%% Gram—Schmidt as Triangular Orthogonalization
clear
M = [1,1,1; 1,1,0; 1,1,9]
[m,n] = size(M);
Q = M;
Rinv = eye(n);
for i = 1:n
Ri = eye(n);
Ri(i,i) = 1 / norm(Q(:,i));
for j = i+1:n
Ri(i,j) = -Q(:,i)'*Q(:,j) / norm(Q(:,i));
end
Q = Q*Ri;
Rinv = Rinv*Ri;
end
Q
R = inv(Rinv)
The question: Q*R gives M butQ is not orthagonal.
This should return the correct result:
clear();
M = [1,1,1; 1,1,0; 1,1,9];
[m,n] = size(M);
Q = zeros(m,n);
R = zeros(n,n);
for j = 1:n
v = M(:,j);
for i = 1:j-1
R(i,j) = Q(:,i).' * M(:,j);
v = v - (R(i,j) * Q(:,i));
end
R(j,j) = norm(v);
Q(:,j) = v / R(j,j);
end
Alternatively, you could give a try to the following implementation I found on Matlab File Exchange (it contains many know variants of the algorithm): https://it.mathworks.com/matlabcentral/fileexchange/55881-gram-schmidt-orthogonalization
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));
I made Clamped Cubic Spline code.
But when I put
f_ = CubicSpline([0,1,2,3],[exp(0),exp(1),exp(2),exp(3)],exp(0),exp(3));
and get answer by sym2poly(f_(1))
result is quite different from my lecture note. And actually, my Cubic Spline result even doesn't match to the prime of bound...
Please I can't understand what is the problem in my code.
This is what I used for my algorithm.
function [f_] = CubicSpline(x0,f0,FPO,FPN)
syms x;
n = length(x0);
h = zeros(n,1);
alpha = zeros(n,1);
l = zeros(n,1);
u = zeros(n,1);
z = zeros(n,1);
a = zeros(n,1);
b = zeros(n,1);
c = zeros(n,1);
d = zeros(n,1);
for iter = 1:n-1
h(iter) = x0(iter+1)-x0(iter);
end
alpha(1) = 3*(f0(2)-f0(1))/h(1)-3*FPO;
alpha(n) = 3*FPN-3*(f0(n)-f0(n-1))/h(n-1);
for iter = 1:n
a(iter) = f0(iter);
end
for iter = 2:n-1
alpha(iter) = 3/h(iter)*(f0(iter+1)-f0(iter))-3/h(iter-1)*(f0(iter)-f0(iter-1));
end
l(1) = 2*h(1);
u(1) = 0.5;
z(1) = f0(1)/l(1);
for iter = 2:n-1
l(iter) = 2*(x0(iter+1)-x0(iter-1)) - h(iter-1)*u(iter-1);
u(iter) = h(iter)/l(iter);
z(iter) = (alpha(iter)-h(iter-1)*z(iter-1))/l(iter);
end
l(n) = h(n-1)*(2-u(n-1));
z(n) = (alpha(n)-h(n-1)*z(n-1))/l(n);
c(n) = z(n);
for iter = (n-1):-1:1
c(iter) = z(iter)-u(iter)*c(iter+1);
b(iter) = (f0(iter+1)-f0(iter))/h(iter)-h(iter)*(c(iter+1)+2*c(iter))/3;
d(iter) = (c(iter+1)-c(iter))/(3*h(iter));
end
for iter = 1:n-1
f_(iter) = a(iter) + b(iter)*(x-x0(iter)) + c(iter)*(x-x0(iter))^2 + d(iter)*(x-x0(iter))^3;
end
end
There is a typo in your code for step 4
z(1) = f0(1)/l(1);
should be
z(1) = alpha(1)/l(1);
My question this time concerns the obtention of the degree distribution of a LDPC matrix through linear programming, under the following statement:
My code is the following:
function [v] = LP_Irr_LDPC(k,Ebn0)
options = optimoptions('fmincon','Display','iter','Algorithm','interior-point','MaxIter', 4000, 'MaxFunEvals', 70000);
fun = #(v) -sum(v(1:k)./(1:k));
A = [];
b = [];
Aeq = [0, ones(1,k-1)];
beq = 1;
lb = zeros(1,k);
ub = [0, ones(1,k-1)];
nonlcon = #(v)DensEv_SP(v,Ebn0);
l0 = [0 rand(1,k-1)];
l0 = l0./sum(l0);
v = fmincon(fun,l0,A,b,Aeq,beq,lb,ub,nonlcon,options)
end
Definition of nonlinear constraints:
function [c, ceq] = DensEv_SP(v,Ebn0)
% It is also needed to modify this function, as you cannot pass parameters from others to it.
