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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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  • 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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  • 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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  • Intersection of line and rectangle with maximum segment length

    - by Aarkan
    I have a vector represented by the slope m. Then there is rectangle (assume axis aligned), which is represented by top-left and bottom-right corner. Of course, there may be many lines with slope m and intersecting the given rectangle. The problem is to find out the line whose length of line intercept inside the rectangle is maximum among all such lines. i.e., if the line intersects rectangle at P1 and P2, then the problem is to find the equation of line for which length of P1P2 is maximum. I proceeded like this. Let the line is: y = m*x + c. Then find out the intersection with each side of rectangle and finding out the maxima for distance function between each pair of points. But it will only give me the length of line segment and there seem to be many corner cases to handle. Could anyone please suggest a better way to do this. Thanks in advance.

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  • Mathmatic errors in basic C++ program

    - by Heather
    I am working with a basic C++ program to determine the area and perimeter of a rectangle. My program works fine for whole numbers but falls apart when I use any number with a decimal. I get the impression that I am leaving something out, but since I'm a complete beginner, I have no idea what. Below is the source: #include <iostream> using namespace std; int main() { // Declared variables int length; // declares variable for length int width; // declares variable for width int area; // declares variable for area int perimeter; // declares variable for perimeter // Statements cout << "Enter the length and the width of the rectangle: "; // states what information to enter cin >> length >> width; // user input of length and width cout << endl; // closes the input area = length * width; // calculates area of rectangle perimeter = 2 * (length + width); //calculates perimeter of rectangle cout << "The area of the rectangle = " << area << " square units." <<endl; // displays the calculation of the area cout << "The perimeter of the rectangle = " << perimeter << " units." << endl; // displays the calculation of the perimeter system ("pause"); // REMOVE BEFORE RELEASE - testing purposes only return 0; }

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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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  • Are doubles faster than floats in c#?

    - by Trap
    I'm writing an application which reads large arrays of floats and performs some simple operations with them. I'm using floats because I thought it'd be faster than doubles, but after doing some research I've found that there's some confusion about this topic. Can anyone elaborate on this? Thanks.

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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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  • 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.

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  • 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

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  • Draw arrow on line

    - by Pete
    Hi, I have this code: CGPoint arrowMiddle = CGPointMake((arrowOne.x + arrowTo.x)/2, (arrowOne.y + arrowTo.y)/2); CGPoint arrowLeft = CGPointMake(arrowMiddle.x-40, arrowMiddle.y); CGPoint arrowRight = CGPointMake(arrowMiddle.x, arrowMiddle.y + 40); [arrowPath addLineToScreenPoint:arrowLeft]; [arrowPath addLineToScreenPoint:arrowMiddle]; [arrowPath addLineToScreenPoint:arrowRight]; [[mapContents overlay] addSublayer:arrowPath]; [arrowPath release]; with this output: http://yfrog.com/edschermafbeelding2010032p What have i to add to get the left and right the at same degree of the line + 30°. If someone has the algorithm of drawing an arrow on a line, pleas give it. It doesn't matter what programming language it is... Thanks

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  • 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; }

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  • 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?

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  • Modify a given number to find the required sum?

    - by Gaurav
    A friend of mine sent me this question. I haven't really been able to come up with any kind of algorithm to solve this problem. You are provided with a no. say 123456789 and two operators * and +. Now without changing the sequence of the provided no. and using these operators as many times as you wish, evaluate the given value: eg: given value 2097 Solution: 1+2+345*6+7+8+9 Any ideas on how to approach problems like these?

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  • 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.

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  • 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.

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  • 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.

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  • 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?

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  • Is there a transformation matrix that can scale the x and/or y axis logarithmically?

    - by Dave M
    I'm using .net WPF geometry classes to graph waveforms. I've been using the matrix transformations to convert from the screen coordinate space to my coordinate space for the waveform. Everything works great and it's really simple to keep track of my window and scaling, etc. I can even use the inverse transform to calculate the mouse position in terms of the coordinate space. I use the built in Scaling and Translation classes and then a custom matrix to do the y-axis flipping (there's not a prefab matrix for flipping). I want to be able to graph these waveforms on a log scale as well (either x axis or y axis or both), but I'm not sure if this is even possible to do with a matrix transformation. Does anyone know if this is possible, and if it is, what is the matrix?

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  • 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.

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