Question

我正在制作的程序是从文本文件中验证数独。到目前为止，我有这个。

file=input("Enter a filename: ")
with open(file) as data:
    rowt=[0 for i in range(9)]
    colt=[0 for i in range(9)]
    for a,b in enumerate(data):
        sb_rowt = 0
        print (a,b)
        for c, d in enumerate(b.split()):
            print (c,d)
            sb_rowt+=int(d)
            colt[c]+=int(d)
        rowt[a]=sb_rowt
    rows=all(i==45 for i in rowt)
    cols=all(i==45 for i in colt)
    print(rows and cols)

这个问题是，如果我有一个像这样的文本文件

5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5 
5 5 5 5 5 5 5 5 5

程序仍将输出为true。我知道我需要删除任何重复的条目，但我不确定如何。

Answer 1

通过你的提问，我认为你没有走上正轨。无论是否有效，电路板上应始终有9 * 9个数字。您获得了正确的输入，但无法验证它。 考虑改进验证算法。

Answer 2

您应该将阅读和验证操作分开：

# read first
file = input("Enter a filename: ")
with open(file) as data:
    array = [ [ int(i) for i in line.split() ] for line in data ]

然后才进行验证。

# first create a set of numbers 1...9
all_numbers = set(range(1, 10))

# validate rows and columns:
valid = True

# we validate ith row and column in the same loop
for i in range(9):
    row_i    = array[i]
    column_i = [ array[rownum][i] for rownum in range(9) ]
    valid   &= set(row_i) == all_numbers
    valid   &= set(column_i) == all_numbers

# valid is True if all constraints succeeded.

# TODO: in sudoku you also need to validate the 
# 3x3 subsquares.

Answer 3

您不能只使用数字的总和来验证网格。相反，您必须检查每行和每列是否包含1到9之间的所有数字。

您可以尝试这样的事情：

all_nums = set(range(1, 10))
with open("data") as data:
    field = [[int(x) for x in line.split()] for line in data]
    rows = map(set, field)
    cols = map(set, zip(*field))
    rows_ok = all(row == all_nums) for row in rows)
    cols_ok = all(col == all_nums) for col in cols)

首先，我从数据输入中创建一个整数列表。然后我提取行和列（zip(*field)基本上翻转field）并将它们转换为set s，因为顺序并不重要。最后，我只是将all行和列与all_nums的集合进行比较。

Answer 4

import csv
import collections

with open('path/to/file') as infile:
    board = [[int(i) for i in row] for row in csv.reader(infile)]
for r,row in enumerate(board):
    counts = collections.Counter(row)
    if all(row[k]==1 for k in range(1,10)):
        continue
    raise ValueError("failed on row %d" %r)

for c,col in enumerate(zip(*rows)):
    if all(col[k]==1 for k in range(1,10)):
        continue
    raise ValueError("failed on col %d" %c)

# todo: check each 3x3 square

Answer 5

将您的董事会作为清单提供：

def validate(board):
    valid = set(range(1, 10))
    return (all(valid == set(r) for r in board) and
            all(valid == set(c) for c in zip(*board)) and
            all(valid == set(board[i][j:j+3] + board[i+1][j:j+3] + board[i+2][j:j+3])
                for i in (0, 3, 6) for j in (0, 3, 6)))

从字典值中删除重复项

5 个答案: