算法导论答案(1~20chapter)-1part
算法导论习题答案 (5)
Three-hole punch your paper on submissions. You will often be called upon to “give an algorithm” to solve a certain problem. Your write-up should take the form of a short essay. A topic paragraph should summarize the problem you are solving and what your results are. The body of the essay should provide the following:
(a) Argue that this problem exhibits optimal substructure.
Solution: First, notice that linecost(i, j) is defined to be � if the words i through j do not fit on a line to guarantee that no lines in the optimal solution overflow. (This relies on the assumption that the length of each word is not more than M .) Second, notice that linecost(i, j) is defined to be 0 when j = n, where n is the total number of words; only the actual last line has zero cost, not the recursive last lines of subprob lems, which, since they are not the last line overall, have the same cost formula as any other line.
算法导论课程作业答案
算法导论课程作业答案Introduction to AlgorithmsMassachusetts Institute of Technology 6.046J/18.410J Singapore-MIT Alliance SMA5503 Professors Erik Demaine,Lee Wee Sun,and Charles E.Leiserson Handout10Diagnostic Test SolutionsProblem1Consider the following pseudocode:R OUTINE(n)1if n=12then return13else 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.Problem2Give a short(1–2-sentence)description of each of the following data structures:(a)FIFO queueSolution:A dynamic set where the element removed is always the one that has been in the set for the longest time.(b)Priority queueSolution:A dynamic set where each element has anassociated priority value.The element removed is the element with the highest(or lowest)priority.(c)Hash tableSolution:A dynamic set where the location of an element is computed using a function of the ele ment’s key.Problem3UsingΘ-notation,describe the worst-case running time of the best algorithm that you know for each of the following:(a)Finding an element in a sorted array.Solution:Θ(log n)(b)Finding an element in a sorted linked-list.Solution:Θ(n)(c)Inserting an element in a sorted array,once the position is found.Solution:Θ(n)(d)Inserting an element in a sorted linked-list,once the position is found.Solution:Θ(1)Problem4Describe an algorithm that locates the?rst occurrence of the largest element in a?nite list of integers,where the integers are not necessarily distinct.What is the worst-case running time of your algorithm?Solution:Idea is as follows:go through list,keeping track of the largest element found so far and its index.Update whenever necessary.Running time isΘ(n).Problem5How does the height h of a balanced binary search tree relate to the number of nodes n in the tree? Solution:h=O(lg n) Problem 6Does an undirected graph with 5vertices,each of degree 3,exist?If so,draw such a graph.If not,explain why no such graph exists.Solution:No such graph exists by the Handshaking Lemma.Every edge adds 2to the sum of the degrees.Consequently,the sum of the degrees must be even.Problem 7It is known that if a solution to Problem A exists,then a solution to Problem B exists also.(a)Professor Goldbach has just produced a 1,000-page proof that Problem A is unsolvable.If his proof turns out to be valid,can we conclude that Problem B is also unsolvable?Answer yes or no (or don’t know).Solution:No(b)Professor Wiles has just produced a 10,000-page proof that Problem B is unsolvable.If the proof turns out to be valid,can we conclude that problem A is unsolvable as well?Answer yes or no (or don’t know).Solution:YesProblem 8Consider the following statement:If 5points are placed anywhere on or inside a unit square,then there must exist two that are no more than √2/2units apart.Here are two attempts to prove this statement.Proof (a):Place 4of the points on the vertices of the square;that way they are maximally sepa-rated from one another.The 5th point must then lie within √2/2units of one of the other points,since the furthest from the corners it can be is the center,which is exactly √2/2units fromeach of the four corners.Proof (b):Partition the square into 4squares,each with a side of 1/2unit.If any two points areon or inside one of these smaller squares,the distance between these two points will be at most √2/2units.Since there are 5points and only 4squares,at least two points must fall on or inside one of the smaller squares,giving a set of points that are no more than √2/2apart.Which of the proofs are correct:(a),(b),both,or neither (or don’t know)?Solution:(b)onlyProblem9Give an inductive proof of the following statement:For every natural number n>3,we have n!>2n.Solution:Base case:True for n=4.Inductive step:Assume n!>2n.Then,multiplying both sides by(n+1),we get(n+1)n!> (n+1)2n>2?2n=2n+1.Problem10We want to line up6out of10children.Which of the following expresses the number of possible line-ups?(Circle the right answer.)(a)10!/6!(b)10!/4!(c) 106(d) 104 ·6!(e)None of the above(f)Don’t knowSolution:(b),(d)are both correctProblem11A deck of52cards is shuf?ed thoroughly.What is the probability that the4aces are all next to each other?(Circle theright answer.)(a)4!49!/52!(b)1/52!(c)4!/52!(d)4!48!/52!