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w32zhong / BTreeimplementationCSC 541 Assignment 4 B-Trees Introduction The goals of this assignment are two-fold: To introduce you to searching data on disk using B-trees. To investigate how changing the order of a B-tree affects its performance. Index File During this assignment you will create, search, and manage a binary index file of integer key values. The values stored in the file will be specified by the user. You will structure the file as a B-tree. Program Execution Your program will be named assn_4 and it will run from the command line. Two command line arguments will be specified: the name of the index file, and a B-tree order. assn_4 index-file order For example, executing your program as follows assn_4 index.bin 4 would open an index file called index.bin that holds integer keys stored in an order-4 B-tree. You can assume order will always be ≥ 3. For convenience, we refer to the index file as index.bin throughout the remainder of the assignment. Note. If you are asked open an existing index file, you can assume the B-tree order specified on the command line matches the order that was used when the index file was first created. B-Tree Nodes Your program is allowed to hold individual B-tree nodes in memory—but not the entire tree—at any given time. Your B-tree node should have a structure and usage similar to the following. #include <stdlib.h> int order = 4; /* B-tree order */ typedef struct { /* B-tree node */ int n; /* Number of keys in node */ int *key; /* Node's keys */ long *child; /* Node's child subtree offsets */ } btree_node; btree_node node; /* Single B-tree node */ node.n = 0; node.key = (int *) calloc( order - 1, sizeof( int ) ); node.child = (long *) calloc( order, sizeof( long ) ); Note. Be careful when you're reading and writing data structures with dynamically allocated memory. For example, trying to write node like this fwrite( &node, sizeof( btree_node ), 1, fp ); will write node's key count, the pointer value for its key array, and the pointer value for its child offset array, but it will not write the contents of the key and child offset arrays. The arrays' contents and not pointers to their contents need to be written explicitly instead. fwrite( &node.n, sizeof( int ), 1, fp ); fwrite( node.key, sizeof( int ), order - 1, fp ); fwrite( node.child, sizeof( long ), order, fp ); Reading node structures from disk would use a similar strategy. Root Node Offset In order to manage any tree, you need to locate its root node. Initially the root node will be stored near the front of index.bin. If the root node splits, however, a new root will be appended to the end of index.bin. The root node's offset will be maintained persistently by storing it at the front of index.bin when the file is closed, and reading it when the file is opened. #include <stdio.h> FILE *fp; /* Input file stream */ long root; /* Offset of B-tree root node */ fp = fopen( "index.bin", "r+b" ); /* If file doesn't exist, set root offset to unknown and create * file, otherwise read the root offset at the front of the file */ if ( fp == NULL ) { root = -1; fp = fopen( "index.bin", "w+b" ); fwrite( &root, sizeof( long ), 1, fp ); } else { fread( &root, sizeof( long ), 1, fp ); } User Interface The user will communicate with your program through a set of commands typed at the keyboard. Your program must support four simple commands: add k Add a new integer key with value k to index.bin. find k Find an entry with a key value of k in index.bin, if it exists. print Print the contents and the structure of the B-tree on-screen. end Update the root node's offset at the front of the index.bin, and close the index file, and end the program. Add Use a standard B-tree algorithm to add a new key k to the index file. Search the B-tree for the leaf node L responsible for k. If k is stored in L's key list, print Entry with key=k already exists on-screen and stop, since duplicate keys are not allowed. Create a new key list K that contains L's keys, plus k, sorted in ascending order. If L's key list is not full, replace it with K, update L's child offsets, write L back to disk, and stop. Otherwise, split K about its median key value km into left and right key lists KL = (k0, ... , km-1) and KR = (km+1, ... , kn-1). Use ceiling to calculate m = ⌈(n-1)/2⌉. For example, if n = 3, m = 1. If n = 4, m = 2. Save KL and its associated child offsets in L, then write L back to disk. Save KR and its associated child offsets in a new node R, then append R to the end of the index file. Promote km , L's offset, and R's offset and insert them in L's parent node. If the parent's key list is full, recursively split its list and promote the median to its parent. If a promotion is made to a root node with a full key list, split and create a new root node