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  • Finding the largest subtree in a BST

    - by rakeshr
    Given a binary tree, I want to find out the largest subtree which is a BST in it. Naive approach: I have a naive approach in mind where I visit every node of the tree and pass this node to a isBST function. I will also keep track of the number of nodes in a sub-tree if it is a BST. Is there a better approach than this ?

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  • What are the most difficult aspects of project management in Software Engineering?

    - by Jamie Chapman
    I have been asked to provide a brief summary of the what the most difficult aspects of being a project manager of a software engineering project. However, I have no experience of this as I'm still at University and have no "hands on" experience of project management. I was hoping that someone on SO would be able to provide some insight based on their experience. What are the most difficult aspects of project management in Software Engineering?

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  • Spatial domain to frequency domain

    - by John Elway
    I know about Fourier Transforms, but I don't know how to apply it here, and I think that is over the top. I gave my ideas of the responses, but I really don't know what I'm looking for... Supposed that you form a low-pass spatial filter h(x,y) that averages all the eight immediate neighbors of a pixel (x,y) but excludes itself. a. Find the equivalent frequency domain filter H(u,v): My answer is to (a): 1/8*H(u-1, v-1) + 1/8*H(u-1, v) + 1/8*H(u-1, v+1) + 1/8*H(u, v-1) + 0 + 1/8*H(u, v+1) + 1/8*H(u+1, v-1) + 1/8*H(u+1, v) + 1/8*H(u-1, v-1) is this the frequency domain? b. Show that your result is again a low-pass filter. does this have to do with the coefficients being positive?

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  • Why isn't this file reading/writing program working?

    - by user320950
    This program is supposed to read files and write them. I took the file open checks out because they kept causing errors. The problem is that the files open like they are supposed to and the names are correct but nothing is on any of the text screens. Do you know what is wrong? #include<iostream> #include<fstream> #include<cstdlib> #include<iomanip> using namespace std; int main() { ifstream in_stream; // reads itemlist.txt ofstream out_stream1; // writes in items.txt ifstream in_stream2; // reads pricelist.txt ofstream out_stream3;// writes in plist.txt ifstream in_stream4;// read recipt.txt ofstream out_stream5;// write display.txt float price=' ',curr_total=0.0; int wrong=0; int itemnum=' '; char next; in_stream.open("ITEMLIST.txt", ios::in); // list of avaliable items out_stream1.open("listWititems.txt", ios::out); // list of avaliable items in_stream2.open("PRICELIST.txt", ios::in); out_stream3.open("listWitdollars.txt", ios::out); in_stream4.open("display.txt", ios::in); out_stream5.open("showitems.txt", ios::out); in_stream.close(); // closing files. out_stream1.close(); in_stream2.close(); out_stream3.close(); in_stream4.close(); out_stream5.close(); system("pause"); in_stream.setf(ios::fixed); while(in_stream.eof()) { in_stream >> itemnum; cin.clear(); cin >> next; } out_stream1.setf(ios::fixed); while (out_stream1.eof()) { out_stream1 << itemnum; cin.clear(); cin >> next; } in_stream2.setf(ios::fixed); in_stream2.setf(ios::showpoint); in_stream2.precision(2); while((price== (price*1.00)) && (itemnum == (itemnum*1))) { while (in_stream2 >> itemnum >> price) // gets itemnum and price { while (in_stream2.eof()) // reads file to end of file { in_stream2 >> itemnum; in_stream2 >> price; price++; curr_total= price++; in_stream2 >> curr_total; cin.clear(); // allows more reading cin >> next; } } } out_stream3.setf(ios::fixed); out_stream3.setf(ios::showpoint); out_stream3.precision(2); while((price== (price*1.00)) && (itemnum == (itemnum*1))) { while (out_stream3 << itemnum << price) { while (out_stream3.eof()) // reads file to end of file { out_stream3 << itemnum; out_stream3 << price; price++; curr_total= price++; out_stream3 << curr_total; cin.clear(); // allows more reading cin >> next; } return itemnum, price; } } in_stream4.setf(ios::fixed); in_stream4.setf(ios::showpoint); in_stream4.precision(2); while ( in_stream4.eof()) { in_stream4 >> itemnum >> price >> curr_total; cin.clear(); cin >> next; } out_stream5.setf(ios::fixed); out_stream5.setf(ios::showpoint); out_stream5.precision(2); out_stream5 <<setw(5)<< " itemnum " <<setw(5)<<" price "<<setw(5)<<" curr_total " <<endl; // sends items and prices to receipt.txt out_stream5 << setw(5) << itemnum << setw(5) <<price << setw(5)<< curr_total; // sends items and prices to receipt.txt out_stream5 << " You have a total of " << wrong++ << " errors " << endl; }

