山东建筑大学计算机学院算法分析算法复习题(Yuconan翻译)上课讲义

山东建筑大学计算机学院算法分析算法复习题(Y u c o n a n翻译)1.The O-notation provides an asymptotic upper bound. The Ω-notation provides anasymptotic lower bound. The Θ-notation asymptotically a function form above2.To represent a heap as an array，the root of tree is A[1], and given the index i ofa node, the indices of its parent Parent(i) { return ⎣i/2⎦; }，left child, Left(i){ return 2*i; }，right child, right(i) { return 2*i + 1; }.代表一个堆中的一个数组，树的根节点是A[1]，并且给出一个节点i，那么该节点的父节点是左孩子右孩子3.Because the heap of n elements is a binary tree, the height of any node is at mostΘ(lg n).因为n个元素的堆是一个二叉树，任意节点的树高最多是4.In optimization problems, there can be many possible solutions. Each solutionhas a value, and we wish to find a solution with the optimal (minimum ormaximum) value. We call such a solution an optimal solution to the problem.在最优化问题中，有很多可能的解，每个解都有一个值，我们希望找到一个最优解（最大或最小），我们称这个解为最优解问题。

5.optimal substructure if an optimal solution to the problem contains within itoptimal solutions to subproblems.最优子结构中问题的最优解，至少包含它的最优解的子问题。

6. A subsequence of X if there exists a strictly increasing sequence <i1,i2, ..., i k> ofindices of X such that for all j = 1, 2, ..., k, we have x i j= z j .Let X = <x1, x2, ..., x m> and Y = <y1, y2, ..., y n> be sequences, and let Z = <z1,z2, ..., z k> be any LCS of X and Y.(1). If x m = y n, then z k = x m = y n and Z k-1 is an LCS of X m-1 and Y n-1.(2). If x m ≠ y n, then z k ≠ x m implies that Z is an LCS of X m-1 and Y.(3). If x m ≠ y n, then z k ≠ y n implies that Z is an LCS of X and Y n-1.7. A greedy algorithm always makes the choice that looks best at the moment. Thatis, it makes a locally optimal choice in the hope that this choice will lead to aglobally optimal solution.贪心算法经常需要在某个时刻寻找最好的选择。

正因如此，它在当下找到希望中的最优选择，以便引导出一个全局的最优解。

8.The greedy-choice property and optimal sub-structure are the two key ingredientsof greedy algorithm.贪心选择和最优子结构是贪心算法的两个重要组成部分。

9.When a recursive algorithm revisits the same problem over and over again, wesay that the optimization problem has overlapping subproblems.当一个递归算法一遍一遍的遍历同一个问题时，我们说这个最优化问题是重叠子问题。

10.greedy-choice property is a globally optimal solution can be arrived at by makinga locally optimal (greedy) choice.贪心选择性质是一个全局的最优解，这个最优解可以做一个全局的最优选择。

11.An approach of Matrix multiplication can develope a Θ(V4)-time algorithm forthe all-pairs shortest-paths problem and then improve its running time to Θ(V3lgV).一个矩阵相乘问题的解决可以一个时间复杂度算法的所有路径的最短路径问题，改进后的时间复杂度是。

12.Floyd-Warshall algorithm, runs in Θ(V3) time to solve the all-pairs shortest-pathsproblem.FW算法在时间复杂度下可以解决最短路径问题。

13.The running time of Quick Sort is O(n2) in the worst case, and O(n lg n) in theaverage case.快速排序的平均时间复杂度是O(n lg n)，最坏时间复杂度是O(n2)。

14.The MERGE(A,p,q,r) procedure in merge sort takes time Θ(n).MERGE在归并排序中所花费的时间是。

15.Given a weighted, directed graph G = (V, E) with source s and weight function w :E →R, the Bellman-Ford algorithm makes |V| - 1 passes over the edges of thegraph.给一个带权重的有向图G = (V, E)，权重关系w : E →R,则the Bellman-Ford算法需经过条边。

