NumPy equivalent of Matlab's magic()

Question

In Ocatave / Matlab, I can use magic() to get a magic square, eg,

magic(4)

  16    2    3   13
   5   11   10    8
   9    7    6   12
   4   14   15    1

Definition: A magic square is an N×N grid of numbers in which the entries in each row, column and main diagonal sum to the same number (equal to N(N^2+1)/2 ).

How can I generate the same using NumPy?

Answer 1

This implementation follows Matlab's and should give exactly the same results with the following exception: it throws an error if n < 3 rather than return a non-magic square [[1, 3], [4, 2]] when n=2 like Matlab does.

As usual, there are three cases: odd, divisible by 4, and even but not divisible by 4, the last one being the most complicated.

def magic(n):
  n = int(n)
  if n < 3:
    raise ValueError("Size must be at least 3")
  if n % 2 == 1:
    p = np.arange(1, n+1)
    return n*np.mod(p[:, None] + p - (n+3)//2, n) + np.mod(p[:, None] + 2*p-2, n) + 1
  elif n % 4 == 0:
    J = np.mod(np.arange(1, n+1), 4) // 2
    K = J[:, None] == J
    M = np.arange(1, n*n+1, n)[:, None] + np.arange(n)
    M[K] = n*n + 1 - M[K]
  else:
    p = n//2
    M = magic(p)
    M = np.block([[M, M+2*p*p], [M+3*p*p, M+p*p]])
    i = np.arange(p)
    k = (n-2)//4
    j = np.concatenate((np.arange(k), np.arange(n-k+1, n)))
    M[np.ix_(np.concatenate((i, i+p)), j)] = M[np.ix_(np.concatenate((i+p, i)), j)]
    M[np.ix_([k, k+p], [0, k])] = M[np.ix_([k+p, k], [0, k])]
  return M 

I also wrote a function to test this:

def test_magic(ms):
  n = ms.shape[0]
  s = n*(n**2+1)//2 
  columns = np.all(ms.sum(axis=0) == s)
  rows = np.all(ms.sum(axis=1) == s)
  diag1 = np.diag(ms).sum() == s 
  diag2 = np.diag(ms[::-1, :]).sum() == s
  return columns and rows and diag1 and diag2 

Try [test_magic(magic(n)) for n in range(3, 20)] to check the correctness.

Answer 2

Here are a quick implementation for odd and doubly even cases.

def magic_odd(n):
    if n % 2 == 0:
        raise ValueError('n must be odd')
    return np.mod((np.arange(n)[:, None] + np.arange(n)) + (n-1)//2+1, n)*n + \
          np.mod((np.arange(1, n+1)[:, None] + 2*np.arange(n)), n) + 1


def magic_double_even(n):
    if n % 4 != 0:
        raise ValueError('n must be a multiple of 4')
    M = np.empty([n, n], dtype=int)
    M[:, :n//2] = np.arange(1, n**2//2+1).reshape(-1, n).T
    M[:, n//2:] = np.flipud(M[:, :n//2]) + (n**2//2)
    M[1:n//2:2, :] = np.fliplr(M[1:n//2:2, :])
    M[n//2::2, :] = np.fliplr(M[n//2::2, :])
    return M

Odd case is from here and I got the rest from How to construct magic squares of even order . Then I got lazy for the single even case but the idea is the similar.

Answer 3

I had the same issue, this is what I used:

import numpy as np

matrix = np.random.random((15,15))
for x in range(15):
    for y in range(15):
        matrix[x][y] = int(matrix[x][y]*10)

I needed integers between 0 and 10, but you get the idea...