I am trying to implement my own LU decomposition with partial pivoting. My code is below and apparently is working fine, but for some matrices it gives different results when comparing with the built-in [L, U, P] = lu(A) function in matlab
Can anyone spot where is it wrong?
function [L, U, P] = lu_decomposition_pivot(A)
n = size(A,1);
Ak = A;
L = zeros(n);
U = zeros(n);
P = eye(n);
for k = 1:n-1
for i = k+1:n
[~,r] = max(abs(Ak(:,k)));
Ak([k r],:) = Ak([r k],:);
P([k r],:) = P([r k],:);
L(i,k) = Ak(i,k) / Ak(k,k);
for j = k+1:n
U(k,j-1) = Ak(k,j-1);
Ak(i,j) = Ak(i,j) - L(i,k)*Ak(k,j);
end
end
end
L(1:n+1:end) = 1;
U(:,end) = Ak(:,end);
return
Here are the two matrices I've tested with. The first one is correct, whereas the second has some elements inverted.
A = [1 2 0; 2 4 8; 3 -1 2];
A = [0.8443 0.1707 0.3111;
0.1948 0.2277 0.9234;
0.2259 0.4357 0.4302];
UPDATE
I have checked my code and corrected some bugs, but still there's something missing with the partial pivoting. In the first column the last two rows are always inverted (compared with the result of lu() in matlab)
function [L, U, P] = lu_decomposition_pivot(A)
n = size(A,1);
Ak = A;
L = eye(n);
U = zeros(n);
P = eye(n);
for k = 1:n-1
[~,r] = max(abs(Ak(k:end,k)));
r = n-(n-k+1)+r;
Ak([k r],:) = Ak([r k],:);
P([k r],:) = P([r k],:);
for i = k+1:n
L(i,k) = Ak(i,k) / Ak(k,k);
for j = 1:n
U(k,j) = Ak(k,j);
Ak(i,j) = Ak(i,j) - L(i,k)*Ak(k,j);
end
end
end
U(:,end) = Ak(:,end);
return
I forgot that If there was a swap in matrix P I had to swap also the matrix L. So just add the next line after after swapping P and everything will work excellent.
L([k r],:) = L([r k],:);
Both functions are not correct.
Here is the correct one.
function [L, U, P] = LU_pivot(A)
[m, n] = size(A); L=eye(n); P=eye(n); U=A;
for k=1:m-1
pivot=max(abs(U(k:m,k)))
for j=k:m
if(abs(U(j,k))==pivot)
ind=j
break;
end
end
U([k,ind],k:m)=U([ind,k],k:m)
L([k,ind],1:k-1)=L([ind,k],1:k-1)
P([k,ind],:)=P([ind,k],:)
for j=k+1:m
L(j,k)=U(j,k)/U(k,k)
U(j,k:m)=U(j,k:m)-L(j,k)*U(k,k:m)
end
pause;
end
end
My answer is here:
function [L, U, P] = LU_pivot(A)
[n, n1] = size(A); L=eye(n); P=eye(n); U=A;
for j = 1:n
[pivot m] = max(abs(U(j:n, j)));
m = m+j-1;
if m ~= j
U([m,j],:) = U([j,m], :); % interchange rows m and j in U
P([m,j],:) = P([j,m], :); % interchange rows m and j in P
if j >= 2; % very_important_point
L([m,j],1:j-1) = L([j,m], 1:j-1); % interchange rows m and j in columns 1:j-1 of L
end;
end
for i = j+1:n
L(i, j) = U(i, j) / U(j, j);
U(i, :) = U(i, :) - L(i, j)*U(j, :);
end
end
Related
I am trying to solve this problem. But I keep getting an error.
This is my First Code.
% Program 3.3
function [L, U, P] = lufact(A)
[N, N] = size(A);
X = zeros(N, 1);
Y = zeros(N, 1);
C = zeros(1, N);
R = 1:N;
for p = 1: N-1
[max1, j] = max(abs(A(p:N, p)));
C = A(p,:);
A(p,:) = A(j + p - 1,:);
A(j + p -1, :) = C;
d = R(p);
R(p) = R(j + p -1);
R(j + p - 1) = d;
if A(p,p) == 0
'A is Singular. No unique Solution'
break
end
for k = p + 1:N
mult = A(k,p)/A(p,p);
A(k,p) = mult;
A(k,p + 1:N) = A(k, p + 1:N) - mult *A(p, p + 1:N);
end
I=(1:N)'*ones(1,N,1); J=I';
L = (I>J).*A + eye(N);
U = (J>=I).*A;
P = zeros(N);
for k=1:N
P(k,R(k))=1;
end
end
X(N) = Y(N)/A(N,N);
for k = N-1: -1: 1
X(k) = (Y(k) - A(k, k+1:N)*X(k+1:N))/A(k,k);
end
And This is my 2nd Code which I'm using to solve this problem.