h = [0 rand(1,19)];
h = h./sum(h); % This is where h comes from
syms x;
X = x.^(0:(length(h)-1));
R = h*transpose(X);
ebn0 = 10^(Ebn0/10);
Rm = 1;
LLR = (-50:50);
p03 = 0.3;
LLR03 = log((1-p03)/p03);
r03 = 1 - p03;
noise03 = (2*r03*Rm*ebn0)^-1;
pf03 = normpdf(LLR, LLR03, noise03);
sumpf03 = sum(pf03(1:length(pf03)/2));
divisions = 100;
Aj = zeros(1, divisions);
rho = zeros(1, divisions);
xj = zeros(1, divisions);
k = 10; % Length(v) -> Same value as in 'Complete.m'
for j=1:1:divisions
xj(j) = sumpf03*j/divisions;
rho(j) = subs(R,x,1-xj(j));
Aj(j) = 1 - rho(j);
end
c = zeros(1, length(xj));
lambda = zeros(1, length(Aj));
for j = 1:1:length(xj)
lambda(j) = sum(v(2:k).*(Aj(j).^(1:(k-1))));
c(j) = sumpf03*lambda(j) - xj(j);
end
save Almacen
ceq = [];
%ceq = sum(v)-1;
end
This question is linked to the one posted here. My problem is that I need that each element from vectors v and h resulting from this optimization problem is a fraction of x/N and x/(N(1-r) respectively.
How could I ensure that condition without losing convergence capability?
I came up with one possible solution, at least for vector h, within the function DensEv_SP:
function [c, ceq] = DensEv_SP(v,Ebn0)
% It is also needed to modify this function, as you cannot pass parameters from others to it.
k = 10; % Same as in Complete.m, desired sum of h
M = 19; % Number of integers
h = [0 diff([0,sort(randperm(k+M-1,M-1)),k+M])-ones(1,M)];
h = h./sum(h);
syms x;
X = x.^(0:(length(h)-1));
R = h*transpose(X);
ebn0 = 10^(Ebn0/10);
Rm = 1;
LLR = (-50:50);
p03 = 0.3;
LLR03 = log((1-p03)/p03);
r03 = 1 - p03;
noise03 = (2*r03*Rm*ebn0)^-1;
pf03 = normpdf(LLR, LLR03, noise03);
sumpf03 = sum(pf03(1:length(pf03)/2));
divisions = 100;
Aj = zeros(1, divisions);
rho = zeros(1, divisions);
xj = zeros(1, divisions);
N = 20; % Length(v) -> Same value as in 'Complete.m'
for j=1:1:divisions
xj(j) = sumpf03*j/divisions;
rho(j) = subs(R,x,1-xj(j));
Aj(j) = 1 - rho(j);
end
c = zeros(1, length(xj));
lambda = zeros(1, length(Aj));
for j = 1:1:length(xj)
lambda(j) = sum(v(2:k).*(Aj(j).^(1:(k-1))));
c(j) = sumpf03*lambda(j) - xj(j);
end
save Almacen
ceq = (N*v)-floor(N*v);
%ceq = sum(v)-1;
end
As above stated, there is no longer any problem with vector h; nevertheless the way I defined ceq value seemed to be insufficient to make the optimization work out (the problems with v have not diminished at all). Does anybody know how to find the solution?
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
I'm trying to vectorize the 2 inner nested for loops, but I can't come up with a way to do this. The FS1 and FS2 functions have been written to accept argument for N_theta and N_e, which is what the loops are iterating over
%% generate regions
for raw_r=1:visual_field_width
for raw_c=1:visual_field_width
r = raw_r - center_r;
c = raw_c - center_c;
% convert (r,c) to polar: (eccentricity, angle)
e = sqrt(r^2+c^2)*deg_per_pixel;
a = mod(atan2(r,c),2*pi);
for nt=1:N_theta
for ne=1:N_e
regions(raw_r, raw_c, nt, ne) = ...
