131
\$\begingroup\$

You are to write a program that will output source code that is

  1. Larger than the original program (character wise)
  2. Will print another program larger than itself when run (i.e. the new program is also a valid answer to this challenge) indefinitely

This is code-golf, so shortest answer wins.

\$\endgroup\$
11
  • 5
    \$\begingroup\$ @Kevin, The definition is recursive. The output should be a program whose output is larger than itself, and an answer to this question. So the output's output should be larger than the output, and the output's output's output should be even larger, etc. \$\endgroup\$
    – ugoren
    Commented Feb 23, 2014 at 10:55
  • 8
    \$\begingroup\$ I think you should clarify your rules. On one hand, any additional code output by such a program is "obviously useless"; on the other hand, all additional code in the output is "useful" in that it furthers the goal of answering this challenge. \$\endgroup\$
    – Jason C
    Commented Feb 23, 2014 at 22:29
  • 6
    \$\begingroup\$ Sorry for destroying your challenge. :^) \$\endgroup\$
    – Justin
    Commented Feb 24, 2014 at 5:03
  • 44
    \$\begingroup\$ Naturally, such a program should be known as a quinine. \$\endgroup\$ Commented Feb 25, 2014 at 5:05
  • 2
    \$\begingroup\$ @facepalm42 Quine challenge answers are forbidden from reading their own source code, which makes comments tricky to use here. \$\endgroup\$ Commented Nov 9, 2019 at 1:02

64 Answers 64

328
\$\begingroup\$

H9+ : 1 char

9

That's right. One character. Outputs the lyrics to 99 bottles of beer, which is a valid program. All the extraneous data does not count, but there are plenty of 9s in there.

The output of the outputted program is the lyrics to 99 bottles of beer 59 times.

This function gives the number of times the lyrics are outputted if you run the program n times (if my calculation is correct):

f(n) = 59n-1
\$\endgroup\$
16
  • 40
    \$\begingroup\$ A 0 char program in a given language, given that it's valid, will likely output 0 characters. And since 0*x=0 for any finite value x, I can claim that my outputted program is actually 9999999999999999 times bigger than the source code. \$\endgroup\$
    – nitro2k01
    Commented Feb 23, 2014 at 18:33
  • 48
    \$\begingroup\$ @JohannesH. For all(solvable) problems there exists a programming language that solves the problem in 0 chars. \$\endgroup\$
    – Cruncher
    Commented Feb 24, 2014 at 18:48
  • 21
    \$\begingroup\$ I still move that we name that "Cruncher's Theorem", unless it's already been discovered. \$\endgroup\$ Commented Feb 25, 2014 at 18:31
  • 30
    \$\begingroup\$ One interesting feature is that it will also contain 2*59^n-2 instances of "Hello, world!", due to the "h" in "the". \$\endgroup\$ Commented Mar 11, 2015 at 18:37
  • 10
    \$\begingroup\$ Looks like you've been beat. \$\endgroup\$ Commented Oct 17, 2016 at 22:03
128
\$\begingroup\$

Java 7: 0 chars


Save as file Blank.java. If you save it as any other file, replace any instance of Blank with the appropriate file name.

Then, run in command line via first compiling, then running. If compiling fails, stop.

I list this as Java 7 because it might output differently for different versions of Java.

First few outputs (outputted to stderr):

