Python - IndexError：リストの索引が範囲外にある（sudokuソルバープログラム）

Question

このコードが従うのが一番簡単ではない場合は申し訳ありません。だから私は9x9次元（スドク板）と二次元のリストboardを持っています。 solveの行のいずれかで、if row <= 8 and column <=8:を確認してからadd1関数を呼び出します。リストされたテスト値を使用すると、最後のセルの永続的な値は9であり、ボードがまだ解決している間にadd1が呼び出されたときにはまだIndexError: list index out of rangeが発生します。 add1関数を呼び出す場合、インデックス値が有効な場合は、と表示されます。

class Cell: 
    def __init__(self, value, isPermanent): 
     self.value = 0 
     self.isPermanent = False 


def solve(board): # solves the board 
    row = 0 # begin in the first cell in the first row 
    column = 0 
    while row != 9: # while there are still empty spaces in the 9 rows, 
     if row <= 8 and column <= 8: # HOW DOES A LIST INDEX ERROR COME UP IN 
            # THE NEXT LINE IF I CHECK THESE VALUES HERE? 
      row, column = add1(board, row, column) # add 1 to the current cell 
     printBoard(board) 
     print(), print() 
     if valid(board) is True: # if the board is valid, 
      if column == 8: # if in the last cell of a row, 
       row += 1 # move up one row, and begin in the first cell 
       column = 0 
      else: # otherwise, move on to the next cell 
       column += 1 
      continue # restart 
     if valid(board) is False: # if the board is invalid, 
      if board[row][column].value == 9: # if the value of the current cell is equal to 9, 
       board[row][column].value = 0 # set it equal to 0 
       if column == 0: # if in the first cell of a row, 
        row -= 1 # go back a row, and into the last cell 
        column = 8 
       else: # otherwise, move back to the previous cell 
        column -= 1 
      continue # restart 


def add1(board, row, column): # increments each cell 
    while True: 
     if board[row][column].isPermanent: 
      if column == 8: 
       row += 1 
       column = 0 
      else: 
       column += 1 
      continue 
     if board[row][column].value == 9: # if the value of the cell is equal to 9, 
      board[row][column].value = 0 # set it equal to 0 
      if column == 0: # if in the first cell of a row, 
       row -= 1 # go back a row, to the last cell 
       column = 8 
      else: # if not in the first cell, 
       column -= 1 # go to the previous cell 
     board[row][column].value += 1 # add 1 to the current cell 
     return row, column # return the new coordinate of the current cell 


def valid(board): 
    for i in range(1, 10): 
     if checkRowsForDuplicate(board, i) is False: 
      return False 
     if checkColumnsForDuplicate(board, i) is False: 
      return False 
     if checkZonesForDuplicate(board, i) is False: 
      return False 
    return True 


def checkRowsForDuplicate(board, x): 
    for row in board: 
     line = [] 
     for cell in row: 
      line.append(cell.value) 
     if line.count(x) > 1: 
      return False 


def checkColumnsForDuplicate(board, x): 
    for i in range(0, 9): 
     column = [] 
     for row in board: 
      column.append(row[i].value) 
     if column.count(x) > 1: 
      return False 


def checkZonesForDuplicate(board, x): 
    y = [0, 3, 6] 
    z = [3, 6, 9] 
    for i in range(3): 
     for j in range(3): 
      if checkSingleZone(board, x, y[i], z[i], y[j], z[j]) is False: 
       return False 
    return True 


def checkSingleZone(board, x, rowStart, rowEnd, columnStart, columnEnd): 
    zoneValues = [] 
    for row in board[rowStart:rowEnd]: 
     for column in row[columnStart:columnEnd]: 
      zoneValues.append(column.value) 
    if zoneValues.count(x) > 1: 
     return False 


def printBoard(board): 
    for row in board: 
     line = [] 
     for cell in row: 
      line.append(cell.value) 
     print(line) 


def initializeBoard(): 
    board = [[], [], [], [], [], [], [], [], []] 
    for row in board: 
     for i in range(0, 9): 
      row.append(Cell(0, False)) 
    return board 

board = initializeBoard() 

# test values 
board[0][0].value = 9 
board[0][0].isPermanent = True 

board[8][8].value = 9 
board[8][8].isPermanent = True 

board[2][7].value = 9 
board[2][7].isPermanent = True 

board[4][1].value = 9 
board[4][1].isPermanent = True 

solve(board) 

Answer 1

私は簡単にトレースバックを再現することができます：あなたはあなたの境界をチェックせずに、このループを打つため

>>> row, column = 8, 8 
>>> board[row][column].isPermanent = True 
>>> add1(board, row, column) 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "<stdin>", line 3, in add1 
IndexError: list index out of range 

です：

def add1(board, row, column): # increments each cell 
    while True: 
     if board[row][column].isPermanent: 
      if column == 8: 
       row += 1 
       column = 0 
      else: 
       column += 1 
      continue 

のでcolumn == 8はあなたが9にrowをインクリメント、真でありますboard[9]を次の反復に当ててください。

皮肉なことに、あなたのboardには有効な解決済みの数独が含まれていますが、これは検出されません。