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  • Segmentation fault in my C program

    - by user233542
    I don't understand why this would give me a seg fault. Any ideas? This is the function that returns the signal to stop the program (plus the other function that is called within this): double bisect(double A0,double A1,double Sol[N],double tol,double c) { double Amid,shot; while (A1-A0 > tol) { Amid = 0.5*(A0+A1); shot = shoot(Sol, Amid, c); if (shot==2.*Pi) { return Amid; } if (shot > 2.*Pi){ A1 = Amid; } else if (shot < 2.*Pi){ A0 = Amid; } } return 0.5*(A1+A0); } double shoot(double Sol[N],double A,double c) { int i,j; /*Initial Conditions*/ for (i=0;i<buff;i++) { Sol[i] = 0.; } for (i=buff+l;i<N;i++) { Sol[i] = 2.*Pi; } Sol[buff]= 0; Sol[buff+1]= A*exp(sqrt(1+3*c)*dx); for (i=buff+2;i<buff+l;i++) { Sol[i] = (dx*dx)*( sin(Sol[i-1]) + c*sin(3.*(Sol[i-1])) ) - Sol[i-2] + 2.*Sol[i-1]; } return Sol[i-1]; } The values buff, l, N are defined using a #define statement. l = 401, buff = 50, N = 2000 Here is the full code: #include <stdio.h> #include <stdlib.h> #include <math.h> #define w 10 /*characteristic width of a soliton*/ #define dx 0.05 /*distance between lattice sites*/ #define s (2*w)/dx /*size of soliton shape*/ #define l (int)(s+1) /*array length for soliton*/ #define N (int)2000 /*length of field array--lattice sites*/ #define Pi (double)4*atan(1) #define buff (int)50 double shoot(double Sol[N],double A,double c); double bisect(double A0,double A1,double Sol[N],double tol,double c); void super_pos(double antiSol[N],double Sol[N],double phi[][N]); void vel_ver(double phi[][N],double v,double c,int tsteps,double dt); int main(int argc, char **argv) { double c,Sol[N],antiSol[N],A,A0,A1,tol,v,dt; int tsteps,i; FILE *fp1,*fp2,*fp3; fp1 = fopen("soliton.dat","w"); fp2 = fopen("final-phi.dat","w"); fp3 = fopen("energy.dat","w"); printf("Please input the number of time steps:"); scanf("%d",&tsteps); printf("Also, enter the time step size:"); scanf("%lf",&dt); do{ printf("Please input the parameter c in the interval [-1/3,1]:"); scanf("%lf",&c);} while(c < (-1./3.) || c > 1.); printf("Please input the inital speed of eiter soliton:"); scanf("%lf",&v); double phi[tsteps+1][N]; tol = 0.0000001; A0 = 0.; A1 = 2.*Pi; A = bisect(A0,A1,Sol,tol,c); shoot(Sol,A,c); for (i=0;i<N;i++) { fprintf(fp1,"%d\t",i); fprintf(fp1,"%lf\n",Sol[i]); } fclose(fp1); super_pos(antiSol,Sol,phi); /*vel_ver(phi,v,c,tsteps,dt); for (i=0;i<N;i++){ fprintf(fp2,"%d\t",i); fprintf(fp2,"%lf\n",phi[tsteps][i]); }*/ } double shoot(double Sol[N],double A,double c) { int i,j; /*Initial Conditions*/ for (i=0;i<buff;i++) { Sol[i] = 0.; } for (i=buff+l;i<N;i++) { Sol[i] = 2.*Pi; } Sol[buff]= 0; Sol[buff+1]= A*exp(sqrt(1+3*c)*dx); for (i=buff+2;i<buff+l;i++) { Sol[i] = (dx*dx)*( sin(Sol[i-1]) + c*sin(3.*(Sol[i-1])) ) - Sol[i-2] + 2.*Sol[i-1]; } return Sol[i-1]; } double bisect(double A0,double A1,double Sol[N],double tol,double c) { double Amid,shot; while (A1-A0 > tol) { Amid = 0.5*(A0+A1); shot = shoot(Sol, Amid, c); if (shot==2.*Pi) { return Amid; } if (shot > 2.*Pi){ A1 = Amid; } else if (shot < 2.*Pi){ A0 = Amid; } } return 0.5*(A1+A0); } void super_pos(double antiSol[N],double Sol[N],double phi[][N]) { int i; /*for (i=0;i<N;i++) { phi[i]=0; } for (i=buffer+s;i<1950-s;i++) { phi[i]=2*Pi; }*/ for (i=0;i<N;i++) { antiSol[i] = Sol[N-i]; } /*for (i=0;i<s+1;i++) { phi[buffer+j] = Sol[j]; phi[1549+j] = antiSol[j]; }*/ for (i=0;i<N;i++) { phi[0][i] = antiSol[i] + Sol[i] - 2.*Pi; } } /* This funciton will set the 2nd input array to the derivative at the time t, for all points x in the lattice */ void deriv2(double phi[][N],double DphiDx2[][N],int t) { //double SolDer2[s+1]; int x; for (x=0;x<N;x++) { DphiDx2[t][x] = (phi[buff+x+1][t] + phi[buff+x-1][t] - 2.*phi[x][t])/(dx*dx); } /*for (i=0;i<N;i++) { ptr[i] = &SolDer2[i]; }*/ //return DphiDx2[x]; } void vel_ver(double phi[][N],double v,double c,int tsteps,double dt) { int t,x; double d1,d2,dp,DphiDx1[tsteps+1][N],DphiDx2[tsteps+1][N],dpdt[tsteps+1][N],p[tsteps+1][N]; for (t=0;t<tsteps;t++){ if (t==0){ for (x=0;x<N;x++){//inital conditions deriv2(phi,DphiDx2,t); dpdt[t][x] = DphiDx2[t][x] - sin(phi[t][x]) - sin(3.*phi[t][x]); DphiDx1[t][x] = (phi[t][x+1] - phi[t][x])/dx; p[t][x] = -v*DphiDx1[t][x]; } } for (x=0;x<N;x++){//velocity-verlet phi[t+1][x] = phi[t][x] + dt*p[t][x] + (dt*dt/2)*dpdt[t][x]; p[t+1][x] = p[t][x] + (dt/2)*dpdt[t][x]; deriv2(phi,DphiDx2,t+1); dpdt[t][x] = DphiDx2[t][x] - sin(phi[t+1][x]) - sin(3.*phi[t+1][x]); p[t+1][x] += (dt/2)*dpdt[t+1][x]; } } } So, this really isn't due to my overwriting the end of the Sol array. I've commented out both functions that I suspected of causing the problem (bisect or shoot) and inserted a print function. Two things happen. When I have code like below: double A,Pi,B,c; c=0; Pi = 4.*atan(1.); A = Pi; B = 1./4.; printf("%lf",B); B = shoot(Sol,A,c); printf("%lf",B); I get a segfault from the function, shoot. However, if I take away the shoot function so that I have: double A,Pi,B,c; c=0; Pi = 4.*atan(1.); A = Pi; B = 1./4.; printf("%lf",B); it gives me a segfault at the printf... Why!?

