Your task is to write a non-empty computer program comprised of some sequence of bytes. If we choose a particular byte in the program and remove all instances of it from the program, the modified program should output the removed byte.

For example if our program were


Then bcb would output a, aaaca would need to output b and aababa would output c.

It does not matter what the unmodified program does.

Answers will be scored in bytes with the goal being to minimize the number of bytes.

  • 4
    \$\begingroup\$ Since this challenge isn't tagged quine, may we read our own source code? \$\endgroup\$
    – Dennis
    Sep 5, 2018 at 22:56
  • 1
    \$\begingroup\$ @Dennis Sure. Be my guest \$\endgroup\$
    – Wheat Wizard
    Sep 5, 2018 at 23:12
  • 2
    \$\begingroup\$ If all the bytes in our program represent digits, may we output via exit code? \$\endgroup\$
    – Mr. Xcoder
    Sep 6, 2018 at 6:04
  • 15
    \$\begingroup\$ I think this would be better as a code challenge where you have to maximise the number of discrete characters used. \$\endgroup\$
    – Notts90
    Sep 6, 2018 at 11:07
  • 2
    \$\begingroup\$ Should've specified more than 1 byte instead of non-empty :P. Or what Notts90 said. \$\endgroup\$ Sep 6, 2018 at 13:54

8 Answers 8


zsh, 603 594 566 561 548 440 415 399 378 370 bytes

ho \\n;ca t<<<$'\x20';exi t
d$c -e8BC6P
d0c -eKp
$'\172\163\150' $'\055\143' $'\146\157\162 v \151\156 \173\043\056\056\134\175\175\073\173 \146\147\162\145\160 \055\161 $\166 '$0$'\174\174\074\074\074$\166\073\175'
$'\145v\141\154' $':\073\072\046\046\145\170\151\164';#%&()*+,/9=>?@ADEFGHIJLMNOQRSTUVWXYZ[]^_`jklmsuwy
0# $#;for b in {$..z};{ fgrep -q $b $0||<<<$b;}

Depends on coreutils + dc.

Try it online!

That was... a journey.

This answer has three parts. The first 4 lines handle certain special cases to simplify the code that follows. The next 2 lines and the last line both accomplish essentially the same thing, but exactly one is run with any given character removal. They are written with mostly complementary character sets, so that removing any character breaks only one at most, allowing the other to continue to function.

Looking at the first part, we first handle

  • newline removal with ec\nho \\n
  • space removal with ca t<<<$'\x20' (followed by exi t to avoid running later code, which would result in extraneous output)
  • $ removal with d$c -e8BC6P (8BC6 = 9226 is 36*256 + 10, and 36 and 10 are the byte values of the $ and newline characters respectively; we use hex digits in decimal to avoid having to include them in the large comment in line 6)
  • 0 removal with d0c -eKp (K gets the decimal precision, which is 0 by default)

In the next part, the only characters used (aside from the garbage at the end of the second line) are $'\01234567v;, space, and newline. Of these, four have been accounted for, so the remainder ('\1234567v) cannot occur in the last line. Expanding the octal escapes ($'\123' represents the ASCII character with value 1238), we get:

zsh -c 'for v in {#..\}};{ fgrep -q $v '$0'||<<<$v;}'
eval ':;:&&exit'

The first line loops through all characters used in the program and searches for each one in its own source code ($0 is the filename of the script being run), printing any character that is not found.

The second line looks a little strange, and appears to do the same thing as exit with a bunch of nops. However, encoding exit as octal directly results in $'\145\170\151\164', which does not contain 2 or 3. We actually need to make this less resilient to removals. This is because if any of '\014567v are removed, breaking the first line, the second line also breaks, allowing the remainder of the code to execute. However, we need it to also break if 2 or 3 are removed so that lines 3 and 4 can run. This is accomplished by shoehorning in : and ;, which have a 2 and 3 in their octal representation respectively.

The junk at the end of line 2 is simply there to ensure every printable ASCII character appears at least once, as the way the checking is done by looping through each one requires this.

If exit was not called in the first section (i.e. it was mangled by the removal of one of '\01234567v), we move on to the second, in which we must accomplish the same thing without using any of these characters. The last line is similar to the decoded first line, except that we can contract the range of the loop to save a few bytes, because we already know that all characters except for '\01234567v have been covered. It also has 0# $# before it, which comments it out and prevents it from producing extraneous output if 0 or $ were removed.

  • 5
    \$\begingroup\$ Wow, very impressive considering the amount of distinct characters involved! Definitely looking forward seeing that explanation. \$\endgroup\$ Sep 6, 2018 at 7:28
  • 3
    \$\begingroup\$ @KevinCruijssen here you go :) \$\endgroup\$
    – Doorknob
    Sep 6, 2018 at 13:35
  • 1
    \$\begingroup\$ @Doorknob if this doesn't win you 548 internets, I don't know what does. Honestly, the 603 byte version is just as impressive hah! \$\endgroup\$ Sep 6, 2018 at 16:09
  • 3
    \$\begingroup\$ The only interesting answer so far. \$\endgroup\$ Sep 7, 2018 at 19:00

Retina, 1 byte


Try it online!

