Sum in a sudoku with sum total of 15 and 45
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
$begingroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have created a sudoku puzzle with the following restrictions:
- Each row and column sum to $45$.
- Each row and column in the nine $3$ by $3$ sub-grids sum to $15$.
Is such a sudoku unique?
grid-deduction sudoku
grid-deduction sudoku
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hugh
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hugh
1,6961721
edited 1 hour ago
Hugh
1,6961721
edited 1 hour ago
Hugh
1,6961721
1,6961721
asked 2 hours ago
Aorica IresonAorica Ireson
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Aorica IresonAorica Ireson
61
asked 2 hours ago
Aorica IresonAorica Ireson
61
61
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aorica Ireson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 2 mins ago
NautilusNautilus
3,741523
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
add a comment |
$begingroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
$endgroup$
No.
Of course, there's rotation and reflection, but even beyond that you can swap any two rows within it's group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
answered 1 hour ago
Dr XorileDr Xorile
12.1k22567
12.1k22567
add a comment |
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 2 mins ago
NautilusNautilus
3,741523
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 2 mins ago
NautilusNautilus
3,741523
$endgroup$
add a comment |
$begingroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 2 mins ago
NautilusNautilus
3,741523
$endgroup$
I was beaten to a very similar answer, but you can create such a square based on the magic square below:
abc
def
ghi
Turn it into this:
abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde
Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).
answered 2 mins ago
NautilusNautilus
3,741523
answered 2 mins ago
NautilusNautilus
3,741523
answered 2 mins ago
NautilusNautilus
3,741523
answered 2 mins ago
NautilusNautilus
3,741523
3,741523
add a comment |
add a comment |
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
Aorica Ireson is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
"...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$.
$endgroup$
– Hugh
2 hours ago
$begingroup$
"where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45?
$endgroup$
– Kryesec
1 hour ago
$begingroup$
@Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15.
$endgroup$
– Hugh
1 hour ago