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  • VU meter implementaion in iphone

    - by Sreelal
    Hi, I am developing an aplication for iphone which records audio and save that audio file .I need to create a UI similar to that in Voice Memo app with VU meter .I implemented codes to record audio,but i have no idea about VU meter implementation.Looking forward for a reply ......Thanks in advance

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  • matlab's phi symbol

    - by ldigas
    Not that significant, but annoying to no end. Why does matlab has no small phi (\varphi) symbol ? It has pretty much all other symbols LaTeX offers, but not this one. Why ? I may be wrong of course, in which case would be delighted if someone could prove me wrong...

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  • Microsoft va renouveler l'intégralité de sa gamme en un an, « du jamais vu » pour le PDG de la filiale française de l'éditeur

    Microsoft va renouveler l'intégralité de sa gamme en un an « Du jamais vu » pour le PDG de l'éditeur « Du jamais vu ». Voilà comment Alain Crozier, nouveau PDG de Microsoft France depuis début juillet, prévoit l'année 2012/2013. Il est vrai que sa feuille de route est assez impressionnante. Du développement au grand public, c'est « l'intégralité de la gamme de Microsoft qui sera renouvelée » en seulement 12 mois. « Microsoft entre dans une nouvelle ère » a ainsi lancé le dirigeant aux quelques centaines de personnes invitées pour la conférence de rentrée de Microsoft France. Si l'on fait la liste, l'année 2012/2013 verra trois nouveaux OS -

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  • LaTex, align alignment characters between align blocks

    - by ccook
    I would like to align two alignment characters between two align blocks so that I can have some text in the middle of a derivation with equations maintaining the horizontal alignment. For example the following excerpt of latex using align \begin{align*} \frac{\delta \phi}{\delta x_1} = {} &\frac{9}{8}\frac{\delta_1\phi}{\delta_1x_1}-\frac{1}{8}\frac{\delta_3\phi}{\delta_3x_1} \\ & \frac{9}{8}\frac{1}{h_1}\left[\phi(x_1+h_1/2)-\phi(x_i-h_1/2)\right]-\frac{1}{8}\frac{1}{3h_1}\left[\phi(x_i+3h_1/2)-\phi(x_1-3h_1/2)\right] \end{align*} some text in the middle \begin{align*} & \frac{9}{8}\frac{1}{h_1}\left[\phi(x_1+h_1/2)-\phi(x_i-h_1/2)\right]-\frac{1}{8}\frac{1}{3h_1}\left[\phi(x_i+3h_1/2)-\phi(x_1-3h_1/2)\right] \end{align*} Ideally I would like the left of the equation in the second block to line up with that of the second equation in the first block. I could do a workaround by not having text in the middle, however, I would like this functionality. EDIT I would like to have a good amount of text between. Say three to four lines that line up as normal paragraphs. Adding text in the alignment block is the workaround I poorly alluded to.

