SQLServerCentral.com / Article Discussions / Article Discussions by Author / Discuss content posted by Kevin Duan / A set based T-SQL Solution for Sudoku Puzzle / Latest PostsInstantForum.NET v2.9.0SQLServerCentral.comhttp://www.sqlservercentral.com/Forums/notifications@sqlservercentral.comWed, 01 Apr 2015 07:55:23 GMT20RE: A set based T-SQL Solution for Sudoku Puzzlehttp://www.sqlservercentral.com/Forums/Topic761296-1264-1.aspxHi Dave,Sorry for my overly simplied comments. Yes, you are right. The input string is composed by the contents in the cells, from left to right, from top to the bottom, it should be exactly 81 characters in length. For non-predefined (empty) cells, use 0's. You can jump to the output script at the bottom anywhere after the input is read, to see the puzzle values.Thanks,KevinTue, 04 Aug 2009 12:37:19 GMTKevin DuanRE: A set based T-SQL Solution for Sudoku Puzzlehttp://www.sqlservercentral.com/Forums/Topic761296-1264-1.aspxHi Ian,Thanks for your posting and your detailed suggestion. It’s my pleasure to have discussion on the “ugly” script that I would never have expected anyone be patient reading through ;-) This script was created to be used in my SQL Server class to illustrate: • The new T-SQL syntax in SQL Server 2005. • The set-based approach for solving problems that have unique constraint on each set (specifically in Soduko puzzles, the rows, columns and blocks). • The performance comparison.For Soduko puzzles, there are two main types of solutions. One is iteration approach, guessing and verifying, as you suggested; the other one is set-based approach, eliminating and exclusion. Because RDBMS are very good at set-based approach, and the uniqueness constraint (in row, column and block) is a perfect scenario for using bitwise operators (using logical AND and OR to find and subtract all duplicates in one shot), I used this as a coaching example. For the purpose of illustration, I didn’t use any of the iteration methods in the script. Actually, I have the scripts using iteration approach in T-SQL, Java and JavaScript, they are posted below. In terms of bitwise operation, it’s not mandatory for the set based approach. I have a script using character string values instead of binary values(script attached below), but I needed an extra loop to implement the elimination rules. I am a big fan of binary data structure, and it's always been my first choice when applicable, say flag related problems(but unfortunately, SQL Server does not support bitmap index unless in the OLAP cube).A couple of extra notes on the set-based script: • There should be only one solution for the puzzle; otherwise the eliminations will be confused and stopped at certain point. But you can find out all solutions by using iteration approach easily. • The elimination rule has to be complete (which is not in my script, see the exception I gave in the posting. But I believe there ARE some elimination rules that can be added to make it perfect). On the performance side, SQL is faster in set-based approach than iteration approach in a lot of cases (BTW, Itzik Ben-gan's SQL books are very good resources on the in-depth SQL skill, especially in this type of comparison). But using iteration approach in procedure languages such as Java, C++ or even JavaScript is even much faster. That is why I usually strongly recommend implementing the business logic at the application (Middle) tier other than doing it inside the databases, which is the most expensive resource in the application. Scalability is also a big concern (you could have tens of application servers running the iteration code).Yes, I should have made the scripts perfect even though I just used it for teaching. If you have any good ideas or suggestions, I am more than happy to learn.Thanks,KevinFollowing is the iteration approach in T-SQL.