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4-dimensional Knight's Tour

The knight's gameLet's revisit this later (a two player knight's game)Knight's Tour QuestionClosed knight tour on 7x7 boardA Knight's KuromasuKnight's Tour on a cube surfaceA Pawn is riding this Knight on his TourKnight's Tour on a 7x7 Board starting from D5Four-dimensional light bulbs

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

11

1

$begingroup$

Imagine a 4-dimensional chess board, but with only 4 squares a side. So, $4times4times4times4=256$ sub-hypercubes.

Your mission - should you choose to accept - is to:

Find a Knight's Tour

Four dimensional spaces are actually quite easy to visualize:

The orange squares represent the possible destinations of a knight that starts from the red square - it moves $1$ in a dimension and $2$ in another.

If the coordinates of a knight are $(x,y,z,t)$, then we can get to $(xpm1,ypm2,z,t)$ or $(x,ypm1,z,tpm2)$, or any other combination, as long as we remain on the board.

The coordinates of the knight shows are $(1,1,1,1)$, with depth representing the $z$-axis, and the $t$-axis being represented by the multiple grid show left to right.

Therefore the first few coordinates of the orange squares, read left-right, top-bottom, are:

$(3,1,0,1), (1,1,0,3), (1,3,0,1), (3,0,1,1), (1,0,1,3), (3,1,1,0), dots$

knight-moves four-dimensional

share|improve this question

edited Aug 8 at 10:31

JonMark Perry

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

$endgroup$

add a comment | 

11

1

$begingroup$

Imagine a 4-dimensional chess board, but with only 4 squares a side. So, $4times4times4times4=256$ sub-hypercubes.

Your mission - should you choose to accept - is to:

Find a Knight's Tour

Four dimensional spaces are actually quite easy to visualize:

The orange squares represent the possible destinations of a knight that starts from the red square - it moves $1$ in a dimension and $2$ in another.

If the coordinates of a knight are $(x,y,z,t)$, then we can get to $(xpm1,ypm2,z,t)$ or $(x,ypm1,z,tpm2)$, or any other combination, as long as we remain on the board.

The coordinates of the knight shows are $(1,1,1,1)$, with depth representing the $z$-axis, and the $t$-axis being represented by the multiple grid show left to right.

Therefore the first few coordinates of the orange squares, read left-right, top-bottom, are:

$(3,1,0,1), (1,1,0,3), (1,3,0,1), (3,0,1,1), (1,0,1,3), (3,1,1,0), dots$

knight-moves four-dimensional

share|improve this question

edited Aug 8 at 10:31

JonMark Perry

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

$endgroup$

add a comment | 

$begingroup$

Imagine a 4-dimensional chess board, but with only 4 squares a side. So, $4times4times4times4=256$ sub-hypercubes.

Your mission - should you choose to accept - is to:

Find a Knight's Tour

Four dimensional spaces are actually quite easy to visualize:

The orange squares represent the possible destinations of a knight that starts from the red square - it moves $1$ in a dimension and $2$ in another.

If the coordinates of a knight are $(x,y,z,t)$, then we can get to $(xpm1,ypm2,z,t)$ or $(x,ypm1,z,tpm2)$, or any other combination, as long as we remain on the board.

The coordinates of the knight shows are $(1,1,1,1)$, with depth representing the $z$-axis, and the $t$-axis being represented by the multiple grid show left to right.

Therefore the first few coordinates of the orange squares, read left-right, top-bottom, are:

$(3,1,0,1), (1,1,0,3), (1,3,0,1), (3,0,1,1), (1,0,1,3), (3,1,1,0), dots$

knight-moves four-dimensional

share|improve this question

edited Aug 8 at 10:31

JonMark Perry

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

$endgroup$

share|improve this question

edited Aug 8 at 10:31

JonMark Perry

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

share|improve this question

edited Aug 8 at 10:31

JonMark Perry

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

asked Aug 8 at 10:24

JonMark PerryJonMark Perry

24.8k66 gold badges4646 silver badges109109 bronze badges

1 Answer
1

active

oldest

votes

10

$begingroup$

This puzzle can be broken down relatively easily.

Start with

A 3-dimensional 4x4x2. This can be done as follows:


This is not as daunting as it looks. There are essentially just two patterns of four and we switch back and forth between them. The patterns are corner-middle-corner-middle, and side-side-side-side.


Note that we're starting from a corner and ending adjacent to that corner. And, of course, that we can do that in reverse.

