Equation to locate a square in a square
$begingroup$
Good evening,
I have been experimenting with different Sudoku checker and have come across a problem:
For a nxn Sudoku where n is a square number (4,6,19,25 etcc.), there would be an n number of sub-squares in a square. For example for a 9 by 9 there are 9 3x3 equal sized sub squares:
let column = j
let row = i
Now what I am trying to do is to come up with an equation in terms of n and sub-square number needed such that it would equal the first row number.
e.g for sub square number 3 the first row is number 3, for sub square 0 first row is 0 etc.
the issue with this is that the sub squares 0,1 and 2 all start at the same row 0 and therefore does this mean it is impossible to find an equation/algorithm or any other way to do this?
would the same apply for columns for sub squares?
recreational-mathematics problem-solving sudoku
asked Jan 1 at 23:16
fredfred
162
$endgroup$
add a comment |
$begingroup$
Good evening,
I have been experimenting with different Sudoku checker and have come across a problem:
For a nxn Sudoku where n is a square number (4,6,19,25 etcc.), there would be an n number of sub-squares in a square. For example for a 9 by 9 there are 9 3x3 equal sized sub squares:
let column = j
let row = i
Now what I am trying to do is to come up with an equation in terms of n and sub-square number needed such that it would equal the first row number.
e.g for sub square number 3 the first row is number 3, for sub square 0 first row is 0 etc.
the issue with this is that the sub squares 0,1 and 2 all start at the same row 0 and therefore does this mean it is impossible to find an equation/algorithm or any other way to do this?
would the same apply for columns for sub squares?
recreational-mathematics problem-solving sudoku
asked Jan 1 at 23:16
fredfred
162
$endgroup$
add a comment |
$begingroup$
Good evening,
I have been experimenting with different Sudoku checker and have come across a problem:
For a nxn Sudoku where n is a square number (4,6,19,25 etcc.), there would be an n number of sub-squares in a square. For example for a 9 by 9 there are 9 3x3 equal sized sub squares:
let column = j
let row = i
Now what I am trying to do is to come up with an equation in terms of n and sub-square number needed such that it would equal the first row number.
e.g for sub square number 3 the first row is number 3, for sub square 0 first row is 0 etc.
the issue with this is that the sub squares 0,1 and 2 all start at the same row 0 and therefore does this mean it is impossible to find an equation/algorithm or any other way to do this?
would the same apply for columns for sub squares?
recreational-mathematics problem-solving sudoku
asked Jan 1 at 23:16
fredfred
162
$endgroup$
Good evening,
I have been experimenting with different Sudoku checker and have come across a problem:
For a nxn Sudoku where n is a square number (4,6,19,25 etcc.), there would be an n number of sub-squares in a square. For example for a 9 by 9 there are 9 3x3 equal sized sub squares:
let column = j
let row = i
Now what I am trying to do is to come up with an equation in terms of n and sub-square number needed such that it would equal the first row number.
e.g for sub square number 3 the first row is number 3, for sub square 0 first row is 0 etc.
the issue with this is that the sub squares 0,1 and 2 all start at the same row 0 and therefore does this mean it is impossible to find an equation/algorithm or any other way to do this?
would the same apply for columns for sub squares?
recreational-mathematics problem-solving sudoku
recreational-mathematics problem-solving sudoku
asked Jan 1 at 23:16
fredfred
162
asked Jan 1 at 23:16
fredfred
162
asked Jan 1 at 23:16
fredfred
162
asked Jan 1 at 23:16
fredfred
162
asked Jan 1 at 23:16
fredfred
162
162
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In your example, square $k$ is in row $lfloor frac k3 rfloor$ and in column $k bmod 3$. For general $n$, replace $3$ by $n$. This shows the nice side of counting starting with zero.
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
|
show 1 more comment
$begingroup$
Assuming the rows and columns are the numbers you wrote outside the square: If $n=m^2$ then the first row in the $k$'th sub-square would be row $m cdotlfloor frac{k}{m} rfloor$ (notice the floor function) and the first column would be $m cdot (k bmod m)$.
So for $n=9$ the first row in e.g. the $7$'th sub-square would be row $3 cdotlfloor frac{7}{3} rfloor = 6$ and the first column would be $3 cdot (7 bmod 3)=3$.
answered Jan 2 at 1:53
JensJens
3,71521030
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your example, square $k$ is in row $lfloor frac k3 rfloor$ and in column $k bmod 3$. For general $n$, replace $3$ by $n$. This shows the nice side of counting starting with zero.
