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

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.

Answer 2

C, 29 bytes (Emacs) / 40 bytes (VI)

Before creating program, execute the shell command

stty -ixon      ### 11 bytes, with the newline

N.B this is required for most terminal editors, but is not required for emacs

then write this code:

main(){printf("%d",'j'*'^S');}

N.B. '^S' is a single XOFF character e.g. in VI use control-V, control-S; in emacs use control-Q, control-S; emacs turns off flow control by default.

I assume the comment [You can't type ^S in a Linux terminal] below left off an implied [without the stty command or equivalent e.g. emacs default behavior upon startup], otherwise that comment is inaccurate.

And here is the requested hexdump:

$ od -a -t x1 x.c
0000000   m   a   i   n   (   )   {   p   r   i   n   t   f   (   "   %
         6d  61  69  6e  28  29  7b  70  72  69  6e  74  66  28  22  25
0000020   d   "   ,   '   j   '   *   ' dc3   '   )   ;   }  nl
         64  22  2c  27  6a  27  2a  27  13  27  29  3b  7d  0a

Answer 3

Python 3, 16 Bytes :

print(ord('ߞ'))

2017 version, 16 Bytes :

print(ord('ߡ'))

Answer 4

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.

Answer 5

Common Lisp, 15 characters

(char-code #\ߞ)

Try it online!

Answer 6

Clojure - 35 characters ASCII only

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

Based on True Soft's answer

Answer 7

Ruby (8 bytes, 7 chars)

'ߞ'.ord

The question mark is ߞ represented in two bytes UTF8.

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Answer 8

Implicit, 3 bytes

`ߞ

Try it online!

     implicit push command
`    character
 ߞ   U+07DE (2014 in decimal)
     implicit integer output

Answer 9

Triangularity, 18 bytes

.. ..
."ߞ".
o    

Try it online!

Answer 10

Pyt, 16 bytes

áŁ!⁺²Đ⁺⁺!₀⁻⁻⇹ᴇ+₅

Explanation:

á                               Push stack into list (pushes empty list)
 Ł                              Get length of top of stack (0)
  !                             0!=1
   ⁺²                           (1+1)^2=4
     Đ                          Duplicate top of stack
      ⁺⁺!                       (4+2)!=720
         ₀⁻⁻                    720/10-2=70
            ⇹                   Swap top two items on stack (4 is on top)
             ᴇ                  10^4=10000
              +₅                (10000+70)/5=2014

Try it online!

Answer 11

Brainf*ck

>++++++++++[<+++++>-]<.--.+.+++.

More readable version:

> Move pointer pos over
++++++++++ Add 10
[ Start loop
<+++++>- Add 5, move over & subtract 1
] End loop
<.--.+.+++. Do some other stuff

Yes, I know this is an old post, but I'm really bored and want to get enough rep to comment because I don't have access to my other account.

Answer 12

Brainfuck, 23 Bytes

-[>+<-----]>-.--.+.+++.

Try it online!

Answer 13

Aceto, 10 bytes

IIppIpPiIp

Try it online!

Uses int(Pi) to it's advantage

11 bytes ^{(previous answer)}

IIppIpIIIIp

Try it online!

Simply increments from zero 2, 0, 1, and 4 times, then prints each time

Fun Version

Piddd++DsJiPidIIJi*p

Try it online!

Uses int(Pi) to base equations off of

Lame Version

'ߞop

Try it online!

Converts the codepoint to number, which is 2014

Answer 14

SNOBOL4 (CSNOBOL4), 42 bytes

	&UCASE 'N' @X ARB 'T' @Y
	OUTPUT =Y X
END

Try it online!

@ assigns the index of the match and (space) concatenates them.

I know, it's 2018, so I should go home...

Answer 15

Python 3, 16 bytes

lambda:ord('ߞ')

Try it online!

Answer 16

Objective-C, 27

NSLog(@"%i",'&'*('V'-'!'));

Answer 17

x86, 15 bytes

No numbers in the source code. Returns in eax. Uses (252*4 - 1) * 2.

   0:   31 c0                   xor    %eax,%eax
   2:   fe c8                   dec    %al
   4:   48                      dec    %eax
   5:   48                      dec    %eax
   6:   48                      dec    %eax
   7:   d1 e0                   shl    %eax
   9:   d1 e0                   shl    %eax
   b:   48                      dec    %eax
   c:   d1 e0                   shl    %eax
   e:   c3                      ret  

If we permit numbers in the source but not binary, we have mov $2014,%ax/ret for 5 bytes.

Answer 18

Ahead, 5 bytes

ezpz

'ߞO@

' push next cell to stack
ߞ U+7DE (2014 decimal)
O output as number
@ die

Try it online!

Answer 19

Z80Golf, 10 bytes

00000000: 080a 0b0e 767e 23e5 ee3a                 ....v~#..:

Try it online!

Disassembly

start:
  ex af,af'  ; db $3a ^ '2' ($08)
  ld a,(bc)  ; db $3a ^ '0' ($0a)
  dec bc     ; db $3a ^ '1' ($0b)
  ld c,$76   ; db $3a ^ '4' ($0e)
             ; halt         ($76)
  ld a,(hl)
  inc hl
  push hl
  xor $3a

Essentially the Hello World trick in Z80Golf.

In short, hl serves two purposes: data address for a and return address for putchar (which is at $8000 and accessed by going all the way through zeroed memory, instead of call $8000).

