title: Python Sudoku Solver with multiple solutions
body: <p>i am new to programing an trying to learn it. As I am always looking for new intressting python tutorials i have come accross an tutorial for a sudoku solver using backtracking technique.</p>
  <p><a href="https://www.techwithtim.net/tutorials/python-programming/sudoku-solver-backtracking/" rel="nofollow noreferrer">https://www.techwithtim.net/tutorials/python-programming/sudoku-solver-backtracking/</a></p>
  <p>Since this sudoku solver only shows one single solution, I was wondering if there was a way to set a count on how many solutions of a sudoku can be found. Or how I can modify the code so that it doesn't stop after finding a solution but keeps on looking for other possible solutions. I don't know how i can get it to backtrack again to find other solutions. I was already looking this problem up on the internet but most of the solutions i found didn't really fit to my problem. If someone has an idea on how to do this I would be really happy about any help i can get, since i am worrying about this problem for a long time now. Already thank you for helping me, looking forward to maybe being able to solving my problem:)</p>
  <pre><code>board = [
      [7,8,0,4,0,0,1,2,0],
      [6,0,0,0,7,5,0,0,9],
      [0,0,0,6,0,1,0,7,8],
      [0,0,7,0,4,0,2,6,0],
      [0,0,1,0,5,0,9,3,0],
      [9,0,4,0,6,0,0,0,5],
      [0,7,0,3,0,0,0,1,2],
      [1,2,0,0,0,7,4,0,0],
      [0,4,9,2,0,6,0,0,7]
  ]
  
  
  def solve(bo):
      find = find_empty(bo)
      if not find:
          return True
      else:
          row, col = find
  
      for num in range(1,10):
          if valid(bo, num, (row, col)):
              bo[row][col] = num          
  
              if solve(bo):                 
                  return True
  
              bo[row][col] = 0              
  
      return False
  
  
  def valid(bo, num, pos):
      # Check row
      for field in range(len(bo[0])):                     
          if bo[pos[0]][field] == num and pos[1] != field:
              return False
  
      # Check column
      for line in range(len(bo)):
          if bo[line][pos[1]] == num and pos[0] != line:
              return False
  
      # Check box
      box_x = pos[1] // 3
      box_y = pos[0] // 3
  
      for i in range(box_y*3, box_y*3 + 3):
          for j in range(box_x * 3, box_x*3 + 3):
              if bo[i][j] == num and (i,j) != pos:
                  return False
  
      return True
  
  
  def print_board(bo):
      for i in range(len(bo)):
          if i % 3 == 0 and i != 0:
              print(&quot;- - - - - - - - - - - - - &quot;)
  
          for j in range(len(bo[0])):
              if j % 3 == 0 and j != 0:
                  print(&quot; | &quot;, end=&quot;&quot;)
  
              if j == 8:
                  print(bo[i][j])
              else:
                  print(str(bo[i][j]) + &quot; &quot;, end=&quot;&quot;)
  
  
  def find_empty(bo):
      for i in range(len(bo)):
          for j in range(len(bo[0])):
              if bo[i][j] == 0:
                  return (i, j)  # row, col
  
      return None
  if __name__ == &quot;__main__&quot;:
      print_board(board)
      solve(board)
      print(&quot;___________________&quot;)
      print(&quot;&quot;)
      print_board(board)
  </code></pre>