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  • Moving from Linear Probing to Quadratic Probing (hash collisons)

    - by Nazgulled
    Hi, My current implementation of an Hash Table is using Linear Probing and now I want to move to Quadratic Probing (and later to chaining and maybe double hashing too). I've read a few articles, tutorials, wikipedia, etc... But I still don't know exactly what I should do. Linear Probing, basically, has a step of 1 and that's easy to do. When searching, inserting or removing an element from the Hash Table, I need to calculate an hash and for that I do this: index = hash_function(key) % table_size; Then, while searching, inserting or removing I loop through the table until I find a free bucket, like this: do { if(/* CHECK IF IT'S THE ELEMENT WE WANT */) { // FOUND ELEMENT return; } else { index = (index + 1) % table_size; } while(/* LOOP UNTIL IT'S NECESSARY */); As for Quadratic Probing, I think what I need to do is change how the "index" step size is calculated but that's what I don't understand how I should do it. I've seen various pieces of code, and all of them are somewhat different. Also, I've seen some implementations of Quadratic Probing where the hash function is changed to accommodated that (but not all of them). Is that change really needed or can I avoid modifying the hash function and still use Quadratic Probing? EDIT: After reading everything pointed out by Eli Bendersky below I think I got the general idea. Here's part of the code at http://eternallyconfuzzled.com/tuts/datastructures/jsw_tut_hashtable.aspx: 15 for ( step = 1; table->table[h] != EMPTY; step++ ) { 16 if ( compare ( key, table->table[h] ) == 0 ) 17 return 1; 18 19 /* Move forward by quadratically, wrap if necessary */ 20 h = ( h + ( step * step - step ) / 2 ) % table->size; 21 } There's 2 things I don't get... They say that quadratic probing is usually done using c(i)=i^2. However, in the code above, it's doing something more like c(i)=(i^2-i)/2 I was ready to implement this on my code but I would simply do: index = (index + (index^index)) % table_size; ...and not: index = (index + (index^index - index)/2) % table_size; If anything, I would do: index = (index + (index^index)/2) % table_size; ...cause I've seen other code examples diving by two. Although I don't understand why... 1) Why is it subtracting the step? 2) Why is it diving it by 2?

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  • moore's law and quadratic algorithm

    - by damon
    I was going thru a video (from coursera - by sedgewick) in which he argues that you cannot sustain Moore's law using a quadratic algorithm.He elaborates like this In year 197* you build a computer of power X ,and need to count N objects.This takes M days According to Moore's law,you have a computer of power 2X after 1.5 years.But now you have 2N objects to count. If you use a quadratic algorithm, In year 197*+1.5 ,it takes (4M)/2 = 2M days 4M because the algorithm is quadratic,and division by 2 because of doubling computer power. I find this hard to understand.I tried to work thru this as below To count N objects using comp=X , it takes M days. -> N/X = M After 1.5 yrs ,you need to count 2N objects using comp=2X -> 2N/(2X) -> N/X -> M days where do I go wrong? can someone please help me understand?

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  • Smoothing Small Data Set With Second Order Quadratic Curve

    - by Rev316
    I'm doing some specific signal analysis, and I am in need of a method that would smooth out a given bell-shaped distribution curve. A running average approach isn't producing the results I desire. I want to keep the min/max, and general shape of my fitted curve intact, but resolve the inconsistencies in sampling. In short: if given a set of data that models a simple quadratic curve, what statistical smoothing method would you recommend? If possible, please reference an implementation, library, or framework. Thanks SO!

