12
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In this challenge, your past self comes back to help or hurt someone else!

The task is simple: write a full program, not a function, that prints the first code submission on this site submitted by the previous user, in the language of your first answer. (In the case that your first answer was purely prose, for example pxeger's first answer to a "Tips for Golfing" thread, use the first code-based answer.)

My first answer was


*/```

That's a newline followed by */```. Suppose that Doorknob was the first to answer this question. His first answer was in JavaScript, so he might respond to this challenge

alert("\n*/```")

And the next answerer here would have to print his first answer,

s='';c=5;r=Math.random;while(c--)s+='bcdfghjklmnpqrstvwxz'[r()*20|0]+'aeiouy'[r()*6|0]

Only accounts that submitted an answer before this question was posted are eligible to participate. Each user may submit only one answer.

This is code golf, so the shortest answer wins. The fun is that who can win depends on who answers!

\$\endgroup\$
15
  • 5
    \$\begingroup\$ The idea looks interesting on paper, but seeing as most people's first submissions were uninteresting and non-golfy, it would be hard to find suitable patterns there. Also, first submissions are usually to easier challenges. So far, every answer was just of the form print(<code>) with print being replaced with an output command. Seeing as my first submission was extremely non-golfy and undesirable by my standards (spaces, curly braces, and a for(i=0;i<Infinity;i++) loop (way too long) this challenge is a little boring, is it not? Unless someone has an interesting submission to print. \$\endgroup\$ – Recursive Co. May 30 at 9:25
  • 3
    \$\begingroup\$ @ophact - I agree. And, even worse, there's a FGITW (fastest gun in the west) element to it, too, which forces one to rush-out an answer, pretty-much guaranteeing that every answer will be (at least initially) boring. I was thinking for a bit how to nicely golf hyper-neutrino's 1055-byte first submission (this would have involved golfing a text compression in R), but now it's been answered (as well as several more), so I've missed that chance to submit something non-trivial... \$\endgroup\$ – Dominic van Essen May 30 at 15:39
  • 4
    \$\begingroup\$ I consider all of the print(<code>)-type answers to be low-quality, and have downvoted them. No malice is intended, and I would be happy to rescind these downvotes if any of the authors can explain why they are non-trivial or otherwise useful. \$\endgroup\$ – Dominic van Essen May 30 at 17:16
  • 2
    \$\begingroup\$ How to address proprietary languages (like Matlab or Excel) that the user can't access anymore? My first answer was in Matlab and I don't have the license now. \$\endgroup\$ – pajonk May 31 at 6:36
  • 2
    \$\begingroup\$ My first answer was a polyglot. What language should I use? Should I create a polyglot in all the languages my first answer was in? \$\endgroup\$ – Command Master May 31 at 7:26

17 Answers 17

5
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5. user, Java (OpenJDK 8), 75 bytes

interface T{static void main(String[]a){System.out.print("print(\"N\")");}}

Try it online!

My first language was Java, lol. This is my first answer. It's 1055 bytes. Have fun.

interface T{static void main(String[]a){System.out.print("The basic idea of a [tag:kolmogorov-complexity] challenge is to print a set output in the shortest code (though that meaning has changed now). We have had many of them, from [Borromean Rings](_%s/53417/ascii-borromean-rings) to [The Gettysburg Address_%s/15395/how-random-is-the-gettysburg-address). \n##Your Task\nThis % is similar, except that it requires printing of a special text - the text of this %. Specifically, the very Markdown code that I am typing right now. \nTo prevent an infinite recursion in the %, the exact text you have to print can be found [here_revisions/1f731ea3-0950-4b03-ae95-24fa812a9a28/view-source). \n##Clarifications\n- You are not allowed to use any external sources like the internet. \n- This means that the purpose of this % is not to download this % text and parse it, but instead to store it in you program. \n- This is [tag:code-golf], so shortest code in **bytes** wins!".replaceAll("_","])http://codegolf.stackexchange.com/").replaceAll("%","question"));}}
\$\endgroup\$
2
  • 6
    \$\begingroup\$ I...I think I'll wait for the next one... \$\endgroup\$ – EasyasPi May 29 at 22:29
  • 1
    \$\begingroup\$ I'm not going to post this as answer, because it would be quite boring, but _=>`<code>` is valid in JavaScript, where <code> is hyper-neutrino's first submission \$\endgroup\$ – Recursive Co. May 30 at 8:48
3
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10. Neil - GolfScript and Funge-98, 154 153 151 147 146 bytes