function B = Ques3(A)
% Computes the inverse of a matrix A
[L,U,P] = lufact(A);
N = max(size(A));
I = eye(N);
B = zeros(N);
for j = 1:N
Y = forsub(L,P*I(:,j));
B(:,j) = backsub(U,Y);
end
But I keep getting an error in MATLAB,
>> Ques3(A)
Unrecognized function or variable 'forsub'.
Error in Ques3 (line 12)
Y = forsub(L,P*I(:,j));
Consider the following calculation of the tangent tangent correlation which is performed in a for loop
v1=rand(25,1);
v2=rand(25,1);
n=25;
nSteps=10;
mean_theta = zeros(nSteps,1);
for j=1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j)=mean(theta);
end
plot(mean_theta)
How can matlab matrix calculations be utilized to make this performance better?
There are several things you can do to speed up your code. First, always preallocate. This converts:
theta = [];
for i = 1:(n-j)
%...
theta = [theta acosd(d/n1/n2)];
end
into:
theta = zeros(1,n-j);
for i = 1:(n-j)
%...
theta(i) = acosd(d/n1/n2);
end
Next, move the normalization out of the loops. There is no need to normalize over and over again, just normalize the input:
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
%...
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
This does change the output very slightly, within numerical precision, because the different order of operations leads to different floating-point rounding error.
Finally, you can remove the inner loop by vectorizing that computation:
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
Timings (n=25):
Original: 0.0019 s
Preallocate: 0.0013 s
Normalize once: 0.0011 s
Vectorize: 1.4176e-04 s
Timings (n=250):
Original: 0.0185 s
Preallocate: 0.0146 s
Normalize once: 0.0118 s
Vectorize: 2.5694e-04 s
Note how the vectorized code is the only one whose timing doesn't grow linearly with n.
Timing code:
function so
n = 25;
v1 = rand(n,1);
v2 = rand(n,1);
nSteps = 10;
mean_theta1 = method1(v1,v2,nSteps);
mean_theta2 = method2(v1,v2,nSteps);
fprintf('diff method1 vs method2: %g\n',max(abs(mean_theta1(:)-mean_theta2(:))));
mean_theta3 = method3(v1,v2,nSteps);
fprintf('diff method1 vs method3: %g\n',max(abs(mean_theta1(:)-mean_theta3(:))));
mean_theta4 = method4(v1,v2,nSteps);
fprintf('diff method1 vs method4: %g\n',max(abs(mean_theta1(:)-mean_theta4(:))));
timeit(#()method1(v1,v2,nSteps))
timeit(#()method2(v1,v2,nSteps))
timeit(#()method3(v1,v2,nSteps))
timeit(#()method4(v1,v2,nSteps))
function mean_theta = method1(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta=[];
for i=1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta = [theta acosd(d/n1/n2)];
end
mean_theta(j) = mean(theta);
end
function mean_theta = method2(v1,v2,nSteps)
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
d = dot([v1(i) v2(i)],[v1(i+j) v2(i+j)]);
n1 = norm([v1(i) v2(i)]);
n2 = norm([v1(i+j) v2(i+j)]);
theta(i) = acosd(d/n1/n2);
end
mean_theta(j) = mean(theta);
end
function mean_theta = method3(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
theta = zeros(1,n-j);
for i = 1:(n-j)
theta(i) = acosd(dot(v(i,:),v(i+j,:)));
end
mean_theta(j) = mean(theta);
end
function mean_theta = method4(v1,v2,nSteps)
v = [v1,v2];
v = v./sqrt(sum(v.^2,2)); % Can use VECNORM in newest MATLAB
n = numel(v1);
mean_theta = zeros(nSteps,1);
for j = 1:nSteps
i = 1:(n-j);
theta = acosd(dot(v(i,:),v(i+j,:),2));
mean_theta(j) = mean(theta);
end
Here is a full vectorized solution:
i = 1:n-1;
j = (1:nSteps).';
ij= min(i+j,n);
a = cat(3, v1(i).', v2(i).');
b = cat(3, v1(ij), v2(ij));
d = sum(a .* b, 3);
n1 = sum(a .^ 2, 3);
n2 = sum(b .^ 2, 3);
