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 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.

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Leaderboard:

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Answer 1

Hexagony, 18 bytes

g{A*'-"'-'"Av<@!}/

   g { A
  * ' - "
 ' - ' " A
  v < @ !
   } / .

I had trouble getting it to fit in 19 bytes (side length 3) because I was always 1 short, then I rearranged my memory accessing to be 1 shorter, which also allowed me to use a very efficient layout. Then I was able to shift a no-op somewhere in the code to the very end saving a byte.

Try it online!

Answer 2

Husk, 4 bytes

c'ߞ

Try it online!

Answer 3

MAWP, 16 bytes

!!M:!!A:!:!!M!M:

chr:pos:stack
! : 1 : [1,1]
! : 2 : [1,1,1]
M : 3 : [1,2]
: : 4 : [1]
! : 5 : [1,1]
! : 6 : [1,1,1]
A : 7 : [1,0]
: : 8 : [1]
! : 9 : [1,1]
: : 10 : [1]
! : 11 : [1,1]
! : 12 : [1,1,1]
M : 13 : [1,2]
! : 14 : [1,2,2]
M : 15 : [1,4]
: : 16 : [1]

Try it!

Answer 4

Python 3, 49 bytes

for i in['..','','.','....']:print(len(i),end='')

Python 3, 15 bytes

print(ord('ߞ'))

Answer 5

Python 3, 16 bytes (15 characters)

print(ord('ߞ'))

Try it online!

Explanation

ord returns the decimal Unicode of a character, and the decimal Unicode of ߞ happens to be 2014.

Python 2?

I tried doing the same thing in Python 2 (print ord('ߞ')), which would be 1 byte less, but this doesn't work. Why? Well, in Python 3, len('ߞ') returns 1, so everything is fine. However, in Python 2, it returns 2. And since ord only takes a string of length 1, Python 2 doesn't really like that: TypeError: ord() expected a character, but string of length 2 found

Answer 6

Scratch, 82 bytes

when gf clicked
say(join(length of[The year two thousnd])(length of[and   fourteen

Answer 7

CSASM v2.4.0.2, 83 bytes

func main:
push "  "
len
print
push ""
len
print
push " "
len
print
push "    "
len
print
ret
end

The only way to push numbers to the stack without using numbers is to create str instances and then get their lengths.

Answer 8

Alphabetti spaghetti, 15 bytes

aiioaoaioaiiiio

Try it online!

Answer 9

APOL, 38 bytes

-(⒏ l(*("eeeeeeeeeeeeeeeee" l("ee"))))

⒏ is a built-in constant that equals 2048, which is only 34 off from 2014. I use the length function to take the length of a 17-character string multiplied by 2 (which is shorter than a 34-character string) and implicitly print the result.

Answer 10

not sure if others have used this one already :

from any UTF-8 aware shell

• printf %d \'ߞ
  

13-chars spanning 14-bytes. the following are equivalent forms :

printf %d \'$'\xDF\x9E'
printf %d \'$'\737\636'

Answer 11

ACCUMULATOR, 2015 28 bytes

-1987 bytes thanks to a near-escape from the Orwellian thought police!

AAAAAAAAAAAAAAAAAAAACMMMMMMO

This is pretty simple:

• 20 As increment the accumulator: 20
• C concatenates it to itself: 2020
• 6 Ms decrement: 2014
• O outputs

Old version here

Answer 12

Python 3, 27 bytes

I went through the effort of writing this script:

def to_base(n: int, b: int) -> str | None:
    if not 1 < b < 37:
        raise "Impossible base, try using a base between 1 and 37 (beginning and end not included)"

    if n == 0:
        return '0'

    possible_digs = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    res = []

    while n > 0:
        res += possible_digs[n % b]
        n //= b
    
    return ''.join(reversed(res))


def to_base_letters_only(n: int) -> str:
    bases_found = []
    for b in range(11, 37):
        c_n = to_base(n, b)
        digs_l_10 = False

        for d in c_n:
            if d in "0123456789":
                digs_l_10 = True
                break

        if not digs_l_10:
            bases_found.append([c_n, b])
    
    print(bases_found)

    if bases_found == []:
        return "No conversion found"
    return f"Conversion found: {(min_base := min(bases_found, key=lambda c: len(c[0])))[0]} in base {min_base[1]}"

print(to_base_letters_only(2014))

Which found BBC in base 13, which is 2014, without using arabic numerals.

So, my answer is this:

print(int("BBC",ord("\r")))

Try it online!

\r is Unicode 13.

Fun fact: This answer is 1 byte behind the Python 2 answer, because of print needing parentheses.

That script can also be used to find solutions for other numbers.

Answer 13

Desmos, 40 26 25 Bytes

b=\ln ee
k=bb
k^kkb-kkb-b

Desmos graph link

Now smaller by a whole 1 more byte! Yaaay...

Version that needs each line to be pasted individually, 24 23 bytes

b=lnee
k=bb
k^kkb-kkb-b

Answer 14

ANSI C - 95 47 52 characters

#include <stdio.h>
main() { printf("%i", (('a' + 'a')/'a') * ('\a' + '\f') * ('<' - '\a') ); }

This program uses characters to initialise integers and multiplies: 2 * 19 * 53.

#include main(){printf("%i",'\aÞ');}

This program initialises an integer using charaterbytes and prints it. '\aÞ' is the bitpattern 00000111 11011110 this is also the bitpattern of 2014.

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Disclaimer: this was made on a windows system with visual studio. This code depends on a lot of things, including - How your compiler endodes the characters you input. Þ has an ascii value of 222 (or its negative equivalent), this may vary depending on your system. The notation int a = 'abcd'; is in itself evil and depends on how memory is handled on your system - this includes endian issues. int a = '\0A'; a is 65 on my system but may be 16640 on your system.

main(){printf("%i",('C'-'A')*('T'-'A')*('v'-'A'));}

I went back to Version one and multiplied 2 * 19 * 53. This version uses only one byte at a time so it is endian compatible. Also it uses only characters in the range of [0 - 127] to be compatible to all systems.

Answer 15

Solution 1

Octave/Matlab (55 chars)

a=pi;b=a*a;disp(ceil(a^a^a/a/a/a-b*b*a-a^a*b+b*b-b-b));

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Solution 2

PHP (9 chars without tags, 12 with them Actually 2022 because of the new lines involved)

<!--Comment
  previous
  2013 lines -->
<?=__LINE__; <!-- This should be on line 2014 -->

Answer 16

Game Maker Language, 22

show_message(ord("ߞ"))

Answer 17

C - 44 characters (85 with headers)

What, no one is abusing strings yet?

#include<stdio.h>
#include<netinet/in.h>
main(){printf("%u",ntohs(*(int*)"\a\xde"));}

Interestingly, this is a special case where neither character is printable, but their special code doesn't involve a number.

If we want no warnings, it needs to become 55 (96) characters:

#include<stdio.h>
#include<netinet/in.h>
int main(){return!printf("%u",ntohs(*(int*)"\a\xde"));}

Answer 18

C/C++ 39

main(){printf("%d%d",':'-'&',':'-',');}

ASCII for: ':' = 58, '&' = 38, ',' = 44. Using that, 58-38 = 20 and 58-44 = 14.

Answer 19

120 characters in Squeak Smalltalk trunk (4.5).
I did not search the shortest, but kind of graphical solution:

((Text string:'Happy\New year'withCRs attribute:TextEmphasis narrow)asMorph borderWidth:Float one+Float one)bounds area

It depends on font, margins, and so is quite fragile, but at least for me it worked.
In Squeak 4.4, it works with lowercase 'happy\new year'.