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  • Microsoft se défend des accusations de Google et dit "apprendre de ses consommateurs", Google persiste et dis n'avoir "jamais rien vu de pareil"

    Microsoft se défend des accusations de Google et dit "apprendre de ses consommateurs", Google persiste et dis n'avoir "jamais rien vu de pareil" Mise à jour du 02.02.2011 par Katleen Il y a quelques heures, de hauts responsables des moteurs de recherche en ligne étaient réunis lors d'une table ronde. D'un côté, Matt Cutts (Google) et de l'autre, Harry Shum (Bing). Le sujet des accusations de Mountain View portées hier envers Microsoft (lire news précédente) a évidement été abordé, et pas vraiment dans le calme. Il faut savoir que dans la nuit (heure française), Redmond avait publié un démenti assurant que jamais les résultats de son concurrents n'avaient été...

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  • OpenGL texture on sphere

    - by Cilenco
    I want to create a rolling, textured ball in OpenGL ES 1.0 for Android. With this function I can create a sphere: public Ball(GL10 gl, float radius) { ByteBuffer bb = ByteBuffer.allocateDirect(40000); bb.order(ByteOrder.nativeOrder()); sphereVertex = bb.asFloatBuffer(); points = build(); } private int build() { double dTheta = STEP * Math.PI / 180; double dPhi = dTheta; int points = 0; for(double phi = -(Math.PI/2); phi <= Math.PI/2; phi+=dPhi) { for(double theta = 0.0; theta <= (Math.PI * 2); theta+=dTheta) { sphereVertex.put((float) (raduis * Math.sin(phi) * Math.cos(theta))); sphereVertex.put((float) (raduis * Math.sin(phi) * Math.sin(theta))); sphereVertex.put((float) (raduis * Math.cos(phi))); points++; } } sphereVertex.position(0); return points; } public void draw() { texture.bind(); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, sphereVertex); gl.glDrawArrays(GL10.GL_TRIANGLE_FAN, 0, points); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); } My problem now is that I want to use this texture for the sphere but then only a black ball is created (of course because the top right corner s black). I use this texture coordinates because I want to use the whole texture: 0|0 0|1 1|1 1|0 That's what I learned from texturing a triangle. Is that incorrect if I want to use it with a sphere? What do I have to do to use the texture correctly?

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  • Arbitrary Rotation about a Sphere

    - by Der
    I'm coding a mechanic which allows a user to move around the surface of a sphere. The position on the sphere is currently stored as theta and phi, where theta is the angle between the z-axis and the xz projection of the current position (i.e. rotation about the y axis), and phi is the angle from the y-axis to the position. I explained that poorly, but it is essentially theta = yaw, phi = pitch Vector3 position = new Vector3(0,0,1); position.X = (float)Math.Sin(phi) * (float)Math.Sin(theta); position.Y = (float)Math.Sin(phi) * (float)Math.Cos(theta); position.Z = (float)Math.Cos(phi); position *= r; I believe this is accurate, however I could be wrong. I need to be able to move in an arbitrary pseudo two dimensional direction around the surface of a sphere at the origin of world space with radius r. For example, holding W should move around the sphere in an upwards direction relative to the orientation of the player. I believe I should be using a Quaternion to represent the position/orientation on the sphere, but I can't think of the correct way of doing it. Spherical geometry is not my strong suit. Essentially, I need to fill the following block: public void Move(Direction dir) { switch (dir) { case Direction.Left: // update quaternion to rotate left break; case Direction.Right: // update quaternion to rotate right break; case Direction.Up: // update quaternion to rotate upward break; case Direction.Down: // update quaternion to rotate downward break; } }

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  • Different results when applying function to equal values

    - by Johannes Stiehler
    I'm just digging a bit into Haskell and I started by trying to compute the Phi-Coefficient of two words in a text. However, I ran into some very strange behaviour that I cannot explain. After stripping everything down, I ended up with this code to reproduce the problem: let sumTup = (sumTuples°concat) frequencyLists let sumFixTup = (138, 136, 17, 204) putStrLn (show ((138, 136, 17, 204) == sumTup)) putStrLn (show (phi sumTup)) putStrLn (show (phi sumFixTup)) This outputs: True NaN 0.4574206676616167 So although the sumTupand sumFixTup show as equal, they behave differently when passed to phi. The definition of phi is: phi (a, b, c, d) = let dividend = fromIntegral(a * d - b * c) divisor = sqrt(fromIntegral((a + b) * (c + d) * (a + c) * (b + d))) in dividend / divisor