holding km and offsets to L and R. Find To find key value k in the index file, search the root node for k. If k is found, the search succeeds. Otherwise, determine the child subtree S that is responsible for k, then recursively search S. If k is found during the recursive search, print Entry with key=k exists on-screen. If at any point in the recursion S does not exist, print Entry with key=k does not exist on-screen. Print This command prints the contents of the B-tree on-screen, level by level. Begin by considering a single B-tree node. To print the contents of the node on-screen, print its key values separated by commas. int i; /* Loop counter */ btree_node node; /* Node to print */ long off; /* Node's offset */ for( i = 0; i < node.n - 1; i++ ) { printf( "%d,", node.key[ i ] ); } printf( "%d", node.key[ node.n - 1 ] ); To print the entire tree, start by printing the root node. Next, print the root node's children on a new line, separating each child node's output by a space character. Then, print their children on a new line, and so on until all the nodes in the tree are printed. This approach prints the nodes on each level of the B-tree left-to-right on a common line. For example, inserting the integers 1 through 13 inclusive into an order-4 B-tree would produce the following output. 1: 9 2: 3,6 12 3: 1,2 4,5 7,8 10,11 13 To support trees with more than 9 levels, we leave space for two characters to print the level at the beginning of each line, that is, using printf( "%2d: ", lvl )" or something similar. Hint. To process nodes left-to-right level-by-level, do not use recursion. Instead, create a queue containing the root node's offset. Remove the offset at the front of the queue (initially the root's offset) and read the corresponding node from disk. Append the node's non-empty subtree offsets to the end of the queue, then print the node's key values. Continue until the queue is empty. End This command ends the program by writing the root node's offset to the front of index.bin, then closing the index file. Programming Environment All programs must be written in C, and compiled to run on the remote.eos.ncsu.edu Linux server. Any ssh client can be used to access your Unity account and AFS storage space on this machine. Your assignment will be run automatically, and the output it produces will be compared to known, correct output using diff. Because of this, your output must conform to the print command's description. If it doesn't, diff will report your output as incorrect, and it will be marked accordingly. Supplemental Material In order to help you test your program, we provide example input and output files. input-01.txt, an input file of commands applied to an initially empty index file saved as an order-4 B-tree, and input-02.txt, an input file of commands applied to the index file produced by input-01.txt. The output files show what your program should print after each input file is processed. output-01.txt, the output your program should produce after it processes input-01.txt. output-02.txt, the output your program should produce after it processes input-02.txt. To test your program, you would issue the following commands: % rm index.bin % assn_4 index.bin 4 < input-01.txt > my-output-01.txt % assn_4 index.bin 4 < input-02.txt > my-output-02.txt You can use diff to compare output from your program to our output files. If your program is running properly and your output is formatted correctly, your program should produce output identical to what is in these files. Please remember, the files we're providing here are meant to serve as examples only. Apart from holding valid commands, you cannot make any assumptions about the size or the content of the input files we will use to test your program. Test Files The following files were used to test your program. Order 3 Test Case. input-03.txt output-03-first.txt Order 4 Test Case. input-04.txt output-04-first.txt Order 10 Test Case. input-10-01.txt, input-10-02.txt output-10-01.txt, output-10-02.txt Order 20 Test Case. input-20.txt output-20-first.txt Your program was run on all test cases using order-3, order-4, and order-20 B-trees. % rm index.bin % assn_4 index.bin 3 < input-03.txt > my-output-03.txt % rm index.bin % assn_4 index.bin 4 < input-04.txt > my-output-04.txt % rm index.bin % assn_4 index.bin 20 < input-20.txt > my-output-20.txt Your program was also run twice using an order-10 B-tree, to test its ability to re-use an existing index file. % rm index.bin % assn_4 index.bin 10 < input-10-01.txt > my-output-10-01.txt % assn_4 index.bin 10 < input-10-02.txt > my-output-10-02.txt Hand-In Requirements Use Moodle (the online assignment submission software) to submit the following files: assn_4, a Linux executable of your finished assignment, and all associated source code files (these can be called