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  • 3D points to quaternions

    - by Hubrus
    For the simplicity, we'll consider two 3D points, that moves one relatively to other, in time. Let's say: at moment t0, we have P1(0,0,0) and P2(0,2,0) at moment t1, P1 is still (0,0,0) but P2 changed to (0,2,2). From what I've understood reading about quaternions, is that, at moment t0, Q1 (representing P1) and Q2 (representing P2) will be both (0, 0, 0, 0). But at the moment t1, Q2 will become something else (w, x, y, z). How do I calculate the Q2 at t1 moment? I've googled a lot on this subject, but I was able to find only rotation between quaternions. I will appreciate any guidance. Thanks!

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  • How do I find the next multiple of 10 of any integer?

    - by Tommy
    Dynamic integer will be any number from 0 to 150. i.e. - number returns 41, need to return 50. If number is 10 need to return 10. Number is 1 need to return 10. Was thinking I could use the ceiling function if I modify the integer as a decimal...? then use ceiling function, and put back to decimal? Only thing is would also have to know if the number is 1, 2 or 3 digits (i.e. - 7 vs 94 vs 136) Is there a better way to achieve this? Thank You,

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  • simple dividing

    - by dontoo
    Lets say I am dividing 2592 / 324 = 8 Practically I must guess that 324 goes 8 times in 2592 without reminder. For example, first I try with 7 ( 7*324 = 2268, reminder 324, so 7 is wrong, so I try with 8 and 8 is correct) Is there anyway to do dividing without guessing( trying ), like multiplication ( all mechanically, digit by digit )?

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  • Inter-rater agreement (Fleiss' Kappa, Krippendorff's Alpha etc) Java API?

    - by adam
    I am working on building a Question Classification/Answering corpus as a part of my masters thesis. I'm looking at evaluating my expected answer type taxonomy with respect to inter-rater agreement/reliability, and I was wondering: Does anybody know of any decent (preferably free) Java API(s) that can do this? I'm reasonably certain all I need is Fleiss' Kappa and Krippendorff's Alpha at this point. Weka provides a kappa statistic in it's evaluation package, but I think it can only evaluate a classifier and I'm not at that stage yet (because I'm still building the data set and classes). Thanks.

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  • How to make simple mathematical operations and displaying the result in a textbox

    - by Audel
    Hi, I know is a rather simple question but I just can't find an appropriate example in google or anywhere. I've got this piece int numberOfPlays = int.Parse(textBox2.Text); numberOfPlays = (numberOfPlays++); textBox2.Text = (numberOfPlays.ToString()); MessageBox.Show(numberOfPlays.ToString()); So basically what I want to do is to get the value of the textBox2, make it an integer and then add 1 to it. I can't think of any more details right now, so if i'm not clear enough please ask Thanks in advance

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  • groovy connect to proxy then download files

    - by senzacionale
    i want to grab the grapes but i am behind proxy so i can not download anything. How can i connect to proxy before downloading? import groovy.text.SimpleTemplateEngine import java.security.MessageDigest import org.apache.commons.cli.OptionBuilder import org.apache.commons.cli.Options import org.apache.commons.cli.PosixParser import org.apache.commons.io.FileUtils import org.apache.ivy.core.settings.IvySettings import org.apache.ivy.plugins.parser.m2.PomModuleDescriptorParser import org.apache.tools.ant.Project import org.apache.tools.ant.ProjectHelper import org.apache.tools.ant.types.Path import org.apache.commons.cli.HelpFormatter //First grab the grapes we need for the script and create a few beans to hold some values @Grab(group = 'org.apache.ant', module = 'ant', version = '1.7.1') @Grab(group = 'commons-io', module = 'commons-io', version = '1.4') @Grab(group = 'commons-cli', module = 'commons-cli', version = '1.2') @Grab(group = 'org.apache.ivy', module = 'ivy', version = '2.1.0')

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  • How do I calculate the average direction of two vectors

    - by Mike Broughton
    Hi, I am writing and opengl based iphone app and would like to allow a user to translate around a view based on the direction that they move two fingers on the screen. For one finger I know I could just calculate the vector from the start position to the current position of the users finger and then find the unit vector of this to get just the direction, but I don't know how I would do this for two fingers, I don't think adding the components of the vectors and calculating the average would work so I'm pretty much stuck... thanks in advance

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  • What are the core mathematical concepts a good developer should know?

