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We have a simplified 3x3 magic square that can be filled with numbers from 0 to 9 with repetitions. The sum of the elements in each row must be equal the sum of the elements in each column, diagonals don't matter.

We need to count the number of ways to fill this square so the sum in each row and column is equal to N.

Oliver Wayne
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    Are you looking for the formula for it, or are you looking for the brute-force approach? Also, what have you tried? – Lasse V. Karlsen Dec 19 '21 at 19:42
  • @LasseV.Karlsen , I've tried only naive brute-force solution. I'm looking for a formula or an optimized and efficient brute-force approach. – Oliver Wayne Dec 19 '21 at 19:51
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    A formula would be more for a mathematics site. Here it should be about code. So what have you tried? Can you share your code and show what the problem is? – trincot Dec 19 '21 at 20:00

1 Answers1

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Let our matrix be

[[a0, a1, a2],
 [a3, a4, a5],
 [a6, a7, a8]]

The sum of the rows would then be:

a0 + a1 + a2 = N
a3 + a4 + a5 = N
a6 + a7 + a8 = N

And columns:

a0 + a3 + a6 = N
a1 + a4 + a7 = N
a2 + a5 + a8 = N

For each constant N, we have 6 equations of 9 unknowns. As it turns out, the last equation is dependent on the others, so we really only have 5 independent equations, and thus end up with 4 free variables.

If you solve the system of equations for a given N, you can brute force these 4 variables, and then just verify that the rest resolve to integer values in the range [0, 8]. Here is an example for N=11:

a1  =    1 a5   +1 a6   +1 a8   +1 a9   -11
a2  =   -1 a5   -1 a8   +11
a3  =   -1 a6   -1 a9   +11
a4  =   -1 a5   -1 a6   +11
a5  =   arbitrary
a6  =   arbitrary
a7  =   -1 a8   -1 a9   +11
a8  =   arbitrary
a9  =   arbitrary
Dillon Davis
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