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  • Quadratic Programming with Oracle R Enterprise

    - by Jeff Taylor-Oracle
         I wanted to use quadprog with ORE on a server running Oracle Solaris 11.2 on a Oracle SPARC T-4 server For background, see: Oracle SPARC T4-2 http://docs.oracle.com/cd/E23075_01/ Oracle Solaris 11.2 http://www.oracle.com/technetwork/server-storage/solaris11/overview/index.html quadprog: Functions to solve Quadratic Programming Problems http://cran.r-project.org/web/packages/quadprog/index.html Oracle R Enterprise 1.4 ("ORE") 1.4 http://www.oracle.com/technetwork/database/options/advanced-analytics/r-enterprise/ore-downloads-1502823.html Problem: path to Solaris Studio doesn't match my installation: $ ORE CMD INSTALL quadprog_1.5-5.tar.gz * installing to library \u2018/u01/app/oracle/product/12.1.0/dbhome_1/R/library\u2019 * installing *source* package \u2018quadprog\u2019 ... ** package \u2018quadprog\u2019 successfully unpacked and MD5 sums checked ** libs /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95 -m64   -PIC  -g  -c aind.f -o aind.o bash: /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95: No such file or directory *** Error code 1 make: Fatal error: Command failed for target `aind.o' ERROR: compilation failed for package \u2018quadprog\u2019 * removing \u2018/u01/app/oracle/product/12.1.0/dbhome_1/R/library/quadprog\u2019 $ ls -l /opt/solarisstudio12.3/bin/f95 lrwxrwxrwx   1 root     root          15 Aug 19 17:36 /opt/solarisstudio12.3/bin/f95 -> ../prod/bin/f90 Solution: a symbolic link: $ sudo mkdir -p /opt/SunProd/studio12u3 $ sudo ln -s /opt/solarisstudio12.3 /opt/SunProd/studio12u3/ Now, it is all good: $ ORE CMD INSTALL quadprog_1.5-5.tar.gz * installing to library \u2018/u01/app/oracle/product/12.1.0/dbhome_1/R/library\u2019 * installing *source* package \u2018quadprog\u2019 ... ** package \u2018quadprog\u2019 successfully unpacked and MD5 sums checked ** libs /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95 -m64   -PIC  -g  -c aind.f -o aind.o /opt/SunProd/studio12u3/solarisstudio12.3/bin/ cc -xc99 -m64 -I/usr/lib/64/R/include -DNDEBUG -KPIC  -xlibmieee  -c init.c -o init.o /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95 -m64  -PIC -g  -c -o solve.QP.compact.o solve.QP.compact.f /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95 -m64  -PIC -g  -c -o solve.QP.o solve.QP.f /opt/SunProd/studio12u3/solarisstudio12.3/bin/f95 -m64   -PIC  -g  -c util.f -o util.o /opt/SunProd/studio12u3/solarisstudio12.3/bin/ cc -xc99 -m64 -G -o quadprog.so aind.o init.o solve.QP.compact.o solve.QP.o util.o -xlic_lib=sunperf -lsunmath -lifai -lsunimath -lfai -lfai2 -lfsumai -lfprodai -lfminlai -lfmaxlai -lfminvai -lfmaxvai -lfui -lfsu -lsunmath -lmtsk -lm -lifai -lsunimath -lfai -lfai2 -lfsumai -lfprodai -lfminlai -lfmaxlai -lfminvai -lfmaxvai -lfui -lfsu -lsunmath -lmtsk -lm -L/usr/lib/64/R/lib -lR installing to /u01/app/oracle/product/12.1.0/dbhome_1/R/library/quadprog/libs ** R ** preparing package for lazy loading ** help *** installing help indices   converting help for package \u2018quadprog\u2019     finding HTML links ... done     solve.QP                                html      solve.QP.compact                        html  ** building package indices ** testing if installed package can be loaded * DONE (quadprog) ====== Here is an example from http://cran.r-project.org/web/packages/quadprog/quadprog.pdf > require(quadprog) > Dmat <- matrix(0,3,3) > diag(Dmat) <- 1 > dvec <- c(0,5,0) > Amat <- matrix(c(-4,-3,0,2,1,0,0,-2,1),3,3) > bvec <- c(-8,2,0) > solve.QP(Dmat,dvec,Amat,bvec=bvec) $solution [1] 0.4761905 1.0476190 2.0952381 $value [1] -2.380952 $unconstrained.solution [1] 0 5 0 $iterations [1] 3 0 $Lagrangian [1] 0.0000000 0.2380952 2.0952381 $iact [1] 3 2 Here, the standard example is modified to work with Oracle R Enterprise require(ORE) ore.connect("my-name", "my-sid", "my-host", "my-pass", 1521) ore.doEval(   function () {     require(quadprog)   } ) ore.doEval(   function () {     Dmat <- matrix(0,3,3)     diag(Dmat) <- 1     dvec <- c(0,5,0)     Amat <- matrix(c(-4,-3,0,2,1,0,0,-2,1),3,3)     bvec <- c(-8,2,0)    solve.QP(Dmat,dvec,Amat,bvec=bvec)   } ) $solution [1] 0.4761905 1.0476190 2.0952381 $value [1] -2.380952 $unconstrained.solution [1] 0 5 0 $iterations [1] 3 0 $Lagrangian [1] 0.0000000 0.2380952 2.0952381 $iact [1] 3 2 Now I can combine the quadprog compute algorithms with the Oracle R Enterprise Database engine functionality: Scale to large datasets Access to tables, views, and external tables in the database, as well as those accessible through database links Use SQL query parallel execution Use in-database statistical and data mining functionality

