Tuesday, November 28, 2017

SUDOKU Solver based on CTE from MySQL 8.0 

SUDOKU is an interesting problem.   Using SQL to solve this problem is not anything magical.



An interesting twitter post from

https://twitter.com/VadimTk/status/916734787557064704

Thanks to  Vadim Tkachenko


The following SQL with MySQL 8.0.3 RC1, the New Feature "Recursive Common Table Expression/CTE" enables the easy way of this SUDOKU solver.



select @myproblem:='..41..2.3........12.....8..82.6.43.....8.9.....67.2.48..5.....64........3.7..69..';

WITH RECURSIVE
        my19(n) AS (
        SELECT 1 AS n
        UNION ALL
        SELECT 1+n FROM my19 WHERE n<9
)
select substr(@myproblem,(n-1)*9+1,9) as sud from my19;


select @t:=sysdate(6);

WITH RECURSIVE
        input(sud) as (
        select @myproblem
),
        digits(z,lp) as (
        select '1', 1
        union all
        select cast(lp+1 as char), lp+1 from digits where lp<9
),
        x(s,ind) as (
        select sud, instr(sud,'.') from input
        union all
        select concat(substr(s,1, ind-1), z, substr(s, ind+1)),
        instr( concat(substr(s,1,ind-1), z, substr(s, ind+1)), '.')
        from x, digits as z
        where ind> 0
        and not exists (
                select 1 from digits as lp
                where z.z = substr(s, ((ind-1) div 9) *9 + lp, 1)
                or z.z = substr(s, ((ind-1)%9) + (lp-1) *9 + 1, 1)
                or z.z = substr(s, (((ind-1) div 3) %3) * 3
                        + ((ind-1) div 27 ) * 27 + lp
                        + ((lp-1) div 3) * 6, 1)
                )
),
        my19(n) AS (
        SELECT 1 AS n
        UNION ALL
        SELECT 1+n FROM my19 WHERE n<9
)

select substr(s,(n-1)*9 + 1,9) as ans from x,my19 where ind=0 ;

select @t as start_time, timediff(sysdate(6),@t);



The Output from the above SQL is shown as follows :




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