(e)None of the above(f)Don’t knowSolution:(a)Problem12The weather forecaster says that the probability of rain on Saturday is25%and that the probability of rain on Sunday is25%.Consider the following statement:The probability of rain during the weekend is50%.Which of the following best describes the validity of this statement?(a)If the two events(rain on Sat/rain on Sun)are independent,then we can add up the twoprobabilities,and the statement is true.Without independence,we can’t tell.(b)True,whether the two events are independent or not.(c)If the events are independent,the statement is false,because the the probability of no rainduring the weekend is9/16.If they are not independent,we can’t tell.(d)False,no matter what.(e)None of the above.(f)Don’t know.Solution:(c)Problem13A player throws darts at a target.On each trial,independentlyof the other trials,he hits the bull’s-eye with probability1/4.How many times should he throw so that his probability is75%of hitting the bull’s-eye at least once?(a)3(b)4(c)5(d)75%can’t be achieved.(e)Don’t know.Solution:(c),assuming that we want the probability to be≥0.75,not necessarily exactly0.75.Problem14Let X be an indicator random variable.Which of the following statements are true?(Circle all that apply.)(a)Pr{X=0}=Pr{X=1}=1/2(b)Pr{X=1}=E[X](c)E[X]=E[X2](d)E[X]=(E[X])2Solution:(b)and(c)only。
《算法导论(第二版)》(中文版)课后答案
5
《算法导论（第二版） 》参考答案 do z←y 调用之前保存结果 y←INTERVAL-SEARCH-SUBTREE(y, i) 如果循环是由于y没有左子树，那我们返回y 否则我们返回z，这时意味着没有在z的左子树找到重叠区间 7 if y≠ nil[T] and i overlap int[y] 8 then return y 9 else return z 5 6 15.1-5 由 FASTEST-WAY 算法知：
15
lg n
2 lg n1 1 2cn 2 cn (n 2 ) 2 1
4.3-1 a) n2 b) n2lgn c) n3 4.3-4
2
《算法导论（第二版） 》参考答案 n2lg2n 7.1-2 (1)使用 P146 的 PARTION 函数可以得到 q=r 注意每循环一次 i 加 1，i 的初始值为 p 1 ，循环总共运行 (r 1) p 1次，最 终返回的 i 1 p 1 (r 1) p 1 1 r (2)由题目要求 q=(p+r)/2 可知，PARTITION 函数中的 i，j 变量应该在循环中同 时变化。 Partition(A, p, r) x = A[p]; i = p - 1; j = r + 1; while (TRUE) repeat j--; until A[j] <= x; repeat i++; until A[i] >= x; if (i < j) Swap(A, i, j); else return j; 7.3-2 （1）由 QuickSort 算法最坏情况分析得知：n 个元素每次都划 n-1 和 1 个，因 为是 p<r 的时候才调用，所以为Θ （n） （2）最好情况是每次都在最中间的位置分，所以递推式是： N(n)= 1+ 2*N(n/2) 不难得到：N(n) =Θ （n） 7.4-2 T(n)=2*T(n/2)+ Θ （n） 可以得到 T(n) =Θ （n lgn） 由 P46 Theorem3.1 可得：Ω (n lgn)
中科大算法导论第一,二次和第四次作业答案
2.2-3 再次考虑线性查找问题 （见练习2.1-3）。在平均情况 下，需要检查输入序列中的多 少个元素？假定待查找的元素 是数组中任何一个元素的可能 性是相等的。在最坏情况下有 怎样呢？用Θ形式表示的话，线 性查找的平均情况和最坏情况 运行时间怎样？对你的答案加 以说明。 • 线性查找问题 • 输入：一列数A=<a1,a2,…,an>和一 个值v。 • 输出：下标i，使得v=A[i]，或者当 v不在A中出现时为NIL。 • 平均情况下需要查找 (1+2+…+n)/n=(n+1)/2 • 最坏情况下即最后一个元素为待 查找元素，需要查找n个。 • 故平均情况和最坏情况的运行时 间都为Θ(n)。
• 2.3-2改写MERGE过程，使之不使 用哨兵元素，而是在一旦数组L或R 中的所有元素都被复制回数组A后， 就立即停止，再将另一个数组中 余下的元素复制回数组A中。 • MERGE(A,p,q,r) 1. n1←q-p+1 2. n2 ←r-q 3. create arrays L[1..n1] and R[1..n2] 4. for i ←1 to n1 5. do L*i+ ←A*p+i-1] 6. for j ←1 to n2 7. do R*j+ ←A*q+j+ 8. i ←1 9. j ←1
10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.
k ←p while((i<=n1) and (j<=n2)) do if L[i]<=R[j] do A[k]=L[i] i++ else do A[k]=R[j] j++ k++ while(i<=n1) do A[k++]=L[i++] while(j<=n2) do A[k++]=R[j++]
算法导论复习资料
算法导论复习资料一、选择题：第一章的概念、术语。
二、考点分析：1、复杂度的渐进表示，复杂度分析。
2、正确性证明。
考点：1）正确性分析（冒泡，归并，选择）；2）复杂度分析（渐进表示O，Ｑ，©，替换法证明，先猜想，然后给出递归方程）。
循环不变性的三个性质：1）初始化：它在循环的第一轮迭代开始之前，应该是正确的；2）保持：如果在循环的某一次迭代开始之前它是正确的，那么，在下一次迭代开始之前，它也应该保持正确；3）当循环结束时，不变式给了我们一个有用的性质，它有助于表明算法是正确的。
插入排序算法：【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 - 15 while i > 0 and A[i] > key6 do A[i + 1] ←A[i]7 i ←i - 18 A[i + 1] ←key插入排序的正确性证明：课本11页。
》归并排序算法：课本17页及19页。
归并排序的正确性分析：课本20页。
3、分治法（基本步骤，复杂度分析）。
——许多问题都可以递归求解考点：快速排序，归并排序，渐进排序，例如：12球里面有一个坏球，怎样用最少的次数找出来。
（解：共有24种状态，至少称重3次可以找出不同的球）不是考点：线性时间选择，最接近点对，斯特拉算法求解。
解：基本步骤：一、分解：将原问题分解成一系列的子问题；二、解决：递归地解各子问题。
若子问题足够小，则直接求解；三、合并：将子问题的结果合并成原问题的解。
复杂度分析：分分治算法中的递归式是基于基本模式中的三个步骤的，T(n)为一个规模为n的运行时间，得到递归式、T(n)=Q(1) n<=cT(n)=aT(n/b)+D(n)+C(n) n>c附加习题：请给出一个运行时间为Q(nlgn)的算法，使之能在给定的一个由n个整数构成的集合S 和另一个整数x时，判断出S中是否存在有两个其和等于x的元素。
算法导论参考答案
算法导论参考答案算法导论参考答案算法导论是计算机科学领域中一本经典的教材，被广泛应用于计算机科学和工程的教学和研究中。
它涵盖了算法设计和分析的基本概念，以及各种常见算法的实现和应用。
本文将为读者提供一些算法导论中常见问题的参考答案，以帮助读者更好地理解和掌握这门课程。
1. 什么是算法？算法是一系列解决问题的步骤和规则。
它描述了如何将输入转换为输出，并在有限的时间内完成。
算法应具备正确性、可读性、健壮性和高效性等特点。
2. 如何分析算法的效率？算法的效率可以通过时间复杂度和空间复杂度来衡量。
时间复杂度表示算法执行所需的时间量级，常用的时间复杂度有O(1)、O(n)、O(logn)、O(nlogn)和O(n^2)等。
空间复杂度表示算法执行所需的额外空间量级，通常以字节为单位。
3. 什么是渐进符号？渐进符号用于表示算法的时间复杂度或空间复杂度的增长趋势。
常见的渐进符号有大O符号、Ω符号和Θ符号。
大O符号表示算法的上界，Ω符号表示算法的下界，Θ符号表示算法的平均情况。
4. 什么是分治法？分治法是一种算法设计策略，将问题分解为若干个子问题，并对子问题进行独立求解，最后将子问题的解合并得到原问题的解。
典型的分治算法有归并排序和快速排序。
5. 什么是动态规划？动态规划是一种通过将问题分解为相互重叠的子问题来求解的方法。
它通常用于求解具有重叠子问题和最优子结构性质的问题。
典型的动态规划算法有背包问题和最短路径问题。
6. 什么是贪心算法？贪心算法是一种通过每一步选择局部最优解来求解整体最优解的方法。
贪心算法通常不能保证得到全局最优解，但在某些问题上能够得到近似最优解。
典型的贪心算法有霍夫曼编码和最小生成树算法。
7. 什么是图算法？图算法是一类用于解决图结构相关问题的算法。
图由节点和边组成，节点表示对象，边表示对象之间的关系。
图算法包括图的遍历、最短路径、最小生成树和网络流等问题的求解。
8. 什么是NP完全问题？NP完全问题是一类在多项式时间内无法求解的问题。
中科大算法导论作业标准标准答案