Error: Could not find or load main class Blank
Blank.java:1: error: class, interface, or enum expected
Error: Could not find or load main class Blank
^
Blank.java:1: error: reached end of file while parsing
Error: Could not find or load main class Blank
                                              ^
2 errors
Blank.java:1: error: class, interface, or enum expected
Blank.java:1: error: class, interface, or enum expected
^
Blank.java:1: error:  expected
Blank.java:1: error: class, interface, or enum expected
                          ^
Blank.java:1: error:  expected
Blank.java:1: error: class, interface, or enum expected
                                     ^
Blank.java:1: error: as of release 5, 'enum' is a keyword, and may not be used as an identifier
Blank.java:1: error: class, interface, or enum expected
                                          ^
  (use -source 1.4 or lower to use 'enum' as an identifier)
Blank.java:1: error: = expected
Blank.java:1: error: class, interface, or enum expected
                                               ^
Blank.java:2: error:  expected
Error: Could not find or load main class Blank
     ^
Blank.java:2: error: ';' expected
Error: Could not find or load main class Blank
      ^
Blank.java:2: error: = expected
Error: Could not find or load main class Blank
                      ^
Blank.java:2: error: = expected
Error: Could not find or load main class Blank
                                   ^
Blank.java:2: error:  expected
Error: Could not find or load main class Blank
                                              ^
Blank.java:3: error: = expected
^
^
Blank.java:3: error: ';' expected
^
 ^
Blank.java:4: error: illegal start of type
Blank.java:1: error: reached end of file while parsing
     ^
Blank.java:4: error: = expected
Blank.java:1: error: reached end of file while parsing
          ^
Blank.java:4: error: illegal start of type
Blank.java:1: error: reached end of file while parsing
           ^
Blank.java:4: error:  expected
Blank.java:1: error: reached end of file while parsing
            ^
Blank.java:4: error: = expected
Blank.java:1: error: reached end of file while parsing
          ^
Blank.java:4: error: illegal start of type
Blank.java:1: error: reached end of file while parsing
           ^
Blank.java:4: error:  expected
Blank.java:1: error: reached end of file while parsing
            ^
Blank.java:4: error: = expected
Blank.java:1: error: reached end of file while parsing
              ^
Blank.java:4: error: ';' expected
Blank.java:1: error: reached end of file while parsing
                   ^
Blank.java:4: error: = expected
Blank.java:1: error: reached end of file while parsing
                                 ^
Blank.java:4: error:  expected
Blank.java:1: error: reached end of file while parsing
                                        ^
Blank.java:4: error: = expected
Blank.java:1: error: reached end of file while parsing
                                         ^
Blank.java:4: error: ';' expected
Blank.java:1: error: reached end of file while parsing
                                              ^
Blank.java:5: error:  expected
Error: Could not find or load main class Blank
     ^
Blank.java:5: error: ';' expected
Error: Could not find or load main class Blank
      ^
Blank.java:5: error: = expected
Error: Could not find or load main class Blank
                      ^
Blank.java:5: error: = expected
Error: Could not find or load main class Blank
                                   ^
Blank.java:5: error:  expected
Error: Could not find or load main class Blank
                                              ^
Blank.java:6: error: = expected
                                              ^
                                              ^
Blank.java:6: error: ';' expected
                                              ^
                                               ^
Blank.java:7: error: reached end of file while parsing
2 errors
        ^
30 errors
\$\endgroup\$
14
  • 26
    \$\begingroup\$ "You are to write a program that will output source code that is" The output is arguably not source code. And furthermore, depending on the definition of output (whether output must come specifically from stdout) nothing is actually output. \$\endgroup\$
    – nitro2k01
    Commented Feb 23, 2014 at 18:48
  • 54
    \$\begingroup\$ @nitro2k01 I quote the same: "You are to write a program that will output source code". Anything is source code. But only some things are valid source code. Additionally, it was not specified what is considered output, so I am allowed freedom there. \$\endgroup\$
    – Justin
    Commented Feb 23, 2014 at 18:50
  • 23
    \$\begingroup\$ @PyRulez That's right. I exploited your rules. You don't need to accept this answer. When I posted this, I knew that a controversy would follow, complete with many up and down votes. I decided that it would be worth it, though. \$\endgroup\$
    – Justin
    Commented Feb 23, 2014 at 21:40
  • 34
    \$\begingroup\$ Are you sure that the program itself printed the output? Wasn't that only the compiler? The program gets never compiled and itself is not interpreted nor executed, thus it wasn't able to produce any output. \$\endgroup\$
    – V-X
    Commented Feb 24, 2014 at 8:25
  • 18
    \$\begingroup\$ @blabla999 I disagree - people don't try these challenges because someone else does something crazy. We just challenge ourselves to do it some other way. As Mark Rosewater says, "Restrictions breed creativity" \$\endgroup\$
    – corsiKa
    Commented Feb 24, 2014 at 23:40
127
\$\begingroup\$

GolfScript, 9 chars

{.'.~'}.~

This code outputs:

{.'.~'}{.'.~'}.~

which outputs:

{.'.~'}{.'.~'}{.'.~'}.~

which outputs:

{.'.~'}{.'.~'}{.'.~'}{.'.~'}.~

and so on.

