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- Magic squares. How to fill a magic square of order n in a symmetrical and logical way by analyzing the possible ways to achieve a given sum of numbers.

I have questions about how to count number of ways to fill a magic square of order n.

Hi there. Happy new year.

I am interested in magic squares. I am particularly interested in how to fill a square of order n in a symmetrical and logical way by analyzing the possible ways to achieve a given sum of numbers.

My question is about combinatorics analyses.

For example for a square of order n=4, the sum of rows, columns and diagonals must be equal to S=n *(n²+1)/2 =34.

There are C4,16 = 1,820 combinations of 4 different numbers between 1 and 16. Only 86 of these combinations are equal to 34, among those possible. And among these 86 combinations, there are several ways to assemble them to form such a magic square of order 4 which require a total of 10 different combined sums (4 rows + 4 columns + 2 diagonals).

My questions are:

a) How to compute this number of combinations whose sum makes 34 (equal to 86 possibilities in the quoted case) knowing that i) we have to consider all combinations of 4 different digits taken among n=1 to n=16, that is to say 1,820 (apply also the remark point iii); ii) in a sum, one must have only different digits used once; iii) all permutations of a sequence are counted only once (so for example, the permuted sums 1+3+14+16 = 3+14+1+16=14+3+16+1=. ...etc...=34 are counted only once; d) the possible sums range from the smallest 1+2+3+4 = 10 to the largest 13+14+15+16=58; e) the sum we are interested in are that equal to S=n *(n²+1)/2 =34. It is thus a question of counting under these conditions, how many of these combinations among 1820 make a sum S=34. What are the combinatorics equations ?

b) How many possibilities are there to form such a magic square except for rotation or mirror symmetries?

How to establish these different combinatorics computations?

The combinatoric analysis, taking into account multiple partitions without permutation, does not seem simple to determine at first sight.

Thank you very much.

New4M.

I am interested in magic squares. I am particularly interested in how to fill a square of order n in a symmetrical and logical way by analyzing the possible ways to achieve a given sum of numbers.

My question is about combinatorics analyses.

For example for a square of order n=4, the sum of rows, columns and diagonals must be equal to S=n *(n²+1)/2 =34.

There are C4,16 = 1,820 combinations of 4 different numbers between 1 and 16. Only 86 of these combinations are equal to 34, among those possible. And among these 86 combinations, there are several ways to assemble them to form such a magic square of order 4 which require a total of 10 different combined sums (4 rows + 4 columns + 2 diagonals).

My questions are:

a) How to compute this number of combinations whose sum makes 34 (equal to 86 possibilities in the quoted case) knowing that i) we have to consider all combinations of 4 different digits taken among n=1 to n=16, that is to say 1,820 (apply also the remark point iii); ii) in a sum, one must have only different digits used once; iii) all permutations of a sequence are counted only once (so for example, the permuted sums 1+3+14+16 = 3+14+1+16=14+3+16+1=. ...etc...=34 are counted only once; d) the possible sums range from the smallest 1+2+3+4 = 10 to the largest 13+14+15+16=58; e) the sum we are interested in are that equal to S=n *(n²+1)/2 =34. It is thus a question of counting under these conditions, how many of these combinations among 1820 make a sum S=34. What are the combinatorics equations ?

b) How many possibilities are there to form such a magic square except for rotation or mirror symmetries?

How to establish these different combinatorics computations?

The combinatoric analysis, taking into account multiple partitions without permutation, does not seem simple to determine at first sight.

Thank you very much.

New4M.