[pure-lang-svn] SF.net SVN: pure-lang:[859] pure/trunk/examples/gauss.pure

[<ag...@us...>]

Revision: 859
          http://pure-lang.svn.sourceforge.net/pure-lang/?rev=859&view=rev
Author:   agraef
Date:     2008-09-25 09:37:07 +0000 (Thu, 25 Sep 2008)

Log Message:
-----------
Comment change.

Modified Paths:
--------------
    pure/trunk/examples/gauss.pure

Modified: pure/trunk/examples/gauss.pure
===================================================================
--- pure/trunk/examples/gauss.pure	2008-09-25 09:36:44 UTC (rev 858)
+++ pure/trunk/examples/gauss.pure	2008-09-25 09:37:07 UTC (rev 859)
@@ -17,12 +17,13 @@
 
 /* Here is a brief rundown of the algorithm: First we find the pivot element
    in column j of the matrix. (We're doing partial pivoting here, i.e., we
-   only look for the pivot in column j, starting at row i.) If the pivot is
-   zero then we're done (the pivot column is already zeroed out). Otherwise,
-   we bring it into the pivot position (swapping row i and the pivot row),
-   divide the picot row by the pivot, and subtract suitable multiples of the
-   pivot row to eliminate the pivot column in all subsequent rows. Finally we
-   update i and p accordingly and return the result. */
+   only look for the pivot in column j, starting at row i. That's usually good
+   enough to achieve numerical stability.) If the pivot is zero then we're
+   done (the pivot column is already zeroed out). Otherwise, we bring it into
+   the pivot position (swapping row i and the pivot row), divide the pivot row
+   by the pivot, and subtract suitable multiples of the pivot row to eliminate
+   the elements of the pivot column in all subsequent rows. Finally we update
+   i and p accordingly and return the result. */
 
 step (p,i,x) j
 = if max_x>0 then


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