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  • Quadratic Bezier Curve: Calculate Tangent

    - by stefan.at.wpf
    I have a quadratic bezier curve and I want to calculate the slope of the tangent in a given point. For example, let it be the middlepoint of the quadratic bezier curve, therefore t=0.5 (please see the link below for a picture of this). I've calculated the first derivative of the formula for the quadratic bezier curve; however I get 400 as value for the slope, though it should be 0. Maybe I'm using the first derivative in a wrong way? I know I could also calculate the tangents using trigonometric functions; however I'd like to do it using the first derivative, shouldn't this be possible? Thanks for any hint! For clarification / please note: I'm interested in a general way to get the slope in a arbitrary given point on a quadratic bezier curve, not only to get the tangent in the start- and end point. A picture of my problem including the text above: http://cid-0432ee4cfe9c26a0.skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf Thank you very much for any hint!

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  • Help with hash tables and quadratic probing in Java

    - by user313458
    I really need help with inserting into a hash table. I'm just not totally getting it right now. Could someone explain quadratic and linear probing in layman's terms? public void insert(String key) { int homeLocation = 0; int location = 0; int count = 0; if (find(key).getLocation() == -1) // make sure key is not already in the table { //****** ADD YOUR CODE HERE FOR QUADRATIC PROBING ******** } } This is the code I'm working on. I'm not asking anyone to do it, I just really need help with learning the whole concept Any help would be greatly appreciated.

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  • python - returns incorrect positive #

    - by tekknolagi
    what i'm trying to do is write a quadratic equation solver but when the solution should be -1, as in quadratic(2, 4, 2) it returns 1 what am i doing wrong? #!/usr/bin/python import math def quadratic(a, b, c): #a = raw_input("What\'s your `a` value?\t") #b = raw_input("What\'s your `b` value?\t") #c = raw_input("What\'s your `c` value?\t") a, b, c = float(a), float(b), float(c) disc = (b*b)-(4*a*c) print "Discriminant is:\n" + str(disc) if disc = 0: root = math.sqrt(disc) top1 = b + root top2 = b - root sol1 = top1/(2*a) sol2 = top2/(2*a) if sol1 != sol2: print "Solution 1:\n" + str(sol1) + "\nSolution 2:\n" + str(sol2) if sol1 == sol2: print "One solution:\n" + str(sol1) else: print "No solution!" EDIT: it returns the following... import mathmodules mathmodules.quadratic(2, 4, 2) Discriminant is: 0.0 One solution: 1.0

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  • Why is my view controllers view not quadratic?

