算法导论习题答案
算法导论15-16-答案.pdf
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15-4 (5-19) Planning a company party. Professor Stewart is consulting for the president of a corporation that is planning a company party. The company has a hierarchical structure; that is, the supervisor relation forms tree rooted at the president. The personal office has ranked each employee with a conviviality rating, which is a real number. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.
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中科大算法导论期末试卷及答案
解：存在������1(n) + ������2(������) = ������(max(������1(������), ������2(������) 证明： ������1(������) = ������(������1(n)) 则存在 a1>0，a2>0, n1>0 使得 n>n1 时有 ������1 ∗ ������1(n) < ������1(������) < ������2 ∗ ������1(n) ������2(������) = ������(������2(n)) 则存在 b1>0，b2>0, n2>0 使得 n>n2 时有 ������1 ∗ ������2(n) < ������2(������) < ������2 ∗ ������2(n) 取 c1=min(a1,b1) >0, c2=2*max(a2,b2) >0, n0=max(n1,n2)>0 当 n>n0 时，有 ������1(n) + ������2(������) > ������1 ∗ ������1(n) + ������1 ∗ ������2(n)
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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
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4.3-1 a) n2 b) n2lgn c) n3 4.3-4
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《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++]
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。
算法导论习题答案 (4)
The father now has to decide how to procure skis for his daughter for every weekly session (until she quits). One can buy skis at a one-time cost of B dollars, or rent skis at a weekly cost of R dollars. (Note that one can buy skis at any time—e.g., rent for two weeks, then buy.)
(a) Suppose that (starting from an empty queue) we do 3 insertions, then 2 removals,
then 3 more insertions, and then 2 more removals. What is the total cost of these 10
Problem Set 4 Solutions
Reading: Chapters 17, 21.1–21.3 Both exercises and problems should be solved, but only the problems should be turned in.
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.。
算法导论第八章答案
算法导论第八章答案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。
#includeusing namespace std;//通过改编计数排序而来，因为有些部分需要注释掉void counting_sort(int*&a, int length, int k, int*&b, int*&c);int main(){ const int LEN =100;int*a =newint[LEN];for(int i =0; i < LEN; i++)a[i] = (i -50)*(i -50) +4;int* b =new int[LEN];const int k =2504;int* c =new int[k +1];counting_sort(a, LEN, k, b, c);//这里需要注释掉//for(int i = 0; i < LEN; i++)//cout<<b[i]<<endl;< p="">int m; int n;while(cin>>m>>n){ if(m >n)cout<<"区间输入不对"<<endl;< p="">else { if(n <0) cout<<"个数为"<<0<<endl;< p="">else if(m <=0&& n <= k) cout<<"个数为"<<c[n]<<endl;< p="">else if(n > k && m >0) cout<<"个数为"<<<endl;<="" c[m=""p="">else if(n > k && m <=0) cout<<"个数为"<<c[k]<<endl;< p="">else cout<<"个数为"<<<endl;<="" c[m="" p="">}} return 0; }void counting_sort(int*&a, int length, int k, int*&b, int*&c){ for(int i =0; i < k +1; i++) c[i] =0;for(int i =0; i < length; i++) c[a[i]]++;for(int i =1; i < k +1; i++) c[i] = c[i] + c[i-1];//这里需注释，因为对c数组内的元素进行减减操作会使其改变/*for(int i = length - 1; i >= 0; i--){b[c[a[i]] - 1] = a[i];c[a[i]]--;}*/ }PS:计数排序的总时间为O(k+n)，在实践中，如果当k = O(n)时，我们常常采用计数排序，这时其运行时间为O(n)8.3-4 :说明如何在O(n)时间内，对0到n^2 - 1之间的n个整数进行排序。
