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算法导论答案

算法导论答案
算法导论答案

算法导论 第三版 第21章 答案 英

Chapter21 Michelle Bodnar,Andrew Lohr April12,2016 Exercise21.1-1 EdgeP rocessed initial{a}{b}{c}{d}{e}{f}{g}{h}{i}{j}{k} (d,i){a}{b}{c}{d,i}{e}{f}{g}{h}{j}{k} (f,k){a}{b}{c}{d,i}{e}{f,k}{g}{h}{j} (g,i){a}{b}{c}{d,i,g}{e}{f,k}{h}{j} (b,g){a}{b,d,i,g}{c}{e}{f,k}{h}{j} (a,h){a,h}{b,d,i,g}{c}{e}{f,k}{j} (i,j){a,h}{b,d,i,g,j}{c}{e}{f,k} (d,k){a,h}{b,d,i,g,j,f,k}{c}{e} (b,j){a,h}{b,d,i,g,j,f,k}{c}{e} (d,f){a,h}{b,d,i,g,j,f,k}{c}{e} (g,j){a,h}{b,d,i,g,j,f,k}{c}{e} (a,e){a,h,e}{b,d,i,g,j,f,k}{c} So,the connected that we are left with are{a,h,e},{b,d,i,g,j,f,k}, and{c}. Exercise21.1-2 First suppose that two vertices are in the same connected component.Then there exists a path of edges connecting them.If two vertices are connected by a single edge,then they are put into the same set when that edge is processed. At some point during the algorithm every edge of the path will be processed,so all vertices on the path will be in the same set,including the endpoints.Now suppose two vertices u and v wind up in the same set.Since every vertex starts o?in its own set,some sequence of edges in G must have resulted in eventually combining the sets containing u and v.From among these,there must be a path of edges from u to v,implying that u and v are in the same connected component. Exercise21.1-3 Find set is called twice on line4,this is run once per edge in the graph,so, we have that?nd set is run2|E|times.Since we start with|V|sets,at the end 1

算法导论 第三版 第六章 答案 英

Chapter6 Michelle Bodnar,Andrew Lohr December31,2016 Exercise6.1-1 At least2h and at most2h+1?1.Can be seen because a complete binary tree of depth h?1hasΣh?1 i=02i=2h?1elements,and the number of elements in a heap of depth h is between the number for a complete binary tree of depth h?1exclusive and the number in a complete binary tree of depth h inclusive. Exercise6.1-2 Write n=2m?1+k where m is as large as possible.Then the heap consists of a complete binary tree of height m?1,along with k additional leaves along the bottom.The height of the root is the length of the longest simple path to one of these k leaves,which must have length m.It is clear from the way we de?ned m that m= lg n . Exercise6.1-3 If there largest element in the subtee were somewhere other than the root, it has a parent that is in the subtree.So,it is larger than it’s parent,so,the heap property is violated at the parent of the maximum element in the subtree Exercise6.1-4 The smallest element must be a a leaf node.Suppose that node x contains the smallest element and x is not a leaf.Let y denote a child node of x.By the max-heap property,the value of x is greater than or equal to the value of y.Since the elements of the heap are distinct,the inequality is strict.This contradicts the assumption that x contains the smallest element in the heap. Exercise6.1-5 Yes,it is.The index of a child is always greater than the index of the parent, so the heap property is satis?ed at each vertex. 1

算法导论第二章答案

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-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(T[] Input) where T:IComparable { 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-1

