Produce the number 2014 without any numbers in your source code

Question

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It's 2017 2018 2019 2020 2021 2022 2023 2024 already, folks, go home.

Woo, 10 years of this challenge!

So, now that it's 2014, it's time for a code question involving the number 2014.

Your task is to make a program that prints the number 2014, without using any of the characters 0123456789 in your code, and independently of any external variables such as the date or time or a random seed.

The shortest code (counting in bytes) to do so in any language in which numbers are valid tokens wins.

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Leaderboard:

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Answer 1

GolfScript, 10 8 7 chars

This solution contains non-printable characters. As xxd output:

0000000: 2714 0e27 7b7d 2f                        '..'{}/

As base 64:

JxQOJ3t9Lw==

Unpacks the ASCII codes for characters 20 and 14 and prints those numbers without any separation.

To actually generate the number 2014, I'm currently at 8 chars:

0000000: 2713 6a27 7b2a 7d2a                      '.j'{*}*

JxNqJ3sqfSo=

Takes a string containing characters with ASCII values 19 and 106 and multiplies them to get 2014.

Answer 2

Haskell, 23 characters, 24 bytes

main=print$fromEnum 'ߞ'

That's U+07DE N'Ko letter KA

Answer 3

EXCEL, 24 characters:

=COLUMN(T:T)&COLUMN(N:N)

Answer 4

Mathematica, 47 bytes

Total@ToCharacterCode@"Happy New Year for all!"

2014

Answer 5

Mathematica (24)

Found by randomly generating expression trees. TraditionalForm chops 5 characters for the floor symbol but I think rich text doesn't count.

Floor[π E(Sin@E+E^E)E^E]

Answer 6

Mathematica, 23 bytes

Tr@ToCharacterCode@"ߞ"

(Using the 2014 unicode char. Credit for Tr goes to alephalpha)

Mathematica, 46 bytes (pure math)

a=⌊E⌋;b=⌈E⌉;c=⌈Pi⌉;a^(b*c-(c-b))-a*a^c-a

idea taken from here.

Answer 7

I and others have posted shorter Mathematica solutions to this challenge, but here are two methods that I don't believe have been tried in any of the other examples, in any languages.

Mathematica, 27 characters / 27 bytes

FromDigits["MMXIV", "Roman"]

Mathematica, 37 characters / 37 bytes

(ToString /@ FromDigits /@ {"K", "E"}) <> ""

and the following 58 character variation

FromDigits[Flatten[IntegerDigits /@ FromDigits /@ {"K", "E"}]]

Answer 8

D

ϯ has the numerical value 1007 in Unicode encoding.

as compiler message during compilation (20 chars)

pragma(msg,'ϯ'+'ϯ');

runtime version (45 chars)

import std.stdio;void main(){write('ϯ'+'ϯ');}

Answer 9

Java (68)

Solution 4: 68 chars (Thanks @radiodef))

class C{public static void main(String[]a){System.out.print(+'ߞ');}}

Solution 3: 69 chars (Thanks @Kamran)

class C{public static void main(String[]a){System.out.print(~~'ߞ');}}

Solution 2: 71 chars

class C{public static void main(String[]a){System.out.print('޾'+' ');}}

Solution 1: 72 chars

class C{public static void main(String[]a){System.out.print((int)'ߞ');}}

<3 Unicode.

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Solution Derivation Technique: Copy the output from the following line and convert that character back to int for your a working solution.

System.out.println((char)2014));

Answer 10

J - 11 bytes - Base arithmetic

+:_bj*p:_bf

Explanation

_bf and _bj are 15 and 19 in infinite base, therefore p:_bf is 15th prime, which is 53. The result is then 19*53 doubled, which is 2014.

J - 31 28 bytes - Math

(((!>:&+:)-(!>:))+:&+:)>.^*_

Explanation

The implementaion of A004126 sequence from OEIS where I searched for the ways to calculate 2014. Basically it's C(2*n+1,3)-C(n+1,3) where n=12.

This means ceiling(exp(signum(infinity))), which is 3

>.^*_

To get 12 we double it two times.