So now we can put it together to finish the whole thing:

Do the red ones first, then the orange and then repeat the whole process for each column (hyper-column?)


Here I only show the first and last position of each 2x4x4 section. So we start at A and go to B, covering all the red squares. Then we jump to the next 2x4x4 section (B->C). Then we do C to D, covering all the orange squares. Similar logic takes us to E and the whole thing repeats.


So the overall loop is A---B-C---D-E---F-G---H-A'---B'-C'---D'-E'---F'-G'---H', where the "-" is a direct jump, and the "---" is the move that covers the corresponding 2x4x4.

share|improve this answer

edited Aug 8 at 12:35

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  And $H^'to A^''$ in the 5-th dimension, and so on... Well done!
  $endgroup$
  – JonMark Perry
  Aug 8 at 12:54
  
  

  
  
  
  

add a comment | 

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10

$begingroup$

This puzzle can be broken down relatively easily.

Start with

A 3-dimensional 4x4x2. This can be done as follows:


This is not as daunting as it looks. There are essentially just two patterns of four and we switch back and forth between them. The patterns are corner-middle-corner-middle, and side-side-side-side.


Note that we're starting from a corner and ending adjacent to that corner. And, of course, that we can do that in reverse.

So now we can put it together to finish the whole thing:

Do the red ones first, then the orange and then repeat the whole process for each column (hyper-column?)


Here I only show the first and last position of each 2x4x4 section. So we start at A and go to B, covering all the red squares. Then we jump to the next 2x4x4 section (B->C). Then we do C to D, covering all the orange squares. Similar logic takes us to E and the whole thing repeats.


So the overall loop is A---B-C---D-E---F-G---H-A'---B'-C'---D'-E'---F'-G'---H', where the "-" is a direct jump, and the "---" is the move that covers the corresponding 2x4x4.

share|improve this answer

edited Aug 8 at 12:35

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  And $H^'to A^''$ in the 5-th dimension, and so on... Well done!
  $endgroup$
  – JonMark Perry
  Aug 8 at 12:54
  
  

  
  
  
  

add a comment | 

10

$begingroup$

This puzzle can be broken down relatively easily.

Start with

A 3-dimensional 4x4x2. This can be done as follows:


This is not as daunting as it looks. There are essentially just two patterns of four and we switch back and forth between them. The patterns are corner-middle-corner-middle, and side-side-side-side.


Note that we're starting from a corner and ending adjacent to that corner. And, of course, that we can do that in reverse.

So now we can put it together to finish the whole thing:

Do the red ones first, then the orange and then repeat the whole process for each column (hyper-column?)


Here I only show the first and last position of each 2x4x4 section. So we start at A and go to B, covering all the red squares. Then we jump to the next 2x4x4 section (B->C). Then we do C to D, covering all the orange squares. Similar logic takes us to E and the whole thing repeats.


So the overall loop is A---B-C---D-E---F-G---H-A'---B'-C'---D'-E'---F'-G'---H', where the "-" is a direct jump, and the "---" is the move that covers the corresponding 2x4x4.

share|improve this answer

edited Aug 8 at 12:35

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  And $H^'to A^''$ in the 5-th dimension, and so on... Well done!
  $endgroup$
  – JonMark Perry
  Aug 8 at 12:54
  
  

  
  
  
  

add a comment | 

$begingroup$

This puzzle can be broken down relatively easily.

Start with

A 3-dimensional 4x4x2. This can be done as follows:


This is not as daunting as it looks. There are essentially just two patterns of four and we switch back and forth between them. The patterns are corner-middle-corner-middle, and side-side-side-side.


Note that we're starting from a corner and ending adjacent to that corner. And, of course, that we can do that in reverse.

So now we can put it together to finish the whole thing:

Do the red ones first, then the orange and then repeat the whole process for each column (hyper-column?)


Here I only show the first and last position of each 2x4x4 section. So we start at A and go to B, covering all the red squares. Then we jump to the next 2x4x4 section (B->C). Then we do C to D, covering all the orange squares. Similar logic takes us to E and the whole thing repeats.


So the overall loop is A---B-C---D-E---F-G---H-A'---B'-C'---D'-E'---F'-G'---H', where the "-" is a direct jump, and the "---" is the move that covers the corresponding 2x4x4.

share|improve this answer

edited Aug 8 at 12:35

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

$endgroup$

share|improve this answer

edited Aug 8 at 12:35

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges

answered Aug 8 at 12:29

Dr XorileDr Xorile

14.7k33 gold badges3232 silver badges9090 bronze badges