Consider the following code.
Wx = zeros(N, N);
for ii = 1 : 1 : N
x_ref = X(ii); y_ref = Y(ii);
nghlst_Local = nghlst(ii, find(nghlst(ii, :))); Nl = length(nghlst_Local);
x_Local = X(nghlst_Local, 1); y_Local = Y(nghlst_Local, 1);
PhiU = ones(Nl+1, Nl+1); PhiU(end, end) = 0;
Phi = ones(Nl+1, Nl+1); Phi(end, end) = 0;
Bx = zeros(Nl+1,1);
for jj = 1 : 1 : Nl
for kk = 1 : 1 : Nl
rx = x_Local(jj,1) - x_Local(kk,1);
ry = y_Local(jj,1) - y_Local(kk,1);
PhiU(jj, kk) = (1 - U(1,1))) / sqrt(rx^2 + ry^2 + c^2);
end
rx = x_ref - x_Local(jj);
ry = y_ref - y_Local(jj);
Bx(jj, 1) = ( (Beta * pi * U(1,1)/(2*r_0*norm(U))) * cos( (pi/2) * (-rx * U(1,1) - ry * U(2,1)) / (r_0 * norm(U)) ) ) / sqrt(rx^2 + ry^2 + c^2) - rx * (1 - Beta * sin( (pi/2) * (-rx * U(1,1) - ry * U(2,1)) / (r_0 * norm(U)) ))/ (rx^2 + ry^2 + c^2)^(3/2);
end
invPhiU = inv(PhiU);
CX = Bx' * invPhiU; CX = CX (1, 1:end-1); Wx (ii, nghlst_Local) = CX;
end
I want to convert the first for loop into parfor loop. The rest of the code works fine, but the following assignment statement does not work when I change for to parfor.
Wx (ii, nghlst_Local) = CX;
I want to know what is this is wrong and how to remove such errors. Thank you.
Related
T = 0.2;
A = [1 T; 0 1];
B = [T^2 / 2 T];
H = [1 0];
G = [0 1]';
Q = 0.00005;
R = 0.006;
x1(1) = 0;
x2(1) = 0;
x1e(1) = 0;
x2e(1) = 0;
xest = [x1e(1) x2e(1)]';
x1p(1) = 0;
x2p(1) = 0;
PE = [R 0; 0 0];
PP = A * PE(1) * A' + Q;
for i= 1:25
if i < 10
u = 0.25;
else
u = 0;
end
x1(i+1) = x1(i) + T * x2(i) + (T^2 / 2) * u;
x2(i+1) = x2(i) + T * u + sqrt(Q) * randn;
y(i+1) = x1(i+1) + sqrt(R) * randn;
PP = A * PE * A' + G * Q * G';
K = PP * H' * inv(H * PP * H' + R);
PE = [eye(2) - K * H] * PP;
xpredict = A * xest + B * u;
xest = xpredict + K * (y(i+1) -H * xpredict);
x1e(i+1) = [1 0] * xest;
x2e(i+1) = [0 1] * xest;
end
Unable to perform assignment because the left and right sides have a different number of elements.
Error in rrrr (line 34)
x1e(i+1) = [1 0] * xest;
how i can solve the error
xpredict must be a 2x1 vector. To solve it, you need to transpose B in line 5 i.e., B = [T^2 / 2 T]',
Since from Newton's laws of motion with constant velocity, we have
I have an equation which goes like this:
Here, I_L(lambdap) is the modified bessel function. This and product with exponential function can be written in matlab as besseli(L,lambdap,1). "i" stands for square root of -1. I want to solve:
1+pt+it=0
where I have to vary 'k' and find values of 'w'. I had posted similar problem at mathematica stack exchange, but I couldn't solve the problem fully, though i have got a clue (please go through the comments at mathematica stack exchange site). I could not convert my equation to the code that has been posted in clue. Any help in this regards will be highly appreciated.
Thanks in advance...