    - by mystify
    I created an UIViewController subclass, and figured out that the default implementation of -loadView in UIViewController will ignore my frame size settings in a strange way. To simplify it and to make sure it's really not the fault of my code, I did a clean test with a plain instance of UIViewController directly, rather than making a subclass. The result is the same. I try to make an exactly quadratic view of 320 x 320, but the view appears like 320 x 200. iPhone OS 3.0, please check this out: UIViewController *ts = [[UIViewController alloc] initWithNibName:nil bundle:nil]; ts.view.frame = CGRectMake(0.0f, 0.0f, 320.0f, 320.0f); ts.view.backgroundColor = [UIColor cyanColor]; [self.view addSubview:ts.view]; like you can see, I do this: 1) Create a UIViewController instance 2) Set the frame of the view to a quadratic dimension of 320 x 320 3) Give it a color, so I can see it 4) Added it as a subview. Now the part, that's even more strange: When I make my own implementation of -loadView, i.e. if I put this code in there like this: - (void)loadView { UIView *v = [[UIView alloc] initWithFrame:CGRectMake(0.0f, 0.0f, 320.0f, 320.0f)]; v.backgroundColor = [UIColor cyanColor]; self.view = v; [v release]; } then it looks right. Now lets think about that: In the first example, I do pretty much exactly the same, just that I let UIViewController create the view on it's own, and then take it over in order to change it's frame. Right? So why do I get this strange error? Right now I see no other way of messing around like that to correct this wrong behavior. I did not activate anything like clipsToBounds and there's no other code touching this.

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  • Revision, Quadratic Time

    - by stan
    I am not sure if you can post revision programming questions in here but i am stuck with some algorithms revision If an algorithm is quadratic it takes time proportional to the number of n^2 ? So if the slides say its almost 1/2 the square of n records is this the same as saying (n^2 * 0.5) Thanks

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  • Testing a quadratic equation

    - by user1201587
    I'm doing a code testing for a program that calculate the results for a quadratic equation I need to have test data for the following situation, when a is not zero and d positive there is two possibilities which are in the code below, I need to find an example for the first satiation when Math.abs(b / a - 200.0) < 1.0e-4 , all the values that I have tried, excute the second one caption= "Two roots"; if (Math.abs(b / a - 200.0) < 1.0e-4) { System.out.println("first one"); x1 = (-100.0 * (1.0 + Math.sqrt(1.0 - 1.0 / (10000.0 * a)))); x2 = (-100.0 * (1.0 - Math.sqrt(1.0 - 1.0 / (10000.0 * a)))); } else { System.out.println("secrst one"); x1 = (-b - Math.sqrt(d)) / (2.0 * a); x2 = (-b + Math.sqrt(d)) / (2.0 * a); } } }

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  • Drawing Quadratic Bezier circles with a given radius: how to determine control points

    - by Casey
    Just to clarify; the code below works, but I don't understand where the formula for the variable "controlRadius" comes from. I wrote this function by dissecting an example I found elsewhere, but I can't find any explanation and the original code comments were not able to be translated. Thanks in advance //returns an array of quadratic Bezier segments public static function generateCircularQuadraticBezierSegments(radius:Number, numControlPoints:uint, centerX:Number, centerY:Number):Array { var segments:Array = []; var arcLength:Number = 2 * Math.PI / numControlPoints; var controlRadius:Number; var segment:QuadraticBezierSegment; for (var i:int = 0; i < numControlPoints; i++) { var startX:Number = centerX + radius * Math.cos(arcLength * i); var startY:Number = centerY + radius * Math.sin(arcLength * i); //control radius formula //where does it come from, why does it work? controlRadius = radius / Math.cos(arcLength * .5); //the control point is plotted halfway between the arcLength and uses the control radius var controlX:Number = centerX + controlRadius * Math.cos(arcLength * (i + 1) - arcLength * .5); var controlY:Number = centerY + controlRadius * Math.sin(arcLength * (i + 1) - arcLength * .5); var endX:Number = centerX + radius * Math.cos(arcLength * (i + 1)); var endY:Number = centerY + radius * Math.sin(arcLength * (i + 1)); segment = new QuadraticBezierSegment(new Point(startX, startY), new Point(controlX, controlY), new Point(endX, endY)); segments.push(segment); } return segments; }

