**Tags**

easter egg, foo.bar, Google, Java, jobs, number sequence, Open source, Python

Recently found some time to attempt Google’s foo.bar challenges. They are really interesting and managed to find the time since fasting and stopped my-self from other works than the work that pays.Hance, I accidentally found the link for foobar or perhaps, the first Easter egg I found in Google.

My impression on them is, they are mainly, the things to do with “Numbers” and there are only two language options Java 1.7 and Python 2.7 (for now), to submit your solutions, so you need to know the basic data-structure APIs in the respective languages (I had to learn about Java Collections for one of the puzzles). No third party packages allowed for Java and for Python also there is a restriction. The attempts I made are kind of militarily related fantasies yet interesting. I thought to share a very simple one I solved, for which I took around 30 mins to crack it, find a formula and program it. Google gave me 72 hours to solve it, it’s plenty.

**The Problem – as Google Says**

Keeping track of Commander Lambda’s many bunny prisoners is starting to get tricky. You’ve been tasked with writing a program to match bunny prisoner IDs to cell locations.

The LAMBCHOP doomsday device takes up much of the interior of Commander Lambda’s space station, and as a result, the prison blocks have an unusual layout. They are stacked in a triangular shape, and the bunny prisoners are given numerical IDs starting from the corner, as follows:

| 7

| 4 8

| 2 5 9

| 1 3 6 10

Each cell can be represented as points (x, y), with x being the distance from the vertical wall, and y is the height from the ground.

For example, the bunny prisoner at (1, 1) has ID 1, the bunny prisoner at (3, 2) has ID 9, and the bunny prisoner at (2,3) has ID 8. This pattern of numbering continues indefinitely (Commander Lambda has been taking a LOT of prisoners).

Write a function answer(x, y) which returns the prisoner ID of the bunny at location (x, y). Each value of x and y will be at least 1 and no greater than 100,000. Since the prisoner ID can be very large, return your answer as a string representation of the number.

**Solution**

So, what they have given here is a number sequence. So if you do some work on this

| 7 *12 18 25*

| 4 8 *13 * *19*

| 2 5 9 *14*

| 1 3 6 10

You’d get a 2D array.

Now if you take the first slice of it, I get 1,2,4,7 – This can be formulated for *[n (n-1) / 2] + 1* where n is the place of the number starting with 1.

Let’s take the second slice, 3,5,8,12 – This also can be formulated to *[(n+1)(n) / 2] + 2* again where n is the place of the number starting with 1.

If you see both of the above formulas, there is a pattern, where the counts related to “n” is getting increased by one with the last constant we add. To make our observation on this pattern is correct, let’s analyze the third slice, 6,9,13,18 – and that can be formulated for *[(n+2)(n+1) / 2] + 3* where n starts with 1. So we can come to the conclusion that the observed pattern is the one we are looking for.

With this we can get a general formula for the 2D array as below

*{[(x + [n – 1])] [([x – 1] + [n – 1])]} / 2 + x*

if we normalize this

*[(x + n – 1)(x + n – 2)] / 2 + x*

Here *x* is the horizontal position of an element and *n* is the vertical one, both start from 1. Also we can say *n* as* y* since they have mentioned it as *y* in the problem.

Finally,

**[(x + y – 1)(x + y – 2)] / 2 + x**

So we’ve derived the formula now and convert this to python.

def answer(x, y): z = ((x+y-1)*(x+y-2))/2 + x return str(z);

That’s it! This is the solution for the puzzle. Very simple, yes?

**Conclusion **

foo.bars from Google are really interesting, you’ll get plenty of time to solve them varying from 48 hours to 72 hours. Find them and solve them – Good Luck – You might land a job at Google Inc.

**Link **

Google for it – Good luck again.