A. V. GERBESSIOTISCS610-101Fall 2019 Aug 20, 2019PrP: Programming ProjectPage 1 Handout(c) Copyright A. Gerbessiotis. All rights reserved.Posting this document on the Web, other than its original location at NJIT, is strictly prohibited unless thereis an explicit written authorization by its author (name on top line of this first page of this document).1 Programming Project (PrP) LogisticsWarning: Besides the algorithmic-related programming, for some or all of the options you need to providecommand-line processing or file-based input/output. If you are not familiar with those topics andrequirements we urge you to become so as soon as possible. Thus start familiarizing with them early in thesemester. The more you procrastinate the more problems you will potentially face when you integrate thesecomponents with the algorithmic-related material.STEP-1. Read carefully Handout 2 and follow all the requirements specified there; moreover, observenaming conventions, comments and testing as specified in sections 2-4 of Handout 2.STEP-2. When the archive per Handout 2 guidelines is ready, upload it to moodle and do not forget to pressthe submit button; the submission timestamp depends on this (submission button).BEFORE NOON-time of Wed 4 December, 2019If you want to get 0 pts ignore Handout 2.You may do one or two options (or none at all). You may utilize the same language or not. We providedescriptions that are to the extent possible language independent: thus a reference in Java is a pointerin C or C++ for example.OPTION 1 (Hash Table related). Do the programming related to the building of a Hash Table that canmaintain arbitrarily long strings in C, C++, or Java; it is similar to that used by Google around 1997-1998.OPTION 2. Do the programming part related to Kleinberg’s HITS, and Google’s PageRank algorithms inJava, C, or C++.Either implementation should be optimized enough for a test execution to take no more than few seconds,maximum 15 seconds forgivingly. (We are a bit more explicit about this in the 8th line or so of OPTION 1.)A. V. GERBESSIOTISCS610-101Fall 2019 Aug 20, 2019PrP: Programming ProjectPage 2 Handout2 OPTION 1: Hashing ( 134 points)We are asking you to implement a Lexicon structure realized by Google in 1997-1998 to store words (akaarbitrarily long strings of characters) in main memory. Lexicon L uses a Hash Table T structure along withan Array A of NULL separated strings . In our case words are going to be English character words only(upper-case or lower case). Table T will be organized as a hash-table using collision-resolution by openaddressingas specified in class. You are going to use quadratic probing for h(k,i) and keep the choice ofthe quadratic function simple: i2so that h(k,i) = (h0(k) +i2) mod m. The keys that you will hash are goingto be English words. Thus function h0(k) is also going to be kept simple: the sum of the ASCII/Unicodevalues of the characters mod m, where m is the slot-size of the hash table. Thus ’alex’ (the string is betweenthe quotation marks) is mapped to 97+108+101+120 mod m whatever m is. In the example below, form = 11, h(alex,0) = 8. Table T however won’t store key values k in it. This is because the keys are stringsof arbitrary length. Instead, T will store pointers/references in the form of an index to another array A. Thesecond table, array A will be a character array and will store the words maintained in T separated by NULvalues \0. This is not 2 characters a backslash followed by a zero. It is 1B (ASCII), 2B (UNICODE) whoseall bits are set to 0, the NUL value. If you don’t know what B is, it is a byte; never use b for a bit, writeinstead bit or bits i.e. read Handout 3.An insertion operation affects T and A. A word w is hashed, an available slot in T is computed andlet that slot be t. In T[t] we store an index to table A. This index is the first location that stores the firstcharacter of w. The ending location is the \0 following w in A. New words that do not exist (never inserted,or inserted but subsequently deleted) are appended in A. Thus originally you need to be wise enough inchoosing the appropriate size of A. If at some point you run-out of space, you need to increase the size ofA accordingly, even if T remains the same. Doubling it, is an option. Likewise the size of T might alsohave to be increased. This causes more problems that you also need to attend to. A deletion will modify Tas needed but will not erase w from A. Let it be there. So A might get dirty (i.e. it contains garbage) afterseveral deletions. If several operations later you end up inserting w after deleting it previously, you do it theinsertion way and you reinsert w, even if a dirty copy of it might still be around. You DO NOT DO a linearsearch to find out if it exists arleady in A; it is inefficient. There is not much to say for a search.You need to support few more operations: Print , Create, Empty/Full/Batch with the last of thosechecking for an empty or full table/array and a mechanism to perform multiple operations in batch form.Print prints nicely T and its contents i.e. index values to A. In addition it prints nicely (linear-wise in oneline) the contents of