# Codefights: alienShip

Immortal aliens from another galaxy invented new spaceship. This spaceship can travel nearly same of speed of light but never exceed it. Some of them travel in that spaceship to find another planet for them.

When they return to home planet, they aged `y`

years.

Suppose they left home planet in `2017`

, your goal is to find out in which year they will return to home planet.

Speed of spaceship is some fraction of speed of light (`v`

).

Use precision of `10`

digits after decimal.

**Example**

For `v = 0.2`

and `y = 10`

, the output should be`alienShip(v, y) = 2027`

.

Aliens travel for `10`

years in spaceship with `20%`

speed of light hence, so they will return to home planet in year `2027`

.

**Input/Output**

**[time limit] 4000ms (js)**

**[input] float v**`v · <speed of light>`

is velocity of the spaceship.*Guaranteed constraints:*`0.01 ≤ v < 1`

.**[input] integer y**Alien aged when they reach home planet.

*Guaranteed constraints:*`1 ≤ y ≤ 100`

.**[output] integer**Year when they return to home planet.

## Solution

From the Special Theory of Relativity we know that traveling with speed \(s\) relates to the time passed \(y\) with

\[y = y_0\sqrt{1-\frac{s^2}{c^2}}\]

It follows that

\[\begin{array}{rrl}& y_0 =& \frac{y}{\sqrt{1-\frac{s^2}{c^2}}}\\& = & \frac{y}{\sqrt{1-\frac{(v c)^2}{c^2}}}\\& = & \frac{y}{\sqrt{1-v^2}}\\\end{array}\]

When we add 2017 as the start date, we find the solution:

alienShip = (v, y) => 2017 + y / (1 - v * v)**.5 | 0