+:&+:

Everything else is a combo of a dyadic fork and monadic hook. You can read about them here.

Other's solution in J

+/(>.^*_)&u:'Happy new year to you!'

Which normally is +/3&u:'Happy new year to you!', it's a shame we need a number to work with strings.

It was fun to remember J :).

Answer 11

JavaScript 59, 44, 36, 29

a=+'';b=''+a+++a++;a+++b+ ++a

_{Assuming alert can be skipped due to implicit return}

---

a=+'';b=''+a+++a++;alert(a+++b+ ++a)

---

alert([-~-~'',+'',-~'',-~-~-~-~''].join(''))

---

_{Those are l characters not 1}

l='length';alert([[,,][l],[][l],[,][l],[,,,,][l]].join(''))

Answer 12

Python 3, 9 bytes

license()

Try it online!

unfortunately this does not work on tio.run but it does work on my device

license() stdouts text which luckily contains 2014 :)

...
2. Subject to the terms and conditions of this License Agreement, PSF hereby
grants Licensee a nonexclusive, royalty-free, world-wide license to reproduce,
analyze, test, perform and/or display publicly, prepare derivative works,
distribute, and otherwise use Python alone or in any derivative version,
provided, however, that PSF's License Agreement and PSF's notice of copyright,
i.e., "Copyright (c) 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010,
2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, 2020, 2021, 2022 Python Software Foundation;

All Rights Reserved" are retained in Python alone or in any derivative version
prepared by Licensee.
...

Answer 13

Perl 6, 37 bytes

This is not my actual solution (the real solution is in PHP, and is shorter), but I decided to make shorter version of Python's solution, because it's a really interesting idea.

say [+] ords 'Happy new year to you!'

Try it online!

So, what does it do? Well, it gets list of character positions (ords function), sums them together ([] is a reduce operator, which takes the exact operator between square brackets, in this case +), and say prints the result with new line. Could be WAY shorter by abusing ~^ prefix operator (which isn't implemented in Rakudo Perl) or Unicode characters, but this is just for fun solution.

Perl 6 is a quite an interesting language, even if it's not as good as Perl 5 in most code golf tasks (because of mandatory whitespace in many situations, and generally less DWIM). However, in this case, because of builtin sum and ords, it wins with Perl 5.

Answer 14

Golfscript - 11 10 chars

'&~I'{}/-*

computes 38 * 53 (ie an expansion of the prime factorization of 2014: 2 * 19 * 53)

---

Old Version

'f.'{}/.*\-

This computes 46²-102.

'f.'{}/ puts the ascii values of f and . on the stack.
.*      squares the value of the .
\-      swaps the two values, then subtracts

Answer 15

AutoHotkey 31/13

send % ++(z:=true) z-z true z+z

This year special solution

send % A_year

Answer 16

R using roman numbers, 29 characters

as.numeric(as.roman('MMXIV'))

Answer 17

Raku (Perl 6), 12 bytes

say 'ߞ'.ord

Answer 18

GolfScript, 10 bytes

I can't beat Quincunx's 10-char plain ascii solution but I can match it:

'N,,'{}/*+

Try it online!

(it calculates 44*44+78)

GolfScript, 26 bytes

And now a different approach :)

'codegolfing is yummy'{+}*

Try it online!

Answer 19

JavaScript:

alert(+!![]+!![]+[+[]]+ +!![]+(+!![]+!![]+!![]+!![]));

---

each +!![] is 1 (![] is false, !false is true and +true is 1)

+[] is 0, and [0] is "0"

so this turns into 1+1+"0"+1+(1+1+1+1)

Answer 20

Python 3 (17 bytes)

print(ord("ߞ"))

Python 2 (50 bytes)

s,t=str,True;print s(t+t)+s(t-t)+s(t/t)+s(-~t<<t)

Answer 21

guess this is already beaten but I like the approach:

<?php
$three=round(pi());
$one=pi()/pi();
$two=$three-$one;
$four=$three+$one;
$zero=$one-$one;
echo $two.$zero.$one.$four;
?>

Answer 22

Ruby — 8 characters

p ?ߞ.ord

where ߞ is Unicode character U+07DE — N'Ko letter ka.

