Solving a Suduku with one SQL-statement, is that possible? A lot of people won’t believe it, but yes, it can be done.
I did already a blog on Solving a Sudoku with Collections, but for this blog I used another aproach: the model clause. The model clause is introduced in Oracle 10g and, according to the documentation “brings a new level of power and flexibility to SQL calculations”. And that isn’t too much said! But how can you use it for solving Sudokus? Quit simple in fact , select it as a string from 81 characters from dual

select '   56  2 ' ||
' 63      ' ||
'   2   37' ||
' 5    173' ||
'327  14  ' ||
'  1  9   ' ||
'6   7    ' ||
'    2 381' ||
'83       ' s
from dual

and add the model clause.

model
reference xxx on
( select i, j, r
from dual
model
dimension by ( 1 i, 1 j )
measures ( 1 x, 1 y, 1 r )
rules
( x[for i from 1 to 81 increment 1, 1] = trunc( ( cv(i) - 1 ) / 9 ) * 9
, y[for i from 1 to 81 increment 1, 1] = mod( cv(i) - 1, 9 ) + 1
, r[for i from 1 to 81 increment 1, for j from 1 to 8 increment 1] = case when x[ cv(i), 1 ] + cv(j) < cv(i)
then x[ cv(i), 1 ] + cv(j)
else x[ cv(i), 1 ] + cv(j) + 1
end
, r[for i from 1 to 81 increment 1, for j from 9 to 16 increment 1] = case when y[ cv(i), 1 ] + ( cv(j) - 9 ) * 9 < cv(i)
then y[ cv(i), 1 ] + ( cv(j) - 9 ) * 9
else y[ cv(i), 1 ] + ( cv(j) - 8 ) * 9
end
, r[for i from 1 to 81 increment 1, 17] = case mod( x[ cv(i), 1 ] / 9, 3 )
when 0 then x[ cv(i), 1 ] + 9
when 1 then x[ cv(i), 1 ] - 9
when 2 then x[ cv(i), 1 ] - 18
end + mod( y[ cv(i), 1 ], 3 ) + trunc( (y[ cv(i), 1 ] - 1) / 3 ) * 3 + 1
, r[for i from 1 to 81 increment 1, 18] = case mod( x[ cv(i), 1 ] / 9, 3 )
when 0 then x[ cv(i), 1 ] + 18
when 1 then x[ cv(i), 1 ] + 9
when 2 then x[ cv(i), 1 ] - 9
end + mod( y[ cv(i), 1 ], 3 ) + trunc( (y[ cv(i), 1 ] - 1) / 3 ) * 3 + 1
, r[for i from 1 to 81 increment 1, 19] = case mod( x[ cv(i), 1 ] / 9, 3 )
when 0 then x[ cv(i), 1 ] + 9
when 1 then x[ cv(i), 1 ] - 9
when 2 then x[ cv(i), 1 ] - 18
end + mod( y[ cv(i), 1 ] + 1, 3 ) + trunc( (y[ cv(i), 1 ] - 1) / 3 ) * 3 + 1
, r[for i from 1 to 81 increment 1, 20] = case mod( x[ cv(i), 1 ] / 9, 3 )
when 0 then x[ cv(i), 1 ] + 18
when 1 then x[ cv(i), 1 ] + 9
when 2 then x[ cv(i), 1 ] - 9
end + mod( y[ cv(i), 1 ] + 1, 3 ) + trunc( (y[ cv(i), 1 ] - 1) / 3 ) * 3 + 1
)
) dimension by ( i, j ) measures ( r )

I use a reference model, which in turn is a SQL-query with a model clause. In a Sudoku can a certain number occur only once on every row, every column and every block. This reference model gives me for every cell in the Sudoku the numbers, that is the index in my Sudoku string, of the others cells on the same row, the same column and the same block. And that is all we need for the main model.

