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5

1

$begingroup$

This isn't something I read in a book or anything, it's more of a puzzle I thought up for myself.

However, I am unable to find a solution.

Here's my problem:

If I create a 9x9 checkerboard, and want to walk a path across it, where each block is walked upon only once, and there are only 90° turns, it's easy enough to create any path. (See first image example:)

However, if I want to create an entrance and exit point NEXT TO EACH OTHER, I am stumped. I can't do it:

Can someone help me think of a path ac cross a 9x9 area, where the entrance and exit points are directly next to each other, and yet each block is used only once and only 90° turns are used?

Thanks again.

pattern geometry

share|improve this question

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  From a glance I believe that it is impossible to do, but I have no proof other then just my intuition.
  $endgroup$
  – Rewan Demontay
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also think that it is impossible to do. From the appearance of the problem I think that there will be an elegant mathematical proof
  $endgroup$
  – Adam
  5 hours ago
  

  
  
  
  

add a comment | 

5

1

$begingroup$

This isn't something I read in a book or anything, it's more of a puzzle I thought up for myself.

However, I am unable to find a solution.

Here's my problem:

If I create a 9x9 checkerboard, and want to walk a path across it, where each block is walked upon only once, and there are only 90° turns, it's easy enough to create any path. (See first image example:)

However, if I want to create an entrance and exit point NEXT TO EACH OTHER, I am stumped. I can't do it:

Can someone help me think of a path ac cross a 9x9 area, where the entrance and exit points are directly next to each other, and yet each block is used only once and only 90° turns are used?

Thanks again.

pattern geometry

share|improve this question

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  From a glance I believe that it is impossible to do, but I have no proof other then just my intuition.
  $endgroup$
  – Rewan Demontay
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also think that it is impossible to do. From the appearance of the problem I think that there will be an elegant mathematical proof
  $endgroup$
  – Adam
  5 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This isn't something I read in a book or anything, it's more of a puzzle I thought up for myself.

However, I am unable to find a solution.

Here's my problem:

If I create a 9x9 checkerboard, and want to walk a path across it, where each block is walked upon only once, and there are only 90° turns, it's easy enough to create any path. (See first image example:)

However, if I want to create an entrance and exit point NEXT TO EACH OTHER, I am stumped. I can't do it:

Can someone help me think of a path ac cross a 9x9 area, where the entrance and exit points are directly next to each other, and yet each block is used only once and only 90° turns are used?

Thanks again.

pattern geometry

share|improve this question

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

etsnymanetsnyman

1283

New contributor

etsnyman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

etsnymanetsnyman

1283

1 Answer
1

active

oldest

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6

$begingroup$

It’s not possible.

Moving (stepping) only horizontally or vertically
(never moving diagonally or jumping over squares),
successive steps are always on squares of alternate colors. 
For example, in your first illustration, the first step is blue,
the second is pink, the third is blue again, and so on. 
In general, the odd-numbered steps are blue
and the even-numbered steps are pink.

Since both dimensions of your board are odd (9),
the total size is odd (9×9=81),
and so the last step, the 81st, is an odd number. 
Therefore, it must be the same color as the first square
(as seen in your first illustration). 
And, since adjacent (next-to-each-other) squares
are always different colors,
the exit square on an odd-sized board
cannot be next to the entrance square.

share|improve this answer

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I guessed so, but I wasn't sure. Thanks for confirming it!
  $endgroup$
  – etsnyman
  4 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

It’s not possible.

Moving (stepping) only horizontally or vertically
(never moving diagonally or jumping over squares),
successive steps are always on squares of alternate colors. 
For example, in your first illustration, the first step is blue,
the second is pink, the third is blue again, and so on. 
In general, the odd-numbered steps are blue
and the even-numbered steps are pink.

Since both dimensions of your board are odd (9),
the total size is odd (9×9=81),
and so the last step, the 81st, is an odd number. 
Therefore, it must be the same color as the first square
(as seen in your first illustration). 
And, since adjacent (next-to-each-other) squares
are always different colors,
the exit square on an odd-sized board
cannot be next to the entrance square.

share|improve this answer

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I guessed so, but I wasn't sure. Thanks for confirming it!
  $endgroup$
  – etsnyman
  4 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

It’s not possible.

Moving (stepping) only horizontally or vertically
(never moving diagonally or jumping over squares),
successive steps are always on squares of alternate colors. 
For example, in your first illustration, the first step is blue,
the second is pink, the third is blue again, and so on. 
In general, the odd-numbered steps are blue
and the even-numbered steps are pink.

Since both dimensions of your board are odd (9),
the total size is odd (9×9=81),
and so the last step, the 81st, is an odd number. 
Therefore, it must be the same color as the first square
(as seen in your first illustration). 
And, since adjacent (next-to-each-other) squares
are always different colors,
the exit square on an odd-sized board
cannot be next to the entrance square.

share|improve this answer

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I guessed so, but I wasn't sure. Thanks for confirming it!
  $endgroup$
  – etsnyman
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It’s not possible.

Moving (stepping) only horizontally or vertically
(never moving diagonally or jumping over squares),
successive steps are always on squares of alternate colors. 
For example, in your first illustration, the first step is blue,
the second is pink, the third is blue again, and so on. 
In general, the odd-numbered steps are blue
and the even-numbered steps are pink.

Since both dimensions of your board are odd (9),
the total size is odd (9×9=81),
and so the last step, the 81st, is an odd number. 
Therefore, it must be the same color as the first square
(as seen in your first illustration). 
And, since adjacent (next-to-each-other) squares
are always different colors,
the exit square on an odd-sized board
cannot be next to the entrance square.

share|improve this answer

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

$endgroup$

share|improve this answer

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939

answered 4 hours ago

Peregrine RookPeregrine Rook

4,79821939