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• How to Refresh / Reload a KML layer in OpenLayers. Dynamic KML Layer.

  - by Ozaki

  TLDR See my answer below on how to refresh the layer. So far I have tried action function as follows: function RefreshKMLData(layer) { layer.loaded = false; layer.setVisibility(true); layer.redraw({ force: true }); } set interval of the function: window.setInterval(RefreshKMLData, 5000, KMLLAYER); the layer itself: var KMLLAYER = new OpenLayers.Layer.Vector("MYKMLLAYER", { projection: new OpenLayers.Projection("EPSG:4326"), strategies: [new OpenLayers.Strategy.Fixed()], protocol: new OpenLayers.Protocol.HTTP({ url: MYKMLURL, format: new OpenLayers.Format.KML({ extractStyles: true, extractAttributes: true }) }) }); the url for KMLLAYER with Math random so it doesnt cache: var MYKMLURL = var currentanchorpositionurl = 'http://' + host + '/data?_salt=' + Math.random(); I would have thought that this would Refresh the layer. As by setting its loaded to false unloads it. Visibility to true reloads it and with the Math random shouldn't allow it to cache? So has anyone done this before or know how I can get this to work? TLDR See my answer below on how to refresh the layer.

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• C++ BigInt multiplication conceptual problem

  - by Kapo

  I'm building a small BigInt library in C++ for use in my programming language. The structure is like the following: short digits[ 1000 ]; int len; I have a function that converts a string into a bigint by splitting it up into single chars and putting them into digits. The numbers in digits are all reversed, so the number 123 would look like the following: digits[0]=3 digits[1]=3 digits[2]=1 I have already managed to code the adding function, which works perfectly. It works somewhat like this: overflow = 0 for i ++ until length of both numbers exceeded: add numberA[ i ] to numberB[ i ] add overflow to the result set overflow to 0 if the result is bigger than 10: substract 10 from the result overflow = 1 put the result into numberReturn[ i ] (Overflow is in this case what happens when I add 1 to 9: Substract 10 from 10, add 1 to overflow, overflow gets added to the next digit) So think of how two numbers are stored, like those: 0 | 1 | 2 --------- A 2 - - B 0 0 1 The above represents the digits of the bigints 2 (A) and 100 (B). - means uninitialized digits, they aren't accessed. So adding the above number works fine: start at 0, add 2 + 0, go to 1, add 0, go to 2, add 1 But: When I want to do multiplication with the above structure, my program ends up doing the following: Start at 0, multiply 2 with 0 (eek), go to 1, ... So it is obvious that, for multiplication, I have to get an order like this: 0 | 1 | 2 --------- A - - 2 B 0 0 1 Then, everything would be clear: Start at 0, multiply 0 with 0, go to 1, multiply 0 with 0, go to 2, multiply 1 with 2 How can I manage to get digits into the correct form for multiplication? I don't want to do any array moving/flipping - I need performance!

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• Question on Split usage in Perl

  - by Nano HE

  Hello. I wrote an small script to use Split() as this, use strict; use warnings; use Data::Dumper; my $fh = \*DATA; while(my $line = <$fh>) { my @values = split(':', $line); foreach my $val (@values) { print "$val\n"; } } __DATA__ 1 : Hello World String10 : NO : A1B2,B3 11 : Hello World String11 : YES : A11B2,B3,B14,B25 A1B2,B3 and A11B2,B3 are characters form like Only One Letter A and One or Two Number 2, 3, 14,25 etc then concatenated with Only One Letter B and one or two Numbers like 2, 3, 14,25. etc Now out put as this 1 Hello World String10 NO A1B2,B3 11 Hello World String11 YES A11B2,B3,B14,B25 How can I hold the last array member from @values and made concatenation and out put as this. 1 Hello World String1 NO A1B2,A1B3 11 Hello World String11 YES A11B2,A11B3,A11B14,A11B25 Appreciated for your comments and replies. [update] My out put need the concatenation followed the rule. A and one or two numbers and joined by B and one or two numbers split by ,