I believe this is the shortest answer in a "real" Turing-complete programming language so far.

Explanation:

Basically, the original code above is a "quine-layer": it outputs a normal quine followed by itself.

In GolfScript, any code block literal (e.g. {foo}), if left undisturbed on the stack, is a quine. Thus, on its own, {.'.~'} simply outputs itself, just like any other code block would.

The .~ at the end of the code takes the last code block on the stack, duplicates it, and executes the copy. When executed, the code .'.~' inside the code block duplicates the topmost item on the stack (i.e. the copy of itself) and appends the string .~.

At the end of the program, the GolfScript interpreter stringifies and outputs everything on the stack, which, in this case, consists of one more {.'.~'} block than in the input, plus the string .~.

Bonus:

Adding a ] before the first . (to collect all the code blocks on the stack into an array before they're duplicated) makes it grow exponentially:

{].'.~'}.~

outputs:

{].'.~'}{].'.~'}.~

which outputs:

{].'.~'}{].'.~'}{].'.~'}{].'.~'}.~

which outputs:

{].'.~'}{].'.~'}{].'.~'}{].'.~'}{].'.~'}{].'.~'}{].'.~'}{].'.~'}.~

and so on.

\$\endgroup\$
6
  • 229
    \$\begingroup\$ {].'.~'}.~︵ ┻━┻ \$\endgroup\$
    – nitro2k01
    Commented Feb 23, 2014 at 18:34
  • 2
    \$\begingroup\$ @nitro2k01 my first thoughts exactly :D \$\endgroup\$
    – Songo
    Commented Feb 27, 2014 at 12:50
  • 1
    \$\begingroup\$ Sorry I don't think I get the joke about the table can someone please explain? \$\endgroup\$ Commented Mar 27, 2016 at 3:22
  • 1
    \$\begingroup\$ Is "real" in quotes because HQ9+ is not real or because GolfScript is not real, or both? \$\endgroup\$
    – None
    Commented Sep 13, 2016 at 8:37
  • 1
    \$\begingroup\$ @thepiercingarrow the source code resemble this \$\endgroup\$
    – Rod
    Commented Nov 16, 2016 at 14:47
107
+500
\$\begingroup\$

GS2 (8636bd8e), 0 bytes



This prints a single newline, which prints two newlines, which prints three newlines, et cetera.

Try it online!

\$\endgroup\$
7
  • 3
    \$\begingroup\$ Isn't this technically non-competing since gs2 is newer than the challenge? \$\endgroup\$
    – DJMcMayhem
    Commented Oct 17, 2016 at 22:22
  • 30
    \$\begingroup\$ This commit predates the challenge by a month and a half. I just tested it, and it works as intended. \$\endgroup\$
    – Dennis
    Commented Oct 17, 2016 at 22:24
  • 17
    \$\begingroup\$ @MamaFunRoll Why would it be? This isn't different from an empty Retina program printing 1, an empty Jelly program printing 0, or an empty Stuck program printing Hello, World!. \$\endgroup\$
    – Dennis
    Commented Nov 12, 2016 at 17:21
  • 3
    \$\begingroup\$ @Dennis well it is slightly different, in that the Jelly program 0 doesn't print 00, the stuck program Hello, World! doesn't print Hello, World!Hello, World!, etc. \$\endgroup\$
    – Cyoce
    Commented Dec 17, 2016 at 21:12
  • 3
    \$\begingroup\$ @Cyoce It just happens that a newline outputs another newline in GS2. \$\endgroup\$ Commented May 31, 2017 at 18:51
84
\$\begingroup\$

HQ9+, HQ9++ and similars, 2 characters

QQ

This is the output:

