Givens rotation QR decomposition

조회 수: 39 (최근 30일)
Duc Anh Le
Duc Anh Le 2020년 2월 11일
댓글: MmO 2021년 9월 30일
Screen Shot 2020-02-11 at 2.08.39 PM.png
I'm trying to create a function that computes the Givens Rotation QR decomposition, following this pseudo-code.
function [Q,R] = givens(A)
[m,n] = size(A);
indexI = zeros(m,n);
indexJ = zeros(m,n);
C = zeros(m,n);
S = zeros(m,n);
for i = 1:n
for j = i+1:m
c = A(i,i)/((A(i,i))^2 + (A(j,i)^2))^0.5;
s = A(j,i)/((A(i,i))^2 + (A(j,i)^2))^0.5;
A(i,:) = c*A(i,:) + s*A(j,:);
A(j,:) = -s*A(i,:) + c*A(j,:);
indexI(j,i) = i;
indexJ(j,i) = j;
C(j,i) = c;
S(j,i) = s;
end
end
R = A;
Q = eye(m);
for i = 1:n
for j= j+1:m
Q(:,i) = c*Q(:,i) + s*Q(:,j);
Q(:,j) = -s*Q(:,i) + c*Q(:,j);
end
end
However, the R matrix, that I get, is not upper triangular. I can't seem to find the mistake here. Any help would be highly appreciated. Thanks in advance.
  댓글 수: 2
Benjamin Ellis
Benjamin Ellis 2020년 3월 10일
Hi! I'm in this class too. I added the lines c = C(j,i) and s = S(j,i) within the second for loop. Also the second for loop should iterate j = (i+1):m.
With these changes I got Q and R to agree with qr(A) up to a sign.
MmO
MmO 2021년 9월 30일
Hello, Where you able to find the mistake? Where did you take this algorithm from? Best regards

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답변 (1개)

Jon
Jon 2020년 2월 11일
It looks to me like your code reproduces what is in the pseudocode. Are you sure that the psuedocode that you based your code on is correct?
A few things look perhaps questionable about the psuedocode.
Why do we compute C, S, idxI idxJ and never use them?
In the second double loop you use the values c and s which are just the last values from the double loop in the first part of the algorithm. So they never change value as i and j change. Maybe that is ok but it looks strange.

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