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• SQL dealing with rubbish in a phone number field

  - by DoctaJonez

  Hello stackers! I've got a wonderfully fun little SQL problem to solve today and thought I'd ask the community to see what solutions you come up with. We've got a really cool email to text service that we use, you just need to send an email to [email protected] and it will send a text message to the desired phone number. For example to send a text to 0790 0006006, you need to send an email to [email protected], pretty neat huh? The problem is with the phone numbers in our database. Most of the phone numbers are fine, but some of them have "rubbish" mixed in with the phone number. Take these wonderful examples of the rubbish you need to deal with (I've anonymised the phone numbers by placing zeroes in): 07800 000647(mobile) 07500 000189 USE 1ST SEE NOTES 07900 000415 HO ONLY try 1st 0770 0000694 then home 07500 000465 Cannot Requirements The solution needs to be in SQL (for MS SQL server). So the challenge is as follows, we need to get the phone number without spaces, and without any of the rubbish seen in the samples. For example: This: try 1st 0770 0000694 then home Should become this: 07700000694 Anything without a phone number in it (e.g. "SEE NOTES") should be null.

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• Cannot Find Symbol Method?

  - by Aaron

  In my driver program, This line gives me cannot find symbol error and I don't know why, the method is clearly defined in the SavingsAccount class, and I can refer to all other methods in my driver program but just not that one, I tried changing the type to double, and etc but still not working. Refer to ** Account acct2 = new SavingsAccount (name); acct2.calculateBalance(); SavingsAccount Class Inherited from Account Class: public class SavingsAccount extends Account { private final short minBalance = 0; private double overdraftFee; private double yearlyInterestRate = 0.02; private double interestAmount; public SavingsAccount (String name) { super(name); } public double withdraw (double amount) { if (accountBalance - amount >= minBalance) { accountBalance -= amount; System.out.print ("Withdraw Successful"); } else { accountBalance -= amount; overdraftFee = accountBalance * (0.10); accountBalance += overdraftFee; System.out.print ("Withdraw Succesful, however overdraft fee of 10% has been applied to your account"); } return accountBalance; } **public void calculateBalance () { interestAmount = (accountBalance * yearlyInterestRate); accountBalance += interestAmount; }** public String toString() { return super.toString() + " Interest Received: " + interestAmount; } } Account class, if needed import java.util.Random; import java.text.NumberFormat; public abstract class Account { protected double accountBalance; protected long accountNumber; protected String accountHolder; public Account (String name) { accountHolder = name; accountBalance = 0; Random accountNo = new Random(); accountNumber = accountNo.nextInt(100000); } public double deposit (double amount) { accountBalance += amount; return accountBalance; } public String toString() { NumberFormat accountBal = NumberFormat.getCurrencyInstance(); return "Account Balance: " + accountBal.format(accountBalance) + "\nAccount Number: " + accountNumber; } public String getAccountHolder() { return accountHolder; } public double getAccountBalance() { return accountBalance; } public abstract double withdraw (double amount); }

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• Building an array out of values from another array

  - by George

  This is a follow up from a question of mine that was just answered concerning parsing numbers in an array. I have an array, data[], with numbers that I'd like to use in a calculation and then put the resulting values into another array. So say data[0] = 100. I'd like to find a percentage using the calculatin, (data[0]/dataSum*100).toFixed(2) where dataSum is the sum of all the numbers in data[]. I've tried: dataPercentage = []; for (var i=0; i < data.length; i++) { data[i] = parseFloat(data[i]); dataSum += data[i]; // looping through data[i] and setting it equal to dataPercentage. dataPercentage[] = (data[i]/dataSum*100).toFixed(2); // thought maybe I was overriding dataPercentage everytime I looped? dataPercentage[] += (data[i]/dataSum*100).toFixed(2); } I also tried just setting dataPercentage = [(data/dataSum*100).toFixed(2)], but I think this creates a nested array, which I don't think is what I need.