算法导论 第三版 第七章 答案 英

Chapter7 Michelle Bodnar,Andrew Lohr April12,2016 Exercise7.1-1 13199512874212611 13199512874212611 13199512874212611 91913512874212611 95131912874212611 95131912874212611 95819121374212611 95871213194212611 95874131912212611 95874131912212611 95874219122113611 95874261221131911 95874261121131912 Exercise7.1-2 If all elements in the array have the same value,PARTITION returns r.To make PARTITION return q= (p+r)/2 when all elements have the same value,modify line4of the algorithm to say this:if A[j]≤x and j(mod2)= (p+1)(mod2).This causes the algorithm to treat half of the instances of the same value to count as less than,and the other half to count as greater than. Exercise7.1-3 The for loop makes exactly r?p iterations,each of which takes at most constant time.The part outside the for loop takes at most constant time.Since r?p is the size of the subarray,PARTITION takes at most time proportional to the size of the subarray it is called on. Exercise7.1-4 To modify QUICKSORT to run in non-increasing order we need only modify line4of PARTITION,changing≤to≥. 1

算法导论 第三版 第二章 答案 英

Chapter2 Michelle Bodnar,Andrew Lohr April12,2016 Exercise2.1-1 314159264158 314159264158 314159264158 263141594158 263141415958 263141415859 Exercise2.1-2 Algorithm1Non-increasing Insertion-Sort(A) 1:for j=2to A.length do 2:key=A[j] 3://Insert A[j]into the sorted sequence A[1..j?1]. 4:i=j?1 5:while i>0and A[i]

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)

算法导论 第三版 第十九章 答案 英

Chapter19 Michelle Bodnar,Andrew Lohr April12,2016 Exercise19.2-1 First,we take the subtrees rooted at24,17,and23and add them to the root list.Then,we set H.min to18.Then,we run consolidate.First this has its degree2set to the subtree rooted at18.Then the degree1is the subtree rooted at38.Then,we get a repeated subtree of degree2when we consider the one rooted at24.So,we make it a subheap by placing the24node under18. Then,we consider the heap rooted at17.This is a repeat for heaps of degree1, so we place the heap rooted https://www.sodocs.net/doc/a215249897.html,stly we consider the heap rooted at23,and then we have that all the di?erent heaps have distinct degrees and are done,setting H.min to the smallest,that is,the one rooted at17. The three heaps that we end up with in our root list are: 23 17 38 30 41 and 1

《算法导论2版》课后答案_加强版2

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

Ch10算法导论 第三版 第十章 答案 英

Chapter10 Michelle Bodnar,Andrew Lohr April12,2016 Exercise10.1-1 4 41 413 41 418 41 Exercise10.1-2 We will call the stacks T and R.Initially,set T.top=0and R.top=n+1. Essentially,stack T uses the?rst part of the array and stack R uses the last part of the array.In stack T,the top is the rightmost element of T.In stack R, the top is the leftmost element of R. Algorithm1PUSH(S,x) 1:if S==T then 2:if T.top+1==R.top then 3:error“over?ow” 4:else 5:T.top=T.top+1 6:T[T.top]=x 7:end if 8:end if 9:if S==R then 10:if R.top?1==T.top then 11:error“over?ow” 12:else 13:R.top=R.top?1 14:T[T.top]=x 15:end if 16:end if 1

Algorithm2POP(S) if S==T then if T.top==0then error“under?ow” else T.top=T.top?1. return T[T.top+1] end if end if if S==R then if R.top==n+1then error“under?ow” else R.top=R.top+1. return R[R.top?1] end if end if Exercise10.1-3 4 41 413 13 138 38 Exercise10.1-4 Algorithm3ENQUEUE if Q.head==Q.tail+1,or Q.head==1and Q.tail==Q.length then error“over?ow” end if Q[Q.tail]=x if Q.tail==Q.length then Q.tail=1 else Q.tail=Q.head+1 end if Exercise10.1-5 As in the example code given in the section,we will neglect to check for over?ow and under?ow errors. 2

算法导论习题答案

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,()()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 +<

