Write a program that sees the New Year in itself

Question

Write the smallest program that maps strings injectively to outputs and maps itself to 2020.

To be clear, your program \$p\$ must have the following properties:

1. every possible input string \$s\$ has a well-defined output \$p(s)\$,
2. for every pair of strings \$s\$ and \$t\$ with \$s\neq t\$, it holds that \$p(s)\neq p(t)\$, and
3. \$p(p)=2020\$.

Be sure to explain why your submission satisfies properties 1 and 2. For property 3, the output can be a string or any type of number.

Edit 1: Programs need to be proper quine variants. Reading the source code is not allowed.

Edit 2: Note that as a consequence of property 2, it holds that \$p(s)\neq p(p)\$ for every \$s\neq p\$.

Answer 1

awk, 4 bytes

The structure of an awk program is pattern { action }. If pattern evaluates to true, { action } is performed. If { action } is omitted, the default is to output the current input record.

Code:

2020

As 2020 always evaluates to true, the program always prints it's input, therefore satisfying properties 1. and 2. When the input is the program itself it satisfies property 3. \$p(p)=2020\$ — even \$p(p)=p\$.

Some test cases:

$ echo 2020 | awk '2020'
2020
$ echo 0 | awk '2020'
0
$ echo "" | awk '2020'

$ echo foo | awk '2020'
foo

Would've been more glorious back in the good old days:

$ echo 1 | awk '1'
1

Answer 2

JavaScript (ES6), 23 bytes

Returns either the input string with a leading \$0\$, or \$2020\$ if the input is the program.

f=s=>s=='f='+f?2020:0+s

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JavaScript (Node.js), 76 bytes

Generates a unique BigInt by using each byte of the input string, then applies a XOR to the result in such a way that \$f(f)=2020\$.

f=s=>(g=s=>Buffer(s).every(b=>t=t<<8n|BigInt(b),t=1n)&&t)(s)^g('f='+f)^2020n

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Answer 3

Python 3.8 (pre-release), 81 60 59 57 56 bytes

-21 Thanks to Jo King

Outputs 2020 if the input matches the program, otherwise the input concatenated to itself 5 time (eg. abc becomes abcabcabcabcabc)

exec(a:='print(5*[i:=input(),404][i=="exec(a:=%r)"%a])')

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Python 2, 56 bytes

Outputs 2020 if the input matches the program, otherwise the input concatenated to itself 5 time (eg. abc becomes abcabcabcabcabc)

a='i=input();print 5*[i,404][i=="a=%r;exec a"%a]';exec a

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Python 3.8 (pre-release), 77 67 65 bytes

-10 Thanks to Jo King

Outputs 2020 if the input matches the program, otherwise the input concatenated to itself 5 time (eg. abc becomes abcabcabcabcabc)

lambda s:5*[s,404][s==(a:='lambda s:5*[s,404][s==(a:=%r)%%a]')%a]

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Answer 4

Ruby -n, 45 bytes

Based on one of the classic Ruby quines, _="_=%p;puts _%%_";puts _%_. Prints 2020 if the input is the source, otherwise prints the Ruby representation of the input (encapsulated in quotes, various characters like " and \ are escaped, etc.)

_="_=%p;p$_==_%%_ ?2020:$_";p$_==_%_ ?2020:$_

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Answer 5

Jelly, 16 bytes

“Ṿ;⁾v`³⁽¥ŒṾ⁼?”v`

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A full program that takes a string as its input. It outputs an unevaluated Jelly version of the input unless provided with itself, in which case it outputs 2020.

Thanks to @JonathanAllan for pointing out a flaw in my original version!

Explanation

 “Ṿ;⁾v`³⁽¥ŒṾ⁼?”v` | Evaluate as Jelly code the following, using itself as the argument:
  Ṿ               | - Unevaluate (effectively wraps string back into “”
   ;⁾v`           | - Append "v`"
       ³    ⁼?    | - If equal to the program’s argument:
        ⁽¥Œ       | - Then: 2020 (compressed integer)
           Ṿ      | - Else: Unevaluated version of the program’s original argument

Answer 6

Raku, 47 bytes

<say "<$_>~~.EVAL"eq($!=get)??2020!!@$!>~~.EVAL

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Outputs 2020 if the input matches the program, otherwise the input surrounded in brackets.

Answer 7

C (gcc), 190 196 194 192 180 176 bytes

Thanks to Jo King for pointing out a bug.

Trailing newline seems accepted in some other answers, so will use that to fix a bug pointed out by NieDzejkob. (Also -12 bytes thanks to the same.)

-4 bytes thanks to @ceilingcat.

Assumes little-endian, ASCII, sizeof(int) >= 4, all the good stuff.

Outputs the input string preceded by a dollar sign, unless given its own source code, in which case it outputs 2020.

*s="*s=%c%s%1$c,u[99]={'0202'};f(int*t){puts(bcmp(u+2,t,sprintf(u+2,s,34,s))?putchar(36),t:u);}",u[99]={'0202'};f(int*t){puts(bcmp(u+2,t,sprintf(u+2,s,34,s))?putchar(36),t:u);}

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Answer 8

05AB1E, 28 bytes

"34çìD«QiŽ7ìëû"34çìD«QiŽ7ìëû

Try it online or verify a few more test cases.

Explanation:

"34çìD«QiŽ7ìëû"            # Push this string
               34ç         # Push 34, and convert it to a character: '"'
                  ì        # Prepend it in front of the string
                   D«      # Append a copy of the string to itself
                     Qi    # If it's equal to the (implicit) input:
                       Ž7ì #  Push compressed integer 2020
                      ë    # Else:
                       û   #  Palindromize the (implicit) input
                           # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ž7ì is 2020.

Answer 9

J, 56 bytes

".a=.}:0 :0
echo@>2020[^:(]-:'".a=.}:0 :0',LF,a&[)1!:1[3

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A full program. Unless provided with itself prints each character on a separate line.

Answer 10

Java 10, 128 bytes

s->{var t="s->{var t=%c%s%1$c;t=t.format(t,34,t);return t.equals(s)?2020:0+s;}";t=t.format(t,34,t);return t.equals(s)?2020:0+s;}

Try it online.

Explanation:

quine part of the explanation:

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

Challenge part of the explanation:

• t.equals(s) checks if the input-String is equal to the source code
• ?2020: if it is, return 2020 as result
• :0+s: if not, return the input-String with a prepended 0 as result instead