When all instances of the single byte (1) are removed, the output is 1. Simple enough.

  • 6
    \$\begingroup\$ I was browsing TIO to find something like this - you beat me to it. Btw this is a polyglot, works with Snails \$\endgroup\$
    – JayCe
    Sep 5, 2018 at 23:43
  • \$\begingroup\$ IMO, this answer should be upgraded to a polyglot answer as the first one (possibly with a forever-incomplete list of languages), and the other two downvoted to oblivion. Oh, and this also works in C. \$\endgroup\$
    – user77406
    Sep 11, 2018 at 21:20
  • \$\begingroup\$ @Rogem I'm not sure what you mean by "this works in C." do you have a C compiler which outputs 1 for the empty program? Regardless, I think the answers in question utilize different approaches and behaviours. IMO a polyglot answer is only warranted if the approach remains the same. (Objectively, this isn't a polyglot as the actual code is different, for the answers below.) Feel free to vote how you want, but a valid answer is a valid answer. I will keep my answer as it is, I don't wish to house a collection of answers on it. \$\endgroup\$ Sep 11, 2018 at 21:40

Lenguage, 216173027061157310 bytes

216173027061157310 = (144115617572598740 + 144115241762960340 + 144115194786755540) / 2. There are 216173027061157310 - 144115617572598740 $s, 216173027061157310 - 144115241762960340 #s and 216173027061157310 - 144115194786755540 spaces.

The 144115617572598740 #s and spaces encode the following BF program:


Try it online!

The 144115241762960340 $s and spaces encode the following BF program:


Try it online!

The 144115194786755540 $s and #s encode the following BF program:


Try it online!

Edit: Saved 72057832274401770 bytes thanks to @Nitrodon.

  • \$\begingroup\$ Why not use U and byte 127? Try it online! Or even just the nul byte and soh? \$\endgroup\$
    – Jo King
    Sep 7, 2018 at 9:47
  • \$\begingroup\$ @JoKing I didn't know that U was the shortest printable ASCII byte that could be output. I didn't want to use unprintable bytes. \$\endgroup\$
    – Neil
    Sep 7, 2018 at 10:51
  • \$\begingroup\$ Even without taking advantage of wrapping cells or unprintable characters, you can get this down to 216173027061157310 bytes by including the space character as a third distinct byte. \$\endgroup\$
    – Nitrodon
    Sep 7, 2018 at 18:46
  • 9
    \$\begingroup\$ I can't help but upvote because of "Edit: Saved 72057832274401770 bytes..." \$\endgroup\$
    – Mr Lister
    Sep 8, 2018 at 11:41

Jelly, 1 byte


Completely different from the Retina answer. whistles

Try it online!

  • 7
    \$\begingroup\$ Polyglot with M and Enlist. \$\endgroup\$
    – Mr. Xcoder
    Sep 6, 2018 at 5:59

Polyglot*, 1 byte (awaiting confirmation)


Try it online! (using Triangularity)

*: This works in a (rather wide) variety of languages (except for esolangs like 4, ><> and the like and some other exceptions). Identical to the Jelly answer in source code, but the method of I/O is different – Output is via exit code. When one removes 0 from the source code, they're left with an empty program, which often doesn't error and yields exit code 0 in the majority of languages.


sed, 1 byte

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Completely different from the Retina answer, or the Jelly answer.

  • \$\begingroup\$ I don't see any code. Wouldn't that make it a 0 byte answer? How does this work? \$\endgroup\$
    – Mast
    Sep 6, 2018 at 10:06
  • 15
    \$\begingroup\$ @Mast There is a newline..... you will have difficulty reading programs written in Whitespace if you keep thinking like that. \$\endgroup\$
    Sep 6, 2018 at 10:14

Unary (non-competitive), 96 bytes

00000000: 0000 0000 0000 0000 0000 0000 0000 0000  ................
00000010: 0000 0000 0000 0000 0000 0000 0000 0000  ................
00000020: 0000 0000 0000 0000 0000 0000 0000 0000  ................
00000030: 0000 0000 0000 0000 0000 0000 0000 0000  ................
00000040: 0000 0000 0000 0000 0000 0000 0000 0000  ................
00000050: 0000 0000 0101 0101 0101 0101 0101 0101  ................

Here is xxd dump.

A wider definition of Unary language allows any characters in its source code. But I havn't find a compiler or interpreter which would work for this. So I marked this answer as non-competitive. If you can find one which posted before this question asked, I will link to it.


Gol><>, 143 bytes


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This contains the characters <\Hs+:PM;o and newline

I wanted to solve this in a 2D language, but most 2D languages are ruled out from this challenge. Gol><> works because of the large number of operations including operations with overlapping functionality, this ensures that we can write a meaningful program with any 1 character banned.

We use the standard technique for 2D languages here. The removal of characters causes the length of a line to change which creates a new alignment and sends the pointer down a new path untouched by the radiation.

Each character gets its own line of the program and which line is run is triggered by the removal. We could possibly choose a smaller character set to work with <\Ho;P1+ and newline should work, however the lines would be very long making this overall more costly.


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