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• NoClassDefFoundError when running Netbeans 6.8 application

  - by Malachi

  My application has recently started throwing NoClassDefFoundError errors when I am running my application from within NetBeans. It never used to do this. And when I run this outside of NetBeans using the command line argument that it provides once built, these errors are not produced Is this a bug with Netbeans? I have reinstalled NetBeans as I recently had another that was resolved by reinstalling, however the problem still persists. http://stackoverflow.com/questions/2654892/symbol-error-in-java-application-using-netbeans-6-8-when-adding-a-shared-project Within Netbeans Outside NetBeans

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• To Plot a x-y graph in c# in Visual Studio 2008?

  - by Nanda

  Using two parameters to plot a line graph in C# using Visual Studio ( Using GUI).

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• javascript too much recursion?

  - by Ken

  Hi, I'm trying to make a script that automatically starts uploading after the data has been enter in the database(I need the autoId that the database makes to upload the file). When I run the javascript the scripts runs the php file but it fails calling the other php to upload the file. too much recursion setTimeout(testIfToegevoegd(),500); the script that gives the error send("/projects/backend/nieuwDeeltaak.php",'deeltaakNaam='+f.deeltaaknaam.value+'&beschrijving='+ f.beschrijving.value+'&startDatum='+f.startDatum.value+'&eindDatum='+f.eindDatum.value +'&deeltaakLeider='+f.leiderID.value+'&projectID='+f.projectID.value,id); function testIfToegevoegd(){ if(document.getElementById('resultaat').innerHTML == "<b>De deeltaak werd toegevoegd</b>"){ //stop met testen + upload file document.getElementById('nieuwDeeltaak').target = 'upload_target'; document.forms["nieuwDeeltaak"].submit() }else{ setTimeout(testIfToegevoegd(),500); } } testIfToegevoegd(); sorry for the dutch names we have to use them it is a school project. when I click the button that calls all this for a second time (after the error) it works fine.

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• How to draw the "trail" in a maze solving application

  - by snow-spur

  Hello i have designed a maze and i want to draw a path between the cells as the 'person' moves from one cell to the next. So each time i move the cell a line is drawn Also i am using the graphics module The graphics module is an object oriented library Im importing from graphics import* from maze import* my circle which is my cell center = Point(15, 15) c = Circle(center, 12) c.setFill('blue') c.setOutline('yellow') c.draw(win) p1 = Point(c.getCenter().getX(), c.getCenter().getY()) this is my loop if mazez.blockedCount(cloc)> 2: mazez.addDecoration(cloc, "grey") mazez[cloc].deadend = True c.move(-25, 0) p2 = Point(getX(), getY()) line = graphics.Line(p1, p2) cloc.col = cloc.col - 1 Now it says getX not defined every time i press a key is this because of p2???

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• Generalizing Fibonacci sequence with SICStus Prolog

  - by Christophe Herreman

  I'm trying to find a solution for a query on a generalized Fibonacci sequence (GFS). The query is: are there any GFS that have 885 as their 12th number? The initial 2 numbers may be restricted between 1 and 10. I already found the solution to find the Nth number in a sequence that starts at (1, 1) in which I explicitly define the initial numbers. Here is what I have for this: fib(1, 1). fib(2, 1). fib(N, X) :- N #> 1, Nmin1 #= N - 1, Nmin2 #= N - 2, fib(Nmin1, Xmin1), fib(Nmin2, Xmin2), X #= Xmin1 + Xmin2. For the query mentioned I thought the following would do the trick, in which I reuse the fib method without defining the initial numbers explicitly since this now needs to be done dynamically: fib2 :- X1 in 1..10, X2 in 1..10, fib(1, X1), fib(2, X2), fib(12, 885). ... but this does not seem to work. Is it not possible this way to define the initial numbers, or am I doing something terribly wrong? I'm not asking for the solution, but any advice that could help me solve this would be greatly appreciated.

