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 ' || <br /> ' 63 ' || <br /> ' 2 37' || <br /> ' 5 173' || <br /> '327 14 ' || <br /> ' 1 9 ' || <br /> '6 7 ' || <br /> ' 2 381' || <br /> '83 ' s<br />from dual<br />

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[1], ' ' ) ) >= 81 or d[1] + 81 < iteration_number ) ( i[1] = instr( s[1], ' ', j[1] ) , v[1] = translate( '123456789' , '#' || substr( s[1], xxx.r[ i[1], 1], 1 ) || substr( s[1], xxx.r[ i[1], 2], 1 ) || substr( s[1], xxx.r[ i[1], 3], 1 ) || substr( s[1], xxx.r[ i[1], 4], 1 ) || substr( s[1], xxx.r[ i[1], 5], 1 ) || substr( s[1], xxx.r[ i[1], 6], 1 ) || substr( s[1], xxx.r[ i[1], 7], 1 ) || substr( s[1], xxx.r[ i[1], 8], 1 ) || substr( s[1], xxx.r[ i[1], 9], 1 ) || substr( s[1], xxx.r[ i[1],10], 1 ) || substr( s[1], xxx.r[ i[1],11], 1 ) || substr( s[1], xxx.r[ i[1],12], 1 ) || substr( s[1], xxx.r[ i[1],13], 1 ) || substr( s[1], xxx.r[ i[1],14], 1 ) || substr( s[1], xxx.r[ i[1],15], 1 ) || substr( s[1], xxx.r[ i[1],16], 1 ) || substr( s[1], xxx.r[ i[1],17], 1 ) || substr( s[1], xxx.r[ i[1],18], 1 ) || substr( s[1], xxx.r[ i[1],19], 1 ) || substr( s[1], xxx.r[ i[1],20], 1 ) , '#' ) , j[1] = case when j[1] >= 81 then 1 else j[1] + 1 end , s[1] = case when length( v[1] ) = 1 then substr( s[1], 1, i[1] - 1 ) || v[1] || substr( s[1], i[1] + 1 ) else s[1] end , d[1] = case when length( v[1] ) = 1 then iteration_number else d[1] end )

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

( i[1] = instr( s[1], ' ', j[1] )

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

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

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

## 16 Comments

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?

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

My solver can solve any solvable sudoku.

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!

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

At the end there’s a link named “code” to http://technology.amis.nl/blog/wp-content/images/sudomod.zip, the zip-file contains the SQL

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

-Mark

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

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

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

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

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).

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

respect!

As always; Brilliant.

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.