I have just uploaded a program I wrote a couple of weeks ago to GitHub . The program is a sudoku solver which I wrote in Java. I am also posting the source code below.

Usage
The program reads the sudoku from a text file called “Sudoku.txt”. Which should be formatted like so:

860020000
00070005 9
000000000
000060 800
040000000
005300007
000000000
020000600
00750 9000

Where the 0s are the blanks in the given puzzle. The program will then print the solution to the command line.

When input the puzzle above:

Here is the source code:

import java.io.File ;
public class SudokuSolver
{
public static void main ( String [] args ) throws Exception
{
Scanner fileScanner = new Scanner ( new File ( "Sudoku.txt" ));
int [][] sudoku = new int [ 9 ][ 9 ]; //Will hold the puzzle in a 2D array.
//We will use this variable to hold the next line of the puzzle, then parse
//each digit on that line.
String line = fileScanner . nextLine ();
//Loops round placing the digits into the correct place of the array.
for ( int y = 0 ; y < 9 ; y ++)
{
for ( int x = 0 ; x < 9 ; x ++)
{
//Gets the digit converts it from a char to an in and places it in the array.
sudoku [ y ][ x ] = Character . getNumericValue ( line . charAt ( x ));
//At the end of each line we load the next line, but check there is a next
//line before trying to load it so the program does not crash.
if ( x == 8 && fileScanner . hasNextLine ())
{
line = fileScanner . nextLine ();
}
} //for x
} //for y
//The recursive function that actually solves the sudoku (starting at 0,0)
solve ( sudoku , 0 , 0 );
} //main
//This is a recursive function used to go through each cell and place a valid number.
//Until they are filled in.
private static void solve ( int [][] sudoku , int cellX , int cellY )
{
//If the y value is 9 then the sudoku has been solved.
if ( cellY > 8 )
{
printSudoku ( sudoku );
System . out . println ();
//System.exit(1); // = This will end the program quicker as it does not have to
//"go back up" through the levels of recursion, but means the main
//routine will not continue running.
//Also if there is more than one solution to the sudoku using this will only print
//the first solution found.
}
else
{
//Here we calculate the next digit for the solve routine to try.
int nextX = cellX ;
int nextY = cellY ;
if ( cellX == 8 )
{
//When at the end of a row add 1 to the row and reset the "column" to 0.
nextX = 0 ;
nextY ++;
}
else
{
nextX ++;
}
//If the digit was already given to us, we can move onto the next one.
if ( sudoku [ cellY ][ cellX ] != 0 )
{
solve ( sudoku , nextX , nextY );
}
else
{
//Otherwise, starting at 1 through 9 we check if the number is "legal"
//and if so place that number, and move on to the next cell.
for ( int checkNum = 1 ; checkNum < 10 ; checkNum ++)
{
if ( checkSquare ( sudoku , cellX , cellY , checkNum )
&& checkRow ( sudoku , cellY , checkNum )
&& checkCol ( sudoku , cellX , checkNum ))
{
sudoku [ cellY ][ cellX ] = checkNum ;
solve ( sudoku , nextX , nextY );
}
}
//If we get to here it means in it's current state the sudoku is impossible
//which means that one of the numbers we "placed" earlier is incorrect.
sudoku [ cellY ][ cellX ] = 0 ;
}
}
}
//This method is given a cell location and a number to check, it then checks
//whether that number is already in the 3x3 square and returns false if so.
private static boolean checkSquare ( int [][] sudoku , int reqX , int reqY , int toCheck )
{
int rowY ;
int colX ;
//First we work out which column the "square" belongs to.
//We take the given x value and if it is below 3 then that means
//the square we need is in the first column (out of 3). etc.
if ( reqX < 3 )
{
colX = 0 ;
}
else if ( reqX < 6 )
{
colX = 3 ;
}
else
{
colX = 6 ;
}
//We do the same but for the rows. For example if the y value is 5 then
//the related square would be on the second row.
if ( reqY < 3 )
{
rowY = 0 ;
}
else if ( reqY < 6 )
{
rowY = 3 ;
}
else
{
rowY = 6 ;
}
//We have now defined the square we need to check and have the top left
//co-ordinate stored in the variables rowY and colX.
//We now loop round and check each digit in the square, and if a digit matches
//we return false.
for ( int y = rowY ; y < rowY + 3 ; y ++)
{
for ( int x = colX ; x < colX + 3 ; x ++)
{
if ( sudoku [ y ][ x ] == toCheck )
{
return false ;
}
}
}
return true ; //number not in the square.
}
//Checks if a given number is in a given row and returns false if it is.
private static boolean checkRow ( int [][] sudoku , int rowY , int toCheck )
{
//loops round each digit in a row.
for ( int x = 0 ; x < 9 ; x ++)
{
//Checks if the given number is the same as the current digit
//and returns false if so.
if ( toCheck == sudoku [ rowY ][ x ])
{
return false ;
}
}
return true ; //the number is not in the row.
}
//Checks if a given number is in a given column and returns false if it is.
private static boolean checkCol ( int [][] sudoku , int colX , int toCheck )
{
//Loops round each digit in a column.
for ( int y = 0 ; y < 9 ; y ++)
{
//Checks if the current digit is the given digit and returns false if so.
if ( toCheck == sudoku [ y ][ colX ])
{
return false ;
}
}
return true ; //the number is not in the column.
}
//Prints the sudoku to the screen.
private static void printSudoku ( int sudoku [][])
{
//Loops round each digit and prints it.
for ( int y = 0 ; y < 9 ; y ++)
{
for ( int x = 0 ; x < 9 ; x ++)
{
System . out . print ( sudoku [ y ][ x ]);
//Starts a new line when at the end of a row.
if ( x == 8 )
{
System . out . println ();
}
} //for x
} //for y
} //printSudoku
} //SudokuSolver