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• html in do_GET() method of a simple Python webserver

  - by Meeri_Peeri

  I am relatively new to Python but have been doing a lot of different things with it recently and I am liking it a lot. However, I ran into trouble/block with the following code. import http.server import socketserver import glob import random class Server(http.server.SimpleHTTPRequestHandler): def do_GET(self): self.send_response(200, 'OK') self.send_header('Content-type', 'html') self.end_headers() self.wfile.write(bytes("<html> <head><title> Hello World </title> </head> <body>", 'UTF-8')) images = glob.glob('*.jpg') rand = random.randint(0,len(images)-1) imagestring = "<img src = \"" + images[rand] + "\" height = 1028 width = 786 align = \"right\"/> </body> </html>" self.wfile.write(bytes(imagestring, 'UTF-8')) def serve_forever(port): socketserver.TCPServer(('', port), Server).serve_forever() if __name__ == "__main__": Server.serve_forever(8000) What I am trying to do here is grab a random image from multiple images in the directory and add it into the response to a web request. The code works fine but when I access the server via browser, the image is not displayed. The html of the page is as intended though. The permissions on the files are 755. Also I tried to create an index.html file in the do_GET method. That didn't work either. I mean the index.html was generated fine, but the response in the browser this time did not show anything (not even the hello world in the title). Am I missing anything very simple here? I was thinking should I overload the handle_request of the underlying SocketServer.BaseServer as the documentation says you should never override BaseHTTPServer's handle() method and should rather override the corresponding do_* method?

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• Confusion with a while statement evaluating if a number is triangular

  - by Darkkurama

  I've been having troubles trying to figure out how to solve a function. I've been assigned the development of a little programme which tells if a number is "triangular" (a number is triangular when the addition of certain consecutive numbers in the [1,n] interval is n. Following the definition, the number 10 is triangular, because in the [1,10] interval, 1+2+3+4=10). I've coded this so far: class TriangularNumber{ boolean numTriangular(int n) { boolean triangular = false; int i = n; while(n>=0 && triangular){ //UE06 is a class which contains the function "f0", which makes the addition of all the numbers in a determined interval UE06 p = new UE06(); if ((p.f0(1, i))==n) triangular = true; else i=i-1; } return triangular; } boolean testTriangular = numTriangular(10) == true && numTriangular(7) == false && numTriangular(6) == true; public static void main(String[] args){ TriangularNumber p = new TriangularNumber(); System.out.println("testTriangular = " + p.testTriangular); } } According to those boolean tests I made, the function is wrong. As I see the function, it goes like this: I state that the input number in the initial state isn't triangular (triangular=false) and i=n (determining the interval [1,i] where the function is going to be evaluated While n is greater or equals 0 and the number isn't triangular, the loop starts The loop goes like this: if the addition of all the numbers in the [1,i] interval is n, the number is triangular, causing the loop to end. If that statement is false, i goes from i to (i-1), starting the loop again with that particular interval, and so on till the addition is n. I can't spot the error in my "algorithm", any advice? Thanks!

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• Algorithm to determine if array contains n...n+m?

  - by Kyle Cronin

  I saw this question on Reddit, and there were no positive solutions presented, and I thought it would be a perfect question to ask here. This was in a thread about interview questions: Write a method that takes an int array of size m, and returns (True/False) if the array consists of the numbers n...n+m-1, all numbers in that range and only numbers in that range. The array is not guaranteed to be sorted. (For instance, {2,3,4} would return true. {1,3,1} would return false, {1,2,4} would return false. The problem I had with this one is that my interviewer kept asking me to optimize (faster O(n), less memory, etc), to the point where he claimed you could do it in one pass of the array using a constant amount of memory. Never figured that one out. Along with your solutions please indicate if they assume that the array contains unique items. Also indicate if your solution assumes the sequence starts at 1. (I've modified the question slightly to allow cases where it goes 2, 3, 4...) edit: I am now of the opinion that there does not exist a linear in time and constant in space algorithm that handles duplicates. Can anyone verify this? The duplicate problem boils down to testing to see if the array contains duplicates in O(n) time, O(1) space. If this can be done you can simply test first and if there are no duplicates run the algorithms posted. So can you test for dupes in O(n) time O(1) space?

