n-level Ouroboros Quine

Question

This 128-language quine ouroboros (a program which outputs a program in another language, which outputs a program in yet another language, (125 languages later), which outputs the original program) is pretty impressive. But unfortunately, it has a static number of iterations.

Write a program which outputs a program (not necessarily in another language, but it can be), which outputs a program, which outputs a program, etc., which after n iterations, outputs the original program for the first time (i.e. no intermediate program should be the same as the original, because otherwise a quine which ignores its input would work), where n is a non-negative integer provided as input. The input can't be just a number in the original source code (e.g. putting x = <the value of n> at the start of your program), it should be one of the following:

1. Passed as a command-line argument
2. Read from standard input
3. Passed as an argument to a function, which returns/outputs the new program.

For intermediate stages in the ouroboros, your program can either be a fully-functioning program, or a function with no arguments, which, when called, will return/output the next one.

You may not read from the source file itself, or use any quine-like builtins (I don't think there are any which would do this, but there might be)

To be clear, if n = 0, the program should output its own source code.

If n = 1, the program should output a different program, which outputs the original source code.

And so on...

Fewest bytes wins!

Edit:

I should have written "For intermediate stages in the ouroboros, your program can either be a fully-functioning program with no input, or a function with no arguments". If your program outputs the next one in the chain, then waits for input, that's fine, but your program shouldn't need the original value of n.

Answer 1

05AB1E, 28 bytes

-4 bytes + fix thanks to Kevin Cruijssen

"34çìD«s<©di®ì"34çìD«s<©di®ì

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Explanation

This works by appending the remaining number to print to the front of the code, which means it's added to the stack the same way an input would be. In the base case of input of "0" it will not concatenate the 0 on the front.

Answer 2

Runic Enchantments, 39 bytes

"3X4+kSql1=c*?S:1-}C'LA-}!i:0)2*?~!'´@

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Would be 4 bytes shorter than 05AB1E using the same tactic if inputs where less than or equal to 10. But as we need to support arbitrarily large values, it gets more complicated.

Numerical value of n gets placed at the front and parsed as a continuous number literal using ´. "3X4+kSq is the basic quine code. If no value is present at the front, the stack length will only be 1 (the quine) otherwise 2 letting l1=d*? determine how to handle things from there.

• If there is a value, S:1-}'LA2+-} runs: swap the value to the top, subtract 1, duplicate it, leaving a copy on the bottom of the stack, get the Log₁₀ of that value times 100 (resulting in the value's own character length, plus 1 for the ´), remove that many characters from the end of the string (effectively truncating itself off the end, both where it isn't needed and because it has the wrong value). C is one byte shorter than 2+ and results in the same value.

• If there is not a value, read one from input.

Regardless: :0)2*? duplicate and compare with zero.

• If non-zero push ´.

• If zero, pop the value. We can cheat using ! instead of 2? and save a byte because when ´ tries to execute the first byte it sees is non-numeric and immediately drops out of number mode at the same position.

Print entire stack top to bottom.

Answer 3

Java 10, 145 bytes

n->{Long N=n;var s="n->{Long N=%s;var s=%c%s%2$c;return s.format(s,N>0?N-1+%2$cL%2$c:%2$cn%2$c,34,s);}";return s.format(s,N>0?N-1+"L":"n",34,s);}

Try it online and see the outputs of some iterations.

Explanation:

quine explanation:

• The var s contains the unformatted source code
• %s is used to put this String into itself with s.format(...)
• %c, %1$c, and 34 are used to format the double-quotes
• s.format(s,34,s) puts it all together

Challenge part:

The first lambda function takes a long input as parameter.

• This is set in variable Long N=n; in the first iteration. Or in Long N=%s; for next iterations.
• The ternary check N>0?N-1+"L":"n" will fill this %s with value N-1, appended with L since it's a long and to convert it to a String for the %s, if N is larger than 1. If N is 0 instead (initial input was 0 or this is the last iteration of the interquine-'loop'), it will fill this %s with the initial n instead.

Answer 4

Haskell, 195 164 bytes

main=putStr(x++show x++"\na=")>>(getLine:cycle[pure$show$a-1])!!a>>=putStr
x="main=putStr(x++show x++\"\\na=\")>>(getLine:cycle[pure$show$a-1])!!a>>=putStr\nx="
a=0

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This uses a pretty simple quine technique. We modify it with a variable a which is set to a number. If that number is zero (which it is at the start) we take input and output our source with a set to the input number. If a is not zero we output our source with a set to one less. This way a counts down to zero before outputting the original source.

Answer 5

Jelly, 22 20 bytes

“Ṫ_1xẸƲ;Ṿ⁾v,⁾¥ ”v,¥ 

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Try repeated calls of code

A full program that takes a single argument \$n\$. Main version outputs correctly to stdout but cannot be used as a function with a function return. The repeated calls version is 2 bytes longer but returns programmatically usable output. L

Answer 6

R, 92 bytes

f=function(n=-1){gsub("\\s","",paste("f=",sub("-?\\d+",n-1,paste(deparse(f),collapse=""))))}

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If deparse is deemed cheating, here’s an alternative:

R, 108 bytes

function(n=-1){s="function(n=%d){s=%s%s%2$s;sprintf(s,n-1,intToUtf8(34),s)}";sprintf(s,n-1,intToUtf8(34),s)}

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Answer 7

Python 3.8 (pre-release), 60 56 55 53 bytes

lambda n=-1,s='lambda n=%d,s=%r:s%%(~-n,s)':s%(~-n,s)

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-2 bytes thanks to Jo King

As of the 53 byte version, also works in Python 2 and Python 3.

Answer 8

Perl 6, 44 bytes

<say '<',S{\d+}=get||0-1,'>~~.EVAL'>~~.EVAL

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Takes input via standard input and returns a program where the only thing changed is the first number. Each subsequent program will then output the same program, except with that number decremented.

Answer 9

C# (Visual C# Interactive Compiler), 112 bytes

void f(int n=-1){var s="void f(int n={2}){{var s={0}{1}{0};Write(s,(char)34,s,n-1);}}";Write(s,(char)34,s,n-1);}

Saved a lot of bytes thanks to @NickKennedy!

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