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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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  • 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?

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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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  • Convert Decimal number into Fraction

    - by alankrita
    I am trying to convert decimal number into its fraction. Decimal numbers will be having a maximum 4 digits after the decimal place. example:- 12.34 = 1234/100 12.3456 = 123456/10000 my code :- #include <stdio.h> int main(void) { double a=12.34; int c=10000; double b=(a-floor(a))*c; int d=(int)floor(a)*c+(int)b; while(1) { if(d%10==0) { d=d/10; c=c/10; } else break; } printf("%d/%d",d,c); return 0; } but I am not getting correct output, Decimal numbers will be of double precision only.Please guide me what I should do.

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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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  • 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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  • 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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  • 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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  • Distributing players to tables

    - by IVlad
    Consider N = 4k players, k tables and a number of clans such that each member can belong to one clan. A clan can contain at most k players. We want to organize 3 rounds of a game such that, for each table that seats exactly 4 players, no 2 players sitting there are part of the same clan, and, for the later rounds, no 2 players sitting there have sat at the same table before. All players play all rounds. How can we do this efficiently if N can be about ~80 large? I thought of this: for each table T: repeat until 4 players have been seated at T: pick a random player X that is not currently seated anywhere if X has not sat at the same table as anyone currently at T AND X is not from the same clan as anyone currently at T seat X at T break I am not sure if this will always finish or if it can get stuck even if there is a valid assignment. Even if this works, is there a better way to do it?

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  • Separating merged array of arithmetic and geometric series

    - by user1814037
    My friend asked me an interseting question. Given an array of positive integers in increasing order. Seperate them in two series, an arithmetic sequence and geometric sequence. The given array is such that a solution do exist. The union of numbers of the two sequence must be the given array. Both series can have common elements i.e. series need not to be disjoint. The ratio of the geometric series can be fractional. Example: Given series : 2,4,6,8,10,12,25 AP: 2,4,6,8,10,12 GP: 4,10,25 I tried taking few examples but could not reach a general way. Even tried some graph implementation by introducing edges if they follow a particular sequence but could not reach solution.

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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?

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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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  • 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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  • Recursively determine average value

    - by theva
    I have to calculate an average value of my simulation. The simulation is ongoing and I want (for each iteration) to print the current average value. How do I do that? I tried the code below (in the loop), but I do not think that the right value is calculated... int average = 0; int newValue; // Continuously updated value. if(average == 0) { average = newValue; } average = (average + newValue)/2; I also taught about store each newValue in an array and for each iteration summarize the whole array and do the calculation. However, I don't think that's a good solution, because the loop is an infinity loop so I can't really determine the size of the array. There is also a possibility that I am thinking too much and that the code above is actually correct, but I don't think so...

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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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  • 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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  • Convert arbitrary length to a value between -1.0 a 1.0?

    - by TheDarkIn1978
    How can I convert a length into a value in the range -1.0 to 1.0? Example: my stage is 440px in length and accepts mouse events. I would like to click in the middle of the stage, and rather than an output of X = 220, I'd like it to be X = 0. Similarly, I'd like the real X = 0 to become X = -1.0 and the real X = 440 to become X = 1.0. I don't have access to the stage, so i can't simply center-register it, which would make this process a lot easier. Also, it's not possible to dynamically change the actual size of my stage, so I'm looking for a formula that will translate the mouse's real X coordinate of the stage to evenly fit within a range from -1 to 1.

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