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  • Off center projection

    - by N0xus
    I'm trying to implement the code that was freely given by a very kind developer at the following link: http://forum.unity3d.com/threads/142383-Code-sample-Off-Center-Projection-Code-for-VR-CAVE-or-just-for-fun Right now, all I'm trying to do is bring it in on one camera, but I have a few issues. My class, looks as follows: using UnityEngine; using System.Collections; public class PerspectiveOffCenter : MonoBehaviour { // Use this for initialization void Start () { } // Update is called once per frame void Update () { } public static Matrix4x4 GeneralizedPerspectiveProjection(Vector3 pa, Vector3 pb, Vector3 pc, Vector3 pe, float near, float far) { Vector3 va, vb, vc; Vector3 vr, vu, vn; float left, right, bottom, top, eyedistance; Matrix4x4 transformMatrix; Matrix4x4 projectionM; Matrix4x4 eyeTranslateM; Matrix4x4 finalProjection; ///Calculate the orthonormal for the screen (the screen coordinate system vr = pb - pa; vr.Normalize(); vu = pc - pa; vu.Normalize(); vn = Vector3.Cross(vr, vu); vn.Normalize(); //Calculate the vector from eye (pe) to screen corners (pa, pb, pc) va = pa-pe; vb = pb-pe; vc = pc-pe; //Get the distance;; from the eye to the screen plane eyedistance = -(Vector3.Dot(va, vn)); //Get the varaibles for the off center projection left = (Vector3.Dot(vr, va)*near)/eyedistance; right = (Vector3.Dot(vr, vb)*near)/eyedistance; bottom = (Vector3.Dot(vu, va)*near)/eyedistance; top = (Vector3.Dot(vu, vc)*near)/eyedistance; //Get this projection projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); //Fill in the transform matrix transformMatrix = new Matrix4x4(); transformMatrix[0, 0] = vr.x; transformMatrix[0, 1] = vr.y; transformMatrix[0, 2] = vr.z; transformMatrix[0, 3] = 0; transformMatrix[1, 0] = vu.x; transformMatrix[1, 1] = vu.y; transformMatrix[1, 2] = vu.z; transformMatrix[1, 3] = 0; transformMatrix[2, 0] = vn.x; transformMatrix[2, 1] = vn.y; transformMatrix[2, 2] = vn.z; transformMatrix[2, 3] = 0; transformMatrix[3, 0] = 0; transformMatrix[3, 1] = 0; transformMatrix[3, 2] = 0; transformMatrix[3, 3] = 1; //Now for the eye transform eyeTranslateM = new Matrix4x4(); eyeTranslateM[0, 0] = 1; eyeTranslateM[0, 1] = 0; eyeTranslateM[0, 2] = 0; eyeTranslateM[0, 3] = -pe.x; eyeTranslateM[1, 0] = 0; eyeTranslateM[1, 1] = 1; eyeTranslateM[1, 2] = 0; eyeTranslateM[1, 3] = -pe.y; eyeTranslateM[2, 0] = 0; eyeTranslateM[2, 1] = 0; eyeTranslateM[2, 2] = 1; eyeTranslateM[2, 3] = -pe.z; eyeTranslateM[3, 0] = 0; eyeTranslateM[3, 1] = 0; eyeTranslateM[3, 2] = 0; eyeTranslateM[3, 3] = 1f; //Multiply all together finalProjection = new Matrix4x4(); finalProjection = Matrix4x4.identity * projectionM*transformMatrix*eyeTranslateM; //finally return return finalProjection; } // Update is called once per frame public void FixedUpdate () { Camera cam = camera; //calculate projection Matrix4x4 genProjection = GeneralizedPerspectiveProjection( new Vector3(0,1,0), new Vector3(1,1,0), new Vector3(0,0,0), new Vector3(0,0,0), cam.nearClipPlane, cam.farClipPlane); //(BottomLeftCorner, BottomRightCorner, TopLeftCorner, trackerPosition, cam.nearClipPlane, cam.farClipPlane); cam.projectionMatrix = genProjection; } } My error lies in projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); The debugger states: Expression denotes a `type', where a 'variable', 'value' or 'method group' was expected. Thus, I changed the line to read: projectionM = new PerspectiveOffCenter(left, right, bottom, top, near, far); But then the error is changed to: The type 'PerspectiveOffCenter' does not contain a constructor that takes '6' arguments. For reasons that are obvious. So, finally, I changed the line to read: projectionM = new GeneralizedPerspectiveProjection(left, right, bottom, top, near, far); And the error I get is: is a 'method' but a 'type' was expected. With this last error, I'm not sure what it is I should do / missing. Can anyone see what it is that I'm missing to fix this error?

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  • What does this piece of code in C++ mean? [closed]

    - by user1838418
    const double pi = 3.141592653589793; const angle rightangle = pi/2; inline angle deg2rad(angle degree) { return degree * rightangle / 90.; } angle function1() { return rightangle * ( ((double) rand()) / ((double) RAND_MAX) - .5 ); } bool setmargin(angle theta, angle phi, angle margin) { return (theta > phi-margin && theta < phi+margin); } Please help me. I'm new to C++

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  • MATLAB intersection of 2 surfaces

    - by caglarozdag
    Hi everyone, I consider myself a beginner in MATLAB so bear with me if the answer to my question is an obvious one. Phi=0:pi/100:2*pi; Theta=0:pi/100:2*pi; [PHI,THETA]=meshgrid(Phi,Theta); R=(1 + cos(PHI).*cos(PHI)).*(1 + cos(THETA).*cos(THETA)); [X,Y,Z]=sph2cart(THETA,PHI,R); surf(X,Y,Z); %display hold on; x1=-4:.1:4; [X1,Y1] = meshgrid(x1); a=1.8; b=0; c=3; d=0; Z1=(d- a * X1 - b * Y1)/c; shading flat; surf(X1,Y1,Z1); I have written this code which plots a 3d cartesian plot of a plane intersecting a peanut shaped object at an angle. I need to get the intersection of these on 2D (going to be the outline of a peanut, but a bit skewed since the intersection happens at an angle), but don't know how. Thanks

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  • Project Euler problem 214, How can i make it more efficient?

