Writing a function that returns boolean true if matrix forms a magic square without importing numpy

Question

I already wrote one part of the program which is below:

def matrix_is_square(matrix):
   for i in range(len(matrix)):
      if len(matrix[i]) != len(matrix):
            


        

Answer 1

Complex solution using sum, range, any functions and set object:

def magic(m):
    length = len(m)
    first_sum = set()
    if length <= 2 \
            or any(len(row) != length for row in m) \
            or any(i > (length * length) for row in m for i in row):
        return False

    for r in m:
        s = sum(r)
        if len(first_sum) == 0:
            first_sum.add(sum(r))
        if s not in first_sum:
            return False

    for i in range(length):
        s = sum([r[i] for r in m])
        if s not in first_sum:
            return False

    if sum(m[i][i] for i in range(length)) not in first_sum \
        or sum(m[i][i] for i in range(length - 1, -1, -1)) not in first_sum:
        return False

    return True

m = [[2,7,6],[9,5,1],[4,3,8]]
print(magic(m))

m = [[16,3,2,13], [5,10,11,8],[9,6,7,12], [4,15,14,1]]
print(magic(m))

m = [[1,2,3,4], [5,6,7,8],[9,10,11,12], [13,14,15,16]]
print(magic(m))

m = [[1,1],[1,1]]
print(magic(m))

m = [[1,1],[1,1],[1,2]]
print(magic(m))

The output(sequentially):

True
True
False
False
False