Sudoku Solver

This code snippet is from a program we had to create during our AI class, that solved a Sudoku puzzle

      
    void set(uint l, uint c, uint val) 
    { 
    // Sets the cell at line l and row c to hold the value val
	  if (isValidPos(l, c) && ((0 < val) && (val <= MAXVAL))) 
	  {
		if (table[l * C + c] == 0) ++filledIn;
		table[l * C + c] = val;
		for (uint i = 0; i < C; ++i) // Update lines
		usedDigits[lineElements[l][i]] |= 1< 0)
		++d;
		set(i / C, i % C, d); // Fill it in
		changed = true; // The board has been changed so this step must be rerun
		} 
		else if (bitcount(usedDigits[i]) == MAXVAL)
		throw 666; // Speed boost	
	  }
    }
    

    void goBruteForce() 
    {
    int max(-1); // Find the cell with the _minimum_ number of posibilities (i.e. the one with the largest number of /used/ digits)
	  for (uint i(0); i < S; ++i)
		if (table[i] == 0) // Is there a digit already written?
		if ((max == -1) || (bitcount(usedDigits[i]) > bitcount(usedDigits[max])))
		max = i;
	  

    if (max != -1) 
    {
		for (uint i(1); i <= MAXVAL; ++i) // Go through each possible digit
		if ((usedDigits[max] & 1<

    { 
  	// If it can be placed in this cell, do
		SudokuBoard temp(*this); // Create a new board
		temp.set(max / C, max % C, i); // Complete the attempt
		temp.solve(); // Solve it
		if (temp.getFilledIn() == S) 
		{ 
		// If the board was completely solved (i.e. the number of filled in cells is S)
		for (uint j(0); j < S; ++j) // Copy the board into this one
		set(j / C, j % C, temp.at(j / C, j % C));
		return; // Break the recursive cascade
		}
		}
	  }
    }