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  • Finding the average of two number using classes and methods

    - by Have alook
    I want to use methods inside class. Q: find the average of two number using classes and methods. import java.util.*; class aaa { int a,b,sum,avrg; void average() { System.out.println("The average is ="+avrg); avrg=(sum/2); } } class ave { public static void main(String args[]){ aaa n=new aaa(); Scanner m=new Scanner(System.in); System.out.println("write two number"); n.a=m.nextInt(); n.b=m.nextInt(); n.average(); } }

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  • Please help me in creating a update query

    - by Rajesh Rolen- DotNet Developer
    I have got a table which contains 5 column and query requirements: update row no 8 (or id=8) set its column 2, column 3's value from id 9th column 2, column 3 value. means all value of column 2, 3 should be shifted to column 2, 3 of upper row (start from row no 8) and value of last row's 2, 3 will be null For example, with just 3 rows, the first row is untouched, the second to N-1th rows are shifted once, and the Nth row has nulls. id math science sst hindi english 1 11 12 13 14 15 2 21 22 23 24 25 3 31 32 33 34 35 The result of query of id=2 should be: id math science sst hindi english 1 11 12 13 14 15 2 31 32 23 24 25 //value of 3rd row (col 2,3) shifted to row 2 3 null null 33 34 35 This process should run for all rows whose id 2 Please help me to create this update query I am using MS sqlserver 2005

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  • Quickest way to write to file in java

    - by user1097772
    I'm writing an application which compares directory structure. First I wrote an application which writes gets info about files - one line about each file or directory. My soulution is: calling method toFile Static PrintWriter pw = new PrintWriter(new BufferedWriter( new FileWriter("DirStructure.dlis")), true); String line; // info about file or directory public void toFile(String line) { pw.println(line); } and of course pw.close(), at the end. My question is, can I do it quicker? What is the quickest way? Edit: quickest way = quickest writing in the file

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  • how to return a list using SwingWorker

    - by Ender
    I have an assignment where i have to create an Image Gallery which uses a SwingWorker to load the images froma a file, once the image is load you can flip threw the image and have a slideshow play. I am having trouble getting the list of loaded images using SwingWorker. This is what happens in the background it just publishes the results to a TextArea // In a thread @Override public List<Image> doInBackground() { List<Image> images = new ArrayList<Image>(); for (File filename : filenames) { try { //File file = new File(filename); System.out.println("Attempting to add: " + filename.getAbsolutePath()); images.add(ImageIO.read(filename)); publish("Loaded " + filename); System.out.println("Added file" + filename.getAbsolutePath()); } catch (IOException ioe) { publish("Error loading " + filename); } } return images; } } when it is done I just insert the images in a List<Image> and that is all it does. // In the EDT @Override protected void done() { try { for (Image image : get()) { list.add(image); } } catch (Exception e) { } } Also I created an method that returns the list called getImages() what I need to get is the list from getImages() but doesn't seam to work when I call execute() for example MySwingWorkerClass swingworker = new MySwingWorkerClass(log,list,filenames); swingworker.execute(); imageList = swingworker.getImage() Once it reaches the imageList it doesn't return anything the only way I was able to get the list was when i used the run() instead of the execute() is there another way to get the list or is the run() method the only way?. or perhaps i am not understanding the Swing Worker Class.

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  • Find the shortest path in a graph which visits certain nodes.

    - by dmd
    I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'. I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order). ( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)

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  • C++ Memory Allocation & Linked List Implementation

    - by pws5068
    I'm writing software to simulate the "first-fit" memory allocation schema. Basically, I allocate a large X megabyte chunk of memory and subdivide it into blocks when chunks are requested according to the schema. I'm using a linked list called "node" as a header for each block of memory (so that we can find the next block without tediously looping through every address value. head_ptr = (char*) malloc(total_size + sizeof(node)); if(head_ptr == NULL) return -1; // Malloc Error .. :-( node* head_node = new node; // Build block header head_node->next = NULL; head_node->previous = NULL; // Header points to next block (which doesn't exist yet) memset(head_ptr,head_node, sizeof(node)); ` But this last line returns: error: invalid conversion from 'node*' to 'int' I understand why this is invalid.. but how can I place my node into the pointer location of my newly allocated memory?

