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May 28, 2022 at 7:11 answer added Sextus Empiricus timeline score: 1
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S Jun 15, 2019 at 11:42 history suggested burning_question CC BY-SA 4.0
Fixed a typo: it should be Nx1 columns, not 1xM columns
Jun 15, 2019 at 11:40 comment added gung - Reinstate Monica Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question.
Jun 15, 2019 at 10:41 review Suggested edits
S Jun 15, 2019 at 11:42
Jun 15, 2019 at 10:38 answer added burning_question timeline score: 0
S Sep 7, 2018 at 13:13 history suggested burning_question CC BY-SA 4.0
added more information to address questions raised in the comments
Sep 7, 2018 at 12:01 review Suggested edits
S Sep 7, 2018 at 13:13
Sep 7, 2018 at 2:37 comment added jbowman It's not clear to me what you mean by "I'm only able to load $1 \times M$ columns of $A$"? Evidently you don't mean "I can load all $M$ columns of $A$"... And why would loading row-by-row not be possible?
Sep 6, 2018 at 19:48 comment added whuber Of course: with a vector $x$ in RAM, accumulate the values of $y_i$ and $y_ix_i$ as you read in the vector $y.$ Repeat for all entries of $A^\prime A.$ Indeed, you only need $O(M^2)$ RAM if you're willing to read the columns an average of $(M+3)/2$ times each.
Sep 6, 2018 at 18:30 review First posts
Sep 6, 2018 at 18:39
Sep 6, 2018 at 18:30 history asked burning_question CC BY-SA 4.0