Answer
code
Work Step by Step
# 11.21 (Game: multiple Sudoku solutions) The complete solution for the Sudoku
# problem is given in Supplement III.A. A Sudoku problem may have multiple
# solutions. Modify Sudoku.py in Supplement III.A to display the total number
# of the solutions. Display two solutions if multiple solutions exist.
"""This not the solution, but the code in supplement III"""
def main():
# Read a Sudoku puzzle
grid = readAPuzzle()
if not isValidGrid(grid):
print("Invalid input")
elif search(grid):
print("The solution is found:")
printGrid(grid)
else:
print("No solution")
return 0
# Read a Sudoku puzzle from the keyboard
def readAPuzzle():
print("Enter a Sudoku puzzle:")
grid = []
for i in range(9):
line = input().split()
grid.append([eval(x) for x in line])
return grid
# Obtain a list of free cells from the puzzle
def getFreeCellList(grid):
freeCellList = []
for i in range(9):
for j in range(9):
if grid[i][j] == 0:
freeCellList.append([i, j])
return freeCellList
# Display the values in the grid
def printGrid(grid):
for i in range(9):
for j in range(9):
print(grid[i][j], end=" ")
print()
# Search for a solution
def search(grid):
freeCellList = getFreeCellList(grid)
numberOfFreeCells = len(freeCellList)
if numberOfFreeCells == 0:
return True # No free cells
k = 0 # Start from the first free cell
while True:
i = freeCellList[k][0]
j = freeCellList[k][1]
if grid[i][j] == 0:
grid[i][j] = 1 # Fill the free cell with number 1
if isValid(i, j, grid):
if k + 1 == numberOfFreeCells:
# No more free cells
return True # A solution is found
else:
# Move to the next free cell
k += 1
elif grid[i][j] < 9:
# Fill the free cell with the next possible value
grid[i][j] = grid[i][j] + 1
else:
# grid[i][j] is 9, backtrack
while grid[i][j] == 9:
if k == 0:
return False # No possible value
grid[i][j] = 0 # Reset to free cell
k -= 1 # Backtrack to the preceding free cell
i = freeCellList[k][0]
j = freeCellList[k][1]
# Fill the free cell with the next possible value,
# search continues from this free cell at k
grid[i][j] = grid[i][j] + 1
return True # A solution is found
# Check whether grid[i][j] is valid in the grid
def isValid(i, j, grid):
# Check whether grid[i][j] is valid at the i's row
for column in range(9):
if column != j and grid[i][column] == grid[i][j]:
return False
# Check whether grid[i][j] is valid at the j's column
for row in range(9):
if row != i and grid[row][j] == grid[i][j]:
return False
# Check whether grid[i][j] is valid in the 3-by-3 box
for row in range((i // 3) * 3, (i // 3) * 3 + 3):
for col in range((j // 3) * 3, (j // 3) * 3 + 3):
if row != i and col != j and grid[row][col] == grid[i][j]:
return False
return True # The current value at grid[i][j] is valid
# Check whether the fixed cells are valid in the grid
def isValidGrid(grid):
for i in range(9):
for j in range(9):
if grid[i][j] < 0 or grid[i][j] > 9 or (grid[i][j] != 0
and not isValid(i, j, grid)):
return False
return True # The fixed cells are valid
main()
