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  • Play a New Random Game Each Day in Chrome

    - by Asian Angel
    Being able to unwind for a few moments each day can make the time pass so much better and help you feel refreshed. If your favorite method for relaxing is playing a quick game, then join us as we take a look at the Random Games from MyGiochi.net extension for Google Chrome. Random Games from MyGiochi.net in Action The really great thing about this extension is that each day you can have a new random game to play. If you love variety this is definitely going to be a perfect match for you. We got “Power Golf” as our random game of the day. Here is a look at things once we got started…this one can be a lot of fun to play. Time to move on to the third hole now… What if you want something different from the game available on any given day? In the upper right corner you will find links for “game categories” that you can look through (clicking on the links will open a new tab). Since the links are in Italian you might need to experiment a little bit to find the category that you want to browse through. We chose the “Games for Girls Category”. With Chrome’s new built in “Translation Bar” you can easily switch the page over to the language of your choice. Note: Translation Bar available in Dev Channel releases. Ready to choose a fun game to play! You really can have a lot of fun with the games available at My Giochi. With our “game of the day” we had a second option for other games to try. More games equals more fun! Conclusion If playing online games is your favorite way to relax then the MyGiochi.net extension will make a great addition to your browser. Have fun with all of those new games each day! Links Download the Random Games from MyGiochi.net extension (Google Chrome Extensions) Similar Articles Productive Geek Tips Geek Fun: Play Alien Arena the Free FPS GamePlay Avalanche!! in Google ChromeFriday Fun: Get Your Mario OnFriday Fun: Play Bubble QuodFriday Fun: 13 Days in Hell TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 PCmover Professional How to Browse Privately in Firefox Kill Processes Quickly with Process Assassin Need to Come Up with a Good Name? Try Wordoid StockFox puts a Lightweight Stock Ticker in your Statusbar Explore Google Public Data Visually The Ultimate Excel Cheatsheet

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  • Truly Random Numbers

    German researchers use a "flip-flop" to improve random number generator Random number generation - Algorithms - Pseudorandom Numbers - Math - Recreation

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  • Random links seo and spam

    - by DoesNotCompute
    I built a mini-forum with social features for a client, to promote user registration, i planned to add a box on the forum pages to display pictures with profile links of random registered users. I managed to make this random selection static for a day, i mean the list will be renewed each day and not change on every page refresh. Could this random list of link could be harmful to seo by being considered as some kind of spam?

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  • In python, is there anyway to have a variable be a different random number everytime?

    - by woah113
    Basically I have this: import random variable1 = random.randint(13, 19) And basically what that does is assign variable1 a random number between 13 and 19. Great. But, what I want it to do is assign a different random number between 13 and 19 to that variable every time it is called. Is there anyway I can do this? If I'm not being clear enough here's an example: import random variable1 = random.randint(13, 19) print(variable1) print(variable1) print(variable1) And the output I want would look something like this: ./script.py 15 19 13 So yeah, anyway I could do this in python? (More specifically python3. but the answer would probably be similar to a python2 answer)

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  • how to generate uncorrelated random numbers in repeated calls in parallel?

    - by user1446948
    I want to write a function which will be repeatedly called by other functions many times. Inside this function it is supposed to generate a lot of random numbers and this part will be treated in parallel. If only for one run, the seed can be chosen differently for each thread, so that the random numbers will be uncorrelated. However, if this function will be called the 2nd time, it seems that the random numbers will repeat unless the seed will be again changed during the later calls. So my question is, is there a good way to generate the random numbers or reset the seed so that the random numbers generated by repeated calls to this function and also by different threads are really random? Thank you.

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  • Quantis Quantum Random Number Generator (QRNG) - any reviews?