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  • Recursive function for a binary search in C++

    - by boomsnack
    Create a recursive function for the binary search. This function accepts a sorted array and a give item being search for and returns the index of the item if this give item in the array or returns -1 if this give item is not in the array. Moreover, write a test program to test your function. Sorry for the bad english but my teacher can not write it or speak it very well. This is for a final project and determines whether I graduate or not I went to the tutor and he did not know how to do it either. Any help is greatly appreicated.

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  • Infinite loop when adding a row to a list in a class in python3

    - by Margaret
    I have a script which contains two classes. (I'm obviously deleting a lot of stuff that I don't believe is relevant to the error I'm dealing with.) The eventual task is to create a decision tree, as I mentioned in this question. Unfortunately, I'm getting an infinite loop, and I'm having difficulty identifying why. I've identified the line of code that's going haywire, but I would have thought the iterator and the list I'm adding to would be different objects. Is there some side effect of list's .append functionality that I'm not aware of? Or am I making some other blindingly obvious mistake? class Dataset: individuals = [] #Becomes a list of dictionaries, in which each dictionary is a row from the CSV with the headers as keys def field_set(self): #Returns a list of the fields in individuals[] that can be used to split the data (i.e. have more than one value amongst the individuals def classified(self, predicted_value): #Returns True if all the individuals have the same value for predicted_value def fields_exhausted(self, predicted_value): #Returns True if all the individuals are identical except for predicted_value def lowest_entropy_value(self, predicted_value): #Returns the field that will reduce <a href="http://en.wikipedia.org/wiki/Entropy_%28information_theory%29">entropy</a> the most def __init__(self, individuals=[]): and class Node: ds = Dataset() #The data that is associated with this Node links = [] #List of Nodes, the offspring Nodes of this node level = 0 #Tree depth of this Node split_value = '' #Field used to split out this Node from the parent node node_value = '' #Value used to split out this Node from the parent Node def split_dataset(self, split_value): fields = [] #List of options for split_value amongst the individuals datasets = {} #Dictionary of Datasets, each one with a value from fields[] as its key for field in self.ds.field_set()[split_value]: #Populates the keys of fields[] fields.append(field) datasets[field] = Dataset() for i in self.ds.individuals: #Adds individuals to the datasets.dataset that matches their result for split_value datasets[i[split_value]].individuals.append(i) #<---Causes an infinite loop on the second hit for field in fields: #Creates subnodes from each of the datasets.Dataset options self.add_subnode(datasets[field],split_value,field) def add_subnode(self, dataset, split_value='', node_value=''): def __init__(self, level, dataset=Dataset()): My initialisation code is currently: if __name__ == '__main__': filename = (sys.argv[1]) #Takes in a CSV file predicted_value = "# class" #Identifies the field from the CSV file that should be predicted base_dataset = parse_csv(filename) #Turns the CSV file into a list of lists parsed_dataset = individual_list(base_dataset) #Turns the list of lists into a list of dictionaries root = Node(0, Dataset(parsed_dataset)) #Creates a root node, passing it the full dataset root.split_dataset(root.ds.lowest_entropy_value(predicted_value)) #Performs the first split, creating multiple subnodes n = root.links[0] n.split_dataset(n.ds.lowest_entropy_value(predicted_value)) #Attempts to split the first subnode.

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  • Optimal sorting algorithm with modified cost... [closed]

    - by David
    The numbers are in a list that is not sorted and supports only one type of operation. The operation is defined as follows: Given a position i and a position j the operation moves the number at position i to position j without altering the relative order of the other numbers. If i j, the positions of the numbers between positions j and i - 1 increment by 1, otherwise if i < j the positions of the numbers between positions i+1 and j decreases by 1. This operation requires i steps to find a number to move and j steps to locate the position to which you want to move it. Then the number of steps required to move a number of position i to position j is i+j. Design an algorithm that given a list of numbers, determine the optimal(in terms of cost) sequence of moves to rearrange the sequence.