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• What data stucture should I use for BigInt class

  - by user1086004

  I would like to implement a BigInt class which will be able to handle really big numbers. I want only to add and multiply numbers, however the class should also handle negative numbers. I wanted to represent the number as a string, but there is a big overhead with converting string to int and back for adding. I want to implement addition as on the high school, add corresponding order and if the result is bigger than 10, add the carry to next order. Then I thought that it would be better to handle it as a array of unsigned long long int and keep the sign separated by bool. With this I'm afraid of size of the int, as C++ standard as far as I know guarantees only that int < float < double. Correct me if I'm wrong. So when I reach some number I should move in array forward and start adding number to the next array position. Is there any data structure that is appropriate or better for this? Thanks in advance.

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• WPF binding comboboxes to parent- child model

  - by PaulB

  I've got a model with a few tiers in it - something along the lines of ... Company Employees Phone numbers So I've got a ListBox showing all the companys in the model. Each ListBoxItem then contains two comboboxes ... one for employees, one for phone numbers. I can successfully get the employee combo to bind correctly and show the right people, but I'd like the phone combo to show the numbers for the selected employee. I'm just setting the DataContext of the ListBox to the model above and using the following data template for each item <DataTemplate x:Key="CompanyBody"> <StackPanel Orientation="Horizontal"> <Label Content="{Binding Path=CompanyName}"></Label> <ComboBox Name="EmployeesCombo" ItemsSource="{Binding Path=Company.Employees}"></ComboBox> <!-- What goes here --> <ComboBox DataContext="???" ItemsSource="??" ></ComboBox> </StackPanel> </DataTemplate> I've tried (naively) <ComboBox ItemsSource="{Binding Path=Company.Employees.PhoneNumbers}" ></ComboBox> and <ComboBox DataContext="EmployeesCombo.SelectedValue" ItemsSource="{Binding Path=PhoneNumbers}" ></ComboBox> and all other manner of combinations ...

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• Read large amount of data from file in Java

  - by Crozin

  Hello I've got text file that contains 1 000 002 numbers in following formation: 123 456 1 2 3 4 5 6 .... 999999 100000 Now I need to read that data and allocate it to int variables (the very first two numbers) and all the rest (1 000 000 numbers) to an array int[]. It's not a hard task, but - it's horrible slow. My first attempt was java.util.Scanner: Scanner stdin = new Scanner(new File("./path")); int n = stdin.nextInt(); int t = stdin.nextInt(); int array[] = new array[n]; for (int i = 0; i < n; i++) { array[i] = stdin.nextInt(); } It works as excepted but it takes about 7500 ms to execute. I need to fetch that data in up to several hundred of milliseconds. Then I tried java.io.BufferedReader: Using BufferedReader.readLine() and String.split() I got the same results in about 1700 ms, but it's still too many. How can I read that amount of data in less that 1 second? The final result should be equal to: int n = 123; int t = 456; int array[] = { 1, 2, 3, 4, ..., 999999, 100000 };

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• What are your best practices for ensuring the correctness of the reports from SQL?

  - by snezmqd4

  Part of my work involves creating reports and data from SQL Server to be used as information for decision. The majority of the data is aggregated, like inventory, sales and costs totals from departments, and other dimensions. When I am creating the reports, and more specifically, I am developing the SELECTs to extract the aggregated data from the OLTP database, I worry about mistaking a JOIN or a GROUP BY, for example, returning incorrect results. I try to use some "best practices" to prevent me for "generating" wrong numbers: When creating an aggregated data set, always explode this data set without the aggregation and look for any obvious error. Export the exploded data set to Excel and compare the SUM(), AVG(), etc, from SQL Server and Excel. Involve the people who would use the information and ask for some validation (ask people to help to identify mistakes on the numbers). Never deploy those things in the afternoon - when possible, try to take a look at the T-SQL on the next morning with a refreshed mind. I had many bugs corrected using this simple procedure. Even with those procedures, I always worry about the numbers. What are your best practices for ensuring the correctness of the reports?