anything you want). There are four important requirements that your assignment must satisfy. Your executable file must be named exactly as shown above. The program will be run and marked electronically using a script file, so using a different name means the executable will not be found, and subsequently will not be marked. Your program must be compiled to run on remote.eos.ncsu.edu. If we cannot run your program, we will not be able to mark it, and we will be forced to assign you a grade of 0. Your program must produce output that exactly matches the format described in the print command section of this assignment. If it doesn't, it will not pass our automatic comparison to known, correct output. You must submit your source code with your executable prior to the submission deadline. If you do not submit your source code, we cannot MOSS it to check for code similarity. Because of this, any assignment that does not include source code will be assigned a grade of 0. Updated 20-Dec-14
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noahjonesx / MarkovModelMarkov Text Generation Problem Description The Infinite Monkey Theorem1 (IFT) says that if a monkey hits keys at random on a typewriter it will almost surely, given an infinite amount of time, produce a chosen text (like the Declaration of Independence, Hamlet, or a script for ... Planet of the Apes). The probability of this actually happening is, of course, very small but the IFT claims that it is still possible. Some people have tested this hypotheis in software and, after billions and billions of simulated years, one virtual monkey was able to type out a sequence of 19 letters that can be found in Shakespeare’s The Two Gentlemen of Verona. (See the April 9, 2007 edition of The New Yorker if you’re interested; but, hypothesis testing with real monkeys2 is far more entertaining.) The IFT might lead to some interesting conversations with Rust Cohle, but the practical applications are few. It does, however, bring up the idea of automated text generation, and there the ideas and applications are not only interesting but also important. Claude Shannon essentially founded the field of information theory with the publication of his landmark paper A Mathematical Theory of Computation3 in 1948. Shannon described a method for using Markov chains to produce a reasonable imitation of a known text with sometimes startling results. For example, here is a sample of text generated from a Markov model of the script for the 1967 movie Planet of the Apes. "PLANET OF THE APES" Screenplay by Michael Wilson Based on Novel By Pierre Boulle DISSOLVE TO: 138 EXT. GROVE OF FRUIT TREES - ESTABLISHING SHOT - DAY Zira run back to the front of Taylor. The President, I believe the prosecutor's charge of this man. ZIRA Well, whoever owned them was in pretty bad shape. He picks up two of the strain. You got what you wanted, kid. How does it taste? Silence. Taylor and cuffs him. Over this we HEAR from a distance is a crude horse-drawn wagon is silhouetted-against the trunks and branches of great trees and bushes on the horse's rump. Taylor lifts his right arm to ward off the blow, and the room and lands at the feet of Cornelius and Lucius are sorting out equipment falls to his knees, buries his head silently at the Ranch). DISSOLVE TO: 197 INT. CAGES - CLOSE SHOT - FEATURING LANDON - FROM TAYLOR'S VOICE (o.s.) I've got a fine veternary surgeons under my direction? ZIRA Taylor! ZIRA There is a small lake, looking like a politician. TAYLOR Dodge takes a pen and notebook from the half-open door of a guard room. Taylor bursts suddenly confronted by his 1https://en.wikipedia.org/wiki/Infinite_monkey_theorem2https://web.archive.org/web/20130120215600/http://www.vivaria.net/experiments/notes/publication/NOTES_ EN.pdf3http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=6773024 1 original pursuer (the dismounted cop coming up with a cigar butt and places it in the drawer beside them. TAYLOR What's the best there is a. loud RAP at the doll was found beside the building. Zira waits at the third table. TAYLOR Good question. Is he a man? CORNELIUS (impatiently. DODGE Blessed are the vegetation. These SHOTS are INTERCUT with: 94 WHAT THE ASTRONAUTS They examine the remnants of the cage. ZIRA (plunging on) Their speech organs are adequate. The flaw lies not in anatomy but in the back of his left sleeve. TAYLOR (taking off his shirt. 