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• Finding specific words in a file (Python language)

  - by Caroline Yi

  I have to write a program in python where the user is given a menu with four different "word games". There is a file called dictionary.txt and one of the games requires the user to input a) the number of letters in a word and b) a letter to exclude from the words being searched in the dictionary (dictionary.txt has the whole dictionary). Then the program prints the words that follow the user's requirements. My question is how on earth do I open the file and search for words with a certain length in that file. I only have a basic code which only asks the user for inputs. I'm am very new at this please help :( this is what I have up to the first option. The others are fine and I know how to break the loop but this specific one is really giving me trouble. I have tried everything and I just keep getting errors. Honestly, I only took this class because someone said it would be fun. It is, but recently I've really been falling behind and I have no idea what to do now. This is an intro level course so please be nice I've never done this before until now :( print print "Choose Which Game You Want to Play" print "a) Find words with only one vowel and excluding a specific letter." print "b) Find words containing all but one of a set of letters." print "c) Find words containing a specific character string." print "d) Find words containing state abbreviations." print "e) Find US state capitals that start with months." print "q) Quit." print choice = raw_input("Enter a choice: ") choice = choice.lower() print choice while choice != "q": if choice == "a": #wordlen = word length user is looking for.s wordlen = raw_input("Please enter the word length you are looking for: ") wordlen = int(wordlen) print wordlen #letterex = letter user wishes to exclude. letterex = raw_input("Please enter the letter you'd like to exclude: ") letterex = letterex.lower() print letterex

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• Implementing Dijkstra's Algorithm

  - by DeadMG

  I've been tasked (coursework @ university) to implement a form of path-finding. Now, in-spec, I could just implement a brute force, since there's a limit on the number of nodes to search (begin, two in the middle, end), but I want to re-use this code and came to implement Dijkstra's algorithm. I've seen the pseudo on Wikipedia and a friend wrote some for me as well, but it flat out doesn't make sense. The algorithm seems pretty simple and it's not a problem for me to understand it, but I just can't for the life of me visualize the code that would realize such a thing. Any suggestions/tips?

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• Available message types in JMS?

  - by Caylem

  This is based on a past exam question. The question is asking to describe the four types of message available using JMS. The problem is it says the four, not just four. So it assumes their is only four, no more no less. However according to this site their seems to be five; streams maps text objects bytes *Another book states that XML is another potential type in future versions of JMS. Is XML already available? Am I missing something or is the question just wrong? Thanks.

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• clearing cin input? cin.ignore not a good way?

  - by igor

  What's a better way to clear cin input? I thought cin.clear and cin.ignore was a good way...? code: void clearInput() { cin.clear(); cin.ignore(1000,'\n'); //cin.ignore( std::numeric_limits<streamsize>::max(), '\n' ); } My teacher gave me this reply... this is basically saying that your clearInput doesn't work FYI: ignore is NEVER a good idea as a way of getting rid of all that remains on a line and your failing this test is exactly the reason why now go clear it the correct way

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• Exception Class: When to Derive from it, In C# (.Net)?

  - by IbrarMumtaz

  I am continuing with my exam revision. I have come across the usage of the Base Exception class and I have seen it on exam papers also. My question is when do you derive from the Base Exception class? I am of the impression if you want a custom class to throw an exception with more meaningful information, then you can create a custom exception class that contains the exact data that is representative of how your custom class is used and what scenario it is designed to be used for? Why can't my custom exception class derive from 'ApplicationException' or 'SecurityException' or the base 'Exception' class? I am of the impression that I should derive from the base Exception class and not the previous two. My question second is, when would you derive from the other two??? Are there any clear-cut distinctions as to when you would derive from either one of these three? Assuming there are no others I have I have missed out? Thanks, Ibrar

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• using compareTo in Binary Search Tree program

  - by Scott Rogener

  I've been working on this program for a few days now and I've implemented a few of the primary methods in my BinarySearchTree class such as insert and delete. Insert seemed to be working fine, but once I try to delete I kept getting errors. So after playing around with the code I wanted to test my compareTo methods. I created two new nodes and tried to compare them and I get this error: Exception in thread "main" java.lang.ClassCastException: TreeNode cannot be cast to java.lang.Integer at java.lang.Integer.compareTo(Unknown Source) at TreeNode.compareTo(TreeNode.java:16) at BinarySearchTree.myComparision(BinarySearchTree.java:177) at main.main(main.java:14) Here is my class for creating the nodes: public class TreeNode<T> implements Comparable { protected TreeNode<T> left, right; protected Object element; public TreeNode(Object obj) { element=obj; left=null; right=null; } public int compareTo(Object node) { return ((Comparable) this.element).compareTo(node); } } Am I doing the compareTo method all wrong? I would like to create trees that can handle integers and strings (seperatly of course)

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• Finding height in Binary Search Tree