[code]USE [tempdb]GOIF OBJECT_ID(N'dbo.uspSolveSudoku_2', N'P') IS NOT NULL DROP PROCEDURE dbo.uspSolveSudoku_2;GOCREATE PROCEDURE [uspSolveSudoku_2] ( @t VARCHAR(81), @Trace INT = 0 ) ASSET NOCOUNT ON DECLARE @datetime DATETIME, @ii BIGINT, @S VARCHAR(max), -- temp variable for building sql query strings @SQL1 VARCHAR(max), -- query for basic elimination rules @SQL2 VARCHAR(max), -- first query for second elimination rules @SQL3 VARCHAR(max) -- second query for second elimination rulesSET @datetime = GETDATE();------------------------------------------------------------------------------------Step 1: Setup working table #SDK.CREATE TABLE #SDK ( i INT, -- Working table, one record per cell, with row/column/block lable and current possible values--DECLARE #SDK TABLE ( i INT, -- Working table, one record per cell, with row/column/block lable and current possible values r INT, -- Row # c INT, -- Column # b INT, -- Block # v INT);WITH n (i, r, c, b, v) AS ( SELECT 1, 1, 1 , 1, CASE WHEN substring(@t,1,1)>0 THEN -CONVERT(int, substring(@t,1,1)) ELSE 0 END UNION ALL SELECT i+1, i/9 +1, (i % 9) +1, (i / 27)*3 + ((i%9)/3+1) , CASE WHEN substring(@t, i+1,1)>0 THEN -CONVERT(INT, substring(@t, i+1,1)) ELSE 0 END FROM n WHERE i < 81) INSERT INTO #SDK (i,r,c,b,v) SELECT * FROM n OPTION (MAXRECURSION 80); DECLARE @i INT, @STEP INT, @V INT, @R INT, @C INT, @B INT, @P TINYINTSELECT @i = 1, @STEP = 1, @ii = 0WHILE @i>0 AND @i<82 BEGIN SET @ii = @ii+1 IF @Step<>0 SELECT @V = V, @R = R, @C = C, @B = B FROM #SDK WHERE i= @i -- Given cell, go forward or backward to skip IF @V <0 BEGIN SELECT @i = @i + @Step END ELSE -- Already hit max, reset to zero and getting backward IF @V=9 BEGIN UPDATE #SDK SET v=0 WHERE i= @i SELECT @Step = -1, @i = @i - 1 END ELSE BEGIN SET @v = @v +1 UPDATE #SDK SET v = @V WHERE i= @i IF EXISTS (select * FROM #SDK WHERE i<>@i AND ABS(v)=@V AND (r=@r OR c= @c OR b= @b)) BEGIN SET @Step = 0 END ELSE BEGIN SELECT @Step = 1, @i=@i+1 END END -- SELECT @step IF @ii = -1 GOTO OutputSudoku ENDSELECT @ii ------------------------------------------------------------------------------------Step 4: Output the SolutionOutputSudoku: -- you can add "GOTO OutputSudoku" anywhere above to stop the calculation and see the interim results!WITH a AS ( SELECT r AS 'Row', c, v FROM #SDK ) SELECT * FROM a pivot ( max(v) for c in ([1],[2],[3],[4],[5],[6],[7],[8],[9]) ) AS pSELECT DATEDIFF(ms, @datetime, GETDATE())GOEXEC uspSolveSudoku_2 '080001000030750000100000027000004301400000006701300000520000004000049070000800090',1 -- an evil level puzzle--EXEC uspSolveSudoku_2 '080700000902000000300090020060800200750109043009004070040050009000000706000007030',1--EXEC uspSolveSudoku_2 '200060000000900871740008006006080030003000100090030400300700018972005000000090002' , 1--EXEC uspSolveSudoku_2 '200060000000900871740008006006080030003000100090030400300700018972005000000090002' ,1--EXEC uspSolveSudoku_2 '029000008030000010000520097070056100000000000006310070760041000050000020800000630' , 1--EXEC uspSolveSudoku_2 '080700000902000000300090020060800200750109043009004070040050009000000706000007030',1--EXEC uspSolveSudoku_2 '060104050008305600200000001800407006006000300700901004500000002007206900040508070',1-- Code Cleaning:[/code]Following is the iteration approach in JavaScript.