[Question]
  [
*Note: This is the **cops'** thread, where one should post the scrambled code. Here is the **[robbers' thread](https://codegolf.stackexchange.com/questions/115034/robbers-square-times-square-root)*** where the cracked source should be posted and linked to the cop's answer.
---
**Task:** Write the shortest *safe* program which multiplies the square root of an integer ***n*** by the square of ***n***
This is [cops-and-robbers](/questions/tagged/cops-and-robbers "show questions tagged 'cops-and-robbers'"), so the rules are:
* In your answer, post a scrambled version of your source code (the characters should be written in any order). The scrambled version **should not** work!
* You can take input in any standard way, the same goes for output. Hardcoding is forbidden
* After the code is cracked by the robbers (if this happens), you must mention that your code has been cracked in your title and add a ***[spoiler](https://meta.stackexchange.com/questions/1191/add-markdown-support-for-hidden-until-you-click-text-aka-spoilers)*** to your answer's body with your exact code
* The same applies to safe answers (mention that it's safe and add the ***[spoiler](https://meta.stackexchange.com/questions/1191/add-markdown-support-for-hidden-until-you-click-text-aka-spoilers)***)
* The code is considered *safe* if nobody has cracked it in 5 days after posting it and you can optionally specify that in the title
* You must specify your programming language
* You should specify your byte count
* You must state the rounding mechanism in your answer (see below)
---
You can assume that the result is lower than 232 and ***n*** is always positive. If the result is an integer, you must return the exact value with or without a decimal point; otherwise the minimum decimal precision will be 3 decimal places with any rounding mechanism of your choice, but can include more. **You must state the rounding mechanism in your answer.** You are not allowed to return as fractions (numerator, denominator pairs - sorry, Bash!)
### Examples:
```
In -> Out
4 -> 32.0 (or 32)
6 -> 88.18163074019441 (or 88.182 following the rules above)
9 -> 243.0
25 -> 3125.0
```
The shortest safe answer by the end of April will be considered the winner.
  
[Answer]
# [Python 3](https://docs.python.org/3/), 44 bytes ([cracked](https://codegolf.stackexchange.com/a/116254/45151))
```
'**:(((paraboloid / rabid,mad,immoral))):**'
```
No rounding. Floating point accuracy.
[Answer]
# [MATL](https://github.com/lmendo/MATL), 12 bytes ([cracked by @tehtmi](https://codegolf.stackexchange.com/a/115451/36398))
```
'Un&0P'/^:+1
```
No rounding; uses floating point.
Intended solution (different from that found by @tehtmi):
> 
> `:&+n10U'P'/^`
> 
> 
> 
### Explanation
> 
> `:&+ % Create a matrix of size n × n, where n is implicit input`  
> 
> `n % Number of elements. Gives n^2`  
> 
> `10U % 10 squared. Gives 100`  
> 
> `'P' % 'P' (ASCII code 80)`  
> 
> `/ % Divide. Gives 1.25`  
> 
> `^ % Power. Implicit display`  
> 
> 
> 
> 
[Answer]
# [Röda](https://github.com/fergusq/roda/tree/roda-0.12), 28 bytes ([Cracked by @tehtmi](https://codegolf.stackexchange.com/a/115449/49958))
```
 (),.025^^cdfhnnnopprstuu{|}
```
Note the space at the beginning. No rounding, but it uses floating point numbers so precision is limited.
[Answer]
## Perl, 42 bytes (Safe)
There are 41 bytes of code and `-p` flag (no other flags).
```
/"/4~~r..rso4r<_$4va=eg1de|i/h0-&$c={}l+"
```
The result isn't rounded (or rather rounded up to the same point Perl would have round up by doing `$_ = (sqrt $_) * ($_ ** 2)`).
**Solution:**
> 
> `$_=eval".i44<4}{|~"=~s/./chr-10+ord$\&/gre`  
> 
>  (without the `\` before the `&` - markdown spoiler seems to dislike `$` followed by `&`)  
> 
> [Try it online!](https://tio.run/nexus/perl5#U1YsSC3KUdAt@K8Sb5talpijpJdpYmJjUltdU6dkW1esr6efnFGka2ignV@UoqKmn16U@v@/JQA "Perl 5 – TIO Nexus")
> 
> 
> 
**Explanation:**
> 
>  `.i44<4}{|~` is `$_**2*sqrt` but with every character replaced by the character with its ascii code + 10. (ascii code of `$` is `36`, so it becomes `.` whose ascii code is `46`, etc.).  