80 DODGE AND LANDON You don't sound happy in your work. GALEN (defensively) Gorilla hunter stands over a dead man, one fo Besides a few spelling errors and some rather odd things that make you wonder about the author, this passage is surprisingly human-like. This is a simple example of natural language generation, a sub-area of natural language processing—a very active area of research in computer science. The particular approach we’re using in this assignment was famously implemented as the fictitious Mark V. Shaney4 and the Emacs command Disassociated Press5. Approach So, here’s the basic idea: Imagine taking a book (say, Tom Sawyer) and determining the probability with which each character occurs. You would probably find that spaces are the most common, that the character ‘e’ is fairly common, and that the character ‘q’ is rather uncommon. After completing this “level 0” analysis, you would be able to produce random Tom Sawyer text based on character probabilities. It wouldn’t have much in common with the real thing, but at least the characters would tend to occur in the proper propor- tion. In fact, here’s an example of what you might produce: Level 0 rla bsht eS ststofo hhfosdsdewno oe wee h .mr ae irii ela iad o r te u t mnyto onmalysnce, ifu en c fDwn oee iteo Now imagine doing a slightly more sophisticated level 1 analysis by determining the probability with which each character follows every other character. You would probably discover that ‘h’ follows ‘t’ more frequently than ‘x’ does, and you would probably discover that a space follows ‘.’ more frequently than ‘,’ does. You could now produce some randomly generated Tom Sawyer text by picking a character to begin with and then always choosing the next character based on the previous one and the probabilities revealed by the analysis. Here’s an example: Level 1 "Shand tucthiney m?" le ollds mind Theybooure He, he s whit Pereg lenigabo Jodind alllld ashanthe ainofevids tre lin-p asto oun theanthadomoere Now imagine doing a level k analysis by determining the probability with which each character follows every possible sequence of characters of length k (kgrams). A level 5 analysis of Tom Sawyer for example, would reveal that ‘r’ follows “Sawye” more frequently than any other character. After a level k analysis, you would be able to produce random Tom Sawyer by always choosing the next character based on the previous k characters (a kgram) and the probabilities revealed by the analysis. 4https://en.wikipedia.org/wiki/Mark_V._Shaney5https://en.wikipedia.org/wiki/Dissociated_press Page 2 of 5 At only a moderate level of analysis (say, levels 5-7), the randomly generated text begins to take on many of the characteristics of the source text. It probably won’t make complete sense, but you’ll be able to tell that it was derived from Tom Sawyer as opposed to, say, The Sound and the Fury. Here are some more examples of text that is generated from increasing levels of analysis of Tom Sawyer. (These “levels of analysis” are called order K Markov models.) K = 2 "Yess been." for gothin, Tome oso; ing, in to weliss of an’te cle - armit. Papper a comeasione, and smomenty, fropeck hinticer, sid, a was Tom, be suck tied. He sis tred a youck to themen K = 4 en themself, Mr. Welshman, but him awoke, the balmy shore. I’ll give him that he couple overy because in the slated snufflindeed structure’s kind was rath. She said that the wound the door a fever eyes that WITH him. K = 6 people had eaten, leaving. Come - didn’t stand it better judgment; His hands and bury it again, tramped herself! She’d never would be. He found her spite of anything the one was a prime feature sunset, and hit upon that of the forever. K = 8 look-a-here - I told you before, Joe. I’ve heard a pin drop. The stillness was complete, how- ever, this is awful crime, beyond the village was sufficient. He would be a good enough to get that night, Tom and Becky. K = 10 you understanding that they don’t come around in the cave should get the word "beauteous" was over-fondled, and that together" and decided that he might as we used to do - it’s nobby fun. I’ll learn you." To create an order K Markov model of a given source text, you would need to identify all kgrams in the source text and associate with each kgram all the individual characters that follow it. This association or mapping must also capture the frequency with which a given character follows a given kgram. For example, suppose that k = 2 and the sample text is: agggcagcgggcg The Markov model would have to represent all the character strings of length two (2-grams) in the source text, and associate with them the characters that follow them, and in the correct proportion. The following table shows one way of representing this information. kgram Characters that follow ag gc gg gcgc gc agg ca g cg g Once you have created an order K Markov model of a given source text, you can generate new text based on this model as follows. Page 3 of 5 1. Randomly pick k consecutive characters that appear in the sample text and use them as the initial kgram. 2. Append the kgram to the output text being generated. 