    - by Tim Post
    I am thinking about getting one of these (PCI) to set up an internal entropy pool similar to this service who incidentally brought us fun captcha challenges. Prior to lightening my wallet, I'm hoping to gather feedback from people who may be using this device. As there is no possible 'correct' answer, I am making this CW and tagging it as subjective. I'm undertaking a project to help write Monte Carlo simulations for a non profit that distributes mosquito nets in Malaria stricken areas. The idea is to model areas to determine the best place to distribute mosquito nets. During development, I expect to consume gigs if not more of the RNG output. We really need our own source. Is this device reliable? Does it have to be re-started often? Is its bandwidth really as advertised? It passes all tests, as far as randomness goes (i.e. NIST/DIEHARD). What I don't want is something in deadlock due to some ioctl in disk sleep that does nothing but radiate heat. This is not a spamvertisement, I'm helping out of pocket and I really, really want to know if such a large purchase will bear fruit. I can't afford to build a HRNG based on radioactive decay, this looks like the next best thing. Any comments are appreciated. I will earn zero rep for this, please do not vote to close. This is no different than questions regarding the utilization of some branded GPU for some odd purpose. Answers pointing to other solutions will be gladly accepted, I'm not married to this idea.

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  • Generating easy-to-remember random identifiers

    - by Carl Seleborg
    Hi all, As all developers do, we constantly deal with some kind of identifiers as part of our daily work. Most of the time, it's about bugs or support tickets. Our software, upon detecting a bug, creates a package that has a name formatted from a timestamp and a version number, which is a cheap way of creating reasonably unique identifiers to avoid mixing packages up. Example: "Bug Report 20101214 174856 6.4b2". My brain just isn't that good at remembering numbers. What I would love to have is a simple way of generating alpha-numeric identifiers that are easy to remember. Examples would be "azil3", "ulmops", "fel2way", etc. I just made these up, but they are much easier to recognize when you see many of them at once. I know of algorithms that perform trigram analysis on text (say you feed them a whole book in German) and that can generate strings that look and feel like German words. This requires lots of data, though, and makes it slightly less suitable for embedding in an application just for this purpose. Do you know of anything else? Thanks! Carl

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  • Best way to get a random number from 1 to 50 which ISN'T x

    - by Cocorico
    Hi guys! So this is probably programming 101 stuff, but I have a problem: I have 2 numbers which are between 0 and 49. Let's call them x and y. Now I want to get a couple of other numbers which are not x or y, but are also between 0 and 49 (I am using Objective C but this is more of a general theory question I think?). Method I thought of is: int a; int b; int c; do { a = arc4random() % 49; } while ((a == x) || (a == y)); do { b = arc4random() % 49; } while ((b == x) || (b == y) || (b == a)); do { c = arc4random() % 49; } while ((c == x) || (c == y) || (c == a) || (c == b)); But it seem kind of bad to me, I don't know, I am just trying to learn to be a better programmer, what would be the most elegant sweet way to do this for best practices? Thanks!

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  • reading a random access file

    - by user1067332
    The file I create has text in it, but when i try to read it, the output is null. here is the file the creates the text file, it creates the file and puts it in the right postion import java.util.*; import java.io.*; public class CreateCustomerFile { public static void main(String[] args) throws IOException { int pos; RandomAccessFile it = new RandomAccessFile("CustomerFile.txt", "rw"); Scanner input = new Scanner(System.in); int id; String name; int zip; final int RECORDSIZE = 100; final int NUMRECORDS = 1000; final int STOP = 99; try { for(int x = 0; x < NUMRECORDS; ++x) { it.writeInt(0); it.writeUTF(""); it.writeInt(0); } } catch(IOException e) { System.out.println("Error: " + e.getMessage()); } finally { it.close(); } it = new RandomAccessFile("CustomerFile.txt","rw"); try { System.out.print("Enter ID number" + " or " + STOP + " to quit "); id = input.nextInt(); while(id != STOP) { input.nextLine(); System.out.print("Enter last name"); name = input.nextLine(); System.out.print("Enter zipcode "); zip = input.nextInt(); pos = id - 1; it.seek(pos * RECORDSIZE); it.writeInt(id); it.writeUTF(name); it.writeInt(zip); System.out.print("Enter ID number" + " or " + STOP + " to quit "); id = input.nextInt(); } } catch(IOException e) { System.out.println("Error: " + e.getMessage()); } finally { it.close(); } } } Here is the file to read, the pos is correct but ouput always goes to the error: null. import javax.swing.*; import java.io.*; public class CustomerItemOrder { public static void main(String[] args) throws IOException { int pos; RandomAccessFile it = new RandomAccessFile("CustomerFile.txt","rw"); String inputString; int id, zip; String name; final int RECORDSIZE = 100; final int STOP = 99; inputString = JOptionPane.showInputDialog(null, "Enter id or "+ STOP + " to quit"); id = Integer.parseInt(inputString); try { while(id != STOP) { pos = id -1 ; it.seek(pos * RECORDSIZE ); id = it.readInt(); name = it.readLine(); zip = it.readInt(); JOptionPane.showMessageDialog(null, "ID:" + id + " last name is " + name + " and zipcode is " + zip); inputString = JOptionPane.showInputDialog(null, "Enter id or " + STOP + " to quit"); id = Integer.parseInt(inputString); } } catch(IOException e) { System.out.println("Error: " + e.getMessage()); } finally { it.close(); } } }