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• Matching up text in 2 different columns in a table?

  - by user297663

  Hey guys, I have recently been working on a pastebin script (for fun) and I've come across a problem that I can't seem to solve in CSS. I have a table with 2 columns. 1 column is used to display the line numbers and the 2nd column is used to display the code. I can't seem to get the numbers match up with the lines in the code so it looks all weird (example: www.zamnproductions.com/paste.php?id=32). Take a look at my code (the snippet): <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> td.num { vertical-align: top; } td.numbers { display:table-cell; padding:1px; vertical-align: top; line-height:25px; } td.code { display:table-cell; vertical-align: top; line-height:20px; } #hide { display:none; } #leftcontent { position: absolute; left:10px; top:119px; width:200px; background:#fff; border:0px solid #000; } #centercontent { background:#fff; margin-left: 199px; margin-right:199px; border:0px solid #000; voice-family: "\"}\""; voice-family: inherit; margin-left: 201px; margin-right:201px; } htmlbody #centercontent { margin-left: 202px; margin-right:201px; } Here is the part where the table is made: listNumbers($_GET['id']); ? viewCode($_GET['id']); ?

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• Is there a more concise regular expression to accomplish this task?

  - by mpminnich

  First off, sorry for the lame title, but I couldn't think of a better one. I need to test a password to ensure the following: Passwords must contain at least 3 of the following: upper case letters lower case letters numbers special characters Here's what I've come up with (it works, but I'm wondering if there is a better way to do this): Dim lowerCase As New Regex("[a-z]") Dim upperCase As New Regex("[A-Z]") Dim numbers As New Regex("\d") Dim special As New Regex("[\\\.\+\*\?\^\$\[\]\(\)\|\{\}\/\'\#]") Dim count As Int16 = 0 If Not lowerCase.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not upperCase.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not numbers.IsMatch(txtUpdatepass.Text) Then count += 1 End If If Not special.IsMatch(txtUpdatepass.Text) Then count += 1 End If If at least 3 of the criteria have not been met, I handle it. I'm not well versed in regular expressions and have been reading numerous tutorials on the web. Is there a way to combine all 4 regexes into one? But I guess doing that would not allow me to check if at least 3 of the criteria are met. On a side note, is there a site that has an exhaustive list of all characters that would need to be escaped in the regex (those that have special meaning - eg. $, ^, etc.)? As always, TIA. I can't express enough how awesome I think this site is.

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• C/C++ Bit Array or Bit Vector

  - by MovieYoda

  Hi, I am learning C/C++ programming & have encountered the usage of 'Bit arrays' or 'Bit Vectors'. Am not able to understand their purpose? here are my doubts - Are they used as boolean flags? Can one use int arrays instead? (more memory of course, but..) What's this concept of Bit-Masking? If bit-masking is simple bit operations to get an appropriate flag, how do one program for them? is it not difficult to do this operation in head to see what the flag would be, as apposed to decimal numbers? I am looking for applications, so that I can understand better. for Eg - Q. You are given a file containing integers in the range (1 to 1 million). There are some duplicates and hence some numbers are missing. Find the fastest way of finding missing numbers? For the above question, I have read solutions telling me to use bit arrays. How would one store each integer in a bit?

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• generate k distinct number less then n

  - by davit-datuashvili

  hi i have following question task is this generate k distinct positive numbers less then n without duplication my method is following first create array size of k where we should write these numbers int a[]=new int[k]; //now i am going to cretae another array where i check if (at given number position is 1 then generate number again else put this number in a array and continue cycle i put here a piece of code and explanations int a[]=new int[k]; int t[]=new int[n+1]; Random r=new Random(); for (int i==0;i<t.length;i++){ t[i]=0;//initialize it to zero } int m=0;//initialize it also for (int i=0;i<a.length;i++){ m=r.nextInt(n);//random element between 0 and n if (t[m]==1){ //i have problem with this i want in case of duplication element occurs repeats this steps afain until there will be different number else{ t[m]=1; x[i]=m; } } so i fill concret my problem if t[m]==1 it means that this element occurs already so i want to generate new number but problem is that number of generated numbers will not be k beacuse if i==0 and occurs duplicate element and we write continue then it will switch at i==1 i need like goto for repeat step or for (int i=0;i<x.length;i++){ loop: m=r.nextInt(n); if ( x[m]==1){ continue loop; } else{ x[m]=1; a[i]=m; continue;//continue next step at i=1 and so on } } i need this code in java please help