  - by mike

  Hey I was wondering if anybody could help me rework this method to find the height of a binary search tree. So far my code looks like this however the answer im getting is larger than the actual height by 1, but when I remove the +1 from my return statements its less than the actual height by 1? I'm still trying to wrap my head around recursion with these BST any help would be much appreciated. public int findHeight(){ if(this.isEmpty()){ return 0; } else{ TreeNode<T> node = root; return findHeight(node); } } private int findHeight(TreeNode<T> aNode){ int heightLeft = 0; int heightRight = 0; if(aNode.left!=null) heightLeft = findHeight(aNode.left); if(aNode.right!=null) heightRight = findHeight(aNode.right); if(heightLeft > heightRight){ return heightLeft+1; } else{ return heightRight+1; } }

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• java interfaces

  - by Codenotguru

  write an interface with one method,two classes that implement the interface, and a main method with an array holding an instance from both classes.Using this array variable call the method in a foreach loop.This is a interview question in java anybody?

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• Python: Beginning problems

  - by Blogger

  ok so basically i very new to programming and have no idea how to go about these problems help if you will ^^ Numerologists claim to be able to determine a person’s character traits based on the “numeric value” of a name. The value of a name is determined by summing up the values of the letters of the name, where ‘a’ is 1, ‘b’ is 2, ‘c’ is 3 etc., up to ‘z’ being 26. For example, the name “Zelle” would have the value 26 + 5 + 12 + 12 + 5 = 60 (which happens to be a very suspicious number, by the way). Write a program that calculates the numeric value of a single name provided as input. Word count. A common utility on Unix/Linux systems is a small program called “wc”. This program counts the number of lines, words (strings of characters separated by blanks, tabs, or new lines), and characters in a file. Write your own version of this program. The program should accept a file name as input and then print three numbers showing the count of lines, words, and characters in the file.

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• Enhanced for loop problem

  - by Antony Kelt

  Why is my enhanced loop not working? Vector<String> v = new Vector<String>(); v.add("one"); v.add("two"); v.add("three"); for(String str : v){ System.out.println(v); }

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• How should I generate the partitions / pairs for the Chinese Postman problem?

  - by Simucal

  I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own. The part that is giving me trouble is generating the partitions of pairings for the odd vertices. For example, if I had the following labeled odd verticies in a graph: 1 2 3 4 5 6 I need to find all the possible pairings / partitions I can make with these vertices. I've figured out I'll have i paritions given: n = num of odd verticies k = n / 2 i = ((2k)(2k-1)(2k-2)...(k+1))/2 So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions. The 15 partions would look like: 1 2 3 4 5 6 1 2 3 5 4 6 1 2 3 6 4 5 ... 1 6 ... Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition). These represent the edges the postman will have to walk twice. At first I thought I had worked out an appropriate algorithm for generating these partitions / pairs but it is flawed. I found it wasn't a simple permutation/combination problem. Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?

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• Help with a sort method

  - by Capsud

  Hi there, If i have an array of strings for example Static final String[] TEST = new String[] { "g","a","b","t","e" }; How would i go about sorting this in alphabetical order please?

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• Problem in finalizing the link list in C#

  - by Yasin

  My code was almost finished that a maddening bug came up! when i nullify the last node to finalize the link list , it actually dumps all the links and make the first node link Null ! when i trace it , its working totally fine creating the list in the loop but after the loop is done that happens and it destroys the rest of the link by doing so, and i don't understand why something so obvious is becoming problematic! (last line) struct poly { public int coef; public int pow; public poly* link;} ; poly* start ; poly* start2; poly* p; poly* second; poly* result; poly* ptr; poly* start3; poly* q; poly* q2; private void button1_Click(object sender, EventArgs e) { string holder = ""; IntPtr newP = Marshal.AllocHGlobal(sizeof(poly)); q = (poly*)newP.ToPointer(); start = q; int i = 0; while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } q->coef = int.Parse(holder); i++; holder = ""; while (this.textBox1.Text[i] != ';') { holder += this.textBox1.Text[i]; i++; } q->pow = int.Parse(holder); holder = ""; p = start; //creation of the first node finished! i++; for (; i < this.textBox1.Text.Length; i++) { newP = Marshal.AllocHGlobal(sizeof(poly)); poly* test = (poly*)newP.ToPointer(); while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } test->coef = int.Parse(holder); holder = ""; i++; while (this.textBox1.Text[i] != ';' && i < this.textBox1.Text.Length - 1) { holder += this.textBox1.Text[i]; if (i < this.textBox1.Text.Length - 1) i++; } test->pow = int.Parse(holder); holder = ""; p->link = test; //the addresses are correct and the list is complete } p->link = null; //only the first node exists now with a null link! }

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• How do I implement a fibonacci sequence in java using try/catch logic?