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
|
show 1 more comment
$begingroup$
In your example, square $k$ is in row $lfloor frac k3 rfloor$ and in column $k bmod 3$. For general $n$, replace $3$ by $n$. This shows the nice side of counting starting with zero.
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
$endgroup$
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
|
show 1 more comment
$begingroup$
In your example, square $k$ is in row $lfloor frac k3 rfloor$ and in column $k bmod 3$. For general $n$, replace $3$ by $n$. This shows the nice side of counting starting with zero.
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
$endgroup$
In your example, square $k$ is in row $lfloor frac k3 rfloor$ and in column $k bmod 3$. For general $n$, replace $3$ by $n$. This shows the nice side of counting starting with zero.
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
answered Jan 1 at 23:19
Ross MillikanRoss Millikan
294k23198371
294k23198371
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
|
show 1 more comment
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Note that calculating $lfloorfrac{k}3rfloor$ in most programming languages simply requires the statement $x:=k/3;$ and $kmbox{ mod }3$ requires $x:=k%3$.
$endgroup$
– SmileyCraft
Jan 1 at 23:20
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
Thank you for your response. To clarify by square do you mean each small square that hold 1 number or a sub-square which holds n numbers?
$endgroup$
– fred
Jan 1 at 23:29
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
For either case, I am not exactly sure if this would work unless I am mistaken (let me know if I am!). For example if we take subsquare 8 and we do 8/3 we get the 2.66..th row.
$endgroup$
– fred
Jan 1 at 23:33
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I may not have made it clear with the picture my apologies! the lines I drew are the outlines of the square and sub squares not the rows and columns.
$endgroup$
– fred
Jan 1 at 23:36
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
$begingroup$
I assumed the rows and columns were numbered $0$ to $2$ in each direction.
$endgroup$
– Ross Millikan
Jan 1 at 23:47
|
show 1 more comment
$begingroup$
Assuming the rows and columns are the numbers you wrote outside the square: If $n=m^2$ then the first row in the $k$'th sub-square would be row $m cdotlfloor frac{k}{m} rfloor$ (notice the floor function) and the first column would be $m cdot (k bmod m)$.
So for $n=9$ the first row in e.g. the $7$'th sub-square would be row $3 cdotlfloor frac{7}{3} rfloor = 6$ and the first column would be $3 cdot (7 bmod 3)=3$.
answered Jan 2 at 1:53
JensJens
3,71521030
$endgroup$
add a comment |
$begingroup$
Assuming the rows and columns are the numbers you wrote outside the square: If $n=m^2$ then the first row in the $k$'th sub-square would be row $m cdotlfloor frac{k}{m} rfloor$ (notice the floor function) and the first column would be $m cdot (k bmod m)$.
So for $n=9$ the first row in e.g. the $7$'th sub-square would be row $3 cdotlfloor frac{7}{3} rfloor = 6$ and the first column would be $3 cdot (7 bmod 3)=3$.
answered Jan 2 at 1:53
JensJens
3,71521030
$endgroup$
add a comment |
$begingroup$
Assuming the rows and columns are the numbers you wrote outside the square: If $n=m^2$ then the first row in the $k$'th sub-square would be row $m cdotlfloor frac{k}{m} rfloor$ (notice the floor function) and the first column would be $m cdot (k bmod m)$.
So for $n=9$ the first row in e.g. the $7$'th sub-square would be row $3 cdotlfloor frac{7}{3} rfloor = 6$ and the first column would be $3 cdot (7 bmod 3)=3$.
answered Jan 2 at 1:53
JensJens
3,71521030
$endgroup$
Assuming the rows and columns are the numbers you wrote outside the square: If $n=m^2$ then the first row in the $k$'th sub-square would be row $m cdotlfloor frac{k}{m} rfloor$ (notice the floor function) and the first column would be $m cdot (k bmod m)$.
So for $n=9$ the first row in e.g. the $7$'th sub-square would be row $3 cdotlfloor frac{7}{3} rfloor = 6$ and the first column would be $3 cdot (7 bmod 3)=3$.
answered Jan 2 at 1:53
JensJens
3,71521030
answered Jan 2 at 1:53
JensJens
3,71521030
answered Jan 2 at 1:53
JensJens
3,71521030
answered Jan 2 at 1:53
JensJens
3,71521030
3,71521030
add a comment |
add a comment |
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