main solve
dimension by ( 1 x )
measures ( cast( s as varchar2(81) ) s
, 1 i
, 1 j
, 0 d
, cast( '' as varchar2(20) ) v
)
rules iterate ( 100000 ) until ( length( replace( s, ' ' ) ) >= 81 or d + 81 < iteration_number )
( i = instr( s, ' ', j )
, v = translate( '123456789'
, '#' ||
substr( s, xxx.r[ i, 1], 1 ) ||
substr( s, xxx.r[ i, 2], 1 ) ||
substr( s, xxx.r[ i, 3], 1 ) ||
substr( s, xxx.r[ i, 4], 1 ) ||
substr( s, xxx.r[ i, 5], 1 ) ||
substr( s, xxx.r[ i, 6], 1 ) ||
substr( s, xxx.r[ i, 7], 1 ) ||
substr( s, xxx.r[ i, 8], 1 ) ||
substr( s, xxx.r[ i, 9], 1 ) ||
substr( s, xxx.r[ i,10], 1 ) ||
substr( s, xxx.r[ i,11], 1 ) ||
substr( s, xxx.r[ i,12], 1 ) ||
substr( s, xxx.r[ i,13], 1 ) ||
substr( s, xxx.r[ i,14], 1 ) ||
substr( s, xxx.r[ i,15], 1 ) ||
substr( s, xxx.r[ i,16], 1 ) ||
substr( s, xxx.r[ i,17], 1 ) ||
substr( s, xxx.r[ i,18], 1 ) ||
substr( s, xxx.r[ i,19], 1 ) ||
substr( s, xxx.r[ i,20], 1 )
, '#' )
, j = case
when j >= 81 then 1
else j + 1
end
, s = case
when length( v ) = 1 then substr( s, 1, i - 1 ) || v || substr( s, i + 1 )
else s
end
, d = case
when length( v ) = 1 then iteration_number
else d
end
)

And that means: find the first unsolved cell, starting from cell j

( i = instr( s, ' ', j )

for this cell, check the values from the other cells on the same row, column, block and find the possible values for this cell with the translate

, v = translate( '123456789'
, '#' ||
substr( s, xxx.r[ i, 1], 1 ) ||
substr( s, xxx.r[ i, 2], 1 ) ||
substr( s, xxx.r[ i, 3], 1 ) ||
substr( s, xxx.r[ i, 4], 1 ) ||
substr( s, xxx.r[ i, 5], 1 ) ||
substr( s, xxx.r[ i, 6], 1 ) ||
substr( s, xxx.r[ i, 7], 1 ) ||
substr( s, xxx.r[ i, 8], 1 ) ||
substr( s, xxx.r[ i, 9], 1 ) ||
substr( s, xxx.r[ i,10], 1 ) ||
substr( s, xxx.r[ i,11], 1 ) ||
substr( s, xxx.r[ i,12], 1 ) ||
substr( s, xxx.r[ i,13], 1 ) ||
substr( s, xxx.r[ i,14], 1 ) ||
substr( s, xxx.r[ i,15], 1 ) ||
substr( s, xxx.r[ i,16], 1 ) ||
substr( s, xxx.r[ i,17], 1 ) ||
substr( s, xxx.r[ i,18], 1 ) ||
substr( s, xxx.r[ i,19], 1 ) ||
substr( s, xxx.r[ i,20], 1 )
, '#' )

and if only one value is possible we have solved this cell

, s = case
when length( v ) = 1 then substr( s, 1, i - 1 ) || v || substr( s, i + 1 )
else s
end

and do this again and again until we have solved the complete Sudoku or don’t find anything to solve

rules iterate ( 100000 ) until ( length( replace( s, ' ' ) ) >= 81 or d + 81 < iteration_number )

Put this all together, and you have a query wich can solve only “simple” Sudokus. A slightly more complex version solves every possible Soduku, by trying every possible value in a cell, and try another possible value if the first gues proves to be wrong.

