[计算机]算法分析与设计课程综合实验

算法分析与设计课程综合实验Design and Analysis of Algorithms1 Map Routing要求：Mandatory.实验目的：Implement the classic Dijkstra's shortest path algorithm and optimize it for maps. Such algorithms are widely used in geographic information systems (GIS) including MapQuest and GPS-based car navigation systems.实验内容及要求：Maps. For this assignment we will be working with maps, or graphs whose vertices are points in the plane and are connected by edges whose weights are Euclidean distances. Think of the vertices as cities and the edges as roads connected to them. To represent a map in a file, we list the number of vertices and edges, then list the vertices (index followed by its x and y coordinates), then list the edges (pairs of vertices), and finally the source and sink vertices. For example, Input6 represents the map below:Dijkstra's algorithm.Dijkstra's algorithm is a classic solution to the shortest path problem. It is described in section 24.3 in CLRS. The basic idea is not difficult to understand. We maintain, for every vertex in the graph, the length of the shortest known path from the source to that vertex, and we maintain these lengths in a priority queue. Initially, we put all the vertices on the queue with an artificially high priority and then assign priority 0.0 to the source. The algorithm proceeds by taking the lowest-priority vertex off the PQ, then checking all the vertices that can be reached from that vertex by one edge to see whether that edge gives a shorter path to the vertex from the source thanthe shortest previously-known path. If so, it lowers the priority to reflect this new information.Here is a step-by-step description that shows how Dijkstra's algorithm finds the shortest path 0-1-2-5 from 0 to 5 in the example above.process 0 (0.0)lower 3 to 3841.9lower 1 to 1897.4process 1 (1897.4)lower 4 to 3776.2lower 2 to 2537.7process 2 (2537.7)lower 5 to 6274.0process 4 (3776.2)process 3 (3841.9)process 5 (6274.0)This method computes the length of the shortest path. To keep track of the path, we also maintain for each vertex, its predecessor on the shortest path from the source to that vertex. The files EuclideanGraph.java, Point.java, IndexPQ.java, IntIterator.java, and Dijkstra.java provide a bare bones implementation of Dijkstra's algorithm for maps, and you should use this as a starting point. The client program ShortestPath.java solves a single shortest path problem and plots the results using turtle graphics. The client program Paths.java solves many shortest path problems and prints the shortest paths to standard output. The client program Distances.java solves many shortest path problems and prints only the distances to standard output.Your goal.Optimize Dijkstra's algorithm so that it can process thousands of shortest path queries for a given map. Once you read in (and optionally preprocess) the map, your program should solve shortest path problems in sublinear time. One method would be to precompute the shortest path for all pairs of vertices; however you cannot afford the quadratic space required to store all of this information. Your goal is to reduce the amount of work involved per shortest path computation, without using excessive space. We suggest a number of potential ideas below which you may choose to implement. Or you can develop and implement your own ideas.Idea 1.The naive implementation of Dijkstra's algorithm examines all V vertices in the graph. An obvious strategy to reduce the number of vertices examined is to stop the search as soon as you discover the shortest path to the destination. With this approach, you can make the running time per shortest path query proportional to E' log V' where E' and V' are the number of edges and vertices examined by Dijkstra's algorithm. However, this requires some care because just re-initializi ng all of the distances to ∞ would take time proportional to V. Since you are doing repeated queries, you can speed things up dramatically by only re-initializing those values that changed in the previous query.Idea 2.You can cut down on the search time further by exploiting the Euclidean geometry of the problem, as described in section 21.5 in the book of Algorithm in C Part V. For general graphs, Dijkstra's relaxes edge v-w by updating d[w] to the sum of d[v] plus the distance from v to w. For maps, we instead update d[w] to be the sum of d[v] plus the distance from v to w plus the Euclidean distance from w to d minus the Euclidean distance from v to d. This is known as the A* algorithm. This heuristics affects performance, but not correctness.Idea e a faster priority queue. There is some room for optimization in the supplied priority queue. You could also consider using a multiway heap as in Sedgewick Program 20.10.Testing.The file usa.txt contains 87,575 intersections and 121,961 roads in the continental United States. The graph is very sparse - the average degree is 2.8. Your main goal should be to answer shortest path queries quickly for pairs of vertices on this network. Your algorithm will likely perform differently depending on whether the two vertices are nearby or far apart. We provide input files that test both cases. You may assume that all of the x and y coordinates are integers between 0 and 10,000.实验类型：Verification.适用对象：Undergraduate for Computer School2 Document Distance Problem要求：Mandatory.实验目的：Design and implement the document distance problem and optimize it for data.