frac{1}{2}$时，$n+a \geq \frac{1}{2}n$。综上，当$n\geq 2|a|$时，$0\leq \frac{1}{2}n \leq (n+a) \leq 2n$。 $\therefore$ 对于$b>0$，有$0\leq (\frac{1}{2}n)^b \leq (n+a)^b \leq (2n)^b$ $\therefore$ 存在$c_{1} = (\frac{1}{2}n)^b$，$c_{2} = (2n)^b$，$n_{0}=2|a|$， 使得$0\leq c_{1}n^b \leq (n+a)^b \leq c_{2}n^b$成立。\ \ \ $\therefore$原命题得证。 \\ \noindent 3.1-3\ 解释：设运行时间为$F(n)$，则$F(n)\geq 0(n^2)$， $\therefore$ 若$F_{1}(n) = 0(n^2)$，则$F(n)\geq F_{1}(n)$， 又$\because$ $\forall$ n，$T(n)=0$时，$T(n)=0(n^2)$，且运行时间都大于0， $\therefore$对于所有的运行时间$F(n)$都有$F(n)\geq 0(n^2)$， $\therefore$这句话是没有意义的。 \\ \noindent 第二章算法入门 由于时间问题有些问题没有写的很仔细，而且估计这里会存在不少不恰当之处。另，思考题2-3 关于霍纳规则，有些部分没有完成，故没把解答写上去，我对其 c 问题有疑问，请有解答方法者提供个意见。 给出的代码目前也仅仅为解决问题，没有做优化，请见谅，等有时间了我再好好修改。 插入排序算法伪代码 INSERTION-SORT(A) 1 for j ← 2 to length[A] 2 do key ←A[j] 3 Insert A[j] into the sorted sequence A[1..j-1] 4 i ←j-1 5 while i > 0 and A[i] > key 6 do A[i+1]←A[i] 7 i ←i ? 1 8 A[i+1]←key C#对揑入排序算法的实现: public static void InsertionSort 1 第一课课程细节；绪论：算法分析，插入排序法(Insertion Sort)，归并排序(Merge Sort) 阅读：1-2章 发测验0 2 演示课1 算法的正确性 发《作业1》 3 第二课渐进记号(Asymptotic Notation)。递归公式(Recurrences)：置换法，迭代法，主方法 阅读：3-4 章，除了§4.4 4 第三课分治法：Strassen 算法，费氏数列，多项式乘法。 阅读：28 章第2 节，30章第1节 5 演示课2 递归公式，松散性 阅读：Akra-Bazzi 的讲义 6 第四课快速排序法，随机化算法 阅读：5 章 1 到 3 节，7 章 收《作业1》发《作业2》 7 演示课3 排序法：堆排序，动态集合，优先队列 阅读：6 章 8 第五课线性时间的排序法：时间下界，计数排序法，基数排序法 阅读：8 章第1 到 3 节 收《作业2》发《作业3》 9 第六课顺序统计学，中位数 阅读：9 章 10 演示课4 中位数的应用，桶排序 阅读：8 章第4 节 11 第七课散列，全域散列 阅读：11 章 1 到 3 节 收《作业3》发《作业4》 12 第八课散列函数，完美散列 阅读：11 章第 5 节 13 演示课5 测验1 复习 收《作业4》 14 评分后的作业4可以在中午拿到 15 测验1 16 演示课6 二叉搜索树，树的遍历 阅读：12 章 1 到 3 节 17 第九课二叉搜索树和快速排序法之间的关系；随机二叉搜索树的分析阅读：12 章 4 节 发《作业5》 18 第十课红黑树，旋转，插入，删除 阅读：13 章 19 演示课7 2-3树，B-树 阅读：18 章 1 到 2 节 20 第十一课高级数据结构，动态顺序统计，线段树（区间树） 阅读：14 章 收《作业5》发《作业6》 21 第十二课计算几何，区间查询 阅读：33 章 1 到 2 节 22 演示课8 凸多边形 阅读：33 章 3 节 23 第十三课van Emde Boas树，优先队列 阅读：van Emde Boas 的讲义 收《作业6》发《作业7》 24 第十四课平摊分析，表的复制，可能法 阅读：17 章 25 演示课9 竞争分析，自我排序列 26 第十五课动态规划，最长公共子序列，最优二叉搜索树 阅读：15 章 收《作业7》发《作业8》 27 第十六课贪婪算法，最小生成树 阅读：16 章 1 到 3 节，23 章 28 演示课10 贪婪算法和动态规划的范例 1 算法导论第三次习题课 2008.12.17 2 19.1-1 如果x 是根节点：degree[sibling[x]] > sibling[x] 如果x 不是根节点：degree[sibling[x]] = sibling[x] –119.1-3 略 3 19.2-2 过程略( union 操作) 19.2-3 （1）decrease-key （2）extract-min 过程略 19.2-6 算法思想：找到堆中最小值min ，用min-1代替-∞. [遍历堆中树的根节点] 4 15.1-1 15.1-2 略P195 或P329 15.1-4 f i [j]值只依赖f i [j-1]的值，从而可以从2n 压缩为2个。再加上f*、l*、l i [j]。 