Mouse joined the chocolate cult?

Question

The (Ω)Mouse starts eating a block of chocolate, consisting of 2007 x 2007 x 2007 small cubes. After eating a small cube, it proceeds to an ajacent (sharing a face) small cube. The plan is to eat the central small cube in the last turn. Can this be done?

Assume that (Ω)Mouse lives long enough and it has a good appetite.

The Mouse can start eating chocolate from any small cube. Not necessarily from one with coordinates (1,1,1).

This is a totally chocolate problem, and the answer can also be formulated in "chocolate" terms.

Remo Aviron, imagine that small cubes are made of white and black chocalate,
and that they are arranged like in a chessboard, only in three dimensions. Suppose that the central cube is made of black chocolate. In the 3 x 3 x 3 case, the central 3 x 3 slice looks like this (X is black chocolate, O is white chocolate):

X 0 X
0 X 0
X 0 X

There are also two side slices:

0 X 0
X 0 X
0 X 0

In total, there are more white cubes (14) , than black cubes (13). This is true for all blocks of dimension (4n-1).

When the (Ω)Mouse eats cubes moving through their faces, the chocolate colors interchange. The chain is like this: 0-X-0-X-0-X-0. Since there are more cubes of white chocolate, the chain should start and finish from 0.
Hence, the central cube cannot be last or first.

This is what Bori Ska wrote, only in terms of parity.

Thanks for answering!

Answer 1

Lets assign integer coordinates X = (i,j k) to each cube.
Also lets define boolean function
parity(X) = even/odd when sum i+j+k is even/odd

When mouse makes first turn he eats chocolate
X1 = (1,1,1) of odd parity(X1) = odd.
At each turn Parity of eaten chocolate changes
from odd to even, and then
from even to odd.

Therefore mouse eats odd chocolate on odd turns and even chocolate on even turns. Central cube has coordiantes
Xc = (1004, 1004, 1004), has even perity and cannot be eaten on odd turn. Last turn is 2007³ and is odd.


***************
> The Mouse can start eating chocolate
> from any small cube. Not necessarily
> from one with coordinates (1,1,1).
***************
Oops!

Lets assign integer coordinates X = (i,j k) to each cube.
Also lets define boolean function
parity(X) = even/odd when sum i+j+k is even/odd

At each turn Parity of eaten chocolate changes
from odd to even, and then
from even to odd.
The sequence of eaten chocolates is therefore balanced ..-e-o-e-o-e-..

Initially there are
4042147171 even chocolates, and
4042147172 odd chocolates.

Odd chocolates are one more, and the mouse must start and end with eating odd chocolates.

Central cube has coordiantes
Xc = (1004, 1004, 1004), has even parity, and cannot be eaten on last turn.


Answer: No, mouse, bad plan.

Answer 2

Being not just any mouse, but an (Ω)Mouse, I would first figure out if I could eat myself out of such a big cube, and if I could, voila, I'd follow that path in.

For a 3x3x3 cube, I'd start in the center, I'd eat one of the middle cubes of any outside layer, then spiral around that layer, spiral around the middle layer (avoiding the previously eaten cube in the center) then sprial around the final layer eating the middle cube of this outside layer last. I would then repeat the pattern for the 5x5x5 layer, spiraling out, on the first layer, eating then eating in a spiral the next three layers (which are all just 1 unit thick), then spiralling in towards the center of the last 5x5 layer.

Repeat for 7, 9, 11 ... 2007 layer. Spiral out, sprial around eating all of the single layers, then sprial back in to the center cube of the last layer before starting on the center cube of the next layer.

Reversing, you would end up starting on the middle cube of an outside layer (1004, 1004, 1) and working inwards, layer by layer.

There may be other solutions, but why make my brain hurt. :-)

>>>>>>>>>>>>>

Just realized that it is even easier than I described. Our mouse just has to eat one layer at a time until he gets to the center-just so long as he doesn't miss a square or finish a layer on an edge. There are plenty more solutions than just layer by layer.