实验内容及要求：Let D be a text document (e.g. the complete works of William Shakespeare).A word is a consecutive sequence of alphanumeric characters, such as "Hamlet" or "2007". We'll treat all upper-case letters as if they are lower-case, so that "Hamlet" and "hamlet" are the same word. Words end at a non-alphanumeric character, so "can't" contains two words: "can" and "t".The word frequency distribution of a document D is a mapping from words w to their frequency count, which we'll denote as D(w).We can view the frequency distribution D as vector, with one component per possible word. Each component will be a non-negative integer (possibly zero).The norm of this vector is defined in the usual way:.The inner-product between two vectors D and D' is defined as usual..Finally, the angle between two vectors D and D' is defined:This angle (in radians) will be a number between 0 and since the vectors are non-negative in each component. Clearly,angle(D,D) = 0.0for all vectors D, andangle(D,D') = π / 2if D and D' have no words in common.Example: The angle between the documents "To be or not to be" and "Doubt truth to be a liar" isWe define the distance between two documents to be the angle between their word frequency vectors.The document distance problem is thus the problem of computing the distance between two given text documents.An instance of the document distance problem is the pair of input text documents.3 Edit Distance要求：Mandatory.实验目的：Many word processors and keyword search engines have a spelling correction feature. If you type in a misspelled word x, the word processor or search engine can suggest a correction y. The correction y should be a word that is close to x. One way to measure the similarity in spelling between two text strings is by “edit distance”. The notion of edit distance is useful in other fields as well. For example, biologists use edit distance to characterize the similarity of DNA or protein sequences.实验内容及要求：The edit distance d(x, y) of two strings of text, x[1..m] and y[1..n], is defined to be the minimum possible cost of a sequence of “transformation operations”(defined below) that transforms string x[1..m] into string y[1..n]. To define the effect of thetransformation operations, we use an auxiliary string z[1..s] that holds the intermediate results. At the beginning of the transformation sequence s = m and z[1..s] = x[1..m] (i.e., we start with string x[1..m]). At the end of the transformation sequence, we should have s = n and z[1..s] = y[1..n](i.e., our goal is to transform into string y[1..n]). Throughout the transformation, we maintain the current length s of string z, as well as a cursor position i, i.e., an index into string z. The invariant 1 ≤i ≤s +1 holds at all times during the transformation. (Notice that the cursor can move one space beyond the end of the string z in order to allow insertion at the end of the string.)Each transformation operation may alter the string z, the size s, and the cursor position i. Each transformation operation also has an associated cost. The cost of a sequence of transformation operations is the sum of the costs of the individual operations on the sequence. The goal of the edit-distance problem is to find a sequence of transformation operation of minimum cost that transforms x[1..m] into y[1..n].There are five transformation operations:Operation Cost Effectleft 0 If i = 1 then do nothing. Otherwise, set i ←i-1right 0 If i = s +1 then do nothing. Otherwise, set i←i-1.replace 4 If i = s +1 then do nothing. Otherwise, replace the character underthe cursor by another character c by setting z[i] ←c, and thenincrementing i.delete 2 If i = s +1 then do nothing. Otherwise, delete the character c underthe cursor by setting z[i..s] ← z[i+1..s+1] and decrementing s. Thecursor position i does not change.insert 3 Insert the character c into string z by incrementing s, setting z[i+1..s] ←z[i..s-1], setting z[i] ←c, and then incrementing index i. As an example, one way to transform the source string algorithm to the target string analysis is to use the sequence of operations shown in Table 1, where the position of the underlined character represents the cursor position i. Many other sequences of transformation operations also transform algorithm to analysis − the solution in Table 1 is not unique − and some other solutions cost more while some others cost less.Operation z Cost Totalinitial string algorithm 0 0right algorithm 0 0right algorithm 0 0replace by y alyorithm 4 4replace by s alysrithm 4 8replace by i alysiithm 4 12replace by s alysisthm 4 16delete alysishm 2 18delete alysism 2 20delete alysis_ 2 22left alysis 0 22left alysis 0 22left alysis 0 22left alysis 0 22left alysis 0 22insert n anlysis 3 25insert a analysis 3 28Table 1: Transforming algorithm into analysis(a)It is possible to transform algorithm to analysis without using the “left” operation.Give a sequence of operations in the style of Table 1 that has the same cost as in Table 1 but does not use the “left” operation.(b)Argue that, for any two strings x and y with edit distance d(x, y), there exists asequence S of transformation operations that transforms x to y with cost d(x, y) where S does not contain any “left” operations.(c)Show that the problem of calculation the edit distance d(x, y) exhibits optimalsubstructure.(Hint: Consider all suffixes of x and y.)(d)Recursively define the value of edit distance d(x, y) in terms of the suffixes of stringsx and y. Indicate how edit distance exhibits overlapping subproblems.(e)Describe a dynamic-programming algorithm that computes the edit distance fromx[1..m] to y[1..n].(Do not use a memoized recursive algorithm. Your algorithm should be a classical, bottom-up, tabular algorithm.) Analyze the running time and space requirements of your algorithm.