chessboard 7x 7 The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraMath and Logic of Infinite ChessCan someone produce a sudoku puzzle where guessing more than one cell's value at a time is required?placing chess knights in a numbered chessboard.How many ways are there to position two black rooks and two white rooks on an 8X8 chessboardPlacing $9$ black and $9$ white rooks on $ 6times6$ board without any attacks between different coloursIn how many ways can the rooks be arranged?Prove that it is not possible to completely cover a 6 × 6 chessboard by tiles which have dimensions 1 × 4.1998 USAMO problem #4coins on chessboard, who has the winning strategyMaximum number of kings on the chessboard subject to some rules
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chessboard 7x 7
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraMath and Logic of Infinite ChessCan someone produce a sudoku puzzle where guessing more than one cell's value at a time is required?placing chess knights in a numbered chessboard.How many ways are there to position two black rooks and two white rooks on an 8X8 chessboardPlacing $9$ black and $9$ white rooks on $ 6times6$ board without any attacks between different coloursIn how many ways can the rooks be arranged?Prove that it is not possible to completely cover a 6 × 6 chessboard by tiles which have dimensions 1 × 4.1998 USAMO problem #4coins on chessboard, who has the winning strategyMaximum number of kings on the chessboard subject to some rules
$begingroup$
A 7 × 7 chessboard that is painted black and white, with the corners painted black;and we have the operator "inverse", which can be applied to a single row or single column in a table that it change the color of the cell from white to black or from black to white ( for example the first cell at row 2 in the chessboard colerd by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)
The goal is to make the whole board colered black. What is the minimum number of steps required to achieve the goal?
there is two soultion for this question, i need the two soultions!!
discrete-mathematics contest-math recreational-mathematics education
asked Mar 24 at 18:54
Nour TahaNour Taha
1
$endgroup$
|
show 2 more comments
$begingroup$
A 7 × 7 chessboard that is painted black and white, with the corners painted black;and we have the operator "inverse", which can be applied to a single row or single column in a table that it change the color of the cell from white to black or from black to white ( for example the first cell at row 2 in the chessboard colerd by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)
The goal is to make the whole board colered black. What is the minimum number of steps required to achieve the goal?
there is two soultion for this question, i need the two soultions!!
discrete-mathematics contest-math recreational-mathematics education
asked Mar 24 at 18:54
Nour TahaNour Taha
1
$endgroup$
$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
1
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
1
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33
|
show 2 more comments
$begingroup$
A 7 × 7 chessboard that is painted black and white, with the corners painted black;and we have the operator "inverse", which can be applied to a single row or single column in a table that it change the color of the cell from white to black or from black to white ( for example the first cell at row 2 in the chessboard colerd by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)
The goal is to make the whole board colered black. What is the minimum number of steps required to achieve the goal?
there is two soultion for this question, i need the two soultions!!
discrete-mathematics contest-math recreational-mathematics education
asked Mar 24 at 18:54
Nour TahaNour Taha
1
$endgroup$
A 7 × 7 chessboard that is painted black and white, with the corners painted black;and we have the operator "inverse", which can be applied to a single row or single column in a table that it change the color of the cell from white to black or from black to white ( for example the first cell at row 2 in the chessboard colerd by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)
The goal is to make the whole board colered black. What is the minimum number of steps required to achieve the goal?
there is two soultion for this question, i need the two soultions!!
discrete-mathematics contest-math recreational-mathematics education
discrete-mathematics contest-math recreational-mathematics education
asked Mar 24 at 18:54
Nour TahaNour Taha
1
asked Mar 24 at 18:54
Nour TahaNour Taha
1
asked Mar 24 at 18:54
Nour TahaNour Taha
1
asked Mar 24 at 18:54
Nour TahaNour Taha
1
asked Mar 24 at 18:54
Nour TahaNour Taha
1
1
$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
1
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
1
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33
|
show 2 more comments
$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
1
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
1
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33
$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
1
1
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
1
1
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If we consider the initial board, we just have to reverse $3$ rows and $3$ columns as indicated by the external stones on the picture above.
Note: $6$ moves is the minimal number since if we consider the stones on the top edge and right edge for instance, there are $6$ of them and they need to be reversed anyway. This cannot be done in less than $6$ operations.
The order has no importance, you can start by the rows $2,4,6$ and then all the columns, but you can also select any row or column that has not been already reversed.
For instance here is the status after $1$ column was reversed.
And after $2$ rows and $1$ column were reversed.
If you continue the process, with the remaining $3$ marked available moves, you end up with an all black board, no matter the order of the moves.
answered Mar 24 at 19:50
zwimzwim
12.7k832
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If we consider the initial board, we just have to reverse $3$ rows and $3$ columns as indicated by the external stones on the picture above.
