Solving Sudoku using backtracking

Question

This is a solver for Sudoku using backtracking. How can I make it more optimized and clean?

#include <stdio.h>

int isAvailable(int sudoku[][9], int row, int col, int num)
{
    int i, j;
    for(i=0; i<9; ++i)
        if( (sudoku[row][i] == num) || ( sudoku[i][col] == num )  )//checking in row and col
            return 0;

     //checking in the grid
    int rowStart = (row/3) * 3;
    int colStart = (col/3) * 3;

    for(i=rowStart; i<(rowStart+3); ++i)
    {
        for(j=colStart; j<(colStart+3); ++j)
        {
            if( sudoku[i][j] == num )
                return 0;
        }
    }    

    return 1;
}    

int fillsudoku(int sudoku[][9], int row, int col)
{
    int i;
    if( row<9 && col<9 )
    {
        if( sudoku[row][col] != 0 )//pre filled 
        {
            if( (col+1)<9 )
                return fillsudoku(sudoku, row, col+1);
            else if( (row+1)<9 )
                return fillsudoku(sudoku, row+1, 0);
            else
                return 1;
        }
        else
        {
            for(i=0; i<9; ++i)
            {
                if( isAvailable(sudoku, row, col, i+1) )
                {
                    sudoku[row][col] = i+1;

                    if( (col+1)<9 )
                    {
                       if( fillsudoku(sudoku, row, col +1) )
                           return 1;
                        else
                            sudoku[row][col] = 0;
                    }
                    else if( (row+1)<9 )
                    {
                        if( fillsudoku(sudoku, row+1, 0) )    
                            return 1;
                        else
                            sudoku[row][col] = 0;
                    }
                    else
                        return 1;
                }
            }
        }
        return 0;
    }
    else
    {
        return 1;
    }
} 

int main()
{
   int i, j;
    int sudoku[9][9]={{3, 0, 6, 5, 0, 8, 4, 0, 0},
                      {5, 2, 0, 0, 0, 0, 0, 0, 0},
                      {0, 8, 7, 0, 0, 0, 0, 3, 1},
                      {0, 0, 3, 0, 1, 0, 0, 8, 0},
                      {9, 0, 0, 8, 6, 3, 0, 0, 5},
                      {0, 5, 0, 0, 9, 0, 6, 0, 0},
                      {1, 3, 0, 0, 0, 0, 2, 5, 0},
                      {0, 0, 0, 0, 0, 0, 0, 7, 4},
                      {0, 0, 5, 2, 0, 6, 3, 0, 0}};

    if( fillsudoku(sudoku, 0, 0) )
    {

        for(i=0; i<9; ++i)
        {
            for(j=0; j<9; ++j)
                printf("%d ", sudoku[i][j]);
            printf("\n");
        }
    }
    else
    {
        printf("\n\nNO SOLUTION\n\n");
    }

    return 0;
}

Answer 1

Very nice:

Few minor changes I would make.

In isAvailable() I would check row/col/box all at the same time.

int isAvailable(int sudoku[][9], int row, int col, int num)
{
    //checking in the grid
    int rowStart = (row/3) * 3;
    int colStart = (col/3) * 3;

    int i, j;
    for(i=0; i<9; ++i)
    {
        if (sudoku[row][i] == num)                             return 0;
        if (sudoku[i][col] == num)                             return 0;
        if (sudoku[rowStart + (i%3)][colStart + (i/3)] == num) return 0;
    }

    return 1;
}    

Which brings my to my first comment:
I prefer to have sub statments of while/for/if inside block quotes.

if( sudoku[i][j] == num )
            return 0;

// I prefer this:
if( sudoku[i][j] == num )
{   return 0;
}

This way there is not chance of accidentally putting a macro that expands and is not all executed.

Your code for moving to the next square is in multiple places in the code:

        if( (col+1)<9 )
           STUFF(row, col+1);
        else if( (row+1)<9 )
            STUFF(row+1, 0);
        else
            WIN

Because it is multiple places you have redundancy. I would move this check to one location. The easiest way to do that is move it to the first few lines of code in the function. Then you always make a recursive call like this fillsudoku(sudoku, row, col+1).

int fillsudoku(int sudoku[][9], int row, int col)
{
    if (col >= 9)
    {
         col = 0;
         ++row;
         if (row >= 9)
         {
             return 1;
         }
    }

    // Original code.
    // When doing a recursive call use ` fillsudoku(sudoku, row, col+1)`
    // You know the start of the code will test it.

Next I would exit early rather than have nested code.
This is a debatable point on style but I don;t think having to scroll down a long way to see a one lineer for failure helps in readability. And since you already exit early that is not a problem.

        if( sudoku[row][col] != 0) //pre filled 
        {
            return fillsudoku(sudoku, row, col+1);
        }
        else
        {
            for(i=0; i<9; ++i)
            {
                if( isAvailable(sudoku, row, col, i+1) )
                {
                    sudoku[row][col] = i+1;

                    int good = fillsudoku(sudoku, row, col +1);
                    if (good)
                    {   return 1;
                    }
                    sudoku[row][col] = 0;
                }
            }
        }
        return 0;
}