Permutations of NxN matrix with equal summation of any row elements or column elements (N being odd number)
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The matrix NxN has N rows and columns. It has all unique elements starting from 1 to (N^2). The condition is the summation of any row elements should be equal to summation of any other row or column elements.
Example: For 3x3 matrix, one of the possible combination looks like following.
4 8 3
2 6 7
9 1 5
Now the question is how many possible combinations can occur to satisfy the given condition of given NxN matrix where N is any odd number?
Thanks for the help in advance.
Patrick
combinations permutation probability
asked Nov 23 '18 at 23:35
PatrickPatrick
61
add a comment |
The matrix NxN has N rows and columns. It has all unique elements starting from 1 to (N^2). The condition is the summation of any row elements should be equal to summation of any other row or column elements.
Example: For 3x3 matrix, one of the possible combination looks like following.
4 8 3
2 6 7
9 1 5
Now the question is how many possible combinations can occur to satisfy the given condition of given NxN matrix where N is any odd number?
Thanks for the help in advance.
Patrick
combinations permutation probability
asked Nov 23 '18 at 23:35
PatrickPatrick
61
add a comment |
The matrix NxN has N rows and columns. It has all unique elements starting from 1 to (N^2). The condition is the summation of any row elements should be equal to summation of any other row or column elements.
Example: For 3x3 matrix, one of the possible combination looks like following.
4 8 3
2 6 7
9 1 5
Now the question is how many possible combinations can occur to satisfy the given condition of given NxN matrix where N is any odd number?
Thanks for the help in advance.
Patrick
combinations permutation probability
asked Nov 23 '18 at 23:35
PatrickPatrick
61
The matrix NxN has N rows and columns. It has all unique elements starting from 1 to (N^2). The condition is the summation of any row elements should be equal to summation of any other row or column elements.
Example: For 3x3 matrix, one of the possible combination looks like following.
4 8 3
2 6 7
9 1 5
Now the question is how many possible combinations can occur to satisfy the given condition of given NxN matrix where N is any odd number?
Thanks for the help in advance.
Patrick
combinations permutation probability
combinations permutation probability
asked Nov 23 '18 at 23:35
PatrickPatrick
61
asked Nov 23 '18 at 23:35
PatrickPatrick
61
asked Nov 23 '18 at 23:35
PatrickPatrick
61
asked Nov 23 '18 at 23:35
PatrickPatrick
61
asked Nov 23 '18 at 23:35
PatrickPatrick
61
61
add a comment |
add a comment |
1 Answer
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The best available answer is, "A heck of a lot."
If you add the condition "same sum down the diagonal", these are magic squares. As http://oeis.org/A006052 notes, the count of magic squares is known for n = 1, 2, 3, 4, and 5. The exact answer for 6 is not known, but it is in the order of 10**20.
Your counts will be higher still because you lack the diagonal condition. But the computational complexities are the same. Brute force will give you answers for n = 1, 2, 3, and 4 fairly easily. 5 will be doable. 6 will be intractable. 7 will be "no hope".
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
add a comment |
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1 Answer
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1 Answer
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The best available answer is, "A heck of a lot."
If you add the condition "same sum down the diagonal", these are magic squares. As http://oeis.org/A006052 notes, the count of magic squares is known for n = 1, 2, 3, 4, and 5. The exact answer for 6 is not known, but it is in the order of 10**20.
Your counts will be higher still because you lack the diagonal condition. But the computational complexities are the same. Brute force will give you answers for n = 1, 2, 3, and 4 fairly easily. 5 will be doable. 6 will be intractable. 7 will be "no hope".
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
add a comment |
The best available answer is, "A heck of a lot."
If you add the condition "same sum down the diagonal", these are magic squares. As http://oeis.org/A006052 notes, the count of magic squares is known for n = 1, 2, 3, 4, and 5. The exact answer for 6 is not known, but it is in the order of 10**20.
Your counts will be higher still because you lack the diagonal condition. But the computational complexities are the same. Brute force will give you answers for n = 1, 2, 3, and 4 fairly easily. 5 will be doable. 6 will be intractable. 7 will be "no hope".
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
add a comment |
The best available answer is, "A heck of a lot."
If you add the condition "same sum down the diagonal", these are magic squares. As http://oeis.org/A006052 notes, the count of magic squares is known for n = 1, 2, 3, 4, and 5. The exact answer for 6 is not known, but it is in the order of 10**20.
Your counts will be higher still because you lack the diagonal condition. But the computational complexities are the same. Brute force will give you answers for n = 1, 2, 3, and 4 fairly easily. 5 will be doable. 6 will be intractable. 7 will be "no hope".
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
The best available answer is, "A heck of a lot."
If you add the condition "same sum down the diagonal", these are magic squares. As http://oeis.org/A006052 notes, the count of magic squares is known for n = 1, 2, 3, 4, and 5. The exact answer for 6 is not known, but it is in the order of 10**20.
Your counts will be higher still because you lack the diagonal condition. But the computational complexities are the same. Brute force will give you answers for n = 1, 2, 3, and 4 fairly easily. 5 will be doable. 6 will be intractable. 7 will be "no hope".
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
answered Nov 27 '18 at 19:01
btillybtilly
27.1k24053
27.1k24053
add a comment |
add a comment |
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