Search Results

Search found 14397 results on 576 pages for 'math mode'.

Page 89/576 | < Previous Page | 85 86 87 88 89 90 91 92 93 94 95 96  | Next Page >

  • Can a real number "cover" all integers within its range?

    - by macias
    Is there a guarantee that a real number (float, double, etc) can "cover" all integers within its range? By cover I mean, that for every integer within its range there is such real number that this equality holds: real == int Or in another example, let's say I have the biggest real number which is smaller than given integer. When I add "epsilon" will I get this number equal to given integer or bigger than integer? (I know that among real numbers you should not write comparisons as == for equality, I am simply asking for better understanding subject, not for coding comparisons.)

    Read the article

  • Uniform 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.

    Read the article

  • 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

    Read the article

  • 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?)

    Read the article

  • Why do so many mathematicians format code so poorly? [closed]

    - by marcog
    I have done a fair amount of programming together with mathematicians. Now I am even teaching some high school kids coming from a mathematics background how to program. Most of these people format their code so hideously it's hard to believe. I've even worked with and taught mathematicians who will fight the auto-indenter! Why is this so common amongst mathematicians? BTW, this is one reason I have started teaching Python. Yet still they find ways other than indentation to produce whacked coding styles!

    Read the article

  • How do I determine when two moving points become visible to each other?

    - by Devin Jeanpierre
    Suppose I have two points, Point1 and Point2. At any given time, these points may be at different positions-- they are not necessarily static. Point1 is located at some position at time t, and its position is defined by the continuous functions x1(t) and y1(t) giving the x and y coordinates at time t. These functions are not differentiable, they are constructed piecewise from line segments. Point2 is the same, with x2(t) and y2(t), each function having the same properties. The obstacles that might prevent visibility are simple (and immobile) polygons. How can I find the boundary points for visibility? i.e. there are two kinds of boundaries: where the points become visible, and become invisible. For a become-visible boundary i, there exists some ?0, such that for any real number a, a ? (i-?, i) , Point1 and Point2 are not visible (i.e. the line segment that connects (x1(a), y1(a)) to (x2(a), y2(x)) crosses some obstacles). For b ? (i, i+?) they are visible. And it is the other way around for becomes-invisible. But can I find such a precise boundary, and if so, how?

    Read the article

  • How do I find the millionth number in the series: 2 3 4 6 9 13 19 28 42 63 ... ?

    - by HH
    It takes about minute to achieve 3000 in my comp but I need to know the millionth number in the series. The definition is recursive so I cannot see any shortcuts except to calculate everything before the millionth number. How can you fast calculate millionth number in the series? Series Def n_{i+1} = \floor{ 3/2 * n_{i} } and n_{0}=2. Interestingly, only one site list the series according to Google: this one. Too slow Bash code #!/bin/bash function series { n=$( echo "3/2*$n" | bc -l | tr '\n' ' ' | sed -e 's@\\@@g' -e 's@ @@g' ); # bc gives \ at very large numbers, sed-tr for it n=$( echo $n/1 | bc ) #DUMMY FLOOR func } n=2 nth=1 while [ true ]; #$nth -lt 500 ]; do series $n # n gets new value in the function through global value echo $nth $n nth=$( echo $nth + 1 | bc ) #n++ done

    Read the article

  • Basic C++ code for multiplication of 2 matrix or vectors (C++ beginner)

    - by Ice
    I am a new C++ user and I am also doing a major in Maths so thought I would try implement a simple calculator. I got some code off the internet and now I just need help to multiply elements of 2 matrices or vectors. Matrixf multiply(Matrixf const& left, Matrixf const& right) { // error check if (left.ncols() != right.nrows()) { throw std::runtime_error("Unable to multiply: matrix dimensions not agree."); } /* I have all the other part of the code for matrix*/ /** Now I am not sure how to implement multiplication of vector or matrix.**/ Matrixf ret(1, 1); return ret; }

    Read the article

  • Easiest way to calculate amount of even numbers in given range

    - by Fdr
    What is the simplest way to calculate the amount of even numbers in a range of unsigned integers? An example: if range is [0...4] then the answer is 3 (0,2,4) I'm having hard time to think of any simple way. The only solution I came up involved couple of if-statements. Is there a simple line of code that can do this without if-statements or ternary operators?

    Read the article

  • 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?

    Read the article

  • JFileChooser returns incorrect path in OS X (folders only mode)

    - by Virat Kadaru
    I have a problem in java swing where the user has to select a folder, so I am using the code below. JFileChooser fc = new JFileChooser(); fc.setFileSelectionMode(JFileChooser.DIRECTORIES_ONLY); if(fc.showDialog(singleton, SELECT) == JFileChooser.APPROVE_OPTION) { File folder = fc.getSelectedFile(); String path = folder.getPath() + File.separatorChar + MYAPPFOLDER; } Now there are 2 ways a user may select the folder Navigate to the folder and select the folder Navigate to the folder, go into the folder, and click select Both ways work fine on windows but on OS X, I get If I do 1 : path = Users/<username>/Desktop/MYAPPFOLDER If I do 2 : path = Users/<username>/Desktop/Desktop/MYAPPFOLDER How do I avoid this 2nd case? Thanks in advance.

