This is a challenge the Cops' Thread can be found here

Your challenge, as robbers is to take outputs from the cops' thread and find anagrams of the output that when run as a program output the original output provided.

The winner will be the person with the most valid cracks on this question.


  • You may not crack answers marked as safe with a provided program.

  • If a cop provides a language that the intended solution is in you must crack it in that particular language, if they choose not to you may crack it in any competing language.

  • Standard rules for Quines apply.

  • Cracks must not be perfect Quines. i.e. they must not output their exact source but a reordering of it (the order is provided by the cop).


Unspecified language (CJam), 254 bytes, DJMcMayhem

0000000: 3235 362c 583e 3130 2d5b 445d 2f41 612a  256,X>10-[D]/Aa*
0000010: 3a63 6523 0102 0304 0506 0708 090b 0c0e  :ce#............
0000020: 0f10 1112 1314 1516 1718 191a 1b1c 1d1e  ................
0000030: 1f20 2122 2425 2627 2829 2b2e 3334 3738  . !"$%&'()+.3478
0000040: 393b 3c3d 3f40 4243 4546 4748 494a 4b4c  9;<=?@BCEFGHIJKL
0000050: 4d4e 4f50 5152 5354 5556 5759 5a5c 5e5f  MNOPQRSTUVWYZ\^_
0000060: 6062 6466 6768 696a 6b6c 6d6e 6f70 7172  `bdfghijklmnopqr
0000070: 7374 7576 7778 797a 7b7c 7d7e 7f80 8182  stuvwxyz{|}~....
0000080: 8384 8586 8788 898a 8b8c 8d8e 8f90 9192  ................
0000090: 9394 9596 9798 999a 9b9c 9d9e 9fa0 a1a2  ................
00000a0: a3a4 a5a6 a7a8 a9aa abac adae afb0 b1b2  ................
00000b0: b3b4 b5b6 b7b8 b9ba bbbc bdbe bfc0 c1c2  ................
00000c0: c3c4 c5c6 c7c8 c9ca cbcc cdce cfd0 d1d2  ................
00000d0: d3d4 d5d6 d7d8 d9da dbdc ddde dfe0 e1e2  ................
00000e0: e3e4 e5e6 e7e8 e9ea ebec edee eff0 f1f2  ................
00000f0: f3f4 f5f6 f7f8 f9fa fbfc fdfe ff0a       ..............

Try it online!

How it works

256,                Push the range [0 ... 255].
    X>              Remove the first item (0).
      10-           Remove 10 (0x0a).
         [D]/       Split at occurrences of 13 (0x0d).
             Aa*    Join, separating by 10 (0x0a).
                :c  Cast all integers to character.
                e#  Begins a comment, terminated by the linefeed at the very end.

Brain-Flak, Riley


Try it online!


This was a bit easier than I had expected.

I wrote 3 separate loops to print each type of brace:


I used standard Mini-Flak technique to avoid using <...> monad and to be honest I never really felt the need to use the <> nilad.

This must have been extremely similar to what Riley did in the first place because when I was done I had an extra () left over. This is not a problem because () is essentially a no-op at the top level so I stuck it at the front of my program.

  • 1
    \$\begingroup\$ That's exactly what/how I wrote it. I needed the extra () to get everything to match up. Well done! \$\endgroup\$ – Riley Feb 28 '17 at 2:40

Haskell, 107 bytes, Laikoni


Plus a trailing newline.

Try it online! (needs a few seconds to run, because it loops through all unicode characters).

The program has an id (right before ['$'..]) which is not needed for the program logic, but somehow I have to spend the i and d.

  • \$\begingroup\$ The output is no in the same order. \$\endgroup\$ – Laikoni Feb 25 '17 at 12:31
  • 1
    \$\begingroup\$ @Laikoni: I misread the challenge. Now it's fixed. \$\endgroup\$ – nimi Feb 25 '17 at 14:25

Python, 96 bytes, Agargara

f= 'B7ofx)wdr0o{1{6{{r){1x0sD00qF1{{24}wr2wlo)y))xm2'; print''.join([r+chr(ord(r)-9)for r in f])

Try it online!

Once I noticed that some characters repeated in pairs, it was just a matter of finding a suitable ascii map between characters.

  • \$\begingroup\$ Congratulations! That's exactly how I wrote it. \$\endgroup\$ – Agargara Mar 6 '17 at 0:31

Pyth, 32 bytes, Rod

J+"J+J=JJ 1-2#pTN%"N #%2J=T-1pJJ

Try it online!

I don't really know Pyth that well, but I started with printing the required output and went from there.

 +"J+J=JJ 1-2#pTN%"N             # Add a '"' to the string in quotes
J                                # Assign result of above step to J
                     #           # Start an infinite loop
                      %2J=T-1    # Loops one iteration, then breaks
                                 # (I have no idea why this works)
                             pJ  # Explicitly print J
                               J # Implicitly print J

CJam, 46 bytes, Business Cat


Try it online!

93]                                             # Push [93] onto the stack
   _                                            # Copy
    :c                                          # Push ascii char 93 ']'
      58 99 95 93                               # Push these numbers...
        ]  ]  ]  ]                              # Push entire stack as array
                  e__                           # Flatten and copy
                     :c                         # Map ascii values to all elements
                       36]95]95]101]101]        # Same as above
                                        e_      # Flatten
                                          $     # Sort
                                           _    # Copy, implicitly print
                                            :c  # Map ascii values to all elements
                                                # Implicitly print

Octave, 91 bytes, drolex


Try it online!

Very similar to the python solution in that it abuses ascii values to print the characters it needs.

  • \$\begingroup\$ Cracked indeed. Well done. \$\endgroup\$ – drolex Mar 7 '17 at 6:32

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