QQQQ
\$\endgroup\$
6
  • 47
    \$\begingroup\$ I saw this and thought "Oh no, nothing can beat this." I started trying to come up with something, then it hit me. So I posted an answer. \$\endgroup\$
    – Justin
    Commented Feb 23, 2014 at 6:09
  • 1
    \$\begingroup\$ @Quincunx Bravo, upvoted your answer. \$\endgroup\$ Commented Feb 23, 2014 at 6:10
  • 4
    \$\begingroup\$ @Quincunx No, I think it's 2^2^n, where the initial program is generation 0. The length recurses m -> m^2. \$\endgroup\$
    – Vortico
    Commented Feb 23, 2014 at 20:11
  • 3
    \$\begingroup\$ @Vortico Good point. Even though I said it wrong, I meant: At code-length n, the output becomes n^n. However, that is false; when the code-length is n, the output's length is n^2 \$\endgroup\$
    – Justin
    Commented Feb 23, 2014 at 21:00
  • 2
    \$\begingroup\$ It is funny though because even the first two iterations are possible contenders for this question. \$\endgroup\$ Commented Feb 23, 2014 at 21:47
30
\$\begingroup\$

Lambda Calculus - 29

A simple lambda term

(λu.(u u)(u u))(λu.(u u)(u u))

Reducing this term by one beta reduction yields

((λu.(u u)(u u))(λu.(u u)(u u)))((λu.(u u)(u u))(λu.(u u)(u u)))

And so on and so on. It's a simple variant on the classic (λu.u u)(λu.u u) which is a quine in lambda calculus, double self application here means we get twice the output.

\$\endgroup\$
7
  • 11
    \$\begingroup\$ I'd argue that in lambda calculus, "running" a lambda term means computing its normal form, not performing a single beta reduction. Also, a term can contain multiple redexes, therefore "reducing a single redex" is an ambiguous operation. \$\endgroup\$
    – Petr
    Commented Feb 24, 2014 at 19:17
  • 4
    \$\begingroup\$ You're right, but since this term has no normal form it makes sense to talk about this in terms of small step reduction. Since lambda calculus lacks any inherent set evaluation, I can just define "running" it to be reducing a single redex top level redex with call-by-name semantics no? \$\endgroup\$ Commented Feb 24, 2014 at 19:54
  • 6
    \$\begingroup\$ Well, terms with no normal form correspond to non-terminating programs. And the problem with top level redex is that a term can have two redexes, neither one being a sub-term of another. You could pick some criterion which one to reduce, but I'd say that then you're getting very far from the standard lambda calculus semantics. (Anyway I appreciate your novel idea.) \$\endgroup\$
    – Petr
    Commented Feb 24, 2014 at 19:56
  • 3
    \$\begingroup\$ (λx.x x x)(λx.x x x) :20 \$\endgroup\$ Commented Feb 24, 2014 at 22:10
  • 3
    \$\begingroup\$ (λx. x x)(λx. x x x) reduces to that in a single step \$\endgroup\$ Commented Feb 26, 2014 at 21:56
30
\$\begingroup\$

Ruby 27

A very slightly modified version of this (via):

puts <<3*3,3
puts <<3*3,3
3

The number of times that puts-line is printed grows exponentially.

$ ruby quine.rb | ruby | ruby
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
puts <<3*3,3
3

$ ruby quine.rb | ruby | ruby | ruby | ruby | ruby | ruby | ruby | wc -l
    3283
\$\endgroup\$
1
  • 20
    \$\begingroup\$ Answer inspired by the Kaiser Chiefs \$\endgroup\$ Commented Feb 23, 2014 at 20:06
25
\$\begingroup\$

SH script, 9

cat $0 $0

Grows at exponential rate.

Either run as sh whatever.sh or set it as executable.

Windows version is here.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ At the time of me writing this comment, the Quine Wiki page says that reading from the filesystem is cheating. \$\endgroup\$
    – NeRoboto
    Commented Dec 7, 2016 at 14:32
  • 8
    \$\begingroup\$ @LordRatte On the other hand, the challenge wasn't to write a quine. \$\endgroup\$
    – Jason C
    Commented Dec 7, 2016 at 23:53
20
\$\begingroup\$

dc 11

Quite simple:

6579792
dfP

The first line is repeated once every generation:

