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Task: Crack the scrambled code for multiplying the square root of an integer n by the square of it!

You must post a comment in the cops' thread with a link to your working source, mentioning clearly that you have Cracked it. In your answer's title, you must include the link to the original answer.

Rules:

  • You may only change the order of the characters in the original source.
  • Safe answers cannot be cracked anymore.
  • The other rules mentioned in the cops' thread
  • Please edit the answer you crack

WINNER: Emigna - 10 submissons (had some trouble counting)

Honorable mentions: Notjagan, Plannapus, TEHTMI

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49 Answers 49

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05AB1E, 22 bytes, P. Knops

n¹t*qA9¥="'?:@->%#[{!.

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Explanation

n # square of input * # times ¹t # square root of input q # end program

The rest of the operations never get executed.
We could have done it without the q as well by having ? after the calculation and escaping the equality sign for example with '=.

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    \$\begingroup\$ Was just doing it for fun :D \$\endgroup\$ Commented Apr 4, 2017 at 18:12
  • \$\begingroup\$ @P.Knops: That's the best reason :) \$\endgroup\$ Commented Apr 4, 2017 at 18:56
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CJam, 8 bytes, Roman Gräf

My First ever CJam answer.
CnR are great for testing new languages :)

ldYYW#+#

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Explanation

ld % double(input) # % ^ Y % (2 + % + Y % 2 # % ^ W % -1)
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HODOR, 171 bytes, wwj

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!!!

Explanation

  • Increment accumulator to 5
  • Divide accumulator by 2
  • Save a copy of accumulator value
  • Set accumulator to 0
  • Read input
  • Raise input to the value of the stored copy (2.5)
  • Output as number
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  • \$\begingroup\$ back to the drawing board \$\endgroup\$ Commented Apr 5, 2017 at 15:07
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JavaScript, 39 bytes, SLuck49

x=>(l=Math).cbrt(l.exp(l.log(x)*7.5))

or similarly

x=>(l=Math).exp(l.log(l.cbrt(x))*7.5)

Test:

alert(( x=>(l=Math).cbrt(l.exp(l.log(x)*7.5)) )(prompt()));

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C, 115 bytes, user4867444

New solution that works for 0.

o(double n){double l=1;;for(double o=0.;o<=8*8*2e3&&(long)o+n*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Old solution that doesn't work for 0:

o(double n){double l=n;;for(double o=1.0;o<=8*8*2e+3&&(long)o*((long)2*3>>1);o++)l=(l+n/l)/2;printf("%.3f",n*n*l);}

Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.

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  • \$\begingroup\$ Close, but not quite, as it returns nan for 0 (not sure we care about that case, asked the OP in a comment) \$\endgroup\$ Commented Apr 6, 2017 at 15:37
  • \$\begingroup\$ As specified by the OP, our programs must return 0 for 0, so I'm afraid your solution doesn't work. \$\endgroup\$ Commented Apr 6, 2017 at 16:59
  • \$\begingroup\$ @user4867444: Sorry, I didn't understand that it must work for 0. I've provided a modified solution that should work for 0. \$\endgroup\$ Commented Apr 6, 2017 at 17:48
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    \$\begingroup\$ 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 nice magic constant to only do a few Newton iterations as opposed to thousands - but great job! \$\endgroup\$ Commented Apr 6, 2017 at 18:33
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05AB1E, 22 bytes, Emigna

蛹DDLÏ_zTnF©¹®/+;}¹DP

Most probably not what was intended, but works ;)

Everything before T is just byte wasting (but which leaves the stack empty), then it does a 100 Newton iterations, which yields a good enough approximation of the square root, then pushes the input twice and returns the product of the whole stack, thus giving the desired output.