第8次作业答案16.1-116.1-2543316.3-416.2-5参考答案：16.4-1证明中要三点：1.有穷非空集合2.遗传性3.交换性第10次作业参考答案16.5-1题目表格：解法1：使用引理16.12性质（2），按wi单调递减顺序逐次将任务添加至Nt（A），每次添加一个元素后，进行计算，{计算方法：Nt（A）中有i个任务时计算N0 (A),…,Ni(A)，其中如果存在Nj (A)>j,则表示最近添加地元素是需要放弃地，从集合中删除}；最后将未放弃地元素按di递增排序，放弃地任务放在所有未放弃任务后面，放弃任务集合内部排序可随意.解法2：设所有n个时间空位都是空地.然后按罚款地单调递减顺序对各个子任务逐个作贪心选择.在考虑任务j时，如果有一个恰处于或前于dj地时间空位仍空着，则将任务j赋与最近地这样地空位，并填入; 如果不存在这样地空位，表示放弃.答案（a1，a2是放弃地）：<a5, a4, a6, a3, a7,a1, a2>or <a5, a4, a6, a3, a7,a2, a1>划线部分按上表di递增地顺序排即可，答案不唯一16.5-2（直接给个计算例子说地不清不楚地请扣分）题目：本题地意思是在O(|A|)时间内确定性质2（性质2：对t=0，1，2，…，n，有Nt(A)<=t,Nt(A)表示A中期限不超过t地任务个数）是否成立.解答示例：思想：建立数组a[n],a[i]表示截至时间为i地任务个数，对0=<i<n，如果出现a[0]+a[1]+…+a[i]>i,则说明A不独立，否则A独立.伪代码：int temp=0;for(i=0;i<n;i++) a[i]=0; ******O(n)=O(|A|)for(i=0;i<n;i++) a[di]++; ******O(n)=O(|A|)for(i=0;i<n;i++) ******O(n)=O(|A|) {temp+=a[i];//temp就是a[0]+a[1]+…+a[i]if（temp>i）//Ni(A)>iA不独立；}17.1-1（这题有歧义，不扣分）a) 如果Stack Operations包括Push Pop MultiPush,答案是可以保持,解释和书上地Push Pop MultiPop差不多.b) 如果是Stack Operations包括Push Pop MultiPush MultiPop,答案就是不可以保持,因为MultiPush,MultiPop交替地话,平均就是O(K).17.1-2本题目只要证明可能性，只要说明一种情况下结论成立即可17.2-1第11次作业参考答案17.3-1题目：答案：备注：最后一句话展开：采用新地势函数后对i 个操作地平摊代价：)1()())1(())(()()(1''^'-Φ-Φ+=--Φ--Φ+=Φ-Φ+=-Di Di c k Di k Di c D D c c i i i i i i17.3-2题目：答案：第一步：此题关键是定义势能函数Φ,不管定义成什么首先要满足两个条件 对所有操作i ，)(Di Φ>=0且)(Di Φ>=)(0D Φ比如令k j+=2i ，j,k 均为整数且取尽可能大，设势能函数)(Di Φ=2k;第二步：求平摊代价，公式是)1()(^-Φ-Φ+=Di Di c c i i 按上面设置地势函数示例：当k=0，^i c =…=2当k ！=0,^i c =…=3 显然，平摊代价为O(1)17.3-4题目：答案：结合课本p249，p250页对栈操作地分析很容易有下面结果17.4-3题目：答案：αα=(第i次循环之后地表中地entry 假设第i个操作是TABLE_DELETE, 考虑装载因子:inum size数)/(第i次循环后地表地大小)=/i i第12 次参考答案19.1.1题目：答案：如果x不是根，则degree[sibling[x]]=degree[child[x]]=degree[x]-1如果x是根，则sibling为二项堆中下一个二项树地根，因为二项堆中根链是按根地度数递增排序，因此degree[sibling[x]]>degree[x]19.1.2题目：答案：如果x是p[x]地最左子节点，则p[x]为根地子树由两个相同地二项树合并而成，以x为根地子树就是其中一个二项树，另一个以p[x]为根，所以degree[p[x]]=degree[x]+1;如果x不是p[x]地最左子节点，假设x是p[x]地子节点中自左至右地第i个孩子，则去掉p[x]前i-1个孩子，恰好转换成第一种情况，因而degree[p[x]]=degree[x]+1+（i-1）=degree[x]+i;综上,degree[p[x]]>degree[x]19.2.2题目：题目：19.2.519.2.6第13次作业参考答案20.2-1题目：解答：20.2-3 题目：解答：20.3-1 题目：答案：20.3-2 题目：答案：第14次作业参考答案这一次请大家自己看书处理版权申明本文部分内容，包括文字、图片、以及设计等在网上搜集整理.版权为个人所有This article includes some parts, including text, pictures, and design. Copyright is personal ownership.6ewMy。
算法导论参考答案
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-15 while i > 0 and A[i] > key6 do A[i+1]←A[i]7 i ←i − 18 A[i+1]←keyC#对揑入排序算法的实现:public static void InsertionSort<T>(T[] Input) where T:IComparable<T>{T key;int i;for (int j = 1; j < Input.Length; j++){key = Input[j];i = j - 1;for (; i >= 0 && Input[i].CompareTo(key)>0;i-- )Input[i + 1] = Input[i];Input[i+1]=key;}}揑入算法的设计使用的是增量（incremental）方法：在排好子数组A[1..j-1]后，将元素A[ j]揑入，形成排好序的子数组A[1..j]这里需要注意的是由于大部分编程语言的数组都是从0开始算起，这个不伪代码认为的数组的数是第1个有所丌同，一般要注意有几个关键值要比伪代码的小1.如果按照大部分计算机编程语言的思路，修改为：INSERTION-SORT(A)1 for j ← 1 to length[A]2 do key ←A[j]3 i ←j-14 while i ≥ 0 and A[i] > key5 do A[i+1]←A[i]6 i ←i − 17 A[i+1]←key循环丌变式(Loop Invariant)是证明算法正确性的一个重要工具。
算法导论习题答案
算法导论习题答案算法导论习题答案算法导论是一本经典的计算机科学教材，讲述了算法设计与分析的基本原理。
在学习过程中，习题是不可或缺的一部分，通过解答习题可以帮助我们巩固所学的知识。
本文将针对算法导论中的一些习题进行解答，帮助读者更好地理解算法导论的内容。
习题1-1：证明对于任意两个实数a和b，有|a + b| ≤ |a| + |b|。
解答：根据绝对值的定义，我们可以将|a + b|、|a|和|b|分别表示为以下三种情况：1. 当a + b ≥ 0，a ≥ 0，b ≥ 0时，|a + b| = a + b，|a| = a，|b| = b。
此时，|a + b| ≤ |a| + |b| 成立。
2. 当a + b < 0，a < 0，b < 0时，|a + b| = -(a + b)，|a| = -a，|b| = -b。
此时，|a + b| ≤ |a| + |b| 成立。
3. 当a + b ≥ 0，a < 0，b < 0时，|a + b| = a + b，|a| = -a，|b| = -b。
此时，|a + b| ≤ |a| + |b| 成立。
综上所述，无论a和b的取值如何，都有|a + b| ≤ |a| + |b| 成立。
习题2-1：证明插入排序的运行时间是O(n^2)。
解答：插入排序是一种简单直观的排序算法，它的基本思想是将待排序的元素一个个地插入到已排好序的序列中。
在最坏情况下，即待排序的序列是逆序排列时，插入排序的运行时间最长。
假设待排序的序列长度为n，那么第一次插入需要比较1次，第二次插入需要比较2次，依次类推，第n次插入需要比较n-1次。
总的比较次数为1 + 2 + 3+ ... + (n-1) = n(n-1)/2。
因此，插入排序的运行时间是O(n^2)。
习题3-1：证明选择排序的运行时间是O(n^2)。
解答：选择排序是一种简单直观的排序算法，它的基本思想是每次从待排序的序列中选择最小的元素，放到已排序序列的末尾。
算法导论答案
算法导论答案算法导论是计算机科学领域的经典教材，它介绍了算法设计和分析的基本原理和方法。
通过学习算法导论，我们可以深入理解算法的运行机制，并且能够运用这些知识解决实际问题。
本文将介绍一些算法导论的常见问题，并给出相应的答案。
第一部分：算法基础在算法导论中，我们首先学习了算法的基础概念和表达方法。
其中最重要的是时间复杂度和空间复杂度的概念。
时间复杂度衡量了算法运行所需的时间，而空间复杂度则衡量了算法所需要的额外空间。
通过计算复杂度，我们可以估算出算法的效率和资源使用情况。
Q1：什么是时间复杂度和空间复杂度？A1：时间复杂度是指算法解决问题所需要的时间代价，通常以大O表示。
空间复杂度是指算法解决问题所需要的额外空间，通常也以大O表示。
时间复杂度和空间复杂度可以帮助我们评估算法的效率和资源使用情况。
Q2：如何计算时间复杂度？A2：时间复杂度可以通过分析算法中的基本操作的执行次数来计算。
通常，我们可以统计算法中循环、递归和条件判断等操作的执行次数，并根据问题规模n来表示。
然后，我们可以将执行次数与n的关系用大O表示法表示。
第二部分：排序算法算法导论中介绍了多种排序算法，包括插入排序、归并排序、快速排序等等。
不同的排序算法适用于不同的问题场景，并且它们的时间复杂度和稳定性也不同。
Q3：什么是稳定的排序算法？A3：稳定的排序算法是指当原始序列中有两个相等的元素时，排序后它们的相对位置不发生改变。
例如，插入排序和归并排序是稳定的排序算法，而快速排序不是稳定的排序算法。
Q4：如何选择合适的排序算法？A4：选择合适的排序算法需要考虑多个因素，包括数据规模、稳定性要求和系统资源等。
对于小规模数据，可以使用插入排序或者冒泡排序。
对于数据规模较大且对稳定性要求较高的情况，可以选择归并排序。
而快速排序则适用于大规模数据和对稳定性没有要求的场景。
第三部分：动态规划动态规划是算法导论中的重要主题，它是一种解决多阶段决策问题的方法。
《计算机算法-设计与分析导论》课后习题答案
在未排序序列（初始时为整个序列）中选择其中最大的元素max ，然后将该元素同未排序序列中的最后一个元素交换。
这时，max 元素就包含在由每次的最大元素组成的已排序序列之中了，也就说这时的max 已经不在未排序序列之中了。
重复上述过程直到完成整个序列的排序。
(a) 写出Maxsort 算法。
其中待排序序列为E ，含有n 个元素，脚标为范围为0,,1n -。
void Maxsort(Element[] E) { int maxID = 0;for (int i=; i>1; i--) {for (int j=0; j<i; j++) {if (E[j] > E[maxID]) maxID = k;}E[i] <--> E[maxID]; }}最坏情况同平均情况是相同的都是11(1)()2n i n n C n i -=-==∑。
遍浏览序列实现。
排序策略是顺序比较相邻元素，如果这两个元素未排序则交换这两个元素的位置。
也就说，首先比较第一个元素和第二个元素，如果第一个元素大于第二个元素，这交换这两个元素的位置；然后比较第二个元素与第三个元素，按照需要交换两个元素的位置；起泡排序的最坏情况为逆序输入，比较次数为11(1)()2n i n n C n i -=-==∑。
(b) 最好情况为已排序，需要(n-1)次比较。
： (a)归纳法：当n=1时显然成立，当n=2时经过一次起泡后，也显然最大元素位于末尾；现假设当n=k-1是，命题也成立，则当n=k 时，对前k-1个元素经过一次起泡后，根据假设显然第k-1个元素是前k-1个元素中最大的，现在根据起泡定义它要同第k 个元素进行比较，当k 元素大于k-1元素时，它为k 个元素中最大的，命题成立；当k 元素小于k-1元素时，它要同k-1交换，这时处于队列末尾的显然时队列中最大的元素。
综上所述，当n=k 时命题成立。
(b)反正法：假设当没有一对相邻的元素需要交换位置的时候，得到的序列是未排序的，则该未排序队列至少存在一对元素是逆序的，现设这两个元素未E(I)和E(i+k)，其中E(i)>E(i+k)。
算法导论习题答案 (1)