$ dc growing_quine.dc
6579792
6579792
dfP

$ dc growing_quine.dc | dc | dc 
6579792
6579792
6579792
6579792
dfP

The last line consists of the following instructions: d duplicates the last value put on the stack (6579792) (so that we get one more copy each time we run it), f prints the whole stack (which is a bunch of that same number) and P prints the number (6579792) out as a byte stream, which displays as dfP.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ This can be made a little shorter: 6579792pP \$\endgroup\$
    – m90
    Commented Oct 8, 2021 at 14:19
  • 2
    \$\begingroup\$ @m90 6579792pP outputs my original program, which isn't a strict superset of 6579792pP; the minuscule p disappears. Therefore I assume that isn't valid, unfortunately. \$\endgroup\$
    – daniero
    Commented Oct 8, 2021 at 17:16
  • 2
    \$\begingroup\$ The requirement of the problem is only that the succesive programs be "larger", which I understood to mean having greater length, although I now see that it's not entirely clear. \$\endgroup\$
    – m90
    Commented Oct 8, 2021 at 17:32
19
\$\begingroup\$

redcode (recursive solution)

This is the code of the easiest warrior writable in redcode, the famous Imp:

MOV 0, 1

When executed, the code writes a copy of its single instruction at the next address in memory; then executes it, etc.

\$\endgroup\$
15
\$\begingroup\$

Python 3 - 55

print(open(__file__).read())
f=lambda:print('f()')
f()

This could be made shorter by replacing __ file__ with a single character filename and saving the file as that, but I felt this answer was more in the spirit of the question. After one iteration it outputs:

print(open(__file__).read())
f=lambda:print('f()')
f()
f()
\$\endgroup\$
4
  • 3
    \$\begingroup\$ This isn't a quine. See the definition. \$\endgroup\$
    – Petr
    Commented Feb 24, 2014 at 19:13
  • 20
    \$\begingroup\$ @PetrPudlák Nor are any programs in this question. This is a quine-variant problem, not strict quining. \$\endgroup\$ Commented Feb 25, 2014 at 2:22
  • \$\begingroup\$ The lambda isn't needed if the file ends in a newline character: print(open(__file__).read()*2) \$\endgroup\$ Commented Oct 7, 2021 at 9:24
  • 2
    \$\begingroup\$ One that meets the new quine definitions is: s='s=%r;print(s%%s*2);';print(s%s) \$\endgroup\$ Commented Oct 7, 2021 at 9:35
8
\$\begingroup\$

Evoloop, 9×9 rectangle (81 cells)

The Evoloop cellular automaton included with Golly supports patterns which replicate themselves in a "quine-like" way. Specifically, these patterns each contain a "program"; a pattern reproduces itself by first executing the program (which creates the "body" of the daughter), and then by copying the program into the daughter.

The above applies to the more famous "Langton's Loops" cellular automaton as well as Evoloop, but Evoloop has an interesting difference, which is that it's easy to create a pattern which grows in each successive generation.

(Much more interesting, in my opinion, is the fact that Evoloop is a simple cellular automaton which contains patterns which reproduce themselves and evolve in a very life-like manner! I think the only known cellular automata which do this are Evoloop and its descendants. A shortcoming of Evoloop, however, is that there is one particular "best" genome; evolution always eventually converges to this one genome.)

Now, there are two shortcomings to this submission. One is that it's not clear what the "output" of a cellular automaton is. But I think that a self-reproducing automaton is "close enough" to being a quine; it's certainly no less interesting! The other shortcoming is that these patterns don't merely create a single copy of themselves; each copy of the original pattern attempts to create infinitely many copies of itself, and these copies end up interacting with each other in a destructive manner. So, I think I've met the requirements of this challenge in spirit, but not in letter.

Without further ado, the pattern is:

022222220
270170172
212222202
202000212
272000272
212000202
202222212
271041032
022222250

Here's the pattern again, in a format that can be copied and pasted into Golly:

x = 9, y = 9, rule = Evoloop
.7B$BG.AG.AGB$BA5B.B$B.B3.BAB$BGB3.BGB$BAB3.B.B$B.5BAB$BGA.DA.CB$.6BE
!

Okay, but what does it look like? It looks like this:

A pattern in a cellular automaton which creates increasingly large copies of itself.