Example runs:

$ echo 0 | ./05AB1E.py -e '蛹DDLÏ_zTnF©¹®/+;}¹DP' 0.0 $ echo 4 | ./05AB1E.py -e '蛹DDLÏ_zTnF©¹®/+;}¹DP' 32.0 $ echo 6 | ./05AB1E.py -e '蛹DDLÏ_zTnF©¹®/+;}¹DP' 88.18163074019441 $ echo 25 | ./05AB1E.py -e '蛹DDLÏ_zTnF©¹®/+;}¹DP' 3125.0 $ echo 7131 | ./05AB1E.py -e '蛹DDLÏ_zTnF©¹®/+;}¹DP' 4294138928.896773
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  • \$\begingroup\$ Nice one! Didn't think of that :P \$\endgroup\$ Commented Apr 7, 2017 at 16:30
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05AB1E, 15 bytes, user4867444

DMžDFD¹s/+;}01P

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Explanation

D # duplicate input M # push the largest value on the stack (the input) žDF # 4096 times do: D # current value + # added to ¹s/ # input divided by current value ; # divide sum by 2 } # end loop 01 # push 1 P # product of the stack
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05AB1E, 23 bytes, user4867444

$DDžHGD¹s/+;}0@rrzz2+\P

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Explanation

$ # push 1 and input DD # duplicate the input twice žHG # 65536 times do: D # duplicate current value ¹s/ # divide input by current value + # add result of division to current value ; # divide by 2 } # end loop 0@ # get the bottom value of the stack (the 1) rr # reverse the stack twice (no-op) zz # calculate 1/(1/1), a no-op 2+ # add 2 \ # discard top of stack (the 3) P # product of stack
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JavaScript, SIGSEGV

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

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  • \$\begingroup\$ You got it! Kudos for you ;) \$\endgroup\$ Commented Apr 12, 2017 at 11:32
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Python - SparklePony

import math f=lambda x:x*x*math.sqrt(x)
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  • \$\begingroup\$ That was very fast! \$\endgroup\$ Commented Apr 3, 2017 at 18:00
  • \$\begingroup\$ Thanks @SparklePony. It was very similar to mine, done it right away \$\endgroup\$ Commented Apr 3, 2017 at 18:01
  • \$\begingroup\$ @SparklePony please accept the edit... \$\endgroup\$ Commented Apr 3, 2017 at 18:01
  • \$\begingroup\$ I think I have done it... first time accepting an edit! \$\endgroup\$ Commented Apr 3, 2017 at 19:34
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Scala, corvus_192

d=>math.pow(d,5/2d)

Try it here!

Simple lambda expression

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05AB1E - P. Knops

t¹n*

Was quite easy :)

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C, Steadybox

float q(float b){return(b-b>0.*4*1/1)-sqrtf(b)*pow(b,2)*-1;}

Not sure what the original did; I had to discard a lot of random operators and numbers, but once the available words were clear it wasn't too hard to piece together the general idea.

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  • \$\begingroup\$ Oops; noticed I'd accidentally included an extra space than the original; fixed. \$\endgroup\$ Commented Apr 3, 2017 at 20:15
  • \$\begingroup\$ Here's the original function: float q(float b){return-1*pow(b,(b>-12)/sqrt(4.0f))*b*b*-1;} \$\endgroup\$ Commented Apr 3, 2017 at 20:46
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R, Flounderer

scan()**mean(1:4.5)

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Ruby, G B

->a{eval''<<97<<42<<42<<50<<46<<53}
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  • \$\begingroup\$ Nice, I was expecting that, maybe not so fast. :-) \$\endgroup\$ Commented Apr 4, 2017 at 8:13
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C#, 135 bytes, Jan Ivan

This was a lot of fun, I particularly liked the use of using static.

using System;using static System.Math;class P{static void Main(){Console.Write(Pow(long.Parse(Console.ReadLine()),1/Round(PI-E,1)));}}
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JavaScript, SIGSEGV

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

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  • \$\begingroup\$ Well, not the same, but well, that works too. maybeIshouldn'tuseasterisks \$\endgroup\$ Commented Apr 12, 2017 at 11:56
  • \$\begingroup\$ @SIGSEGV: I'd be interested to see what the intended version was then :) \$\endgroup\$ Commented Apr 12, 2017 at 12:03
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Python 2, 174 bytes, cracks penalosa

print str(input()**(5. * .5))or ()/22*2444599.9.HHHWWW_______________aaaaabbbcccccccccddddddeeeeeeeeeeeefffgghhhhiiiiiiikkkkkkkkkllllllllmmnnnooooooooprrrrrrrrrrrt.tttuuuuuu;

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Vyxal, cracks emanresuA's answer

5|2/en[(`ĖĖ:

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Easy. I came up with the solution in the shower that's how easy it is.

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  • \$\begingroup\$ So you have, thanks to an oversight of mine. On to round 2 \$\endgroup\$ Commented Nov 25, 2021 at 7:47
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