Find all possible magic squares (3×3) python

Question

I can find only one magic square how to find all please.

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Answer 1

I’ll leave finding out how to generate a magic square as an exercise. If you’re still having trouble with it, you can find other questions on StackOverflow about how to generate a magic square of a given size in Python.

Once you have your 3×3 magic square magic(3) (as a numpy ndarray), you can obtain all of the possible magic squares of that size by performing all of the possible rotations and reflections on it:

rotations = [np.rot90(magic(3), x) for x in range(4)]
reflections = [np.flip(x, 1) for x in rotations]
all_magic_3x3 = rotations + reflections

This produces a list containing the following 8 magic 3×3 matrices:

[[8 1 6]
 [3 5 7]
 [4 9 2]]

[[6 7 2]
 [1 5 9]
 [8 3 4]]

[[2 9 4]
 [7 5 3]
 [6 1 8]]

[[4 3 8]
 [9 5 1]
 [2 7 6]]

[[6 1 8]
 [7 5 3]
 [2 9 4]]

[[2 7 6]
 [9 5 1]
 [4 3 8]]

[[4 9 2]
 [3 5 7]
 [8 1 6]]

[[8 3 4]
 [1 5 9]
 [6 7 2]]

Answered By: Will Da Silva

Answer 2

from itertools import permutations

def generate_all_magic_squares():
    magic_squares = []
    for seq in permutations(range(1, 10)):
        cand = [seq[i:i+3] for i in range(0, 9, 3)]
        # filter cand whose row sum is a const
        if sum(cand[0]) == sum(cand[1]) == sum(cand[2]):
            cols = list(zip(*cand)) # invert cols to rows to check their sums as well
            if sum(cols[0]) == sum(cols[1]) == sum(cols[2]) == sum(cand[0]):
                pd =  [cand[0][0],cand[1][1],cand[2][2]] # principle diagnol
                od =  [cand[0][2], cand[1][1], cand[2][0]] # other diagnol
                if sum(pd) == sum(od) == sum(cand[0]): # check the sums of the diagnol are equal to other rows/cols
                    magic_squares.append(cand)

    return magic_squares

Answered By: mwikya

Find all possible magic squares (3×3) python

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