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How can I write a function to test whether a matrix magic square

21st Feb 2017, 7:05 PM
aslicann
aslicann - avatar
2 Antworten
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""" Ok, fine, I think I got it. I wrote a function which iterates through rows, columns and diagonals, sequentially. The code can be four times simpler by using four temporary sums instead of one, but I leave it for educational purpose to see what it is doing. Also, I left matrix2 from the previous attempt to show false result, too. """ matrix = [[2, 7, 6], [9, 5, 1], [4, 3, 8]] matrix2 = [[4, 5, 7], [3, 6, 8], [7, 8, 9]] def is_magic(square): magic_sum = 0 check_sum = 0 for x in range(len(square)): check_sum += square[0][x] print(check_sum) # check for the sum of the first row as a control sum for y in range(len(square)): for x in range(len(square)): magic_sum += square[x][y] if magic_sum != check_sum: print(magic_sum) return False else: print(magic_sum) magic_sum = 0 #rows checked for y in range(len(square)): for x in range(len(square)): magic_sum += square[y][x] if magic_sum != check_sum: print(magic_sum) return False else: print(magic_sum) magic_sum = 0 # columns checked for x in range(len(square)): magic_sum += square[x][x] if magic_sum != check_sum: print(magic_sum) return False else: print(magic_sum) magic_sum = 0 # diagonal checked for x in range(len(square)): magic_sum += square[len(square)-x-1][x] if magic_sum != check_sum: print(magic_sum) return False else: print(magic_sum) magic_sum = 0 # contrdiagonal checked return True # if got here, everything is checked positive print(is_magic(matrix)) print(is_magic(matrix2))
21st Feb 2017, 8:36 PM
Kuba Siekierzyński
Kuba Siekierzyński - avatar
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Thank you so much
21st Feb 2017, 7:56 PM
aslicann
aslicann - avatar