3. Repeat the following steps until the output text is sufficiently long. (a) Select a character c that appears in the sample text based on the probability of that character following the current kgram. (b) Append this character to the output text. (c) Update the kgram by removing its first character and adding the character just chosen (c) as its last character. If this process encounters a situation in which there are no characters to choose from (which can happen if the only occurrence of the current kgram is at the exact end of the source), simply pick a new kgram at random and continue. As an example, suppose that k = 2 and the sample text is that from above: agggcagcgggcg Here are four different output text strings of length 10 that could have been the result of the process described above, using the first two characters (’ag’) as the initial kgram. agcggcagcg aggcaggcgg agggcaggcg agcggcggca For another example, suppose that k = 2 and the sample text is: the three pirates charted that course the other day Here is how the first three characters of new text might be generated: •A two-character sequence is chosen at random to become the initial kgram. Let’s suppose that “th” is chosen. So, kgram = th and output = th. •The first character must be chosen based on the probability that it follows the kgram (currently “th”) in the source. The source contains five occurrences of “th”. Three times it is followed by ’e’, once it is followed by ’r’, and once it is followed by ’a’. Thus, the next character must be chosen so that there is a 3/5 chance that an ’e’ will be chosen, a 1/5 chance that an ’r’ will be chosen, and a 1/5 chance that an ’a’ will be chosen. Let’s suppose that we choose an ’e’ this time. So, kgram = he and output = the. •The next character must be chosen based on the probability that it follows the kgram (currently “he”) in the source. The source contains three occurrences of “he”. Twice it is followed by a space and once it is followed by ’r’. Thus, the next character must be chosen so that there is a 2/3 chance that a space will be chosen and a 1/3 chance that an ’r’ will be chosen. Let’s suppose that we choose an ’r’ this time. So, kgram = er and output = ther. •The next character must be chosen based on the probability that it follows the kgram (currently “er”) in the source. The source contains only one occurrence of “er”, and it is followed by a space. Thus, the next character must be a space. So, kgram = r_ and output = ther_, where ’_’ represents a blank space. Page 4 of 5 Implementation Details You are provided with two Java files that you must use to develop your solution: MarkovModel.java and TextGenerator.java. The constructors of MarkovModel build the order-k model of the source text. You are required to represent the model with the provided HashMap field. The main method of TextGenerator must process the following three command line arguments (in the args array): •A non-negative integer k •A non-negative integer length. •The name of an input file source that contains more than k characters. Your program must validate the command line arguments by making sure that k and length are non- negative and that source contains at least k characters and can be opened for reading. If any of the command line arguments are invalid, your program must write an informative error message to System.out and terminate. If there are not enough command line arguments, your program must write an informative error message to System.out and terminate. With valid command line arguments, your program must use the methods of the MarkovModel class to create an order k Markov model of the sample text, select the initial kgram, and make each character selection. You must implement the MarkovModel methods according to description of the Markov modeling process in the section above. A few sample texts have been provided, but Project Gutenberg (http://www.gutenberg.org) maintains a large collection of public domain literary works that you can use as source texts for fun and practice. Acknowledgments This assignment is based on the ideas of many people, Jon Bentley and Owen Astrachan in particular.
fazeelkhalid / Graph Real Time ProblemsPoints: 100 Topics: Graphs, topological sort, freedom to decide how to represent data and organize code (while still reading in a graph and performing topological sort) PLAGIARISM/COLLUSION: You should not read any code (solution) that directly solves this problem (e.g. implements DFS, topological sorting or other component needed for the homework). The graph representation provided on the Code page (which you are allowed to use in your solution) and the pseudocode and algorithm discussed in class provide all the information needed. If anything is unclear in the provided materials check with us. You can read materials on how to read from a file, or read a Unix file or how to tokenize a line of code, BUT not in a sample code that deals with graphs or this