Cops - square times square root

Question

Note: This is the cops' thread, where one should post the scrambled code. Here is the robbers' thread where the cracked source should be posted and linked to the cop's answer.

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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, 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 to your answer's body with your exact code
• The same applies to safe answers (mention that it's safe and add the spoiler)
• 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)

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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 1

05AB1E, 35 bytes

ö£=s‰sr2xöR+RM0`.T"YVBYCDž„¨¨"6£H-L

This one took me a while to make.

There's some double quotes in there, so you can surround as much of the code as you want in them.

No rounding, decimal point precision.

Answer 2

Fireball, 8 bytes (Cracked by Roman Gräf)

♥1Z*^²/♥

Does not round, uses floating point precision.

Code surrounded with hearts ;)

Should pretty easy to crack, once you have a look into Fireball.

Answer 3

C#, 112 bytes (cracked by Emigna)

     ((((())))),.....25;;;=CCLMMPPPRSWaaaaaaabbbccdddeeeeeeeeghiiiiiillllmnnnnnooooooorrrsssssssstttttuuvvwy{{}}

no rounding done

using System;
class P
{
    static void Main()
    {
        var b=double.Parse(Console.ReadLine());
        Console.Write(Math.Pow(b,2.5));
    }
}

Answer 4

NO!, 41 bytes (Cracked by Emigna)

NOOO! NO! can finally compete! Although no-one will be bothered to crack this anyway :(

Anyway sigh here's the code:

NNNOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO??
nnno!

Full list of commands here. Coded in Python so uses Python rounding.

NOOOOOOOOO!!!!!!!! Someone cracked it. (Curse you Emigna!) (just kidding!)

Answer 5

C, 115 bytes, (Cracked by @tehtmi)

""%&&((((((()))))))******++++,..//01122233388;;;;;;<====>>   bbbdddeeeefffggillllllllllnnnnnnnnoooooooooooprrtuuu{}

Notice that it does include 3 spaces.

Always outputs 3 exact decimals, rounded to the closest thousandth.

Should be rather fun to crack as it doesn't involve any built-in to compute powers or square roots, as opposed to most (cracked) solutions so far.

Example run (which also gives you a few small hints):

int main() {
  int n[] = {4, 6, 9, 25};
  for (int i = 0; i < 4; i++) {
    printf("%i => ", n[i]);
    o(n[i]);
    printf("\n");
  }
}

outputs:

4 => 32.000
6 => 88.182
9 => 243.000
25 => 3125.000

Tested with both gcc and clang, with no compilation flag needed.

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It's been cracked; the original solution was

o(double n){long o=2.303e18+(*(long*)&n>>1);double l=*(double*)&o;for(o=2;o++<8;)l=(l+n/l)/2;printf("%.3f",l*n*n);}

which uses a magic constant in the same spirit as the infamous fast inverse square root algorithm to do just a handful of Newton iterations (as opposed to several thousands in the cracked solution)

Answer 6

05AB1E, 23 bytes (Cracked by Emigna)

A totally different approach than my previous answers :)

$++/02;@DDDGHP\rrszz}¹ž

No rounding.

Example runs

In   -> Out
0    -> 0
4    -> 32.0
6    -> 88.18163074019441
25   -> 3125.0
7131 -> 4294138928.8967724

Answer 7

Excel VBA, 59 bytes (Safe)

?(())*..//11124AAAAFPSW[[]]^acceehiiiiklnnnnoooppqrrstttttu

No particular rounding.
Uses the Immediate Window.

Answer 8

Octave, 42 bytes (Safe)

((((()))))+,,,-//01111111@[]^mnooorrssst~~

No rounding. Floating point accuracy.

Intended solution

@(s)norm(roots([-1,~1,s^(10/(1+1))]),1/~1)

Try it online!

Answer 9

Lua 5.3, 110 bytes (safe)

((((()))))**,,,,,-....=====>~             aaaaaaaaacddddeeeffhhhiiiiiiilllmmnnnnnnooopppppprrrrrrrsssssssstttt

Calculations are standard using Lua numbers (ie double in common implementations). Tested using PUC-Rio interpreter. (I hope I didn't do anything bad, but I think it is reasonable...)

This is a full program with input from the command line.

Original solution:

for a,r in pairs(math)do l,s=pcall(r,-math.pi)if s~=s and a>=(i or a)then i,p=a,r end end s=...print(s*s*p(s))

Explanation:

math.sqrt is retrieved from the math library by looping through the math table and looking for functions that return NaN (and don't throw an error) when applied to negative pi. There are only a few functions like this, and sqrt is alphabetically last.

