## Produce the number 2014 without any numbers in your source code

598

206

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 2020 2021 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.

var QUESTION_ID=17005,OVERRIDE_USER=7110;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

15Even though numbers are ignored in brainfuck, I thought I'd post one anyway. 32 Chars: ++++++[>++++++++<-]>++.--.+.+++. – Braden Best – 2015-04-01T21:37:33.103

6Brainfuck isn't a valid language for this challenge. – Joe Z. – 2015-04-01T22:49:22.940

6I know. That's why I posted it as a comment – Braden Best – 2015-04-01T22:51:45.383

10I wonder if this question gets a small spike in popularity around New Year's. – Joe Z. – 2015-12-26T23:28:27.887

3Waiting for "Come on folks, don't you realize it's 2016?" :) – padawan – 2016-01-04T23:35:40.280

@cagirici Done. – Joe Z. – 2016-01-05T04:50:49.897

1@BradenBest It's possible to do it in 31 characters in at least two different ways:

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



and

++++++++++[>+++++<-]>.--.+.+++.
 – Zubin Mukerjee  – 2016-02-21T17:47:16.207

@JoeZ. It's almost that time of the year again. – Ian Limarta – 2016-11-13T01:43:28.127

2possible to do the brainfuck answer in 23 bytes -[>+<-----]>-.--.+.+++. – FinW – 2016-12-05T18:05:54.643

That language is full of language! – cascading-style – 2016-12-12T21:44:04.253

@JoeZ. We're discussing this question on meta.

– xnor – 2016-12-30T00:30:38.963

C++: cout<<to_string(('V'-'B'))+to_string(('V'-'H'))<<"\n"; – abhinav1602 – 2019-09-06T08:34:19.677

(unsigned char)'\xca'*'\xa' for 2020 in C. – S.S. Anne – 2020-04-26T14:42:50.937

Locomotive basic (Amstrad CPC): ?ASC("-")*ASC("-")-&b, 22 bytes. (Cannot asnwer since the question is protected) – Amessihel – 2020-07-24T20:48:28.927

The question text says "any language in which numbers are valid tokens". Since any character that is not an instruction is a comment in BF, numbers are valid therein. Though admittedly you could go into a philosophical debate over whether digits count as tokens or whitespace in such a language. – Stewart – 2021-02-24T10:53:26.483

146

# Mouse-2002, 4 bytes.

That's 4 bytes of pure, sweet ASCII.

In Mouse, the letters of the alphabet are initialised to the values 0-25. ! is the operator for printing integers, thus this prints 20 then 14 (no intermittent newline).

U!O!


There's no online interpreter available, but here you will find an interpreter written in C (needing some tweaks before one can coerce gcc to compile it) and the same compiled interpreter for Win32 but which works perfectly on Linux with wine.

Here you can find the fixed version of the interpreter, which compiles.

8Wow, this is an amazing find so much later after the challenge. And with a language so esoteric nobody here had heard of it yet! – Joe Z. – 2015-12-09T01:19:22.930

1

I just discovered this language today and I already love it. However, it's not purely esoteric, which makes it even better!

– cat – 2015-12-09T01:25:59.000

4BTW: this does the same thing as CJam's KE, but is actually valid because it's a language from before 2014. – Joe Z. – 2015-12-09T01:30:44.070

10The smug look on my face right now resembles that of a Fortran-77 wizard after beating some proponent of Python in a [tag:fastest-code]. – cat – 2015-12-09T01:33:09.633

1Also this solution will work until 2025 at least, until you pass U!Z!. – Joe Z. – 2016-01-11T22:24:04.313

2But by then, somebody will invent a language that prints out the Unicode value of any unrecognized token, and then the high score will be at 2 bytes forevermore. – Joe Z. – 2016-01-11T22:25:12.867

74This a game of @cat and Mouse-2002. – Digital Trauma – 2016-10-07T05:00:35.170

6@DigitalTrauma indeed :3 – cat – 2016-10-07T06:44:15.943

1Running the 2007 binary using Wine on Ubuntu, the program as written produces an error message to the standard output which doesn't include the string 2014. The modified program U!O!$ works, producing 2014 and no trailing newline. – LegionMammal978 – 2017-08-11T01:50:10.613 1@JoeZ, that's... that's kinda what I've done. ;) – Lyxal – 2019-11-09T07:48:57.470 1 336 # Python, 52 bytes print sum(ord(c) for c in 'Happy new year to you!')  Updated for 2015 thanks to @Frg: print sum(ord(c) for c in 'A Happy New Year to You!')  Mouse over to see 2016 version: print sum(ord(c) for c in 'Happy New Year to you!!!') Try it online! 5Updated: print sum(ord(c) for c in 'A Happy New Year to You!') – Frg – 2015-01-08T20:14:25.570 Uh....I get invalid syntax in 3.4? How does one run this? – jvriesem – 2015-09-16T03:57:59.520 For Python 3.x, print needs ( )'s around the expression. – dansalmo – 2015-09-17T04:23:38.823 12Congrats on being the first answer on this site to reach 1,000 upvotes. :) – Joe Z. – 2015-12-06T02:48:18.617 You don't need the space after ord(c) – Cyoce – 2016-04-02T20:52:42.060 2print sum(ord(c) for c in 'Happy New Year to You!!!!') for 2017 :D – fedorqui – 2016-07-20T13:18:46.363 Why were revisions undone for print sum(map(ord,'A Happy New Year to You!'))? It does the same thing. – Sunny Patel – 2016-08-23T16:07:29.503 3@SunnyPatel it's frowned upon to edit in golfs – Cyoce – 2016-12-07T21:59:14.413 When can haz 2017? – Pavel – 2017-01-01T05:30:08.790 2@Phoenix print sum(ord(c) for c in 'Happy New Year to You!!!!') – dansalmo – 2017-05-05T18:34:17.850 1@dansalmo Perhaps you want to add that one to the answer as well? Would be cool to see the entire list in 2025. ;) (Oh, and I took the liberty to add a TIO-link.) – Kevin Cruijssen – 2017-05-19T06:46:49.870 193You deserve the cleverness award – Carter Pape – 2014-01-01T19:49:30.683 You forgot a print somewhere. Nice one otherwise :) – jub0bs – 2014-01-01T19:49:39.370 171wow - this is so in the spirit of this question. <3 – Johannes – 2014-01-01T19:51:18.320 80I was a little surprised that it was only the 4th phrase I tried after "Happy new year!", which would have been perfect for the year 1374. – dansalmo – 2014-01-01T20:14:24.980 3Is this a rip-off of Mikael Mayer's answer in Scala, though? – SimonT – 2014-01-01T21:48:41.493 No, I did not look at anyone else's answers first. – dansalmo – 2014-01-01T21:55:12.983 10I only posted mine two hours before yours. This one is not that hard to find by oneself. – Mikaël Mayer – 2014-01-01T23:05:20.810 1Ugh, ever since I started programming in Befunge, Python looks so wordy. But since this is such an amazing solution (+1), here is the equivalent Befunge: "Happy new year to you!"aa+k+.@ – Justin – 2014-01-02T06:54:31.193 14You want terse? Ruby: p 'Happy new year to you!'.sum – histocrat – 2014-01-02T15:19:57.390 57print sum(ord(c) for c in 'HAPPY NEW YEAR To you too.') Oh wait, I'm a year late. That's 2013. – Warren P – 2014-01-02T15:31:00.903 173sum(map(ord,'Happy new year to you!')) would save 7 characters. – None – 2014-01-02T16:19:49.690 4Even at my best I am a bogey golfer. – dansalmo – 2014-01-02T17:04:38.730 30I created this account just to give +1! – mkoistinen – 2014-01-02T19:32:02.783 38sum(b'Happy new year to you!') in Python 3 – Kabie – 2014-01-04T12:56:14.837 6sum(bytearray('Happy new year to you!')) in Python 2.6+ – martineau – 2014-01-04T14:22:41.990 Awesome timing for awesome answer! – tumchaaditya – 2014-01-05T04:06:04.603 haha good answer, in PHP : echo array_sum(array_map('ord',str_split('Happy new year to you!'))); – Tufan Barış Yıldırım – 2014-01-05T16:32:50.093 @KevinCruijssen That wasn't their solution. fedorqui posted that solution nearly a year prior, up a few comments. – None – 2017-08-11T00:43:43.570 2You guys made my day! Thank you and have a great year wherever you are :) – amyassin – 2014-01-08T15:31:07.840 2Exactly 1234 upvotes at the time of the posting of this comment. – Comrade SparklePony – 2018-02-26T20:30:54.810 One byte can be saved by deleting whitespace before "for". – stackzebra – 2019-05-17T13:58:43.407 @stackzebra and also by deleting whitespace after in – dansalmo – 2019-05-18T14:28:27.680 I offer: print(ord('&')*ord('j')/len('NY')) – Max – 2019-05-23T03:21:18.350 This does not make a serious attempt at solving the question (this is code-golf) – MilkyWay90 – 2019-07-05T04:30:58.910 print(ord('ߞ')) for -36 bytes. – None – 2019-09-22T02:43:37.410 Printable ASCII at 32-127 means about max 62 chars, min 15. You clocked 22. I'd like to see your list of things you considered before picking this one, added to the answer. If it's code golf, how much smaller while being literate could it get...margin is 7 chars. :-) – HostileFork says dont trust SE – 2014-03-02T23:40:29.653 1It's hard to remember exactly, but the first thing I tried was "Happy new year!", then "Happy new year!!!!!", then "Happy new year everyone!", and finally 'Happy new year to you!' – dansalmo – 2014-03-03T21:32:23.540 print(sum(ord(c) for c in 'have a happy new year!!')) – MirroredFate – 2020-01-10T07:22:24.047 2018: print sum(ord(c) for c in 'Happy New Year To You!!!!!') – Seth – 2021-02-23T18:59:37.820 Mind blown! That is very clever of you sir! – rageit – 2014-07-11T14:26:44.227 248 # Ruby, 15 p Time.new.year  Temporary ;) Note that the section of the question independently of any external variables such as the date or time or a random seed was not edited in until long after I posted my answer... Jan Dvorak offers a great alternative in the comments: Happy = Time Happy.new.year  But it's so unenthusiastic. I prefer: Happy = Time class Time; alias year! year; end Happy.new.year!  Or even: class Have; def self.a; A.new; end; end class A; def happy; Time; end; end class Time; alias year! year; end Have.a.happy.new.year!  And here's correct English punctuation: def noop x = nil; end alias a noop alias happy noop alias new noop alias year! noop def Have x p Time.new.year end Have a happy new year!  Okay okay, I couldn't help it: def noop x = nil; end eval %w[we wish you a merry christmas! christmas and a happy new].map{|x|"alias #{x} noop"}*"\n" def year!; p Time.new.year; end we wish you a merry christmas! we wish you a merry christmas! we wish you a merry christmas and a happy new year!  175Happy = Time; Happy.new.year – John Dvorak – 2014-01-01T08:18:20.343 246 # Go, 2 bytes (UTF-16) One unicode character (2 bytes in UTF-16, 3 bytes in UTF-8 format), output 2014 as part of an error —  http://ideone.com/dRgKfk can't load package: package : prog.go:1:1: illegal character U+2014 '—'  7That's very clever, but the question requests a program that prints the number. – HelloGoodbye – 2014-01-04T15:18:37.043 45If printed error codes count, this submission should win. – ToastyMallows – 2014-01-05T15:51:53.140 42+1. The question doesn't say, "prints only the number and nothing else". – Kaz – 2014-01-14T22:31:49.557 2I don't care if its valid or not, it's pretty awesome. – Redwolf Programs – 2019-05-05T02:01:44.023 2 This is a standard loophole – MilkyWay90 – 2019-08-03T04:56:02.083 I joined this community just to like this post. – Gaurav – 2020-01-18T12:53:21.007 244 # Befunge 98 - 17119 8 bytes '-:*b-.@  Similar to the old version, but I remembered about ' '-:* pushes 45, duplicates it, then squares it, producing 2025 b- subtracts 11 from it, resulting in 2014 .@ prints the result, then ends the program  Interestingly, 452-11 is the only pairing of numbers a,b where $$(a,b)∈[32,126]\times[10,15]\land a^2-b=2014$$ The significance of those sets is that [32,126] is the set of printable ascii characters and [10,15] is the set of easily accessible Befunge numbers. I found that pair with this python program: for a in range(32,127): for c in range(10,16): if (a**2-c)==2014: print("%s,%s"%(a,c))  Or, if your interpreter supports unicode, then this works: # Befunge 98 - 5 bytes (4 chars) 'ߞ.@  It at least works on http://www.quirkster.com/iano/js/befunge.html with the following code (Befunge 93 - 6 bytes / 5 chars): "ߞ".@  # Old version cdd**e-.@  computes the number, then prints it: cdd pushes numbers to the stack so that it is this: 12,13,13 ** multiplies top three values of stack, which is now: 2028 e pushes 14 - subtracts the top two values of the stack, resulting in: 2014 . prints the numerical value @ end of program  # Older version: "*'&("#;:aj@a+,;  Pushes the ascii values for 2014, -10. Then prints each after adding 10 to it. Can I edit the question to include my answer? – cat – 2015-12-09T00:13:33.463 79http://en.wikipedia.org/wiki/Befunge ...what is this i don't even... – Plato – 2014-01-02T13:16:37.563 38"The language was originally created by Chris Pressey in 1993 as an attempt to devise a language which is as hard to compile as possible"... – will – 2014-01-02T15:21:41.003 1Wouldn't a:+.e.@ achieve the same? Saves a character by printing 20 and then 14. – MSalters – 2014-01-02T15:38:58.577 @MSalters That would actually print 20 14 – Justin – 2014-01-02T16:15:04.723 @Quincunx Where does it say that isn't allowed? – 11684 – 2014-01-02T16:44:13.987 7@11684 The output is rather well defined in the question. If a program printed: randomstuff2randomstuff0randomstuff1randomstuff4 I wouldn't consider it a valid solution. – Cruncher – 2014-01-02T17:45:34.097 1I agree. But at least in my native language (Dutch) spaces can be used in numbers to make numbers more readable. So I wouldn't categorise a space as 'random stuff'. – 11684 – 2014-01-02T18:20:04.453 7 @Plato Rather than read the wikipedia article (which only includes Befunge 93), if you want to learn about Befunge 98, read the official specs – Justin – 2014-01-03T09:06:21.270 1By the same idea, '&:f+*.@ should work as well. – MvG – 2014-01-05T19:53:30.740 @MvG yeah, I was just manually searching for short solutions, then wrote a program, then tried different methods with that program. Of course there are other short methods. – Justin – 2014-01-06T06:27:15.267 3Wiki: 'The language was originally created by Chris Pressey in 1993 as an attempt to devise a language which is as hard to compile as possible' ...but why? – CousinCocaine – 2014-01-28T20:17:54.290 184 # Python 2, 26 bytes print int('bbc',ord("\r"))  Try it online! 3Woah I almost upvoted your bbc is 2014 in base 13 but then I realized it has 42 upvotes. Perfection – User – 2015-11-01T16:04:16.617 Here is a permalink to Base 13 as of Jan 4 '14 for anyone interested. Be wary that it contains a mild spoiler to "The Restaurant at the End of the Universe". – arekolek – 2015-11-21T13:36:40.557 58bbc is 2014 in base 13 – Darren Stone – 2014-01-01T10:50:32.950 Would've been my approach. Until I saw the accepted answer... – Kiruse – 2014-01-01T18:45:49.733 72I assume the use of the characters 'bbc' and the use of base 13 is a Douglas Adams tribute. "I may be a sorry case, but I don't write jokes in base 13." – Brian Minton – 2014-01-02T15:34:53.857 1013 years ago. RIP. – Darren Stone – 2014-01-02T18:30:38.280 5IMO, this answer is in "True Pythonese" – kmonsoor – 2014-01-04T15:11:32.230 Wikipedia reference for the Douglas Adams and base-13, http://en.wikipedia.org/wiki/Base_13 – Juha Palomäki – 2014-01-04T21:50:38.240 10@BrianMinton: The use of Base 13 is probably because it's the only base between 2 and 36 that allows the number 2014 to be represented without any of the characters 0-9. – dan04 – 2014-01-05T04:25:34.217 1Coincidence? I think not! :-) – Darren Stone – 2014-01-05T04:30:05.400 148 ## MATLAB, Scala (4 characters, 5 bytes) You can take advantage of MATLAB's (and Scala's) relatively weak type system, here. The trick is to apply the unary + operation on a string composed only of the character ߞ (of UTF-8 code point U+07DE, or 2014 in decimal). This operation implicitly converts the string to a double (in MATLAB) and to an Int (in Scala): +'ߞ'  Byte-count details: • + is ASCII and counts for 1 byte • ' is ASCII and counts for 1 byte (but appears twice in the expression) • ߞ is a 2-byte UTF-8 character Total: 5 bytes ## TeX (32 26 characters, as many bytes) \def~{\the\catcode}~}~\\~\%\bye  An even shorter alternative (proposed by Joseph Wright) is \number^^T\number^^N\bye  ## XeTeX/LuaTeX (13 characters, 14 bytes) If XeTeX or LuaTeX are allowed, UTF-8 input can be used directly (as proposed by Joseph Wright): \numberߞ\bye  \let~\number~^^T~^^N\bye 25 chars/bytes. You IMHO count wrong, it's 25 for your solution as well. – yo' – 2014-01-02T19:46:43.473 1@tohecz I think both our solutions are actually 26-byte long. – jub0bs – 2014-01-02T19:55:35.470 1If the current year is wanted, then 13 bytes: \the\year\bye – yo' – 2014-01-02T20:03:03.350 2+'ߞ' also works in Scala (and a few other languages I imagine) – theon – 2014-01-02T21:22:41.203 120 # dc, 6 chars DiBBCp  • D pushes 13 on the stack, even tho the input radix is 10 initially • i changes input radix (to 13 from 10) • BBC is 2014 base 13. • p prints. Console output: $ dc <<< "DiBBCp"
2014