    - by Once
    I am becoming more and more addicted to the Project Euler problems. However since one week I am stuck with the #214. Here is a short version of the problem: PHI() is Euler's totient function, i.e. for any given integer n, PHI(n)=numbers of k<=n for which gcd(k,n)=1. We can iterate PHI() to create a chain. For example starting from 18: PHI(18)=6 = PHI(6)=2 = PHI(2)=1. So starting from 18 we get a chain of length 4 (18,6,2,1) The problem is to calculate the sum of all primes less than 40e6 which generate a chain of length 25. I built a function that calculates the chain length of any number and I tested it for small values: it works well and fast. sum of all primes<=20 which generate a chain of length 4: 12 sum of all primes<=1000 which generate a chain of length 10: 39383 Unfortunately my algorithm doesn't scale well. When I apply it to the problem, it takes several hours to calculate... so I stop it because the Project Euler problems must be solved in less than one minute. I thought that my prime detection function might be slow so I fed the program with a list of primes <40e6 to avoid the primality test... The code runs now a little bit faster, but there is still no way to get a solution in less than a few hours (and I don't want this). So is there any "magic trick" that I am missing here ? I really don't understand how to be more efficient on this one... I am not asking for the solution, because fighting with optimization is all the fun of Project Euler. However, any small hint that could put me on the right track would be welcome. Thanks !

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  • Matlab code works with one version but not the other

