Unscramble the Source Code [robber thread for cracking attempts]

Question

This is the companion thread to the main Unscramble the Source Code challenge. If you think you have managed to unscramble one of the cop answers, you should post your solution as the answer to this thread.

As a reminder, you have one attempt at cracking each submission. Your cracking attempt will be an unscrambled version of the source code. If your guess matches the description (same characters, output, and of course language), and you are the first correct guess, then you win a point. It is important to note that your program does not have to exactly match the original, simply use the same characters and have the same functionality. This means there could be more than one correct answer.

The robber with the most points (successful cracks) wins.

Leaderboard

Too many solves

• Martin Büttner: (Python 3, 16, matsjoyce), (CJam, 15, Ypnypn), (Mathematica, 29, Fox Wilson), (JavaScript, 15, Caridorc), (CJam, 52, user23013), (Pyth, 11, isaacg), (PHP, 22, kenorb), (PHP, 13, kenorb), (Ruby, 17, Doorknob), (PHP, 49, bwoebi), (Mathematica, 35, Tally), (GolfScript, 13, Peter Taylor), (JavaScript, 25, Cris), (JavaScript, 29, Cris), (JavaScript, 26, Cris), (Bash, 33, Debasis), (JavaScript, 10, Cris), (Lua, 28, ChipperNickel), (Mathematica, 18, Arcinde), (JavaScript, 30, Qwertiy), (CJam, 13, user23013), (JavaScript, 41, Ismael Miguel), (Ruby, 38, Doorknob), (Marbelous, 36, es1024), (PHP, 33, Ismael Miguel), (JavaScript, 29, Joe), (JavaScript, 28, Shawn Holzworth), (Ruby, 35, histocrat), (CJam, 19, Ypnypn), (Ruby, 17, Stephen Touset), (JavaScript, 36, MegaTom), (JavaScript, 24, fogcityben), (Python 3, 21, Reticality), (JavaScript, 10, Jamie Barker), (JavaScript, 15, Jamie Barker)

20 solves

• feersum: (Python 3, 44, Sp3000), (C, 70, es1024), (MATLAB, 41, COTO), (Brainfuck, 118, Sp3000), (C, 30, Ethiraric), (C, 28, Mig), (Python 3, 46, hosch250), (Java, 70, Rodolvertice), (C, 29, imallett), (Java, 226, nhahtdh), (Little Man Computer, 63, The Wolf), (Python 2, 89, Beta Decay), (Python 2, 41, muddyfish), (C, 63, es1024), (C++, 192, Arcinde), (Java, 108, durron597), (C#, 604, eshansingh1), (C, 44, Art), (Java, 134, Olavi Mustanoja), (Bash, 47, Vi.)

15 solves

• user23013: (CJam, 18, Ypnypn), (JavaScript, 26, hsl), (CJam, 12, COTO), (PHP, 23, bwoebi), (PHP, 54, Steve Robbins), (CJam, 32, Dennis), (CJam, 19, Martin Büttner), (Bash, 23, The Wolf), (Ruby, 33, Doorknob), (CJam, 34, Dennis), (JavaScript, 82, Ismael Miguel), (PHP, 80, Ismael Miguel), (QBasic, 43, DLosc), (QBasic, 42, DLosc), (ECMAScript, 90, Cris)

10 solves

• squeamish ossifrage: (Python, 44, Caridorc), (PHP, 52, PleaseStand), (Befunge-93, 17, user23013), (BBC BASIC, 187, Beta Decay), (C, 43, Allbeert), (Ruby, 58, Rodolvertice), (JavaScript, 32, hsl), (PHP, 33, kenorb), (Python 2, 16, imallett), (PHP, 53, PleaseStand)
• Sp3000: (Python, 154, Fox Wilson), (Python, 48, kgull), (Python, 68, horns), (Python 2, 38, FryAmTheEggman), (Python 2, 57, FryAmTheEggman), (Python, 74, xnor), (JavaScript, 66, Cris), (JavaScript, 70, Cris), (Python 2, 37, muddyfish), (Python, 55, FireFly)