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  • asking the container to notify your application whenever a session is about to timeout in Java

    - by user136101
    Which method(s) can be used to ask the container to notify your application whenever a session is about to timeout?(choose all that apply) A. HttpSessionListener.sessionDestroyed -- correct B. HttpSessionBindingListener.valueBound C. HttpSessionBindingListener.valueUnbound -- correct this is kind of round-about but if you have an attribute class this is a way to be informed of a timeout D. HttpSessionBindingEvent.sessionDestroyed -- no such method E. HttpSessionAttributeListener.attributeRemoved -- removing an attribute isn’t tightly associated with a session timeout F. HttpSessionActivationListener.sessionWillPassivate -- session passivation is different than timeout I agree with option A. 1) But C is doubtful How can value unbound be tightly coupled with session timeout.It is just the callback method when an attribute gets removed. 2) and if C is correct, E should also be correct. HttpSessionAttributeListener is just a class that wants to know when any type of attribute has been added, removed, or replaced in a session. It is implemented by any class. HttpSessionBindingListener exists so that the attribute itself can find out when it has been added to or removed from a session and the attribute class must implement this interface to achieve it. Any ideas…

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  • File processing-Haskell

    - by Martinas Maria
    How can I implement in haskell the following: I receive an input file from the command line. This input file contains words separated with tabs,new lines and spaces.I have two replace this elements(tabs,new lines and spaces) with comma(,) .Observation:more newlines,tabs,spaces will be replaced with a single comma.The result has to be write in a new file(output.txt). Please help me with this.My haskell skills are very scarse. This is what I have so far: processFile::String->String processFile [] =[] processFile input =input process :: String -> IO String process fileName = do text <- readFile fileName return (processFile text) main :: IO () main = do n <- process "input.txt" print n In processFile function I should process the text from the input file. I'm stuck..Please help.

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  • Ad distribution problem: an optimal solution?

    - by Mokuchan
    I'm asked to find a 2 approximate solution to this problem: You’re consulting for an e-commerce site that receives a large number of visitors each day. For each visitor i, where i € {1, 2 ..... n}, the site has assigned a value v[i], representing the expected revenue that can be obtained from this customer. Each visitor i is shown one of m possible ads A1, A2 ..... An as they enter the site. The site wants a selection of one ad for each customer so that each ad is seen, overall, by a set of customers of reasonably large total weight. Thus, given a selection of one ad for each customer, we will define the spread of this selection to be the minimum, over j = 1, 2 ..... m, of the total weight of all customers who were shown ad Aj. Example Suppose there are six customers with values 3, 4, 12, 2, 4, 6, and there are m = 3 ads. Then, in this instance, one could achieve a spread of 9 by showing ad A1 to customers 1, 2, 4, ad A2 to customer 3, and ad A3 to customers 5 and 6. The ultimate goal is to find a selection of an ad for each customer that maximizes the spread. Unfortunately, this optimization problem is NP-hard (you don’t have to prove this). So instead give a polynomial-time algorithm that approximates the maximum spread within a factor of 2. The solution I found is the following: Order visitors values in descending order Add the next visitor value (i.e. assign the visitor) to the Ad with the current lowest total value Repeat This solution actually seems to always find the optimal solution, or I simply can't find a counterexample. Can you find it? Is this a non-polinomial solution and I just can't see it?

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  • Problem on a Floyd-Warshall implementation using c++

    - by Henrique
    I've got a assignment for my college, already implemented Dijkstra and Bellman-Ford sucessfully, but i'm on trouble on this one. Everything looks fine, but it's not giving me the correct answer. Here's the code: void FloydWarshall() { //Also assume that n is the number of vertices and edgeCost(i,i) = 0 int path[500][500]; /* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path from i to j using intermediate vertices (1..k-1). Each path[i][j] is initialized to edgeCost(i,j) or infinity if there is no edge between i and j. */ for(int i = 0 ; i <= nvertices ; i++) for(int j = 0 ; j <= nvertices ; j++) path[i][j] = INFINITY; for(int j = 0 ; j < narestas ; j++) //narestas = number of edges { path[arestas[j]->v1][arestas[j]->v2] = arestas[j]->peso; //peso = weight of the edge (aresta = edge) path[arestas[j]->v2][arestas[j]->v1] = arestas[j]->peso; } for(int i = 0 ; i <= nvertices ; i++) //path(i, i) = 0 path[i][i] = 0; //test print, it's working fine //printf("\n\n\nResultado FloydWarshall:\n"); //for(int i = 1 ; i <= nvertices ; i++) // printf("distancia ao vertice %d: %d\n", i, path[1][i]); //heres the problem, it messes up, and even a edge who costs 4, and the minimum is 4, it prints 2. //for k = 1 to n for(int k = 1 ; k <= nvertices ; k++) //for i = 1 to n for(int i = 1 ; i <= nvertices ; i++) //for j := 1 to n for(int j = 1 ; j <= nvertices ; j++) if(path[i][j] > path[i][k] + path[k][j]) path[i][j] = path[i][k] + path[k][j]; printf("\n\n\nResultado FloydWarshall:\n"); for(int i = 1 ; i <= nvertices ; i++) printf("distancia ao vertice %d: %d\n", i, path[1][i]); } im using this graph example i've made: 6 7 1 2 4 1 5 1 2 3 1 2 5 2 5 6 3 6 4 6 3 4 2 means we have 6 vertices (1 to 6), and 7 edges (1,2) with weight 4... etc.. If anyone need more info, i'm up to giving it, just tired of looking at this code and not finding an error.

