Cubical Infection

Question

You may remember the C. Coli bacteria, which infects checkerboard squares. There is a similar disease which plagues cubes, known as cubic lice. Once a cube gets infested with cubic lice, it will be infested for the rest of its life!

A refresher on cube biology: cubes live together in colonies, which consist of $1000$ cubes tightly packed together in a $10\times 10 \times 10$ cubical pattern. If a healthy cube is ever adjacent to at least three infested cubes, then that cube will itself become infested the next day.

A bioterrorist wants to infest an entire colony of cubes with cubic lice. He plans to break into a WHO lab and steal several cubic larvae, each of which can cause a single cube to become infested.

What is fewest number of larvae he needs to steal to be able to infect the entire colony?

As usual, you should prove your answer is correct, by showing how he can infect the colony with that many larvae, and why he can't with fewer.

Answer 1

Answer - we need at least 100.

Just as before, after each iteration the surface of the infected part does not increase. Since the surface of a unit cube is 6 and the surface of the big cube is 600, we need at least 600/6=100 infected cubes. This is my pattern for infecting 10x10x10 block with 200 larvae:

1st layer:
X 0 0 0 0 0 0 0 0 0
0 X 0 0 0 0 0 0 0 0
0 0 X 0 0 0 0 0 0 0
0 0 0 X 0 0 0 0 0 0
............................
0 0 0 0 0 0 0 0 0 X

2nd layer:
0 X 0 0 0 0 0 0 0 0
0 0 X 0 0 0 0 0 0 0
0 0 0 X 0 0 0 0 0 0
0 0 0 0 X 0 0 0 0 0
............................
X 0 0 0 0 0 0 0 0 0

10th layer:
0 0 0 0 0 0 0 0 0 X
X 0 0 0 0 0 0 0 0 0
0 X 0 0 0 0 0 0 0 0
0 0 X 0 0 0 0 0 0 0
............................
0 0 0 0 0 0 0 0 X 0

However, there also exist patterns that will work, for which no two infested cubes are in the same line.

Answer 2

Well, i think you only need

3 larvaes, places like a L

Because

!To simplify my demonstration let's imagine a 4*4 cube where each element have a position {x,y,z}

oooo oooo oooo oooo
oooo oooo oooo oooo
oooo oooo oooo oooo
oooo oooo oooo oooo

Let's add 3 infested elements in {3,1,1}, {3,2,1},{4,1,1} (step 1)

ooxx oooo oooo oooo
ooxo oooo oooo oooo
oooo oooo oooo oooo
oooo oooo oooo oooo

The next day obviously the element {4,2,1} will be infected. But there is also the element {3,1,2} (step 2)

ooxx ooxo oooo oooo
ooxx oooo oooo oooo
oooo oooo oooo oooo
oooo oooo oooo oooo

The next day the elements {2,1,1} will be infected, but also the elements {3,2,2}, {4,1,2} and {4,2,2}(step 3)

oxxx ooxx oooo oooo
ooxx ooxx oooo oooo
oooo oooo oooo oooo
oooo oooo oooo oooo

here we can separate each "square" with respect to the z axis : this square
ooxx
ooxx
oooo
oooo
is repeating on every steps ({x,y,2} on the 3rd step is the same as {x,y,1} on the 2nd step, and it will be the same as {x,y,3} on the 4th step). We simplified this problem on a 2-dimentions problem
Let's take {x,y,1} on the last step

oxxx
ooxx
oooo
oooo

It should be easy to proove that every elements will be infected.

Hope my answer isn't too hard to understand ..

EDIT

If a healthy cube is ever adjacent to at least three infested cubes, then that cube will itself become infested the next day.

My answer suppose that 'adjacent' means diagonally to.