[code]//Create two arrays:// a: Holds the puzzle value.// b: Holds the predefined flag(non-zero)a = new Array(0,0);b = "060590004"+"005004302"+"290700100"+"000017238"+"400000006"+"378920000"+"002001049"+"507600800"+"800035010";for (i = 0; i < b.length; i++){ a[i] = parseInt(b[i]);}//Loop through each none predefined cellL: for (k = FindNext(-1); k<b.length; k++){ while (a[k]+= 1 ){ if (a[k] == 10){ if ((k = GoBack(k))== -1){ document.write("Cannot find solution!!"); break L; } } Else{ if (Validate(k)){ if ((k = FindNext(k)) == -1){ // solution found ! document.write("Found solution"); document.write(a); break L; } } } } }// Sets the pointer to the next non-predefined cell after cell i,// returns -1 when goes beyond the boundary.function FindNext(i){ for(j=1;;j++){ if (i+j== b.length ) return -1; if (b[i+j]==0) return i+j; } }//Sets the pointer back to the last increasable non-predefined cell,// meanwhile, sets the values of the passed cells to zero,// returns -1 when goes beyond the boundary.function GoBack(i){ a[i] = 0; while (--i >= 0){ if (b[i]==0){ if (a[i]<9){ return i;} else{ a[i] = 0; } } } return -1;}//Validate the cell xfunction Validate(x){ xr = Math.floor(x/9)*9; //Row offset xc = x % 9; //Column offset xb = (Math.floor(x / 27)) * 27 + ( (Math.floor((x % 9)/3)) *3 ); //Block offset for (i = 0; i < 3; i++){ for (j = 0; j < 3; j++){ if ((((Xr+i*3+j) != x) && (a[Xr+i*3+j] ==a[x])) || (((Xc+(i*3+j)*9)!= x) && (a[Xc+(i*3+j)*9] ==a[x])) || (((Xb+i*9 + j) != x) && (a[Xb+i*9 + j] ==a[x]))) return false; } } return true;} [/code]Following is the iteration approach in Java.[code]public class Sudoku { public int[][] s = { {0,0,9,0,2,8,0,0,0}, {0,0,0,7,0,0,0,0,2}, {0,0,0,0,0,0,1,3,4}, {0,0,0,0,6,0,0,2,8}, {8,0,0,0,4,0,0,0,9}, {3,5,0,0,7,0,0,0,0}, {6,8,4,0,0,0,0,0,0}, {9,0,0,0,0,1,0,0,0}, {0,0,0,5,9,0,7,0,0} }; private int[][] f = new int[9][9]; public int row = 0, col = -1; private String msg = "Still Trying..."; public void solveIt(){ for (int i = 0; i < 9; i++){ for (int j = 0; j < 9; j++){ f[i][j]=s[i][j]; } } while (moveNext()){ while (!this.findNextNumber()) { if (s[row][col]==10){ if (!moveBack()){ msg = "No Solution Found"; return; } } } } msg = "Congratulations! Found Solutions!"; } private boolean findNextNumber(){ while (++s[row][col]<=9) { if (checkCell()) {return true;} } return false; } private boolean moveNext(){ do { if (++col ==9) {row++; col = 0;} } while ( row<9 && f[row][col]!=0); return (row<9?true:false); } private boolean moveBack(){ s[row][col]=0; do { if (--col == -1) {row--; col = 8;} } while ( row>=0 && f[row][col] != 0); return (row < 0 ? false : true); } private boolean checkCell(){ for (int i = 0 ; i < 9; i++){ if (i != col && s[row][i]==s[row][col]) return false; if (i != row && s[i][col]==s[row][col]) return false; } for (int i = row/3*3 ; i < row/3*3+3; i++){ for (int j = col/3*3 ; j < col/3*3+3; j++){ if (!(i == row && j == col) && s[i][j]==s[row][col]) return false; } } output(); return true; } public void output(){ System.out.println(msg); for (int i = 0 ; i < 9; i++){ for (int j = 0 ; j < 9; j++){ System.out.print(s[i][j]); }; System.out.println(""); }; } public static void main(String[] args){ Sudoku mySudoku = new Sudoku (); mySudoku.solveIt(); mySudoku.output(); }}[/code]Following is the T-SQL set-based solution for character flags (vs. bitwise values) [code]IF OBJECT_ID(N'dbo.uspSolveSudoku_C', N'P') IS NOT NULL DROP PROCEDURE dbo.uspSolveSudoku_C;GOCreate procedure