A. (For a \0 you will do the SEMI obvious: print a backslash but not its zero). Theintent of Print is to assist the grader. Print however does not print the words of A for deleted words. Itprints stars for every character of a deleted word instead. (An alternative is that during deletion each suchcharacter has already been turned into a star.) Function Create creates T with m slots, and A with 15m charsand initializes A to spaces. The following is a suggested minimal interface (we try to be language agnostic).We call the class that supports and realizes A and T a lexicon: L is one instance of a lexicon.HashCreate (lexicon L, int m); // Create T, A. T will have m slots; A should be 15mHashEmpty (lexicon L); // Check if L is emptyHashFull (lexicon L); // Check if L can maintain more wordsHashPrint (lexicon L); // Print of LHashInsert (lexicon L, word w); //Insert w into L (and T and A)HashDelete (lexicon L, word w); //Delete w from L (but not necessarily from A)HashSearch (lexicon L, word w); //Search for string in L (and this means T)HashBatch (lexicon L, file filename)The list of functions above is a suggestion of what needs to be implemented: it is too generic and generalto be programming language agnostic.A. V. GERBESSIOTISCS610-101Fall 2019 Aug 20, 2019PrP: Programming ProjectPage 3 HandoutTesting utilizes HashBatch. Its argument filename, an arbitrary filename contains several operationsthat are executed in batch mode. Operation 10 is Insert, Operation 11 is Deletion, and Operation 12 isSearch. Operation 13 is Print, Operation 14 is Create. Operation 15 is Comment; the rest of the line isignored. (Create accepts as its second parameter and that of HashCreate, an integer value next to its code 14;this becomes m.) The HashBatch accepts an arbitrary filename such as command.txt or file.txt that containsa sequence of operations.% java mplexicon filearbitrary.txt% ./mplexicon file.txt15 ready-to-print CAUTION: 15 is a comment string (chars,numbers,-)13 operation 15 is skipped/ignoredThe six-line batch file above will print the following. The T entries for 0, 5, 9 are the indexes (first position)for alex, tom, jerry respectively. Note that the ASCII values for ’alex’ mod 11 give an 8, but for ’tom’and ’jerry’ give 6, i.e. a collision occurs. A minimal output for Print is available below.T A: alex\tom\jerry\0: CAUTION: \ means \01: \t is not a tab character !!!If the following lines were added CS610-101代写、Programming代做、C/C+to the filethey will generate in addition on screenalex found at slot 8tom found at slot 6jerry found at slot 7mary not foundtom deleted from slot 6T A: alex\***\jerry\Deliverables. An archive per Handout 2 guidelines.A. V. GERBESSIOTISCS610-101Fall 2019 Aug 20, 2019PrP: Programming ProjectPage 4 Handout3 OPTION 2: HITS and PageRank implementations ( 134 points)Implement Kleinberg’s HITS Algorithm, and Google’s PageRank algorithm in Java, C, or C++ as explained.(A) Implement the HITS algorithm as explained in class/Subject notes adhering to the guidelines ofHandout 2. Pay attention to the sequence of update operations and the scaling. For an example of theSubject notes, you have output for various initialization vectors. You need to implement class or functionhits (e.g. hitsWXYZ. For an explanation of the arguments see the discussion on PageRank to follow.% java hits iterations initialvalue filename% ./hits iterations initialvalue filename(B) Implement Google’s PageRank algorithm as explained in class/Subject notes adhering also to theguidelines of Handout 2 and this description. The input for this (and the previous) problem would be a filecontaining a graph represented through an adjacency list representation. The command-line interface is asfollows. First we have the class/binary file (eg pgrk). Next we have an argument that denotes the numberof iterations if it is a positive integer or an errorrate for a negative or zero integer value. The nextargument initialvalue indicates the common initial values of the vector(s) used. The final argument is astring indicating the filename that stores the input graph.% ./pgrk iterations initialvalue filename // in fact pgrkWXYZ% java pgrk iterations initialvalue filename // in fact pgrkWXYZThe two algorithms are iterative. In particular, at iteration t all pagerank values are computed using resultsfrom iteration t − 1. The initialvalue helps us to set-up the initial values of iteration 0 as needed.Moreover, in PageRank, parameter d would be set to 0.85. The PageRank PR(A) of vertex A depends onthe PageRanks of vertices T1,...,Tm incident to A, i.e. pointing to A; check Subject 7 section 3.6 for moredetails. The pageranks at iteration t use the pageranks of iteration t −1 (synchronous update). Thus PR(A)on the left in the PageRank equation is for iteration t, but all PR (Ti) values are from the previous iterationt − 1. Be careful and synchronize! In order to run the ’algorithm’ we either run it for a fixed numberof iterations and iterations determines that, or for a fixed errorrate (an alias for iterations); aniterations equal to 0 corresponds to a default errorrate of 10−5. A -1, -2, etc , -6 for iterationsbecomes an errorrate of 