算法导论第十五章习题答案
算法导论第十五章习题答案算法导论第十五章习题答案算法导论是一本经典的计算机科学教材，其中第十五章涵盖了图算法的内容。
本文将针对该章节中的习题进行解答，并对其中一些问题进行深入探讨。
1. 习题15.1-1题目要求证明：对于任意的有向图G=(V,E)，如果图中不存在从节点u到节点v的路径，则在每个强连通分量中，节点u和节点v都在同一个强连通分量中。
证明：假设存在一个强连通分量C，其中节点u在C中，节点v在C'中（C'为除了C之外的其他强连通分量）。
由于不存在从u到v的路径，所以在C中不存在从u到v的路径。
但是根据强连通分量的定义，C中的任意两个节点之间都存在路径。
所以存在一条从v到u的路径。
这与C'中的节点v不在C中矛盾，所以假设不成立，节点u和节点v必定在同一个强连通分量中。
2. 习题15.2-2题目要求证明：在一个有向无环图中，存在一个拓扑排序，使得任意两个非根节点u和v，u在v之前。
证明：假设存在一个有向无环图G=(V,E)，不存在上述所要求的拓扑排序。
即对于任意的拓扑排序，存在两个非根节点u和v，u在v之后。
那么我们可以得到一条从u到v的路径。
由于图中不存在环，所以路径上的节点不会重复。
我们可以将路径上的节点按照拓扑排序的顺序排列，得到一个新的拓扑排序，使得u在v之前。
这与假设矛盾，所以原命题成立。
3. 习题15.3-3题目要求证明：在一个有向图G=(V,E)中，如果存在一条从节点u到节点v的路径，那么在图的转置G^T中，存在一条从节点v到节点u的路径。
证明：假设存在一条从节点u到节点v的路径。
那么在图的转置G^T中，边(u,v)变成了边(v,u)。
所以存在一条从节点v到节点u的路径。
因此，原命题成立。
4. 习题15.4-1题目要求：给出一个算法，判断一个有向图G=(V,E)是否是有向无环图。
算法思路：我们可以使用深度优先搜索（DFS）来判断是否存在环。
具体步骤如下：1. 对于图中的每个节点v，设置一个状态标记visited[v]为false。
算法导论第三版答案
Procedure BINARY-SEARCH takes a sorted array A, a value , and a range Œlow : : high of the array, in which we search for the value . The procedure compares to the array entry at the midpoint of the range and decides to eliminate half the range from further consideration. We give both iterative and recursive versions, each of which returns either an index i such that AŒi D , or NIL if no utions for Chapter 2: Getting Started
AŒlow : : high contains the value . The initial call to either version should have the parameters A; ; 1; n.
Selected Solutions for Chapter 2: Getting Started
2-3
d. We follow the hint and modify merge sort to count the number of inversions in ‚.n lg n/ time.
To start, let us define a merge-inversion as a situation within the execution of merge sort in which the MERGE procedure, after copying AŒp : : q to L and AŒq C 1 : : r to R, has values x in L and y in R such that x > y. Consider an inversion .i; j /, and let x D AŒi and y D AŒj , so that i < j and x > y. We claim that if we were to run merge sort, there would be exactly one mergeinversion involving x and y. To see why, observe that the only way in which array elements change their positions is within the MERGE procedure. Moreover, since MERGE keeps elements within L in the same relative order to each other, and correspondingly for R, the only way in which two elements can change their ordering relative to each other is for the greater one to appear in L and the lesser one to appear in R. Thus, there is at least one merge-inversion involving x and y. To see that there is exactly one such merge-inversion, observe that after any call of MERGE that involves both x and y, they are in the same sorted subarray and will therefore both appear in L or both appear in R in any given call thereafter. Thus, we have proven the claim.