FS_1(nt-1,a,N_theta) * ...
FS_2(ne-1,e,N_e,e0_in_deg, e_max);
end
end
end
end
Ideally, I could replace the two inner nested for loops by:
regions(raw_r,raw_c,:,:) = FS_1(:,a,N_theta) * FS_2(:,N_e,e0_in_deg,e_max);
But this isn't possible. Maybe I'm missing an easy fix or vectorization technique? e0_in_deg and e_max are parameters.
The FS_1 function is
function h = FS_1(n,theta,N,t)
if nargin==2
N = 9;
t=1/2;
elseif nargin==3
t=1/2;
end
w = (2*pi)/N;
theta = theta + w/4;
if n==0 && theta>(3/2)*pi
theta = theta - 2*pi;
end
h = FS_f((theta - (w*n + 0.5*w*(1-t)))/w);
the FS_2 function is
function g = FS_gne(n,e,N,e0, e_max)
if nargin==2
N = 10;
e0 = .5;
elseif nargin==3
e0 = .5;
end
w = (log(e_max) - log(e0))/N;
g = FS_f((log(e)-log(e0)-w*(n+1))/w);
and the FS_f function is
function f = FS_f(x, t)
if nargin<2
t = 0.5;
end
f = zeros(size(x));
% case 1
idx = x>-(1+t)/2 & x<=(t-1)/2;
f(idx) = (cos(0.5*pi*((x(idx)-(t-1)/2)/t))).^2;
% case 2
idx = x>(t-1)/2 & x<=(1-t)/2;
f(idx) = 1;
% case 3
idx = x>(1-t)/2 & x<=(1+t)/2;
f(idx) = -(cos(0.5*pi*((x(idx)-(1+t)/2)/t))).^2+1;
I had to assume values for the constants, and then used ndgrid to find the possible configurations and sub2ind to get the indices. Doing this I removed all loops. Let me know if this produced the correct values.
function RunningFunction
%% generate regions
visual_field_width = 10;
center_r = 2;
center_c = 3;
deg_per_pixel = 17;
N_theta = 2;
N_e = 5;
e0_in_deg = 35;
e_max = 17;
[raw_r, raw_c, nt, ne] = ndgrid(1:visual_field_width, 1:visual_field_width, 1:N_theta, 1:N_e);
ind = sub2ind(size(raw_r), raw_r, raw_c, nt, ne);
r = raw_r - center_r;
c = raw_c - center_c;
% convert (r,c) to polar: (eccentricity, angle)
e = sqrt(r.^2+c.^2)*deg_per_pixel;
a = mod(atan2(r,c),2*pi);
regions(ind) = ...
FS_1(nt-1,a,N_theta) .* ...
FS_2(ne-1,e,N_e,e0_in_deg, e_max);
regions = reshape(regions, size(raw_r));
end
function h = FS_1(n,theta,N,t)
if nargin==2
N = 9;
t=1/2;
elseif nargin==3
t=1/2;
end
w = (2*pi)./N;
theta = theta + w/4;
theta(n==0 & theta>(3/2)*pi) = theta(n==0 & theta>(3/2)*pi) - 2*pi;
h = FS_f((theta - (w*n + 0.5*w*(1-t)))/w);
end
function g = FS_2(n,e,N,e0, e_max)
if nargin==2
N = 10;
e0 = .5;
elseif nargin==3
e0 = .5;
end
w = (log(e_max) - log(e0))/N;
g = FS_f((log(e)-log(e0)-w*(n+1))/w);
end
function f = FS_f(x, t)
if nargin<2
t = 0.5;
end
f = zeros(size(x));
% case 1
idx = x>-(1+t)/2 & x<=(t-1)/2;
f(idx) = (cos(0.5*pi*((x(idx)-(t-1)/2)/t))).^2;
% case 2
idx = x>(t-1)/2 & x<=(1-t)/2;
f(idx) = 1;
% case 3
idx = x>(1-t)/2 & x<=(1+t)/2;
f(idx) = -(cos(0.5*pi*((x(idx)-(1+t)/2)/t))).^2+1;
end