Answer 10

Javascript, 15 17 bytes, Cracked by @Emigna

(x=>x****-125*10)

Solution:

x=>x**(25*10**-1)

Javascript, 18 bytes, Again, Cracked.

x=>x**-(!3.5++)[];

Solution:

x=>x**(3-!+[]+.5);

Answer 11

JavaScript + paper.js, 210 bytes (safe)

This one's fun. No comments for you, and no strings either.

paper.js is a vector graphics framework. This is a full program that takes input via a prompt and outputs graphically, because it's possible. There is no rounding and floating-point accuracy (I think).
Reference
Try code here

Math.tan(prompt(a,0));
leg Rave, ew(0,0),ectoPlasm = Real new qrt(;
marvin atTax M(en) in vr;tent * hat.o = Mat
nottet near(0a,a) w cgee;
brown.cern(lantern) ;( w 0,0.art;vict.o brace
r=v=,h.c.)an=.flClr= Cor0)

Edit: Just noticed (my fault for not reading properly)

The shortest safe answer by the end of April will be considered the winner.

Well, oops.

Original solution:

var a = prompt(0,0);
var r = new Rectangle(0,0,a,a).area;
var t = new Rectangle(0,0,Math.cbrt(a),Math.sqrt(Math.cbrt(a))).area;
var n = new PointText(view.center);
n.content = r * t;
n.fillColor = new Color(0);

Try code here

Answer 12

HODOR, 172 bytes, Cracked by Emigna

Walder
Hodor Hodor Hodor Hodor Hodor Hodor
Hodor Hodor Hodor Hodor Hodor
Hodor Hodor Hodor Hodor
Hodor Hodor Hodor
Hodor Hodor
Hodor
hodor hodor
HODOR HODOR!!!!!!!,,,,,,..?

let me know if I missed anything

Answer 13

Python 3.6 - 15 bytes (Cracked by Kritixi Lithos)

axx 2*d:b.*5mal

Standard rounding precision of Python

Answer 14

C++, 100 bytes (Cracked by @fergusq)

#.leiha2dm"
tuchn"crmueie<linaatt#dso
fameng pcae sli;t>dt nnsiuno;s mnpti;ouic(<){n(a;>w<n>n)co.,5}

Standard C++ rounding precision

Answer 15

Python 2, 44 Bytes (Cracked by @fəˈnɛtɪk)

np5w***i0(0(0r0n0t+iarn00p0)_(0tu00i50.0)) t

Has accuracy up to 10 decimal digits (I think). No rounding involved and there is a space involved.

Answer 16

Scala, 19 bytes (Cracked by @math_junkie)

)>,/2(=addd.hmoptw5

This is a function of type Int=>Double.

Answer 17

05AB1E, 4 bytes (Cracked by @Mr. Xcoder)

Must be really easy to crack, but I'll still post it here

n*t¹

Normal rounding

Answer 18

C, 60 bytes (Cracked by @Dave)

1(root, flat*fat bern**lotr);pw(f{bus(--q.b4/q0*1-2)})(b>1)b

Standard C float accuracy.

Answer 19

05AB1E, 23 bytes (Cracked by Emigna)

Another try

A9n*¥="'q?:@->%t#[{¹!.

Normal rounding

Answer 20

HODOR, 198 bytes (Cracked by wwj)

NB: Had to correct the code!

Uses the brand-new hodor command (Updated Saturday)! Yes, I know I'm not supposed to say but it's pretty obvious anyway and (as far as I know) I'm the only person who can code in Hodor so worth some help :)

This is the code: and the full list of commands is here

H!HDHrlrHrhoooo!h, oo!do ordoH!oHHd
dodHrodorooooHdddoooOOo ddro roorrro.r,H
ooO ooHr,
HdHo oo H!O?Hd, oo,, Hdodhd oor, dd!HHrDo aod
rdroHdH,HrRr
 o o RHrdHoo,oro Hr
oo
oHoer o r oddodo,rr
rdd!WrH d

Answer 21

Stacked, 64 bytes

:u#{`E2t#tv',lnp~.iuP#2qr+t,'3m3em`,#9''.::mluls on\ts r:a\:+.S0

Good luck.

Result for 6: 88.181630740194411535.

Answer 22

Javascript (ES6), 39 bytes (Cracked)

(((())))*....57===>Mabcefghlllloprttxxx

The rounding is normal floating point precision for Javascript.

Answer 23

bash, 158 bytes

This program requires a 64 bit processor to function. Pass input number as argument; result on stdout. I believe rounding is floor function. This program actually does decimal output. If you got integer only keep trying.