specific problem. E.g. you can read tutorials about these topics, but not a solution to this problem (or a problem very similar to it). You should not share your code with any classmate or read another classmate's code. Part 1: Main program requirements (100 pts) Given a list of courses and their prerequisites, compute the order in which courses must be taken so that when taking a courses, all its prerequisites have already been taken. All the files that the program would read from are in Unix format (they have the Unix EOL). Provided files: ● Grading Criteria ● cycle0.txt ● data0.txt ● data0_rev.txt ● data1.txt - like data0.txt but the order of the prerequisite courses is modified on line 2. ● slides.txt (graph image) - courses given in such a way that they produce the same graph as in the image. (The last digit in the course number is the same as the vertex corresponding to it in the drawn graph. You can also see this in the vertex-to-course name correspondence in the sample run for this file.) ● run.html● data0_easy.txt - If you cannot handle the above file format, this is an easier file format that you can use, but there will be 15 points lost in this case. More details about this situation are given in Part 3. ● Unix.zip - zipped folder with all data files. ● For your reference: EOL_Mac_Unix_Windows.png - EOL symbols for Unix/Mac/Windows Specifications: 1. You can use structs, macros, typedef. 2. All the code must be in C (not C++, or any other language) 3. Global or static variables are NOT allowed. The exception is using macros to define constants for the size limits (e.g. instead of using 30 for the max course name size). E.g. #define MAX_ARRAY_LENGTH 20 4. You can use static memory (on the frame stack) or dynamic memory. (Do not confuse static memory with static variables.) 5. The program must read from the user a filename. The filename (as given by the user) will include the extension, but NOT the path. E.g.: data0.txt 6. You can open and close the file however many times you want. 7. File format: 1. Unix file. It will have the Unix EOL (end-of-line). 2. Size limits: 1. The file name will be at most 30 characters. 2. A course name will be at most 30 characters 3. A line in the file will be at most 1000 characters. 3. The file ends with an empty new line. 4. Each line (except for the last empty line) has one or more course names. 5. Each course name is a single word (without any spaces). E.g. CSE1310 (with no space between CSE and 1310). 6. There is no empty space at the end of the line. 7. There is exactly one empty space between any two consecutive courses on the same line. (You do not need to worry about having tabs or more than one empty space between 2 courses.) The first course name on each line is the course being described and the following courses are the prerequisites for it. E.g. CSE2315 CSE1310 MATH1426 ENGL13018. The first line describes course CSE2315 and it indicates that CSE2315 has 2 prerequisite courses, namely: CSE1310 and MATH1426. The second line describes course ENG1301 and it indicates that ENG1301 has no prerequisites. 9. You can assume that there is exactly one line for every course, even for those that do not have prerequisites (see ENGL1301 above). Therefore you can count the number of lines in the file to get the total number of courses. 10.The courses are not given in any specific order in the file. 8. You must create a directed graph corresponding to the data in the file. 1. The graph will have as many vertices as different courses listed in the file. 2. You can represent the vertices and edges however you want. 3. You do NOT have to use a graph struct. If you can do all the work with just the 2D table (the adjacency matrix) that is fine. You HAVE TO implement the topological sorting covered in class (as this assignment is on Graphs), but you can organize, represent and store the data however you want. 4. For the edges, you can use either the adjacency matrix representation or the adjacency list. If you use the adjacency list, keep the nodes in the list sorted in increasing order. 5. For each course that has prerequisites, there is an edge, from each prerequisite to that course. Thus the direction of the edge indicates the dependency. The actual edge will be between the vertices in the graph corresponding to these courses. E.g. file data0.txt has: c100 c300 c200 c100 c200 c100 Meaning: c100-----> c200 \ | \ | \ | \ | \ | \ | V V c300(The above drawing is provided here to give a picture of how the data in the file should be interpreted and the graph that represents this data. Your program should *NOT* print this drawing. See the sample run for expected program output.) From this data you should create the correspondence: vertex 0 - c100 vertex 1 - c300 vertex 2 - c200 and you can represent the graph using adjacency matrix (the row and column indexes are provided for convenience): | 0 1 2 ----------------- 0| 0 1 1 1| 0 0 0 2| 0 1 0 e.g. E[0][1] is 1 because vertex 0 corresponds to c100 and vertex 1 corresponds to c300 and c300 has c100 as a prerequisite. Notice that E[1][0] is not 1. If you use the adjacency list representation, then you can print the adjacency list. The list must be sorted in increasing order (e.g. see the list for 0). It should show the corresponding node numbers. E.g. for the above example the adjacency list will be: 0: 1, 2, 1: 2: 1, 6. 