579
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Note to challenge writers as per meta consensus: This question was well-received when it was posted, but challenges like this, asking answerers to Do X without using Y are likely to be poorly received. Try using the sandbox to get feedback on if you want to post a similar challenge.


It's 2017 2018 2019 already, folks, go home.

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.


Leaderboard:

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  • 10
    \$\begingroup\$ Even though numbers are ignored in brainfuck, I thought I'd post one anyway. 32 Chars: ++++++[>++++++++<-]>++.--.+.+++. \$\endgroup\$ – Braden Best Apr 1 '15 at 21:37
  • 4
    \$\begingroup\$ Brainfuck isn't a valid language for this challenge. \$\endgroup\$ – Joe Z. Apr 1 '15 at 22:49
  • 8
    \$\begingroup\$ I wonder if this question gets a small spike in popularity around New Year's. \$\endgroup\$ – Joe Z. Dec 26 '15 at 23:28
  • 3
    \$\begingroup\$ Waiting for "Come on folks, don't you realize it's 2016?" :) \$\endgroup\$ – padawan Jan 4 '16 at 23:35
  • 1
    \$\begingroup\$ @BradenBest It's possible to do it in 31 characters in at least two different ways: +++++++[>+++++++<-]>+.--.+.+++. and ++++++++++[>+++++<-]>.--.+.+++. \$\endgroup\$ – Zubin Mukerjee Feb 21 '16 at 17:47

274 Answers 274

0
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k 11 & 17 chars

this 11-char one is from a colleague

@:[:']*.(.)

this 17-char one is the best i could come up with on my own; it's a variant on ASCII abuse:

.,/$-/"i"$$`zz`fl

and just for interest, here are a few others of mine:

."c"$"RPQT"-"e"$" " / ascii (only 2.x)
."c"$-/"e"$("RPQT";" ") / ascii (all versions)

-_-(s*(exp acos@-`=`)xexp x)-(s xexp s:x*x)%x:+/``=`` / port of David Carraher's solution above

.,/$#:'(``;();`;````) / another approach
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  • \$\begingroup\$ @David Carraher my latest improvements on my port of your algo--two implementations, both 39 chars: -_-(xpp:exp acos@-#)-f*x*x*x:f*f:#`` f+(x*_p*p:exp acos@-#)-fxxx:ff:#`` (i don't have the rep to comment on your post yet) \$\endgroup\$ – Aaron Davies Jan 10 '14 at 2:02
0
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Delphi (26bytes & 26 chars)

ord('-')*ord('/')-ord('e')

Ascii values
- : 45
/ : 47
e : 101
45*47 = 2115 - 101 = 2014

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0
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You never said we couldn't put it on an external page!

Javascript - 19 Chars

location='//x.vu/u'

PS. It took 2 tries to get a shortened URL without numbers :P

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  • \$\begingroup\$ Yes, but sadly your code is far from being the shortest. :P \$\endgroup\$ – Joe Z. Jan 1 '14 at 18:26
  • 1
    \$\begingroup\$ @Joe Z. Right, you could include an HTML script src instead :P \$\endgroup\$ – Cilan Jan 1 '14 at 23:10
  • \$\begingroup\$ location.href.match(/-(\d+)-/).pop() \$\endgroup\$ – Alf Eaton Jan 3 '14 at 9:18
  • \$\begingroup\$ location='http://goo.gl/miVwHe' would be shorter in my opinion. \$\endgroup\$ – Konrad Borowski Jan 4 '14 at 11:23
  • \$\begingroup\$ @xfix Thanks for the tip, I even removed 'http:' :) \$\endgroup\$ – Cilan Feb 8 '14 at 19:24
0
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e-TeX, 21 bytes

\the\numexpr`*`j\bye

It contains an invisible control character with code 19 (0x13) before the asterisk. A version with printable ASCII characters needs two more bytes:

\the\numexpr`^^S*`j\bye

In TeX ` takes the character code of the next token:

  • [0x13] (^^S): 19
  • j: 106

\numexpr calculates: 19 * 106 = 2014

The result is a DVI file with "2014" on the first page.

Variant with 2014 as page number:

\pageno\numexpr`^^S*`j~\bye

(25 bytes, if ^^S is replaced by the byte with character code 19).

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0
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Java without the weird unicode charas @ 115

enum A{A;public static void main(String[]z){int a=A.ordinal(),b=a++;System.out.print(""+(a<<a)+b+a+(a<<(a<<a)));}}

enumerators are pretty handy :)

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0
\$\begingroup\$

GTB, 38

π/π→A:A++B:A+A→C:C*C→D~B*C*C+B*C-C-A-A

Compile assuming : at front

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  • \$\begingroup\$ π is 2 bytes and is 3, bringing the byte count to 49. \$\endgroup\$ – Joe Z. Jan 7 '14 at 2:18
  • \$\begingroup\$ @JoeZ. All characters at tibasicdev.wikidot.com/83lgfont are 1 byte. \$\endgroup\$ – Timtech Jan 7 '14 at 12:10
0
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Python, 8 characters

ord('ߞ')

I think this ought to be valid :P

Ok, if you want me to use print(),

15 characters

print(ord('ߞ'))
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0
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Cardinal, 14 chars

%n=*+.-d++.
 d

The pointer starts at %, reads in the character ‘d’(ASCII 100), creates a duplicate as inactive value (the bottom of the stack), adds active and inactive values, resulting in 200, adds 1, prints out the result 201, subtracts one, divides by inactive value (100), resulting in 2, adds two, prints out 4. Cardinal pointer stacks can only carry values up to 255 (OEM 437 range), everything above leads to a wrap-around.