    Read the article

  • Who owes who money optimisation problem

    - by Francis
    Say you have n people, each who owe each other money. In general it should be possible to reduce the amount of transactions that need to take place. i.e. if X owes Y £4 and Y owes X £8, then Y only needs to pay X £4 (1 transaction instead of 2). This becomes harder when X owes Y, but Y owes Z who owes X as well. I can see that you can easily calculate one particular cycle. It helps for me when I think of it as a fully connected graph, with the nodes being the amount each person owes. Problem seems to be NP-complete, but what kind of optimisation algorithm could I make, nevertheless, to reduce the total amount of transactions? Doesn't have to be that efficient, as N is quite small for me.

    Read the article

  • Dynamic creation of a pointer function in c++

    - by Liberalkid
    I was working on my advanced calculus homework today and we're doing some iteration methods along the lines of newton's method to find solutions to things like x^2=2. It got me thinking that I could write a function that would take two function pointers, one to the function itself and one to the derivative and automate the process. This wouldn't be too challenging, then I started thinking could I have the user input a function and parse that input (yes I can do that). But can I then dynamically create a pointer to a one-variable function in c++. For instance if x^2+x, can I make a function double function(double x){ return x*x+x;} during run-time. Is this remotely feasible, or is it along the lines of self-modifying code?

    Read the article

  • 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

    Read the article

  • Getting Factors of a Number

    - by Dave
    Hi Problem: I'm trying to refactor this algorithm to make it faster. What would be the first refactoring here for speed? public int GetHowManyFactors(int numberToCheck) { // we know 1 is a factor and the numberToCheck int factorCount = 2; // start from 2 as we know 1 is a factor, and less than as numberToCheck is a factor for (int i = 2; i < numberToCheck; i++) { if (numberToCheck % i == 0) factorCount++; } return factorCount; }

    Read the article

  • A Plot Graph .NET WindowsForm Component Free

    - by user255946
    Hello All, I'm searching for a plot .NET component to plot a 2D line chart, given an array of data. It will be used with WindowsForm (C#) and It will be very helpful if it could be freeware. It is for a scientific application. This is my first asked question in stackoverflow, and excuse me for my terrible English written.

    Read the article

  • Better mode for do a select with group by

    - by Luca Romagnoli
    Hi i've wrote a query that works: SELECT `comments`.* FROM `comments` RIGHT JOIN (SELECT MAX( id ) AS id, core_id, topic_id FROM comments GROUP BY core_id, topic_id order by id desc) comm ON comm.id = comments.id LIMIT 10 I want know if is possible and how rewrite it for get better performance. thanks

    Read the article

  • How would you calculate all possible permutations of 0 through N iteratively?

    - by Bob Aman
    I need to calculate permutations iteratively. The method signature looks like: int[][] permute(int n) For n = 3 for example, the return value would be: [[0,1,2], [0,2,1], [1,0,2], [1,2,0], [2,0,1], [2,1,0]] How would you go about doing this iteratively in the most efficient way possible? I can do this recursively, but I'm interested in seeing lots of alternate ways to doing it iteratively.

    Read the article

  • Special simple random number generator

    - by psihodelia
    How to create a function, which on every call generates a random integer number? This number must be most random as possible (according to uniform distribution). It is only allowed to use one static variable and at most 3 elementary steps, where each step consists of only one basic arithmetic operation of arity 1 or 2. Example: int myrandom(void){ static int x; x = some_step1; x = some_step2; x = some_step3; return x; } Basic arithmetic operations are +,-,%,and, not, xor, or, left shift, right shift, multiplication and division. Of course, no rand(), random() or similar staff is allowed.

    Read the article

  • Symbian textshell mode border problem

    - by MostafaEweda
    I edited epoc.ini added textshell in the first line. When I run from the emulator: NO window border is shown when I run. It prints directly to the screen, so I can't interact with it and the console window is isolated from the emulator window I'm runninbg Windows 7 with Screen Resolution 1280*780

    Read the article

  • Fast ceiling of an integer division in C / C++

    - by andand
    Given integer values x and y, C and C++ returns as the quotient q = x/y the floor of the floating point valued equivalent. I'm interestd in a method of returning the ceiling instead? For example, ceil(10/5) = 2 and ceil(11/5) = 3. The obvious approach involves something like: q = x / y; if (q * y < x) ++q; This requires an extra comparison and multiplication; and other methods I've seen (used in fact) involve casting as a float or double. Is there a more direct method that avoids the additional multiplication (or a second division) and branch, and that also avoids casting as a floating point number?

    Read the article

< Previous Page | 85 86 87 88 89 90 91 92 93 94 95 96  | Next Page >