> 
>  The purpose of `s/./chr-10+ord$\&/gre` is then to undo this transformation: it replaces each character by the character with ascii code 10 lower. (`chr-10+ord$\&` is probably clearer as `chr(ord($\&)-10)` where `chr` returns the character corresponding to an ascii code, and `ord` returns the ascii code corresponding to a character).  
> 
>  finally, `eval` evaluates this string, and thus computes the result, which is stored in `$_`, which is implicitly printed at the end thanks to `-p` flag.
> 
> 
> 
[Answer]
# Octave, 43 bytes (Safe)
```
$'()*+,-/23579:[]aelnouv'*,-23:[]lu',-23]',
```
This is a script that requires input from the command line (it's not a function). It's floating point accuracy (so no rounding).
**Solution:**
> 
> `eval(-[5,-2:3,-3:2]+['nlouu*$$',39,']2/7'])`  
> 
> 
> 
> 
**Explanation:**
> 
> `eval( <string> ) % Evaluated the string inside the brackets and executes it`   
> 
> Everything inside the `eval` call gets evaluated to `input('')^2.5`   
>   
> 
> **How?**  
>   
> 
> `-[5,-2:3,-3:2] % A vector: [-5, 2, 1, 0, -1, -2, -3, 3, 2, 1, 0, -1, -2]`  
> 
> `['nlouu**$$',39,']2/7'] % This is a string: nlouu**$ concatenated with the number`   
> 
> `. % 39 (ASCII ']'), and ']2/7'. Thus, combined: 'nlouu**$$']2/7'`  
>   
> 
> Adding the first vector to this string will convert it to the integer vector:   
> 
> `[105, 110, 112, 117, 116, 40, 39, 39, 41, 94, 50, 46, 53]`  
>   
> 
> `eval` implicitly converts this to a string, and these numbers just so happens to be:
> `input('')^2.5`
> 
> 
> 
[Answer]
# C, 50 bytes ([Cracked by fergusq](https://codegolf.stackexchange.com/a/115176/8927))
```
%(())   ,-12225;>\\aaabbdddeeefffllnoooprrttuuuuw{
```
Uses standard IEEE754 rounding. As noted by fergusq's answer, may require `-lm` depending on your compiler.
[Answer]
# Mathematica, 131 bytes, non-competing?, [cracked](https://codegolf.stackexchange.com/a/115146/66104)
*This has been **cracked by [@lanlock4](https://codegolf.stackexchange.com/a/115146/66104)**! However, I still have internet points to bestow on someone who finds the original solution, where all the characters are actually needed....*
```
f[y_]:=With[{x=@@@@@@#####^^&&&(((()))){{}}111111,,+-/y},Print[#,".",IntegerString[Round@#2,10,3]]&@@QuotientRemainder[1000x,1000]]
```
This is intended as a puzzle. Although you may use the above characters however you want, I certainly intend for the answer to follow the form
```
f[y_]:=With[{x=
    @@@@@@#####^^&&&(((()))){{}}111111,,+-/y
},Print[#,".",IntegerString[Round@#2,10,3]]&@@QuotientRemainder[1000x,1000]]
```
where the first and third lines are just a wrapper to make the rounding and display legal (it writes every output to exactly three decimal places, rounded), and the second line is the scrambled version of the guts of the code. Sample outputs:
```
6 -> 88.182
9 -> 243.000
9999 -> 9997500187.497
```
(Mathematica is non-free software, but there is a [Wolfram sandbox](https://sandbox.open.wolframcloud.com) where it is possible to test modest amounts of code. For example, cutting and pasting the code
```
f[y_]:=With[{x=
    y^2.5
},Print[#,".",IntegerString[Round@#2,10,3]]&@@QuotientRemainder[1000x,1000]]
```
defines a function, which you can subsequently call like `f@6` or `f[9]`, that does the same thing as the unscrambled version of the code above. So does this really have to be non-competing?)
[Answer]
# Swift - 64 bytes (Safe)
```
prot Fdnufi;nooitamunc xetgru(->atl)Ior:n{tFn pg,F(ao.o25t)(w)l}
```
No rounding, and displays a `.0` even if the result is an integer.
[Answer]
# Haskell, 16 bytes ([Cracked by @nimi](https://codegolf.stackexchange.com/a/115104/61405))
```
()*...25=eglopxx
```
No particular rounding
[Answer]
# R, 28 bytes ([Cracked by @Flounderer](https://codegolf.stackexchange.com/a/115132/26905))
```
funny(p1)-tio(^*^)/pc(2)<p2;
```
Standard R floating-point accuracy.
