Blog: Google, Rabbits, and Cellular Automata
A few weeks ago, a friend of mine sent me a link, telling me it's a series of programming questions. When I opened it, it brought me to an odd looking website. Feeling curious, I hopped in the rabbit hole wondering where it will lead me. Two weeks later of on and off coding, I emerged successful, satisfied, and with a renewed love for Python.
That website was foobar, a hidden programming / recruiting website by Google.
The challenge consists of five levels of programming questions in increasing difficulty. Unlike other online judges, the interface has no buttons to click. Instead, it's centered around a virtual terminal with several common shell commands (e.g. ls
, cat
, cd
, etc.). While neither vi
nor emacs
are available, edit
opens a small editor which shares the screen with the terminal.
I was going to write this post about all the questions I've solved, but I found the last one particularly interesting and felt it deserved a whole post.
Here is the problem verbatim, including Google's rabbitthemed storyline:
Expanding Nebula
You've escaped Commander Lambda's exploding space station along with numerous escape pods full of bunnies. But  oh no!  one of the escape pods has flown into a nearby nebula, causing you to lose track of it. You start monitoring the nebula, but unfortunately, just a moment too late to find where the pod went. However, you do find that the gas of the steadily expanding nebula follows a simple pattern, meaning that you should be able to determine the previous state of the gas and narrow down where you might find the pod.
From the scans of the nebula, you have found that it is very flat and distributed in distinct patches, so you can model it as a 2D grid. You find that the current existence of gas in a cell of the grid is determined exactly by its 4 nearby cells, specifically, (1) that cell, (2) the cell below it, (3) the cell to the right of it, and (4) the cell below and to the right of it. If, in the current state, exactly 1 of those 4 cells in the 2x2 block has gas, then it will also have gas in the next state. Otherwise, the cell will be empty in the next state.
For example, let's say the previous state of the grid (p
) was:
.O..
..O.
...O
O...
To see how this grid will change to become the current grid (c
) over the next time step, consider the 2x2 blocks of cells around each cell. Of the 2x2 block of [p[0][0], p[0][1], p[1][0], p[1][1]]
, only p[0][1]
has gas in it, which means this 2x2 block would become cell c[0][0]
with gas in the next time step:
.O > O
..
Likewise, in the next 2x2 block to the right consisting of [p[0][1], p[0][2], p[1][1], p[1][2]]
, two of the containing cells have gas, so in the next state of the grid, c[0][1]
will NOT have gas:
O. > .
.O
Following this pattern to its conclusion, from the previous state p, the current state of the grid c will be:
O.O
.O.
O.O
Note that the resulting output will have 1 fewer row and column, since the bottom and rightmost cells do not have a cell below and to the right of them, respectively.
Write a function answer(g)
where g
is an array of array of bools saying whether there is gas in each cell (the current scan of the nebula), and return an int with the number of possible previous states that could have resulted in that grid after 1 time step. For instance, if the function were given the current state c
above, it would deduce that the possible previous states were p (given above) as well as its horizontal and vertical reflections, and would return 4. The width of the grid will be between 3 and 50 inclusive, and the height of the grid will be between 3 and 9 inclusive. The answer will always be less than one billion (10^9).
Test cases
g = [[true, false, true], [false, true, false], [true, false, true]]
=> 4
g = [[true, false, true, false, false, true, true, true], [true, false, true, false, false, false, true, false], [true, true, true, false, false, false, true, false], [true, false, true, false, false, false, true, false], [true, false, true, false, false, true, true, true]]
=> 254
g = [[true, true, false, true, false, true, false, true, true, false], [true, true, false, false, false, false, true, true, true, false], [true, true, false, false, false, false, false, false, false, true], [false, true, false, false, false, false, true, true, false, false]]
=> 11567
tl;dr
You are given a 2D grid, where each cell is either filled (O
) or not filled (.
). Consider an operation that takes in such a grid, and performs the following steps:
 Take the topleftmost 2x2 subgrid. Among those four cells, if only one of them is filled, output a filled cell. Otherwise, output an empty cell.
 Shift the subgrid one cell to the right. This new subgrid will contain the two rightmost cells from the previous subgrid, along with two new cells to the right of them. If exactly one of these cells is filled, output a filled cell to the right of the previously outputted cell. Otherwise, output an empty cell.
 Repeat this process for every possible 2x2 subgrid, iterating horizontally and vertically. This will create a new 2D grid with one fewer width and height.
The problem is to count the number of distinct 2D grids which, when performed the operation upon, will produce the given grid.
The Game of Life