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  • Why is my /dev/random so slow when using dd?

    - by Mikey
    I am trying to semi-securely erase a bunch of hard drives. The following is working at 20-50Mb/s dd if=/dev/zero of=/dev/sda But dd if=/dev/random of=/dev/sda seems not to work. Also when I type dd if=/dev/random of=stdout It only gives me a few bytes regardless of what I pass it for bs= and count= Am I using /dev/random wrong? What other info should I look for to move this troubleshooting forward? Is there some other way to do this with a script or something like makeMyLifeEasy | dd if=stdin of=/dev/sda Or something like that...

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  • java random percentages

    - by erw
    I need to generate n percentages (integers between 0 and 100) such that the sum of all n numbers adds up to 100. If I just do nextInt() n times, each time ensuring that the parameter is 100 minus the previously accumulated sum, then my percentages are biased (i.e. the first generated number will usually be largest etc.). How do I do this in an unbiased way?

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  • Weighted random numbers in MATLAB

    - by yuk
    How to randomly pick up N numbers from a vector a with weight assigned to each number? Let's say: a = 1:3; % possible numbers weight = [0.3 0.1 0.2]; % corresponding weights In this case probability to pick up 1 should be 3 times higher than to pick up 2. Sum of all weights can be anything.

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  • How do i generate random data with RSA?

    - by acidzombie24
    After loading my RSACryptoServiceProvider rsa object i would like to create a key for my AES object. Since i dont need to store the AES key (i only need it to decrypt on my prv side) i figure i dont need to store it and i can generate it with my public key. I thought doing rsa.Encrypt(byte[] with 4 hardcoded bytes); would generate the data i need. It turns out everytime i call this function even with the same data i get different results. So theres no way for me to recreate the AES key if its different everytime. How can i generate data with RSA in a way that i can recreate anytime i need?

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  • Computationally simple Pseudo-Gaussian Distribution with varying mean and standard deviation?

    - by mstksg
    This picture from wikipedia has a nice example of the sort of functions I'd ideally like to generate http://en.wikipedia.org/wiki/File:Normal_Distribution_PDF.svg Right now I'm using the Irwin-Hall Distribution, which is more or less a Polynomial approximation of the Gaussian distribution...basically, you use uniform random number generator and iterate it x times, and take the average. The more iterations, the more like a Gaussian Distribution it is. It's pretty nice; however I'd like to be able to have one where I can vary the mean. For example, let's say I wanted a number between the range 0 and 10, but around 7. Like, the mean (if I repeated this function multiple times) would turn out to be 7, but the actual range is 0-10. Is there one I should look up, or should I work on doing some fancy maths with standard Gaussian Distributions?

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  • Efficiently generate a 16-character, alphanumeric string

    - by ensnare
    I'm looking for a very quick way to generate an alphanumeric unique id for a primary key in a table. Would something like this work? def genKey(): hash = hashlib.md5(RANDOM_NUMBER).digest().encode("base64") alnum_hash = re.sub(r'[^a-zA-Z0-9]', "", hash) return alnum_hash[:16] What would be a good way to generate random numbers? If I base it on microtime, I have to account for the possibility of several calls of genKey() at the same time from different instances. Or is there a better way to do all this? Thanks.