#.c5*01:$pd-1x
 v "var isSorted = function(array) { return \""'"'" + array == array.sort(function(a, b) { return a - b; }); }"
#>:'\-#^_$
#^@#,  _

My first answer was:

#>&:*.@
~2-1??
\$\endgroup\$
2
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1. Purple P, √ å ı ¥ ® Ï Ø ¿, 9 bytes

'
*/```'W

Try it online!

This was my first answer:

Ißo

Ignore the OUTPUT and Program execution information banners. Those are automatically printed by the interpreter when any program in √ å ı ¥ ® Ï Ø ¿ is executed.

It's been a while since I used √ å ı ¥ ® Ï Ø ¿ (or remembered that it existed). This pushes the string


*/```

to the stack, then outputs it with W

\$\endgroup\$
2
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6. hyper-neutrino, Brain-Flak (BrainHack), 19516 bytes

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Try it online!

This is a bit of a placeholder, since golfing brain-flak is a lot of work and I'd rather not lose all my progress since someone else posted an answer. The current code was generated by this hyper-neutrino answer.

My first answer was:

(<({}<>)>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}<>(({}<{}>)){{}{}(<(())>)}{}

Should present a interesting challenge.

\$\endgroup\$
1
  • \$\begingroup\$ For once, someone has an interesting first submission! \$\endgroup\$ – Recursive Co. May 30 at 11:47
2
\$\begingroup\$

7. Wheat Wizard♦, MATLAB/Octave, 94 bytes

disp('(<({}<>)>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}<>(({}<{}>)){{}{}(<(())>)}{}')

Try it online!
Meh, a no-brainer. Didn't find any smarty way to golf it.

My first answer was:

ezpolar(@(x)r);s=r+5;axis([-s,s,-s,s])

(it was later edited but if I understand correctly we're supposed to do the original first answers)


Alternative solution, trying to golf, but is longer (110 bytes)

e='$&\IIEPW0+,H&X*&Y9.W$&XW)I^\OE WF])V^X@!-/W'-32;c='()[]<>{}';disp(c(reshape([floor(e/8);mod(e,8)],1,[])+1))

Try it online!
Each character in$&\IIEPW0+,H&X*&Y9.W$&XW)I^\OE WF])V^X@!-/W represents two characters to display (more exactly their indices in ()[]<>{}). Needed to add 32 to chars (and subtract them in the decoding process) to avoid using non-printable characters that render the code unrunnable.

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4
  • \$\begingroup\$ I don't know matlab but base 8 encoding it might be able save you if you can write a shortish decoder. \$\endgroup\$ – Wheat Wizard May 30 at 13:35
  • \$\begingroup\$ @WheatWizard I was thinking about it but it probably would end up much longer than straightforward solution. At least I can't think of a way to do it shorter. \$\endgroup\$ – elementiro May 30 at 13:54
  • 1
    \$\begingroup\$ Naively, if you replace each character of the answer with the smallest possible octal digit (10123045450421236157504112315612357552231012312355453230430411230234552232310115545323) you get a number that takes 64 hex digits to represent (0x82989651114f1be884a66e29df6a4c82994ed95a6231094c272d49a64136cad3). Is 30 bytes enough space to "convert to base 8 and replace every character"? \$\endgroup\$ – Purple P May 31 at 2:02
  • \$\begingroup\$ @PurpleP it isn't. You would have to write your own converter. My alternative solution is IMO better than what you propose since it converts Wizard's answer to number in base 64 (each letter holds 2 octal digits) - and converter for this base is much simpler (and thus shorter) since you can convert it letter by letter. \$\endgroup\$ – elementiro May 31 at 18:58
2
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12. tail spark rabbit ear, Vyxal, 6 bytes

kH‛.p+

Try it Online!

