{{ message }}

Instantly share code, notes, and snippets.

Created Jul 10, 2017

bomb_baby

The challenge specification

You're so close to destroying the LAMBCHOP doomsday device you can taste it! But in order to do so, you need to deploy special self-replicating bombs designed for you by the brightest scientists on Bunny Planet. There are two types: Mach bombs (M) and Facula bombs (F). The bombs, once released into the LAMBCHOP's inner workings, will automatically deploy to all the strategic points you've identified and destroy them at the same time.

But there's a few catches. First, the bombs self-replicate via one of two distinct processes: Every Mach bomb retrieves a sync unit from a Facula bomb; for every Mach bomb, a Facula bomb is created; Every Facula bomb spontaneously creates a Mach bomb.

For example, if you had 3 Mach bombs and 2 Facula bombs, they could either produce 3 Mach bombs and 5 Facula bombs, or 5 Mach bombs and 2 Facula bombs. The replication process can be changed each cycle.

Second, you need to ensure that you have exactly the right number of Mach and Facula bombs to destroy the LAMBCHOP device. Too few, and the device might survive. Too many, and you might overload the mass capacitors and create a singularity at the heart of the space station - not good!

And finally, you were only able to smuggle one of each type of bomb - one Mach, one Facula - aboard the ship when you arrived, so that's all you have to start with. (Thus it may be impossible to deploy the bombs to destroy the LAMBCHOP, but that's not going to stop you from trying!)

You need to know how many replication cycles (generations) it will take to generate the correct amount of bombs to destroy the LAMBCHOP. Write a function answer(M, F) where M and F are the number of Mach and Facula bombs needed. Return the fewest number of generations (as a string) that need to pass before you'll have the exact number of bombs necessary to destroy the LAMBCHOP, or the string "impossible" if this can't be done! M and F will be string representations of positive integers no larger than 10^50. For example, if M = "2" and F = "1", one generation would need to pass, so the answer would be "1". However, if M = "2" and F = "4", it would not be possible.

Test cases

Inputs: `(string) M = "2" (string) F = "1"`

Output: `(string) "1"`

Inputs: `(string) M = "4" (string) F = "7"`

Output: `(string) "4"`

Inputs: `(string) M = "2" (string) F = "4"`

Output: `(string) "impossible"`

Understanding

I guess the best way to describe the problem is to imagine yourself starting with two numbers `F` and `M`, both of which have the value of `1`. At the start you are currently in `step 0`, to progress a step you must either;

• Add the value of `F` onto `M` to calculate a new `M`, or,
• Add the value of `M` onto `F` to calculate a new `F`.

Because of the rules of the specification we can create some facts:

• Because of the steps, `F` can never equal `M` if `step n` > `0`.
• If `F` > `M` we know that the last step to take place involved adding `M` to `F`.
• If `M` > `F` we know that the last step to take place involved adding `F` to `M`.

A Solution

One sure way to solve this challenge is to reverse engineer the two numbers they provide. From the rules to get from `step n` to `step n-1` we just do `F` = `F` - `M` or `M` = `M` - `F` depending if `F` or `M` is larger than the other.

For example:

``````F = 7   M = 4
F > M ∴ F = 7 - 4
F = 3   M = 4
M > F ∴ M = 4 - 3
F = 3   M = 1
F > M ∴ F = 3 - 1
F = 2   M = 1
F > M ∴ F = 2 - 1
F = 1   M = 1
``````

Well.. that is fairly easy, we have a solution already - we know that if we do not end up with `1` and `1` there is no solution.

But... It is pretty slow. The specification mentions that the inputs can be up to 10^50 in size. Now imagine if we had the inputs `F = 1` and `M = 10^50` we would have to do our calculation 10^50 - 1 times, and who knows how long you will be waiting for that to compute.

This problem occurs when `F | M` is larger by several multiples than its counterpart. So how about we see how many times `F | M` fits into `M | F`, then we increase the counter for how many times it can be divided into the other one.

For example:

``````F = 31   M = 4
F > M ∴ F = 31 - 4 * (31 / 4) //Rounded down of course
F = 3    M = 4
...
``````

This solution avoids the problem with massive differences in the numbers and greatly optimizes the solution.

I think this challenge is pretty straight forward but the real challenge is making it optimized, I think there is enough in this file for you to go and take a good shot at this problem.

Summary

In summary the steps for this solution are:

• Find out if the current `step` you are on is solvable.

• If it is, find out if `F` or `M` is bigger.

• Divide the smaller one into the bigger one, round down, to find out the multiplier to increase the counter and to subtract the larger one to find the answer faster.

• Repeat steps until you have `1` and `1`, frequently check if solvable using a custom function and finally print out the counter of how many steps it took.

Some tips:

• Write a custom method to check if you can actually solve the problem, i.e. check if > 0, check `F` != `M` ect.

• Expect numbers larger than 10^32! So use `BigInteger` for example.

• If you are testing your solution and the tests are not running and instead you get a generic `error` it is most likely because the solution you have provided is unoptimized and is too slow. Test out your solution with big numbers.

• If you want some good numbers to test edge case, provide a number N and N + 1 as long as they are greater than 1. Good luck!