7. In order for the output to look the same for everyone, use the correspondence given here: vertex 0 for the course on the first line, vertex 1 for the course on the second line, etc. 1. Print the courses in topological sorted order. This should be done using the DFS (Depth First Search) algorithm that we covered in class and the topological sorting based on DFS discussed in class. There is no topological order if there is a cycle in the graph; in this case print an error message. If in DFV-visit when looking at the (u,v) edge, if the color of v is GRAY then there is a cycle in the graph (and therefore topological sorting is not possible). See the Lecture on topological sorting (You can find the date based on the table on the Scans page and then watch the video from that day. I have also updated the pseudocodein the slides to show that. Refresh the slides and check the date on the first page. If it is 11/26/2020, then you have the most recent version.) 8. (6 points) create and submit 1 test file. It must cover a special case. Indicate what special case you are covering (e.g. no course has any prerequisite). At the top of the file indicate what makes it a special case. Save this file as special.txt. It should be in Unix EOL format. Part 2: Suggestions for improvements (not for grade) 1. CSE Advisors also are mindful and point out to students the "longest path through the degree". That is longest chain of course prerequisites (e.g. CSE1310 ---> CSE1320 --> CSE3318 -->...) as this gives a lower bound on the number of semesters needed until graduation. Can you calculate for each course the LONGEST chain ending with it? E.g. in the above example, there are 2 chains ending with c300 (size 2: just c100-->c300, size 3: c100-->c200-->c300) and you want to show longest path 3 for c300. Can you calculate this number for each course? 2. Allow the user the enter a list of courses taken so far (from the user or from file) and print a list of the courses they can take (they have all the prerequisites for). 3. Ask the user to enter a desired number of courses per semester and suggest a schedule (by semester). Part 3: Implementation suggestions 1. Reading from file: (15 points) For each line in the file, the code can extract the first course and the prerequisites for it. If you cannot process each line in the file correctly, you can use a modified input file that shows on each line, the number of courses, but you would lose the 15 points dedicated to line processing. If your program works with the "easy files", in order to make it easy for the TAs to know which file to provide, please name your C program courses_graph_easy.c. Here is the modification shown for a new example. Instead of c100 c300 c200 c100 c200 the file would have: 1 c1003 c300 c200 c100 1 c200 1. that way the first data on each line is a number that tells how many courses (strings) follow after it on that line. Everything is separated by exactly one space. All the other specifications are the same as for the original file (empty line at the end, no space at the end of any line, length of words, etc). Here is data0_easy.txt Make a direct correspondence between vertex numbers and course names. E.g. the **first** course name on the first line corresponds to vertex 0, the **first** course name on the second line corresponds to vertex 1, etc... 2. 3. The vertex numbers are used to refer to vertices. 4. In order to add an edge in the graph you will need to find the vertex number corresponding to a given course name. E.g. find that c300 corresponds to vertex 1 and c200 corresponds to vertex 2. Now you can set E[2][1] to be 1. (With the adjacency list, add node 1 in the adjacency list for 2 keeping the list sorted.) To help with this, write a function that takes as arguments the list/array of [unique] course names and one course name and returns the index of that course in the list. You can use that index as the vertex number. (This is similar to the indexOf method in Java.) 5. To see all the non-printable characters that may be in a file, find an editor that shows them. E.g. in Notepad++ : open the file, go to View -> Show symbol -> Show all characters. YOU SHOULD TRY THIS! In general, not necessarily for this homework, if you make the text editor show the white spaces, you will know if