Print-station(l, I, j ) //i 为线路，j 为station if j>1 then Print-station(l, l i [j], j-1 ) print “line”I, “station”j; 5 15.2-1 略（见课本） 15.2-2 15.2-4 略 MATRIX-CHAIN-MULTIPLY(A, s, i, j) if j>i x= MATRIX-CHAIN-MULTIPLY(A, s, s(i,j), j) y= MATRIX-CHAIN-MULTIPLY(A, s, s(i,j)+1,j) return MATRIX-MULTIPLY(x, y) else return A i 6 15.3-1 （归纳法）递归调用 枚举15.3-2 没有效果，没有重叠子问题 15.3-4 略 (3)n Ο3/2(4/) n n Θ CSCI 303Homework 1 Problem 1(2.1-1): Using Figure 2.2as a model,illustrate the operation of Insertion-Sort on the array A = 31,41,59,26,41,58 . Solution 1: 265841594131Problem 2(2.2-1): Express the function n 3/1000?100n 2?100n +3in terms of Θ-notation. Solution 2: n 3/1000?100n 2?100n +3=Θ(n 3) Problem 3(Derived from 2.2-2): Consider sorting n numbers stored in array A by ?rst ?nding the smallest element of A and exchanging it with the element in A [1].Then ?nd the second smallest element of A ,and exchange it with A [2].Continue in this manner for the ?rst n ?1elements of A .Write pseudocode for this algorithm,which is known as Selection-Sort .Give the worst-case running times of selection sort in Θ-notation. Solution 3: Selection-Sort (A ) for i ←1to length [A ] do min-value ←A [i ] min-index =i for j =i +1to length [A ] do if A [j ]≤min-value min-value =A [j ] min-index =j A [i ]?A [min-index ] The worst-case running time of Selection-Sort is Θ(n 2). Chapter2 Getting Start 2.1 Insertion sort 2.1.2 将Insertion-Sort 重写为按非递减顺序排序 2.1.3 计算两个n 位的二进制数组之和 2.2 Analyzing algorithms 当前n-1个元素排好序后，第n 个元素已经是最大的元素了. 最好时间和最坏时间均为2()n Θ 2.3 Designing algorithms 2.3.3 计算递归方程的解 22()2(/2)2,1k if n T n T n n if n for k =?=?+ = >? （1） 当1k =时，2n =，显然有()lg T n n n = （2） 假设当k i =时公式成立，即()lg 2lg 22i i i T n n n i ===?， 则当1k i =+，即12i n +=时， 2.3.4 给出insertion sort 的递归版本的递归式 2.3-6 使用二分查找来替代insertion-sort 中while 循环内的线性扫描，是否可以将算法的时间提高到(lg )n n Θ？ 虽然用二分查找法可以将查找正确位置的时间复杂度降下来，但 是移位操作的复杂度并没有减少，所以最坏情况下该算法的时间复杂度依然是2()n Θ 2.3-7 给出一个算法，使得其能在(lg )n n Θ的时间内找出在一个n 元素的整数数组内，是否存在两个元素之和为x 首先利用快速排序将数组排序，时间(lg )n n Θ，然后再进行查找： Search(A,n,x) QuickSort(A,n); i←1; j←n; while A[i]+A[j]≠x and i 8.2-4 :在O(1)的时间内，回答出输入的整数中有多少个落在区间[a...b]内。给出的算法的预处理时间为O(n+k) 算法思想：利用计数排序，由于在计数排序中有一个存储数值个数的临时存储区C[0...k]，利用这个数组即可。 #include Chapter35 Michelle Bodnar,Andrew Lohr April12,2016 Exercise35.1-1 We could select the graph that consists of only two vertices and a single edge between them.Then,the approximation algorithm will always select both of the vertices,whereas the minimum vertex cover is only one vertex.more generally,we could pick our graph to be k edges on a graph with2k vertices so that each connected component only has a single edge.In these examples,we will have that the approximate solution is o?by a factor of two from the exact one. Exercise35.1-2 It is clear that the edges picked in line4form a matching,since we can only pick edges from E ,and the edges in E are precisely those which don’t share an endpoint with any vertex already in C,and hence with any already-picked edge. Moreover,this matching is maximal because the only edges we don’t include