Solid tower is constructed of black and white cubic bricks?

Question

Base layer is rectangle nxm, and has arbitrary chosen pattern of brick colors.
The bricks in each next layer nxm are chosen so, that
the color of each brick satisfies the following conditons:

1) Each black brick has even number of black neighbours;
2) Each white brick has odd number of black neighbours.

Prove: there will always appear a layer at certain
height, which pattern is similar to the base layer,
but all colors reversed.

That is if base layer is for exmaple
bwbw
bbbb

then sooner or later layer colored as
wbwb
wwww
will appear.

Neighbours of a given brick are six or less bricks, which share common face.

The tower has bottom, but costruction newer stops on the top, there is no top layer. You can say the tower grows to infinity like tower of Babylon.

You can always construct the next layer n+1, so that layer n becomes valid. After that layer n+2 will make valid layer n-1 below it, and so on...

Sorry about misunderstanding.

The tower has bottom, but costruction newer stops on the top, there is no top layer. You can say the tower grows to infinity like tower of Babylon.

You can always construct the next layer n+1, so that layer n becomes valid. After that layer n+2 will make valid layer n-1 below it, and so on...

Sorry about misunderstanding.

Dr. D:
base floor:
b w
b w
second floor:
b w
b w
third floor
w w
w w
fourth floor
w b
w b

There are at most 2^(mn) possible patterns.
There are at most 2^(2mn) possible pairs of layers k, k+1.

A pair k,k+1 uniquely defines layer k+2, therefore the tower has period P.
Furthermore, a pair k,k+1 uniquely defines layer k-1, so the tower is periodic in downward direction too.

Layer P has the same pattern as ground layer.
Layer P-1 is all white (= zeroth layer right below ground layer).
Layer P-2 is inverse of P (= -1th layer).

Answer 1

The question at first glance is very interesting indeed, but I'm not sure I've understood the conditions correctly. Take a base floor to be 2x1 - 2 white bricks. How to build the 2nd floor?
- - /2nd floor/
w w /base floor/ - these must have odd number of black neighbours to satisfy 2), so the 2nd floor must be b b, but then 1) is violated!
So may I ask for more additional information: is the height of the tower infinite? Of course a Babylon tower would be another story.
Or are 1) and 2) valid ABOVE the base layer only?

Continued (after having read the additional details): the problem is very intriguing, I have no solution yet, but the time is about to expire, I'll ask Dr. Dumbfellow to grant extra time as much as Yahoo answers allow. Let me share with the community what I've done: I decided to begin with the case n x 1 base layer - so the tower in this case is practically a wall. Let us replace every black brick with 0, every white with 1, then each layer will be coded with a n-digit binary number, easily can be seen that we can always add also an underground layer (floor 0) 111..11, then having an arbitrary base (floor 1), every digit on "ith" position in "(k+1)th" layer (floor) can be obtained by the recurrence relationship:
(***) d[i, k+1] = (d[i, k] + d[i-1, k] + d[i+1, k] + d[i, k-1]) mod 2,
here k=2,3, . . ; i=2,3, . . ,n-1; for the leftmost and rightmost digits with obvious adjustments - replace the missing addend with 1, or say it otherwise, imagine the wall is bounded from both sides with columns of 1s. Here are 2 examples /imagine the wall grows DOWN for convenience/:
1 1 1 . . . . 1 1 1 (floor 0)
0 0 1 . . . . 1 0 1 (base floor)
0 0 1 . . . . 1 1 1 (2nd floor)
1 1 1 . . . . 0 1 0
1 1 0 . . . . 1 0 1
0 1 1 . . . . 0 1 0 cycle end
1 1 1 . . . . 1 1 1 new cycle begins /next 6 floors/
1 0 0 . . . . 1 0 1 etc.
1 0 0 . . . .
1 1 1 . . . .
0 1 1 . . . .
1 1 0 cycle end
1 1 1 new cycle begins /next 12 floors/
0 0 1 etc.
Then I wrote a computer program, generating n-bit binary numbers for various values of n and displaying the wall sufficiently far. From more then 100 such trials I noticed a cyclic behaviour - the floors pattern repeats infinitely many times from some point on /look the examples above/. I can't prove for the time being that this will always happen, but the evidence is very strong according the tests I've made /looks like this is a math problem where computers can help/. Now look at the cycle ends: the floor 111..11 acts like the binary bit-wise operation "not": not 0 = 1, not 1 = 0 - it inverts all bits, which follows from (***) above - indeed not 110 = 001, not 010 = 101, so the floor at the end of a cycle is inverted base floor.
Now if we consider a tower n x m as consisting from m vertical walls n x 1 and if the pattern in each wall repeats in such a way /of course, instead of (***) we'll have:
d[i,,j,k+1] = (d[i,j,k-1] + d[i,j,k] + d[i-1,j,k] + d[i+1,j,k] + d[i,j-1,k] + d[i,j+1,k]) mod 2 with the same details/,
we can take the LCM of the cycles lengths - this floor will be a matrix of 1s only, the next will resemble the base floor, the previous will be inversed.

Of course I can't be sure that this approach is most promising, I would appreciate any full solution, my curiosity isn't less than Dr D's.

Answer 2

I was wondering the same thing as Duke. Is the tower infinite or finite? Also could n = m?

Consider a tower 2 floors high (in math a tower can be 2 floors high), with n = m = 2.

Base floor:
b w
b w

Top floor:
b w
b w

Every b has 2 black neighbors, and every w has 1 black neighbor. Yet the pattern is not reversed.

*EDIT*
Yes I realized that if you go up to the 4th floor, the pattern will reverse. That was while I wasn't sure whether the number of floors was finite or not.

Now that the floors are infinite, I can verify it for a few cases, but so far not able to prove it formally - other than saying that over infinite trials, any event has a probability of 1. But these are not random events.

*EDIT*
If you actually have a solution for this, I'd love to see it if no one gets it before the question expires. I just hope it's not something like "XYZ's theorem says such and such, therefore the pattern will be reversed".

Answer 3

Please elaborate - do diagonal neighbours (in the current and preceding layers) count in the enumeration?

Answer 4

challenging issue. look onto google. that can assist!