^=======|       -!
///////......
''(((((((((((())))))))))))
[]$$$$$$$$$****
\\\\\\
#++
000000000000
111111222
bcdddeeeehhhiill
nn
NNNNNNoooOOOOOOO
sssSSSS
tw

Answer 24

Python 2.7, 174 bytes

_cc*gnta(dkree_rtrto._i9kuc_ l5_r_cctbimWka(Wihlkeu_5_eur_rri_keeluc4clrp ..u_)(HohhH)ap5o(trbnord_ir9/*in)_clrikroap.ie.tg_obkfkseeeel*in2Hkdu*ffmo4lWlrncde_22);doodt a49 hu

Answer 25

05AB1E, 15 bytes, (Cracked by Emigna)

+/01;DDDFMPs}¹ž

Normal rounding

Example runs

In   -> Out
0    -> 0
4    -> 32.0
6    -> 88.18163074019441
25   -> 3125.0
7131 -> 4294138928.896773

Answer 26

C#, 135 bytes (cracked by SLuck49)

      (((((()))))),,-..../11;;;CCEILMMMPPPRRSSWaaaaaaacccdddeeeeeeeeggghiiiiiiiillllmmnnnnnnnnoooooooorrsssssssssssttttttttuuuvwyy{{}}

no rounding done
(First try)

Answer 27

05AB1E, 22 bytes, Cracked by user4867444

+/;DDDFLPT_nz}©®¹¹¹Ïè›

Example runs

In -> Out
0  -> 0
4  -> 32.0
6  -> 88.18163074019441

Original solution

LnD¹›_Ï®©èTFD¹/z+;}¹DP

Answer 28

JavaScript (ES7), 24 bytes (Safe)

()+01488M``aabeiijlooqtv

Standard 64 bit floating point rounding.

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Solution

eval(atob`8j0+8ioqMi41`)

Tests

console.log(eval(atob`8j0+8ioqMi41`)(0));    // 0
console.log(eval(atob`8j0+8ioqMi41`)(1));    // 1
console.log(eval(atob`8j0+8ioqMi41`)(2));    // 5.656854249492381
console.log(eval(atob`8j0+8ioqMi41`)(4));    // 32
console.log(eval(atob`8j0+8ioqMi41`)(10));   // 316.22776601683796

Answer 29

Javascript (ES7), 53 bytes (safe)

(((())))*++++,,,/0233346=====>>[[]]aaaabbbccelootttuv

The rounding is normal floating point precision for Javascript.

Tested this in the latest Chrome and FF.

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Solution

u=>(o=(t,c,a=o+c,b=a[t])=>eval(a[42]+b+b+3*30/t))(36)

This works because a=o+c is '(t,c,a=o+c,b=a[t])=>eval(a[42]+b+b+3*30/t)undefined'.
So a[42]='u', a[36]='*', 3*30/36=2.5, making eval('u**2.5').

Tests

var f=u=>(o=(t,c,a=o+c,b=a[t])=>eval(a[42]+b+b+3*30/t))(36);
console.log(f(0));    // 0
console.log(f(1));    // 1
console.log(f(2));    // 5.656854249492381
console.log(f(4));    // 32
console.log(f(10));   // 316.22776601683796

Answer 30

05AB1E, 27 bytes - safe

Yet another totally different approach from my previous answers.

(/1DFFOOPT`ins}}©®¹¹Ðèëž‚‚‹

No rounding.

Example runs

In   -> Out
0    -> 0
4    -> 32.0
6    -> 88.18163074019441
25   -> 3125.0
7131 -> 4294138928.896773

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Solution

Ð(‚žFFDO©n¹‹i`T/ë1è®s}‚}O¹P

This is simply computing the square root with a trivial (and very dumb) trial-and-error strategy...

Detailed explanation

Ð(‚žFFDO©n¹‹i`T/ë1è®s}‚}O¹P
Ð(‚                         # triplicate the input, reverse the sign of the last copy, and wrap the last 2 items in a list, which going forward is going to be the list of our current sqrt approximation and of the next decrement in our dumb search (so it starts at [x, -x])
   žFF                      # 16384(!) times, do
      D                     # duplicate the current [value, decrement] list
       O©                   # compute the sum value + decrement, and also save it in the register
         n                  # compute the square of that potential new approximation of the square root
          ¹‹                # and compare it with the input
            i               # if our new approximation is smaller than the actual square root, then
             `              # push the flattened [value, decrement] list to the stack as value, decrement
              T/            # and divide our decrement by 10
                ë           # else (our new approximation is still bigger than the actual square root, then)
                 1è         # extract the current decrement from the list
                   ®        # retrieve the new approximation from the register
                    s       # and switch them to again have them in the value, decrement order
                     }      # end of the if
                      ‚     # no matter which branch of the if we went into, the last 2 items of the stack are now the new approximation and the new decrement, in order: wrap them in a list for the next loop iteration
                       }    # end of the loop
                        O   # replaces the output of the loop (the final [value, decrement] list) by its sum, which given the very small value of the decrement at that point is pretty much the same as our final approximation of sqrt(x)
                         ¹  # push another copy of x; the stack is now [x, sqrt(x), x] (where the first x comes from the very first copy of the initial Ð)
                          P # computes the product of the whole stack

Try it online!