This is what it looks like when I copy and paste the above line and run it in my Terminal. Notice the character does not display:

Answer 23

C#, 60 characters

class a{static void Main(){System.Console.Write(- -'ߞ');}}

Thanks to mleise for the original answer for D!

Note: space between hyphens is important

Answer 24

T-SQL, 50 40

My original entry seems too straightforward. How's this?

PRINT RIGHT(CHECKSUM(',,.'),LEN('````'))

(Disclaimer: Given Steve Matthews's comment, the above may be dependent on configuration.)

Here's the SELECT version in SQL Server 2012.

Original entry:
This may look obscene in more ways than one, but it's valid.

DECLARE @ INT=\PRINT CONCAT(-~-~@,@,-~@,-~-~-~-~@)

Try it in SQL Server 2012. Here is a version using SELECT.

Explanation: "\" is a valid currency symbol. (Do a find for "String to" here.) If you assign just a currency symbol to an INT variable, you store 0. Also, "@" is a valid variable name. "~" is bitwise NOT, and "-" is negative. If you negate a NOT-ed INT, you get the INT plus 1. So, repeat "-~" until you make @ into the digit you need. Then CONCAT() your digits.

Answer 25

Javascript (29 chars)

t=!'',a=t+t+'';a+!t*t+t*t+a*a

Not the shortest but some boolean logic

Answer 26

PHP: 54 chars

Slightly different approach, albeit long, using concatenation, and not using ordinal values:

<?=floor(M_E).floor(M_EULER).ceil(M_EULER).ceil(M_PI);

Answer 27

PHP - 13 bytes

<?=m_n_^_o_k;

A fairly trivial ASCII-only solution.

Other variants:

<?=Q__Q^cone;
<?=VonV^d__b;
<?=PAAP^bqpd;
<?=BEAU^pupa;

etc.

Answer 28

Bash 28

echo $((`printf %d \':`#yG))

Similar to the Ruby and Python solutions, but uses base62, now without literal 6 and 2 (but pretty long...)! Using base58 as recommended by @DigitalTrauma along with removing quotes around the printf format string, Thanks!

Better 17 (Thanks @user13955 and @DigitalTrauma!)

tr a-l /-:<<<dbcf

"It's shorthand for tr abcdefghijkl /0123456789:. Not very geeky, I'm afraid." -user13955 Also updated to use a heredoc instead of echo, thanks @DigitalTrauma!

Answer 29

Japt, 3 bytes (2 chars)

Japt is newer than this competition (created in 2015), but was not created specifically to answer it.

#ߞ

That's right. 2 chars. Explanation:

#   // Char-code of next character
 ߞ  // Character with char-code of 2014
    // Implicit output

Try it online!

Answer 30

C, 33 30 bytes

Numerical Solution 1

f(){printf("%d",'<'*'#'-'V');}

// 2014 = 60 * 35 - 86

---

C, 86 83 bytes

Numerical Solution 2

#define A ((int)'}')
#define B (((int)'r')-((int)'d'))
f(){printf("%d",A*B+A+A+B);}

// A = 125
// B = 14
// 2014 = 125*14 + 125 + 125 + 14

---

C, 53 50 bytes

ASCII Art 1

f(){printf("┌┐ ┌┐ ┐ ┐┌\n┌┘ ││ │ └┤\n└┘ └┘ ┴  ┴");}

Result

┌┐ ┌┐ ┐ ┐┌
┌┘ ││ │ └┤
└┘ └┘ ┴  ┴

---

C, 94 91 bytes

ASCII Art 2

f(){printf("╔═╗ ╔═╗ ╗ ╦ ╦\n  ║ ║ ║ ║ ║ ║\n╔═╝ ║ ║ ║ ╚═╣\n║   ║ ║ ║   ║\n╚═╝ ╚═╝ ╩   ╩\n");}

Result

╔═╗ ╔═╗ ╗ ╦ ╦
  ║ ║ ║ ║ ║ ║
╔═╝ ║ ║ ║ ╚═╣
║   ║ ║ ║   ║
╚═╝ ╚═╝ ╩   ╩