Introduction to Algorithms September 24, 2004Massachusetts Institute of Technology 6.046J/18.410J Professors Piotr Indyk and Charles E. Leiserson Handout 7Problem Set 1 SolutionsExercise 1-1. Do Exercise 2.3-7 on page 37 in CLRS.Solution:The following algorithm solves the problem:1.Sort the elements in S using mergesort.2.Remove the last element from S. Let y be the value of the removed element.3.If S is nonempty, look for z=x−y in S using binary search.4.If S contains such an element z, then STOP, since we have found y and z such that x=y+z.Otherwise, repeat Step 2.5.If S is empty, then no two elements in S sum to x.Notice that when we consider an element y i of S during i th iteration, we don’t need to look at the elements that have already been considered in previous iterations. Suppose there exists y j∗S, such that x=y i+y j. If j<i, i.e. if y j has been reached prior to y i, then we would have found y i when we were searching for x−y j during j th iteration and the algorithm would have terminated then.Step 1 takes �(n lg n)time. Step 2 takes O(1)time. Step 3 requires at most lg n time. Steps 2–4 are repeated at most n times. Thus, the total running time of this algorithm is �(n lg n). We can do a more precise analysis if we notice that Step 3 actually requires �(lg(n−i))time at i th iteration.However, if we evaluate �n−1lg(n−i), we get lg(n−1)!, which is �(n lg n). So the total runningi=1time is still �(n lg n).Exercise 1-2. Do Exercise 3.1-3 on page 50 in CLRS.Exercise 1-3. Do Exercise 3.2-6 on page 57 in CLRS.Exercise 1-4. Do Problem 3-2 on page 58 of CLRS.Problem 1-1. Properties of Asymptotic NotationProve or disprove each of the following properties related to asymptotic notation. In each of the following assume that f, g, and h are asymptotically nonnegative functions.� (a) f (n ) = O (g (n )) and g (n ) = O (f (n )) implies that f (n ) = �(g (n )).Solution:This Statement is True.Since f (n ) = O (g (n )), then there exists an n 0 and a c such that for all n √ n 0, f (n ) ←Similarly, since g (n )= O (f (n )), there exists an n � 0 and a c such that for allcg (n ). �f (n ). Therefore, for all n √ max(n 0,n Hence, f (n ) = �(g (n )).�()g n ,0← �),0c 1 � g (n ) ← f (n ) ← cg (n ).n √ n c � 0 (b) f (n ) + g (n ) = �(max(f (n ),g (n ))).Solution:This Statement is True.For all n √ 1, f (n ) ← max(f (n ),g (n )) and g (n ) ← max(f (n ),g (n )). Therefore:f (n ) +g (n ) ← max(f (n ),g (n )) + max(f (n ),g (n )) ← 2 max(f (n ),g (n ))and so f (n ) + g (n )= O (max(f (n ),g (n ))). Additionally, for each n , either f (n ) √max(f (n ),g (n )) or else g (n ) √ max(f (n ),g (n )). Therefore, for all n √ 1, f (n ) + g (n ) √ max(f (n ),g (n )) and so f (n ) + g (n ) = �(max(f (n ),g (n ))). Thus, f (n ) + g (n ) = �(max(f (n ),g (n ))).(c) Transitivity: f (n ) = O (g (n )) and g (n ) = O (h (n )) implies that f (n ) = O (h (n )).Solution:This Statement is True.Since f (n )= O (g (n )), then there exists an n 0 and a c such that for all n √ n 0, �)f ()n ,0← �()g n ,0← f (n ) ← cg (n ). Similarly, since g (n ) = O (h (n )), there exists an n �h (n ). Therefore, for all n √ max(n 0,n and a c � such thatfor all n √ n Hence, f (n ) = O (h (n )).cc�h (n ).c (d) f (n ) = O (g (n )) implies that h (f (n )) = O (h (g (n )).Solution:This Statement is False.We disprove this statement by giving a counter-example. Let f (n ) = n and g (n ) = 3n and h (n )=2n . Then h (f (n )) = 2n and h (g (n )) = 8n . Since 2n is not O (8n ), this choice of f , g and h is a counter-example which disproves the theorem.(e) f(n)+o(f(n))=�(f(n)).Solution:This Statement is True.Let h(n)=o(f(n)). We prove that f(n)+o(f(n))=�(f(n)). Since for all n√1, f(n)+h(n)√f(n), then f(n)+h(n)=�(f(n)).Since h(n)=o(f(n)), then there exists an n0such that for all n>n0, h(n)←f(n).Therefore, for all n>n0, f(n)+h(n)←2f(n)and so f(n)+h(n)=O(f(n)).Thus, f(n)+h(n)=�(f(n)).(f) f(n)=o(g(n))and g(n)=o(f(n))implies f(n)=�(g(n)).Solution:This Statement is False.We disprove this statement by giving a counter-example. Consider f(n)=1+cos(�≈n)and g(n)=1−cos(�≈n).For all even values of n, f(n)=2and g(n)=0, and there does not exist a c1for which f(n)←c1g(n). Thus, f(n)is not o(g(n)), because if there does not exist a c1 for which f(n)←c1g(n), then it cannot be the case that for any c1>0and sufficiently large n, f(n)<c1g(n).For all odd values of n, f(n)=0and g(n)=2, and there does not exist a c for which g(n)←cf(n). By the above reasoning, it follows that g(n)is not o(f(n)). Also, there cannot exist c2>0for which c2g(n)←f(n), because we could set c=1/c2if sucha c2existed.We have shown that there do not exist constants c1>0and c2>0such that c2g(n)←f(n)←c1g(n). Thus, f(n)is not �(g(n)).Problem 1-2. Computing Fibonacci NumbersThe Fibonacci numbers are defined on page 56 of CLRS asF0=0,F1=1,F n=F n−1+F n−2for n√2.In Exercise 1-3, of this problem set, you showed that the n th Fibonacci number isF n=�n−� n,�5where �is the golden ratio and �is its conjugate.A fellow 6.046 student comes to you with the following simple recursive algorithm for computing the n th Fibonacci number.F IB(n)1 if n=02 then return 03 elseif n=14 then return 15 return F IB(n−1)+F IB(n−2)This algorithm is correct, since it directly implements the definition of the Fibonacci numbers. Let’s analyze its running time. Let T(n)be the worst-case running time of F IB(n).1(a) Give a recurrence for T(n), and use the substitution method to show that T(n)=O(F n).Solution: The recurrence is: T(n)=T(n−1)+T(n−2)+1.We use the substitution method, inducting on n. Our Induction Hypothesis is: T(n)←cF n−b.To prove the inductive step:T(n)←cF n−1+cF n−2−b−b+1← cF n−2b+1Therefore, T(n)←cF n−b+1provided that b√1. We choose b=2and c=10.