  - by Lars Flyger

  I know how to do it using simple recursion, but in order to complete this particular assignment I need to be able to accumulate on the stack and throw an exception that holds the answer in it. So far I have: public static int fibo(int index) { int sum = 0; try { fibo_aux(index, 1, 1); } catch (IntegerException me) { sum = me.getIntValue(); } return sum; } fibo_aux is supposed to throw an IntegerException (which holds the value of the answer that is retireved via getIntValue) and accumulates the answer on the stack, but so far I can't figure it out. Can anyone help?

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• about visual c++ programming language

  - by amit kumar dey

  give a good program to count the number of ones in a "n"which is 64 bit long.find a solution that does it in log(n) steps.

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• How do I improve my performance with this singly linked list struct within my program?

  - by Jesus

  Hey guys, I have a program that does operations of sets of strings. We have to implement functions such as addition and subtraction of two sets of strings. We are suppose to get it down to the point where performance if of O(N+M), where N,M are sets of strings. Right now, I believe my performance is at O(N*M), since I for each element of N, I go through every element of M. I'm particularly focused on getting the subtraction to the proper performance, as if I can get that down to proper performance, I believe I can carry that knowledge over to the rest of things I have to implement. The '-' operator is suppose to work like this, for example. Declare set1 to be an empty set. Declare set2 to be a set with { a b c } elements Declare set3 to be a set with ( b c d } elements set1 = set2 - set3 And now set1 is suppose to equal { a }. So basically, just remove any element from set3, that is also in set2. For the addition implementation (overloaded '+' operator), I also do the sorting of the strings (since we have to). All the functions work right now btw. So I was wondering if anyone could a) Confirm that currently I'm doing O(N*M) performance b) Give me some ideas/implementations on how to improve the performance to O(N+M) Note: I cannot add any member variables or functions to the class strSet or to the node structure. The implementation of the main program isn't very important, but I will post the code for my class definition and the implementation of the member functions: strSet2.h (Implementation of my class and struct) // Class to implement sets of strings // Implements operators for union, intersection, subtraction, // etc. for sets of strings // V1.1 15 Feb 2011 Added guard (#ifndef), deleted using namespace RCH #ifndef _STRSET_ #define _STRSET_ #include <iostream> #include <vector> #include <string> // Deleted: using namespace std; 15 Feb 2011 RCH struct node { std::string s1; node * next; }; class strSet { private: node * first; public: strSet (); // Create empty set strSet (std::string s); // Create singleton set strSet (const strSet &copy); // Copy constructor ~strSet (); // Destructor int SIZE() const; bool isMember (std::string s) const; strSet operator + (const strSet& rtSide); // Union strSet operator - (const strSet& rtSide); // Set subtraction strSet& operator = (const strSet& rtSide); // Assignment }; // End of strSet class #endif // _STRSET_ strSet2.cpp (implementation of member functions) #include <iostream> #include <vector> #include <string> #include "strset2.h" using namespace std; strSet::strSet() { first = NULL; } strSet::strSet(string s) { node *temp; temp = new node; temp->s1 = s; temp->next = NULL; first = temp; } strSet::strSet(const strSet& copy) { if(copy.first == NULL) { first = NULL; } else { node *n = copy.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } strSet::~strSet() { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } } int strSet::SIZE() const { int size = 0; node *temp = first; while(temp!=NULL) { size++; temp=temp->next; } return size; } bool strSet::isMember(string s) const { node *temp = first; while(temp != NULL) { if(temp->s1 == s) { return true; } temp = temp->next; } return false; } strSet strSet::operator + (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string newEle = temp->s1; if(!isMember(newEle)) { if(newSet.first==NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first = newNode; } else if(newSet.SIZE() == 1) { if(newEle < newSet.first->s1) { node *tempNext = newSet.first; node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = tempNext; newSet.first = newNode; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; newSet.first->next = newNode; } } else { node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if(newEle < curr->s1) { if(prev == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; newSet.first = newNode; break; } else { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = curr; prev->next = newNode; break; } } if(curr->next == NULL) { node *newNode; newNode = new node; newNode->s1 = newEle; newNode->next = NULL; curr->next = newNode; break; } prev = curr; curr = curr->next; } } } temp = temp->next; } return newSet; } strSet strSet::operator - (const strSet& rtSide) { strSet newSet; newSet = *this; node *temp = rtSide.first; while(temp != NULL) { string element = temp->s1; node *prev = NULL; node *curr = newSet.first; while(curr != NULL) { if( element < curr->s1 ) break; if( curr->s1 == element ) { if( prev == NULL) { node *duplicate = curr; newSet.first = newSet.first->next; delete duplicate; break; } else { node *duplicate = curr; prev->next = curr->next; delete duplicate; break; } } prev = curr; curr = curr->next; } temp = temp->next; } return newSet; } strSet& strSet::operator = (const strSet& rtSide) { if(this != &rtSide) { if(first != NULL) { while(first->next != NULL) { node *nextNode = first->next; first->next = nextNode->next; delete nextNode; } } if(rtSide.first == NULL) { first = NULL; } else { node *n = rtSide.first; node *prev = NULL; while (n) { node *newNode = new node; newNode->s1 = n->s1; newNode->next = NULL; if (prev) { prev->next = newNode; } else { first = newNode; } prev = newNode; n = n->next; } } } return *this; }