7 solves

• grc: (Python 2, 61, FryAmTheEggman), (Perl 5, 47, chilemagic), (Python, 69, Sp3000), (Perl, 39, GentlePurpleRain), (Perl, 36, squeamish ossifrage), (Python 3, 110, Sp3000), (C, 53, FireFly)

5 solves

• hsl: (CoffeeScript, 25, Martin Büttner), (Golfscript, 20, Josiah Winslow), (Lua, 18, user3155415), (Python, 54, kgull), (Arithmetic, 31, Cris)
• PleaseStand: (PHP, 14, Tryth), (C++, 56, hosch250), (PHP, 21, kenorb), (QBasic, 37, DLosc), (JavaScript, 46, FireFly)

4 solves

• bwoebi: (PHP, 52, kenorb), (PHP, 30, Tryth), (PHP, 27, kenorb), (PHP, 44, kenorb)
• FireFly: (JavaScript, 94, Shawn Holzworth), (Python, 34, Sp3000), (Brainfuck, 39, Mig), (JavaScript, 69, Shawn Holzworth)

3 solves

• Optimizer: (JavaScript, 61, jsh), (JavaScript, 47, palerdot), (JavaScript, 42, Shawn Holzworth)
• es1024: (PHP, 46, Ismael Miguel), (Python shell, 15, xnor), (Python shell, 21, xnor)
• DLosc: (Python, 41, Sp3000), (Ruby, 37, Rodolvertice), (CJam, 14, User23013)

2 solves

• xnor: (Python 3, 37, matsjoyce), (Python 2, 29, Dennis)
• Daniel Wagner: (Haskell, size 34, Petr Pudlák), (Haskell, 42, proud haskeller)
• nneonneo: (Python, 41, Sp3000), (C, 43, FryAmTheEggman)
• Art: (C, 51, es1024), (C, 48, es1024)
• n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳: (Java 6+, 101, Rodolvertice), (Java, 97, Olavi Mustanoja)

1 solve

• Ypnypn: (Python 2, 50, Geobits)
• matsjoyce: (Python 3, 12, xnor)
• user1354557: (Python, 56, Fox Wilson)
• Dennis: (CJam, 20, Martin Büttner)
• isaacg: (Pyth, 71, FryAmTheEggman)
• ConMan: (SAS, 17, user3490)
• Arcinde: (JavaScript, 285, Beta Decay)
• Shawn Hotazworth: (Javascript, 82, TrungDQ)
• nooodl: (Ruby, 49, Doorknob)
• Bhante Nandiya: (Python 3, 37, Sp3000)
• Geobits: (APL, 17, user23013)
• histocrat: (Ruby, 23, MegaTom)
• ProgramFOX: (Python, 13, Reticality)

Answer 1

Python, 60, f.rodrigues

This post was so bad it had to be stopped.

print(60132)

4 or((odad95oamamarrdsn,nn.m7dtit.rminnrope8ed))

Answer 2

Bash, 47, Vi.

If you saw a $100 bill lying on the ground, wouldn't you pick it up?

Don't know much about bash, but I tested it and it works.

printf pp;o='${cel-sr(xe\     pp/pp/) fec1/; }'

Answer 3

DC, 28 bytes, by MichaelT

[ls+dn11ls1+dss<n]sn1ss0lnx

But I counted 27 bytes. Maybe there is a newline...

Answer 4

CJam, 51, by Optimizer

"JamesBdo,"YZ+/)BA*c+Y*Y%:BS@SB)))[JW:Z____)ci+*]U*

Answer 5

Python, 45, by muddyfish

print(eval((chr(32)+chr(45)).join(str(413))))

Interesting idea, I have to say. Wasn't hard to brute force though.

Answer 6

Brainfuck, 39, by Mig

+++++++[->++++++<]>[->++<]>-.----.++++.

Answer 7

Python shell, size 15, xnor

1215954^1548696

Obtained by brute force in 1.42 seconds.

