[bonanova]

An old favorite asks whether a termite can bore his way from a corner cell of a 3x3x3 Rubik's Cube

[assumed to be wooden] to the center, passing once through the center of every cell.

His path is always parallel to one of the cube edges.

Forget for the sake of the puzzle that a true Rubik's Cube has no center cell.

You may know the answer to this, or want to figure it out.

But let's generalize to other orders of cubes and ask:

For an nxnxn cube, for what values of n is it possible for a termite,

starting from a corner, and passing once through each of the cells,

either to

1. end his journey at the center cell, or
   
2. come back to his starting point?
   

Enjoy.

[unreality]

I calculated it first for a 1x1x1, like a dice. It's possible:

  4

 3 2 5

  1

  6

if you go on the 1 -> 6 path

That means it's possible for all cubes if you can travel that way on the faces from centers to corners to corners on the individual square faces

then I figured out, from the center or rough center on any NxN square, how to emerge in any of the four directions from the center space after touching all spaces. I did that from N=1 through N=5, so I'm pretty sure it's possible for all N (could be wrong of course)

Going from corner to another corner is even easier, just go on long rows... for diagonal corners (only possible every other for adjacent corners depending on odd/even parity). So have to recheck that, and get a more general proof...

but I think, for #1, it's possible for all. I'll be back with a better analysis though ;P

edit - codebox

Edited June 27, 2009 by unreality

[Guest]

unreality, i don't think you understand the puzzle.

for a 1x1x1 cube you would start at the center and stay there.

the goal is to build a tunnel from cube to cube.

for a 3x3x3 cube, i'm fairly certain its impossible to visit every cube and the center cube last,

for an nxnxn cube i have no idea what rule you would use to determine whether it is possible.

my guess is it has something to do with prime numbers but really don't know.

[Guest]

it works for all even n, and for no odd n (except 1, if that counts).

[CaptainEd]

Coloring argument: Color the starting corner cube Black. Color remaining cubes in alternating White, Black, like a 3D checkerboard.

For any odd N, there is one more B than W. Therefore, the path must end in a B. That B is not adjacent to the original B, so no way you can return to the original cell.

Furthermore, for n = 3(mod 4), the center cube is W, so the path cannot end there.

Don't yet see why you can't end in the center of N = 1(mod 4).

[CaptainEd]

There are 5 planes, we're starting at 1b, moving through 1s until 2a. Then from 2b, following 2s to 3a, etc. until 10a, through 10s to the O in the middle.

[font="Courier New"]
1b 1 1 1 1 7a 6 6 5 3 7b 7 7 5 3 7 7 7 5 3 7 7 8a 5 3
1 1 1 1 1 6 6 6 5 3 10 10 10b 5 3 9 9 10a 5 3 8 8 8b 5 3
1 1 1 1 1 6 6 6b 6a 3 10 10 O 5 3 9 9 9b 5 3 8 8 9a 5b 3
1 1 1 1 1 4 4 4 4b 4a 4 4 4 4 3 4 4 4 4 3 4 4 4 5a 3b
1 1 1 1 2a 2 2 2 2 2b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3a [/font]

If this holds up, I'll conjecture that you can start at a corner and wind up in the center for N = 1(mod 4).

[unreality]

edit - codebox

I calculated it first for a 1x1x1, like a dice. It's possible:

  4

 3 2 5

  1

  6

if you go on the 1 -> 6 path

That means it's possible for all cubes if you can travel that way on the faces from centers to corners to corners on the individual square faces

then I figured out, from the center or rough center on any NxN square, how to emerge in any of the four directions from the center space after touching all spaces. I did that from N=1 through N=5, so I'm pretty sure it's possible for all N (could be wrong of course)

Going from corner to another corner is even easier, just go on long rows... for diagonal corners (only possible every other for adjacent corners depending on odd/even parity). So have to recheck that, and get a more general proof...

but I think, for #1, it's possible for all. I'll be back with a better analysis though ;P

lol I was thinking in terms of going across the center of individual squares on the surface area of the cube, not boring through the cube itself... I see now Dinnertime now, but I'll get back to this problem now that I know what it's asking

[bonanova]

Spoiler for In fact, I think this is a successful solution for N=5:

There are 5 planes, we're starting at 1b, moving through 1s until 2a. Then from 2b, following 2s to 3a, etc. until 10a, through 10s to the O in the middle.

 <font face="Courier New">

 1b  1   1   1   1       7a  6   6   5   3       7b  7   7   5   3       7   7   7   5   3       7   7   8a  5   3   

 1   1   1   1   1       6   6   6   5   3       10  10  10b 5   3       9   9   10a 5   3       8   8   8b  5   3   

 1   1   1   1   1       6   6   6b  6a  3       10  10  O   5   3       9   9   9b  5   3       8   8   9a  5b  3   

 1   1   1   1   1       4   4   4   4b  4a      4   4   4   4   3       4   4   4   4   3       4   4   4   5a  3b  

 1   1   1   1   2a      2   2   2   2   2b      2   2   2   2   2       2   2   2   2   2       2   2   2   2   3a  </font>

 [/codebox]

If this holds up, I'll conjecture that you can start at a corner and wind up in the center for N = 1(mod 4).



The Captain has it!

Kudos