算法导论 第三版 第十六章 答案 英

Chapter16 Michelle Bodnar,Andrew Lohr April12,2016 Exercise16.1-1 The given algorithm would just stupidly compute the minimum of the O(n) numbers or return zero depending on the size of S ij.There are a possible number of subproblems that is O(n2)since we are selecting i and j so that 1≤i≤j≤n.So,the runtime would be O(n3). Exercise16.1-2 This becomes exactly the same as the original problem if we imagine time running in reverse,so it produces an optimal solution for essentially the same reasons.It is greedy because we make the best looking choice at each step. Exercise16.1-3 As a counterexample to the optimality of greedily selecting the shortest, suppose our activity times are{(1,9),(8,11),(10,20)}then,picking the shortest ?rst,we have to eliminate the other two,where if we picked the other two instead,we would have two tasks not one. As a counterexample to the optimality of greedily selecting the task that con?icts with the fewest remaining activities,suppose the activity times are {(?1,1),(2,5),(0,3),(0,3),(0,3),(4,7),(6,9),(8,11),(8,11),(8,11),(10,12)}.Then, by this greedy strategy,we would?rst pick(4,7)since it only has a two con- ?icts.However,doing so would mean that we would not be able to pick the only optimal solution of(?1,1),(2,5),(6,9),(10,12). As a counterexample to the optimality of greedily selecting the earliest start times,suppose our activity times are{(1,10),(2,3),(4,5)}.If we pick the ear-liest start time,we will only have a single activity,(1,10),whereas the optimal solution would be to pick the two other activities. Exercise16.1-4 Maintain a set of free(but already used)lecture halls F and currently busy lecture halls B.Sort the classes by start time.For each new start time which you encounter,remove a lecture hall from F,schedule the class in that room, 1

算法导论 第三版 第十七章 答案 英

Chapter17 Michelle Bodnar,Andrew Lohr April12,2016 Exercise17.1-1 It woudn’t because we could make an arbitrary sequence of MULT IP USH(k),MULT IP OP(k). The cost of each will beΘ(k),so the average runtime of each will beΘ(k)not O(1). Exercise17.1-2 Suppose the input is a1followed by k?1zeros.If we call DECREMENT we must change k entries.If we then call INCREMENT on this it reverses these k changes.Thus,by calling them alternately n times,the total time isΘ(nk). Exercise17.1-3 Note that this setup is similar to the dynamic tables discussed in section 17.4.Let n be arbitrary,and have the cost of operation i be c(i).Then, n i=1c(i)= lg(n) i=1 2i+ i≤n not a power of2 1≤ lg(n) i=1 2i+n=21+ lg(n) ?1+n≤4n?1+n≤5n∈O(n) So,since to?nd the average,we divide by n,the average runtime of each com-mand is O(1). Exercise17.2-1 To every stack operation,we charge twice.First we charge the actual cost of the stack operation.Second we charge the cost of copying an element later on.Since we have the size of the stack never exceed k,and there are always k operations between backups,we always overpay by at least enough.So,the ammortized cost of the operation is constant.So,the cost of the n operation is O(n). Exercise17.2-2 1

算法导论 第三版 第35章 答案 英

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

算法导论 第三版 第一章 答案 英

Chapter1 Michelle Bodnar,Andrew Lohr April12,2016 Exercise1.1-1 An example of a real world situation that would require sorting would be if you wanted to keep track of a bunch of people’s?le folders and be able to look up a given name quickly.A convex hull might be needed if you needed to secure a wildlife sanctuary with fencing and had to contain a bunch of speci?c nesting locations. Exercise1.1-2 One might measure memory usage of an algorithm,or number of people required to carry out a single task. Exercise1.1-3 An array.It has the limitation of requiring a lot of copying when re-sizing, inserting,and removing elements. Exercise1.1-4 They are similar since both problems can be modeled by a graph with weighted edges and involve minimizing distance,or weight,of a walk on the graph.They are di?erent because the shortest path problem considers only two vertices,whereas the traveling salesman problem considers minimizing the weight of a path that must include many vertices and end where it began. Exercise1.1-5 If you were for example keeping track of terror watch suspects,it would be unacceptable to have it occasionally bringing up a wrong decision as to whether a person is on the list or not.It would be?ne to only have an approximate solution to the shortest route on which to drive,an extra little bit of driving is not that bad. 1

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