dc<<<DiBBCp - a bit shorter – mpapis – 2014-01-03T17:32:44.003

11True, but the actual dc program is still DiBBCp (6 chars), the rest is just a way to run it. – daniero – 2014-01-03T17:51:17.650

12

I was going to upvote this but it has 42 points! uses base 13 and the word BBC. How cool is that! Seems that this year we will find the question for life, universe and everithing ;-) http://en.wikipedia.org/wiki/Base_13. I am upvoting @daniero's comment instead and leave the answer with this magnificent 42 reputation ;-)

– Pablo Marin-Garcia – 2014-01-12T19:01:29.187

5@PabloMarin-Garcia, unfortunatelly some unaware person broke it... Has 43 votes now. Please go back and downvote! :) – Tomas – 2014-01-14T16:07:49.657

4@Tomas Vogons always the Vogons. Resistance is futile against the intergalactic burocracy. – Pablo Marin-Garcia – 2014-01-15T00:03:07.350

2I cannot even find the D command in the man page. What does it do? Never mind... D is hex for 13. – kzh – 2014-04-28T18:40:54.373

I would upvote but you have 64 votes – None – 2014-09-08T20:10:07.377

@mpapis Or dc -eDiBBCp. Same length, I think. – nyuszika7h – 2014-11-30T15:03:44.707

89

## 72 45 characters in the code; Zero character codes

This is far from the shortest answer posted, but no one has yet posted an answer that

• doesn't use character codes as a substitute for numbers, and
• doesn't call the system date.

Using pure math (okay, and an automatic boolean conversion) in R, from the R console:

x<-(T+T);x+floor(exp(pi)^x)*x*x-(x*x)^(x*x)/x


Prints out the number 2014. T is a pre-defined synonym for true in R. The floor and exp functions are directly available in the base package, as is the pi constant. R doesn't have an increment operator, but repeating the (x*x) turned out to be fewer characters that doing increment and decrement twice each.

## Original version in Javascript (72 characters)

For the simple reason that I could test out in the console, and it doesn't mind a complete lack of whitespace:

m=Math;p=m.pow;t=true;++t+m.floor(p(m.exp(m.PI),t))*t*t++-p(++t,t--)/--t


run in your console and it will print back the number 2014.

Props to xkcd (and also) for getting me to think about exp(pi):

P.S. If you can make the same algorithm shorter in a different language, post a comment with it.

1o=!"";(o+o)+""+(o-o)+(o+o-o)+(o+o+o+o) – Kernel James – 2014-01-02T10:32:28.360

525 bytes: alert('ߞ'.charCodeAt()) – oberhamsi – 2014-01-02T14:13:35.793

1

+1. However, you claim your answer was the first answer that "doesn't use character codes as a substitute for numbers, and doesn't call the system date". That is actually false. My answer has this solution cdd**e-.@ (posted before yours) which does not make use of character codes or system date. It computes the number 2014. c,d, and e are hexadecimal number digits. a,b,...,f push (respectively) 10,11,...15 so 12 * 13 * 13 - 14 is the computation.

– Justin – 2014-01-04T09:19:08.433

This wasn't the first answer to not use character codes; I have several earlier ones. – Timtech – 2014-01-04T12:39:32.350

1@Quincunx: you're right; skimming through I didn't catch that your version was using hex digits, not character codes. – AmeliaBR – 2014-01-04T15:47:02.527

o=+!'';t=o+o;''+t+(+!o)+o+(t+t), a bit smaller than @kernel-james one – Ricardo Panaggio – 2014-01-05T13:08:34.020

R code with 38 characters x=T+T;y=x*x;x+floor(exp(pi)^x)*y-y^y/x – djhurio – 2014-01-07T11:12:28.750

Nice catch, @djhurio. – AmeliaBR – 2014-01-07T16:42:05.293

I also like @KernelJames version (and Ricardo's improvement) though they really qualify as a completely different method rather than an improvement on mine -- I assume the "protected" status wasn't letting you post? – AmeliaBR – 2014-01-07T16:44:51.883

two k implementations of this algo, both 39 chars: 1: -_-(x*p*p:exp acos@-#)-f*x*x*x:f*f:#; 2: f+(x*_p*p:exp acos@-#)-f*x*x*x:f*f:#. I can't seem to squeeze it any further at the moment.... – Aaron Davies – 2014-01-22T20:25:24.240

and here it is in APL in 31 chars (which i'm quite new to, so i'm sure it can be improved--particularly the hacky generation of the 2 constant): ⌈(x×p×p)-f×x×x×x←f×f←⍴⍕⌊p←*○~⊃⍬ – Aaron Davies – 2014-01-22T20:48:53.953

87

### Scala REPL

34 29 characters.

+"Happy new year to you!".sum


Well ok if you really want it golfed with any chars, you can use:

'@'*' '-'"'


or

"{yz}"map(_-'I'toChar)


which have 11 and 22 chars respectively.

4Interesting idea, even if it's not particularly golfed. But the string itself makes it more interesting, than something using high bytes, or something like that. – Konrad Borowski – 2014-01-01T19:32:46.910

I added a golfed version. – Mikaël Mayer – 2014-01-01T20:12:55.343

You could save a couple characters from the toInt like this: +"Happy new year to you!".sum Test

– theon – 2014-01-02T21:26:44.283

Updated. I did not know that + could also be a prefix. – Mikaël Mayer – 2014-01-03T09:49:16.023

69

## C, 33 bytes

main(){printf("%d",'A'*' '-'B');}


@JonathonReinhart multi-character literals are entirely implementation-defind. – FUZxxl – 2015-04-08T12:14:05.213

20How annoying that the prime factorization of 2014 requires ASCII 5! – Ben Jackson – 2014-01-01T08:56:06.520

I got for(int a;YES;a++){NSLog(@"%i",a);} for Cocoa Touch objective C but I can't add yet (not rated enough). It does show a 2014 eventually and it has a compiler error and is possible it may not work at all and compiled is probably about 4.2 meg - but hey. – Recycled Steel – 2014-01-03T17:21:14.703

main(){printf("%d",'\a\xde');} – mjy – 2014-01-05T15:07:12.807

1@mjy Byte order of multi-character literals is not guaranteed. – Jonathon Reinhart – 2014-01-05T22:14:53.000

On specified env. using multichar is good – l4m2 – 2018-03-23T17:16:37.657

68

# PHP, 9 bytes

This requires PHP 7.1 or lower. It will work in PHP 7.2, but it will result in a warning. No guarantees for any future version.

xxd needed because of binary data (so copying and pasting it would be easier). May return E_NOTICE, but it doesn't really matter, does it?