what you see as 4 empty spaces comes from 4 spaces or from one tab or show other hidden characters. This can help when you tokenize. E.g. here I am using Notepad++ to see the EOL for files saved with Unix/Mac/Windows EOL (see the CR/LF/CRLF at the end of each line): EOL_Mac_Unix_Windows.png How to submit Submit courses_graph.c (or courses_graph_easy.c) and special.txt (the special test case you created) in Canvas . (For courses_graph_easy.c you can submit the "easy" files that you created.)Your program should be named courses_graph.c if it reads from the normal/original files. If instead it reads from the 'easy' files, name it courses_graph_easy.c As stated on the course syllabus, programs must be in C, and must run on omega.uta.edu or the VM. IMPORTANT: Pay close attention to all specifications on this page, including file names and submission format. Even in cases where your answers are correct, points will be taken off liberally for non-compliance with the instructions given on this page (such as wrong file names, wrong compression format for the submitted code, and so on). The reason is that non-compliance with the instructions makes the grading process significantly (and unnecessarily) more time consuming. Contact the instructor or TA if you have any questions
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IBM / AI Assisted Chemical SensingThis repository provides chemsense, a package developed for chemical sensor array data processing. chemsense leverages visual encoding of sensor data and pre-trained vision models to extract vector-based measurement representations, i.e. the "digital fingerprints" of the chemical specimens under test
rahulthawal / Finding Object By Robotic ArmDependencies OpenCV The Open Computer Vision library, in the form of python, was used to perform image parsing and analysis. This was used to apply a masking filter to each image, then parse each image for the contours of the marker / target. Numpy The Numerical Python (Numpy) library was used to provide advanced array and matrix functionality for the mathematics used in the program. In particular, the arrays were used for creating and passing the necessary RGB (in this instance, BGR) vectors for creating the appropriate mask. Access The Python program has been designed to run at the allocated hours of each member of Team 4 automatically (See below for a time table). It first starts by retrieving the system time of the executing host and compares it to the hard-coded times for each member. If the system time does not coincide with members' time, the program will inform the user through Standard Output and close. Otherwise, the program will continue with the Execution Procedure. Function Definitions readResponse() Takes the encoded response from the server after issuing a command and parses through the encoded white-space characters for more a more user-friendly report. getCoords(output) Given the output from the server, it parses through the data to find the X, Y, and Z co-ordinates. sendAJMA() Sends the Angular commands to the server to move the arm. After issuing, it acquires the resulting X, Y, and Z co-ordinates from the parsed return output. sendCapture() Sends the capture command to the server and prints the response. find_marker(image) Given an input image, the function parses over each pixel, applying a mask. After massaging the data, it attempts to find contours within the image to find a target marker (for example, a piece of paper). download_image(url, filename) Issues a HTTP request for an image. rawCommand(givenData) Parses through the encoded response from the server and makes it more user-friendly. printData() Does the actual printing of parsed data for the user's benefit. downloadCheckImage() First, it attempts to download the latest image from the server and reports any errors it encounters. Then, it parses the image for the marker if possible and displays the image to the user. testCom() With an initial list of commands for the arm, it proceeds to iterate through them while checking each image for the marker/box. After locating it, it proceeds to center the camera horizontally on the box and then vertically. After having centered the camera on the box, it will proceed to use forward kinematics to calculate the box's location. Execution Procedure The program proceeds to first authenticate with the R12 server. On success, it proceeds to call on the primary driver, testComm, to find the box. This process then instructs the arm for move, capture an image, downloads the image, and parses it for the target. All the while, reporting all pertinent data to the user for diagnostics purposes.
pyfar / SpharpySpherical array processing in python.
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ed741 / CainCain: Cellular Processor Array Convolution filter code Generator