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  • What's going on here? Repeating rows in random list of lists.

    - by Jesse Aldridge
    I expected to get a grid of unique random numbers. Instead each row is the same sequence of numbers. What's going on here? from pprint import pprint from random import random nrows, ncols = 5, 5 grid = [[0] * ncols] * nrows for r in range(nrows): for c in range(ncols): grid[r][c] = int(random() * 100) pprint(grid) Example output: [[64, 82, 90, 69, 36], [64, 82, 90, 69, 36], [64, 82, 90, 69, 36], [64, 82, 90, 69, 36], [64, 82, 90, 69, 36]]

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  • How to generate a number in arbitrary range using random()={0..1} preserving uniformness and density?

    - by psihodelia
    Generate a random number in range [x..y] where x and y are any arbitrary floating point numbers. Use function random(), which returns a random floating point number in range [0..1] from P uniformly distributed numbers (call it "density"). Uniform distribution must be preserved and P must be scaled as well. I think, there is no easy solution for such problem. To simplify it a bit, I ask you how to generate a number in interval [-0.5 .. 0.5], then in [0 .. 2], then in [-2 .. 0], preserving uniformness and density? Thus, for [0 .. 2] it must generate a random number from P*2 uniformly distributed numbers. The obvious simple solution random() * (x - y) + y will generate not all possible numbers because of the lower density for all abs(x-y)>1.0 cases. Many possible values will be missed. Remember, that random() returns only a number from P possible numbers. Then, if you multiply such number by Q, it will give you only one of P possible values, scaled by Q, but you have to scale density P by Q as well.

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  • Software/Application needed to store and organize random content [on hold]

    - by Rami.Shareef
    I have the need to store random contents (on my local hard drive/ laptop hard drive) for personal use so I can look these contents up later on when I need them. What I mean by random content is one of the following: Some sort of code Random fact Css / html / Javascript tips Or anything I might find interesting and want to keep for future reference I want to look up these entries by tags, the software should give me the ability to associate every entry with one tag or more. I have been having difficult time find applications offer this functionality, it is like a local database with search ability and easy data-entry methods. I can build one, but I don't have the time to invest in doing so. Can you refer me to any application/software that can do that, it would be great. It does not matter if it is paid or free.

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  • reservoir sampling problem: correctness of proof

    - by eSKay
    This MSDN article proves the correctness of Reservoir Sampling algorithm as follows: Base case is trivial. For the k+1st case, the probability a given element i with position <= k is in R is s/k. The probability i is replaced is the probability k+1st element is chosen multiplied by i being chosen to be replaced, which is: s/(k+1) * 1/s = 1/(k+1), and prob that i is not replaced is k/k+1. So any given element's probability of lasting after k+1 rounds is: (chosen in k steps, and not removed in k steps) = s/k * k/(k+1), which is s/(k+1). So, when k+1 = n, any element is present with probability s/n. about step 3: What are the k+1 rounds mentioned? What is chosen in k steps, and not removed in k steps? Why are we only calculating this probability for elements that were already in R after the first s steps?

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  • reservoir sampling problem

    - by eSKay
    This MSDN article proves the correctness of Reservoir Sampling algorithm as follows: Base case is trivial. For the k+1st case, the probability a given element i with position <= k is in R is s/k. The probability i is replaced is the probability k+1st element is chosen multiplied by i being chosen to be replaced, which is: s/(k+1) * 1/s = 1/(k+1), and prob that i is not replaced is k/k+1. So any given element's probability of lasting after k+1 rounds is: (chosen in k steps, and not removed in k steps) = s/k * k/(k+1), which is s/(k+1). So, when k+1 = n, any element is present with probability s/n. about step 3: What are the k+1 rounds mentioned? What is chosen in k steps, and not removed in k steps? Why are we only calculating this probability for elements that were already in R after the first s steps?

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