    - by Jose B.
    Since Graduating from a very small school in 2006 with a badly shaped & outdated program (I'm a foreigner & didn't know any better school at the time) I've come to realize that I missed a lot of basic concepts from a mathematical & software perspective that are mostly the foundations of other higher concepts. I.e. I tried to listen/watch the open courseware from MIT on Introduction to Algorithms but quickly realized I was missing several mathematical concepts to better understand the course. So what are the core mathematical concepts a good software engineer should know? And what are the possible books/sites you will recommend me?

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  • Uniforme distance between points

    - by Reonarudo
    Hello, How could I, having a path defined by several points that are not in a uniform distance from each other, redefine along the same path the same number of points but with a uniform distance. I'm trying to do this in Objective-C with NSArrays of CGPoints but so far I haven't had any luck with this. Thank you for any help. EDIT I was wondering if it would help to reduce the number of points, like when detecting if 3 points are collinear we could remove the middle one, but I'm not sure that would help.

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  • Django AND .htaccess rewrites/redirects, is this possible?

    - by Infinity
    Is it possible to have Apache htaccess rewrites take effect before it hits django? For example so I can redirect iPhone users to a completely different domain without even hitting django. We're using apache2 with mod_wsgi and the apache vhost looks like this: <VirtualHost *:80> DocumentRoot /usr/local/www/site/static Alias /css/ /usr/local/www/site/static/css/ Alias /js/ /usr/local/www/site/static/js/ Alias /img/ /usr/local/www/site/static/img/ Alias /flash/ /usr/local/www/site/static/flash/ <Directory /usr/local/www/site/static> AllowOverride All Order allow,deny Allow from all </Directory> WSGIDaemonProcess mallorca-site threads=15 WSGIScriptAlias / /usr/local/www/site/config/dev/dev.wsgi </VirtualHost> Thanks

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  • How do I get points on a curve in PHP with log()?

    - by Erick
    I have a graph I am trying to replicate: I have the following PHP code: $sale_price = 25000; $future_val = 5000; $term = 60; $x = $sale_price / $future_val; $pts = array(); $pts[] = array($x,0); for ($i=1; $i<=$term; $i++) { $y = log($x+0.4)+2.5; $pts[] = array($i,$y); echo $y . " <br>\n"; } How do I make the code work to give me the points along the lower line (between the yellow and blue areas)? It doesn't need to be exact, just somewhat close. The formula is: -ln(x+.4)+2.5 I got that by using the Online Function Grapher at http://www.livephysics.com/ Thanks in advance!!

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  • bcdiv() bcadd() bcsub() with Php

    - by Pieman
    Will this code be 'stressful' for a server? Or is it easy to bcdiv/sub/add to 10000 decimal places? I'm thinking of looping it afew times... Not Sure... $s2 = (bcdiv('1', $test, 10000)); $s = bcsub($s, $s2, 10000); $test += 2; $s3 = (bcdiv('1', $test, 10000)); $s = bcadd($s, $s3, 10000); $test += 2; Any advice? :)

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  • What is the smallest amount of bits you can write twin-prime calculation?

    - by HH
    A succinct example in Python, its source. Explanation about the syntactic sugar here. s=p=1;exec"if s%p*s%~-~p:print`p`+','+`p+2`\ns*=p*p;p+=2\n"*999 The smallest amount of bits is defined by the smallest amount of 4pcs of things you can see with hexdump, it is not that precise measure but well-enough until an ambiguity. $ echo 's=p=1;exec"if s%p*s%~-~p:print`p`+','+`p+2`\ns*=p*p;p+=2\n"*999' > .test $ hexdump .test | wc 5 36 200 $ hexdump .test 0000000 3d73 3d70 3b31 7865 6365 6922 2066 2573 0000010 2a70 2573 2d7e 707e 703a 6972 746e 7060 0000020 2b60 2b2c 7060 322b 5c60 736e 3d2a 2a70 0000030 3b70 2b70 323d 6e5c 2a22 3939 0a39 000003e so in this case it is 31 because the initial parts are removed.