∗{For the base case consider n0,1}and note the running time is no more than10−2=8.(b) Similarly, show that T(n)=�(F n), and hence, that T(n)=�(F n).Solution: Again the recurrence is: T(n)=T(n−1)+T(n−2)+1.We use the substitution method, inducting on n. Our Induction Hypothesis is: T(n)√F n.To prove the inductive step:T(n)√F n−1+F n−2+1√F n+1Therefore, T(n)←F n. For the base case consider n∗{0,1}and note the runningtime is no less than 1.1In this problem, please assume that all operations take unit time. In reality, the time it takes to add two numbers depends on the number of bits in the numbers being added (more precisely, on the number of memory words). However, for the purpose of this problem, the approximation of unit time addition will suffice.Professor Grigori Potemkin has recently published an improved algorithm for computing the n th Fibonacci number which uses a cleverly constructed loop to get rid of one of the recursive calls. Professor Potemkin has staked his reputation on this new algorithm, and his tenure committee has asked you to review his algorithm.F IB�(n)1 if n=02 then return 03 elseif n=14 then return 15 6 7 8 sum �1for k�1to n−2do sum �sum +F IB�(k) return sumSince it is not at all clear that this algorithm actually computes the n th Fibonacci number, let’s prove that the algorithm is correct. We’ll prove this by induction over n, using a loop invariant in the inductive step of the proof.(c) State the induction hypothesis and the base case of your correctness proof.Solution: To prove the algorithm is correct, we are inducting on n. Our inductionhypothesis is that for all n<m, Fib�(n)returns F n, the n th Fibonacci number.Our base case is m=2. We observe that the first four lines of Potemkin guaranteethat Fib�(n)returns the correct value when n<2.(d) State a loop invariant for the loop in lines 6-7. Prove, using induction over k, that your“invariant” is indeed invariant.Solution: Our loop invariant is that after the k=i iteration of the loop,sum=F i+2.We prove this induction using induction over k. We assume that after the k=(i−1)iteration of the loop, sum=F i+1. Our base case is i=1. We observe that after thefirst pass through the loop, sum=2which is the 3rd Fibonacci number.To complete the induction step we observe that if sum=F i+1after the k=(i−1)andif the call to F ib�(i)on Line 7 correctly returns F i(by the induction hypothesis of ourcorrectness proof in the previous part of the problem) then after the k=i iteration ofthe loop sum=F i+2. This follows immediately form the fact that F i+F i+1=F i+2.(e) Use your loop invariant to complete the inductive step of your correctness proof.Solution: To complete the inductive step of our correctness proof, we must show thatif F ib�(n)returns F n for all n<m then F ib�(m)returns m. From the previous partwe know that if F ib�(n)returns F n for all n<m, then at the end of the k=i iterationof the loop sum=F i+2. We can thus conclude that after the k=m−2iteration ofthe loop, sum=F m which completes our correctness proof.(f) What is the asymptotic running time, T�(n), of F IB�(n)? Would you recommendtenure for Professor Potemkin?Solution: We will argue that T�(n)=�(F n)and thus that Potemkin’s algorithm,F ib�does not improve upon the assymptotic performance of the simple recurrsivealgorithm, F ib. Therefore we would not recommend tenure for Professor Potemkin.One way to see that T�(n)=�(F n)is to observe that the only constant in the programis the 1 (in lines 5 and 4). That is, in order for the program to return F n lines 5 and 4must be executed a total of F n times.Another way to see that T�(n)=�(F n)is to use the substitution method with thehypothesis T�(n)√F n and the recurrence T�(n)=cn+�n−2T�(k).k=1Problem 1-3. Polynomial multiplicationOne can represent a polynomial, in a symbolic variable x, with degree-bound n as an array P[0..n] of coefficients. Consider two linear polynomials, A(x)=a1x+a0and B(x)=b1x+b0, where a1, a0, b1, and b0are numerical coefficients, which can be represented by the arrays [a0,a1]and [b0,b1], respectively. We can multiply A and B using the four coefficient multiplicationsm1=a1·b1,m2=a1·b0,m3=a0·b1,m4=a0·b0,as well as one numerical addition, to form the polynomialC(x)=m1x2+(m2+m3)x+m4,which can be represented by the array[c0,c1,c2]=[m4,m3+m2,m1].(a) Give a divide-and-conquer algorithm for multiplying two polynomials of degree-bound n,represented as coefficient arrays, based on this formula.Solution:We can use this idea to recursively multiply polynomials of degree n−1, where n isa power of 2, as follows:Let p(x)and q(x)be polynomials of degree n−1, and divide each into the upper n/2 and lower n/2terms:p(x)=a(x)x n/2+b(x),q(x)=c(x)x n/2+d(x),where a(x), b(x), c(x), and d(x)are polynomials of degree n/2−1. The polynomial product is thenp(x)q(x)=(a(x)x n/2+b(x))(c(x)x n/2+d(x))=a(x)c(x)x n+(a(x)d(x)+b(x)c(x))x n/2+b(x)d(x).The four polynomial products a(x)c(x), a(x)d(x), b(x)c(x), and b(x)d(x)are computed recursively.