(f)Implement your algorithm as a computer program in any language you wish. Yourprogram should calculate the edit distance d(x, y) between two strings x and y using dynamic programming and print out the corresponding sequence of transformation operations in the style of Table 1. Run your program on the stringsx= “electrical engineering”,y= “computer science”.Submit the source code of your program electronically on the class website, and hand in a printout of your source code and your results. Sample input and output text is provided on the class website to help you debug your program. These solutions are not necessarily unique: there may be other sequences of transformation operations that achieve the same cost. As usual, you may collaborate to solve this problem, but you must write the program by yourself.(g) Run your program on the three input files provided on the class website. Each inputfile contains the following four lines:1.The number of characters m in the string x.2.The string x.3.The number of characters n in the string y.4.The string y.Compute the edit distance d(x, y) for each input. Do not hand in a printout of the transformation operations for this problem part. (Extra bonus kudos if you can identify the source of all the texts, without searching the web.)(h) If z is implemented using an array, then the “insert” and “delete” operations requiresΘ(n) time. Design a suitable data structure that allow each of the five transformation operations to be implemented in Ο(1) time.实验类型：Synthesis适用对象：Undergraduate for Computer School4 Global Sequence Alignment要求：Optional实验目的：Write a program to compute the optimal sequence alignment of two DNA strings. This program will introduce you to the emerging field of computational biology in which computers are used to do research on biological systems. Further, you will be introduced to a powerful algorithmic design paradigm known as dynamic programming.实验内容及要求：Biology review. A genetic sequence is a string formed from a four-letter alphabet {Adenosine (A), Thymidine (T), Guanosine (G), Cytidine (C)} of biological macromolecules referred to together as the DNA bases. A gene is a genetic sequence that contains the information needed to construct a protein. All of your genes taken together are referred to as the human genome, a blueprint for the parts needed to construct the proteins that form your cells and, by extension, your body. Each new cell produced by your body receives a copy of the genome. This copying process, as well as natural wear and tear, introduces a small number of changes into the sequences of many genes. Among the most common changes are the substitution of one base for another and the deletion of a substring of bases; such changes are generally referred to as point mutations. As a result of these point mutations, the same gene sequenced from closely related organisms will have slight differences.The problem.Through your research you have found the following sequence of a gene in a previously unstudied organism.A A C A G T T A C CWhat is the function of the protein that this gene encodes? You could immediately begin a series of uninformed experiments in the lab to determine what role this gene plays. However, there is a good chance that it is a variant of a known gene in a previously studied organism. Since biologists and computer scientists have laboriously determined (and published) the genetic sequence of many organisms (including humans), you wouldlike to leverage this information to your advantage. We'll compare the above genetic sequence with one which has already been sequenced and whose function is well understood.T A A G G T C AIf the two genetic sequences are similar enough, we might expect them to have similar functions. We would like a way to quantify "similar enough".Edit-distance.We measure the similarity of two genetic sequences by using a very popular method known as the edit distance, a concept which is also widely used in spell checking, speech recognition, plagiarism detection, file revision, and computational linguistics. We align the two sequences, but we are permitted to insert gaps in either sequence (e.g., to make them have the same length). We pay a penalty for each gap that we insert and also for each pair of characters that mismatch in the final alignment. Intuitively, these penalties model the relative likeliness of point mutations arising from deletion/insertion and substitution. We produce a numerical score according to the following simple rule, which is widely used in biological applications:As an example, two possible alignments for the genetic sequences discussed above are:The first alignment has a score of 8, while the second one has a score of 7. The edit-distance is the score of the best possible alignment between the two genetic sequences over all possible alignments. In this example, the second alignment is in fact optimal, so the edit-distance between the two strings is 7. Computing the edit-distance between two strings is a nontrivial computational problem because we must find the best alignment among exponentially many possibilities. For example, if both strings are 100 characters long, then there are more than 10^75 possible alignments.A recursive solution.Your job is to write a program to compute the edit-distance and the optimal alignment of two genetic sequences. We can compute the edit-distancerecursively by breaking up the sequence alignment problem on the two original strings x and y into many alignment problems on the suffixes of the two strings. We use the notation x[i..M] to refer to the suffix of x consisting of the characters x[i], x[i+1], ..., x[M]. Note that x[M] is the string termination character '\0'. For example, consider the two strings x = "AACAGTTACC" and y = "TAAGGTCA" of length M = 10 and N = 8, respectively. Then, x[2..M] is "CAGTTACC" and y[8..N] is the empty string.Now, we return to our recursive solution technique. Consider the zeroth column in an optimal alignment of x[0..M] with y[0..N]. There are three possibilities:1.The optimal alignment matches x[0] = 'A' up with y[0] = 'T'. In this case, we pay apenalty of 1 for a mismatch and still need to align the string x[1..M] with y[1..N].What is the best way to do this? This subproblem is exactly the same as the original sequence alignment problem, except that the two inputs are each suffixes of the original inputs. We can solve this subproblem recursively.2.The optimal alignment matches the x[0] = 'A' up with a gap. In this case, we pay apenalty of 2 for a gap and still need to align the string x[1..M] with y[0..N]. This subproblem is identical to the original sequence alignment problem, except that the first input is a suffix of the original input.3.The optimal alignment matches the y[0] = 'T' up with a gap. In this case, we pay apenalty of 2 for a gap and still need to align the string x[0..M] with y[1..N]. This subproblem is identical to the original sequence alignment problem, except that the second input is a suffix of the original input.All of the resulting subproblems are sequence alignment problem on suffixes of the original inputs. Thus, we need a function, say opt(i, j) that returns the optimum value of aligning the input string x[i..M] with y[j..N]. Using this notation, the optimal value of our original sequence alignment problem is opt(0, 0). In general, to compute opt(i, j), we need to compute the minimum of the following three quantities:opt(i+1, j+1) + 0/1 by aligning x[i] with y[j], add 0 if a match, 1 if a mismatchopt(i+1, j) + 2 by aligning x[i] with a gapopt(i, j+1) + 2 by aligning y[j] with a gapFor our recursive scheme to bottom-out, we need a base case. Aligning an empty string with another string of length j requires inserting j gaps, for a total cost of 2j. Thus, in general we should set opt(M, j) = 2(N-j) and opt(i, N) = 2(M-i). For our example, the final matrix is:6 T 9 87 5 3 3 5 6 87 A 11 9 7 6 4 2 3 4 68 C 13 11 9 7 5 3 1 3 49 C 14 12 10 8 6 4 2 1 210 - 16 14 12 10 8 6 4 2 0The score of the optimal alignment is opt(0, 0) = 7.A dynamic programming approach. A direct implementation of the above recursive scheme will work, but it is spectacularly inefficient. If both input strings have N characters, then the number of recursive calls will exceed 2^N.To overcome this performance bug, we will use dynamic programming. (Read sections 15.1 for an introduction to this technique and 15.4 for the problem.) Dynamic programming is a powerful algorithmic paradigm that forms the core computational engine of many programs, including BLAST(the sequence alignment program almost universally used by molecular biologist in their experimental work). The key idea of dynamic programming is to break a large computational problem up into smaller subproblems, store the answers to those smaller subproblems, and, eventually, using the stored answers to solve the original problem. This avoids recomputing the same quantity over and over again. Specifically, we maintain an array, say c[i][j], that contains a table (or cache) of the solutions to all of the subproblems that the recursive function has already solved. The recursive function computes and fills in the appropriate table entry when it is presented with a new subproblem. However, when the function is presented with a previously solved subproblem, it simply looks up the appropriate value in the table and returns it.Finding the alignment itself.The above procedure above indicates how to compute the value of the optimal alignment. We now describe how to find the optimal alignment itself. In order to reconstruct the optimal alignment, we maintain a character matrix, say bl[i][j], to keep track of where the minimum value for aligning x[i..M] with y[j..N] came from. For example, if the minimum came from aligning x[i] with y[j] (in which case we solved the subproblem c(i+1, j+1), then we can record this fact by drawing an arrow from (i, j) to (i+1, j+1). We can obtain a crude ASCII picture of such an arrow by storing one of the three characters '\', '-', or '|' into b[i][j]. We interpret the three symbols as arrows emanating from (i, j) and terminating at (i+1, j+1), (i+1, j), and (i, j+1), respectively. For the example above, we get the the following solution matrix:5 T \ \ \ \ \ \ \ | |6 T \ \ \ \ \ \ \ | |7 A - \ \ \ \ \ | \ |8 C \ \ \ \ \ \ \ \ |9 C - - - - - - \ \ |10 - - - - - - - - - .Note that all arrows point from top to bottom and left to right. In order to reconstruct the alignment, we follow the arrows from the upper left corner c[0][0] until we arrive at the lower right corner c[M][N]. When we encounter '-', we insert a gap in x; when we encounter '|', we insert a gap in y; and, when we encounter '\', we align the two characters. The optimal alignment is the second candidate alignment in the edit-distance section.Input and output.Read the two input strings from standard input, one per line. Print the edit-distance to standard output. If both strings have length at most 40, also print out a table of optimal values, optimal choices, and an optimal alignment. Test your program using the example provided above, as well as the genomic data sets from GenBank.Analysis.Estimate the running time and memory usage of your program as a function of the lengths of the two input strings M and N.实验类型：Synthesis适用对象：Undergraduate for Computer School。