算法导论作业3答案
Introduction to Algorithm s Day 14 Massachusetts Institute of Technology 6.046J/18.410J Singapore-MIT Alliance SMA5503 Professors Erik Demaine, Lee Wee Sun, and Charles E. Leiserson Handout 17Problem Set 3 SolutionsMIT students: This problem set is due in lecture on Day 11.Reading: Chapters 8 and 9Both exercises and problems should be solved, but only the problems should be turned in. Exercises are intended to help you master the course material. Even though you should not turn in the exercise solutions, you are responsible for material covered by the exercises.Mark the top of each sheet with your name, the course number, the problem number, your recitation instructor and time, the date, and the names of any students with whom you collaborated. MIT students: Each problem should be done on a separate sheet (or sheets) of three-hole punched paper.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 your essay should provide the following:1. A description of the algorithm in English and, if helpful, pseudocode.2. At least one worked example or diagram to show more precisely how your algorithm works.3. A proof (or indication) of the correctness of the algorithm.4. An analysis of the running time of the algorithm.Remember, your goal is to communicate. Graders will be instructed to take off points for convoluted and obtuse descriptions.Exercise 3-1. Do exercise 8.1-2 on page 167 of CLRS.Exercise 3-2. Do exercise 8.1-3 on page 168 of CLRS.Exercise 3-3. Do exercise 8.2-3 on page 170 of CLRS.Exercise 3-4. Do exercise 8.4-2 on page 177 of CLRS.Exercise 3-5. Do exercise 9.3-1 on page 192 of CLRS.Exercise 3-6. Show that the second smallest of n elements can be found with n+Θ(lg n)comparisons in the worst case. (Hint: Also find the smallest element.)Problem 3-1. Largest i numbers in sorted orderGiven a set of n numbers, we wish to find the i largest in sorted order using a comparison-based algorithm. Find the algorithm that implements each of the following methods with the best asymptotic worst-case running time, and analyze the running times of the algorithms in terms of n and i.(a) Sort the numbers, and list the i largest.Solution:Use any optimal sorting algorithm, such as MergeSort or HeapSort. Then this can bedone in Θ(n lg n).(b) Build a max-priority queue from the numbers, and call E XTRACT-M AX i times.Solution:Call Build-Heap, Θ(n). Then call Extract-Max, Θ(lg i), i times. So, total runningtime is Θ(n+i lg i).(c) Use an order-statistic algorithm to find the i th largest number, partition around thatnumber, and sort the i largest numbers.Solution:Select the i-th largest number using SELECT, Θ(n), call partition, Θ(n), and then sortthe i largest numbers, Θ(i lg i). So our algorithm takes Θ(n+i lg i).Problem 3-2. At the wading poolYou work at a summer camp which holds regular outings for the n children which attend. One of these outings is to a nearby wading pool which always turns out to be something of a nightmare at the end because there are n wet, cranky children and a pile of 2n shoes (n left shoes and n right shoes) and it is not at all clear which kids go with which shoes. Not being particularly picky, all you care about is getting kids into shoes that fit. The only way to determine if a shoe is a match for a child is to try the shoe on the child’s foot. After trying on the shoe, you will know that it either fits, is too big, or is too small. It is important to note that you cannot accurately compare children’s feet directly with each other, nor can you compare the shoes. You know that for every kid, there are at least two shoes (one left shoe and one right shoe) that will fit, and your task is to shoe all of the children efficiently so that you can go home. There are enough shoes that each child will find a pair which fits her. Assume that each comparison (trying a shoe on a foot) takes one time unit.