Explanation

kH       # Hello, World! pre-defined constant
‛        # Two byte string literal
.p       # Capture next two bytes: .p and push to stack
+        # Concatenate top two items on stack
          (implicit output of top of stack)

Prints Hello, World.p, couldn't waste the big chance and I am winner until now!!!

My first answer, 1.."$args"-match"2$"

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3
  • 4
    \$\begingroup\$ Isn't your first answer in powershell and the answer need to be in the same language as your first answer? I think I'm missing something \$\endgroup\$ – Command Master May 31 at 7:33
  • 1
    \$\begingroup\$ same as command's. \$\endgroup\$ – tail spark rabbit ear Jun 1 at 10:49
  • \$\begingroup\$ Also isn't your first submission this, rather than that? \$\endgroup\$ – tail spark rabbit ear Jun 2 at 9:14
2
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9. Nick Kennedy - JavaScript, 136 bytes

function(){return'x=function(){rev(rawToBits(rev(charToRaw(sprintf("x=%s;x()",gsub("\\\\s","",paste(deparse(x),collapse="")))))))};x()'}

My first answer was :

var isSorted = function(array) { return "" + array == array.sort(function(a, b) { return a - b; }); }
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2
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14. SjoerdPennings, JavaScript (Node.js), 310 302 297 bytes

alert(`def f(i):
i=i.split(" ")
print i[0],i[2],
for f in i[0:3]: print f,
print ""
for t in["TT","TF","FT","FF"]:
p,q=t[0],t[1]
y = t[0]+" "+t[1]
if i[1]=="^${a='": r=(False,True)[p'}==q]
if i[1]=="v${a}!=q]
if r: y+="   T"
else: y+="   F"
print y`.replace(/\n/g,(b,i)=>b+' '.repeat(4+(i>90)*4)))

Try it online!

My first answer, 68 bytes

l=(a,b,c)=>b.startsWith(a.slice(c=~~c,-1))?a.slice(0,c)+b:l(a,b,++c)
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2
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16. Aaron Miller, AArch64 machine code, 41 bytes

Machine code dump (xxd):

00000000: 20 00 80 52 c1 00 00 10 a2 01 80 52 08 08 80 52   ..R.......R...R
00000010: 01 00 00 d4 a8 0b 80 52 01 00 00 d4 3a 2e 5c 21  .......R....:.\!
00000020: 2b 3a 22 64 22 60 23 40 5f                       +:"d"`#@_

Assembly source:

        .text
        .globl _start
_start:
        // write(1, .Lstr, strlen(.Lstr))
        mov     w0, #1
        adr     x1, .Lstr
        mov     w2, #13
        mov     w8, #64 // SYS_write
        svc     #0
        // exit(dontcare)
        mov     w8, #93 // SYS_exit
        svc     #0
.Lstr:
        .ascii ":.\\!+:\"d\"`#@_"

Given how all instructions are 4 bytes long, there isn't any point in doing anything but a direct write.

My first submission (xxd)

00000000: e4 03 27 1e 00 44 40 bc 01 58 20 0e 21 38 30 2e  ..'..D@..X .!80.
00000010: 23 3c a4 0e 02 1c a3 2e 24 1c a3 2e 21 04 00 f1  #<......$...!...
00000020: 21 ff ff 54 40 00 26 1e c0 03 5f d6              !..T@.&..._.

You must emit the raw bytes, not the assembler source or the hexdump. 😏

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1
  • \$\begingroup\$ I have officially killed this chain \$\endgroup\$ – EasyasPi Jun 12 at 18:44
1
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2. caird coinheringaahing - Python 3, 13 bytes

print("Ißo")

Try it online!

My first (code-based) answer is long and high-entropy, so good luck :P

import zlib;lambda x:filter(33 .__ne__,zlib.decompress(b'x\x9cKJM-PH\xc2A(\x92\xc7\xa26\x97nb4!\0hm{7')[:x*5])
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1
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4. A username, Scala, 141 bytes

_=>"<?php $w=0;foreach(str_split($argv[1])as$v){if($v=='o'){echo$w.' ';}else{$w=$v=='s'?$w**2:($v=='i'?++$w:--$w);}if($w==256||$w<0){$w=0;}}"

Try it in Scastie!