The instructions in the data section have to be effective no-ops (not touching hl and sp should suffice), and the last one should be 2-byte in order to shadow $76 (halt). I carefully selected the xor value so that the resulting binary has no digits ($30 to $39).

Z80Golf, 10 bytes, 2018 edition

00000000: 1416 171e 767e 23e5 ee26                 ....v~#..&

Try it online!

Disassembly

start:
  inc d      ; db $26 ^ '2' ($14)
  ld d,$17   ; db $26 ^ '0' ($16)
             ; db $26 ^ '1' ($17) ; rla
  ld e,$76   ; db $26 ^ '8' ($1e)
             ; halt         ($76)
  ld a,(hl)
  inc hl
  push hl
  xor $26

Same principle, but this time we can't use ld c,$76 since it gives $36 for the xor value. So I moved to e which starts with 1e. Also, different starting address gives two different interpretations of the data section:

• $16 $17: ld d,$17
• $17: rla (Rotate the register A to the left)

So I had to check both are effective no-ops in this program.

Answer 20

Lua 5.3.1, 31 bytes

It's an old question, but I may as well join in on the fun!

For those not familiar with Lua, the # operator returns the length of an array or string, and the .. operator is for string and number concatenation. The // operator is integer divide, and is used so that there is no trailing decimal in the numbers caused by Lua treating the result of regular divisions as a float.

2014 in 31 bytes: t=#"aa"print(t..t-t..t//t..t*t) = print(2..2-2..2/2..2*2)

2015 in 36 bytes: t=#"aa"print(t..t-t..t//t..t*t+t//t) = print(2..2-2..2//2..2*2+2/2)

2016 in 33 bytes: t=#"aa"print(t..t-t..t//t..t*t+t) = print(2..2-2..2//2..2*2+2)

2017 in 38 bytes: t=#"aa"print(t..t-t..t//t..t*t+t+t//t) = print(2..2-2..2//2..2*2+2+2/2)

2018 in 33 bytes: t=#"aa"print(t..t-t..t//t..t*t*t) = print(2..2-2..2//2..2*2*2)

The solutions for other years are pretty similar. Here's a function that can do any number:

function f(y)
    t=#"a"
    s=""
    for i=t, #(y.."") do
        n=t-t
        for j=t,(y..""):sub(i,i) do
            n=n+t
        end
        s=s..n
    end
    return s
end

And here's the golfed version of the function at 108 bytes:

function f(y)t=#"a"s=""for i=t,#(y.."")do n=t-t for j=t,(y..""):sub(i,i)do n=n+t end s=s..n end return s end

Answer 21

;# - 203

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

Answer 22

K (ngn/k), 10 bytes

"&"*-/"U "

Try it online!

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and 2018 just because this is so late

K (ngn/k), 11 bytes

-"^"-*/"@!"

Try it online!

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EDIT: I'm unsure when this language was created, it's a variant of k which has existed for a long time, but the gitlab link seems to indicate that it may be only a year or so old (judging by commit info), so this answer may be invalid as a result

Answer 23

Aheui (esotope), 8 chars (24 bytes)

반밝따바뱟해망어

Try it online!

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It's too easy for Aheui since it never use any kind of digits at all.

Explanation:

# An Aheui code starts with default stack "아".
반: push 2, move cursor right by 1(→).
밝: push 7, →
따: mul(push 14), →
바: push 0, →
뱟: push 2, move cursor right by 2(→→).
해: end.
망: print as integer, →
어: move cursor left by 1(←).

Note: Print instruction moves cursor in reverse direction if current storage is empty.

Answer 24

TI-BASIC, 7 bytes

int(₁₀^(³√(tan(cosh(cos(π

A significant improvement from the previously winning 12-byte TI-BASIC answer. Using one-byte functions allows the greatest number of chances to achieve a certain value within a given byte count.

Answer 25

Forth, 29 bytes

hex fbc 'B' '!' / / decimal .

Try it online!

Answer 26

C# .NET 75 bytes

public class p{public static void Main(){System.Console.Write((int)'ߞ');}}

Just displaying the integer value of the unicode character ߞ
Try online (also has 2015-2019)

Answer 27

*nix shell (POSIX/bourne) 41 bytes

a=.. c=. d=....;echo ${#a}${#b}${#c}${#d}

Other solutions

bashprintf %x \'— # 13 characters Try it online!
bashecho $[x=++y+y]$?$y$[x+x] # 25 characters Try it online!

Answer 28

HaykamScript, 11 bytes

s[⛉,☄]i

Explanation

s       - string
  [     - executed parameters
    ⛉   - constant for 201
    ,
    ☄   - constant for 4
  ]
i       - convert to integer

Other Years

It's pretty simple to make any other year with the same amount of bytes, Unicode codepoints 0x2600 to 0x26FF represent integers 0 to 255 in my language.

For example, to make the year this was posted:

s[⛉,☈]i

Answer 29

k4, 20 bytes

*/(-)."j"$("Ud";"/")

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      "j"$("Ud";"/") / convert "Ud" and "/" to underlying integer representation -> (85 100;47)
  (-).               / subtract 47 from 85 and 100 -> 38 53
*/                   / multiply over

Answer 30

MAWP 2.1, 15 bytes

!+!:!!-:!!$:!+:

Try it online lol.