0             100             100             200             201
0               0             100             100             100
>n=*+.-d++.    %>=*+.-d++.    %n>*+.-d++.    %n=>+.-d++.    %n=*>.-d++.
 d              d              d              d              d
———————————————————————————————————————————————————————————————————————
   201             200               2               3               4
   100             100             100             100             100
%n=*+>-d++.    %n=*+.>d++.    %n=*+.->++.    %n=*+.-d>+.    %n=*+.-d+>.

print:"201"
———————————————————————————————————————————————————————————————————————
          4
        100  
%n=*+.-d++>

  print:"4"

Result:

Executing program..

2014

Execution complete.
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0
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C, 24 / 68

main(){printf("MMXIV");}

Haha, Roman Numerals For The Win!


Or, for real:

main(){printf("%i%i%i%i",strlen("aa"),nil,strlen("a"),sizeof(int));}
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0
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q, 20 17 bytes

{x+y*z}."j"$"$+."
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0
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TeaScript, 7 bytes

'ߞ'c()

Takes the char code of the character with a char code of 2014

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0
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Groovy, 48 bytes

For 2015 change cabe to cabf...

n={it.each{print(((int)it)-(int)'a')}}
n('cabe')

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0
\$\begingroup\$

Factor, 10 bytes

Not gonna beat my winning answer, but as a follow-on to the other "2014th Unicode char" answers:

CHAR: ߞ .

Prints 2014.

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0
\$\begingroup\$

JavaScript, 17 Bytes

atob("MjAxNA==")
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0
\$\begingroup\$

C (gcc 5.3.1), 49 bytes

Pure arithmetic! \o/

main(a){printf("%d",(++a<<(a<<a|a))-(a<<a+a)-a);}

Undefined behaviours.

Specifically, use this compiler (languages are defined by implementation).

Without undefined behaviour (which would work on all compilers), 51 bytes

main(a){a++;printf("%d",(a<<(a<<a|a))-(a<<a+a)-a);}

Explanation

Basically 2048 - 32 - 2, constructed using powers of 2.

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0
\$\begingroup\$

Jelly, 10 bytes (non-competing)

⁹⁴×H_⁴Ḥ¤’’

Explanation:

⁹            Set the current value to 256.
 ⁴×          Multiply by 16. The current value is now 4096.
   H         Divide by 2. The current value is now 2048.
    _⁴Ḥ¤     Subtract by 16/2. The current value is now 2016.
        ’’   Decrement twice. The current value is now 2014.
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0
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Swift 2.2, 28 bytes

print("\(ENOTDIR)\(EFAULT)")

Falling back on Darwin/glibc error codes from errno.h.

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0
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Mathematica, 16 bytes

FromDigits@"JAE"
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0
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QBIC, 10 bytes - Not Competing

?z^s*r+z+t

Calculates 2014 from the pre-initialised variables q-z (1-10) and prints it. Development of QBIC started some two years after 2014...

Alternative 12-byte version:

?(u*y)^r-z-q
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0
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Pushy, 6 bytes

`��`j#

Contains control characters, so here's a hexdump:

0000-0006:  60 14 0e 60 6a 23

The first unprintable is the literal DC4 byte (\x14), and the second is SHIFT-OUT (\x0e).

First these bytes are pushed as charcodes, so the stack is [20, 14]. The j operator concatenates these and # outputs the result: "2014".


10-byte solution:

`<:;>`KT-"

Uses char-code manipulation and the builtin T (10).

`<:;>`       Push string as char-codes: [60, 58, 59, 62]
     KT-     Take 10 from each: [50, 48, 49, 52]
        "    Interpret as char-codes and print: results in "2014"
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0
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Microscript, 10 bytes

Noncompeting, language postdates the question..

'js'(s';-*

Explanation: 106*(59-40), using the language's equivalent of character literals. I'd do 53*38, but the character corresponding to 53... is the digit 5. And the character corresponding to 19 is, of course, nonprintable, so that wouldn't work.

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0
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Pyke, 3 bytes, noncompetitive

Try it here!

Where ߾ is 0xDFBE

Loads ord(0xDFBD-0x20) as an integer and implicit prints it

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  • \$\begingroup\$ How does it work? \$\endgroup\$ – Pavel Dec 13 '16 at 4:04
  • \$\begingroup\$ @Pavel explanation added \$\endgroup\$ – Blue Dec 13 '16 at 7:35
0
\$\begingroup\$

tcl, 16

puts [scan ߞ %c]

Can be seen on: http://rextester.com/live/SVXB29034

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0
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Groovy, casting to int, 8 bytes

(int)'ߞ'
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0
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C, 25 bytes

f(){printf("%d",L'ߞ');}

24 characters, but one character is UTF-8 encoded. Still the shortest C answer! How it works:

U+07DE ߞ NKO LETTER KA

7DE in decimal is 2014.

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0
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Python 3, 16 Bytes :

print(ord('ߞ'))

2017 version, 16 Bytes :

print(ord('ߡ'))
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0
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Python - 23 Bytes

print ord('&')*ord('5')
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  • 2
    \$\begingroup\$ The 5 is a number. \$\endgroup\$ – pppery Aug 31 '17 at 20:39
0
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Whitespace, 19 bytes

Visible representation:

SSSTTTTTSTTTTSNTNST

Just pushes 2014 onto the stack and prints it. Whitespace's lack of any visible characters makes this pretty easy.

Valid as numbers are completely valid tokens in whitespace, they just don't do anything.

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0
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Common Lisp, 15 characters

(char-code #\ߞ)

Try it online!

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0
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Clojure - 35 characters ASCII only

(print(-(*(int\#)(int\<))(int\V)))

Based on True Soft's answer

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protected by Community Jan 14 at 6:34

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