At first, this may sound familiar to Conway's Game of Life. They're both examples of cellular automata: a model of a grid of "cells" in some number of states that evolves over discrete time steps according to a set of rules. Usually, these cells are in a square grid, can be in one of two states, and the next evolution depends on the states of a cell and its neighbouring cells. Conway's example follow these rules, with the states being "alive" or "dead", along with the following evolution rules:
 Any live cell with less than two neighbouring live cells will die (from underpopulation).
 Any live cell with two or three neighbouring live cells will stay alive.
 Any live cell with more than three neighbouring live cells will die (from overpopulation).
 Any dead cell with three neighbouring live cells will become alive (from reproduction).
As simple as these rules are, the Game of Life produces extremely complex patterns and a wide variety of behaviors. In fact, it is proven to be Turing complete, and people have made Turing machines within it. And once you achieve Turingcompleteness, you can do pretty much anything.
Reversibility
Another characteristic of the Game of Life is that it is not reversible. A reversible cellular automaton has a unique predecessor state for every possible state. This can be easily shown as there are many states that evolve into a blank grid (e.g. a single alive cell).
Although the Game of Life is similar to Google's problem, there's one twist to the evolution rules: with each time step, the size of the grid shrinks by one cell. This makes it trivial to show its nonreversibility as all states will eventually become nothing.
While it is possible to determine whether a 1D automaton is reversible, the same problem is undecidable for higher dimensions.
For nonreversible cellular automata, there must exist some state that has no predecessor. These states are called Garden of Eden patterns as they are impossible to reach by going forwards in time. Finding these states is also a similar conundrum; for 1D cellular automata, there exists an algorithm that takes time polynomial of the number of evolution rules (called a rule table), but for higher dimensions, it remains undecidable. Despite this, many researchers have found Garden of Eden patterns for the Game of Life, with continually smaller sizes.
First Solution
Spoiler Alert: Please try and solve this problem yourself before reading on!
A bruteforce solution (i.e. simply generating all m+1 x n+1 grids and checking their next step) will take O(mn * 2^(mn))
time and definitely take too long, so I started with a better approach.
Rather than testing all possible grids, it would be faster to generate predecessors for each cell and merge the resulting 2x2 grids. This only goes through actual predecessors and skips those that aren't. One other choice I made was hinted at in the problem statement, and was to split the grid into columns. We'll see why later on.
The grid is first split into columns, basically turning it into multiple 1D arrays. Then for each m x 1 column, find all m+1 x 2 grids that evolve into the column. Let's call these 2wide columns "column predecessors". This can be done by using DFS along the column, building up all possible column predecessors. Two cells must be added to the m+1 x 2 grid with every cell in the column. There are several cases to consider, but overall this step should take O(m * 2^n)
time.
If any of these columns don't have any predecessors, then the whole grid also can't have any predecessors and is a Garden of Eden. Otherwise, we continue.
We now have a list of column predecessors for each column in the grid. To complete an entire grid, we need to merge these column predecessors. Since a column predecessor contains two columns, the second column will overlap with a predecessor of the next column. A valid predecessor for the entire grid will consists of multiple column predecessors that each share one column (the overlap) with an adjacent one. If there are two columns without any overlapping column predecessors, then there's is no way to construct a whole grid, and is again a Garden of Eden. A naive bruteforce will take O(mnp^2)
, where p
is the average number of column predecessors per column. Note that p
can be very big, up to 2^n
as shown above, so this step takes considerably long time.
The previously mentioned hint is that the constraints of the grid are very skewed: a grid can be as wide as 50 cells, but only 9 cells tall. Even with an exponential solution, it may be fast enough if it's only exponential of the height of the grid (2^10 = 1024, while 2^50 = 1024^5).
Nevertheless, it was far too slow and solved about 3/10 of the test cases. I would need some improvements to this approach.
Improvements
Currently, checking if two column predecessors share an overlap is essentially an array comparison, and requires O(m)
time. To improve this, each column in a column predecessor can be converted into a binary number with m
bits, since each cell can be in one of two states. Comparing two integers is faster than manually comparing two bit arrays, so we save a bit of time here.
Next, the column predecessors can be found and joined in succession. Each column predecessor is only compared with adjacent ones, so there's no need to find them for all columns, then joined after. This greatly reduces the space needed from all columns (O(np)
) to two columns (O(p)
