Counting Boring Squares

Probability Level 4

A 3 × 3 3 \times 3 boring square is (as defined in this problem ) is a 3 × 3 3 \times 3 grid filled with integers 1 , 2 , . . . , 9 1,2,...,9 such that the sum of the integers is unique in every row, column and diagonal.

How many 3 × 3 3 \times 3 boring squares are there ?

NB : Squares that are rotations or reflections of each other are considered the same.


The answer is 3120.

Romain Bouchard by Romain Bouchard , 34, France

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1 solution

Arjen Vreugdenhil
Dec 18, 2017

Brute force method

The following code counts 24960 24960 boring squares. They can be grouped into sets of eight identical squares (under rotation/reflection), so that the answer is 24960 / 8 = 3120 24960/8 = \boxed{3120} . 

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sq = [1,2,3,4,5,6,7,8,9]
N = 0

while sq > []:
    while len(sq) < 9:
        i = 1
        while i in sq:
            i = i + 1
        sq.append(i)

    s1 = sq[0] + sq[1] + sq[2]
    s2 = sq[3] + sq[4] + sq[5]
    s3 = sq[6] + sq[7] + sq[8]
    s4 = sq[0] + sq[3] + sq[6]
    s5 = sq[1] + sq[4] + sq[7]
    s6 = sq[2] + sq[5] + sq[8]
    s7 = sq[0] + sq[4] + sq[8]
    s8 = sq[2] + sq[4] + sq[6]

    sums = set([s1,s2,s3,s4,s5,s6,s7,s8])
    if len(sums) == 8:
        N = N+1

    i = 10
    while i > 9 and len(sq) > 0:
        i = sq[-1]
        del sq[-1]
        i = i + 1
        while i <= 9 and (i in sq):
            i = i + 1

    if i <= 9:
        sq.append(i)

print(N)

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