theta = acosd(d./sqrt(n1.*n2));
idx = (1:nSteps).' <= (n-1:-1:1);
mean_theta = sum(theta .* idx ,2) ./ sum(idx,2);
Result of Octave timings for my method,method4 from the answer provided by #CrisLuengo and the original method (n=250):
Full vectorized : 0.000864983 seconds
Method4(Vectorize) : 0.002774 seconds
Original(loop) : 0.340693 seconds
Similar problem to this question matrix that forms an orthogonal basis with a given vector but I'm having problems with my implementation. I have a given vector, u, and I want to construct an arbitrary orthonormal matrix V such that V(:,1) = u/norm(u). The problem I'm getting is that V'*V isn't I (but is diagonal). Here's my code:
function V = rand_orth(u)
n = length(u);
V = [u/norm(u) rand(n,n-1)];
for i = 1:n
vi = V(:,i);
rii = norm(vi);
qi = vi/rii;
for j = i+1:n
rij = dot(qi,V(:,j));
V(:,j) = V(:,j) - rij*qi;
end
end
end
Simply normalize columns of V once you've finished the orthogonalization. Note there's also no need to normalize u (first column) initially, since we will normalize the whole matrix at the end:
function V = rand_orth(u)
n = length(u);
V = [u, rand(n,n-1)];
for i = 1:n
vi = V(:,i);
rii = norm(vi);
qi = vi/rii;
for j = i+1:n
rij = dot(qi,V(:,j));
V(:,j) = V(:,j) - rij*qi;
end
end
V = bsxfun(#rdivide, V, sqrt(diag(V'*V))');
end
Is there an algorithm that computes the Drazin inverse of a singular matrix? I would like to apply it either in MATLAB or Mathematica.
Read this article:
Fanbin Bu and Yimin Wei, The algorithm for computing the Drazin inverses of two-variable polynomial matrices, Applied mathematics and computation 147.3 (2004): 805-836.
in appendix there are several MATLAB code. The first one is this:
function DrazinInverse1a = DrazinInverse1(a)
%-----------------------------------------
%Compute the Drazin Inverse of a matrix 'a' using the limited algorithm.
%Need computing the index of 'a'.
global q1 q2 s1 s2
[m,n] = size(a);
if m~= n
disp('Matrix is must be square!')
end
%-----------------------------------------
% Computer the index of A and note r = rank(A^k).
[k,r,a1,a] = index(a);
F = eye(n);
g = -trace(a);
g = collect(g);
for i = 1:r-1
G = g*eye(n);
F = a*F+G;
g = -trace(a*F)/(i+1);
g = collect(g);
end
DrazinInverse1a = a1*F;
DrazinInverse1a = -1/g*DrazinInverse1a;
DrazinInverse1a = simplify(DrazinInverse1a);
end
function [k,r,a1,a] = index(a)
%To compute the index of 'a'.
k = 0;
n = length(a);
r = n;
a0 = a;
r1 = rank(a);
a1 = eye(n);
while r ~= r1
r = r1;
a1 = a;
a = a*a0;
r1 = rank(a);
k = k+1;
end
r = sym2poly(r);
end
I have this code for matlab to multiply matrixes how di i make the user add the matrixes and then use the matrixes in the code?
eg [n,m] = input(user inputs matrix here)
[n,m] = size(A);
[p,q] = size(B);
C = zeros(n,p);
if p~=m
error('Inner Matrix Dimensions Must Agree.')
end
for k = 1:n
for j = 1:q
temp=0;
for i = 1:p
temp = temp+(A(k,i)*B(i,j));
end
C(k,j) = temp;
end
end
You can use in the script:
A = input('input array A ');
B = input('input array B ');
[n,m] = size(A);
[p,q] = size(B);
C = zeros(n,p);
if p~=m
error('Inner Matrix Dimensions Must Agree.')
end
for k = 1:n
for j = 1:q
temp=0;
for i = 1:p
temp = temp+(A(k,i)*B(i,j));
end
C(k,j) = temp;
end
end
or you can write the above as a function:
function C = matrixmultiply(A,B)
[n,m] = size(A);
[p,q] = size(B);
C = zeros(n,p);
if p~=m
error('Inner Matrix Dimensions Must Agree.')
end
for k = 1:n
for j = 1:q
temp=0;
for i = 1:p
temp = temp+(A(k,i)*B(i,j));
end
C(k,j) = temp;
end
end
end