Answer 20

JavaScript, 49 Chars

A mathematical JavaScript version making use of only PI and E as source numbers.

(m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|""

... mmmm PIE.

Oh and just in case implicit returns are vetoed (56 Chars with alert):

alert((m=Math).pow(e=m.E,e*(p=~~m.PI))/m.sqrt(p)+e+e|"")

Answer 21

Clojure - 22

(apply *(map int"j#"))

(note: the # is ASCII character 19, Stack Overflow doesn't seem to like this but it's valid Clojure source...)

Clojure - 36

(dec(reduce +(nnext(range(int\@)))))

Answer 22

C++ - 63 bytes

I'm not sure if this method has been used, but I designed this myself anyway:

#include<iostream>
int main(){std::cout<<int('&'*(','+'\t'));}

Answer 23

Ruby 1.9, 10 bytes 

p 'ߞ'.ord

Answer 24

vba (immediate window), 38 26 13

using regular ascii characters (no funny typing needed)

?&ha+&ha&&&he

26

?val("&hfbc")/-(true+true)

38

?year((cdbl(asc("ê"))*cdbl(asc("²"))))

find a date that can be represented as a number, and select the year from that (in this case, Jan, 13, 2014)

have to use cdbl, as it assumes signed int, and overflows

Answer 25

Clojure, no unicode tricks (49 characters/bytes)

Uses the fact that * called with no args evaluates to 1:

(let[b(inc(*))j(+(* b b b)b)](+(* b j j j)j b b))

Using the same trick and doing string concatenation instead of arithmetic, the lowest I could get was 51 chars:

(let[n(*)t(+ n n)z(+)f(+ t t)](print(str t z n f)))

Answer 26

Python, 23

print ord("<DC3>")*ord("j")

<DC3> should be replaced with ASCII symbol 19 (device control 3).

Answer 27

SAS, 34 characters/bytes

data a;x=put(' ',hex.);put x;run;

That puts it to the log, it's 6 longer if you need it to the output window. Note I'm not seeing the second character there; it is backwards-P, which is hex 14.

There should be a shorter solution with %sysfunc(putc(..., but I can't get that to work properly.

Answer 28

JavaScript 45

alert(parseInt('bbc','twentyonefour'.length))

Answer 29

Bash - 10 (or 8)

Well there have been a couple of answers that have been disqualified because they rely on the year. When golfing, one side goal is to see how close we can get to breaking the rules as currently written without breaking the letter of the rules (I include the clarifications by Joe Z in the 66 existing comments on the rules). The question very specifically states that I can not depend on 2014 being the current year. I instead rely on it being 8:14pm in my timezone.

date +%H%M 

When I ran it, it output 2014 exactly, thus it satisfies it No, it has to be 2014 exactly. comment. (Due to context people seem to misread it as ... 2014 always, but that was not what was written, even if that were perhaps what was intended.) This lets me beat the current Bash record, at least until this loophole is closed. This interpretation may seem too cheaty since all the existing popular answers assume that the rules really meant always. Indeed some of them exploit this and export something that isn't exactly 2014, but instead contains 2014. I am fine with that interpretation too since Bash can do:

cat /*/*

This is a mere 8 characters, which will concatenates a bunch of files including /dev/urandom/, and it generally takes my machine under a minute to find 2014 in /dev/urandom. Although my rule twisting golfing code of honour won't let me pick this solution since it violates the letter of Joe Z's clarification, the only objection Joe Z raised to the random approach in the 66 comments was that it was too long. At 8 characters this answer is actually shorter than my rules-lawyer answer.

Answer 30

Python, 55 bytes (no math import and no char or unicode trickery!)

x=False;a=x**x;b=a+a;c=b+b;print c**c*(c+c)-b**(c+a)-b

Uses the fact that zero to the zeroth power is defined as one and False can be implicitly casted to 0. Hence a, b and c will contain 1, 2 and 4 respectively.