My first answer

print("N")

Simple string replacement can't be done here, as there aren't any patterns. Using GZIPOutputStream and stuff gets really long just from the names of the classes involved, and the string's too short for the compression algorithm to work its magic. In contrast to A username's first answer, mine is really short and easily printed.

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1
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3. pxeger, PHP, 110 bytes

import zlib;lambda x:filter(33 .__ne__,zlib.decompress(b'x\x9cKJM-PH\xc2A(\x92\xc7\xa26\x97nb4!\0hm{7')[:x*5])

Try it online!

My first answer was this:

<?php $w=0;foreach(str_split($argv[1])as$v){if($v=='o'){echo$w.' ';}else{$w=$v=='s'?$w**2:($v=='i'?++$w:--$w);}if($w==256||$w<0){$w=0;}}
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3
  • \$\begingroup\$ :/⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ \$\endgroup\$ – Makonede May 29 at 21:40
  • 1
    \$\begingroup\$ @Makonede What? It's valid PHP, see the Hello world. \$\endgroup\$ – emanresu A May 29 at 21:43
  • \$\begingroup\$ I know. But this is going to end badly for the next person to answer :/ \$\endgroup\$ – Makonede May 29 at 21:44
1
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8. elementiro R, 45 bytes

cat("ezpolar(@(x)r);s=r+5;axis([-s,s,-s,s])")

Try it online!

A rather boring program that just cats the previous person’s first answer. My first answer was this one:

x=function(){rev(rawToBits(rev(charToRaw(sprintf("x=%s;x()",gsub("\\s","",paste(deparse(x),collapse="")))))))};x()
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1
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13.Wasif, Python 2, 27 bytes

EDIT: My sleepy brain didn't notice the "language of your first answer" rule, it has been fixed at the cost of 5 bytes.

Just a simple print. There's not enough of a pattern to try replacements or formatted strings, compressed versions of the strings are longer than the original, and since this is Python 2, using exit('text') won't save me a byte. As far as I know, the best way to golf this is to simply print it, and remove the space between print and the string.

print'1.."$args"-match"2$"'

Try it online!

My first answer is:

def f(i):
    i=i.split(" ")
    print i[0],i[2],
    for f in i[0:3]: print f,
    print ""
    for t in["TT","TF","FT","FF"]:
        p,q=t[0],t[1]
        y = t[0]+" "+t[1]
        if i[1]=="^": r=(False,True)[p==q]
        if i[1]=="v": r=(False,True)[p!=q]
        if r: y+="   T"
        else: y+="   F"
        print y
\$\endgroup\$
1
  • \$\begingroup\$ I didn't notice the "language of your first answer" rule, it has now been fixed. \$\endgroup\$ – SjoerdPennings May 31 at 7:30
1
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11. Ross Long, Pxem, 0 (content) + 17 (filename).

Since it was 15 bytes, terminating LF is also output.

Filename is as follows:

#>&:*.@
~2-1??
.p

Content is empty.

Try it online!

My first post:

  • Filename: Hello, World!.p
  • Content: empty
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1
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Alex bries, Befunge-93, 79 bytes

")c++,b,a(l:b+)c,0(ecils.a?))1-,c~~=c(ecils.a(htiWstrats.b>=)c,b,a(=l">,# :# _@

Try it online!

My first answer:

:.\!+:"d"`#@_
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1
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17. EasyasPi, Python 3 2, 110 bytes

I realised that my first answer was written in Python 2 after I had posted. Here is the updated and golfed version.

import os,base64 as b
os.write(1,b.b64decode("5AMnHgBEQLwBWCAOITgwLiM8pA4CHKMuJByjLiEEAPEh//9UQAAmHsADX9Y="))

This was my first answer:

n='\n'
p,a='p'*8+n,'rnbqkbnr'+n
print a+p+('.'*8+n)*4+(p+a).upper()
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1
  • \$\begingroup\$ I also just realised that this wasn't the python I was looking for. Will have to change that. \$\endgroup\$ – Lord Ratte Jul 19 at 21:45

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