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qr decomposition run-time performance

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Niv Shapira
Niv Shapira on 31 Aug 2020
Commented: Niv Shapira on 1 Sep 2020
I'm comparing the run-time of the "qr" function:
Option1: R = triu(qr(A))
Option2: [~, R]=qr(A) (or [Q, R]=qr(A)),
Only the "R" output is required.
I've tested them on a large number of possible choices of A, a full (not-sparse) matrix with m >> n.
The result was a significant run-time advantage for triu(qr(A)).
To the best of my knowlegde from Matlab documention, option1 use LAPACK xGEQRF Householder Reflectors, but I couldn't find how option2 was implemented.
Can anyone explain what is (or may be) the cause for such performance difference?
(Using Matlab R2019b)
Thank you in advance!

Answers (2)

Bruno Luong
Bruno Luong on 31 Aug 2020
Edited: Bruno Luong on 31 Aug 2020
I'm pretty sure both Q-less-qr and qr use the same algorithm.
What you should try is
Option 3
[~, R] = qr(A,0);
that will return (n x n) R matrix without the bottom part padded with 0.
It's about 2 times slower than q-less qr, but it can be simply explained by Q (m x n) that is built and cleared.
If you do the Option2
[~, R] = qr(A);
the invisible cleared Q is (m x m), totally useless orthogonal much bigger matrix that has been built and then cleared.
Niv Shapira
Niv Shapira on 1 Sep 2020
I'll add here a note to whom it may concern:
The test i've performed above on the "magic" matrix "A" was not sufficient. The qr functionality and run-time may vary dramatically depending on the input matrix properties. It seems that for some matrices the run-time difference between triu(qr(A)) to qr(A,0) (or qr(A)) might be even larger.

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Christine Tobler
Christine Tobler on 1 Sep 2020
When a "~" is used for a return value, this is only for code clarity. MATLAB still needs to compute that value, so the second case is computing the Q, even though you don't intend to use it.

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