Incepand cu versiunea 11 Release 2 Oracle introduce un nou feature in RDBMS lor, Recursive Subquery Factoring. Prima aplicatie practica la care s-a gandit Anton Scheffer este cum sa rezolve SUDOKU

<pre>with x( s, ind ) as
( select sud, instr( sud, ' ' )
from ( select '53  7    6  195    98    6 8   6   34  8 3  17   2   6 6    28    419  5    8  79' sud from dual )
union all
select substr( s, 1, ind - 1 ) || z || substr( s, ind + 1 )
, instr( s, ' ', ind + 1 )
from x
, ( select to_char( rownum ) z
from dual
connect by rownum <= 9
) z
where ind > 0
and not exists ( select null
from ( select rownum lp
from dual
connect by rownum <= 9
)
where z = substr( s, trunc( ( ind - 1 ) / 9 ) * 9 + lp, 1 )
or    z = substr( s, mod( ind - 1, 9 ) - 8 + lp * 9, 1 )
or    z = substr( s, mod( trunc( ( ind - 1 ) / 3 ), 3 ) * 3
+ trunc( ( ind - 1 ) / 27 ) * 27 + lp
+ trunc( ( lp - 1 ) / 3 ) * 6
, 1 )
)
)
select s
from x
where ind = 0
/</pre>

Astfel ca daca vrem sa rezolvam exemplul de mai jos folosim codul de mai sus

Sudoku Example

Practic folosim sirul din exemplu “53 7 6 195 98 6 8 6 34 8 3 17 2 6 6 28 419 5 8 79″ si ii aplicam sintaxa SQL. Rezultatul este rezolvarea la exemplul nostru.

Sudoku Example - Rezolvarea

Rareori vezi cod care sa te impresioneze.

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