Jan 14, 2016

Solve your SUDOKU by SQL!



Oracle Database 11g Release 2 introduces a new feature called Recursive Subquery Factoring with the help of which you can solve your Sudoku.




To solve this Sudoku you first have to transform this to a single string by appending all rows together:(give exact amount of spaces)


“53  7    6  195    98    6 8   6   34  8 3  17   2   6 6    28    419  5    8  79”

Past this string into a Recursive Subquery, run it and you get a new string with your solved Sudoku:


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
/


The output:


534678912672195348198342567859761423426853791713924856961537284287419635345286179


This string can be transformed back to a nice display of the solution.






Adopted from blog

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