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http://www.openoffice.org/issues/show_bug.cgi?id=26022
------- Additional comments from [EMAIL PROTECTED] Sun Jul 17 15:10:56 -0700
2005 -------
The trick I did is this.
In the old algorithm, the poisson variable was computed as follows (in
pseudocode):
exp( -lambda ) * lambda^x / x!
But alas! lambda^x can overflow when lambda or x (or both) is sufficiently
large. So I changed it to the following equivalent formula:
lambda lambda lambda
exp( -lambda ) * ------ * ------ * ... * ------
1 2 x
This yields an equivalent result within a binary float-point rounding error,
without a reduced risk of having an intermediate value that is large enough to
overflow.
Kohei
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