dbo.uspSolveSudoku_C ( @t varchar(100), @Trace bit = 0 ) ASSet nocount ondeclare @datetime datetime; select @datetime = getdate();-----------------------------------------Step 1: Setup working table #s. -- r for Row#, -- c for Column#, -- b for Block#, -- v for Value-- d for current # of possible values---------------------------------------Create table #s (i int, r int, c int, b int, v varchar(9), d as len(v));WITH n (i, r, c, b, v) AS ( SELECT 1, 1, 1 , 1, case when substring(@t,1,1)>0 then substring(@t,1,1) else '123456789' end UNION ALL SELECT i+1, i/9 +1, (i % 9) +1, (i / 27)*3 + ((i%9)/3+1) , case when substring(@t, i+1,1)>0 then substring(@t, i+1,1) else '123456789' end from n where i < 81)insert into #s (i,r,c,b,v) select * from n OPTION (MAXRECURSION 80); -----------------------------------------Step 2: Create sequential number-----------------------------------------Create table #num ( vInt int, vBinary int);WITH n (i ) AS ( SELECT 1 UNION ALL SELECT i+1 from n WHERE i < 9) select convert(char(1), i) as P into #num from n -----------------------------------------Step 3: Solve the puzzle. A "range" means a single row, column or block---------------------------------------Declare @SQL varchar(2000), @SQL1 varchar(2000), @SQL2 varchar(2000), @SQL3 varchar(2000) -- query to eliminate determined values from un-determined cells in the same rangeSet @SQL1 = 'while @@rowcount >0 begin update #s set v = replace(#s.v, b.v, '''') from (select Range, v from #s where d = 1) as b where #s.Range= b.Range and #s.d >1 and charindex(b.v, #s.v)>0; end;'-- query to determin the cell that solely contains a possible value in the same rangeSet @SQL2 = 'Update #s set v = pfrom ( select Range, p from #s cross apply (select P from #num where charindex(#num.p, #s.v)>0) as n where d >1 group by Range, p having count (*) =1 ) as twhere #s.Range = t.Range and #s.d >1 and charindex (p, #s.v) >0; '-- if 2 cells in the same range have the same pair of possible values, eliminate the pair of values from other cells in the same rangeSet @SQL3 = 'Update #s set v = replace(replace(v, substring(v2,1,1), ''''),substring(v2,2,1), '''')from ( select Range, v as v2 from #s where d = 2 group by Range,v having count(*) =2 ) as twhere #s.Range = t.Range and #s.d>1 and (charindex(substring(v2,1,1), #s.v)>0 or charindex(substring(v2,2,1), #s.v)>0) ; ' Declare @Count1 smallint, -- Count of determined cells, before updates @Count2 smallint, -- Count of determined cells, after updates @CountSecondRuleExecTimes int ;Select @Count1 = 0, @CountSecondRuleExecTimes = 0;Select @Count2 = Count(*) from #s where d=1While @Count2 < 81 and @CountSecondRuleExecTimes < 100 Begin While @Count1 <> @Count2 and @Count2 < 81 Begin Select @Count1 = count(*) from #s where d = 1 Set @SQL = replace(@SQL1, 'Range','r'); Exec (@SQL); If @Trace= 1 print Char(13)+Char(10)+@SQL Set @SQL = replace(@SQL1, 'Range','c'); Exec (@SQL); If @Trace= 1 print Char(13)+Char(10)+@SQL Set @SQL = replace(@SQL1, 'Range','b'); Exec (@SQL); If @Trace= 1 print Char(13)+Char(10)+@SQL Select @Count2 = count(*) from #s where d = 1 End If @Count2 < 81 Begin Set @SQL = replace ( case when (@CountSecondRuleExecTimes % 6) <3 then @SQL2 else @SQL3 end, 'Range', case (@CountSecondRuleExecTimes % 3) when 0 then 'r' when 1 then 'c' when 2 then 'b' end) If @Trace= 1 print Char(13)+Char(10)+@SQL; Exec (@SQL); Select @Count2 = count(*) from #s where d = 1 Select @CountSecondRuleExecTimes = @CountSecondRuleExecTimes + 1 if @Count1<> @Count2 AND @Trace = 1 Print replace(@SQL, char(13)+char(10),'') EndEnd ;---------------------------------------OutputSudoku:---------------------------------------with a as ( select r , c, v from #s ) select * from a pivot ( max(v) for c in ([1],[2],[3],[4],[5],[6],[7],[8],[9]) ) as p;If @Trace = 1 select datediff(ms, @datetime, getdate()), @CountSecondRuleExecTimesGOexec uspSolveSudoku_C '200060000000900871740008006006080030003000100090030400300700018972005000000090002' , 1[/code]Tue, 04 Aug 2009 11:56:17 GMTKevin DuanRE: A set based T-SQL Solution for Sudoku Puzzlehttp://www.sqlservercentral.com/Forums/Topic761296-1264-1.aspxHi Kevin,I really enjoyed reading your script, it was good to see TSQL used in a completely different way than usual. I would never have thought of using the bitwise operator to solve this problem. I like to see something different because seeing how other people approach a problem and write code is the best way of learning how to improve your own skills.Could you explain a bit about why you chose to work with the numbers in binary? was it your first choice or did you experiment with other ways first? The problem with sudoku is that for some of the puzzles you can only get so far by counting the numbers, eventually you may be forced to take a best guess at the content of a square. This is something that is missing from your script, do you intend to extend it in the future?I had a go by using recursion, but there is probably a better way, I modified the script to return 3 states: complete, Fail, Try Again. When the Try again state was reached the code would generate a new value for @t with one of the 0's replaced with a guess to see if that would lead to the completion of the puzzle.I identifed a square that had the minimum number of possible values and then created a new @t values for each of the possible values. I limited this to two levels of guesses.Here are the results I got for this sudoku: 080700000902000000300090020060800200750109043009004070040050009000000706000007030_ C1 C2 C3 C4 C5 C6 C7 C8 C9R1 _ 8 _ 7 3 2 9 _ _R2 9 _ 2 _ _ _ 3 _ _R3 3 _ _ _ 9 _ _ 2 _R4 _ 6 _ 8 7 _ 2 9 _R5 7 5 8 1 2 9 6 4 3R6 _ _ 9 _ 6 4 _ 7 _R7 _ 4 7 _ 5 _ _ _ 9R8 _ _ _ _ _ _ 7 5 6R9 _ _ _ _ _ 7 4 3 2_ C1 C2 C3 C4 C5 C6 C7 C8 C9R1 _ 8 _ 7 3 2 9 [b]1[/b] _R2 9 _ 2 _ _ _ 3 6 _R3 3 _ _ _ 9 _ _ 2 _R4 _ 6 _ 8 7 _ 2 9 _R5 7 5 8 1 2 9 6 4 3R6 _ _ 9 _ 6 4 _ 7 _R7 _ 4 7 _ 5 _ 1 8 9R8 _ _ _ _ _ _ 7 5 6R9 _ _ _ _ _ 7 4 3 2_ C1 C2 C3 C4 C5 C6 C7 C8 C9R1 5 8 6 7 3 2 9 1 [b]4[/b]R2 9 7 2 4 8 1 3 6 5R3 3 1 4 5 9 6 8 2 7R4 4 6 3 8 7 5 2 9 1R5 7 5 8 1 2 9 6 4 3R6 1 2 9 3 6 4 5 7 8R7 6 4 7 2 5 3 1 8 9R8 2 3 1 9 4 8 7 5 6R9 8 9 5 6 1 7 4 3 2I would be very interested to see how you would solve this problem.Thanks.Ian.Thu, 30 Jul 2009 03:21:44 GMTmulletcheeseRE: A set based T-SQL Solution for Sudoku Puzzlehttp://www.sqlservercentral.com/Forums/Topic761296-1264-1.aspxJust want to be sure I'm using this right. So, do I enter the entire puzzle number set beginning from the top left corner, reading a whole row, and then moving to the next row and continuing until I've reached the bottom right corner? Though this might seem obvious to the writer, not having this instruction may have people confused as to how to enter their puzzle problem set.Wed, 29 Jul 2009 07:16:12 GMTDave HauzeA set based T-SQL Solution for Sudoku Puzzlehttp://www.sqlservercentral.com/Forums/Topic761296-1264-1.aspxComments posted to this topic are about the item [B]<A HREF="/scripts/67539/">A set based T-SQL Solution for Sudoku Puzzle</A>[/B]Wed, 29 Jul 2009 00:48:45 GMTKevin Duan