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• Fastest method to define whether a number is a triangular number

  - by psihodelia

  A triangular number is the sum of the n natural numbers from 1 to n. What is the fastest method to find whether a given positive integer number is a triangular one? I suppose, there must be a hidden pattern in a binary representation of such numbers (like if you need to find whether a number is even/odd you check its least significant bit). Here is a cut of the first 1200th up to 1300th triangular numbers, you can easily see a bit-pattern here (if not, try to zoom out): (720600, '10101111111011011000') (721801, '10110000001110001001') (723003, '10110000100000111011') (724206, '10110000110011101110') (725410, '10110001000110100010') (726615, '10110001011001010111') (727821, '10110001101100001101') (729028, '10110001111111000100') (730236, '10110010010001111100') (731445, '10110010100100110101') (732655, '10110010110111101111') (733866, '10110011001010101010') (735078, '10110011011101100110') (736291, '10110011110000100011') (737505, '10110100000011100001') (738720, '10110100010110100000') (739936, '10110100101001100000') (741153, '10110100111100100001') (742371, '10110101001111100011') (743590, '10110101100010100110') (744810, '10110101110101101010') (746031, '10110110001000101111') (747253, '10110110011011110101') (748476, '10110110101110111100') (749700, '10110111000010000100') (750925, '10110111010101001101') (752151, '10110111101000010111') (753378, '10110111111011100010') (754606, '10111000001110101110') (755835, '10111000100001111011') (757065, '10111000110101001001') (758296, '10111001001000011000') (759528, '10111001011011101000') (760761, '10111001101110111001') (761995, '10111010000010001011') (763230, '10111010010101011110') (764466, '10111010101000110010') (765703, '10111010111100000111') (766941, '10111011001111011101') (768180, '10111011100010110100') (769420, '10111011110110001100') (770661, '10111100001001100101') (771903, '10111100011100111111') (773146, '10111100110000011010') (774390, '10111101000011110110') (775635, '10111101010111010011') (776881, '10111101101010110001') (778128, '10111101111110010000') (779376, '10111110010001110000') (780625, '10111110100101010001') (781875, '10111110111000110011') (783126, '10111111001100010110') (784378, '10111111011111111010') (785631, '10111111110011011111') (786885, '11000000000111000101') (788140, '11000000011010101100') (789396, '11000000101110010100') (790653, '11000001000001111101') (791911, '11000001010101100111') (793170, '11000001101001010010') (794430, '11000001111100111110') (795691, '11000010010000101011') (796953, '11000010100100011001') (798216, '11000010111000001000') (799480, '11000011001011111000') (800745, '11000011011111101001') (802011, '11000011110011011011') (803278, '11000100000111001110') (804546, '11000100011011000010') (805815, '11000100101110110111') (807085, '11000101000010101101') (808356, '11000101010110100100') (809628, '11000101101010011100') (810901, '11000101111110010101') (812175, '11000110010010001111') (813450, '11000110100110001010') (814726, '11000110111010000110') (816003, '11000111001110000011') (817281, '11000111100010000001') (818560, '11000111110110000000') (819840, '11001000001010000000') (821121, '11001000011110000001') (822403, '11001000110010000011') (823686, '11001001000110000110') (824970, '11001001011010001010') (826255, '11001001101110001111') (827541, '11001010000010010101') (828828, '11001010010110011100') (830116, '11001010101010100100') (831405, '11001010111110101101') (832695, '11001011010010110111') (833986, '11001011100111000010') (835278, '11001011111011001110') (836571, '11001100001111011011') (837865, '11001100100011101001') (839160, '11001100110111111000') (840456, '11001101001100001000') (841753, '11001101100000011001') (843051, '11001101110100101011') (844350, '11001110001000111110') For example, can you also see a rotated normal distribution curve, represented by zeros between 807085 and 831405?