rules iterate ( 999999 ) until ( length( replace( s, ' ' ) ) >= 81 )
( i[ it ] = case
when m = 1 then instr( s, ' ' )
else i[ cv() ]
end
, v[ it ] = case
when m = 1 then
translate( '123456789'
, '#' ||
substr( s, xxx.r[ i[cv()], 1], 1 ) ||
substr( s, xxx.r[ i[cv()], 2], 1 ) ||
substr( s, xxx.r[ i[cv()], 3], 1 ) ||
substr( s, xxx.r[ i[cv()], 4], 1 ) ||
substr( s, xxx.r[ i[cv()], 5], 1 ) ||
substr( s, xxx.r[ i[cv()], 6], 1 ) ||
substr( s, xxx.r[ i[cv()], 7], 1 ) ||
substr( s, xxx.r[ i[cv()], 8], 1 ) ||
substr( s, xxx.r[ i[cv()], 9], 1 ) ||
substr( s, xxx.r[ i[cv()],10], 1 ) ||
substr( s, xxx.r[ i[cv()],11], 1 ) ||
substr( s, xxx.r[ i[cv()],12], 1 ) ||
substr( s, xxx.r[ i[cv()],13], 1 ) ||
substr( s, xxx.r[ i[cv()],14], 1 ) ||
substr( s, xxx.r[ i[cv()],15], 1 ) ||
substr( s, xxx.r[ i[cv()],16], 1 ) ||
substr( s, xxx.r[ i[cv()],17], 1 ) ||
substr( s, xxx.r[ i[cv()],18], 1 ) ||
substr( s, xxx.r[ i[cv()],19], 1 ) ||
substr( s, xxx.r[ i[cv()],20], 1 )
, '#' )
else v[ cv() ]
end
, m = nvl2( v[ it ], m, 0 )
, it = case
when m = 1 then it
else it - 1
end
, j[ it ] = case
when m = 1 then 1
else j[ cv() ] + 1
end
, m = case
when length( v[ it ] ) >= j[ it ] then 1
else m
end
, s = case
when m = 1 then substr( s, 1, i[ it ] - 1 ) || substr( v[ it ], j[ it ], 1 ) || substr( s, i[ it ] + 1 )
else substr( s, 1, i[ it ] - 1 ) || ' ' || substr( s, i[ it ] + 1 )
end
, it = case
when m = 1 then it + 1
else it
end
)

This query can take some more time, up to 60 seconds.

These queries should work on every 10G database, but I have tested it only on 10.2.0.1.0, and I have heard that they won’t work on 10.2.0.3.0

Anton

code Oracle Consultant at AMIS

1. This is a nice way to import a Sudoku puzzle.

Try integrating it with a user interface like this example: http://www.vantasyworld.com/fun/sudoku/sudokusolver.html.

Does it also detect sudoku puzzles that are impossible to solve?

2. Sathya Narayanan on

Great work Anton, I am trying hard to understand this but I have failed every time… Could you please explain theÂ indexing that isÂ Â made in the reference model? Thank You

3. Anton Scheffer on

My solver can solve any solvable sudoku.

4. I am impressed to solve a Sudoku in one statement but how about solving complex Sudokus can it handle all?

http://vbaexcel.eu/vba-macro-code/sudoku-solver

This solver I have been using works fine but it is alot of code compared to yours!

5. Incroyable, trÃ¨s intÃ©ressant et challenge de rÃ©soudre une grille de Sudoku avec SQL, chapeau l’artiste.

6. Never mind….just saw the comment about not working in 10.2.0.3…..sigh……

-Mark

7. Does anyone have this in one simple SQL, without it broken up into pieces? I can’t get it to execute.

I assume I need the first part, select from dual, then the model clause (model reference xxx on …. dimension by (i,j) measures (r), then followed by the “main solve” clause?

I keep getting syntax errors trying to run it, in 10.2.0.3.

-Mark

8. Thanks for this wonderful and amazing sql! It’s really impressive 🙂

9. It is sad that this site thinks Mac OS X Safari is “

10. you need to read this. This guy wrote a high-performance stored-procedure version of the same thing:

http://www.devx.com/dbzone/Article/33551

FULL DISCLOSURE: That guy happens to be my dad 😉

11. Mike Brandt on

Well, that was very impressive. I am still working it out .

By the way – if you like sql pluzzles you might like to try this, which can be solved by a single sql statement ( it will work even with Oracle 8i) , assuming you have a table called nums that holds the values 2..100.
I like this problem , since it is almost impossible to solve by hand. By the way, the answer is the same even if you make the largest value 1000 (and then solving by hand becomes a nightmare).