~ $xxd -r > 2014.php 0000000: 3c3f 3d7e cdcf cecb 3b <?=~....; ~$ php 2014.php
2014


Alternatively, save this using ISO-8859-1 encoding.

<?=~ÍÏÎË;


Why not <?=~ÍÏÎË instead? – Ismael Miguel – 2015-01-03T18:46:06.693

@IsmaelMiguel: I guess (if you save it using ISO-8859-1). – Konrad Borowski – 2015-01-03T20:13:08.417

@Quincunx: Uploading this was relatively tricky, but here is Ideone - http://ideone.com/APHKhn.

1This looks interesting; what does it do? – Justin – 2014-01-01T10:48:24.727

7@Quincunx: It does bitwise not (0x00 changes into 0xFF, and 0x7F changes into 0x80) on every character of the string. As the string is valid identifier (anything with high bit set is an identifier character for PHP, probably to support other encodings), PHP thinks it's a constant, but because it's not defined, it treats it as a string. – Konrad Borowski – 2014-01-01T10:50:08.657

2Since the question was "Now that it is 2014...", how about echo date('Y');? – John – 2014-01-02T19:05:13.307

6@John: Nope. If you would read the comments for the question, you would notice that OP doesn't want the to get current year as an answer. Besides, <?=date(Y); would be 11 characters, and I have solution in 9 characters. – Konrad Borowski – 2014-01-03T12:39:32.380

58

### Python3.4.0b2 (0 bytes)

% python3.4
Python 3.4.0b2 (v3.4.0b2:ba32913eb13e, Jan  5 2014, 11:02:52)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
>>>


2Also works in Python 3.4.2. – Erik the Outgolfer – 2016-09-14T17:56:59.917

This won't work once people update their version of Python 3 a year from now. – Joe Z. – 2014-01-11T22:51:29.427

22That's why I mentioned python3.4.0b2, It wont happen twice. Its release date, not system date. – YOU – 2014-01-12T03:18:35.937

You are not allowed to add extraneous output. – MilkyWay90 – 2019-07-05T04:31:48.937

6This made me smile! – Matt Fletcher – 2014-08-02T11:24:40.330

56

## Mathematica, 14 characters (or 15 if you count the bitmap as a character)

TextRecognize@

52Actually, you should count bytes so it's 14 + the size of the bitmap image. – Sylwester – 2014-01-01T18:46:16.863

23The bitmap would probably be 7,357 characters, really. – Joe Z. – 2014-01-01T18:46:25.163

5Also, I believe some of the bytes in the image fall into the \d range. Namely, if it's the GIF, then the header itself is guaranteed to contain some. – John Dvorak – 2014-01-01T18:49:19.760

15You are all killjoys. Fine, use First@ToCharacterCode@"ߞ" for 25 characters, 26 bytes. – Michael Stern – 2014-01-01T18:59:19.753

@MichaelStern Yup, that's been done allready. – Ajasja – 2014-01-01T19:08:36.677

@Ajasja only after I fixed it. The old solution as given produced a list of answers rather than '2014' as specified by the problem. – Michael Stern – 2014-01-01T19:21:39.630

9

As a bonus, here's a Mathematica solution in 30 characters --URLFetch@"http://goo.gl/miVwHe"

– Michael Stern – 2014-01-01T19:42:52.623

1

I liked this answer so much I made it into its own question: http://codegolf.stackexchange.com/q/17588/2605

– Charles – 2014-01-05T06:17:39.503

48

# Javascript, 23 bytes

Uses Base 64 Conversion

alert(atob("MjAxNA=="))


23 chars (16 with atob by itself)

Or

alert("MMXIV") // ;)


alert("ߞ".charCodeAt()) – programmer5000 – 2017-04-10T12:56:38.923

27MMXIV has no characters in common with 2014, last I checked... // ;) – Joe Z. – 2014-01-01T06:45:20.290

23The ancient Romans might disagree. The first one produces 2014 though, in modern English. – logic8 – 2014-01-01T06:48:01.030

1Your first one doesn't output though... – Doorknob – 2014-01-01T16:14:52.157

1if you open the console in chrome, paste it and hit enter it returns the value - I'll add an alert to it as its longer than others anyway. – logic8 – 2014-01-01T17:54:52.677

5@logic8: Nice one +1. Another JavaScript version which also works: (4 bytes) [!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]] - alert([!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]); Though I'm not sure it meets Joe Z's requirements. – Nope – 2014-01-03T15:03:46.710

@FrançoisWahl, Now that's cool! Nice one. – logic8 – 2014-01-03T19:20:41.640

Or go the other way btoa("ÛMx") for 11 characters – Phil McCullick – 2014-01-03T21:35:49.513

Just so you know, atob("MjAxNA==") returns "2014" (a string). +atob("MjAxNA==") returns a number. Doesn't matter for alert, though. – Pascal – 2014-01-05T15:01:12.400

@logic8 The ancient Romans would probably find it quirky as they didn't use generally use subtractive notation (e.g. IV for 4). – Peter Cooper – 2014-01-05T16:56:30.013

1Here's another javascript base one: alert(parseInt('bbc',+[]+'xd')) – Adam Iley – 2014-01-05T20:03:23.183

@PeterCooper: That's interesting. Would the standard have been "IIII" then? – logic8 – 2014-01-05T21:54:58.940

1@logic8 You can omit the equals signs to get 21 chars – gengkev – 2014-01-12T21:04:22.757

41

# Perl - 10 characters

This solution is courtesy of BrowserUK on PerlMonks, though I've shaved off some unnecessary punctuation and whitespace from the solution he posted. It's a bitwise "not" on a four character binary string.

say~"ÍÏÎË"


The characters displayed above represent the binary octets cd:cf:ce:cb, and are how they appear in ISO-8859-1 and ISO-8859-15.

Here's the entire script in hex, plus an example running it:

$hexcat ~/tmp/ten-stroke.pl 73:61:79:7e:22:cd:cf:ce:cb:22$ perl -M5.010 ~/tmp/ten-stroke.pl
2014


Perl (without high bits) - 14 characters

say''^RPQT


This uses a bitwise "or" on the two four-character strings "RPQT" and "" (that is, four backticks).

$~/tmp/fourteen-stroke.pl 73:61:79:27:60:60:60:60:27:5e:52:50:51:54$ perl -M5.010 ~/tmp/fourteen-stroke.pl
2014


(I initially had the two strings the other way around, which required whitespace between print and RPQT to separate the tokens. @DomHastings pointed out that by switching them around I could save a character.)

Perl (cheating) - 8 characters

This is probably not within the spirit of the competition, but hdb on PerlMonks has pointed out that Perl provides a variable called $0 that contains the name of the current program being executed. If we're allowed to name the file containing the script "2014", then $0 will be equal to 2014. $0 contains a digit, so we can't use it directly, but ${...} containing an expression that evaluates to 0 will be OK; for example:

say${$|}


For consistency, let's do the hexcat-then-perl thing with that:

$hexcat 2014 73:61:79:24:7b:24:7c:7d$ perl -M5.010 2014
2014


I think this is cheating, but it's an interesting solution nonetheless, so worth mentioning.

2Hey there, if you reverse the string and barewords, you can save a char: print""^RPQT. It might be possible to use say too on 5.10+ using -E instead of -e, but I don't know if that incurs a +2 penalty for different command-line args? – Dom Hastings – 2014-01-01T11:54:38.850

2It's considered acceptable to specify that you're using Perl 5 and use say for no penalty. – Peter Taylor – 2014-01-01T12:04:27.927

@DomHastings, personally I think if you're going to do it with "-e" or "-E", I'd say that you should then have to include the entire command, including "perl -E" in your character count. – tobyink – 2014-01-01T13:35:50.170

@PeterTaylor, say is a syntax error in Perl 5 unless you explicitly write a pragma to enable it such as use v5.10 or use feature 'say'. In Perl 6 it's enabled automatically. – tobyink – 2014-01-01T13:37:11.443

1Relevant meta answer – Peter Taylor – 2014-01-01T14:24:22.860

4You may also save another byte by using string literals that don't require quotes, such as say ABCD^srrp. – primo – 2014-01-09T17:56:10.053

40

## Ruby, 20

p 'bbc'.to_i ?\r.ord


Explanation: bbc is 2014 in base 13. Shorter than Python. Not as short as Forth.

(late comment) Can you do i? instead of i ?? – Erik the Outgolfer – 2016-06-18T15:56:43.867

23"Nobody writes jokes in base 13!" – Jean-Michaël Celerier – 2014-01-01T12:21:45.560

1w=?$;"^XA[_AXeMFGIAHJLjKNAEFEJJNHQHNKLAEMINJOJOHLAGKHOJOJ[AG[HQHRFJAH}IHAIGGwIIAHHGwKHAHGHrEUAGQFiGVAGQGfIPAFHKHHbJHAQII]MGASHNSOHATIdIAUJJRLIAWLIQGKAZOFUA]ZAeSAiPAjOAkLA".codepoints{|r|r-=68;$><<(r<0??\n:(w=w==?$?' ':?$)*r)} – Darren Stone – 2014-01-01T17:49:06.293

1p 'ߞ'.ord for a fully functional program, 'ߞ'.ord inside irb. Works on 1.9+ – Ricardo Panaggio – 2014-01-03T02:37:51.457

1Without "strange" unicode characters: ?-*?--?\v (45 * 45 - 11) (Ruby 1.8) – Ricardo Panaggio – 2014-01-03T02:45:54.963

One character smaller: ?.*?.-?f (46 * 46 - 102) (Ruby 1.8 only again) – Ricardo Panaggio – 2014-01-03T02:58:02.487

How is the base 13 indicated in the print? – Recycled Steel – 2014-01-03T17:13:51.193

@RecycledSteel, to_i takes a base parameter. I'm passing it 13 since that's the ordinal value of a carriage return (\r). – Darren Stone – 2014-01-03T17:53:56.633

z=''.to_i;o=Math.cos(z).to_i;t=o+o;"#{t}#{z}#{o}#{t+t}" without relying on charcodes – Ricardo Panaggio – 2014-01-05T13:13:08.290

@RicardoPanaggio, those are some sweet answers. You should submit. I'm going to leave mine as-is, simply for the BBC - Base-13 - Douglas Adams connections. :-) – Darren Stone – 2014-01-05T16:42:45.897

29

## Powershell, 9

+"ߞ"[""]


ߞ (U+07DE NKO LETTER KA) is counted as two bytes according to the code-golf tag info.

[""] returns the first character from the string ("" is converted to 0). The unary plus opeartor (+) converts the character to an integer.

1

According to http://codegolf.stackexchange.com/tags/code-golf/info, it should count as 9 characters.

@GlitchMr, thanks, I've corrected my answer. – Danko Durbić – 2014-01-01T11:22:35.520

3The [''] is a nice trick. I usually used [char] but that's indeed shorter :) – Joey – 2014-01-08T08:57:12.403

27

## Javascript, 18 characters

alert(btoa('ÛMx'))

Update: in ES6, using a template literal can save two characters:

alert(btoaÛMx)

The code above is fairly easy to understand by keeping in mind that btoa converts a string into another string according to a set of well-defined rules (RFC 4648). To see how the conversion works, we're going to write the input string "ÛMx" as a sequence of binary digits, where each character is rendered as its 8-bit character code.

Input character          |        Û |        M |        x
Character code (decimal) |      219 |       77 |      120
Character code (binary)  | 11011011 | 01001101 | 01111000


After reorganizing the binary digits in the last row in groups of 6, we get the binary representation of 4 new numbers, corresponding to the Base64 indices of the 4 characters in the string "2014".

Base64 index (binary)  | 110110 | 110100 | 110101 | 111000
Base64 index (decimal) |     54 |     52 |     53 |     56
Output character       |      2 |      0 |      1 |      4


As per HTML specification, the output characters can be retrieved from their Base64 indices according to this table: http://dev.w3.org/html5/spec-LC/webappapis.html#base64-table.

If you don't care about the details, you could let the browser do the calculations for you and find out that "ÛMx" is the result of evaluating atob('2014') in Javascript.

Or in a function expression (allowed by default) x=>btoa\ÛMx – programmer5000 – 2017-04-10T13:03:21.077

@programmer5000 Times have changed but the consensus in 2014 was to use alert. You can find the discussion in the link in xfix's comment. – GOTO 0 – 2017-04-10T13:24:59.590

the question mentions printing the number. I know that evaluation in a console will lead to the result being returned, but on [CodeGolf.SE] it's generally expected that JS answers will include some form of IO (alert is most common). – zzzzBov – 2014-01-03T16:22:07.573

@zzzzBov Yes, my original answer included an alert call.

– GOTO 0 – 2014-01-03T17:21:37.743

It lists Community as having removed the alert. I wonder if the standards have changed for JS codegolf or whether it's just a change that was adopted by Community.

– zzzzBov – 2014-01-03T17:28:47.633

@zzzzBov: It was a change by anonymous user accepted by SHiNKiROU and luser droog. Anyway, this edit triggered a discussion on Meta (http://meta.codegolf.stackexchange.com/questions/803/javascript-standards-for-io).

@xfix, It's not this edit alone that caused me to ask that question on meta, I've seen enough code golfs where people get into disagreements over whether the dev console counts as printing, that I figured it was worth asking for a standard. – zzzzBov – 2014-01-04T18:12:51.173

I'd love to have an explanation of how this works. – DA. – 2014-01-08T01:23:26.880

@DA. I added a short explanation. – GOTO 0 – 2014-01-09T18:38:02.440

if functions are allowed instead of full program: _=>btoa'ÛMx' which is 13 bytes (the ' symbols should be , but i can't because of the limitations in comments) – Brian H. – 2017-11-15T14:22:35.783

27

### Scala REPL - 6 chars

"?="##


(## is Scala's symbol meaning hashCode, and the Java string "?=" hashes to 2014.)

### Scala REPL - 4 chars, 5 bytes

+'ߞ'


Math on our favorite unicode character produces an Int.

24

### Yet another GolfScript entry, 9 printable characters

I believe this is shorter than any of the printable GolfScript entries so far:

"!="{*}*)


(Peter Taylor's 7-char entry beats it, but includes non-printable control characters.)

I call this the "that's so last year!" entry, because what it actually does is generate the number 2013 in 8 chars, as 33 × 61, and then increments it by one. ;-)

You win (for ascii golfscript) – aditsu quit because SE is EVIL – 2014-01-01T16:39:58.520

21

## C (31)

main(){printf("%o",' b'/'\b');}


## C (32)

main(){printf("%x%o",' ','\f');}


## C (30)

main(){printf("%x",' j'^'~');}


## C (30)

main(){printf("%d",'\a\xde');}


The most recent edit included a variation of code which breaks the rule of including numbers in the code. It uses the number "5". – Albert Renshaw – 2017-01-29T22:22:21.470

How about just "%x" and 'b'*'S' ? That's 8212 dec or 2014 hex. Saves one char one literal and two on format string. – MSalters – 2014-01-02T15:26:11.103

@MSalters No, 'b'*'S' is 8134. The prime factorization of 8212 is 222053, so I don't see an easy way to produce it. – fredoverflow – 2014-01-02T16:14:17.073

Bah, checked with integer division that 8212/'b' == 'S' :( – MSalters – 2014-01-02T16:37:48.373

and octal doesn't work either (02014=1036=37*28, 28 is unprintable) – MSalters – 2014-01-02T16:44:38.807

@MSalters The octal was a nice hint though, see my new 31 solution ;) – fredoverflow – 2014-01-02T17:34:25.427

1@FredOverflow what about my 28 char solution?

main(){puts(DATE+'\a');} – Arunprasad Rajkumar – 2014-01-08T05:18:49.050

@ArunprasadRajkumar Will it work in the next year too? (eg. display "2014") – Spook – 2014-01-08T11:41:30.017

28 characters: main(){printf("%o\b",' a');} A space is 32 in decimal. 'a' is 97. Appending them in a character literal, 32*256+97=8289. 8289 decimal is 20141 in octal. Following that printout with a backspace produces just the required 2014. – durette – 2019-11-11T05:08:08.583

Your last solutions employs implementation defined behavior and is likely invalid. – FUZxxl – 2014-09-22T12:21:50.593

20

# GolfScript, 14

'-+,/'{)))))}%


How it works: ASCII goes like this:

...
+
,
-
.
/
0
1
2
3
4
...


So, this takes the ASCII codes of each character, subtracts five, and sticks it in a string.

{...}% yields an array of the characters of a string mapped, when given a string as an argument. So, it increments each character by 5 () means increment).

The CJam version of this: "-+,/"_,)f+. :) (Then again, CJam has KD.) – Martin Ender – 2014-12-18T18:38:07.483

The + at the end is unnecessary. Also, rather than increment 5 times, just add five. Total savings: 4 chars. '-+,/'{5+}% – Justin – 2014-01-01T10:42:14.760

19@Quincunx, 5 is a number. – Peter Taylor – 2014-01-01T10:48:18.453

8@PeterTaylor I keep forgetting. – Justin – 2014-01-01T16:43:40.767

Clever solution – Sumurai8 – 2014-01-02T07:38:38.173

20

## Forth, 14

'> '" * '^ - .


1Can you explain how this works? – ProgramFOX – 2014-01-01T11:04:58.047

1I'm expressing integer constants as character literals using their ordinal (ASCII) values. So this is: 62 34 * 94 - . If you don't speak Forth, this means print (62 * 34 - 94). – Darren Stone – 2014-01-01T11:14:09.887

Forth (Gforth 0.7.0), 5 bytes, 4 chars: 'ߞ . It prints the character. – 18446744073709551615 – 2014-01-09T11:03:33.333

19

# Python, 32 chars

print ord(',')*ord('-')+ord('"')


Probably possible to reduce it using the 2014 th Unicode char ߞ, but I didn't try.

Quincunx notes that

a=ord('.');print a*a-ord('f')


is shorter by three chars.

The unicode version works, at least in Python 3 with IDLE: print(ord("ߞ")). It does not appear to work in Python 2; Python 2 probably does not support unicode. – Justin – 2014-01-01T09:59:19.137

3

Also, with your version, a=ord('.');print(a*a-ord('f')) is 2 chars shorter.

– Justin – 2014-01-01T10:06:58.723

@Quincunx It does support, however you need to have a specific encoding set to your file and append u at the end of the string (would be "ߞ"u) – Kroltan – 2014-01-01T16:23:41.703

2@Kroltan Actually, I'm pretty sure the second part is not true. In Python 2, you had to prepend strings with u to say that they were unicode, but in Python 3, all strings are automatically unicode – murgatroid99 – 2014-01-02T15:41:38.903

@murgatroid99 Ah, sorry. It's been a while since I used python, and my brain remembered about C-style float literals (1453f) – Kroltan – 2014-01-02T18:38:45.143

1Similar as commented above, without unicode print ord('.')**2-ord('f') is slightly shorter. – MLS – 2014-01-07T16:45:31.720

19

## Python 32 10 bytes, 91 85 (66 with math imported by default) bytes pure math

my_lst = []
for i in range(33, 126):
for j in range(i, 126):
if 2014 - 126 < i * j < 2014 - 33:
if j not in range(48, 58):
my_lst.append("ord('" + unichr(i) +
"')*ord('" + unichr(j) +
"')+ord('" + unichr(2014 - i * j) + "')")

for val in my_lst:
print val, '->', eval(val)


Prints all the possible ways I can write 2014 using Bruno Le Floch's method (32 chars):

ord('!')*ord(':')+ord('d') -> 2014
ord('!')*ord(';')+ord('C') -> 2014
ord('!')*ord('<')+ord('"') -> 2014
ord('"')*ord(':')+ord('*') -> 2014
ord(')')*ord('/')+ord('W') -> 2014
ord('*')*ord('-')+ord('|') -> 2014
ord('*')*ord('.')+ord('R') -> 2014
ord('*')*ord('/')+ord('(') -> 2014
ord('+')*ord(',')+ord('z') -> 2014
ord('+')*ord('-')+ord('O') -> 2014
ord('+')*ord('.')+ord('$') -> 2014 ord(',')*ord(',')+ord('N') -> 2014 ord(',')*ord('-')+ord('"') -> 2014  But this is obviously redundant, so if your interpreter is set to utf-8 by default, then all it takes is: >>> ord(u'ߞ') 2014  Also, thanks to AmeliaBR (for the idea), I tried my best to implement a pure math version: from math import* a,b,c=int(e),round(e),ceil(pi);print int(a**(b*c-(c-b))-a*a**c-a)  @JoachimIsaksson A digit is no number ;) – univalence – 2016-08-26T18:50:20.023 Here is an optimized math version (42 characters) a=True;a+=a;print(a<<a*a*a)*a*a-(a<<a*a|a) – Phylliida – 2017-03-30T06:12:54.673 ord(',')**2+ord('N') – MLS – 2014-01-07T16:42:31.893 6@MLS Well, that has a digit in it :) – Joachim Isaksson – 2014-01-07T23:00:54.293 2I optimized your math version a bit: a,b=int(e),int(pi);c=a+a;print a**(b*c-c+b)-a*a**c-a, and you can eliminate the math import altogether by making use of the fact that True in Python 2.x is identical to the integer 1 in operation, bringing it down to 50 characters: o=True;a=o+o;b=a+o;c=b+o;print a**(b*c-o)-a*a**c-a – Ponkadoodle – 2014-01-08T02:19:09.820 You can save a character if you use Python 3: ord('ߞ') – asmeurer – 2014-01-11T20:24:24.870 18 ## APL (6 bytes, 4 chars) ⊃⎕TS  Only works this year though. Why it works:  ⎕TS 2014 1 1 11 58 5 811 ⊃⎕TS 2014  Without relying on the system date, it's 10 bytes (7 characters): ⎕UCS'ߞ'  6Unfortunately, answers that only work for this year are invalid. – Joe Z. – 2014-01-01T15:34:19.607 I don't know APL: is that square box an APL char, or am I missing a font representation (Chrome) ? – Carl Witthoft – 2014-01-01T16:24:33.457 @JoeZ.: The question doesn't say that but I've added one that doesn't rely on it being 2014. – marinus – 2014-01-01T17:12:18.843 4@CarlWitthoft: It's called a quad (⎕), it's supposed to look like that. – marinus – 2014-01-01T17:12:48.470 17 ## R (20 character) @popojan (he is not allowed to post an answer here yet) has provided the solution within 20 characters. sum(T+T:exp(T+pi))-T  Output: > sum(T+T:exp(T+pi))-T [1] 2014  ## R (22 character) Anonymous user has suggested shorter solution. strtoi("bbc",pi*pi+pi)  2014 is BBC in base 13. pi*pi+pi (=13.0112) is treated by R in this context as the integer 13. Output: > strtoi("bbc",pi*pi+pi) 2014  ## R (30 character) Thanks to @Tomas! cat(a<-T+T,T-T,T/T,a*a,sep="")  Output: > cat(a<-T+T,T-T,T/T,a*a,sep="") 2014  ## R (31 character) cat(T+T,T-T,T/T,T+T+T+T,sep="")  Inspired from the answer by AmeliaBR. Output: > cat(T+T,T-T,T/T,T+T+T+T,sep="") 2014  Wow, +1! And I hoped my roman numerals solution will be the best... :( ;-) – Tomas – 2014-01-14T15:35:38.607 The variants below can be made shorter: cat(a<-T+T,T-T,T/T,a^a,sep="") (30 chars), paste0(aT+T,T-T,T/T,a^a) (26 chars) – Tomas – 2014-01-15T10:16:19.117 @Tomas, the function paste0 is not allowed, as it contains the symbol 0. Your solution makes 30 characters if using cat. – djhurio – 2014-01-21T10:47:43.640 create file 2014 in current directory and write list.files() - 12 characters – Zbynek – 2014-01-31T17:00:16.177 @Zbynek, I expect your solution not to be valid in the general scope of codegolf rules. – djhurio – 2014-02-01T09:19:04.050 15 ## Java - 77 75 characters 75 characters if print is added in a class with the main method: class C{public static void main(String[]a){System.out.print('#'*'<'-'V');}}  It means 35*60-86 which is equal to 2014 2I like your formula. – Johannes – 2014-01-01T23:30:21.640 7This actually needs to be longer, because the class definition needs to be in it to run. Shortest way I know (with Java 6):class a{static{System.out.println('#'*'<'-'V');}} – Justin – 2014-01-02T00:29:03.230 1@Quincunx, with Java 7, a main method is needed – True Soft – 2014-01-02T20:53:12.023 2@TrueSoft Note I said "with Java 6". You should choose the shortest Java possible. – Justin – 2014-01-03T00:43:51.107 2You can remove 'ln' in println to reduce chars... – laksys – 2014-01-03T10:22:21.343 @laksys, thanks for the tip :) – True Soft – 2014-01-03T12:01:09.530 @Quincunx, a class without main() would throw an error at runtime, and I don't think it's ok to put that as an answer. – True Soft – 2014-01-03T21:33:02.203 Another fun way would have been value of "?=".hashCode(). Not shorter, just fun. – MickJ – 2014-01-03T22:26:58.027 @TrueSoft see http://codegolf.stackexchange.com/a/16099/9498 – Justin – 2014-01-03T22:40:38.700 I can improve this answer (to 72 chars) but I can't find the answer button on this StackExchange.. weird :S – javatarz – 2014-01-08T10:56:26.543 You can shave 4 bytes off the expression if you're willing to use non-printables: 'j'*'' (106*19). – Boann – 2014-01-08T19:30:53.063 Read it and weep: class C{public static void main(String[]a){System.out.print(0xFBCA/' ');}} – Hot Licks – 2014-01-09T01:28:24.660 My solution (http://codegolf.stackexchange.com/a/18051/14156) comes at 69 chars. Beats @HotLicks' solution by 5 chars :) – javatarz – 2014-01-09T05:17:32.453 @JAnderton - Of course, that's a non-ASCII solution. Different category. – Hot Licks – 2014-01-09T12:10:56.963 1@HotLicks 0x has a digit. – Boann – 2014-01-09T18:28:37.350 @Boann - Drat!!! Forgot about that. – Hot Licks – 2014-01-09T18:58:02.357 13 # CJam - 2 KE  K and E are variables preset to 20 and 14. I created CJam in 2014 so it's ok if it doesn't qualify. 12 ## GolfScript, 108 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. Does it really take 3 characters to output the ascii codes for a string? That's sad :/ – aditsu quit because SE is EVIL – 2014-01-01T12:08:20.400 Printing 20 and 14 without separation is acceptable. – Joe Z. – 2014-01-07T01:37:36.127 3I guess this can be actually shortened to 6 chars with {xy}.* where x and y represent 20 and 14 in ASCII. – Vereos – 2014-01-08T17:42:28.863 @Vereos, I somehow missed seeing your comment earlier. Very nice, and sufficiently different that it's worth posting as a separate answer. – Peter Taylor – 2014-03-05T14:24:17.860 Alright, I will :) – Vereos – 2014-03-05T15:05:15.673 11 ## Haskell, 23 characters, 24 bytes main=print$fromEnum 'ߞ'


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

According to http://codegolf.stackexchange.com/tags/code-golf/info, it has 24 characters.

2Methinks that's 23 characters, @GlitchMr, though it may be 24 bytes. – TRiG – 2014-01-01T18:18:14.330

@TRiG: Well, yes. The info page says that "If you use Unicode, byte count should use UTF-8." – Konrad Borowski – 2014-01-01T19:06:36.030

11

# Mathematica, 47 bytes

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


2014

If you remove the space between All and !, you can make All lowercase. – Joe Z. – 2014-01-03T00:33:11.340

@JoeZ. Yes!.. tks – Murta – 2014-01-03T00:35:40.177

1That's pretty awesome. – BrainSlugs83 – 2014-01-13T04:17:19.753

11

## EXCEL, 24 characters:

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


Why not =COLUMN(T1)&COLUMN(N1)? – Neil – 2016-03-04T09:47:37.910

@Neil because that has digits. – Titus – 2016-12-05T21:01:55.463

1you can shorten to 16 bytes by using =COLUMN(BYL:BYL) – Taylor Scott – 2017-07-12T10:06:40.813

1How does this even work? – Timtech – 2014-01-07T23:13:34.820

2@Timtech T:T and N:N define an entire column, and COLUMN returns the column number of the reference passed in (of its top left cell, IIRC). & is string concatenation. – John Dvorak – 2014-01-08T03:15:11.760

Good job there :) – Timtech – 2014-01-08T11:50:51.537

How about the sheet references to shorten it even more? – Konrad Viltersten – 2014-01-19T18:19:34.333

10

## 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.

2 bytes "=[LongDash]" where "[LongDash]" is replaced to character and wolfram alpha returns 2014. – Margus – 2016-03-03T16:31:31.013

1That returns a list. To return only the desired string without using any numbers, try First@ToCharacterCode@"ߞ". That's 25 characters, 26 bytes. – Michael Stern – 2014-01-01T19:01:26.010

2@MichaelStern Hmm I would think a List is OK. – Ajasja – 2014-01-01T20:39:59.370

Nope, has to be the number itself. – Joe Z. – 2014-01-02T15:00:31.773

@MichaelStern Tr@ToCharacterCode@"ߞ". – alephalpha – 2014-01-02T15:59:10.963

@MichaelStern Thanks, great idea! JoeZ: Oh well:) – Ajasja – 2014-01-02T17:34:46.950

10

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"}]]


37 chars:(ToString /@ FromDigits /@ {"K", "E"}) <> "" By the way, why does FromDigits["K"] return 20?? – DavidC – 2014-09-23T02:25:03.523

@DavidCarraher Ah, 37 characters indeed, thank you. As for FromDigits["K"], "0"->0, "1"->1, "2"->2, "3"->3, "4"->4, "5"->5, "6"->6, "7"->7, "8"->8, "9"->9, "A"->10 etc. – Michael Stern – 2014-09-23T10:16:45.777

10

## 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]


1+1 Very nice! Pi counts as two chars (takes two bytes in UTF8, so it's 25 bytes). You can get it to 22 bytes by ⌊π*E(Sin@E+E^E)E^E⌋ (as you mentioned). – Ajasja – 2014-01-03T13:53:20.010

9

# 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');}


(late comment) Can you do pragma(msg,'ߞ'); and import std.stdio;void main(){write('ߞ');}? NOTE: ߞ IS U+07DE (2014). – Erik the Outgolfer – 2016-06-18T15:53:53.337

Arithmetical results are always widened to at least 32-bit integers, which is why the two character values add up and print as "2014". A single character retains its type and should be printed verbatim. – mleise – 2016-08-03T21:25:37.717

1Couldn't you improve this by ditching the import and just qualifiying the write call (e.g., void main(){std.stdio.write('ϯ'+'ϯ');}) – Travis – 2014-01-10T02:14:15.833

No, the compiler still requires the import. – mleise – 2014-05-06T00:17:00.550

9

# 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.

+:&+:


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 :).

2Another 11 char J solution: *./a.I.'&j'. – randomra – 2014-12-27T21:28:35.293

it's a shame we need a number to work with strings. You don't. You could do +/a.i.'Happy new year to you!' – ɐɔıʇǝɥʇuʎs – 2014-07-19T14:58:22.470

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.

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));


1n->+'ߞ' for 8 bytes (7 characters), since lambdas are appropriate equivalents of a program. – Olivier Grégoire – 2017-08-09T08:15:56.280

I'm pretty sure that 1 is a digit and hence forbidden. – Peter Taylor – 2014-01-09T10:14:29.143

2Mine is better, without any numbers, and has 69 character.

class C{public static void main(String[]a){System.out.print(~~'ߞ');}} – Kamran – 2014-01-09T10:29:32.043

Didn't think of that. Nice solution @Kamran – javatarz – 2014-01-09T11:16:53.513

4By the way instead of the double complement, you can basically cheat and use the useless "positive" operator. System.out.print(+'ߞ'); All it will do is promote to int. – Radiodef – 2014-01-09T17:01:12.847

1Note: This current answer produces text, not a Number object, as the question requested. – djangofan – 2014-01-10T00:55:47.067

@Radiodef Updated the answer! Thanks for the suggestion. – javatarz – 2014-01-10T05:34:06.963

@djangofan not sure if @JoeZ. meant it to be a number specifically. Either way, I don't think your statement is completely true. The only way Java can bit compliment the value is to consider it an Integer (or int). Here's a SSCCE that proves my point: Object o = ~~'ߞ'; System.out.println(o.getClass()); the output to which is class java.lang.Integer – javatarz – 2014-01-10T05:38:35.153

@PeterTaylor Thanks for the note. I removed that solution. The current solution, you'll find, is compliant to requirements (to the best of my knowledge ;)) – javatarz – 2014-01-10T05:39:36.830

@OlivierGrégoire In that case you no longer need the +. Simply o->'ߞ' will be enough, if the return-type is an int. Try it online. But the challenge states "Your task is to make a program", so I would just leave it as is. Maybe changing class C{public static void main to interface C(static void main to save a byte, by going from Java 7 to 8.

– Kevin Cruijssen – 2017-09-19T12:11:10.033

1Tip: you can use enum C{ instead of Class C{ Some IDEs think its a compiler error but it compiles flawlessly. that can result in another byte shaved off – masterX244 – 2014-03-05T15:40:11.537

8

### 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(''))


this is nice. +1 – gion_13 – 2014-01-05T17:28:51.637

7

## AutoHotkey 31/13

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


### This year special solution

send % A_year


27 bytes: a:=Asc("")Asc(""), Send %a%. The characters in the Asc functions don't show correctly SE for me but they're the ASCII characters 20 and 14 copied and pasted into AHK. They show up as DC4 and SO (or maybe S0) in SciTE. – Engineer Toast – 2017-02-24T18:53:35.917

7

# 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


Interestingly, the Befunge version of this Golfscript ('&~I'{}/-*), "&~I"-*.@, is one char longer than my Befunge answer and one character shorter than Golfscript. – Justin – 2014-01-01T10:33:03.587

1Note that 2*9=18; what you meant was 2*19=38. – Kyle Kanos – 2014-01-01T18:23:07.727

7

## GolfScript, 10

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

'N,,'{}/*+


(it calculates 44*44+78)

And now a different approach :)

'codegolfing is yummy'{+}*


5Yet another 10-char solution: "xr"{(;}% (free bonus smiley included!) – Ilmari Karonen – 2014-01-01T14:58:01.903

7

# Perl 6, 37 characters

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!'


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.

say ord "ߞ" is shorter but certainly less whimsical. – Brad Gilbert b2gills – 2015-01-13T17:02:14.193

At first I thought you meant your solution was in Perl and was 6 characters long. And then I was like, "no, that can't be right." – Joe Z. – 2014-01-07T02:03:36.617

7

# R using roman numbers, 29 characters

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


@plannapus yeah I know crazy idea :) Well, but quite normal in ancient times :) – Tomas – 2014-01-14T15:34:40.450

probably a "legacy" function :) – plannapus – 2014-01-14T15:35:38.857

4Nice! There are 28 characters in as.double(as.roman('MMXIV')). – djhurio – 2014-01-15T07:11:15.343

6

# 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

1 is not allowed – Random – 2014-01-07T05:31:36.747

I came up with 'ϯ'+'ϯ' but I believe this is 1 char shorter. – Jodrell – 2014-01-07T11:23:53.123

2if an exit code is acceptable class a{static int Main(){return - -'ߞ';}} is just 42 chars. – Jodrell – 2014-01-07T12:04:05.167

7No need for "- -'ߞ'", could just do +'ߞ'. So: class a{static void Main(){System.Console.Write(+'ߞ');}} which is 56 bytes – David_001 – 2014-01-07T13:00:41.077

1If we want to get without using the ASCII characters this is simple, int i; i = " ".Length; Console.WriteLine(i + "" + ((i-i)) + "" + --i + "" + ++i*i); – Vara Prasad.M – 2014-01-08T06:33:32.717

System.Console.Write(DateTime.Now.Year) has the added advantage that it will solve similar future contests as well :) – Julia Hayward – 2014-01-08T11:31:44.393

@JuliaHayward But it will fail this contest next year... – Spook – 2014-01-08T11:46:44.803

1@JuliaHayward and it will also fail this contest this year... (Reread the OP). – BrainSlugs83 – 2014-01-13T03:43:07.353

1OK, tongue-in-cheek comment failed utterly – Julia Hayward – 2014-01-13T09:38:56.827

1VS 2010 crashes when I try this with a fresh new Console Application... – Otiel – 2014-01-13T10:27:22.830

6

### 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:

I understand what you're trying to do, but help me out... How do I test this? Copy & pasting your code doesn't work. If I copy & paste the little "07DE" Unicode char then I actually see \U+FFDF in irb, but that char has an ord of 85. – Darren Stone – 2014-01-01T20:30:59.693

I tested it by copying the symbol from the site listed above and running it in pry. I don't think Stack Exchange recognizes the symbol by default (so simple copy & paste of the above won't work), but Ruby 2.0+ should. – O-I – 2014-01-01T21:54:48.453

2This script prints nothing. – tobyink – 2014-01-02T15:36:31.407

1How did you enter it? It works for me. Also, what version of Ruby are you using? – O-I – 2014-01-02T16:16:34.280

Interesting. The N'Ko alphabet reads right to left. Interspersing it with Roman characters seems to shuffle characters depending on where you edit/paste. Has anyone else been able to get this to run, or am I the only one? – O-I – 2014-01-02T16:29:06.863

1p ?ߞ.ord would meet the requirement of printing. – steenslag – 2014-01-03T18:55:07.260

1this works for me – mpapis – 2014-01-03T20:04:04.470

@steenslag I must have been snoozing to miss the print requirement. I thought returning 2014 was sufficient. Corrected. Thanks for pointing that out! – O-I – 2014-01-03T20:39:52.670

6

Perl 6, 11 characters:

say 'ߞ'.ord


6

# 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)


1Those are actually 16 and 49 bytes respectively. The trailing newline doesn't count. – nyuszika7h – 2015-03-15T20:04:24.513

Python 2 version can be shortened to t=True;print't+t'+'t-t'+'t/t'+'-~t<<t', where the ' have to be changed to backticks. (I don't know how the enter real backticks that are shown as such). – Reinstate Monica – 2014-01-15T00:18:44.607

1@WolframH wrap the code-block in an extra backtick. Like so: t=True;printt+t+t-t+t/t+-~t<<t where the initialization and end of the code block are done by use of: \. Alternatively, replace every \ in the code with\: t=True;print\t+t+t-t+t/t+-~t<<t – Justin – 2014-03-05T06:06:52.980

6

## 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)

6

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.$four; ?>  5 # Vim 7.4, 18 bytes :h os_v<CR>$bbbyiwZZp


Here, <CR> means a literal newline. Uses the help page os_vms.txt which contains the number 2014 on the first line.

How it works:

:h os_v<CR>            This opens the help page for os_vms.txt, the line the cursor lands on contains '2014'.
$bbb Move the cursor to the '2014' (end-back-back-back) yiw Copy it (yank-in-word) ZZ Close the help page p Paste the copied word  If numbers were allowed we could do: i2014  It's 2010 for me instead – user41805 – 2016-12-18T13:17:28.157 @KritixiLithos Weird, it works for me. What help page does :h os_v open for you? – Loovjo – 2016-12-18T13:18:34.357 1It opens os_vms.txt says Vim version 7.3 and last change was made in 2010 Jul 8 – user41805 – 2016-12-18T13:24:48.117 What version of Vim are you using? – Loovjo – 2016-12-18T13:26:23.177 I'm using Vim version 7.3 – user41805 – 2016-12-18T13:44:48.127 Perhaps it is the difference in OS (I'm using macOS El Capitan) – user41805 – 2016-12-18T13:50:52.960 @KritixiLithos I'm using 7.4 (also Capitan), that's probably why. I'll update the answer accordingly. – Loovjo – 2016-12-18T13:51:16.240 2If numbers were allowed you could just do i2014. :P – Joe Z. – 2016-12-18T21:01:26.090 @JoeZ. Oh right. I'm stupid :P – Loovjo – 2016-12-18T21:02:16.727 5 # TI-BASIC, 12 bytes A completely new approach. iPart(π+iPart(e)sinh(π+√(π  This could probably be golfed further, using a 1-byte token in place of e, but I have spent far too long on this today... 5 # Javascript (29 chars) t=!'',a=t+t+'';a+!t*t+t*t+a*a  Not the shortest but some boolean logic That was my first idea too haha. This is the best I could get: a=!'';a+a+""+(+!a)+(+a)+(a+a+a+a) (33 chars) – Benno – 2014-01-11T12:44:05.063 @Benno (+!a) is pretty clever, didn't know it works like this – Rob Fox – 2014-01-11T16:37:56.620 Yeah, it seems to treat it as an integer, only when in brackets though! Without the brackets, it reverts to 2falsetruetruetruetruetrue ;) – Benno – 2014-01-13T05:56:28.977 5 ## 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. Running PRINT RIGHT(CHECKSUM(',,.'),LEN('')) on MS SQL Server 2008 R2 produces the response "1200" – Steve Matthews – 2015-12-01T10:15:47.557 Can I suggest instead something like PRINT ASCII('')*ASCII('j') which is just 27 bytes (note that the character that isn't rendered here is the DC3 CHAR(19) in the first set of quote marks. – Steve Matthews – 2015-12-01T10:34:52.533 @Steve I actually ran the checksum one on the same version, too, but you touch on a good point that maybe the checksum algorithm could vary based on configuration. I should add a disclaimer. Also, I like your better 27-byte solution. You should post it as an answer so people can vote it up. – Muqo – 2015-12-01T18:59:22.267 PRINT UNICODE(N'ߢ') has 20 bytes – Jan Drozen – 2018-11-09T13:33:18.907 @JanDrozen Very nice, but you must've meant 'ߞ'. – Muqo – 2018-11-13T00:15:07.427 5 # Javascript, 37 bytes ((_=+!![])<<''+_+''+_)-(_<<_+++_+_)-_  It's not code golf until someone bitshifts. $ node
> ((_=+!![])<<''+_+''+_)-(_<<_+++_+_)-_
2014


It is actually Javascript, I promise.

((_=+!![])    // create 1 and save in _
<<          // left bitshift by 11 to get 2048
''+_+''+_ // create 11 from '1' + '1'
)
-           // subtract (2048 - 32) === 2016
(
_<<       // bitshift by 5 to get 32
_+++_+_     // create 5 from 1 + (++1) + 2
)-_           // subtract 2 to get 2014


2That looks so much like Brainfuck. – Joe Z. – 2014-10-19T14:59:13.530

4

# Rockstar, 34 32 bytes

X was up equalizing a word
Say X


Say hello to rockstar! (No I didn't make this one)

## Explanation:

This first line:

X was up equalizing a word


Uses Rockstar's Poetic number literals. (As opposed to regular literals, which use numbers).

This means that the length of every word after was indicates the digit in that position. So here we have a 2 length (up), a 10 length (equalizing) a 1 length (a) and a 4 length (word). The length is % 10, so equalizing becomes a 0, and the result is that the variable X has the value 2014.

Then of course we print it with Say

1Save 2 bytes by using X's instead of X was. – Shaggy – 2020-09-24T14:44:58.537

4

echo $((printf %d \':#yG))  Similar to the Ruby and Python solutions, but uses base62, now without literal 6 and 2 (but pretty long...)! Using base58 as recommended by @DigitalTrauma along with removing quotes around the printf format string, Thanks! ## Better 17 (Thanks @user13955 and @DigitalTrauma!) tr a-l /-:<<<dbcf  "It's shorthand for tr abcdefghijkl /0123456789:. Not very geeky, I'm afraid." -user13955 Also updated to use a heredoc instead of echo, thanks @DigitalTrauma! 3Not a valid solution since 2 and 6 are forbidden. – jub0bs – 2014-01-01T12:32:28.110 haha, Oh dear... – Dom Hastings – 2014-01-01T12:41:26.390 1user13955 suggested this edit, might be worth considering: echo dbcf|tr a-l /-: It's shorthand for tr abcdefghijkl /0123456789:. Not very geeky, I'm afraid. – Justin – 2014-01-07T08:18:14.620 I have a solution in pure bash/ksh. i.e. it does not use any external program like tr. However, I lack the 10 reputation to post this. ;-) – Henk Langeveld – 2014-01-08T02:10:11.397 1Pure bash|ksh 110 bytes: sum(){ (IFS=+;echo$(($*)));}&&d(){ echo$((n=n+n));}&&false||n=$?&&sum$(d;d;d;d;echo $((n=n+n+n+n));d;d;d;d) – Henk Langeveld – 2014-01-08T02:38:36.727 1I verified that the unicode solution works in ksh93. – Henk Langeveld – 2014-01-08T02:46:37.000 As the question doesn't seem to preempt showing any other characters in addition to 2014, how about 'date' for 4 characters? HNY – ICL1901 – 2014-01-08T23:55:28.220 @DavidDelMonte, I think that answer has been suggested a few times, but it has to be 2014 specifically, otherwise, you'd be on to a winner! – Dom Hastings – 2014-01-09T06:19:28.687 1Use a bash here-string to shave 3 chars of the tr solution: tr a-l /-:<<<dbcf – Digital Trauma – 2014-02-01T02:30:04.147 1And no need for the quotes around printf format string: echo$((\printf %d "'>"#wu)) – Digital Trauma – 2014-02-01T02:34:16.463

1And use base 58 and drop one more char: echo $((\printf %d ':#yG)) – Digital Trauma – 2014-02-01T05:06:50.293 @DigitalTrauma Amazing! Updated, thank you! – Dom Hastings – 2014-02-01T08:26:48.637 2@DigitalTrauma Pure bash 18: echo$[$[$#xd]#bbc] ... without fork! – F. Hauri – 2014-04-16T22:23:49.347

@F.Hauri Good one! – Digital Trauma – 2014-04-16T22:27:10.107

4

## Javascript: 47 characters

Not the shortest, but another solution:

$node > [!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]] "2014"  Source ## Javascript: 6 characters Your task is to make a program that prints the number 2014 without using any of the characters 0123456789 Ok, here you go: 6 characters. $ node
> "²⁰¹⁴"
'²⁰¹⁴'


Yeah, only for ASCII lowers. :-)

Will I win? :-D

4

Ruby:

p eval("#{""=~//}x#{"\a".ord}de")


Here is how it works inside IRB shell:

>> p eval("#{""=~//}x#{"\a".ord}de")
2014


4

# TI-Basic, 4941 31

π/π+π/π→B:B^B^B→C:C^C(C/B)-CB-B


Explanation

• π/π+π/π→B Store 2 in B
• :B^B^B→C Store 16 in C
• :C^C(C/B)-CB-B Display 2014

I'm up voting on the assumption this works; I have no way to test it. – Michael Stern – 2014-01-03T12:59:03.667

@MichaelStern Explanation added :) – Timtech – 2014-01-03T15:22:56.807

:π/π+π/π→B:B^B^B→C:Disp C^C*(C/B)-C*B-B would be 2 characters shorter. – Michael Madsen – 2014-01-08T11:02:35.380

@MichaelMadsen Thanks, updated! – Timtech – 2014-01-08T11:49:22.263

8 bytes can be saved with various tweaks: iPart(e→B:BBBB:Ans^B(Ans/B)-AnsB-B – Khuldraeseth na'Barya – 2017-11-18T20:12:00.367

Another 23-byter: iPart(e→B:BBBB^(BBB)-BBBBB-B – Khuldraeseth na'Barya – 2017-11-18T20:17:50.730

That can be 21 without assignment to B, using Ans instead. I would advise at least fixing your current solution, as it currently outputs a number well over 2014. That should be C^B instead of C^C. – Khuldraeseth na'Barya – 2017-11-19T17:13:20.850

4

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);


<?= echo date('Y') ?> – Surabhil Sergy – 2014-01-07T11:37:18.527

1@SurabhilSergy that is not a valid answer, the output will change in 2015. – zamnuts – 2014-01-07T22:44:28.323

4

## J (18)

This one doesn't use any character codes. Uses the idea that the sum of the sequence of natural numbers from 1 -> 63 = 2016.

<:<:+/i.*:+~+~p:%_


You could save two bytes <:<:+/i.*:+~dyad – FrownyFrog – 2017-11-18T16:50:13.423

4

Multiboot x86 Assembly (250 bytes):

[BITS N]
%define L(x) mov dword [eax],'zaza'-x
O equ 'A'<<\xe
F equ 'bab'-'aaa'
M equ \xFF\xAF\xAA\xFA-\xFD\xFF\xFC\xDE
C equ -(M+F)
h:
dd M
dd F
dd C
dd h+O
dd h+O
dq 'a'-'a'
dd s+O
s:
mov eax,'\'<<\r
L('HBJB')
L('IBFB')
hlt


Compile with:

nasm src.S -o 2014.bin -f bin -DN=32


(If someone has a suggestion for replacing the N macro, I'm all ears)

1You can also submit the machine code as its own answer, provided final compiled binary doesn't have any bytes from 0x30 to 0x39 in it. – Joe Z. – 2014-01-07T02:48:44.820

4

## 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.

BEAU^pupa LOL. – Joe Z. – 2014-01-09T18:59:18.343

<?=date('Y'); in any year ) – Dmitry Dubovitsky – 2014-01-13T14:50:39.260

@DmitryDubovitsky <?=date(Y); would also work, but it won't always print 2014, as the problem requires. – primo – 2014-01-13T23:32:33.747

4

# 05AB1E, 4 bytes

ToRÀ


Try it online!

T        # 10
o       # 2**x (1024)
R      # reverse (4201)
À     # rotate left (2014)
# implicit output


1Boring alternative: ŽGMR. – Kevin Cruijssen – 2019-11-19T15:42:43.193

4

# Golfscript, 6 chars

{xy}.*


This solution involves non-printable characters too.

Where x and y represent 20 and 14 in ASCII (the non-printable characters we were talking about before).

4

# Windows Calculator - 5 characters

Inspired by this answer to a different question.

Open the Windows Calculator in Programmer View (Hex mode) and type:

DFEC±

This is what the result looks like.

5If you change the size to Word, it will be an almost-valid answer that will output only 2014 – Ismael Miguel – 2015-01-09T11:49:10.283

I don't think this is a programming language... – cat – 2016-04-07T14:40:20.807

4

# 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!

Does this mean I can undelete my Microscript answer? – SuperJedi224 – 2015-12-09T02:25:14.237

@SuperJedi224 There's a list in the main post of "invalid but interesting answers". I think newer languages fall into this category. – ETHproductions – 2015-12-09T02:31:50.690

4

# Emotinomicon, 14 bytes

ߞ


Try it online!

Explanation:

  ߞ       explanation
begin quote string
ߞ
end quote string
pop N and output as a number


ߞ is U+7DE. 0x7DE is 2014.

Okay, thanks! In your code, do you convert the quote string to a number when it's outputting? – MilkyWay90 – 2019-05-25T15:05:39.507

@MilkyWay90 Emotinomicon pushes the character code of each character in the string. – Erik the Outgolfer – 2019-05-25T15:07:27.653

4

# The Shakespeare Programming Language, 219 bytes

I am using drsam94's compiler.

.
Ajax,.
Puck,.
Act I:
Scene I:
[Enter Ajax and Puck]
Ajax:
You is the difference between a fat fat fat fat fat fat fat fat fat fat fat cat and the sum of a fat fat fat fat fat cat and a fat cat!Open thy heart!
[Exeunt]


## Explanation

.


Everything from the first line to the first period is the title, parsed as a comment.

Ajax,.
Puck,.


These are characters from Shakespeare's plays, and descriptions (also comments)

Act I:
Scene I:


Act I and Scene I, used for gotos (not used here)

Ajax:


The character speaking.

You is the difference between


Assigns the difference of the next two values to the character being spoken to.

a fat fat fat fat fat fat fat fat fat fat fat cat


Every adjective multiplies by 2, a noun is either 1 or -1 depending on its connotation (in this case it's 1). 2^11 * 1 = 2048.

the sum of a fat fat fat fat fat cat and a fat cat


The sum of the next two values. 32 + 2 = 34, so 2048 - 34 = 2014

Open thy heart!


Prints the value of the character being spoken to as a number.

3

# PHP, 27 bytes

not in any way competitive to xfix´s solution, but here are a few versions with 27 bytes each:

<?=hexdec($c=hexdec(e)),$c;   # e->14->20, e->14
<?=ord(U)-ord(A),hexdec(e);   # 85-65,14
<?=($c=ord(","))*$c+ord(N);   # 44*44+78
<?=($c=ord("."))*$c-ord(f);   # 46*46-102


The first one already has been posted by brother Filip; but the others have not.

This one has 37 bytes, but I like it (inspired by the Calculator solution)

<?=dechex(hexdec(dfeb)^hexdec(ffff));


or 13 bytes with

<?=sqpu^AAAA;


But that idea has already been used by primo

3

## EXCEL: 148 bytes

=POWER(ROW()+ROW(),(ROW()+ROW()+ROW())*(ROW()+ROW()+ROW())+ROW()+ROW())-(POWER(ROW()+ROW(),ROW()+ROW()+ROW())*(ROW()+ROW()+ROW()+ROW())+ROW())-ROW()

only works in A1.

>

• It works anywhere in row 1.
• Use the ^ operator instead.
• You can also save a few using row 2 instead and a little algebra: =ROW()^((ROW()+ROW()/ROW())^ROW()+ROW())-ROW()-ROW()^(ROW()+ROW()+ROW()/ROW(. But then again, you can also save some by going all the way to Row 2014 too.
• < – Calculuswhiz – 2020-07-10T12:23:33.763

Of course, concatenation is also an option. – Calculuswhiz – 2020-07-10T12:28:59.923

3

## R, 39 31 bytes:

x=T+T;x^(x*x*x+x)*x-x^(x*x)*x-x


### R, also 39 31 bytes:

x=T+T;z=x*x;x^(z*x+x)*x-x^z*x-x


Thanks Scrooble!

### More entertaining version: 46 bytes

z=pi;x=z*z;y=exp;j=z/y(z);floor(y(x)/(x-j-j))


Not especially efficient, but I had a lot of fun messing around with this. I'm sure there's a shorter way using just those two numbers

Long-form, subbing in the variables: floor(exp(pi*pi)/((pi*pi) - pi/exp(pi) - pi/exp(pi))

In real-person numbers: floor(19333.69 / (9.869604 - 0.1357605 - 0.1357605)) = floor(2014.328)

1Save 8 bytes by getting 2 a shorter way. – Khuldraeseth na'Barya – 2018-03-15T21:58:18.137

Save a few in the entertaining one by eliminating some assignment. – Khuldraeseth na'Barya – 2018-03-19T18:52:32.580

And some more – Khuldraeseth na'Barya – 2018-03-19T18:59:28.907

3

# Julia 0.6, 9 bytes

Int('ߞ')


Try it online!

Just for completeness' sake. Here's 2018 (same trick, different character):

Int('ߢ')


Try it online!

And just for fun, here's a function using bit shifting and arithmetic instead of using character codepoints (depends on this being Julia version 0.6, which seems an appropriately golf-y hack):

# Julia 0.6, 47 bytes

(l=VERSION.minor,o=true)->o<<(l+l-o)-o<<~-l-o-o


Try it online!

Here, o=true evaluates as 1 during arithmetic. VERSION is an inbuilt constant containing the current Julia version, and VERSION.minor is 6 in this case. We left shift 1 by 6+6-1=11, giving 2048, then subtract 1<<(6-1)=32 and 1 and 1 from it, to give 2014.

2018 version would be:

(l=VERSION.minor,o=true)->o<<(l+l-o)-o<<~-l+o+o


-1 byte : '߾'-' ' – MarcMush – 2021-02-23T13:06:13.060

3

# C++, 50 bytes

#include<iostream>
int main(){std::cout<<'U'^'A';}


You should provide compile-ready code in C++. – Joe Z. – 2014-01-01T05:24:31.243

Also, use four spaces before each line for blocks of code. – Joe Z. – 2014-01-01T05:29:30.457

i added "iostrean but it disappeared, actually am new so don't know how to" – zeeshan mughal – 2014-01-01T05:29:48.790

5This is supposed to be code golf, i.e. shortest code possible. Remove unnecessary whitespace, etc – Doorknob – 2014-01-01T05:32:47.690

@DoorknobofSnow thanx, after your edition i can still see few extra spaces – zeeshan mughal – 2014-01-01T05:34:53.547

Rather verbose. int('U'-'A') prints 20, int('O'-'A') 14. Even shorter: ~~'U'-'A' also is 20. Result: #include <iostream> int main(){std::cout<<~~'U'-'A'<<~~'O'-'A';}. – MSalters – 2014-01-02T15:13:43.213

Even more shorter: 'U'^'A'. – MSalters – 2014-01-02T15:18:51.140

@MSalters Did you try compiling this? g++ on all standard versions gives an error: no match for ‘operator^’. – LegionMammal978 – 2017-08-11T02:03:40.800

Doesn't work! – NieDzejkob – 2017-12-17T17:16:50.873

3

## LiveScript, 18 bytes

The temporary solution

new Date!.getYear!


Unicode

\ߞ .charCodeAt!


Over Excitement

x=!Happy
Happy = -> console.log it
New = -> +it
Year = ->++x and Year
Year.valueOf = -> x

Happy New Year!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


4I initially assumed that you mean LiveScript, as in, JavaScript in first Netscape 2 beta. – Konrad Borowski – 2014-01-02T13:51:11.083

1@GlitchMr, that's where the name came from :-) – Brigand – 2014-01-02T20:06:55.153

Only the "Unicode" solution is valid. The "temporary" solution violates the rule "... independently of any external variables such as the date or time" – pppery – 2019-09-06T02:25:27.337

... and the "Over Excitement" solution is not a serious contender, – pppery – 2019-09-06T02:35:16.723

3

### C, 37 Characters

main(){printf("%d%d",'d'-'P','\xE');}


You could knock off 3 bytes by renaming main to f since we don't require main to be used. – MD XF – 2017-05-12T19:51:18.847

3

## MATLAB (no char codes), 47 chars

p=pi,q=p^p;e=exp(p);floor(q*q+q*e-q-q-q-p-e-e)


I played with a similar approach in Mathematica (which has a much larger set of mathematical constants built-in). How long did it take you to develop that solution. – Michael Stern – 2014-01-02T22:46:59.363

Just playing around, maybe five minutes, probably less. I’m pretty sure this is far from the most compact formula. – Christopher Creutzig – 2014-01-03T13:13:01.737

3

# JSFuck, 1267 bytes

In Javascript, here is the alert(2014) ! (Try in browser Console).

[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]][([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]+(![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]]+[!+[]+!+[]]+[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]])()


This sample uses only six different characters to write and execute code. This was generated by https://github.com/aemkei/jsfuck.

1Even though it's the longest answer rather than the shortest, +1 for JSFuck. – Joe Z. – 2014-01-01T21:04:05.993

3

# Python 2, 68 bytes

Not really short, but it abuses the fact that Python accepts booleans where an integer is required, because bool is a subclass of int.

import string;d=string.digits;T=True;print d[T+T]+d[:T+T]+d[T+T+T+T]


+1 for the first non-cheating program in this question that doesn't depend on ASCII or UTF-8 (unless I missed something) – Konrad Borowski – 2014-01-01T21:54:41.903

False can be shortened to d>d. On the other hand, d[d>d]+d[T] can be shortened to d[:T+T], so that optimization is not really necessary ;-). Also, from string import*;d=digits;saves one character. – Reinstate Monica – 2014-01-15T00:12:41.163

@WolframH Noted. I'll update it later. – nyuszika7h – 2014-01-21T21:49:29.547

3

# MathGolf, 2 bytes

ID


Try it online!

## Explanation

I   Pushes 20
D  Pushes 14


The stack is printed in full on termination.