(b) Give and solve a recurrence for the worst-case running time of your algorithm.Solution:Since we can perform the dividing and combining of polynomials in time �(n), recursive polynomial multiplication gives us a running time ofT(n)=4T(n/2)+�(n)=�(n2).(c) Show how to multiply two linear polynomials A(x)=a1x+a0and B(x)=b1x+b0using only three coefficient multiplications.Solution:We can use the following 3 multiplications:m1=(a+b)(c+d)=ac+ad+bc+bd,m2=ac,m3=bd,so the polynomial product is(ax+b)(cx+d)=m2x2+(m1−m2−m3)x+m3.� (d) Give a divide-and-conquer algorithm for multiplying two polynomials of degree-bound nbased on your formula from part (c).Solution:The algorithm is the same as in part (a), except for the fact that we need only compute three products of polynomials of degree n/2 to get the polynomial product.(e) Give and solve a recurrence for the worst-case running time of your algorithm.Solution:Similar to part (b):T (n )=3T (n/2) + �(n )lg 3)= �(n �(n 1.585)Alternative solution Instead of breaking a polynomial p (x ) into two smaller polynomials a (x ) and b (x ) such that p (x )= a (x ) + x n/2b (x ), as we did above, we could do the following:Collect all the even powers of p (x ) and substitute y = x 2 to create the polynomial a (y ). Then collect all the odd powers of p (x ), factor out x and substitute y = x 2 to create the second polynomial b (y ). Then we can see thatp (x ) = a (y ) + x b (y )· Both a (y ) and b (y ) are polynomials of (roughly) half the original size and degree, and we can proceed with our multiplications in a way analogous to what was done above.Notice that, at each level k , we need to compute y k = y 2 (where y 0 = x ), whichk −1 takes time �(1) per level and does not affect the asymptotic running time.。
算法导论作业1答案
Introduction to Algorithm s Day 10 Massachusetts Institute of Technology 6.046J/18.410J Singapore-MIT Alliance SMA5503 Professors Erik Demaine,Lee Wee Sun,and Charles E.Leiserson Handout11Exercise1-1.Do Exercise2.3-5on page37in CLRS.Solution:Procedure B INARY-S EARCH takes a sorted array,a value,and a range low high of the array, in which the value should be searched for.The procedure compares to the midpoint of the range and decides to eliminate half the range from further consideration.Both iterative and recursive versions are given.These versions should be initially called with the range length.I TERATIVE-B INARY-S EARCH low high1while low high2do mid low high3if mid4then return mid5if mid6then low mid7else high mid8return NILR ECURSIVE-B INARY-S EARCH low high1if low high2then return NIL3mid low high4if mid5then return mid6if mid7then return R ECURSIVE-B INARY-S EARCH mid high8else return R ECURSIVE-B INARY-S EARCH low midBoth procedures terminate the search unsuccessfully when the range is empty(i.e.,low high) and terminate 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,whose solution is.Exercise1-2.Do Exercise2.3-7on page37in CLRS.Solution:The following algorithm solves the problem:1.Sort the elements in using mergesort.2.Remove the last element from.Let be the value of the removed element.3.If is nonempty,look for in using binary search.4.If contains such an element,then STOP,since we have found and such that.Otherwise,repeat Step2.5.If is empty,then no two elements in sum to.Notice that when we consider an element of during th iteration,we don’t need to look at the elements that have already been considered in previous iterations.Suppose there exists, such that.If,i.e.if has been reached prior to,then we would have found when we were searching for during th iteration and the algorithm would have terminated then.Step1takes time.Step2takes time.Step3requires at most time.Steps2–4 are repeated at most times.Thus,the total running time of this algorithm is.We can do a more precise analysis if we notice that Step3actually requires time at th iteration. However,if we evaluate,we get,which is.So the total running time is still.Exercise1-3.Do Exercise3.1-1on page50in CLRS.Solution:By the definition of-notation(CLRS p.42)we must show that there exist positive constants, ,and such that for,Without loss of generality,let max(.Clearly,.Also, since,.Thus,selecting and and satisfies the definition.Exercise1-4.Do Exercise4.1-6on page67in CLRS.Solution:Let or,equivalently,.The recurrence becomesWe will need one more substitution:Let.The recurrence then becomes:By the master method,.Equivalently,in terms of we have. Going back to(),we getExercise1-5.Rank the following functions by order of growth;that is,find an arrangement of the functions satisfying,,...,.Parti-tion your list into equivalence classes such that and are in the same class if and only if .)5.6.Finally Stirling’s approximation bounds are useful in ranking expression with factorials:So here’s the ranking(listed from left to right by row)The oscillating function does notfit in the ranking because although and ,it is not-related to.The equivalence classes determined by the relationship are:1.Problem1-1.Asymptotic notation for multivariate functionsThe generalization of asymptotic notation from one variable to multiple variables is surprisingly tricky.One proper generalization of-notation for two variables is the following: Definition1there exist positive constants,,and such thatfor all orConsider the following alternative definition:Definition2there exist