# Square CTF 2018 - C10: fixed point
## Description
```If you are a Charvis then this is a simple application of Xkrith’s First and Sixth Theorems.  If you are not a Charvis then you should familiarize yourself with Xkrith’s works.  A primer can be found in Room 100034B.```Note: For the humans out there, this system is built on an oscillator. It’ll disable itself when you find the oscillator’s fixed point.
## Preparation
The challenge was to find the ```x```, and f(x) should be equal x ( ```f(x) == x```).
We have fixed_point.html, where f(x) was declared:
```function f(x) {  if ((x.substr(0, 2) == '?') && (x.slice(-2) == '?')) {    return x.slice(2, -2);  }  if (x.substr(0, 2) == '?') {    return '?' + f(x.slice(2));  }  if (x.substr(0, 2) == '?') {    return f(x.slice(2)).match(/..|/g).reverse().join("");  }  if (x.substr(0, 2) == '?') {    return f(x.slice(2)).repeat(5);  }  if (x.substr(0, 2) == '?') {    var t = f(x.slice(2));    return t.substr(0, t.length/2);  }
  return "";}```
So.. At first sight, it is not difficult to determine that ```x``` - it is a string, that contains emoji symbols of rockets, unos, moons, go on. And ```f(x)``` - it is a function that interpreting input one by one symbol.
Let's to dive in it. 
1) ```?x?```
Rockets.
Rockets are key players of this game. Because when ```f(x)``` gets into ```if(rockets)```, function just print ```x``` ( input in rockets, WITHOUT rockets).When another emojis pass substring after one of their to another emoji, rockets just print it, and the current recursion spins in the opposite direction!
I call rockets as BRACKETS. Because they do nothing, they don't call ```f(x)``` again. And they disappears.. It is the hardest part of this game because rockets disappears. 
For example, when you print , you will see:
```?1122334455?``` -> ```1122334455```
P.S. I use plain numbers (by 2 symbols) to test this challenge, it is easy to see what happens
2) Another emojis - are functions. And I use next formula, I use any count of function, that are following each other, and one BRACKETS:
FuncFuncFuncFuncFunc......?some_emojis_that_just_will_be_print?
Where Func - one of ?, ?, ?, ?
3) ```?```
This emoji just reverse string, after this one.
```??1122334455?``` -> ```5544332211```
4) ```?```
White moon gets string after this and multiples it by 5.
```??11?``` -> ```1111111111```
5) ```?```
Black moon gets string after this and divides it by 2.
```??11223344?``` -> ```1122```
6) ```?```
Uno - is solution for rockets. Because uno creates one rocket and continue to send substring to ```f(x)``` again
## Solution
This challenge is solved from the end. Because a output of any function - it is a input for another one.
So, let's start with BRACKETS(rockets)!!!
```?11223344?``` -> ```11223344```
Poorly... :) Okay, let's add uno to this.
```??11223344?``` -> ```?11223344```
Good, but output contains the rocket at first position, and it is unpossible with my rules. O, I can reverse it! 
```???11223344?``` -> ```44332211?```
Good output, I have rocket at the last position, but my string into rockets are reversed. Okay, solve it in future.
Also I have only one rocket it output, but moons can help me to multiple rocket. At first, multiple it with 1 white moon. And then divide with 2 moons.
```??????11223344?``` - > ```44332211?44```
Let's increment count of white and black moons, until you will see the following output .
```????????????11223344?``` - > ```44332211?44332211� ```
Fine! What I look here? I look that my input string in BRACKETS is reflected two times ( but it is reversed now), and I see one moon, and another one.. almost.
So, let's clear bad rocket at the end and reverse input at the same time.
```?????????????11223344?``` -> ```11223344?11223344```
Will you see it? It is almost the end. You can paste  ```????????????```  instead of ```11223344``` and you see :). But it is only a one rocket, where is another one.. last one..
Okay, what I know.. I know that with emoji-functions, I get ```11223344``` * 2 and one rocket. I have 4 symbols ( 11,22,33,44) * 2 and another one symbol (rocket). 
Can I do that I will have two rockets, but only (4 * 2 - 1 ) symbols? Exactly! I did a mistake ( but really not), in my previous steps. Just replace ? and ? symbols in the middle.
```?????????????11223344?``` -> ```223344?11223344?```
And it is the end :))) In input I have unnecessary symbol ```11```.
Replace ```223344``` with ????????????
```?????????????11?????????????``` -> Good!!!
## Finally
Finally , this solution is not unique :) Maybe, you have another one? :)