(a) Describe a deterministic algorithm that uses Θ(n 2) comparisons to pair children withshoes.Solution:For each child, try on all the shoes until you find the two shoes that fit. T (n )= T (n − 2) + O (n ) = Θ(n 2).(b) Prove a lower bound of Ω(n lg n ) for the number of comparisons that must be madeby an algorithm solving this problem. (Hint: How many leaves does the decision tree have?)Solution:There are n ! ways that left shoes can be assigned to children and n ! ways that right shoes can be assigned to children. So the decision tree should have n !2 leaves. n !2 ≥ n ! h ≥ lg(n !)h ≥ lg ( ne )n by Stirling’s Approximation = n lg n − n lg e= Ω(n lg n )(c) How might you partition the children into those with a smaller shoe size than a “pivot”child, and those with a larger shoe size than the pivot child?Solution:Take the pivot child and try on all the shoes until you find one that fits. This should take Θ(n ) time as there are 2n shoes. Then try the shoe on all the children. If the shoe is too small, then they have larger feet than the pivot child. If the shoe is too big, then they have smaller feet than the pivot child. This should also take Θ(n ) time making our partition algorithm run in linear time.(d) Give a randomized algorithm whose expected number of comparisons is O (n lg n ),and prove that this bound is correct. What is the worst-case number of comparisons for your algorithm?Solution:This is similar to quicksort. Pick a random child. Partition the children around that child as in part (c). Then take the shoe you used to partition the children and partition the shoes around it. Take the two shoes and pivot child and put them in the group of paired children. Then recurse on the two groups of shoes and children. This should have the same analysis as randomized quicksort because we have only added an extra call to partition which will still make the work done at each level Θ(n ).。
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
算法导论答案Ch3
f (n) =∞ g (n)
Ω(g (n, m)) = {f (n, m) : there exist positive constants c, n0 , and m0 such that f (n, m) ≥ cg (n, m) for all n ≥ n0 or m ≥ m0 }
Θ(g (n, m)) = {f (n, m) : there exist positive constants c1 , c2 , n0 , and m0 such that c1 g (n, m) ≤ f (n, m) ≤ c2 g (n, m) for all n ≥ n0 or m ≥ m0 } Exercise 3.2-1 Let n1 < n2 be arbitrary. From f and g being monatonic increasing, we know f (n1 ) < f (n2 ) and g (n1 ) < g (n2 ). So f (n1 ) + g (n1 ) < f (n2 ) + g (n1 ) < f (n2 ) + g (n2 ) 2
n→∞
2e n
n
≤ lim
n→∞
2e n
n
1 =0 2n
nn 1 en n −.5 n √ = lim √ e = lim O ( n ) e ≥ lim 1 n→∞ n! n→∞ n→∞ c1 n n→∞ 2πn(1 + Θ( n )) lim ≥ lim en en = lim =∞ n→∞ c1 n n→∞ c1
n a/b n if and only if if and only if n − 21 /b ≥ −a if and only if n + a ≥ (1/2) (n + a)b ≥ cnb . Therefore (n + a)b = Ω(nb ). By Theorem 3.1, (n + a)b = Θ(nb ).
《算法导论》习题答案