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• How can I put back a character that I've read when I detect it's the start of a new row?

  - by gcc

  char nm; int i=0; double thelow, theupp; double numbers[200]; for(i=0;i<4;++i) { { char nm; double thelow,theupp; /*after erased ,created again*/ scanf("%c %lf %lf", &nm, &thelow, &theupp); for (k = 0; ; ++k) ; { scanf("%lf",numbers[k]); if(numbers[k]=='\n') break; } /*calling function and sending data(nm,..) to it*/ } /*after } is seen (nm ..) is erased*/ ; } I want say compiler : hey my dear code read only i-th row,dont touch characters at placed in next line. because characters at placed in next line is token after i increased by 1 and nm ,thelow,theupp is being zero or erased after then again created. how can I do ? input; D -1.5 0.5 .012 .025 .05 .1 .1 .1 .025 .012 0 0 0 .012 .025 .1 .2 .1 .05 .039 .025 .025 B 1 3 .117 .058 .029 .015 .007 .007 .007 .015 .022 .029 .036 .044 .051 .058 .066 .073 .080 .088 .095 .103

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• Iterating over a String to check for a number and printing out the String value if it doesn't have a number

  - by wheelerlc64

  I have set up my function for checking for a number in a String, and printing out that String if it has no numbers, and putting up an error message if it does. Here is my code: public class NumberFunction { public boolean containsNbr(String str) { boolean containsNbr = false; if(str != null && !str.isEmpty()) { for(char c : str.toCharArray()) { if(containsNbr = Character.isDigit(c)) { System.out.println("Can't contain numbers in the word."); break; } else { System.out.println(str); } } } return containsNbr; } } import com.imports.validationexample.function.NumberFunction; public class Main { public static void main(String[] args) { NumberFunction nf = new NumberFunction(); System.out.println(nf.containsNbr("bill4")); } } I am trying to get it to print out the result to the console, but the result keeps printing multiple times and prints the boolean value, which I do not want, something like this: bill4 bill4 bill4 bill4 Can't contain numbers in the word. true Why is this happening? I've tried casting but that hasn't worked out either. Any help would be much appreciated.

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• question about permut-by-sorting

  - by davit-datuashvili

  hi i have following question from book introduction in algorithms second edition there is such problem suppose we have some array A int a[]={1,2,3,4} and we have some random priorities array P={36,3,97,19} we shoud permut array a randomly using this priorities array here is pseudo code P ERMUTE -B Y-S ORTING ( A) 1 n ? length[A] 2 for i ? 1 to n do P[i] = R ANDOM(1, n 3 ) 3 4 sort A, using P as sort keys 5 return A and result will be permuted array B={2, 4, 1, 3}; please help any ideas i have done this code and need aideas how continue import java.util.*; public class Permut { public static void main(String[]args){ Random r=new Random(); int a[]=new int[]{1,2,3,4}; int n=a.length; int b[]=new int[a.length]; int p[]=new int[a.length]; for (int i=0;i<p.length;i++){ p[i]=r.nextInt(n*n*n)+1; } // for (int i=0;i<p.length;i++){ // System.out.println(p[i]); //} } } please help

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• Find existence of number in a sorted list in constant time? (Interview question)

