On a 2-dimensional Van der Waerden-like theorem on 2-coloured square grids

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Let us consider a $ntimes n$ grid. We colour each point of the grid using two colours.



We say that a grid is bad is there exists a set of 4 points on the grid with same color which are the vertices a square whose sides are parallel to the sides of the grid.



A grid which is not bad is said to be good.



Example: let say we colour using Red (R) or Blue (B).



The $3times 3$ grid $beginmatrix R & B & Rcr B & R & B cr R & B & B endmatrix$ is good.



A theorem of Furstenberg and Weiss states that there exists $ngeq 0$ such that any $(n+1)times (n+1)$ grid is bad.



The natural question which arises is :



Question. What is the minimal value of $n$ ? In other words, what is the value of the unique integer such that there is a $ntimes n$ good grid, but there is no $(n+1)times (n+1)$ good grid ?



I don't know the answer, nor any estimate on the size of $n$. For the moment, the only method I coud think of to solve the problem is to use a computer.



The only small thing I know is that there is a $5times 5$ good grid, so $ngeq 5.$



For example, the following grid is good:



$$beginmatrix
R & R & R & R & Rcr
B & R & B & R & B cr
B & R & B & B & R cr
B & B & R & R & B cr
R & B & B & R & B
endmatrix$$



Thanks in advance for your answers!







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  • This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
    – Peter Kagey
    Aug 18 at 0:15







  • 1




    And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
    – Peter Kagey
    Aug 18 at 0:16











  • Thank you very much!
    – GreginGre
    Aug 18 at 7:35














up vote
0
down vote

favorite












Let us consider a $ntimes n$ grid. We colour each point of the grid using two colours.



We say that a grid is bad is there exists a set of 4 points on the grid with same color which are the vertices a square whose sides are parallel to the sides of the grid.



A grid which is not bad is said to be good.



Example: let say we colour using Red (R) or Blue (B).



The $3times 3$ grid $beginmatrix R & B & Rcr B & R & B cr R & B & B endmatrix$ is good.



A theorem of Furstenberg and Weiss states that there exists $ngeq 0$ such that any $(n+1)times (n+1)$ grid is bad.



The natural question which arises is :



Question. What is the minimal value of $n$ ? In other words, what is the value of the unique integer such that there is a $ntimes n$ good grid, but there is no $(n+1)times (n+1)$ good grid ?



I don't know the answer, nor any estimate on the size of $n$. For the moment, the only method I coud think of to solve the problem is to use a computer.



The only small thing I know is that there is a $5times 5$ good grid, so $ngeq 5.$



For example, the following grid is good:



$$beginmatrix
R & R & R & R & Rcr
B & R & B & R & B cr
B & R & B & B & R cr
B & B & R & R & B cr
R & B & B & R & B
endmatrix$$



Thanks in advance for your answers!







share|cite|improve this question




















  • This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
    – Peter Kagey
    Aug 18 at 0:15







  • 1




    And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
    – Peter Kagey
    Aug 18 at 0:16











  • Thank you very much!
    – GreginGre
    Aug 18 at 7:35












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let us consider a $ntimes n$ grid. We colour each point of the grid using two colours.



We say that a grid is bad is there exists a set of 4 points on the grid with same color which are the vertices a square whose sides are parallel to the sides of the grid.



A grid which is not bad is said to be good.



Example: let say we colour using Red (R) or Blue (B).



The $3times 3$ grid $beginmatrix R & B & Rcr B & R & B cr R & B & B endmatrix$ is good.



A theorem of Furstenberg and Weiss states that there exists $ngeq 0$ such that any $(n+1)times (n+1)$ grid is bad.



The natural question which arises is :



Question. What is the minimal value of $n$ ? In other words, what is the value of the unique integer such that there is a $ntimes n$ good grid, but there is no $(n+1)times (n+1)$ good grid ?



I don't know the answer, nor any estimate on the size of $n$. For the moment, the only method I coud think of to solve the problem is to use a computer.



The only small thing I know is that there is a $5times 5$ good grid, so $ngeq 5.$



For example, the following grid is good:



$$beginmatrix
R & R & R & R & Rcr
B & R & B & R & B cr
B & R & B & B & R cr
B & B & R & R & B cr
R & B & B & R & B
endmatrix$$



Thanks in advance for your answers!







share|cite|improve this question












Let us consider a $ntimes n$ grid. We colour each point of the grid using two colours.



We say that a grid is bad is there exists a set of 4 points on the grid with same color which are the vertices a square whose sides are parallel to the sides of the grid.



A grid which is not bad is said to be good.



Example: let say we colour using Red (R) or Blue (B).



The $3times 3$ grid $beginmatrix R & B & Rcr B & R & B cr R & B & B endmatrix$ is good.



A theorem of Furstenberg and Weiss states that there exists $ngeq 0$ such that any $(n+1)times (n+1)$ grid is bad.



The natural question which arises is :



Question. What is the minimal value of $n$ ? In other words, what is the value of the unique integer such that there is a $ntimes n$ good grid, but there is no $(n+1)times (n+1)$ good grid ?



I don't know the answer, nor any estimate on the size of $n$. For the moment, the only method I coud think of to solve the problem is to use a computer.



The only small thing I know is that there is a $5times 5$ good grid, so $ngeq 5.$



For example, the following grid is good:



$$beginmatrix
R & R & R & R & Rcr
B & R & B & R & B cr
B & R & B & B & R cr
B & B & R & R & B cr
R & B & B & R & B
endmatrix$$



Thanks in advance for your answers!









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 18 at 0:02









GreginGre

1,27328




1,27328











  • This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
    – Peter Kagey
    Aug 18 at 0:15







  • 1




    And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
    – Peter Kagey
    Aug 18 at 0:16











  • Thank you very much!
    – GreginGre
    Aug 18 at 7:35
















  • This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
    – Peter Kagey
    Aug 18 at 0:15







  • 1




    And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
    – Peter Kagey
    Aug 18 at 0:16











  • Thank you very much!
    – GreginGre
    Aug 18 at 7:35















This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
– Peter Kagey
Aug 18 at 0:15





This question, Size of a 3-colored square grid to produce a monochrome rectangle, might be related.
– Peter Kagey
Aug 18 at 0:15





1




1




And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
– Peter Kagey
Aug 18 at 0:16





And this question on MathOverflow, The Problem about 2-coloring finite plane, says the answer is $n=14$. (That is, there is a good $14 times 14$ grid, but no good $15 times 15$ grid.)
– Peter Kagey
Aug 18 at 0:16













Thank you very much!
– GreginGre
Aug 18 at 7:35




Thank you very much!
– GreginGre
Aug 18 at 7:35















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