<div class="gmail_extra"><div class="gmail_quote">On Sat, Apr 28, 2012 at 04:56, Umut Tabak <span dir="ltr"><<a href="mailto:u.tabak@tudelft.nl" target="_blank">u.tabak@tudelft.nl</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div class="im"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">All you will get are asymptotics, not sharp operation counts. For 2D, the bounds are O(n^{1.5} flops and O(n log n) space; for 3D they are O(n^2) flops and O(n^{4/3}) space. This is covered in Alan George's book, <a href="http://www.cse.illinois.edu/courses/cs598mh/george_liu.pdf" target="_blank">http://www.cse.illinois.edu/<u></u>courses/cs598mh/george_liu.pdf</a><br>
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Dear Jed,<br>
I am not good at these cost computations, is 'n' the number of nonzeros in the matrices?</blockquote></div><br></div><div class="gmail_extra">No, n is the number of degrees of freedom in the model.</div><div class="gmail_extra">
<br></div><div class="gmail_extra">There is no estimate just in terms of the number of nonzeros because ultimately, the cost is limited by the size of minimal vertex separators.</div>