I never attempted this before, but... is this returning a suitable result?
syms w k;
fun = 1 + pt(w,k) + it(w,k);
sol = vpasolve(fun == 0,w,k);
disp(sol.w);
disp(sol.k);
function res = pt(w,k)
eps_l0 = w / (1.22 * k);
lam_k = 0.25 * k^2;
res = sym('res',[5 1]);
res_off = 1;
for L = -2:2
gam = besseli(L,lam_k) * exp(-lam_k);
eps_z = (w - L) / (1.22 * k);
zeta = 1i * sqrt(pi()) * exp(-eps_z^2) * (1 + erfc(1i * eps_z));
res(res_off,:) = ((25000 * gam) / k^2) * (1 + (eps_l0 * zeta));
res_off = res_off + 1;
end
res = sum(res);
end
function res = it(w,k)
eps_l0 = (w - (0.86 * k)) / (3.46 * k);
lam_k = 0.03 * k^2;
res = sym('res',[5 1]);
res_off = 1;
for L = -2:2
gam = besseli(L,lam_k) * exp(-lam_k);
eps_z = (w - (8 * L) - (0.86 * k)) / (3.46 * k);
zeta = 1i * sqrt(pi()) * exp(-eps_z^2) * (1 + erfc(1i * eps_z));
res(res_off,:) = ((2000000 * gam) / k^2) * (1 + (eps_l0 * zeta));
res_off = res_off + 1;
end
res = sum(res);
end
EDIT
For numeric k and symbolic w:
syms w;
for k = -3:3
fun = 1 + pt(w,k) + it(w,k);
sol = vpasolve(fun == 0,w);
disp(sol.w);
end
I have seen that there was an interest in custom interpolation kernels for resize (MATLAB imresize with a custom interpolation kernel). Did anyone implemented the parametric Mitchell-Netravali kernel [1] that is used as default in ImageMagick and is willing to share the Matlab code? Thank you very much!
[1] http://developer.download.nvidia.com/books/HTML/gpugems/gpugems_ch24.html
// Mitchell Netravali Reconstruction Filter
// B = 0 C = 0 - Hermite B-Spline interpolator
// B = 1, C = 0 - cubic B-spline
// B = 0, C = 1/2 - Catmull-Rom spline
// B = 1/3, C = 1/3 - recommended
float MitchellNetravali(float x, float B, float C)
{
float ax = fabs(x);
if (ax < 1) {
return ((12 - 9 * B - 6 * C) * ax * ax * ax +
(-18 + 12 * B + 6 * C) * ax * ax + (6 - 2 * B)) / 6;
} else if ((ax >= 1) && (ax < 2)) {
return ((-B - 6 * C) * ax * ax * ax +
(6 * B + 30 * C) * ax * ax + (-12 * B - 48 * C) *
ax + (8 * B + 24 * C)) / 6;
} else {
return 0;
}
}
Here I got another approach with vectorization; according to my tests with upscaling (1000x1000 -> 3000x3000) this is faster than the standard bicubic even with a large Mitchell radius = 6:
function [outputs] = Mitchell_vect(x,M_B,M_C)
outputs= zeros(size(x,1),size(x,2));
ax = abs(x);
temp = ((12-9*M_B-6*M_C) .* ax.^3 + (-18+12*M_B+6*M_C) .* ax.^2 + (6-2*M_B))./6;
temp2 = ((-M_B-6*M_C) .* ax.^3 + (6*M_B+30*M_C) .* ax.^2 + (-12*M_B-48*M_C) .* ax + (8*M_B + 24*M_C))./6;
index = find(ax<1);
outputs(index)=temp(index);
index = find(ax>=1 & ax<2);
outputs(index)=temp2(index);
end
I got the following proposal for the Mitchel kernel called by imresize with the parameters B and C and a kernel radius using for-loops (and preallocation):
img_resize = imresize(img, [h w], {#(x)Mitchell(x,B,C),radius});
function [outputs] = Mitchell(x,B,C)
outputs= zeros(size(x,1),size(x,2));
for i = 1 : size(x,1)
for j = 1 : size(x,2)
ax = abs(x(i,j));
if ax < 1
outputs(i,j) = ((12-9*B-6*C) * ax^3 + (-18+12*B+6*C) * ax^2 + (6-2*B))/6;
elseif (ax >= 1) && (ax < 2)
outputs(i,j) = ((-B-6*C) * ax^3 + (6*B+30*C) * ax^2 + (-12*B-48*C) * ax + (8*B + 24*C))/6;
else
outputs(i,j) = 0;
end
end
end
end
I am right now stuck on a problem in matlab. What I have done is that I have an equation that is passed on into another function which works by the bisection-method.
But I have a multiplier that I am trying to implement which somehow leads to the function crashing.