[Answer]
# C#, 172 bytes ([Cracked by SLuck49](https://codegolf.stackexchange.com/questions/115034/robbers-square-times-square-root/115332#115332))
```
       (((((())))))***,,,,......1225;;;;;;<====>CFLMMMMMPPPRSSSSSWaaaaaaabbbbcccddddddeeeeeeeeeeegghiiiiiiiillllllmmnnnnnnnooooooooqqqqrrrssssssssstttttttttuuuuuuuvvwyy{{}}
```
This code is a full program.
There are seven space characters at the start.
The input is read form STDIN and printed to STDOUT. The result is in `double`, no rounding done.
Original Code ungolfed:
```
using System;
using S = System.Console;
class P
{
    static void Main()
    {
        var t = S.ReadLine();
        double q = int.Parse(t);
        Func<double, double, double> M = Math.Pow;
        S.Write(M(q, 2 * .25) * M(q * q, 1));
    }
}
```
[Answer]
## Ohm, 11 bytes
```
M ⁿ¡D¼½;+1I
```
Use with `-c` flag. Uses CP-437 encoding.
[Answer]
## JavaScript (ES7), 20 bytes ([Cracked by @IlmariKaronen](//codegolf.stackexchange.com/a/115142))
```
****..22255666=>____
```
Standard JavaScript precision.
[Answer]
# Python 2, 60 Bytes ([Cracked by @notjagan](https://codegolf.stackexchange.com/a/115140/63641))
```
 3n0)4  5)594p3(p5*5i9t4542)0/*((8(t.84- 90945 u)i*48/95n8r8
```
No rounding involved. Accurate up to 10 decimal digits.
[Answer]
# Python 3.6, 59 bytes
```
ba(,b5,d' (,a/([m:'-)oa)(bl*aadplma dba](r)d )l*d,:)*m:-mml
```
No rounding. Floating point accuracy.
[Answer]
## Haskell, 64 bytes, ([cracked by Laikoni](https://codegolf.stackexchange.com/a/115223/34531))
```
$$$$$$(((((())))))**,...0<<<>>>[]cccccdddeffiiiiilloopppprrsstuu
```
Standard Haskell floating point operations.
My original version is:
> 
>  product.((($succ$cos$0)(flip(\*\*).)[id,recip])).flip(id)
> 
> 
> 
[Answer]
# [Fourier](https://esolangs.org/wiki/Fourier), 124  119 Bytes
```
((()))*******--011111<=>>>HHINNNN^^^eeehhhhkkkkmmmmmmmmmmmmmmmmossuuuuuuuuuuuuuuuuu{{{{{{{{{{}}}}}}}}}}~~~~~~~~~~~~~~~~
```
There are no whitespaces or newline characters.
Square root is rounded to the nearest whole number because Fourier doesn't seem to handle anything other than integers (and since @ATaco got permission, I hope this is ok)
fixed an editing mistake, if you were already cracking this, the previous was functional
Realized that I had misunderstood part of the code, and was using more characters than I needed to
If I missed anything let me know
[Answer]
# [Inform 7](https://en.wikipedia.org/wiki/Inform#Inform_7), 71 bytes ([Cracked by @Ilmari Karonen](https://codegolf.stackexchange.com/questions/115034/robbers-square-times-square-root/115135#115135))
```
""()**-..:[]
                 RT
aaaabeeeffilmmnnnnnooooooqrrrrrssstuuy
```
The code includes 17 spaces and 2 new lines.
This is a full Infrom 7 program defining a function that prints the result with a precision of 5 decimal places.
[Answer]
## R, 19 bytes [(Cracked by @Steadybox)](https://codegolf.stackexchange.com/a/115136/26905)
```
mensana(4*5*c(.1)):
```
Standard rounding
## R, 33 bytes [(Cracked by @plannapus)](https://codegolf.stackexchange.com/a/115230/26905)
```
(rofl(17)^coins(2*e)/pisan(10))--
```
## R, 31 bytes [(Cracked by @plannapus)](https://codegolf.stackexchange.com/a/115993/26905)
`h=f`l`u`n`c`t`i`o`n([],[])^(.9)`
[Answer]
# Octave, 30 bytes (Safe)
```
(((((())))))**++/:@eeeiiijmsu~
```
A bit simpler than my first one. Shouldn't be too hard, but it's hopefully a fun puzzle.
[Answer]
# 05AB1E, 20 bytes - safe
Another totally different approach from my previous answers.
```
****++/133DDFPTs}¹¹Ð
```
No rounding.