Code used to find solution:

#include <iostream>
#include <algorithm>

int main(){
    int digits[] = {1,1,1,2,4,4,5,5,5,6,6,8,9,9};
    int ndigits = sizeof(digits)/sizeof(digits[0]);

    do{
        int n1 = 0, n2 = 0;
        for(int i = 0; i < ndigits - 1; ++i){
            n1 = (n1 * 10) + digits[i];
            n2 = 0;
            for(int j = i+1; j < ndigits; ++j)
                n2 = (n2 * 10) + digits[j];
            if((n1 ^ n2) == 339018){
                std::cout << n1 << ' ' << n2 << std::endl;
                return 0;
            }
        }
    }while(std::next_permutation(&digits[0], &digits[ndigits]));
}

Answer 8

Python shell, size 21, xnor

6446781078^9350512293

Obtained by brute force in ~27 minutes.

Code used:

#include <iostream>
#include <cinttypes>
#include <cmath>

inline bool check(int64_t a, int64_t b){
    int used[10] = {};
    do{
        auto r = std::div(a,10ll);
        ++used[r.rem];
        a = r.quot;
    }while(a);
    do{
        auto r = std::div(b,10ll);
        ++used[r.rem];
        b = r.quot;
    }while(b);
    if(used[0] > 2) return false;
    for(int i = 1; i < 10; ++i)
        if(used[i] != 2) 
            return false;
    return true;
}

int main(int argc, char **argv){
    int64_t i = 1ll << 32;
    while(!check(i,i ^ 15788895283ll)) ++i;
    std::cout << i << ' ' << (i^15788895283ll) << std::endl;
}

This checks every integer starting from 2^32 until it and its matching integer use up all the characters provided.

Answer 9

C++, 244 bytes^?, by Gabe Evans

#include<iostream>
int main(){int(CCDD)='%';char(a)[14]={'h','e','l','l','o',22+20/2,'w','o','r','l','d','!'};CCDD+=CCDD<=a[0];{{}};aa:adffillllllnnnnnnnnnnnnooooooooooooorrrrrrrrrssstttttttttttttttuuuuuuuuuuuuwwwwww:cout<<a;}

It was a bit of a struggle getting all those spare characters out of the way. You can see the code working here.

Answer 10

Insomnia, 14, by n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳

e9-i|yig9s pre

Answer 11

C, size 62, by Olavi Mustanoja

#include<stdio.h>
#define _ +8
int*main(){printf("%d",1 _ _);}

Answer 12

Python, 28, by user3863227

print(int(bin(1234)[2:])**2)

Factorizing the output gave this one away quite easily:

100220321420320200100 = 2² × 5² × 457² × 2190593²
sqrt(100220321420320200100) = 10011010010
int('10011010010',2) = 1234

Answer 13

Perl - 26 bytes by primo

f:print po=>glob q$t{a,o}$

The q was easy to recognise. I took the potato letters away and shuffled the rest to discover glob suddenly. Then, I realised $$ will be used instead of {} and that => can be created. The last thing was f:, it took me a while to think of a label.

Answer 14

CJam, 14, by User23013

92e8.17 5e6.3%

I started by making some assumptions: that the exponent in each case was a floating point number, and that it had at least one digit after the dot (otherwise it's just the same as using an integer) and one or two before it (otherwise the result would be too small or too large). Then I used Python's itertools library to do some permutations. I got a result in under a minute.

Here's my cracking code:

from itertools import permutations
target = 108183.05975155532

def e(mantissa, exponent):
    return mantissa * 10 ** exponent

def ePartition(digits):
    """Generator function: gives every way to partition digits into an
    integer mantissa and a floating point exponent that is > 1 and < 100."""
    for eIndex in range(1, len(digits) - 1):
        for dotIndex in range(eIndex + 1, min(len(digits), eIndex + 3)):
            yield (digits[:eIndex],
                   digits[eIndex:dotIndex] + "." + digits[dotIndex:])


for p in permutations("12356789"):
    for spIndex in range(3,6):
        digList1 = "".join(p[:spIndex])
        digList2 = "".join(p[spIndex:])
        for mant1, expt1 in ePartition(digList1):
            for mant2, expt2 in ePartition(digList2):
                value = eval("e(%s,%s)%%e(%s,%s)" % (mant1,
                                                     expt1,
                                                     mant2,
                                                     expt2))
                if value == target:
                    print("%se%s %se%s%% -> %s" % (mant1,
                                                   expt1,
                                                   mant2,
                                                   expt2,
                                                   value))

Answer 15

JavaScript, 36, by MegaTom

x=alert;x(("x"+x)["substring"](6,9))

Tested in Chrome's console.