Before I introduced the multiplier it all worked, I tried breaking it down by entering the multiplier value manually and it didn't work
P_{1} = 0.6;
P_{2} = 0.2;
P_{3} = 0.2;
a_1 = 4/3;
a_2 = -7/3;
b_1 = -1/3;
b_2 = 4/3;
persistent multiplier
multiplier = exp(a_1 * 44 + a_2 * 14 + 0);
eqn = #(x) ((a_1 * x + b_1)^a_1) * ((a_2 * x + b_2)^a_2) * x ...
-(P_{1}^a_1) * (P_{2}^a_2) * P_{3} * multiplier;
Q_{3} = Bisectionmethod(a_1, a_2, b_1, b_2, eqn);
Here is the calculating part of the bisection method.
x_lower = max(0, -b_1 / a_1);
x_upper = -b_2 / a_2;
x_mid = (x_lower + x_upper)/2;
Conditional statement encompassing the method of bisection
while abs(eqn(x_mid)) > 10^(-10)
if (eqn(x_mid) * eqn(x_upper)) < 0
x_lower = x_mid;
else
x_upper = x_mid;
end
x_mid = (x_lower + x_upper)/2;
end
Based on the information you provided this is what I came up with
function Q = Stackoverflow
persistent multiplier
P{1} = 0.6;
P{2} = 0.2;
P{3} = 0.2;
a1 = 4/3;
a2 = -7/3;
b1 = -1/3;
b2 = 4/3;
multiplier = exp(a1 * 44 + a2 * 14 + 0);
eqn = #(x) ((a1 .* x + b1).^a1) .* ((a2 .* x + b2).^a2) .* x -(P{1}.^a1) .* (P{2}.^a2) .* P{3} .* multiplier;
Q{3} = Bisectionmethod(eqn, max([0, -b1/a1]), -b2/a2, 1E-10);
end
function XOut = Bisectionmethod(f, xL, xH, EPS)
if sign(f(xL)) == sign(f(xH))
XOut = [];
error('Cannot bisect interval because can''t ensure the function crosses 0.')
end
x = [xL, xH];
while abs(diff(x)) > EPS
x(sign(f(mean(x))) == sign(f(x))) = mean(x);
end
XOut = mean(x);
end
I'm trying to solve some ODEs in MatLab and seeing as the variables in the equations are populations they need to be constrained to being positive. So I tried using odeset() before calling the equation solver to make them non-negative but on plotting the values afterwards they are actually negative at times (in the code below it is the magenta line). What am I doing wrong?
Here's some code:
%Lots of variables
includeJ=1;
cullLIRate=1/2000;
cullDIRate=1/2000;
N = 16800;
beta = 2e-7;
delta = 0.5;
gamma = 1/50;
sigma = 1/400;
mu = 1/365;
maxTime = 30*365;
kappa = N;
gR = 0.05;
mJ = 1/3650;
initJPerAdult = 10;
numInitE = 1000;
TSpan = [0,maxTime];
initState = [N-numInitE,numInitE,0,0,0,initJPerAdult*N];
%IMPORTANT BIT HERE
options = odeset('NonNegative', 1:6)
scirSoln = ode45(#equation,TSpan,initState,[],beta,delta,gamma,sigma,mu,kappa,gR,mJ,cullLIRate,cullDIRate,includeJ);
scirVals = deval(scirSoln,timeToPlot);
plot(timeToPlot,scirVals(1,:));
hold on;
plot(timeToPlot,scirVals(3,:),'k');
plot(timeToPlot,scirVals(4,:),'g');
plot(timeToPlot,scirVals(6,:),'m');
timeToPlot = [0:max(TSpan)/1000:max(TSpan)];
The code for equation(...) is:
function retVal = equation(t,y,beta,delta,gamma,sigma,mu,kappa,gR,mJ,cullLIRate,cullDIRate,includeJ)
retVal = zeros(6,1);
S = y(1);
E = y(2);
LI = y(3);
DI = y(4);
R = y(5);
J = y(6);
retVal(1)= mJ * J - beta * S * (delta * LI + DI);
retVal(2) = beta * S * (delta * LI + DI) - gamma * E;
retVal(3) = gamma * E - (cullLIRate + sigma) * LI;
retVal(4) = sigma * LI - (mu + cullDIRate) * DI;
retVal(5) = mu * DI + cullLIRate* LI + cullDIRate * DI;
retVal(6) = gR * S * (1 - S / kappa) - mJ * J;
end
You are not passing your defined odeset (options variable) to the ODE45 - solver.
The syntax for the ODE45 is: [T,Y] = ODE45(ODEFUN,TSPAN,Y0,OPTIONS,P1,P2...)
Glad it worked! :)