are the ones we removed from E .We did this because they shared an endpoint with an edge we already picked,so if we added it to the matching it would no longer be a matching. Exercise35.1-3 We will construct a bipartite graph with V=R∪L.We will try to construct it so that R is uniform,not that R is a vertex cover.However,we will make it so that the heuristic that the professor(professor who?)suggests will cause us to select all the vertices in L,and show that|L|>2|R|. Initially start o?with|R|=n?xed,and L empty.Then,for each i from 2up to n,we do the following.Let k= n i .Select S a subset of the vertices of R of size ki,and so that all the vertices in R?S have a greater or equal degree.Then,we will add k vertices to L,each of degree i,so that the union of their neighborhoods is S.Note that throughout this process,the furthest apart the degrees of the vertices in R can be is1,because each time we are picking the smallest degree vertices and increasing their degrees by1.So,once this has been done for i=n,we can pick a subset of R whose degree is one less than the rest of R(or all of R if the degrees are all equal),and for each vertex in 1 Partial Solutions for Introduction to algorithms second edition Professor: Song You TA: Shao Wen ACKNOWLEDGEMENT CLASS ONE: JINZI CLASS TWO: LIUHAO, SONGDINMIN, SUNBOSHAN, SUNYANG CLASS FOUR:DONGYANHAO, FANSHENGBO, LULU, XIAODONG, CLASS FIVE:GAOCHEN, WANGXIAOCHUAN, LIUZHENHUA, WANGJIAN, YINGYING CLASS SIX: ZHANGZHAOYU, XUXIAOPENG, PENGYUN, HOULAN CLASS: LIKANG,JIANGZHOU, ANONYMITY The collator of this Answer Set, SHAOWen, takes absolutely no responsibility for the contents. This is merely a vague suggestion to a solution to some of the exercises posed in the book Introduction to algorithms by Cormen, Leiserson and Rivest. It is very likely that there are many errors and that the solutions are wrong. If you have found an error, have a better solution or wish to contribute in some constructive way please send an Email to shao_wen_buaa@https://www.360docs.net/doc/a41654970.html, It is important that you try hard to solve the exercises on your own. Use this document only as a last resort or to check if your instructor got it all wrong. Have fun with your algorithms and get a satisfactory result in this course. Best regards, SHAOWen 麻省理工学院-算法导论 关于课本的介绍如下： 本书自第一版出版以来，已经成为世界范围内广泛使用的大学教材和专业人员的标准参考手册。本书全面论述了算法的内容，从一定深度上涵盖了算法的诸多方面，同时其讲授和分析方法又兼顾了各个层次读者的接受能力。各章内容自成体系，可作为独立单元学习。所有算法都用英文和伪码描述，使具备初步编程经验的人也可读懂。全书讲解通俗易懂，且不失深度和数学上的严谨性。第二版增加了新的章节，如算法作用、概率分析与随机算法、线性编程等，几乎对第一版的各个部分都作了大量修订。 学过计算机的都知道，这本书是全世界最权威的算法课程的大学课本了，基本上全世界的名牌大学用的教材都是它。