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  • Solving quadratic programming using R

    - by user702846
    I would like to solve the following quadratic programming equation using ipop function from kernlab : min 0.5*x'*H*x + f'*x subject to: A*x <= b Aeq*x = beq LB <= x <= UB where in our example H 3x3 matrix, f is 3x1, A is 2x3, b is 2x1, LB and UB are both 3x1. edit 1 My R code is : library(kernlab) H <- rbind(c(1,0,0),c(0,1,0),c(0,0,1)) f = rbind(0,0,0) A = rbind(c(1,1,1), c(-1,-1,-1)) b = rbind(4.26, -1.73) LB = rbind(0,0,0) UB = rbind(100,100,100) > ipop(f,H,A,b,LB,UB,0) Error in crossprod(r, q) : non-conformable arguments I know from matlab that is something like this : H = eye(3); f = [0,0,0]; nsamples=3; eps = (sqrt(nsamples)-1)/sqrt(nsamples); A=ones(1,nsamples); A(2,:)=-ones(1,nsamples); b=[nsamples*(eps+1); nsamples*(eps-1)]; Aeq = []; beq = []; LB = zeros(nsamples,1); UB = ones(nsamples,1).*1000; [beta,FVAL,EXITFLAG] = quadprog(H,f,A,b,Aeq,beq,LB,UB); and the answer is a vector of 3x1 equals to [0.57,0.57,0.57]; However when I try it on R, using ipop function from kernlab library ipop(f,H,A,b,LB,UB,0)) and I am facing Error in crossprod(r, q) : non-conformable arguments I appreciate any comment

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  • How do I draw part of parabola using iText ? Or how do I create quadratic bezier curves from cubic b

    - by drasto
    I need to draw a shape whose boundaries are parts of parabola (that is quadratic bezier curves) using iText. I have found only method for drawing cubic bezier curves in PdfContentByte class. So how do I draw quadratic bezier curves using iText ? One way would be to use method for cubic bezier curves. Is it possible to draw quadratic bezier curves as a cubic bezier curves (with 2 control points). I gues it is but I cannot make up the formula. If somebody states the formula tu "translate" cubic bezier curves to quadratic that would solve the problem. Any other ways to draw quadratic bezier(parts of parabola) curves in iText (and filled shapes made of them) is also the solution. Thanks

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  • Linear complexity and quadratic complexity

    - by jasonline
    I'm just not sure... If you have a code that can be executed in either of the following complexities: A sequence of O(n), like for example: two O(n) in sequence O(n²) The preferred version would be the one that can be executed in linear time. Would there be a time such that the sequence of O(n) would be too much and that O(n²) would be preferred? In other words, is the statement C x O(n) < O(n²) always true for any constant C? Why or why not? What are the factors that would affect the condition such that it would be better to choose the O(n²) complexity?

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  • Sparse quadratic program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an QP solver for large sparse matrices. Also, A is huge (around 2e6x2e6) but very sparse with <=4 elements per row. Any ideas/recommendations?

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  • Linear time and quadratic time

    - by jasonline
    I'm just not sure... If you have a code that can be executed in either of the following complexities: (1) A sequence of O(n), like for example: two O(n) in sequence (2) O(n²) The preferred version would be the one that can be executed in linear time. Would there be a time such that the sequence of O(n) would be too much and that O(n²) would be preferred? In other words, is the statement C x O(n) < O(n²) always true for any constant C? If no, what are the factors that would affect the condition such that it would be better to choose the O(n²) complexity?

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  • Attempting my first fortran 95 program, to solve quadratic eqn. Getting weird errors.