    - by user1325655
    I have a code that works in Matlab version R2010a but shows errors in matlab R2008a. I am trying to implement a self organizing fuzzy neural network with extended kalman filter. I have the code running but it only works in matlab version R2010a. It doesn't work with other versions. Any help? Code attach function [ c, sigma , W_output ] = SOFNN( X, d, Kd ) %SOFNN Self-Organizing Fuzzy Neural Networks %Input Parameters % X(r,n) - rth traning data from nth observation % d(n) - the desired output of the network (must be a row vector) % Kd(r) - predefined distance threshold for the rth input %Output Parameters % c(IndexInputVariable,IndexNeuron) % sigma(IndexInputVariable,IndexNeuron) % W_output is a vector %Setting up Parameters for SOFNN SigmaZero=4; delta=0.12; threshold=0.1354; k_sigma=1.12; %For more accurate results uncomment the following %format long; %Implementation of a SOFNN model [size_R,size_N]=size(X); %size_R - the number of input variables c=[]; sigma=[]; W_output=[]; u=0; % the number of neurons in the structure Q=[]; O=[]; Psi=[]; for n=1:size_N x=X(:,n); if u==0 % No neuron in the structure? c=x; sigma=SigmaZero*ones(size_R,1); u=1; Psi=GetMePsi(X,c,sigma); [Q,O] = UpdateStructure(X,Psi,d); pT_n=GetMeGreatPsi(x,Psi(n,:))'; else [Q,O,pT_n] = UpdateStructureRecursively(X,Psi,Q,O,d,n); end; KeepSpinning=true; while KeepSpinning %Calculate the error and if-part criteria ae=abs(d(n)-pT_n*O); %approximation error [phi,~]=GetMePhi(x,c,sigma); [maxphi,maxindex]=max(phi); % maxindex refers to the neuron's index if ae>delta if maxphi<threshold %enlarge width [minsigma,minindex]=min(sigma(:,maxindex)); sigma(minindex,maxindex)=k_sigma*minsigma; Psi=GetMePsi(X,c,sigma); [Q,O] = UpdateStructure(X,Psi,d); pT_n=GetMeGreatPsi(x,Psi(n,:))'; else %Add a new neuron and update structure ctemp=[]; sigmatemp=[]; dist=0; for r=1:size_R dist=abs(x(r)-c(r,1)); distIndex=1; for j=2:u if abs(x(r)-c(r,j))<dist distIndex=j; dist=abs(x(r)-c(r,j)); end; end; if dist<=Kd(r) ctemp=[ctemp; c(r,distIndex)]; sigmatemp=[sigmatemp ; sigma(r,distIndex)]; else ctemp=[ctemp; x(r)]; sigmatemp=[sigmatemp ; dist]; end; end; c=[c ctemp]; sigma=[sigma sigmatemp]; Psi=GetMePsi(X,c,sigma); [Q,O] = UpdateStructure(X,Psi,d); KeepSpinning=false; u=u+1; end; else if maxphi<threshold %enlarge width [minsigma,minindex]=min(sigma(:,maxindex)); sigma(minindex,maxindex)=k_sigma*minsigma; Psi=GetMePsi(X,c,sigma); [Q,O] = UpdateStructure(X,Psi,d); pT_n=GetMeGreatPsi(x,Psi(n,:))'; else %Do nothing and exit the while KeepSpinning=false; end; end; end; end; W_output=O; end function [Q_next, O_next,pT_n] = UpdateStructureRecursively(X,Psi,Q,O,d,n) %O=O(t-1) O_next=O(t) p_n=GetMeGreatPsi(X(:,n),Psi(n,:)); pT_n=p_n'; ee=abs(d(n)-pT_n*O); %|e(t)| temp=1+pT_n*Q*p_n; ae=abs(ee/temp); if ee>=ae L=Q*p_n*(temp)^(-1); Q_next=(eye(length(Q))-L*pT_n)*Q; O_next=O + L*ee; else Q_next=eye(length(Q))*Q; O_next=O; end; end function [ Q , O ] = UpdateStructure(X,Psi,d) GreatPsiBig = GetMeGreatPsi(X,Psi); %M=u*(r+1) %n - the number of observations [M,~]=size(GreatPsiBig); %Others Ways of getting Q=[P^T(t)*P(t)]^-1 %************************************************************************** %opts.SYM = true; %Q = linsolve(GreatPsiBig*GreatPsiBig',eye(M),opts); % %Q = inv(GreatPsiBig*GreatPsiBig'); %Q = pinv(GreatPsiBig*GreatPsiBig'); %************************************************************************** Y=GreatPsiBig\eye(M); Q=GreatPsiBig'\Y; O=Q*GreatPsiBig*d'; end %This function works too with x % (X=X and Psi is a Matrix) - Gets you the whole GreatPsi % (X=x and Psi is the row related to x) - Gets you just the column related with the observation function [GreatPsi] = GetMeGreatPsi(X,Psi) %Psi - In a row you go through the neurons and in a column you go through number of %observations **** Psi(#obs,IndexNeuron) **** GreatPsi=[]; [N,U]=size(Psi); for n=1:N x=X(:,n); GreatPsiCol=[]; for u=1:U GreatPsiCol=[ GreatPsiCol ; Psi(n,u)*[1; x] ]; end; GreatPsi=[GreatPsi GreatPsiCol]; end; end function [phi, SumPhi]=GetMePhi(x,c,sigma) [r,u]=size(c); %u - the number of neurons in the structure %r - the number of input variables phi=[]; SumPhi=0; for j=1:u % moving through the neurons S=0; for i=1:r % moving through the input variables S = S + ((x(i) - c(i,j))^2) / (2*sigma(i,j)^2); end; phi = [phi exp(-S)]; SumPhi = SumPhi + phi(j); %phi(u)=exp(-S) end; end %This function works too with x, it will give you the row related to x function [Psi] = GetMePsi(X,c,sigma) [~,u]=size(c); [~,size_N]=size(X); %u - the number of neurons in the structure %size_N - the number of observations Psi=[]; for n=1:size_N [phi, SumPhi]=GetMePhi(X(:,n),c,sigma); PsiTemp=[]; for j=1:u %PsiTemp is a row vector ex: [1 2 3] PsiTemp(j)=phi(j)/SumPhi; end; Psi=[Psi; PsiTemp]; %Psi - In a row you go through the neurons and in a column you go through number of %observations **** Psi(#obs,IndexNeuron) **** end; end

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  • How do I make this nested for loop, testing sums of cubes, more efficient?

    - by Brian J. Fink
    I'm trying to iterate through all the combinations of pairs of positive long integers in Java and testing the sum of their cubes to discover if it's a Fibonacci number. I'm currently doing this by using the value of the outer loop variable as the inner loop's upper limit, with the effect being that the outer loop runs a little slower each time. Initially it appeared to run very quickly--I was up to 10 digits within minutes. But now after 2 full days of continuous execution, I'm only somewhere in the middle range of 15 digits. At this rate it may end up taking a whole year just to finish running this program. The code for the program is below: import java.lang.*; import java.math.*; public class FindFib { public static void main(String args[]) { long uLimit=9223372036854775807L; //long maximum value BigDecimal PHI=new BigDecimal(1D+Math.sqrt(5D)/2D); //Golden Ratio for(long a=1;a<=uLimit;a++) //Outer Loop, 1 to maximum for(long b=1;b<=a;b++) //Inner Loop, 1 to current outer { //Cube the numbers and add BigDecimal c=BigDecimal.valueOf(a).pow(3).add(BigDecimal.valueOf(b).pow(3)); System.out.print(c+" "); //Output result //Upper and lower limits of interval for Mobius test: [c*PHI-1/c,c*PHI+1/c] BigDecimal d=c.multiply(PHI).subtract(BigDecimal.ONE.divide(c,BigDecimal.ROUND_HALF_UP)), e=c.multiply(PHI).add(BigDecimal.ONE.divide(c,BigDecimal.ROUND_HALF_UP)); //Mobius test: if integer in interval (floor values unequal) Fibonacci number! if (d.toBigInteger().compareTo(e.toBigInteger())!=0) System.out.println(); //Line feed else System.out.print("\r"); //Carriage return instead } //Display final message System.out.println("\rDone. "); } } Now the use of BigDecimal and BigInteger was delibrate; I need them to get the necessary precision. Is there anything other than my variable types that I could change to gain better efficiency?