Note: $6$ moves is the minimal number since if we consider the stones on the top edge and right edge for instance, there are $6$ of them and they need to be reversed anyway. This cannot be done in less than $6$ operations.
The order has no importance, you can start by the rows $2,4,6$ and then all the columns, but you can also select any row or column that has not been already reversed.
For instance here is the status after $1$ column was reversed.
And after $2$ rows and $1$ column were reversed.
If you continue the process, with the remaining $3$ marked available moves, you end up with an all black board, no matter the order of the moves.
answered Mar 24 at 19:50
zwimzwim
12.7k832
$endgroup$
add a comment |
$begingroup$
If we consider the initial board, we just have to reverse $3$ rows and $3$ columns as indicated by the external stones on the picture above.
Note: $6$ moves is the minimal number since if we consider the stones on the top edge and right edge for instance, there are $6$ of them and they need to be reversed anyway. This cannot be done in less than $6$ operations.
The order has no importance, you can start by the rows $2,4,6$ and then all the columns, but you can also select any row or column that has not been already reversed.
For instance here is the status after $1$ column was reversed.
And after $2$ rows and $1$ column were reversed.
If you continue the process, with the remaining $3$ marked available moves, you end up with an all black board, no matter the order of the moves.
answered Mar 24 at 19:50
zwimzwim
12.7k832
$endgroup$
add a comment |
$begingroup$
If we consider the initial board, we just have to reverse $3$ rows and $3$ columns as indicated by the external stones on the picture above.
Note: $6$ moves is the minimal number since if we consider the stones on the top edge and right edge for instance, there are $6$ of them and they need to be reversed anyway. This cannot be done in less than $6$ operations.
The order has no importance, you can start by the rows $2,4,6$ and then all the columns, but you can also select any row or column that has not been already reversed.
For instance here is the status after $1$ column was reversed.
And after $2$ rows and $1$ column were reversed.
If you continue the process, with the remaining $3$ marked available moves, you end up with an all black board, no matter the order of the moves.
answered Mar 24 at 19:50
zwimzwim
12.7k832
$endgroup$
If we consider the initial board, we just have to reverse $3$ rows and $3$ columns as indicated by the external stones on the picture above.
Note: $6$ moves is the minimal number since if we consider the stones on the top edge and right edge for instance, there are $6$ of them and they need to be reversed anyway. This cannot be done in less than $6$ operations.
The order has no importance, you can start by the rows $2,4,6$ and then all the columns, but you can also select any row or column that has not been already reversed.
For instance here is the status after $1$ column was reversed.
And after $2$ rows and $1$ column were reversed.
If you continue the process, with the remaining $3$ marked available moves, you end up with an all black board, no matter the order of the moves.
answered Mar 24 at 19:50
zwimzwim
12.7k832
edited Mar 24 at 19:56
answered Mar 24 at 19:50
zwimzwim
12.7k832
answered Mar 24 at 19:50
zwimzwim
12.7k832
answered Mar 24 at 19:50
zwimzwim
12.7k832
12.7k832
add a comment |
add a comment |
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$begingroup$
i have one answer now: at the begining i will "inverse" row 2 alone then row 4 then row 6 , now i have colmun 2, 4, and 6 is colered by white i will " inverse then but each one separately, but i need another solution:(
$endgroup$
– Nour Taha
Mar 24 at 19:04
1
$begingroup$
"which can be applied to a single row or single column ... by white and the second is black.... if i will use "inverse" to this row it will change the first cell to black and the second to white ...)" These are three inconsistent statements. The first says something undefined happens to an entire row or column. The second says it only applies to a single cell. The third says it applies only to the first two cells of a row or column. Why not just say "it reverses the colors of all cells in a row or column"?
$endgroup$
– fleablood
Mar 24 at 19:08
$begingroup$
i can say it reverses the colors of all cells in a row or in a colmun, im not that good in english😜
$endgroup$
– Nour Taha
Mar 24 at 19:11
$begingroup$
Well, it's easy to see you can do it in 6 turns by inverting rows 2,4 and 6. That leaves the board completely striped. Then invert colums 2,4,6 to make them all black. The question is is there any better.
$endgroup$
– fleablood
Mar 24 at 19:13
1
$begingroup$
I guess the other way is to start by columns 2,4,6 so then the stripped pattern is rotated compared to your solution. Does that count for another solution ? You can also alternate rows and columns to have a mixed pattern, in the end after 6 moves it is still all black.
$endgroup$
– zwim
Mar 24 at 19:33