这本书一共四位作者，Thomas H. Cormen，Charles E. Leiserson和Ronald L. Rivest是来自MIT的教授，Clifford Stein是MIT出来的博士，现在哥伦比亚大学做教授，四人姓氏的首字母联在一起即是此书的英文简称(CLRS 2e)，其中的第三作者Ronald L. Rivest是RSA算法的老大(算法名字里面的R即是指他)，四个超级大牛出的一本书，此书不看人生不能算完整。 再介绍一下课堂录像里面授课的两位MIT的老师，第一位，外表“绝顶聪明”的，是本书的第二作者Charles E. Leiserson，以逻辑严密，风趣幽默享誉MIT。第二位，留着金黄色的络腮胡子和马尾发的酷哥是Erik Demaine，21岁即取得MIT教授资格的天才，1981出生，今年才25岁，业余爱好是俄罗斯方块、演戏、琉璃、折纸、杂耍、魔术和结绳游戏。 另外，附上该书的中文电子版，pdg转pdf格式，中文版翻译自该书的第一版，中文书名没有使用《算法导论》，而使用的是《现代计算机常用数据结构和算法》，1994年出版时没有得到国外的授权，属于“私自翻译出版”，译者是南京大学计算机系的潘金贵。 课程重点 算法导论是麻省理工学院电机工程与计算机科学系“理论计算机科学”集中选修课程的先导科目。课程几乎将所有资料放到线上，包括了完整的课堂讲义和习题。课本“算法导论，第二版”，是由Charles Leiserson 教授副笔。 课程描述 本课程教授高效率算法的设计及分析技巧，并着重在有实用价值的方法上。课程主题包含了：排序、搜寻树、堆积及散列；各个击破法、动态编程、偿还分析、图论算法、最短路径、网络流、计算几何、数字理论性算法；多项式及矩阵的运算；高速缓存技术及并行运算。 一般资讯 讲师： Erik Demaine Solution to Exercise2.2-2 S ELECTION-S ORT.A/ n D A:length for j D1to n 1 smallest D j for i D j C1to n if A?i 2-2Selected Solutions for Chapter2:Getting Started A?low::high contains the value .The initial call to either version should have the parameters A; ;1;n. I TERATIVE-B INARY-S EARCH.A; ;low;high/ while low high mid D b.low C high/=2c if ==A?mid return mid elseif >A?mid low D mid C1 else high D mid 1 return NIL R ECURSIVE-B INARY-S EARCH.A; ;low;high/ if low>high return NIL mid D b.low C high/=2c if ==A?mid return mid elseif >A?mid return R ECURSIVE-B INARY-S EARCH.A; ;mid C1;high/ else return R ECURSIVE-B INARY-S EARCH.A; ;low;mid 1/ Both procedures terminate the search unsuccessfully when the range is empty(i.e., low>high)and terminate it successfully if the value has been found.Based on the comparison of to the middle element in the searched range,the search continues with the range halved.The recurrence for these procedures is therefore T.n/D T.n=2/C?.1/,whose solution is T.n/D?.lg n/. Introduction to Algorithms Massachusetts Institute of Technology 6.046J/18.410J Singapore-MIT Alliance SMA5503 Professors Erik Demaine,Lee Wee Sun,and Charles E.Leiserson Handout10 Diagnostic Test Solutions Problem1 Consider the following pseudocode: R OUTINE(n) 1if n=1 2then return1 3else return n+R OUTINE(n?1) (a)Give a one-sentence description of what R OUTINE(n)does.(Remember,don’t guess.) Solution:The routine gives the sum from1to n. (b)Give a precondition for the routine to work correctly. Solution:The value n must be greater than0;otherwise,the routine loops forever. (c)Give a one-sentence description of a faster implementation of the same routine. Solution:Return the value n(n+1)/2. Problem2 Give a short(1–2-sentence)description of each of the following data structures: (a)FIFO queue Solution:A dynamic set where the element removed is always the one that has been in the set for the longest time. 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