positive constants,,and such thatfor all and(a)Explain why Definition2is a“bogus”definition.That is,what anomalies does thedefinition of permit that are counterintuitive?You mayfind it helpful to illustrateyour answer with a diagram of relevant regions of the plane.Solution:The distinction between these two interpretations can best be illustrated with a di-agram of the space parameterized by and;see Figure1.The definition ofrequires that the inequality hold in the shaded rectangle in Figure1(a),defined by and,leaving the strips and uncovered.In contrast, the definition of requires in addition that the inequality hold for sufficiently largevalues in those strips,i.e.,for the shaded region in Figure1(b).Ideally,we would alsohope for the inequality to hold for all values of and,as in Figure1(c);we call thisthe unrestricted interpretation.(a)and(b)or(c)UnrestrictedFigure1:Three candidate regions in which a statement about a two-variable function should hold.The definition of is bogus because it allows to be outsidefor infinitely many pairs of values.Recall that for univariate functions,anequivalent interpretation of what means is the following:there existsa constant such that for all butfinitely many1values of.We shouldthus expect the definition of notation in multivariate functions to allow for onlyfinitely many points(tuples)to be outside the stated range.1or equivalently,“for infinitely many values of”Definition3A two-variable function is monotonically increasingifandfor all nonnegative and.(b)Explain this definition in plain English.Solution:A function is monotonically increasing if whenever either(or both)of thefunction’s arguments increase,the function’s value either increases or remains con-stant,but never decreases.ii...However,this is true only if we assume that and,iii...v...vi...(d)Prove that the following two functions are not multiplicatively separable:i.ii.Solution:i.Proof by contradiction:Suppose the function was multiplicatively separable.Then we would have:And so would not be bounded in terms of the argument itself,independent of the other argumentIndeed,in this case,would not necessarily hold under the“or”interpretation if it holds under the“and”interpretation.For example, consider the functionforforforLet.Then holds whenever and but is not necessarily true under the“or”definition.ii.Similarly,2notice thatforforwith.Then holds whenever and but is not necessarily true under the“or”definition.(The’s are necessary to deal with the possibility that takes on values less than.)Thus,suffices in this case.(g)Conclude that.Solution:Let.By parts(a)and(b),for all and,Therefore,holds under the unrestricted interpretation and thusin particular the‘or’definition.(h)(Extra credit.)Give a proper generalization of to two variables.Justify your defini-tion.Solution:Awaiting ideas from students...Problem1-2.Tree TraversalThe following pseudocode is a standard recursive tree-traversal algorithm for counting the number of nodes in a tree.The initial call is C OUNT-N ODES.C OUNT-N ODES1if NIL2then return3else return C OUNT-N ODES leftC OUNT-N ODES rightDefine size to be the number of nodes in the subtree rooted at node,and let denote the worst-case running time of C OUNT-N ODES.(a)Give a recurrence for in terms of left and right.Solution:left right(b)Use the substitution method to prove that size.Solution:For convenience,let denote,that is,the size of the tree whose root is node .In order to prove that,we need to show that there exists a constant such that.Proof.Let be an upper bound on the term3.Assume that there exists some constant such that:for all trees with,.That is,we assume that the statement holds for all trees whose size is less than.We want to prove that.From part(a),we have:left right.Since right and left are smaller than,we haveleft rightleft rightleft right3notice that the term does not depend on the size of the treeConsider the loop invariantC OUNT-N ODES-T AIL left right(1) where is the number of times the while loop(lines2–4)in C OUNT-N ODES-T AIL has been executed.(c)Prove that if Equation(1)holds for,then it holds for.Solution:Let be the new value of after one more execution of the loop.Since afterexecutions we haveC OUNT-N ODES-T AIL left rightat the end of the-st execution we will haveC OUNT-N ODES-T AIL left rightC OUNT-N ODES-T AIL left right C OUNT-N ODES-T AIL left rightC OUNT-N ODES-T AIL left rightSo it holds forC OUNT-N ODES-T AIL left rightHowever,we need to prove that Equation(1)holds.Proof.Base Case:Consider a tree of size1.The loop will execute only once and at the end of the loop,as predicted by Equation(1).So the base case holds.Inductive step:The inductive step was taken care of in part(c).(e)Prove by induction that C OUNT-N ODES-T AIL correctly computes size.Solution:Base case:tree of size.As shown above,the algorithm returns,which is the right answer.Inductive step:assume that the algorithm returns the right result for all trees up to.We want to prove that it will return the right result for any tree ofas well.Consider a tree.As shown in part(d),the algorithm returnsC OUNT-N ODES-T AIL left rightSince all left subtrees have size at most,we know thatC OUNT-N ODES-T AIL left rightcorrectly counts the number of leaves of all the left subtrees.Since the tree consists of right-most nodes plus all their left subtrees,thealgorithm returns the right result for trees of as well.Problem1-3.Polynomial multiplicationIf we have two linear polynomials and,we can multiply them using the four coefficient multiplicationsto form the polynomial(a)Give a divide-and-conquer algorithm for multiplying two polynomials of degree-boundbased on this formula.Solution:We can use this idea to recursively multiply polynomials of degree,where isa power of2,as follows:Let and be polynomials of degree,and divide each into the upperand lower terms:where,,,and are polynomials of degree.The polynomial product is thenThe four polynomial products,,,and are com-puted recursively.(b)Give and solve a recurrence for the worst-case running time of your algorithm.Solution:Since we can perform the dividing and combining of polynomials in time,re-cursive polynomial multiplication gives us a running time of(c)Show how to multiply two linear polynomials and using only threecoefficient multiplications.Solution:We can use the following3multiplications:so the polynomial product is(d)Give a divide-and-conquer algorithm for multiplying two polynomials of degree-boundbased on your formula from part(c).Solution:The algorithm is the same as in part(a),except for the fact that we need only compute three products of polynomials of degree to get the polynomial product.(e)Give and solve a recurrence for the worst-case running time of your algorithm.Solution:Similar to part(b):Alternative solution Instead of breaking a polynomial into two smaller poly-nomials and such that,as we did above,we could do the following:Collect all the even powers of and substitute to create the polynomial .Then collect all the odd powers of,factor out and substitute to create the second polynomial.Then we can see thatBoth and are polynomials of(roughly)half the original size and degree,and we can proceed with our multiplications in a way analogous to what was done above.Notice that,at each level,we need to compute(where),which takes time per level and does not affect the asymptotic running time.。
算法导论(第二版)课后习题解答
Θ
i=1
i
= Θ(n2 )
This holds for both the best- and worst-case running time. 2.2 − 3 Given that each element is equally likely to be the one searched for and the element searched for is present in the array, a linear search will on the average have to search through half the elements. This is because half the time the wanted element will be in the first half and half the time it will be in the second half. Both the worst-case and average-case of L INEAR -S EARCH is Θ(n). 3
Solutions for Introduction to algorithms second edition
Philip Bille
The author of this document 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 a message to beetle@it.dk. 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. Please note that the document is under construction and is updated only sporadically. Have fun with your algorithms. Best regards, Philip Bille
算法导论中文版答案
24.2-3
24.2-4
24.3-1 见图 24-6 24.3-2
24.3-3
24.3-4 24.3-5 24.3-6
24.3-7
24.3-8 这种情况下不会破坏已经更新的点的距离。 24.4**** 24.5****
25.1-1 见图 25-1 25.1-2 为了保证递归定义式 25.2 的正确性 25.1-3
8.3-3 8.3-4
8.3-5(*) 8.4-1 见图 8-4 8.4-2
8.4-3 3/2，1/2 8.4-4(*) 8.4-5(*)
9.1-1
9.1-2 9.2-1 9.3-1
第九章
9.3-2 9.3-3
9.3-4 9.3-5
9.3-6 9.3-7
9.3-8
9.3-9
15.1-1
6.4-4
6.4-5
6.5-1 据图 6-5 6.5-2
6.5-3 6.5-4 6.5-5
6.5-6 6.5-7
6.5-8
7.1-1 见图 7-1 7.1-2
7.1-3 7.1-4 7.2-1 7.2-2
7.2-3 7.2-4 7.2-5
第七章
7.2-6 7.3-1
7.3-2
7.4-1 7.4-2
5.3-6
6.1-1 6.1-2 6.1-3 6.1-4 6.1-5 6.1-6
第6章
6.1-7
6.2-1 见图 6-2 6.2-2
6.2-3
6.2-4
6.2-5 对以 i 为根结点的子树上每个点用循环语句实现 6.2-6
6.3-1
见图 6-3 6.3-2
6.3-3
6.4-1 见图 6-4 6.4-2 HEAPSORT 仍然正确，因为每次循环的过程中还是会运行 MAX-HEAP 的过程。 6.4-3
算法导论中文版答案
} cout<<len<<endl; } return 0; } 15.5-1
15.5-2 15.5-3
15.5-4
16.1-1
第 16 章
16.1-2 16.1-3
16.1-4 16.2-1 16.2-2
16.2-3
16.2-4
16.2-5 16.2-6
16.2-7
25.3-6
5.3-6
6.1-1 6.1-2 6.1-3 6.1-4 6.1-5 6.1-6
第6章
6.1-7
6.2-1 见图 6-2 6.2-2
6.2-3
6.2-4
6.2-5 对以 i 为根结点的子树上每个点用循环语句实现 6.2-6
6.3-1
见图 6-3 6.3-2
6.3-3
6.4-1 见图 6-4 6.4-2 HEAPSORT 仍然正确，因为每次循环的过程中还是会运行 MAX-HEAP 的过程。 6.4-3
6.4-4
6.4-5
6.5-1 据图 6-5 6.5-2
6.5-3 6.5-4 6.5-5
6.5-6 6.5-7
6.5-8
7.1-1 见图 7-1 7.1-2
7.1-3 7.1-4 7.2-1 7.2-2
7.2-3 7.2-4 7.2-5
第七章
7.2-6 7.3-1
7.3-2
7.4-1 7.4-2
16.3-1 16.3-2
16.3-3 16.3-4
16.3-5
16.3-6 那就推广到树的结点有三个孩子结点，证明过程同引理 16.3 的证明。 16.3-7 16.3-8
第 24 章
24.1-1 同源顶点 s 的运行过程，见图 24-4 24.1-2
算法导论附录a习题答案
算法导论附录a习题答案算法导论附录A习题答案算法导论是一本经典的计算机科学教材，被广泛应用于计算机科学和工程的教育和研究领域。
附录A是该书中的一个重要部分，提供了一系列习题，用于帮助读者巩固和应用书中所讲述的算法概念和理论。
本文将回答附录A中的一些习题，旨在帮助读者更好地理解和运用算法导论中的知识。
习题1：给定一个由n个整数构成的数组A，设计一个算法，找到数组中的两个元素，使得它们的和等于给定的值x。
解答：可以使用哈希表来解决这个问题。
首先，遍历一遍数组A，将每个元素插入哈希表中。
然后，再次遍历数组A，对于每个元素a，计算目标值x与a的差值diff。
检查哈希表中是否存在diff，如果存在，则找到了满足条件的两个元素。
习题2：给定一个由n个整数构成的数组A，设计一个算法，找到数组中的一个元素，使得它出现的次数超过数组长度的一半。
解答：可以使用摩尔投票算法来解决这个问题。
首先，假设数组中的第一个元素为候选元素，初始化计数器count为1。
然后，遍历数组，对于每个元素a，如果a与候选元素相同，则计数器加1；否则，计数器减1。
如果计数器减到0，则将当前元素设为候选元素，并将计数器重置为1。
最后，返回候选元素即可。
习题3：给定一个由n个整数构成的数组A，设计一个算法，找到数组中的一个元素，使得它出现的次数严格大于n/3。
解答：可以使用摩尔投票算法的变种来解决这个问题。
首先，假设数组中的前两个元素为候选元素1和候选元素2，初始化计数器count1和count2为0。
然后，遍历数组，对于每个元素a，如果a与候选元素1相同，则计数器count1加1；如果a与候选元素2相同，则计数器count2加1；如果count1为0，则将当前元素设为候选元素1，并将计数器count1重置为1；如果count2为0，则将当前元素设为候选元素2，并将计数器count2重置为1。
最后，再次遍历数组，统计候选元素1和候选元素2的出现次数，如果其中一个元素的出现次数严格大于n/3，则返回该元素。