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  • Complexity of subset product

    - by threenplusone
    I have a set of numbers produced using the following formula with integers 0 < x < a. f(x) = f(x-1)^2 % a For example starting at 2 with a = 649. {2, 4, 16, 256, 636, 169, 5, 25, 649, 576, 137, ...} I am after a subset of these numbers that when multiplied together equals 1 mod N. I believe this problem by itself to be NP-complete (based on similaries to Subset-Sum problem). However starting with any integer (x) gives the same solution pattern. Eg. a = 649 {2, 4, 16, 256, 636, 169, 5, 25, 649, 576, 137, ...} = 16 * 5 * 576 = 1 % 649 {3, 9, 81, 71, 498, 86, 257, 500, 135, 53, 213, ...} = 81 * 257 * 53 = 1 % 649 {4, 16, 256, 636, 169, 5, 25, 649, 576, 137, 597, ...} = 256 * 25 * 137 = 1 % 649 I am wondering if this additional fact makes this problem solvable faster? Or if anyone has run into this problem previously or has any advice?

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  • How to calculate the state of a graph?

    - by zcb
    Given a graph G=(V,E), each node i is associated with 'Ci' number of objects. At each step, for every node i, the Ci objects will be taken away by the neighbors of i equally. After K steps, output the number of objects of the top five nodes which has the most objects. Some Constrains: |V|<10^5, |E|<2*10^5, K<10^7, Ci<1000 My current idea is: represent the transformation in each step with a matrix. This problem is converted to the calculation of the power of matrix. But this solution is much too slow considering |V| can be 10^5. Is there any faster way to do it?

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  • How do I point a domain name to a Django url?

    - by username2
    I have a subdomain m.example.com that I want to point to the same location as example.com/mobile running on an apache2/django1.3 installation. example.com is the landing page, and I have the urls.py configured such that urls that match /^mobile$/ will be served the mobile version of the page. I looked into <VirtualHost>, but I think it requires a physical location for me to point m.example.com at and with the django urls there is no physical location except for the root of the project directory. I am unsure if the configuration change is made on the apache side or the django side. I've also looked into the mod_rewrite module for Apache, but I would prefer if I didnt have to redirect m.example.com to example.com/mobile

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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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  • m-estimate for continuous values

    - by Null
    I'm building a custom regression tree and want to use m-estimate for pruning. Does anyone know how to calculate that. http://www.ailab.si/blaz/predavanja/UISP/slides/uisp07-RegTrees.ppt might help (slide 12, how should Em look like?)

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  • C - Rounding number up

    - by Dave
    Hi all, I was curious to know how I can round a number to the nearest tenth. For instance If I had int a = 59 / 4 /* which would be 14.75 and how can i Store the number as 15 in "a"*/ Thanks, Dave

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  • Figuring out QuadCurveTo's parameters

    - by Fev
    Could you guys help me figuring out QuadCurveTo's 4 parameters , I tried to find information on http://docs.oracle.com/javafx/2/api/javafx/scene/shape/QuadCurveTo.html, but it's hard for me to understand without picture , I search on google about 'Quadratic Bezier' but it shows me more than 2 coordinates, I'm confused and blind now. I know those 4 parameters draw 2 lines to control the path , but how we know/count exactly which coordinates the object will throught by only knowing those 2 path-controller. Are there some formulas? import javafx.animation.PathTransition; import javafx.animation.PathTransition.OrientationType; import javafx.application.Application; import static javafx.application.Application.launch; import javafx.scene.Group; import javafx.scene.Scene; import javafx.scene.paint.Color; import javafx.scene.shape.MoveTo; import javafx.scene.shape.Path; import javafx.scene.shape.QuadCurveTo; import javafx.scene.shape.Rectangle; import javafx.stage.Stage; import javafx.util.Duration; public class _6 extends Application { public Rectangle r; @Override public void start(final Stage stage) { r = new Rectangle(50, 80, 80, 90); r.setFill(javafx.scene.paint.Color.ORANGE); r.setStrokeWidth(5); r.setStroke(Color.ANTIQUEWHITE); Path path = new Path(); path.getElements().add(new MoveTo(100.0f, 400.0f)); path.getElements().add(new QuadCurveTo(150.0f, 60.0f, 100.0f, 20.0f)); PathTransition pt = new PathTransition(Duration.millis(1000), path); pt.setDuration(Duration.millis(10000)); pt.setNode(r); pt.setPath(path); pt.setOrientation(OrientationType.ORTHOGONAL_TO_TANGENT); pt.setCycleCount(4000); pt.setAutoReverse(true); pt.play(); stage.setScene(new Scene(new Group(r), 500, 700)); stage.show(); } public static void main(String[] args) { launch(args); } } You can find those coordinates on this new QuadCurveTo(150.0f, 60.0f, 100.0f, 20.0f) line, and below is the picture of Quadratic Bezier

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