10−1,10−2,...,10−6respectively. At iteration t when all authority/hub/PageRankvalues have been computed (and auth/hub values scaled) we compare for every vertex the current and theprevious iteration values. If the difference is less than errorrate for EVERY VERTEX, then and only thencan we stop at iteration t.Argument initialvalue sets the initial vector values. If it is 0 they are initialized to 0, if it is 1 theyare initialized to 1. If it is -1 they are initialized to 1/N, where N is the number of web-pages (vertices ofthe graph). If it is -2 they are initialized to 1/√N. filename first.)Argument filename describes the input (directed) graph and it has the following form. The firstline contains two numbers: the number of vertices followed by the number of edges which is also thenumber of remaining lines. PAY ATTENTION THAT NUMBER of VERTICES comes first. The samplegraph is treacherous: n and m are the same! All vertices are labeled 0,...,N − 1. Expect N to be lessthan 1,000,000. In each line an edge (i, j) is represented by i j. Thus our graph has (directed) edges(0,2),(0,3),(1,0),(2,1). Vector values are printed to 7 decimal digits. If the graph has N GREATER than10, then the values for iterations, initialvalue are automatically set to 0 and -1 respectively. Insuch a case the hub/authority/pageranks at the stopping iteration (i.e t) are ONLY shown, one per line. Thegraph below will be referred to as samplegraph.txtA. V. GERBESSIOTISCS610-101Fall 2019 Aug 20, 2019PrP: Programming ProjectPage 5 HandoutThe following invocations relate to samplegraph.txt, with a fixed number of iterations and the fixederror rate that determines how many iterations will run. Your code should compute for this graph the samerank values (intermediate and final). A sample of the output for the case of N > 10 is shown (output truncatedto first 4 lines of it).% ./pgrk 15 -1 samplegraph.txtBase : 0 :P[ 0]=0.2500000 P[ 1]=0.2500000 P[ 2]=0.2500000 P[ 3]=0.2500000Iter : 1 :P[ 0]=0.2500000 P[ 1]=0.2500000 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 2 :P[ 0]=0.2500000 P[ 1]=0.1596875 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 3 :P[ 0]=0.1732344 P[ 1]=0.1596875 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 4 :P[ 0]=0.1732344 P[ 1]=0.1596875 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 5 :P[ 0]=0.1732344 P[ 1]=0.1319559 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 6 :P[ 0]=0.1496625 P[ 1]=0.1319559 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 7 :P[ 0]=0.1496625 P[ 1]=0.1319559 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 8 :P[ 0]=0.1496625 P[ 1]=0.1234406 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 9 :P[ 0]=0.1424245 P[ 1]=0.1234406 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 10 :P[ 0]=0.1424245 P[ 1]=0.1234406 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 11 :P[ 0]=0.1424245 P[ 1]=0.1208259 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 12 :P[ 0]=0.1402020 P[ 1]=0.1208259 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 13 :P[ 0]=0.1402020 P[ 1]=0.1208259 P[ 2]=0.0970858 P[ 3]=0.0970858Iter : 14 :P[ 0]=0.1402020 P[ 1]=0.1200230 P[ 2]=0.0970858 P[ 3]=0.0970858Iter : 15 :P[ 0]=0.1395195 P[ 1]=0.1200230 P[ 2]=0.0970858 P[ 3]=0.0970858% ./pgrk -3 -1 samplegraph.txtBase : 0 :P[ 0]=0.2500000 P[ 1]=0.2500000 P[ 2]=0.2500000 P[ 3]=0.2500000Iter : 1 :P[ 0]=0.2500000 P[ 1]=0.2500000 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 2 :P[ 0]=0.2500000 P[ 1]=0.1596875 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 3 :P[ 0]=0.1732344 P[ 1]=0.1596875 P[ 2]=0.1437500 P[ 3]=0.1437500Iter : 4 :P[ 0]=0.1732344 P[ 1]=0.1596875 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 5 :P[ 0]=0.1732344 P[ 1]=0.1319559 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 6 :P[ 0]=0.1496625 P[ 1]=0.1319559 P[ 2]=0.1111246 P[ 3]=0.1111246Iter : 7 :P[ 0]=0.1496625 P[ 1]=0.1319559 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 8 :P[ 0]=0.1496625 P[ 1]=0.1234406 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 9 :P[ 0]=0.1424245 P[ 1]=0.1234406 P[ 2]=0.1011066 P[ 3]=0.1011066Iter : 10 :P[ 0]=0.1424245 P[ 1]=0.1234406 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 11 :P[ 0]=0.1424245 P[ 1]=0.1208259 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 12 :P[ 0]=0.1402020 P[ 1]=0.1208259 P[ 2]=0.0980304 P[ 3]=0.0980304Iter : 13 :P[ 0]=0.1402020 P[ 1]=0.1208259 P[ 2]=0.0970858 P[ 3]=0.0970858% ./pgrk 0 -1 verylargegraph.txtIter : 4P[ 0]=0.0136364P[ 1]=0.0194318P[ 2]=0.0310227... other vertices omittedFor the HITS algorithm, you need to print two values not one. Follow the convention of the Subject notesBase : 0 :A/H[ 0]=0.3333333/0.3333333 A/H[ 1]=0.3333333/0.3333333 A/H[ 2]=0.3333333/0.3333333Iter : 1 :A/H[ 0]=0.0000000/0.8320503 A/H[ 1]=0.4472136/0.5547002 A/H[ 2]=0.8944272/0.0000000or for large graphsIter : 37A/H[ 0]=0.0000000/0.0000002A/H[ 1]=0.0000001/0.0000238A/H[ 2]=0.0000002/1.0000000A/H[ 3]=0.0000159/0.0000000...Deliverables. Include source code of all implemented functions or classes in an archive per Handout 2guidelines. Document bugs; no bug report no partial points.转自：http://www.daixie0.com/contents/9/4331.html