Re: sudoku in an sql statement - Mailing list pgsql-general

From Richard Broersma
Subject Re: sudoku in an sql statement
Date
Msg-id 396486430911041531r1596e2aek401766e7f55aabbb@mail.gmail.com
Whole thread Raw
In response to Re: sudoku in an sql statement  (marcin mank <marcin.mank@gmail.com>)
Responses Re: sudoku in an sql statement  (Andrej <andrej.groups@gmail.com>)
Re: sudoku in an sql statement  (marcin mank <marcin.mank@gmail.com>)
List pgsql-general
On Wed, Nov 4, 2009 at 3:18 PM, marcin mank <marcin.mank@gmail.com> wrote:

>                                         s
> -----------------------------------------------------------------------------------
>  534678912672195348198342567859761423426853791713924856961537284287419635345286179
> (1 row)


I don't get the same results:

broersr=> with recursive x( s, ind ) as
broersr-> ( select sud, position( ' ' in sud )
broersr(>  from  (select '53  7    6  195    98    6 8   6   34  8 3  17   2
broersr'> 6 6    28    419  5    8  79'::text as sud) xx
broersr(>  union all
broersr(>  select substr( s, 1, ind - 1 ) || z || substr( s, ind + 1 )
broersr(>       , position(' ' in repeat('x',ind) || substr( s, ind + 1 ) )
broersr(>  from x
broersr(>     ,  (select gs::text as z from generate_series(1,9) gs)z
broersr(>  where ind > 0
broersr(>  and not exists ( select null
broersr(>                   from generate_series(1,9) lp
broersr(>                   where z.z = substr( s, ( (ind - 1 ) / 9 )
* 9 + lp, 1 )
broersr(>                   or    z.z = substr( s, mod( ind - 1, 9 ) -
8 + lp * 9, 1 )
broersr(>                   or    z.z = substr( s, mod( ( ( ind - 1 )
/ 3 ), 3 ) * 3
broersr(>                                      + ( ( ind - 1 ) / 27 ) * 27 + lp
broersr(>                                      + ( ( lp - 1 ) / 3 ) * 6
broersr(>                                   , 1 )
broersr(>                 )
broersr(> )
broersr-> select s
broersr-> from x
broersr-> where ind = 0;
 s
---
(0 rows)


--
Regards,
Richard Broersma Jr.

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