How would you code this in MATLAB?
This is what I've tried, but it doesn't work quite right.
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
for k = 1:tot_it
for j = 1:n
for i = 1:n
if (j ~= i)
x(i) = -((A(i,j)/A(i,i)) * x(j) + (b(i)/A(i,i)));
else
continue;
end
end
end
end
end
j is an iterator of a sum over each i, so you need to change their order. Also the formula has a sum and in your code you're not adding anything so that's another thing to consider. The last thing I see that you're omitting is that you should save the previous state of xbecause the right side of the formula needs it. You should try something like this:
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
s = 0; %Auxiliar var to store the sum.
xold = x
for k = 1:tot_it
for i = 1:n
for j = 1:n
if (j ~= i)
s = s + (A(i,j)/A(i,i)) * xold(j);
else
continue;
end
end
x(i) = -s + b(i)/A(i,i);
s = 0;
end
xold = x;
end
end
Related
Given an nxn matrix A_k and a nx1 vector x, is there any smart way to compute
using Matlab? x_i are the elements of the vector x, therefore J is a sum of matrices. So far I have used a for loop, but I was wondering if there was a smarter way.
Short answer: you can use the builtin matlab function polyvalm for matrix polynomial evaluation as follows:
x = x(end:-1:1); % flip the order of the elements
x(end+1) = 0; % append 0
J = polyvalm(x, A);
Long answer: Matlab uses a loop internally. So, you didn't gain that much or you perform even worse if you optimise your own implementation (see my calcJ_loopOptimised function):
% construct random input
n = 100;
A = rand(n);
x = rand(n, 1);
% calculate the result using different methods
Jbuiltin = calcJ_builtin(A, x);
Jloop = calcJ_loop(A, x);
JloopOptimised = calcJ_loopOptimised(A, x);
% check if the functions are mathematically equivalent (should be in the order of `eps`)
relativeError1 = max(max(abs(Jbuiltin - Jloop)))/max(max(Jbuiltin))
relativeError2 = max(max(abs(Jloop - JloopOptimised)))/max(max(Jloop))
% measure the execution time
t_loopOptimised = timeit(#() calcJ_loopOptimised(A, x))
t_builtin = timeit(#() calcJ_builtin(A, x))
t_loop = timeit(#() calcJ_loop(A, x))
% check if builtin function is faster
builtinFaster = t_builtin < t_loopOptimised
% calculate J using Matlab builtin function
function J = calcJ_builtin(A, x)
x = x(end:-1:1);
x(end+1) = 0;
J = polyvalm(x, A);
end
% naive loop implementation
function J = calcJ_loop(A, x)
n = size(A, 1);
J = zeros(n,n);
for i=1:n
J = J + A^i * x(i);
end
end
% optimised loop implementation (cache result of matrix power)
function J = calcJ_loopOptimised(A, x)
n = size(A, 1);
J = zeros(n,n);
A_ = eye(n);
for i=1:n
A_ = A_*A;
J = J + A_ * x(i);
end
end
For n=100, I get the following:
t_loopOptimised = 0.0077
t_builtin = 0.0084
t_loop = 0.0295
For n=5, I get the following:
t_loopOptimised = 7.4425e-06
t_builtin = 4.7399e-05
t_loop = 1.0496e-04
Note that my timings fluctuates somewhat between different runs, but the optimised loop is almost always faster (up to 6x for small n) than the builtin function.
For my studies I had to write a PDE solver for the Poisson equation on a disc shaped domain using the finite difference method.
I already passed the Lab exercise. There is one issue in my code I couldn't fix. Function fun1 with the boundary value problem gun2 is somehow oscillating at the boundary. When I use fun2 everything seems fine...
Both functions use at the boundary gun2. What is the problem?
function z = fun1(x,y)
r = sqrt(x.^2+y.^2);
z = zeros(size(x));
if( r < 0.25)
z = -10^8*exp(1./(r.^2-1/16));
end
end
function z = fun2(x,y)
z = 100*sin(2*pi*x).*sin(2*pi*y);
end
function z = gun2(x,y)
z = x.^2+y.^2;
end
function [u,A] = poisson2(funame,guname,M)
if nargin < 3
M = 50;
end
%Mesh Grid Generation
h = 2/(M + 1);
x = -1:h:1;
y = -1:h:1;
[X,Y] = meshgrid(x,y);
CI = ((X.^2 +Y.^2) < 1);
%Boundary Elements
Sum= zeros(size(CI));
%Sum over the neighbours
for i = -1:1
Sum = Sum + circshift(CI,[i,0]) + circshift(CI,[0,i]) ;
end
%if sum of neighbours larger 3 -> inner note!
