創建幻方

Question

所以我創建一個幻方，這裏是我的代碼

高清is_magic_square（S）： 「」」 返回一個二維整數數組S是否是魔術，那就是： 1）s的維數爲nxn 2）[1,2，...，n * n]中的每個整數都出現在s中，只有一次。 3）s中所有行的總和等於s中所有 列的總和，與s中對角線 元素的總和相同。

:param s: A two dimensional integer array represented as a nested list. 
:return: True if square is magic, False otherwise 

QUESTION 1: Write DocTest for 

TEST Matricies that are magic a few different sizes, special cases seem to be, 1x1 matrix, 2x2 matrix. 
>>> is_magic_square([[1]]) 
True 
>>> is_magic_square([[8, 3, 4], [1, 5, 9], [6, 7, 2]]) 
True 

YOUR TEST CASES GO HERE 

NOTE:!!! LEAVE A BLANK LINE AFTER THE LAST TEST RESULT, BEFORE A COMMENT !!! 
TEST Matricies that are not magic. 

TEST NOT 1) The dimensions of s is nxn 
>>> is_magic_square([[1,2],[3,4],[5,6]]) 
False 
>>> is_magic_square([[1,2],[3,4,5],[6,7]]) 
False 

YOUR TEST CASES GO HERE 
>>>is_magic_square([[8, 3, 4], [1, 5, 9], [6, 7, 2]]) 
True 

TEST NOT 2) Every integer in [1,2,...,n*n] appears in s, exactly once. 

YOUR TEST CASES GO HERE 
>>> is_magic_square([8, 3, 4],[9,3,3],[6,7,2]) 
False 

TEST NOT 3) The sum of all rows in s is the same as the sum of all 
    columns in s, is the same as the sum of the diagonal 
    elements in s. 

YOUR TEST CASES GO HERE 
>>> is_magic_square([8,3,4], [1, 5, 9], [6,7,1]) 
False 

nrows = len(s) 
ncols = 0 
if nrows != 0: 
    ncols = len(s[0]) 
if nrows != ncols: 
    return False 
l = [] 
for row in s: 
    for elem in row: 
     if elem in l: 
      return False 
     l.append(elem) 
m = sum(s[0]) 
for row in s: 
    sums = 0 
    for elem in row: 
     sums+=elem 
    if sums != m: 
     return False 


return True 

我測試基本上一切都那麼遠，如果N×N的平方的長度是相同的長度，同樣的金額，等我的問題是，現在我試着去計算，如果對角和相同的行和柱。我會怎麼做？

Answer 1

您可以通過執行 sum([ s[i][i] for i in range(len(s))]) 和 sum([ s[len(s)-i-1][i] for i in range(len(s))]) ，其中s變量列表要測試

Answer 2

我解決了類似的問題，不久前爲編程拼圖網站列表計算對角和。我的問題在範圍上稍大一些 - 請確認NxN Sudoku難題已妥善解決。

你將有兩個對角線在一個NxN的正方形。

import itertools 

square = [ [1, 2, 3], 
      [4, 5, 6], 
      [7, 8, 9] ] # not a valid square! 

n = len(square) # nxn square 

rows = iter(square) 
columns = zip(*square) # this is a beautiful idiom! Learn it! 
diagonals = zip(*[(square[i][i], square[-i-1][i]) for i in range(n)]) 

然後，你可以做一個簡單的sum檢查

sum(rows) == sum(columns) == sum(diagonals) 

代碼的唯一怪異的一部分是列表理解的拉鍊明星。它基本上：

diagonal1 = [] 
diagonal2 = [] 
for i in range(n): 
    diagonal1.append(square[i][i]) 
    diagonal2.append(square[-i-1][i]) 
diagonals = [diagonal1, diagonal2]