《算法导论》习题答案Chapter2 Getting Start2.1 Insertion sort2.1.2 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 algorithms2.2.1将函数用符号表示2.2.2写出选择排序算法selection-sort 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2.3 Designing algorithms计算递归方程的解(1) 当时，，显然有T((2) 假设当时公式成立，即，则当，即时，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j?n;if A[i]+A[j]<xi?i+1elsej?j-1if A[i]+A[j]=xreturn trueelsereturn false时间复杂度为。
或者也可以先固定一个元素然后去二分查找x减去元素的差，复杂度为。
Chapter3 Growth of functions3.1Asymptotic notation3.1.2证明对于b时，对于，时，存在，当时，对于，3.1-4 判断与22n是否等于O(2n)3.1.6 证明如果算法的运行时间为，如果其最坏运行时间为O(g(n))，最佳运行时间为。
最坏时间O(g(n))，即;最佳时间，即3.1.7:证明定义3.2 Standard notation and common functions 3.2.2 证明证明当n>4时，，是否多项式有界～与设lgn=m，则?lgn～不是多项式有界的。
mememmmm2设，，是多项式有界的3.2.5比较lg(lg*n)与lg*(lgn)lg*(lgn)= lg*n-1设lg*n=x，lgx<x-1较大。
算法导论答案(经典)
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
16.3-1 16.3-2Leabharlann 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
24.1-3
24.1-4 24.1-5* 24.1-6 修改 Bellman-Ford 算法，先找到负环上的一个节点，再依次找到负环上的一个节点，再依 次找到负环上的其他节点。 24.2-1 见图 24-5 24.2-2 最后一次不影响结果
len=j;//更新 len
} 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
第 25 章
25.1-4 25.1-5
25.1-6 25.1-7 25.1-8
25.1-9 25.1-10 25.2-1 见图 25-4 25.2-2 25.2-3 25.2-4
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
算法导论习题解答2.3-7
算法导论习题解答2.3-7
•2.3-7 请给出⼀个运⾏为Θ(nlgn)的算法（伪码），使之能在给定⼀个由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。
解：解题思路：先对集合S进⾏归并排序，然后新建⼀个数组S1，使得S1[i] = x – S[i]，再将两个数组并起来。
如果在并的过程中发现有两个元素相等且两个元素⼀个来⾃S，⼀个来⾃S1，则可以确定S中存在有两个其和等于x的元素。
Find whether x exits
1、Sort(S)
2、for i <- 0 to Length(S) – 1
3、 do S1[i] <- x - S[i]
4、for i <- 0 to Length(S) – 1
5、 do Merge( S,S1 )
6、 if S[p] > S1[q]
7、 S0[k] <- S[p] p++ k++
8、 if S[p] < S1[q]
9、 S0[k] <- S[q] q++ k++
10、 if S[p] == S1[q]
11、 return true
12、return false
在第⼀⾏进⾏排序时，时间代价为Θ(nlgn)，后来的合并过程时间代价为Θ(n)，总的时间代价为Θ(nlgn)。
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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<j
if A[i]+A[j]<x
i←i+1
else
j←j -1
if A[i]+A[j]=x
return true
else
return false
时间复杂度为)(n Θ。
或者也可以先固定一个元素然后去二分查找x 减去元素的差，复杂度为)lg (n n Θ。
Chapter3 Growth of functions
3.1Asymptotic notation
0b >,()()b b n a n +=Θ
0a >时，()()2b b b b n a n n n +<+=
对于121,2b c c ==，12()b b b c n n a c n <+<
0a <时，()b b n a n +<
存在02n a >-，当0n n >时，()(/2)b b n a n +> 对于122,1b c c -==，12()b b b c n n a c n <+<
3.1-4 判断122()n n n O +与2是否等于2
3.1.6 证明如果算法的运行时间为(())g n Θ，如果其最坏运行时间为(())O g n ，最佳运行时间为(())g n Ω。
最坏时间(())O g n ，即2()T c g n <；
最佳时间(())g n Ω，即1()T c g n >
定义
3.2 Standard notation and common functions
3.2.2 证明log log b b
c a a c = 设lg n=m ，则2(ln 1)ln 1ln ln !()()m m m m m m n m m m e e m n e e
--=>=>>
∴lg n ！不是多项式有界的。
lg*(lgn)= lg*n-1
设lg*n=x ，lgx<x-1
∴ lg*(lgn)较大。
Chapter4 Recurrences
4.1 The recursion-tree
4.1.1 证明T()()1n n T =⎡⎤+2的解为(lg )O n
4.1.2 证明()2()T n T n n =⎣/2⎦+的解为(lg )O n n
设()lg T n c n n /2≥⎣/2⎦⎣/2+1⎦
lg 1lg(1)n n +≥+，当1/3,(lg 2)c n c n n <>+
4.1.3
将假设改为T(n) = cn lgn +b ，只需要在T(1)=1的情况下成立。
4.1.6 计算
()21T n T =+的解 令/2lg ,(2)2(2)1m m m n T T ==+
令T(n)=S(m),则()2(/2)1S m S m =+
其解为()(),()()(lg )S m m T n S m n =Θ∴==Θ
4.2 The recursion-tree method
4.2.1 lg3()()T n n =Θ
4.2.2 略
4.2.3
2()()T n n =Θ 4.2.5 ()(lg )T n n n =Θ
4.3 The master method
4.3.1 略
4.3.4 主方法是否适用方程
2()4(/2)T n T n n lgn =+，给出该方程解的一个上界
不适用
使用递归树方法可以求得其一个上界为22
(lg )O n n 1,1a b ≥>()f n ，使得其满足主定理case3中除了(/)()af n b cf n ≤外的所有条件。
()(sin 2)f n n n =+，1,2log =0b a b a == ，
找不到这样的1c <，总是满足上面的条件，如sin 1n =-时，sin(/2)/2n ±为，此时sin(/2)212(sin 2)
n n +>+，所以(/)()af n b cf n ≤不满足。