  - by Rich

  I'm studying for upcoming interviews and have encountered this question several times (written verbatim) Find or determine non existence of a number in a sorted list of N numbers where the numbers range over M, M N and N large enough to span multiple disks. Algorithm to beat O(log n); bonus points for constant time algorithm. First of all, I'm not sure if this is a question with a real solution. My colleagues and I have mused over this problem for weeks and it seems ill formed (of course, just because we can't think of a solution doesn't mean there isn't one). A few questions I would have asked the interviewer are: Are there repeats in the sorted list? What's the relationship to the number of disks and N? One approach I considered was to binary search the min/max of each disk to determine the disk that should hold that number, if it exists, then binary search on the disk itself. Of course this is only an order of magnitude speedup if the number of disks is large and you also have a sorted list of disks. I think this would yield some sort of O(log log n) time. As for the M N hint, perhaps if you know how many numbers are on a disk and what the range is, you could use the pigeonhole principle to rule out some cases some of the time, but I can't figure out an order of magnitude improvement. Also, "bonus points for constant time algorithm" makes me a bit suspicious. Any thoughts, solutions, or relevant history of this problem?

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• Calculating percent "x/y * 100" always results in 0?

  - by Patrick Beninga

  In my assignment i have to make a simple version of Craps, for some reason the percentage assignments always produce 0 even when both variables are non 0, here is the code. import java.util.Random; Header, note the variables public class Craps { private int die1, die2,myRoll ,myBet,point,myWins,myLosses; private double winPercent,lossPercent; private Random r = new Random(); Just rolls two dies and produces their some. public int roll(){ die1 = r.nextInt(6)+1; die2 = r.nextInt(6)+1; return(die1 + die2); } The Play method, this just loops through the game. public void play(){ myRoll = roll(); point = 0; if(myRoll == 2 ||myRoll == 3 || myRoll == 12){ System.out.println("You lose!"); myLosses++; }else if(myRoll == 7 || myRoll == 11){ System.out.println("You win!"); myWins++; }else{ point = myRoll; do { myRoll = roll(); }while(myRoll != 7 && myRoll != point); if(myRoll == point){ System.out.println("You win!"); myWins++; }else{ System.out.println("You lose!"); myLosses++; } } } This is where the bug is, this is the tester method. public void tester(int howMany){ int i = 0; while(i < howMany){ play(); i++; } bug is right here in these assignments statements winPercent = myWins/i * 100; lossPercent = myLosses/i* 100; System.out.println("program ran "+i+" times "+winPercent+"% wins "+ lossPercent+"% losses with "+myWins+" wins and "+myLosses+" losses"); } }

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• Printing the results in the original order

  - by Sam

  String[] numbers = new String[] {"3", "4", "s", "a", "c", "h", "i", "n", "t", "e", "n", "d", "u", "l", "k"}; Map<String, Integer> map = new HashMap<String, Integer>(); for (int i = 0; i < numbers.length; i++) { String key = numbers[i]; if (map.containsKey(key)) { int occurrence = map.get(key); occurrence++; map.put(key, occurrence); } else { map.put(key, 1); }// end of if else }// end of for loop Iterator<String> iterator = map.keySet().iterator(); while (iterator.hasNext()) { String key = iterator.next(); int occurrence = map.get(key); System.out.println(key + " occur " + occurrence + " time(s)."); } This program tries to count the number of occurrences of a string. When I execute it I am getting the answer, but the output is not in the original order, it is shuffled. How can I output the strings in the original order?

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• Python - creating a list with 2 characteristics bug

  - by user2733911

  The goal is to create a list of 99 elements. All elements must be 1s or 0s. The first element must be a 1. There must be 7 1s in total. import random import math import time # constants determined through testing generation_constant = 0.96 def generate_candidate(): coin_vector = [] coin_vector.append(1) for i in range(0, 99): random_value = random.random() if (random_value > generation_constant): coin_vector.append(1) else: coin_vector.append(0) return coin_vector def validate_candidate(vector): vector_sum = sum(vector) sum_test = False if (vector_sum == 7): sum_test = True first_slot = vector[0] first_test = False if (first_slot == 1): first_test = True return (sum_test and first_test) vector1 = generate_candidate() while (validate_candidate(vector1) == False): vector1 = generate_candidate() print vector1, sum(vector1), validate_candidate(vector1) Most of the time, the output is correct, saying something like [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0] 7 True but sometimes, the output is: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 2 False What exactly am I doing wrong?

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