).
Finally, we can represent the column predecessors as a simple graph. Each column predecessor can be visualized as a node with two values, the binary values for each column. Grouping and ordering the processors into their original column, the number of predecessors of the grid is equal to the number of paths from the first column to the last column.
Even though many column predecessors can share a column, their previous columns are not important in finding the next one; only the number of column predecessors that share that column is needed. Since each predecessor's nodes are distinct and can't be backtracked, a simple DP solution can solve for the answer in O(np + nm^2) = O(n * 2^n)
.
This step also only requires two column predecessors to calculate each step. Hence, this step can be done iteratively as well to save memory.
In the end, this method runs in O(n * 2^n)
time and O(2^n)
space, and is fast enough to solve Google's problem.
((`\
____ \\ \.
_/ `` 0 }
/ \ '. __/
( /_ \ \_\_
{_\______)~~\__\_)
Example
Let's run through the first test case as an example to visualize how the solution works:
O.O
.O.
O.O
We first need to generate all the column predecessors of O / . / O
. To do so, let's focus on the first cell: O
. Here are the four predecessors of O
:
.. .. O. .O
.O O. .. ..
Notice how the two lefthand cells are either . / .
, . / O
, or O / .
. They can't be O / O
since then there would be more than one filled cell in the 2x2 grid. Instead, if the original cell was .
, O / O
would be a valid start.
Next, we consider the second cell in the column .
. These predecessors can be generated by adding onto the previous four that we found for O
:
.. .. O. .O
.O O. .. ..
\/ \/ \/ \/
.. .. O. .O
.O O. .. ..
.O .O .. ..
.. .. O. .O
.O O. .. ..
O. O. OO OO
.. ..
.O O.
OO OO
The left three grids are created from the first 2x2 predecessor, and the other grids are created similarly. Finally, with the last cell, we'll have all of the column predecessors of O / . / O
.
.. .. .. ..
.O O. O. .O
.O .O O. O.
.. .. .. ..
O. O. .O .O
.. .. .. ..
.. .. .. ..
O. .O O. .O
We've eliminated some the choices since their third row is O O
. This can't be possible as the last cell in the original column is O
. Now that we have the column predecessors of O / . / O
, we can convert them into binary, and continue to the next column.
.. .. .. ..
.O O. O. .O
.O .O O. O.
.. .. .. ..
06 42 60 24
O. O. .O .O
.. .. .. ..
.. .. .. ..
O. .O O. .O
90 81 18 09
= [0,6], [4,2], [6,0], [2,4], [9,0], [8,1], [1,8], [0,9]
A small thing to note is for every column predecessor, there is a reversed one (e.g. [0,6] => [6,0]). This is because swapping the columns of a column predecessor doesn't change the number of O
's in any 2x2 block.
To save space, here are the column predecessors of the second column . / O / .
:
[13,9], [12,8], [9,13], [8,12], [11,9], [11,8], [10,9], [9,11], [9,10], [8,11], [13,1], [12,0], [9,5], [8,4], [5,9], [4,8], [1,13], [0,12], [3,1], [3,0], [2,1], [1,3], [1,2], [0,3]
Since the first column is the same as the third column, we're done finding the column predecessors.
If we combine the column predecessors of the first two columns, we'll get 16 different predecessors (finding them is left as an exercise for the reader).
Repeating the step with the column predecessors of the last column, we get the following four predecessors:
[1,8,4,2], [2,4,8,1], [4,2,1,8], [8,1,2,4]
. O . . . . O . . . . O O . . .
. . O . . O . . O . . . . . . O
. . . O O . . . . O . . . . O .
O . . . . . . O . . O . . O . .
Improving the Example
Since we don't actually need the full predecessors (only how many there are), we can improve the speed of the algorithm (as mentioned in the previous section).
Let's represent the column predecessors by creating a map, where the key is the right column, and the value is the number of predecessors with that right column.
0: [9, 6], 1: [8], 2: [4], 4: [2], 6: [0], 8: [1], 9: [0]
=> {0: 2, 1: 1, 2: 1, 4: 1, 6: 1, 8: 1, 9: 1}
Here, there are two column predecessors with a right column of 0, one with 8, etc. The sum of these values is 2 + 1 + 1 + 1 + 1 + 1 + 1 = 8, representing the number of predecessors for the first column. This matches what we found before, except this time we ignored what the actual predecessors are.
Similarly, here are the predecessors for the second column:
0: [12, 3], 1: [13, 3, 2], 2: [1], 3: [1, 0], 4: [8], 5: [9], 8: [12, 11, 4], 9: [13, 11, 10, 5], 10: [9], 11: [9, 8], 12: [8, 0], 13: [9, 1]
There are two column predecessors with the right column 0: [12,0] and [3,0]. Since there aren't any column predecessors for the first column that have a right column of 12 or 3, neither [12,0] nor [3,0] can be part of a grid predecessor. For 1, there is one that ends in 2 but none for 3 or 13, so there's only one predecessor (namely [4,2,1]). Continuing for the rest of the column predecessors yields this:
{0: 0, 1: 1, 2: 1, 3: 3, 4: 1, 5: 1, 8: 1, 9: 0, 10: 1, 11: 2, 12: 3, 13: 2}
Finally, we add the column predecessors of the last column.
It is easy to see how this method breaks the problem down into subproblems: it counts the number of predecessors of a section of the grid (e.g. the first three columns), then adds in the next column. This iterative method allows us to only store the column predecessors of two columns at any time, which saves space.
In short:

Predecessors:
0: [9, 6], 1: [8], 2: [4], 4: [2], 6: [0], 8: [1], 9: [0]

Ending with right column:
{0: 2, 1: 1, 2: 1, 4: 1, 6: 1, 8: 1, 9: 1}

8 predecessors for
O / . / O

Predecessors:
0: [12, 3], 1: [13, 3, 2], 2: [1], 3: [1, 0], 4: [8], 5: [9], 8: [12, 11, 4], 9: [13, 11, 10, 5], 10: [9], 11: [9, 8], 12: [8, 0], 13: [9, 1]

Ending with right column:
{0: 0, 1: 1, 2: 1, 3: 3, 4: 1, 5: 1, 8: 1, 9: 0, 10: 1, 11: 2, 12: 3, 13: 2}

16 predecessors for
O. / .O / O.

Predecessors:
0: [9, 6], 1: [8], 2: [4], 4: [2], 6: [0], 8: [1], 9: [0]

Ending with right column:
{0: 0, 1: 1, 2: 1, 4: 1, 6: 0, 8: 1, 9: 0}

4 predecessors for
O.O / .O. / O.O
Hence, there is one predecessor that ends in 1, one in 2, one in 4, and one in 8. Summing these values, the answer is four possible predecessors. This matches the previous calculation!
Trivia
After some research, I came across a 1974 paper by Jean HardouinDuparc titled Paradis terrestre dans l'automate cellulaire de Conway (Garden of Eden in Conway's cellular automata). In it, he describes a similar method to find Garden of Edens in Conway's Game of Life. Since Garden of Edens have no predecessors, this problem and his work go hand in hand with each other. Although his work is more complex than my solution, he had the same idea of splitting a grid into rows to help find predecessors of nearby rows. Using finite automaton, he took these possible predecessors and found their complementary set, resulting in patterns that don't have any predecessors. Eventually, this will yield a Garden of Eden split into rows.
Like this solution, his method only takes time exponential to the grid's width or height rather than its area. With his research, HarduoinDuparc attempted to search for the narrowest possible Garden of Eden. In his same paper, he proved that no Garden of Edens of height 1 exist. For height 2, he complains how his automaton cannot be developed due to the memory constraints of his computer (a whopping 131,072 bytes!). Adapting his method to be heuristic, he continued his search and failed to find any up to height 5. He did find two very narrow ones with bounding boxes of 122 × 6 and 117 × 6.
Final Thoughts
I've had a lot of fun going through this problem, and it introduced me to a new branch of math. The most I've delved before in cellular automata was creating small Game of Life simulations for school assignments, and my research broadened my view on Conway's creation and what lies underneath it.
Although this problem is over, I'm still curious about several questions:
 As long as the given grid is narrow, its height doesn't significantly affect the running time. Is there an algorithm that runs faster for square grids?
 Splitting the grid into rows can be thought of several 1D cellular automata. Since the bottleneck of this algorithm is finding predecessors of the rows, can techniques used for 1D automata be applied to this step?
 Instead of splitting the grid into columns or rows, will other splitting methods improve the algorithm (e.g. halfcolumns or 2x2 blocks)?
 Google's problem asked for the number of predecessors of a grid. How much harder is it to list those predecessors?