CI = (Sum > 3);
%else boundary
CB = (Sum < 3 & Sum ~= 0);
Sum= zeros(size(CI));
%Sum over the boundary neighbour nodes....
for i = -1:1
Sum = Sum + circshift(CB,[i,0]) + circshift(CB,[0,i]);
end
%If the sum is equal 2 -> Diagonal boundary
CB = CB + (Sum == 2 & CB == 0 & CI == 0);
%Converting X Y to polar coordinates
Phi = atan(Y./X);
%Converting Phi R back to cartesian coordinates, only at the boundarys
for j = 1:M+2
for i = 1:M+2
if (CB(i,j)~=0)
if j > (M+2)/2
sig = 1;
else
sig = -1;
end
X(i,j) = sig*1*cos(Phi(i,j));
Y(i,j) = sig*1*sin(Phi(i,j));
end
end
end
%Numberize the internal notes u1,u2,......,un
CI = CI.*reshape(cumsum(CI(:)),size(CI));
%Number of internal notes
Ni = nnz(CI);
f = zeros(Ni,1);
k = 1;
A = spalloc(Ni,Ni,5*Ni);
%Create matix A!
for j=2:M+1
for i =2:M+1
if(CI(i,j) ~= 0)
hN = h;hS = h; hW = h; hE = h;
f(k) = fun(X(i,j),Y(i,j));
if(CB(i+1,j) ~= 0)
hN = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i+1,j))*2/(hN^2+hN*h);
A(k,CI(i-1,j)) = -2/(h^2+h*hN);
else
if(CB(i-1,j) ~= 0) %in negative y is a boundry
hS = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i-1,j))*2/(hS^2+h*hS);
A(k,CI(i+1,j)) = -2/(h^2+h*hS);
else
A(k,CI(i-1,j)) = -1/h^2;
A(k,CI(i+1,j)) = -1/h^2;
end
end
if(CB(i,j+1) ~= 0)
hE = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j+1),Y(i,j))*2/(hE^2+hE*h);
A(k,CI(i,j-1)) = -2/(h^2+h*hE);
else
if(CB(i,j-1) ~= 0)
hW = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j-1),Y(i,j))*2/(hW^2+h*hW);
A(k,CI(i,j+1)) = -2/(h^2+h*hW);
else
A(k,CI(i,j-1)) = -1/h^2;
A(k,CI(i,j+1)) = -1/h^2;
end
end
A(k,k) = (2/(hE*hW)+2/(hN*hS));
k = k + 1;
end
end
end
%Solve linear system
u = A\f;
U = zeros(M+2,M+2);
p = 1;
%re-arange u
for j = 1:M+2
for i = 1:M+2
if ( CI(i,j) ~= 0)
U(i,j) = u(p);
p = p+1;
else
if ( CB(i,j) ~= 0)
U(i,j) = gun(X(i,j),Y(i,j));
else
U(i,j) = NaN;
end
end
end
end
surf(X,Y,U);
end
I'm keeping this answer short for now, but may extend when the question contains more info.
My first guess is that what you are seeing is just numerical errors. Looking at the scales of the two graphs, the peaks in the first graph are relatively small compared to the signal in the second graph. Maybe there is a similar issue in the second that is just not visible because the signal is much bigger. You could try to increase the number of nodes and observe what happens with the result.
You should always expect to see numerical errors in such simulations. It's only a matter of trying to get their magnitude as small as possible (or as small as needed).
I only have about two weeks of experience with programming/Matlab, so I'm just a beginner. In my code I would like to plot mu as a function of alpha. When I display mu it shows the 10 values of mu for each value of alpha. However, when I plot the graph it gives the values of mu as seperate points. But I want the points to be connected with just one line. How can I solve this problem?
n=40; %number of months
p=0.23; %probability of success
num_of_simulations=100;
s=rng; x = rand(1,n)<p;
rng(s);
hold on;
for alpha=0.01:0.01:0.1;
for i=1:num_of_simulations
x = rand(1,n)<p;
S0=5000; %initial value
Y(1)=S0*alpha; %deposit
for k=1
if x(1,1)==1;
S(1, i)=S0+2*Y(1);
else
S(1, i)=S0-Y(1);
end
end
for k=2:n
Y(k)=S(k-1, i)*alpha;
if x(1,k)==1;
S(k, i)=S(k-1, i)+2*Y(k);
else
S(k, i)=S(k-1, i)-Y(k);
end
end
Sn(i)=S(n,i); %end value for each simulations
end
mu=mean(Sn);
disp(mu);
plot(alpha,mu);
end
The reason your points aren't connected is because you plot each point separately. If we take a different approach and take alpha = 0.01:0.01:0.1; out of the for loop definition and then change the for loop definition to for j=1:numel(alpha) we can still loop over every element of alpha. Now we need to change each use of alpha in the loop to alpha(j) so that we are referring to the current element of alpha and not every element. Following on from this we need to change mu to mu(j). What this means is that when the entire loop has finished we have all of the values of alpha and mu stored and 1 call to plot(alpha, mu) will plot the data with the points connected as in
This also enables us to remove hold on; too as we only plot once.