    - by Damon
    So, I'm attempting my first program in Fortran, trying to solve quadratic eqn. I have double and triple checked my code and don't see anything wrong. I keep getting "Invalid character in name at (1)" and "Unclassifiable statement at (1)" at various locations. Any help would be greatly appreciated... ! This program solves quadratic equations ! of the form ax^2 + bx + c = 0. ! Record: ! Name: Date: Notes: ! Damon Robles 4/3/10 Original Code PROGRAM quad_solv IMPLICIT NONE ! Variables REAL :: a, b, c REAL :: discrim, root1, root2, COMPLEX :: comp1, comp2 CHARACTER(len=1) :: correct ! Prompt user for coefficients. WRITE(*,*) "This program solves quadratic equations " WRITE(*,*) "of the form ax^2 + bx + c = 0. " WRITE(*,*) "Please enter the coefficients a, b, and " WRITE(*,*) "c, separated by commas:" READ(*,*) a, b, c WRITE(*,*) "Is this correct: a = ", a, " b = ", b WRITE(*,*) " c = ", c, " [Y/N]? " READ(*,*) correct IF correct = N STOP IF correct = Y THEN ! Definition discrim = b**2 - 4*a*c ! Calculations IF discrim > 0 THEN root1 = (-b + sqrt(discrim))/(2*a) root2 = (-b - sqrt(discrim))/(2*a) WRITE(*,*) "This equation has two real roots. " WRITE(*,*) "x1 = ", root1 WRITE(*,*) "x2 = ", root2 IF discrim = 0 THEN root1 = -b/(2*a) WRITE(*,*) "This equation has a double root. " WRITE(*,*) "x1 = ", root1 IF discrim < 0 THEN comp1 = (-b + sqrt(discrim))/(2*a) comp2 = (-b - sqrt(discrim))/(2*a) WRITE(*,*) "x1 = ", comp1 WRITE(*,*) "x2 = ", comp2 PROGRAM END quad_solv Thanks in advance!

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  • The most efficient method of drawing multiple quads in OpenGL

    - by CPatton
    I'm not very keen with OpenGL and I was wondering if someone could give me some insight on this. I'm a 'seasoned' programmer, I've read the redbook about VBOs and the like, but I was wondering from a more experienced person about the best/most efficient way of achieving this. I've been producing this 2d tile-based game engine to be used in several projects. I have a class called "ScreenObject" which is mainly composed of a Dictionary<Point, Tile> The Point key is to show where to render the Tile on the screen, and the Tile contains one or more textures to be drawn at that point. This ScreenObject is where the tiles will be modified, deleted, added, etc.. My original method of drawing the tiles in the testing I've done was to iterate through the ScreenObject and draw each quad at each location separately. From what I've read, this is a massive waste of resources. It wasn't horribly slow in the testing, but after I've completed the animation classes and effect classes, I'm sure it would be extremely slow. And one last thing, if you wouldn't mind.. As I said before, the Tile class can contain multiple textures to be drawn at the Point location on the screen. I recognize possibly two options for me here. Either add a quad at that location for each texture to be drawn, or, somehow.. use a multiple texture for the same quad (if it's possible). Even if each tile contained one texture only, that would be 64 quads to be drawn on the screen. Most of the tiles will contain 2-5 textures, so the number of total quads would increase dramatically with this method. Would it be feasible to add a quad for each new texture, or am I ignoring a better way to do this? Just need some help understanding this if you don't mind :) I've tried to be as concise as possible, and I'd greatly appreciate any responses.. and even some criticism. Programming is often a learning process and one who develops seems to never stops learning. Thanks for your time.

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  • Quadratic bezier curve: Y coordinate for a given X ?

    - by stefan.at.wpf
    Hello, I have a quadratic bezier curve and I need the Y coordinate of a point on the bezier curve for a given X coordinate. I know that in pure maths this can be easily done, but I'm wondering is there's a simple / another way for this in C# / WPF? Is it possible to get the single points used by C# / WPF for drawing the bezier curve and then maybe just loop them and compare the X coordinate of each point with the given X coordinate? BTW for the mathematical way it would be good to know which step for the parameter t of the bezier curve has been choosen by C# / WPF? Any chance to find this out? Probably t is just scaled by / steps for t are 1/(distance of P0 and P2) ? Thank you very much for any hint!