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• python programme.

  - by siva

  hi, i am siva this is frist time taken the python programming language i have a small problem please help me the question is **Write two functions, called countSubStringMatch and countSubStringMatchRecursive that take two arguments, a key string and a target string. These functions iteratively and recursively count the number of instances of the key in the target string. You should complete definitions for def countSubStringMatch(target,key): and def countSubStringMatchRecursive (target, key): **

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• Validating an integer or String without try-catch

  - by Phil

  Ok, I'm lost. I am required to figure out how to validate an integer and String, but for some stupid reason, I can't use the Try-Catch method. I know this is the easiest way and so all the solutions on the internet are using it. I'm writing in Java. The deal is this, I need someone to put in an numerical ID and String name. If either one of the two inputs are invalid I must tell them they made a mistake. Can someone help me?

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• finding "distance" between two pixel's colors.

  - by igor

  Once more something relatively simple, but confused as to what they want. the method to find distance on cartesian coordinate system is distance=sqrt[(x2-x1)^2 + (y2-y1)^2] but how do i apply it here? //Requires: testColor to be a valid Color //Effects: returns the "distance" between the current Pixel's color and // the passed color // uses the standard method to calculate "distance" // uses the same formula as finding distance on a // Cartesian coordinate system double colorDistance(Color testColor) const;

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• Information Gain and Entropy

  - by dhorn

  I recently read this question regarding information gain and entropy. I think I have a semi-decent grasp on the main idea, but I'm curious as what to do with situations such as follows: If we have a bag of 7 coins, 1 of which is heavier than the others, and 1 of which is lighter than the others, and we know the heavier coin + the lighter coin is the same as 2 normal coins, what is the information gain associated with picking two random coins and weighing them against each other? Our goal here is to identify the two odd coins. I've been thinking this problem over for a while, and can't frame it correctly in a decision tree, or any other way for that matter. Any help? EDIT: I understand the formula for entropy and the formula for information gain. What I don't understand is how to frame this problem in a decision tree format. EDIT 2: Here is where I'm at so far: Assuming we pick two coins and they both end up weighing the same, we can assume our new chances of picking H+L come out to 1/5 * 1/4 = 1/20 , easy enough. Assuming we pick two coins and the left side is heavier. There are three different cases where this can occur: HM: Which gives us 1/2 chance of picking H and a 1/4 chance of picking L: 1/8 HL: 1/2 chance of picking high, 1/1 chance of picking low: 1/1 ML: 1/2 chance of picking low, 1/4 chance of picking high: 1/8 However, the odds of us picking HM are 1/7 * 5/6 which is 5/42 The odds of us picking HL are 1/7 * 1/6 which is 1/42 And the odds of us picking ML are 1/7 * 5/6 which is 5/42 If we weight the overall probabilities with these odds, we are given: (1/8) * (5/42) + (1/1) * (1/42) + (1/8) * (5/42) = 3/56. The same holds true for option B. option A = 3/56 option B = 3/56 option C = 1/20 However, option C should be weighted heavier because there is a 5/7 * 4/6 chance to pick two mediums. So I'm assuming from here I weight THOSE odds. I am pretty sure I've messed up somewhere along the way, but I think I'm on the right path! EDIT 3: More stuff. Assuming the scale is unbalanced, the odds are (10/11) that only one of the coins is the H or L coin, and (1/11) that both coins are H/L Therefore we can conclude: (10 / 11) * (1/2 * 1/5) and (1 / 11) * (1/2) EDIT 4: Going to go ahead and say that it is a total 4/42 increase.

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