I've included the complete edited code here for you to see. The changes are minuscule and should make sense.
clear all
close all
n = 40; %number of months
p = 0.23; %probability of success
num_of_simulations = 100;
s = rng;
x = rand(1, n) < p;
rng(s);
alpha = 0.01:0.01:0.1;
for j = 1:numel(alpha)
for i = 1:num_of_simulations
x = rand(1, n) < p;
S0 = 5000; %initial value
Y(1) = S0*alpha(j); %deposit
for k = 1
if x(1, 1) == 1;
S(1, i) = S0 + 2*Y(1);
else
S(1, i) = S0 - Y(1);
end
end
for k = 2:n
Y(k) = S(k-1, i)*alpha(j);
if x(1, k) == 1;
S(k, i) = S(k-1, i) + 2*Y(k);
else
S(k, i) = S(k-1, i) - Y(k);
end
end
Sn(i) = S(n, i); %end value for each simulations
end
mu(j) = mean(Sn);
disp(mu(j));
end
plot(alpha, mu);
I'm trying to build make a code where an equation is not calculated for some certain values. I have a meshgrid with several values for x and y and I want to include a for loop that will calculate some values for most of the points in the meshgrid but I'm trying to include in that loop a condition that if the points have a specified index, the value will not be calculated. In my second group of for/if loops, I want to say that for all values of i and k (row and column), the value for z and phi are calculated with the exception of the specified i and k values (in the if loop). What I'm doing at the moment is not working...
The error I'm getting is:
The expression to the left of the equals sign is not a valid target for an assignment.
Here is my code at the moment. I'd really appreciate any advice on this! Thanks in advance
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
for i = 1:length(x)
for k = 1:length(x)
% Zeroing values where cylinder is
if sqrt(x(i,k).^2 + y(i,k).^2) < a
x(i,k) = 0;
y(i,k) = 0;
end
end
end
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
z = zeros(length(x));
phi = zeros(length(x));
for i = 1:length(x)
for k = 1:length(x)
if (i > 16 && i < 24 && k > 16 && k <= length(x))
z = 0;
phi = 0;
else
z = U_i.*r.*(1-a^2./r.^2).*sin(theta); % Stream function
phi = U_i*r.*(1+a^2./r.^2).*cos(theta); % Velocity potential
end
end
end
The original code in the question can be rewritten as seen below. Pay attention in the line with ind(17:24,:) since your edit now excludes 24 and you original question included 24.
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
ind = find(sqrt(x.^2 + y.^2) < a);
x(ind) = 0;
y(ind) = 0;
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
ind = true(size(x));
ind(17:24,17:length(x)) = false;
z = zeros(size(x));
phi = zeros(size(x));
z(ind) = U_i.*r(ind).*(1-a^2./r(ind).^2).*sin(theta(ind)); % Stream function
phi(ind) = U_i.*r(ind).*(1+a^2./r(ind).^2).*cos(theta(ind)); % Velocity potential
I can't seem to find a fix to my infinite loop. I have coded a Jacobi solver to solve a system of linear equations.
Here is my code:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
i =1;
for r=1:m
sum = 0;
for c=1:n
if r~=c
sum = sum + A(r,c)*x(c);
else
x(r) = (-sum + b(r))/A(r,c);
end
x(r) = (-sum + b(r))/A(r,c);
xxx end xxx
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
When I terminate the code it ends at the line with xxx
The reason why your code isn't working is due to the logic of your if statements inside your for loops. Specifically, you need to accumulate all values for a particular row that don't belong to the diagonal of that row first. Once that's done, you then perform the division. You also need to make sure that you're dividing by the diagonal coefficient of A for that row you're concentrating on, which corresponds to the component of x you're trying to solve for. You also need to remove the i=1 statement at the beginning of your loop. You're resetting i each time.
In other words:
function [x, i] = Jacobi(A, b, x0, TOL)
[m n] = size(A);
i = 0;
x = [0;0;0];
while (true)
for r=1:m
sum = 0;
for c=1:n
if r==c %// NEW
continue;
end
sum = sum + A(r,c)*x(c); %// NEW
end
x(r) = (-sum + b(r))/A(r,r); %// CHANGE
end
if abs(norm(x) - norm(x0)) < TOL;
break
end
x0 = x;
i = i + 1;
end
Example use:
A = [6 1 1; 1 5 3; 0 2 4]
b = [1 2 3].';
[x,i] = Jacobi(A, b, [0;0;0], 1e-10)
x =
0.048780487792648
-0.085365853612062
0.792682926806031
i =
20
This means it took 20 iterations to achieve a solution with tolerance 1e-10. Compare this with MATLAB's built-in inverse:
x2 = A \ b
x2 =
0.048780487804878
-0.085365853658537
0.792682926829268
As you can see, I specified a tolerance of 1e-10, which means we are guaranteed to have 10 decimal places of accuracy. We can certainly see 10 decimal places of accuracy between what Jacobi gives us with what MATLAB gives us built-in.