efine minval=2
define maxval=100
set numwidth 6
set pages 0
set feedback off
set verify off
column a3 format a3
column a15 format a15
column num format 9999
prompt
prompt Solve the problem set by:
prompt
prompt There are 2 integers a and b between &amp;&amp;minval and &amp;maxval, with a= a.num;

create view sums as select a+b s,count(*) cc
from pairs
group by a+b;

create view products as select a*b p,count(*) cc
from pairs
group by a*b;

prompt
prompt Paul: I don't know what the numbers are
prompt Sam : I knew that. I don't know what the numbers are either
prompt
prompt Paul's first statement is more helpful than it first appears
prompt The pair of numbers cannot be uniquely determined by their product
prompt This includes not just 2 primes such as 7 and 11,
prompt but also pairs like 19 and 38, or 4 and 53, 77 and 17
prompt (assuming we are only looking at 2 digit numbers)
prompt
prompt Sam's first statement is even more helpful.
prompt Sam knows that Paul can't possibly know the numbers
prompt So whatever his sum is, it cannot
prompt be formed by a pair that uniquely determines their product
prompt For example : the sum cannot be 18, or 57 or 94 using the above examples
prompt
prompt generate poss_sums as
prompt the only sums that cannot possibly be made by distinct product pairs
prompt There are surprisingly few

create view poss_sums as
select s from sums where cc &gt; 1
minus
select pairs.a+pairs.b s from pairs where pairs.a*pairs.b in
(select p from products where cc=1);

prompt
prompt From Sam's first statement the sum must be one of
prompt
select s from poss_sums
order by s;

prompt
prompt Paul's last statement:
prompt Paul: I do now!
prompt
prompt So Paul knows that Sam knew that he did not know the numbers and so
prompt must have a sum matching the above criterion
prompt This enables him to deduce the pairs.
prompt So Paul must have a product that decomposes to one of the sums
prompt in the above list
prompt in exactly one way (2 or more would not give him a solution and 0
prompt would violate Sam's previous statement)
prompt find all products that can have exactly one sum matching the list of sums above
prompt

create view poss_prods as
select products.p,max(poss_sums.s) s
from products,pairs,poss_sums
where products.cc &gt; 1
and pairs.a*pairs.b = products.p
and pairs.a+pairs.b = poss_sums.s
group by products.p
having count(poss_sums.s) = 1;

select 'product is ', p, 'sum is', s from poss_prods
order by p,s;

-- now find the sum that has only one possible product fitting that criterion

prompt
prompt Sam's last statement:
prompt Sam : So do I!
prompt
prompt So we need to find a sum that only occurs once in the above list
prompt Otherwise Sam could still not deduce the pair.
prompt and as we know both the product and the sum we can find the pair
prompt

col plan_plus_exp for a120
set lines 160

select 'In the range' a15,min(num) num,'to' a3,max(num) num from nums;

set autotrace on explain

select 'The numbers are' a15, a num,'and' a3, b num
from pairs,
( select max(p) p,s from poss_prods group by s having count(*) = 1) pp
where pairs.a*pairs.b = pp.p
and pairs.a+pairs.b= pp.s
order by a,b;

set autotrace off

drop view poss_prods;
drop view poss_sums;

drop view products;
drop view sums;
drop view pairs;

prompt
prompt or to be perverse, and use only in-line views
prompt

-- using the with clause is much too slow in 10.1
-- with nums as (select rownum+&amp;&amp;minval-1 num from sys.tab\$
-- where rownum = a.num
) pairs,
( select max(p) p,s
from ( select products.p,max(poss_sums.s) s
from ( select a.num*b.num p,count(*) cc
from nums a, nums b
where b.num&gt;= a.num
group by a.num*b.num
having count(*) &gt; 1
) products,
( select a.num a, b.num b
from nums a, nums b
where b.num &gt;= a.num
) prs,
(
select s
from ( select a.num+b.num s
from nums a, nums b
where b.num &gt;= a.num
group by a.num+b.num
having count(*) &gt; 1
) sums
minus
select prs2.a+prs2.b s
from ( select a.num a, b.num b
from nums a, nums b
where b.num &gt;= a.num
) prs2
where prs2.a*prs2.b in
( select a.num*b.num p
from nums a, nums b
where b.num &gt;= a.num
group by a.num*b.num
having count(*) = 1
)
) poss_sums
where prs.a*prs.b = products.p
and prs.a+prs.b = poss_sums.s
group by products.p
having count(poss_sums.s) = 1
)
group by s having count(*) = 1
) pp
where pairs.a*pairs.b = pp.p
and pairs.a+pairs.b= pp.s
order by a,b;

set autotrace off
set timing off

drop table nums;

12. I lost you directly after the ‘select string from dual’ I’m afraid. I hope I never have to write anything like this 😉

13. respect!

14. As always; Brilliant.

15. I’m filing this one away for posterity.

I hope it will work on Oracle 10XE because I would like to have this available to me on my laptop.