Answer 16

Python 2, 13, by mbomb007

71e0//36/0x92

The output is 0.00684931506849315 on my machine. But it worked on http://repl.it/.

Answer 17

Python, 39 bytes, Reticality

print("{0}{2}{1}{1}{1}".format(0,5,6,))

You could also swap the {0} & {2} and the 0 and 6.

Answer 18

JavaScript, 15, by Jamie Barker

Math.sin(1)+'O'

EZ PZ, as you said. ;)

Answer 19

Python, 13 bytes, by Reticality

To be run on an interactive Python console:

len("printi")

Outputs 6.

Answer 20

CMD, 9, by unclemeat

set/a!!!0

There are too few possibilities here. The 4 alphabet characters are most likely command name, so I attack in that direction.

Answer 21

Bash, size 23, by The Wolf

q=echo;_(){ $q $q;};""_

I think the original solution (or red herring) should be:

q(){ q=;};q ""echo;$_ $_

which is invalid because it has one more space character.

Answer 22

C, size 63, es1024

main(_){int __=_;_<<=_;__=_<<_<<__;printf("%d",_<<__|__);_|__;}

What I'm trying to get is 2 << 16 | 16. Three leftover underscores, woohoo.                                    

Answer 23

Rust, size 123, by Vi.

fn x(r:int)->int{for i in range(0,2){if i==r/r{{}}else if i==0{}}r%222*912}fn main(){print!("{}",x(993+2))}                

This shouldn't be the intended solution because of the no-op for loop, but it produces no warning in play-pen anyway.

(I'll leave the Clojure part for others.)

Answer 24

PHP, 62, aks.

for($i=1;$i<3.;)echo"$i. ".((($i++==2))?'Tues':'Mon').'day ';

Quite easy compared to other crazy PHP submissions.

Answer 25

JavaScript, 24 bytes (by fogcityben)

'.+';console.log( 'hi' )

Not sure what he really did with that . and +.

Answer 26

Python 3, 21 bytes, Reticality

print(str(...)[4::2])

Prints the p and final i from Ellipsis.

Answer 27

Python, size 26, by Oliver

print 2 * 3 * 37037,
print

The 2 prints threw me off a bit.

Answer 28

Python, size 15, by mbomb007

There are at least 5 permutations that work. This is one of them:

`3*.7/50`[::~4]

It was pretty obvious that the original code was something of the form `«expression»`[«range»], so it wasn't difficult to place the backticks, colons and square brackets. On a hunch, I assumed that the range specifier was [::~i] for some value of i, leaving only 8 characters to permute. This took only a fraction of a second. Here's the code I used:

from itertools import permutations

perms = [''.join(p) for p in permutations('03457*./')]
for p in perms:
    try:
        e = '`'+p[:-1]+'`[::~'+p[-1:]+']'
        if eval(e)=='6991':
            print e
    except:
        continue

(The original code is actually a bit flakey, since the result of 3*.7/50 isn't guaranteed to be 0.041999999999999996 all of the time.)

Answer 29

Python, size 30, by mbomb007

Didn't need much unscrambling:

' rsin[~(ddf) 0]rot13'[9+6: :]

Answer 30

CMD, 21, by unclemeat

set /a!!"echo>>tfile"

Repeating the set /a trick and ! operator is reducing the search space for otherwise a tough nut to crack.

I'm not too sure how set /a works, but when I evaluate set /a "tfile", it returns 0. Not sure if all strings evaluates to 0, but in this case, we end up evaluating (!!0) >> 0, which is 0.