In the above animation, you can see the initial pattern create a larger daughter, which creates a larger granddaughter, then a larger great-granddaughter, and finally an even larger great-great-granddaughter which starts to construct a yet larger third-great-granddaughter. If you ran this pattern for a long time, it would keep going on like this forever (or perhaps they would eventually be overtaken by the evolved organisms, which are capable of reproducing much faster; I'm not sure).

\$\endgroup\$
2
  • \$\begingroup\$ When you run it long enough you get genomes which create smaller offspring, which eventually self-destruct. The 'mother' at the center stays alive. \$\endgroup\$ Commented Jun 12, 2020 at 23:09
  • \$\begingroup\$ @MarkJeronimus Yup. Those were the inspiration for this answer: I saw that there were organisms that get smaller with each generation, so I figured it would probably be possible to create one that would get larger with each generation. And sure enough. \$\endgroup\$ Commented Jun 13, 2020 at 2:46
7
\$\begingroup\$

Smalltalk, 125 61 57

The golf version looks almost unreadable so I'll explain first (and use real identifiers).

This is a variant of the "weirdest-way-to-produce-a-stack-overflow" self modifying method.

The method prints out a hello message, and its current source (for the demonstration only). Then, the code is modified to output a longer string and installed. Finally, the new code is called recursively.

In order to protect myself from an immediate runaway, it lets the user confirm in each cycle.

compile in Object:

eatMe_alice
   |msg mySource|

   mySource := thisContext method source.

   '**** Hello Alice' printCR.
   '  ---- my current code is:' printCR.
   mySource printCR.
   '  ---------------' printCR.

   (UserConfirmation confirm:'Again? ') ifTrue:[
       Object compile:
            (mySource
                copyReplaceString:'Hello ','Alice'
                withString:'Hello ','Alice !').
       self eatMe_alice
   ]

start the show by sending "eatMe_alice" to any Object; nil will do:

nil eatMe_alice

A nice variant is to not call the new code recursively, but instead iteratively, by unwindig the call stack and reentering into the new method. This has the advantage of not leading to a recursion exception. To do this, replace the recursive call ("self eatMe_alice") by:

thisContext resend

Golfing:

Obviously, printing and self calling was not asked for, so the shortest (for golf) is to simply append a comment to my own source and return it. As a side effect, it also gets installed for the next call...

x|s|Object compile:(s:=thisContext method source,'""').^s
\$\endgroup\$
5
\$\begingroup\$

SH script, 12 8 7

Store a file with

sed p *

in its own, empty directory and run from this directory using sh [file] or set executable.


Old alternative with 8 characters, but doesn't need its own directory. Store a file with

sed p $0

and run using sh [file] or set executable.

Old alternative with 12 characters:

sed -i- p $0

This will actually output to the program file itself, but where to output was not specified. Replicates itself at an exponential rate.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ You might still have chance using ed, didn't want to look through its man page... \$\endgroup\$
    – Jens Erat
    Commented Feb 24, 2014 at 10:49
5
\$\begingroup\$

ECMAScript 6 (38 Characters)

(f=_=>'(f='+f+')();(f='+f+')();')();

Which outputs:

(f=_=>'(f='+f+')();(f='+f+')();')();(f=_=>'(f='+f+')();(f='+f+')();')();

Edit

You could do (28 characters):

(f=_=>'(f='+f+')();'+f())();

However it will recurse infinitely and never return anything... but this can be solved by doing something like this (42 characters):

(f=_=>_?'(f='+f+')('+_+');'+f(_-1):'')(3);

Which will output:

(f=_=>_?'(f='+f+')('+_+');'+f(_-1):'')(3);(f=_=>_?'(f='+f+')('+_+');'+f(_-1):'')(2);(f=_=>_?'(f='+f+')('+_+');'+f(_-1):'')(1);
\$\endgroup\$
2
  • 5
    \$\begingroup\$ +_+ =_= +1 for emoticons \$\endgroup\$ Commented Oct 18, 2016 at 12:21
  • 2
    \$\begingroup\$ 33: (f=_=>(f=${f})();.repeat(2))(); \$\endgroup\$
    – emanresu A
    Commented Apr 8, 2021 at 21:09
5
\$\begingroup\$

JavaScript, 41, 40 chars

function f(){console.log(f+"f(f())")}f()

The first time you run it it outputs itself with another ;f() at the end. Subsequent runs of the output results in each "input" source printed twice.

alert would be shorter than console.log but I don't consider multiple alert dialogs to be "the" output while it seems reasonable to call multiple lines in the console as an output.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ You can save one byte with "f(f())" instead of "f();f()" \$\endgroup\$
    – Hedi
    Commented Nov 12, 2016 at 15:40
  • 1
    \$\begingroup\$ This doesn't seem to work for me... \$\endgroup\$ Commented Nov 12, 2016 at 22:27
5
\$\begingroup\$

Runic Enchantments, 6 bytes

"'<S@>

Try it online!

