Re: SQL (was: Why using "Group By")

>>Do you have an example problem requiring recursion for which
transitive closure provides no solution? <<

The Ackermann function, for one:

 A(0, m) = m+1
 A(n+1, 0) = A(n, 1)
 A(n+1, m+1) = A(n, A(n+1, m))