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  • How can I prevent segmentation faults in my program?

    - by worlds-apart89
    I have a C assignment. It is a lot longer than the code shown below, and we are given the function prototypes and instructions only. I have done my best at writing code, but I am stuck with segmentation faults. When I compile and run the program below on Linux, at "735 NaN" it will terminate, indicating a segfault occurred. Why? What am I doing wrong? Basically, the program does not let me access table-list_array[735]-value and table-list_array[735]-key. This is of course the first segfault. There might be more following index 735. #include <stdio.h> #include <stdlib.h> typedef struct list_node list_node_t; struct list_node { char *key; int value; list_node_t *next; }; typedef struct count_table count_table_t; struct count_table { int size; list_node_t **list_array; }; count_table_t* table_allocate(int size) { count_table_t *ptr = malloc(sizeof(count_table_t)); ptr->size = size; list_node_t *nodes[size]; int k; for(k=0; k<size; k++){ nodes[k] = NULL; } ptr->list_array = nodes; return ptr; } void table_addvalue(count_table_t *table) { int i; for(i=0; i<table->size; i++) { table->list_array[i] = malloc(sizeof(list_node_t)); table->list_array[i]->value = i; table->list_array[i]->key = "NaN"; table->list_array[i]->next = NULL; } } int main() { count_table_t *table = table_allocate(1000); table_addvalue(table); int i; for(i=0; i<table->size; i++) printf("%d %s\n", table->list_array[i]->value, table->list_array[i]->key); return 0; }

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  • Sample/Example needed for a table/field setup

    - by acctman
    Can someone explain the statement below to me with a working sample/example. thanks in advance. You can not create duplicate fields, but simply add a single extra field, "coupleId", which would have a unique id for each couple; and two rows (one for each person) per couple; then JOIN the table against itself with a constraint like a.coupleId = b.coupleId AND a.id < b.id so that you can condense the data into a single result row for a given couple.

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  • how to fix my error saying expected expression before 'else'

    - by user292489
    this program intended to read a .txt, a set of numbers, file and wwrite to another two .txt files called even amd odd as follows: #include <stdio.h> #include <stdlib.h> int main(int argc, char *argv[]) { int i=0,even,odd; int number[i]; // check to make sure that all the file names are entered if (argc != 3) { printf("Usage: executable in_file output_file\n"); exit(0); } FILE *dog = fopen(argv[1], "r"); FILE *feven= fopen(argv[2], "w"); FILE *fodd= fopen (argv[3], "w"); // check whether the file has been opened successfully if (dog == NULL) { printf("File %s cannot open!\n", argv[1]); exit(0); } //odd = fopen(argv[2], "w"); { if (i%2!=1) i++;} fprintf(feven, "%d", even); fscanf(dog, "%d", &number[i]); else { i%2==1; i++;} fprintf(fodd, "%d", odd); fscanf(dog, "%d", &number[i]); fclose(feven); fclose(fodd);

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  • Regex Question ...

    - by kate
    Hi, Could someone help me with the following RegEx query: based on the following rules: 1) 1 letter followed by 4 letters or numbers, then 2) 5 letters or numbers, then 3) 3 letters or numbers followed by a number and one of the following signs: ! & @ ? You will have to allow customers to input the fidelity card code as a 15-character string, or as 3 groups of 5 chars, separated by one space.

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  • Finding an available network port on the machine

    - by Tomer Vromen
    I'm trying to implement a simple FTP server (a variation of the EFTP protocol) in linux. When a client connects and sends the PASV command, the server should respond with a port number, so the client can connect to that port to transmit the file. How can the server choose a port number? Do I need to iterate through all the ports from 1024 to 65535 until I find a port that the process can bind to? I know that calling bind() with 0 as the port automatically chooses the port to bind to, but then how can I know which port was chosen? Many thanks.

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