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  • Bracketing algorithm when root finding. Single root in "quadratic" function

    - by Ander Biguri
    I am trying to implement a root finding algorithm. I am using the hybrid Newton-Raphson algorithm found in numerical recipes that works pretty nicely. But I have a problem in bracketing the root. While implementing the root finding algorithm I realised that in several cases my functions have 1 real root and all the other imaginary (several of them, usually 6 or 9). The only root I am interested is in the real one so the problem is not there. The thing is that the function approaches the root like a cubic function, touching with the point the y=0 axis... Newton-Rapson method needs some brackets of different sign and all the bracketing methods I found don't work for this specific case. What can I do? It is pretty important to find that root in my program... EDIT: more problems: sometimes due to reaaaaaally small numerical errors, say a variation of 1e-6 in some value the "cubic" function does NOT have that real root, it is just imaginary with a neglectable imaginary part... (checked with matlab) EDIT 2: Much more information about the problem. Ok, I need root finding algorithm. Info I have: The root I need to find is between [0-1] , if there are more roots outside that part I am not interested in them. The root is real, there may be imaginary roots, but I don't want them. Probably all the rest of the roots will be imaginary The root may be double in that point, but I think that actually doesn't mater in numerical analysis problems I need to use the root finding algorithm several times during the overall calculations, but the function will always be a polynomial In one of the particular cases of the root finding, my polynomial will be similar to a quadratic function that touches Y=0 with the point. Example of a real case: The coefficient may not be 100% precise and that really slight imprecision may make the function not to touch the Y=0 axis. I cannot solve for this specific case because in other cases it may be that the polynomial is pretty normal and doesn't make any "strange" thing. The method I am actually using is NewtonRaphson hybrid, where if the derivative is really small it makes a bisection instead of NewRaph (found in numerical recipes). Matlab's answer to the function on the image: roots: 0.853553390593276 + 0.353553390593278i 0.853553390593276 - 0.353553390593278i 0.146446609406726 + 0.353553390593273i 0.146446609406726 - 0.353553390593273i 0.499999999999996 + 0.000000040142134i 0.499999999999996 - 0.000000040142134i The function is a real example I prepared where I know that the answer I want is 0.5 Note: I still haven't check completely some of the answers I you people have give me (Thank you!), I am just trying to give al the information I already have to complete the question.

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  • Quaddratic Bezier Curve: Calculate Tangent

    - by stefan.at.wpf
    Hello, I have have a quadratic bezier curve and I want to calculate the slope of the tangent in a given point. For example, let it be the middlepoint of the quadratic bezier curve, therefore t=0.5 (please see the link below for a picture of this). I've calculated the first derivation of the formula for the quadratic bezier curve, however I get 400 as value for the slope, though it should be 0. Maybe I'm using the first derivation in a wrong way? I know I could also calculate the tangents using trigonometric functions, however I'd like to do it using the first derivation, shouldn't this be possible? Thanks for any hint! For clarification / please note: I'm interested in a general way to get the slope in a arbitrary given point on a quadratic bezier curve, not only to get the tangent in the start- and end point. A picture of my problem including the text above: http://cid-0432ee4cfe9c26a0.skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf Thank you very much for any hint!

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  • Faster projected-norm (quadratic-form, metric-matrix...) style computations

    - by thekindamzkyoulike
    I need to perform lots of evaluations of the form X(:,i)' * A * X(:,i) i = 1...n where X(:,i) is a vector and A is a symmetric matrix. Ostensibly, I can either do this in a loop for i=1:n z(i) = X(:,i)' * A * X(:,i) end which is slow, or vectorise it as z = diag(X' * A * X) which wastes RAM unacceptably when X has a lot of columns. Currently I am compromising on Y = A * X for i=1:n z(i) = Y(:,i)' * X(:,i) end which is a little faster/lighter but still seems unsatisfactory. I was hoping there might be some matlab/scilab idiom or trick to achieve this result more efficiently?

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