16.The Bellman-Ford algorithm runs in time O(V E).Bellman ford 算法的时间复杂度是。

17.A decision tree represents the comparisons made by a comparison sort.Theasymptotic height of any decision tree for sorting n elements is (n lg n).一个决策树代表一个比较类型，通过比较排序。

N个元素的任意决策树的渐进高度是。

True-false questions1.An algorithm is said to be correct if, for some input instance, it halts with thecorrect output F如果给一个算法输入一些实例，并且它给力正确的输出，则认识这个算法是正确的。

2.Insertion sort always best merge sort F插入排序总是优越与归并排序。

3.Θ(n lg n) grows more slowly than Θ(n2). Therefore, merge sort asymptoticallybeats insertion sort in the worst case. TΘ(n lg n)4.Currently computers are fast and computer memory is very cheap, we have noreason to study algorithms. F5.In RAM (Random-Access Machine) model, instructions are executed withconcurrent operations. F6.The running time of an algorithm on a particular input is the number of primitiveoperations or “steps” executed. T7.Quick sorts, have no combining step: two subarrays form an already-sorted array.T8.The running time of Counting sort is O(n + k). But the running time of sorting isΩ(n lg n). So this is contradiction. F9.The Counting sort is stable. T10.In the selection problem,there is a algorithm of theoretical interest only with O(n)worst-case running time. T11.Divide-and-conquer algorithms partition the problem into independentsubproblems, solve the subproblems recursively, and then combine their solutions to solve the original problem. In contrast, dynamic programming is applicablewhen the subproblems are not independent, that is, when subproblems sharesubsubproblems. T12.In dynamic programming, we build an optimal solution to the problem fromoptimal solutions to subproblems. T13.The best-case running time is the longest running time for any input of size n. F14.When we analyze the running time of an algorithm, we actually interested on therate of growth (order of growth). T15.The dynamic programming approach means that it break the problem into severalsubproblems that are similar to the original problem but smaller in size, solve the subproblems recursively, and then combine these solutions to create a solution to the original problem. T16.Insertion sort and merge sort both use divide-and-conquer approach. F17.Θ(g(n)) = { f (n) : there exist positive constants c1, c2, and n0 such that 0 ≤ c1 g(n)≤ f (n) ≤ c2 g(n) for all n ≥ n0 }18.Min-Heaps satisfy the heap property: A[Parent(i)] ≥ A[i] for allnodes i > 1. F19.For array of length n, all elements in range A[⎣n/2⎦ + 1 .. n] are heaps. T20.The tighter bound of the running time to build a max-heap from an unorderedarray isn’t in linear time. F21.The call to BuildHeap() takes O(n) time, Each of the n - 1 calls to Heapify()takes O(lg n) time, Thus the total time taken by HeapSort() = O(n) + (n - 1) O(lg n)= O(n) + O(n lg n)= O(n lg n). T22.Quick Sort is a dynamic programming algorithm. The array A[p..r] is partitionedinto two non-empty subarrays A[p..q] and A[q+1..r], All elements in A[p..q] are less than all elements in A[q+1..r], the subarrays are recursively sorted by calls to quicksort. F23.Assume that we have a connected, undirected graph G = (V, E) with a weightfunction w : E→R, and we wish to find a minimum spanning tree for G. BothKruskal and Prim algorithms use a dynamic programming approach to theproblem. F24.A cut (S, V - S) of an undirected graph G = (V, E) is a partition of E. F25.An edge is a light edge crossing a cut if its weight is the maximum of any edgecrossing the cut. F26.Kruskal's algorithm uses a disjoint-set data structure to maintain several disjointsets of elements. T27.Optimal-substructure property is a hallmark of the applicability of both dynamicprogramming. T28.Dijkstra's algorithm is a dynamic programming algorithm. F29.Floyd-Warshall algorithm, which finds shortest paths between all pairs of vertices ,is a greedy algorithm. F30.Given