<p>No, a different ordering will be computed, and the ordering with the sparser matrix could even lead to more fill.</p>
<div class="gmail_quote">On Dec 9, 2011 4:13 PM, "Xiangdong Liang" <<a href="mailto:xdliang@gmail.com">xdliang@gmail.com</a>> wrote:<br type="attribution"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
On Fri, Dec 9, 2011 at 7:02 PM, Barry Smith <<a href="mailto:bsmith@mcs.anl.gov">bsmith@mcs.anl.gov</a>> wrote:<br>
><br>
> On Dec 9, 2011, at 5:55 PM, Xiangdong Liang wrote:<br>
><br>
>> Hello everyone,<br>
>><br>
>> I am solving Ax=b with sparse direct solver Pastix. I have two<br>
>> equivalent A's (upto these zero entries): A1 and A2. A1 is generated<br>
>> with ignor_zero_entries and A2 is without this option. For example A1<br>
>> has 9 millions nonzeros, while A2 has 10 millions zeros. When I solve<br>
>> them with Pastix, I found the time for solving sparser A1 actually is<br>
>> longer (10%--20% worse) than A2. Does anyone have thoughts on this?<br>
><br>
> Completely possible and not particularly surprising. The amount of work required for sparse LU depends in an incrediably complicated way on the nonzero structure of the matrix, it is only very very minorly related to the number of nonzeros in the matrix.<br>
> One fun thing to check would be the number of nonzeros in the factor of A1 and the number of nonzeros in the factor of A2. (I'm not sure if Pastix has a way to check this).<br>
<br>
Should the nnz in the factor of A1 be the same as nnz in the factor of<br>
A2 since A1 and A2 are the same matrices except some zeros (due to<br>
ignore_zero_entries in the matrix assembling)?<br>
<br>
Xiangdong<br>
<br>
><br>
> Barry<br>
><br>
>> Thanks.<br>
>><br>
>> Xiangdong<br>
>><br>
>> P.S. The time I count is only for Spare LU solving (not including the<br>
>> matrix assembling time).<br>
><br>
</blockquote></div>