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  • error in encryption program

    - by Raja
    #include<iostream> #include<math.h> #include<string> using namespace std; int gcd(int n,int m) { if(m<=n && n%m ==0) return m; if(n<m) return gcd(m,n); else return gcd(m,n%m); } int REncryptText(char m) { int p = 11, q = 3; int e = 3; int n = p * q; int phi = (p - 1) * (q - 1); int check1 = gcd(e, p - 1); int check2 = gcd(e, q - 1); int check3 = gcd(e, phi); // // Compute d such that ed = 1 (mod phi) //i.e. compute d = e-1 mod phi = 3-1 mod 20 //i.e. find a value for d such that phi divides (ed-1) //i.e. find d such that 20 divides 3d-1. //Simple testing (d = 1, 2, ...) gives d = 7 // double d = Math.Pow(e, -1) % phi; int d = 7; // public key = (n,e) // (33,3) //private key = (n,d) //(33 ,7) double g = pow(m,e); int ciphertext = g %n; // Now say we want to encrypt the message m = 7, c = me mod n = 73 mod 33 = 343 mod 33 = 13. Hence the ciphertext c = 13. //double decrypt = Math.Pow(ciphertext, d) % n; return ciphertext; } int main() { char plaintext[80],str[80]; cout<<" enter the text you want to encrpt"; cin.get(plaintext,79); int l =strlen(plaintext); for ( int i =0 ; i<l ; i++) { char s = plaintext[i]; str[i]=REncryptText(s); } for ( int i =0 ; i<l ; i++) { cout<<"the encryption of string"<<endl; cout<<str[i]; } return 0; }

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  • Taking fixed direction on hemisphere and project to normal (openGL)

    - by Maik Xhani
    I am trying to perform sampling using hemisphere around a surface normal. I want to experiment with fixed directions (and maybe jitter slightly between frames). So I have those directions: vec3 sampleDirections[6] = {vec3(0.0f, 1.0f, 0.0f), vec3(0.0f, 0.5f, 0.866025f), vec3(0.823639f, 0.5f, 0.267617f), vec3(0.509037f, 0.5f, -0.700629f), vec3(-0.509037f, 0.5f, -0.700629), vec3(-0.823639f, 0.5f, 0.267617f)}; now I want the first direction to be projected on the normal and the others accordingly. I tried these 2 codes, both failing. This is what I used for random sampling (it doesn't seem to work well, the samples seem to be biased towards a certain direction) and I just used one of the fixed directions instead of s (here is the code of the random sample, when i used it with the fixed direction i didn't use theta and phi). vec3 CosWeightedRandomHemisphereDirection( vec3 n, float rand1, float rand2 ) float theta = acos(sqrt(1.0f-rand1)); float phi = 6.283185f * rand2; vec3 s = vec3(sin(theta) * cos(phi), sin(theta) * sin(phi), cos(theta)); vec3 v = normalize(cross(n,vec3(0.0072, 1.0, 0.0034))); vec3 u = cross(v, n); u = s.x*u; v = s.y*v; vec3 w = s.z*n; vec3 direction = u+v+w; return normalize(direction); } ** EDIT ** This is the new code vec3 FixedHemisphereDirection( vec3 n, vec3 sampleDir) { vec3 x; vec3 z; if(abs(n.x) < abs(n.y)){ if(abs(n.x) < abs(n.z)){ x = vec3(1.0f,0.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } }else{ if(abs(n.y) < abs(n.z)){ x = vec3(0.0f,1.0f,0.0f); }else{ x = vec3(0.0f,0.0f,1.0f); } } z = normalize(cross(x,n)); x = cross(n,z); mat3 M = mat3( x.x, n.x, z.x, x.y, n.y, z.y, x.z, n.z, z.z); return M*sampleDir; } So if my n = (0,0,1); and my sampleDir = (0,1,0); shouldn't the M*sampleDir be (0,0,1)? Cause that is what I was expecting.

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  • installing Oracle Database 10g XE Server in Ubuntu 11.04, "E: Unable to locate package oracle-xe"

    - by Nguyen Phi Vu
    I have read many posts for installing Oracle Database 10g XE Server in Ubuntu, such as this But I get an error: E: Unable to locate package oracle-xe when execute the command sudo apt-get install oracle-xe At the previous step (sudo apt-get update), it also notices that E: Some index files failed to download. They have been ignored, or old ones used instead. Did any one meet and solve this problem? I have searched for this problem but got no proper answer.