a weighted, directed graph G = (V, E) with weight function w : E →R, let p= <v1,v2,..., v k_>be a shortest path from vertex v1 to vertex v k and, for any i and j such that 1 ≤i ≤j ≤k, let p ij = <v i, v i+1,..., v j> be the subpath of p from vertex v i to vertex v j . Then, p ij is a shortest path from v i to v j. T31.Given a weighted, directed graph G = (V, E) with weight function w : E →R,Ifthere is a negative-weight cycle on some path from s to v , there exists a shortest-path from s to v. F32.Since any acyclic path in a graph G = (V, E) contains at most |V| distinct vertices,it also contains at most |V| - 1 edges. Thus, we can restrict our attention to shortest paths of at most |V| - 1 edges. T33.The process of relaxing an edge (u, v) tests whether we can improve the shortestpath to v found so far by going through u. T34.In Dijkstra's algorithm and the shortest-paths algorithm for directed acyclic graphs,each edge is relaxed exactly once. In the Bellman-Ford algorithm, each edge is also relaxed exactly once . F35.The Bellman-Ford algorithm solves the single-source shortest-paths problem inthe general case in which edge weights must be negative. F36.Given a weighted, directed graph G = (V, E) with source s and weight function w :E →R, the Bellman-Ford algorithm can not return a Boolean value indicatingwhether or not there is a negative-weight cycle that is reachable from the source.F37.Given a weighted, directed graph G = (V, E) with source s and weight function w :E →R, for the Bellman-Ford algorithm, if there is such a cycle, the algorithmindicates that no solution exists. If there is no such cycle, the algorithm produces the shortest paths and their weights. F38.Dijkstra's algorithm solves the single-source shortest-paths problem on a weighted,directed graph G = (V, E) for the case in which all edge weights are negative. F 39.Dijkstra's algorithm solves the single-source shortest-paths problem on a weighted,directed graph G = (V, E) for the case in which all edge weights are nonnegative.Bellman-Ford algorithm solves the single-source shortest-paths problem on aweighted, directed graph G = (V , E ), the running time of Dijkstra's algorithm is lower than that of the Bellman-Ford algorithm. T40. The steps for developing a dynamic-programming algorithm:1. Characterize thestructure of an optimal solution. 2. Recursively define the value of an optimalsolution. 3. Compute the value of an optimal solution in a bottom-up fashion. 4. Construct an optimal solution from computed information. T三 Each of n input elements is an integer in the range 0 to k , Design a linear running time algorithm to sort n elements.四Design a expected linear running time algorithm to find the i th smallest element of n elements using divide and conquer strategy.五Write the INSERT-SORT procedure to sort into non-decreasing order. Analyze the running time of it with RAM Model. What ’s the best-case running time, the worst-case running time and average case running time. Write the MERGE-SORTprocedure to sort into non-decreasing order. Give the recurrence for the worst-case running time T(n) of Merge sort and find the solution to the recurrence.六 What is an optimal Huffman code for the following set of frequencies, <a:45, b:13, c:12,d:16,e:9,f:5>七 The traveling-salesman problem (TSP): in the traveling-salesman problem, we are given a complete undirected graph G=(V,E) that has a nonnegative integer cost c(u,v) associated with each edge (u,v)∈E , and we must find a tour of G with minimum cost. The following is an instance TSP. Please compute a tour with minimum cost with greedy algorithm.∞∞∞∞∞69326911699252312514216214八Given items of different values and weights, find the most valuable set of items that fit in a knapsack of fixed weight C .For an instance of knapsack problem, n =8, C =110,value V ={11,21,31,33,43,53,55,65} weight W ={1,11,21,23,33,43,45,55}. Use greedy algorithms to solve knapsack problem.