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Consecutive numbers that are Manhattan distance 5 apart

Consecutive numbers that are Manhattan distance 3 apartPulling Apart a Jigsaw PuzzleThirteen Minesweeper mines that make seven 5sNeighbouring numbers summing to a prime on a 4x4Generating numbers with cubes3x3 self-descriptive squaresConsecutive numbers that are Manhattan distance 3 apartTransferring 9 pegs on a 9x9 gridA curious 5x5 squarePeaceable Bishops on an 8x8 grid

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5

$begingroup$

Can you place numbers from 1 to 36 on a 6x6 grid, such that the distance between any two consecutive numbers ($a$ and $a+1$) is Manhattan distance 5?

Bonus question: can you also make 1 and 36 be separated by Manhattan distance 5, thus making it a closed tour?

Note that the Manhattan distance between two locations is the distance between their row locations plus the distance between their column locations.

Here is a similar question for a 4x4 grid: Consecutive numbers that are Manhattan distance 3 apart

Good luck!

mathematics logical-deduction combinatorics

share|improve this question

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Can you place numbers from 1 to 36 on a 6x6 grid, such that the distance between any two consecutive numbers ($a$ and $a+1$) is Manhattan distance 5?

Bonus question: can you also make 1 and 36 be separated by Manhattan distance 5, thus making it a closed tour?

Note that the Manhattan distance between two locations is the distance between their row locations plus the distance between their column locations.

Here is a similar question for a 4x4 grid: Consecutive numbers that are Manhattan distance 3 apart

Good luck!

mathematics logical-deduction combinatorics

share|improve this question

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Can you place numbers from 1 to 36 on a 6x6 grid, such that the distance between any two consecutive numbers ($a$ and $a+1$) is Manhattan distance 5?

Bonus question: can you also make 1 and 36 be separated by Manhattan distance 5, thus making it a closed tour?

Note that the Manhattan distance between two locations is the distance between their row locations plus the distance between their column locations.

Here is a similar question for a 4x4 grid: Consecutive numbers that are Manhattan distance 3 apart

Good luck!

mathematics logical-deduction combinatorics

share|improve this question

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

$endgroup$

share|improve this question

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

share|improve this question

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

asked Oct 2 at 6:16

Dmitry KamenetskyDmitry Kamenetsky

5,75699 silver badges5555 bronze badges

1 Answer
1

active

oldest

votes

6

$begingroup$

Give a non-loop solution firstly:

 04 29 16 33 36 05

 13 26 23 20 11 30

 32 35 08 01 14 17

 19 10 03 06 27 24

 22 15 12 31 34 09

 07 28 25 18 21 02

Update:

Give another solution with loop:

 01 26 23 04 07 10

 16 29 12 21 02 17

 31 06 09 36 15 32

 14 33 18 27 24 13

 35 20 03 30 11 34

 28 25 22 05 08 19

Strategy: We may categorize cells by distance of center:

 A B C C B A

 B C D D C B

 C D E E D C

 C D E E D C

 B C D D C B

 A B C C B A

Consider that if we could make a route from A to E, and contains 1 A, 2 Bs, 3 Cs, 2 Ds and 1 E, then this is the $1/4$ sub-route and we could copy to another $3/4$ sub-routes by point symmetry.


Also E has 5 distance to A, thus we could finally connect those 4 sub-routes, forming a loop.

share|improve this answer

edited Oct 2 at 8:59

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't mind the edit.. I was having a hard time seeing the numbers before.
  $endgroup$
  – JS1
  Oct 2 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's fine~ thanks! (My current environment can't upload images so I need to represent tables by all characters :( )
  $endgroup$
  – Conifers
  Oct 2 at 7:35
  

  
  
  
  

add a comment
 | 

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6

$begingroup$

Give a non-loop solution firstly:

 04 29 16 33 36 05

 13 26 23 20 11 30

 32 35 08 01 14 17

 19 10 03 06 27 24

 22 15 12 31 34 09

 07 28 25 18 21 02

Update:

Give another solution with loop:

 01 26 23 04 07 10

 16 29 12 21 02 17

 31 06 09 36 15 32

 14 33 18 27 24 13

 35 20 03 30 11 34

 28 25 22 05 08 19

Strategy: We may categorize cells by distance of center:

 A B C C B A

 B C D D C B

 C D E E D C

 C D E E D C

 B C D D C B

 A B C C B A

Consider that if we could make a route from A to E, and contains 1 A, 2 Bs, 3 Cs, 2 Ds and 1 E, then this is the $1/4$ sub-route and we could copy to another $3/4$ sub-routes by point symmetry.


Also E has 5 distance to A, thus we could finally connect those 4 sub-routes, forming a loop.

share|improve this answer

edited Oct 2 at 8:59

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't mind the edit.. I was having a hard time seeing the numbers before.
  $endgroup$
  – JS1
  Oct 2 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's fine~ thanks! (My current environment can't upload images so I need to represent tables by all characters :( )
  $endgroup$
  – Conifers
  Oct 2 at 7:35
  

  
  
  
  

add a comment
 | 

6

$begingroup$

Give a non-loop solution firstly:

 04 29 16 33 36 05

 13 26 23 20 11 30

 32 35 08 01 14 17

 19 10 03 06 27 24

 22 15 12 31 34 09

 07 28 25 18 21 02

Update:

Give another solution with loop:

 01 26 23 04 07 10

 16 29 12 21 02 17

 31 06 09 36 15 32

 14 33 18 27 24 13

 35 20 03 30 11 34

 28 25 22 05 08 19

Strategy: We may categorize cells by distance of center:

 A B C C B A

 B C D D C B

 C D E E D C

 C D E E D C

 B C D D C B

 A B C C B A

Consider that if we could make a route from A to E, and contains 1 A, 2 Bs, 3 Cs, 2 Ds and 1 E, then this is the $1/4$ sub-route and we could copy to another $3/4$ sub-routes by point symmetry.


Also E has 5 distance to A, thus we could finally connect those 4 sub-routes, forming a loop.

share|improve this answer

edited Oct 2 at 8:59

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't mind the edit.. I was having a hard time seeing the numbers before.
  $endgroup$
  – JS1
  Oct 2 at 7:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's fine~ thanks! (My current environment can't upload images so I need to represent tables by all characters :( )
  $endgroup$
  – Conifers
  Oct 2 at 7:35
  

  
  
  
  

add a comment
 | 

$begingroup$

Give a non-loop solution firstly:

 04 29 16 33 36 05

 13 26 23 20 11 30

 32 35 08 01 14 17

 19 10 03 06 27 24

 22 15 12 31 34 09

 07 28 25 18 21 02

Update:

Give another solution with loop:

 01 26 23 04 07 10

 16 29 12 21 02 17

 31 06 09 36 15 32

 14 33 18 27 24 13

 35 20 03 30 11 34

 28 25 22 05 08 19

Strategy: We may categorize cells by distance of center:

 A B C C B A

 B C D D C B

 C D E E D C

 C D E E D C

 B C D D C B

 A B C C B A

Consider that if we could make a route from A to E, and contains 1 A, 2 Bs, 3 Cs, 2 Ds and 1 E, then this is the $1/4$ sub-route and we could copy to another $3/4$ sub-routes by point symmetry.


Also E has 5 distance to A, thus we could finally connect those 4 sub-routes, forming a loop.

share|improve this answer

edited Oct 2 at 8:59

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

$endgroup$

share|improve this answer

edited Oct 2 at 8:59

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges

answered Oct 2 at 6:53

ConifersConifers

5,19411 gold badge88 silver badges5555 bronze badges