**Example runs**
```
In   -> Out
0    -> 0
4    -> 32.0
6    -> 88.18163074019441
25   -> 3125.0
7131 -> 4294138928.896773
```
I have no doubt @Emigna is going to crack it in a jiffy, but eh, one has to try! :-D
---
**Solution**
```
D1TFÐ*D¹3*+s3*¹+/*}P
```
This is using the fact that this sequence:
u\_0 = 1, u\_{n+1} = u\_n \* (u\_n ^ 2 + 3 x) / (3 u\_n ^ 2 + x)
converges to sqrt(x), and cubically fast at that (sorry, didn't find how to format math equations in PCG).
**Detailed explanation**
```
D1TFÐ*D¹3*+s3*¹+/*}P
D1                   # Duplicate the input, then push a 1: stack is now [x, x, 1] (where x is the input)
  TF                 # 10 times (enough because of the cubic convergence) do
    Ð                # triplicate u_n
     *               # compute u_n ^ 2
      D              # and duplicate it
       ¹3*+          # compute u_n ^ 2 + 3 x
           s         # switch that last term with the second copy of u_n ^ 2
            3*¹+     # compute 3 u_n ^ 2 + x
                /    # compute the ratio (u_n ^ 2 + 3 x) / (3 u_n ^ 2 + x)
                 *   # compute u_n * (u_n ^ 2 + 3 x) / (3 u_n ^ 2 + x), i.e. u_{n+1}, the next term of the sequence
                  }  # end of the loop
                   P # compute the product of the whole stack, which at that point contains u_10 (a sufficiently good approximation of sqrt(x)), and the 2 copies of the input from the initial D: we get x ^ 2 * sqrt(x)
```
[Try it online!](https://tio.run/nexus/05ab1e#@@9iGOJ2eIKWy6GdxlraxcZah3Zq62vVBvz/bwYA)
[Answer]
## [OCaml](http://www.ocaml.org/), 13 bytes ([Cracked by @Dada](https://codegolf.stackexchange.com/questions/115034/robbers-square-times-square-root/115078#115078))
```
2*fn-5f>f*u .
```
No rounding (within IEEE 754 scope).
[Answer]
# Javascript, 123 bytes, [Cracked by notjagan](https://codegolf.stackexchange.com/a/115108/64505)
```
 """"""((((((((()))))))))********,--.....//2;;======>>Seeeeeeegggggggggggghhhhhhhhhhhilllllnnnnnnnnnnorrrsstttttttttttttu{}
```
This code is a full function 
There is one space character at the very start of the list of characters 
The rounding of this answer is the floating point precision for Javascript, accuracy is within 10^-6 for every answer. 
Got shorter because the precision didn't need to be maintained quite as high as I thought it did.
I had realized that it would be much easier to solve than I initially had made it but it was already there :P
**Initial code:**
```
g=t=>t-(t*t-n)/("le".length*t);e=h=>{n=h*h*h*h*h,s=2**(n.toString("ng".length).length/"th".length);return g(g(g(g(g(s)))))}
```
Newtons method, applied 5 times from the closest power of 2
[Answer]
# Python 3.6 - 52 bytes ([Cracked by @xnor](https://codegolf.stackexchange.com/a/115161/67047))
```
f=lambda x:x**125*77*8+8/5/((('aafoort.hipie.xml')))
```
Standard Python rounding
[Answer]
## Ruby, 35 bytes (cracked by [xsot](https://codegolf.stackexchange.com/a/115171/18535))
```
'a'0-a<2<e<2<l<3<v<4<4<4<5<5<6>7{9}
```
No rounding. Floating point accuracy.
[Answer]
# [05AB1E](https://github.com/Adriandmen/05AB1E), 47 bytes
```
)*.2555BFHIJJKKKPQRST``cgghilnstwx}«¹¹Áöž‚„…………
```
Does not round, uses floating point accuracy.
[Answer]
## CJam, 8 bytes ([Cracked by E~~n~~mig~~m~~na](https://codegolf.stackexchange.com/a/115203/56341))
```
WYdYl##+
```
No rounding. Uses double precision.
[Answer]
# R, 32 bytes ([Cracked by @plannapus](https://codegolf.stackexchange.com/a/115227/61405))
```
i=na*0.5f*n(2*s*cos(t))*22*s*12u
```
Standard floating-point accuracy.
[Answer]
## Excel, 26 bytes
```
=(())*//11122AAAIINPQRSST^
```
No rounding.
*Note: As Excel is paid software, this works also in free LibreOffice*
[Answer]
# [RProgN 2](https://github.com/TehFlaminTaco/RProgN-2), 6 Bytes ([Cracked by @notjagan](https://codegolf.stackexchange.com/a/115286/63641))
```
š2]^*\
```
No rounding, displays many decimal places.
Does not display any for an integer solution.
]