九Use dynamic programming to solve Assembly-line scheduling problem: A Motors Corporation produces automobiles that has two assembly lines, numbered i =1,2. Eachline has n stations, numbered j=1,2…n. We denote the j th station on line i by S ij. The following figure is an instance of the assembly-line problem with costs entry time e i, exit time x i, the assembly time required at station S ij by a ij, the time to transfer a chassis away from assembly line I after having gone through station S ij is t ij. Please compute the fastest time and construct the fastest way through the factory of the instance.十. The matrix-chain multiplication problem can be stated as follows: given a chain <A1,A2,…,An>of matrices, where for i=1，2…，n, matrix A i has dimensionP i-1 P i, fully parenthesize the product A1,A2,…,A n in a way that minimizes the number of scalar multiplication. We pick as our subproblems the problems of determining the minimum cost of a parenthesization of A i A i+1 A j for 1 ≤i ≤j ≤n. Let m[i, j] be the minimum number of scalar multiplications needed to compute the matrix A i..j; for the full problem, the cost of a cheapest way to compute A1..n would thus be m[1, n]. Can you define m[i, j] recursively? Find an optimal parenthesization of a matrix-chain product whose sequence of dimensions is <4,3,5,2,3>十一 In the longest-common-subsequence (LCS) problem, we are given two sequences X = <x1, x2, ...,xm> and Y = <y1, y2, ..., yn> and wish to find a maximum-length common subsequence of X and Y. Please write its recursive formula and determine an LSC of Sequence S1=ACTGATCG and sequence S2=CATGC. Please fill in the blanks in the table below.十二 Proof: Any comparison sort algorithm requires Ω(nlgn) comparisons in the worst case.How many leaves does the tree have? （叶节点的数目）–At least n! (each of the n!permutations if the input appears as some leaf) ⇒n! ≤l （至少n! 个，排列）–At most 2hleaves （引理，至多2h个）⇒n! ≤l ≤2h⇒h ≥lg(n!) = Ω(nlgn)十三Proof: Subpaths of shortest paths are shortest paths.Given a weighted, directed graph G = (V, E) with weight function w : E → R, let p=<v1,v2,..., v k> be a shortest path from vertex v1to vertex v k and, for any i and j such that 1 ≤i ≤j ≤k, let p ij= <v i, v i+1,..., v j>be the subpath of p from vertex v i to vertex v j . Then, p ij is a shortest path from v i to v j.十四Proof : The worst case running time of quicksort is Θ(n2)十五Compute shortest paths with matrix multiplication and the Floyd-Warshall algorithm for the following graph.十六 Write the MAX-Heapify() procedure to for manipulating max-heaps. And analyze the running time of MAX-Heapify().三（10分)1 CountingSort(A, B, k)2 for i=1 to k3 C[i]= 0;4 for j=1 to n5 C[A[j]] += 1;6 for i=2 to k7 C[i] = C[i] + C[i-1];8 for j=n downto 19 B[C[A[j]]] = A[j];10 C[A[j]] -= 1;四算法描述3分The best-case running time is T(n) = c1n + c2(n - 1) + c4(n - 1) + c5(n - 1) + c8(n - 1) = (c1 + c2 + c4 + c5 + c8)n - (c2+ c4 + c5 + c8). This running time can be expressed as an + b for constants a and b that depend on the statement costs c i; it is thus a linear function of n.This worst-case running time can be expressed as an2 + bn + c for constants a, b, and c that again depend on the statement costs c i; it is thus a quadratic function of n.分析2分算法描述2分Θ(1) if n = 1T(n) =2T(n/2) + Θ(n) if n > 1.递归方程和求解3分五7 RAND-SELECT(A, p, r, i) (5分)if p = r then return A[p]q ← RAND-PARTITION(A, p, r)k ← q – p + 1if i = k then return A[q]if i < kthen return RAND-SELECT(A, p, q – 1, i )else return RAND-SELECT(A, q + 1, r, i – k ) Randomized RANDOMIZED-PARTITION(A; p; r) (5分) { i ←RANDOM(p, r)exchange A[r] ← A[i]return PARTITION(A; p; r)}PARTITION(A; p; r){ x← A[r]i ←p-1for j ← p to r-1do if A[j] ≤ xthen i ←i+1exchange A[i] ↔A[j]exchange A[i+1] ↔ A[r]return i+1}六首先画出它对应的图，加上标号，假设从1出发，每次贪心选择一个权重最小的顶点作为下一个要去的城市。