This one was weird. All I had to do was remove a ~ from the original quine found by Jo King.

Every additional run appends another < to the end, e.g.:

"'<S@><<<<<<<<<

All of which do nothing.

\$\endgroup\$
5
\$\begingroup\$

Microscript II, 6 bytes

"qp"qp

The first iteration adds an extra qp to the end, and each successive iteration adds an extra copy of this original program to the beginning.

\$\endgroup\$
4
\$\begingroup\$

Windows .BAT, 25

@COPY %~nx0+%~nx0 CON>NUL

Grows at exponential rate.

Equivalent SH version here.

\$\endgroup\$
4
\$\begingroup\$

Julia, 66 chars

x="print(\"x=\$(repr(x))\\n\$x;\"^2)";print("x=$(repr(x))\n$x;"^2)

Output (134 chars):

x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);

Result of executing the result (268 chars):

x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);

next result (536 chars):

x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);

Next result (1072 chars):

x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);x="print(\"x=\$(repr(x))\\n\$x;\"^2)"
print("x=$(repr(x))\n$x;"^2);

I hope this is according to the rules.

It produces larger output, and the output itself is valid source code that produces larger output again.

\$\endgroup\$
6
  • \$\begingroup\$ Will the output, when ran, produce a valid entry? \$\endgroup\$ Commented Jun 30, 2015 at 1:11
  • \$\begingroup\$ In the first example the output is valid code. It’s a string literal containing 5 periods. In the second example, entering the julia prompt itself wouldn’t do anything. I just thought that just giving the repl actually a newline character by hitting enter produces a longer output than the input. \$\endgroup\$
    – M L
    Commented Jun 30, 2015 at 1:24
  • \$\begingroup\$ Will "....." produce a larger version of itself? \$\endgroup\$ Commented Jun 30, 2015 at 1:25
  • \$\begingroup\$ Ah, now I understand. I just noticed it had to be recursive... time to correct my solution. Thanks for the hint. \$\endgroup\$
    – M L
    Commented Jun 30, 2015 at 1:28
  • 1
    \$\begingroup\$ I guess the code is according to the rules now. \$\endgroup\$
    – M L
    Commented Jul 12, 2015 at 2:26
4
\$\begingroup\$

Husk, 15 11 bytes

-4 bytes thanks to Unrelated String

S+s"So+uswg

Try it online!

Outputs So+uswg"So+uswg" then S o+uswg"S o + u s w g" then S o+uswg"S o + u s w g"... This is a variant of the usual quine S+s"S+s" but with increasing spaces between each character in the string. Currently there's a bug in the parser that prevents double spaces in the code itself, otherwise this could forgo the unique stripping out spaces in the code section and could be 9 bytes.

\$\endgroup\$
1
3
\$\begingroup\$

Common Lisp, 16 Characters

(print `(or ,-))

Granted, it's interactive-only, but being able to reference the current top-level form is probably the single best way to minimize a non-trivial program that meets the specification.

What would be really interesting is what blows up the fastest. Maybe something like

(print `(progn ,@(loop repeat (length -) collect -)))

\$\endgroup\$
2
  • \$\begingroup\$ +1 For Common Lisp. Looks to me like you can skip ` and ,? \$\endgroup\$
    – daniero
    Commented Feb 26, 2014 at 13:55
  • \$\begingroup\$ @daniero Without the quotation, it just prints the original program, never growing. At that point, you may as well omit the or, too: (print -) \$\endgroup\$ Commented Feb 26, 2014 at 20:14
3
\$\begingroup\$

reticular, 11 bytes, noncompeting

"'34'coo1o;

This is the standard quine framework, except an extra 1 is printed after each iteration. Try it online!

