On Tue, Apr 8, 2008 at 9:54 AM, Brendan Jurd <direvus@gmail.com> wrote:
> If I understand your design correctly, this claim isn't true. If the
> topmost node reports 9 bytes free, could you not have seven pages each
> with 1 byte free, and an eighth page with 2 bytes free?
>
> 9
> 4 5
> 2 2 2 3
> 1 1 1 1 1 1 1 2
>
> So you'd actually end up walking two levels down the left hand side of
> the tree, discovering not enough space, and then walking two levels
> down the right hand side to again discover not enough space.
As I read it, each node takes the value of the largest child, not the
sum of the children.
--
Dave Page
EnterpriseDB UK: http://www.enterprisedb.com