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  • Matlab: Why is '1' + 1 == 50? [migrated]

    - by phi
    Matlab has weak dynamic typing, which is what causes this weird behaviour. What I do not understand is what exactly happens, as this result really surprises me. Edit: To clarify, what I'm describing is clearly a result of Matlab storing chars in ASCII-format, which was also mentioned in the comments. I'm more interested in the way Matlab handles its variables, and specifically, how and when it assigns a type/tag to the values. Thanks. '1' is a 1-by-1 matrix of chars in matlab and '123' is a 1-by-3 matrix of chars. As expected, 1 returns a 1-by-1 double. Now if I enter '1' + 1 I get 50 as a 1-by-1 double, and if I enter '123' + 1 I get a 1-by-3 double [ 50 51 52 ] Furthermore, if I type 'a' + 1 the result is 98 in a 1-by-1 double. I assume this has to do with how Matlab stores char-variables in ascii form, but how exactly is it handling these? Are the data actually unityped and tagged, or how does it work? Thanks.

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  • How do I assign a number value to a non-numerical value in Excel

    - by Keyslinger
    Greetings I have an some survey responses with values like "VU" for "Very Unlikely" and "S" for Sometimes. Each survey response occupies a cell. For each cell containing a survey response, I want to fill another cell with a corresponding number. For example, for every cell containing "VU" I want to fill a corresponding cell with the number 1. How is this done?

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  • VBA + Polymorphism: Override worksheet functions from 3rd party

    - by phi
    my company makes extensive use of a data provider using a (closed source) VBA plugin. In principal, every query follows follows a certain structure: Fill one cell with a formula, where arguments to the formula specify the query the range of that formula is extended (not an arrray formula!) and cells below/right are filled with data For this to work, however, a user has to have a terminal program installed on the machine, as well as a com-plugin referenced in VBA/Excel. My Problem These Excelsheets are used and extended by multiple users, and not all of them have access to the data provider. While they can open the sheet, it will recalculate and the data will be gone. However, frequent recalculation is required. I would like every user to be able to use the sheets, without executing a very specific set of formulas. Attempts remove the reference on those computers where I do not have terminal access. This generates a NAME error i the cell containing the query (acceptable), but this query overrides parts of the data (not acceptable) If you allow the program to refresh, all data will be gone after a failed query Replace all formulas with the plain-text result in the respective cells (press a button and loop over every cell...). Obviously destroys any refresh-capabilities the querys offer for all subsequent users, so pretty bad, too. A theoretical idea, and I'm not sure how to implement it: Replace the functions offered by the plugin with something that will be called either first (and relay the query through to the original function, if thats available) or instead of the original function (by only deploying the solution on non-terminal machines), which just returns the original value. More specifically, if my query function is used like this: =GETALLDATA(Startdate, Enddate, Stockticker, etc) I would like to transparently swap the function behind the call. Do you see any hope, or am I lost? I appreciate your help. PS: Of course I'm talking about Bloomberg... Some additional points to clarify issues raise by Frank: The formula in the sheets may not be changed. This is mission-critical software, and its way too complex for any sane person to try and touch it. Only excel and VBA may be used (which is the reason for the previous point...) It would be sufficient to prevent execution of these few specific formulas/functions on a specific machine for all excel sheets to come This looks more and more like a problem for stackoverflow ;-)

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  • Meaning of tcp_delack_min

    - by Phi
    Hi, the current Linux Kernel (e.g. 2.6.36) uses Delayed Acknowledgments (delack). In /include/net/tcp.h it says: define TCP_DELACK_MIN ((unsigned)(HZ/25)) So, for a Kernel using a HZ value of 1000, an ACK should be delayed by a minimum of 40 ms. However, RFC 2581 says a TCP implementation should acknowledge every second full sized segment without further delay. Does anybody know whether the Linux Kernel follows that 'should' or whether the TCP_DELACK_MIN value means that even after a full sized segment was received, the ACK continues to be delayed until 40 ms have passed?

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  • La publicité plus efficace sur l'iAd d'Apple que sur la télévision ? Oui d'après une étude de Nielsen

    La publicité plus efficace sur l'iAd d'Apple que sur la télévision ? Oui d'après une étude de Nielsen Les annonces sur iAd (la plate-forme de publicité mobile d'Apple) seraient deux fois plus efficaces que celles diffusées sur les écrans de télévision. C'est en tout cas la conclusion d'une étude menée par le cabinet Nielsen sur une campagne publicitaire de cinq semaines pour les produits de la firme Campbell. L'étude aurait constaté que les personnes ayant vu l'annonce sur iAd étaient deux fois plus susceptibles de se rappeler le produit que celles qui l'ont vu sur un écran de télévision. En outre, cinq fois plus de personnes auraient pris contact avec la branche iAd de Camp...

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  • How to right align the search box [closed]

    - by Hai Vu
    I am a newbie in HTML, CSS. I am working on a simple web app using CherryPy and ran into the following problem. The page I serve has a h1 title and a form for search box. I would like to lay them out on the same line with the h1 left aligned and the form right align, like this: My H1 Title here [ ] Search But instead, I got them in two separate lines: My H1 Title here [ ] Search My HTML code: <span class="header"> <h1>${pagetitle}</h1> </span> <span class="searchbox"> <form> <input type="text" name="searchterm"> <input type="submit" value="Search"> </form> </span> My CSS: span.header { text-align: left; } span.searchbox { text-align: right; } I have tried to work around using table, but was told to use span to do the right thing. I appreciate your help to set it right.

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