First few outputs:

"'34'coo1o;
"'34'coo1o;1
"'34'coo1o;11
"'34'coo1o;111
\$\endgroup\$
3
\$\begingroup\$

05AB1E, 15 bytes, noncompeting

0"DÐ34çý"DÐ34çý

Try it online!

Prints 0"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý,
which prints 0"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý"DÐ34çý,
etc.

\$\endgroup\$
2
  • \$\begingroup\$ Both instances of DD can be replaced by Ð. \$\endgroup\$
    – Emigna
    Commented Nov 11, 2016 at 9:30
  • 4
    \$\begingroup\$ Why noncompeting? \$\endgroup\$
    – MD XF
    Commented May 25, 2017 at 20:10
3
\$\begingroup\$

J, 1 byte

'

Try it online!

The open quote gives, obviously, the open quote error:

|open quote
|   '
|   ^
|[-1] /home/runner/.code.tio

Note that, by the nature of J interpreter, the errors are printed to STDOUT, not STDERR.

When the above is run as code, it prints:

|open quote
|   |   '
|       ^
|[-2] /home/runner/.code.tio

Then

|open quote
|   |   |   '
|           ^
|[-2] /home/runner/.code.tio

and so on. Each time the code runs, the second line is left-padded with four bytes |   , fulfilling the requirement of this challenge.

J, proper quine variant, 25 bytes

echo,~'echo,:~(,quote)'''

Try it online!

Outputs

echo,:~(,quote)'echo,:~(,quote)'

And then it outputs itself twice, on separate lines:

echo,:~(,quote)'echo,:~(,quote)'
echo,:~(,quote)'echo,:~(,quote)'

then

echo,:~(,quote)'echo,:~(,quote)'
echo,:~(,quote)'echo,:~(,quote)'
echo,:~(,quote)'echo,:~(,quote)'
echo,:~(,quote)'echo,:~(,quote)'

and so on.

The first output is a simple variant of standard J quine. The added ,:~ concatenates itself vertically, where the resulting 2D array is printed as two rows of the same string.

\$\endgroup\$
3
\$\begingroup\$

Python 2, 38 37 36 34 bytes

-1 bytes thanks to Jo King

s='print"s=%r;exec s;"%s+s';exec s

Try it online!


Python 3.8 (pre-release), 39 38 36 bytes

exec(s:='print("exec(s:=%r);"%s+s)')

Try it online!

\$\endgroup\$
0
2
\$\begingroup\$

EcmaScript 6 (51 bytes):

(_=x=>'(_='+_+Array(x++).join(','+_)+')('+x+')')(2)

It produces a longer version of itself, which can produce a longer version of itself, which can produce a longer version of itself, etc. ...

\$\endgroup\$
2
\$\begingroup\$

Underload, 9 bytes

(:::aSSS):^

Modification on the standard underload quine, which is (:aSS):aSS. First, I changed the second :aSS into :^ since it still runs the same code (by duplicating and then running the top item of the stack, which is the stuff inside the bracket) whilst saving bytes. Then I added another S to make it so the program gets longer, and added two more :s to make it so the program it produces doesn't error.

Try it Online!

\$\endgroup\$
2
\$\begingroup\$

Japt, 12 bytes

"iQ ²Ä"iQ ²Ä

Try it online!

Based off the Standard Japt quine

Explanation

For first iteration

"iQ ²Ä"       // Take this string.        iQ ²Ä
       iQ     // Insert a quote.          "iQ ²Ä
          ²   // Double.                  "iQ ²Ä"iQ ²Ä
           Ä  // Concatenate 1 to end     "iQ ²Ä"iQ ²Ä1
              // Implicitly output.

On Each iteration it will concatenate another 1 to the end of the program

\$\endgroup\$
2
\$\begingroup\$

Jelly, 4 bytes

⁾ṘȮv

Try it online!

Originally written for a CMC in JHT.

⁾ṘȮ     "ṘȮ"
   v    evaluated with the argument
⁾ṘȮ     "ṘȮ".

 Ṙ      Print a Jelly string representation of the argument,
  Ȯ     print the argument,
        then implicitly print it again.
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.