3

## bash 35

I want something that doesn't use ASCII or UNICODE.

[ ];x=$?;y=$[x+x];echo $y$?$x$[y+y]


If you don't mind some error messages mixed in (28 characters):

[;x=$?;echo$x$?$[x/x]$[x+x]  You can get rid of the error messages if you don't mind corrupting a file (32 characters): [ 2>x;x=$?;echo $x$?$[x/x]$[x+x]


1That doesn't use ASCII? That's a bit hard. – Joe Z. – 2014-01-07T02:04:40.073

@JoeZ. My program would work just as well if the system used ebcdic. And as you know, a multitude of ebcdic systems use bash. :-) – Stephen Montgomery-Smith – 2014-01-07T03:20:57.527

Maybe 34 bytes: :;y=$[x=!$?,x+x];echo $y$?$x$[y+y] Try it online!

– Isaac – 2019-08-29T20:32:38.243

Hmmm, Just 29 bytes: echo $[x=y=!x,y+=y]$?$x$[y+y] Try it online!

– Isaac – 2019-08-29T21:04:11.163

3

# VB.NET, 59 bytes

MsgBox(((Asc(vbTab) + Asc(vbTab)) & Asc("~")) / Asc(vbTab))


takes the ascii values of a Tab twice (18) concats the ascii value of "~" (126), giving "18126" and then divides the lot by ascii of a Tab (9) = 2014

Alternatively, you can do

MsgBox Asc("j") * vbKeyPause


i.e. ascii of "j" (106) * value of the constant vbKeyPause (19), for a total of 28 characters (less than half the original).

The exact same code works for VB6 too. – Rob – 2014-01-04T00:06:26.347

user14566 suggested this edit: 27 bytes: MsgBox(Asc("") & Asc(""))

=20 =14 – Justin – 2014-01-13T07:05:42.283

2You can run this in the immediate window of VBA as ?Asc("j")*vbKeyPause, which shortens it up a bit. – Gaffi – 2014-03-05T16:46:10.420

3

### newLISP - 61 characters

I can't seem to play golf in Lisp -- is it just too wordy?

(int(apply string(map(fn(c)(find c'(b c a e d)))'(a b c d))))


3

# Javascript, 6 characters (8 bytes)

I never saw any rule saying we had to produce the number 2014 in the absence of any other output.
(Nor anything about not outputting to an error, but that's more obvious.)

new—


For me, on Firefox Nightly, this produces TypeError: "\u2014" is not a constructor, which contains the number 2014.

(If it isn't obvious, this happens because the em dash, —, is U+2014. Or, in other terms, it's the unicode character that can be represented by the hexadecimal number 2014.)

3

# C# - 64 chars

class P{static void Main(){System.Console.Write('!'*'>'-' ');}}


pretty printed

class P
{
static void Main()
{
System.Console.Write('!' * '>' - ' ');
}
}


What if we use class P { static void Main() { System.Console.Write(2014); } } – Amit Ranjan – 2014-01-07T06:55:16.800

@Amit, you can't use numbers in this challenge... – shamp00 – 2014-01-07T09:10:44.703

2if an exit code is acceptable class a{static int Main(){return '!'*'>'-' ';}} is just 47 chars. – Jodrell – 2014-01-07T12:05:54.847

1I like this one, my countdown numbers skills sucked so I ended up with ('d' + 'e').ToString() + Math.Ceiling(Math.PI); – NibblyPig – 2014-01-08T15:02:19.160

3

# Clojure, 9 bytes

Inspired by the Matlab answer, converts char \u075e to an int:

(int \ߞ)


3

How do you get a number without using any digits in the source? Lots of people had already done it with characters or strings, so I decided to use pi, predefined in most languages. From pi, you can get the numbers 3 and 4 easily using the ceiling and floor functions. Then you can use some combination of addition, subtraction, multiplication, and maybe division to get 2014. Just by experimenting around, it would be easy to figure out a function that takes in 3 and 4 and returns 2014 (such as 4^4 * 4 + 3^3*3^3 + 4*(3^4) - 4*4*4 + 4 - 3 = 2014). This one's 70 characters:

main=print$(\x y->y^y*y+x^x*x^x+y*x^y-y*y*y+y-x)(floor pi)$ceiling pi


Now, that's fine, but writing a function like that isn't much different than just repeatedly writing floor(pi) and ceiling(pi). Is it doable with only one pi? Well, in Haskell, functions can be treated as Monads with an instance defined in Control.Monad.Instances:

instance Monad ((->) r) where
return = const
f >>= k = \ r -> k (f r) r


So you can use the bind function to pass one value into two different functions: g (f x) (h x) can be rewritten f >>= flip (g.h). id >>= f can be used to pass the one value twice into the same function: id >>= (^) for example is a function that returns x to the x power. The resulting program at 207 characters is more obfuscated than golfed, but it was fun to write:

import Control.Monad.Instances
main=print.((id>>=(^)>>=flip((+).(id>>=(+)>>=flip((+).(id>>=(-)>>=flip((+).(id>>=div))))))).floor>>=flip((-).(id>>=(^)>>=flip((*).(round.sqrt.fromInteger>>=(*)))).ceiling))$pi  3 ## x86 - 16 bytes (Assembled) This little snippet moves the stack so it overlaps the video memory, to then push the individual numbers straight to the screen (They even blink! Surely a sign of good times to come). Code: (Note: '0'-'9' -> 0x30-0x39) B890A8 8ED0 2C5C 50 2C04 50 48 50 40 40 50  Assembled from: mov ax, 0xA890 mov ss, ax sub al, 0x5c push ax sub al, 3 push ax dec ax push ax inc ax inc ax push ax  Build'n run: nasm -f bin -o 2014.com 2014.asm dosbox 2014.com  @TeunPronk but that's the generation code, not the submission -- the submission is the binary – cat – 2016-04-15T23:47:22.047 1I think it shouldn't have any characters [0..9] in your code – Teun Pronk – 2014-01-31T14:56:26.113 3 # Keg, 2 bytes ߞ  Keg auto pushes any characters that aren't instructions to the stack, and ߞ has a unicode value of 2014, which then gets printed. Try it Online! 3 # Sclipting, 3 characters (6 bytes) This outputs the string "2014". 꼣갱꽀  That's not 3 bytes though. – daniero – 2014-01-23T19:52:52.037 I didn’t say it was. – Timwi – 2014-01-23T20:02:06.143 Well I can see that obviously, but the question actually asks for a byte count. – daniero – 2014-01-23T20:30:30.677 Byte count stands at 12, I believe. – cjfaure – 2014-02-08T19:34:54.163 1No, it’s 6 bytes. – Timwi – 2014-02-08T20:35:59.150 3 # Raku (previously Perl6) Raku interpret Unicode numbers as usual numbers, so: ## In REPL – 7 bytes (3 chars): ⑳~⑭  ## Without REPL – 11 bytes (7 chars): say ⑳~⑭  you can run it as rakudo -e "say ⑳~⑭" 3 ## Pure bash 18 Without fork! echo$[$[$#xd]#bbc]
2014


1How about: printf %x \'— Just 13 characters (15 bytes). – Isaac – 2019-08-30T22:18:01.153

@Isaac Awesome! I learn today this printf %d \'A newer seen before! – F. Hauri – 2019-08-31T07:41:22.273

3

# Pyth, 4 bytes

C"ߞ


Pretty straightforwards, just convert that character to an integer and print.

While this answer is now technically the shortest, I can't accept it because it was written in a language that didn't exist at the time of writing the question. – Joe Z. – 2014-07-08T11:08:44.847

@JoeZ. I understand. No worries, the rules make sense. – isaacg – 2014-07-08T15:50:17.623

3

# awk (28)

There's definitely a need for an "awky" answer... ;-)

BEGIN{print++I+I--I++I++I*I}


...oookaaayyy... the last * may be a + too. But please don't call it an "awkf*ck" solution then... ;-)

BEGIN{print++I+I--I++I++I+I}


I think, I prefer the later version now because of less different characters...

(tested with gawk and mawk)

A different use of variables that saves one byte: BEGIN{print++I+I J++I++I*I} – Pedro Maimere – 2021-01-19T00:57:06.717

3

# Racket, 18 bytes

(~a(+ #xa #xa)#xe)


3

### Excel VBA, 51 bytes

MsgBox Len("aa") & Len("") & Len("a") & Len("four")


As it's 2015 you could add an extra character onto "four" for an extra byte.

3

# 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

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


You could shorten all your mains to f, saving 3 bytes each, and your submission will still be valid. – MD XF – 2017-05-12T19:53:31.647

3

# BASIC v2.0 (Commodore 64), 24 bytes

PRINT ASC("&")*(ASC("V")-ASC("!"))


The Commodore BASIC tokens are single-byte representations of BASIC keywords. This should result in storing the above line as 24 bytes instead of 34.

I was aiming to construct an equation based on character codes, so I was looking for ideal characters in PETSCII. To minimize the number of operations in the equation, it seemed like a good idea picking a pair of divisors of 2014.

However, none of the combinations were perfect, because either one of the divisors were too large, too small or the character code of a number.

So I ended up using 38 which is CHR$("&") and 53, but instead of directly using the latter which is CHR$("5"), I used the difference of 86 and 33. These are the character codes for CHR$("V") and CHR$("!") respectively.

Tested in VICE and FC64.

3

# Java 8, 33 bytes

()->Integer.parseInt("bbc",'\r');


3

# Jelly, 3 bytes

⁽¥Æ


Try it online!

– Stephen – 2017-08-11T01:14:51.170

2

# ><>, 7 4 bytes

"nߞ


Try it online!

## Explanation

"nߞ      : Put the string nߞ onto the stack.            Stack: [110, 2014]
n       : Print the top item of the stack as a number. Stack: [110]
ߞ      : Error out.


Huh, this worked really well in Runic too. One byte shorter than the solution I came up with a couple weeks ago. Not sure why I didn't try that before.

– Draco18s no longer trusts SE – 2019-02-06T02:30:33.543

2

# 05AB1E, 6 bytes (non-competing)

T·žvÍ«


Uses the CP-1252 encoding. Try it online!

Explanation:

T       # Push 10
·      # Multiply by 2
žv    # Push 16
Í   # Subtract 2
«  # Concatenate
# Implicit output


+1 from me. Here a 6-byte alternative: Txs4+J (Push 10; Duplicate and double copy (10,20); swap; Add 4; Join together).

– Kevin Cruijssen – 2018-09-10T13:31:54.257

5-byter 'ߞÇ. Uses default ASCII encoding (so both ߞ and Ç are 2 bytes each), because ߞ doesn't exist in 05AB1E's codepage.

– Kevin Cruijssen – 2019-01-14T12:51:20.300

2

# Brain-Flak, 50 bytes

((((((()()()()()){}){})){}{}()){()()({}[()])}{}())


Try it online!

Brain-flak is great for restricted source challenges because there are only 8 valid character to begin with: brackets. (e.g. []{}()<>).

This was made possible with the help of @ASCII-only's integer metagolfer, which is currently hosted online at brain-flak.github.io/integer

1

Using the integer metagolfer in WheatWizard's brain-flak optimizer I found ((((((()()()()()){}){})){}{}()){()()({}[()])}{}()) which is 4 bytes shorter.

– 0 ' – 2016-12-15T07:30:55.973

The challenge requires languages in which 0123456789 are valid tokens. Does Brain-Flak satisfy this requirement? – Dennis – 2017-01-07T17:17:15.563

2

# Brain-Flak, 46 + 3 = 49

Try it online

(([()]([(()()())]((((({}){}){}()){}){})))()())


This one uses the -A flag for +3 bytes. It pushes the ASCII values for 2014 to the stack which outputs as 2014 in ASCII mode.

The challenge requires languages in which 0123456789 are valid tokens. Does Brain-Flak satisfy this requirement? – Dennis – 2017-01-07T17:17:03.453

2@Dennis yes @lt flags use decimal literals – Wheat Wizard – 2017-01-07T21:59:15.817

2

# Sinclair ZX81 15 bytes 10 bytes

 PRINT CODE "=";CODE ":"


As the ZX81 has a non-ASCII compatible character set, the character code for = is 20 and for : it is 14 - simples.

1You've used a 1 in your answer. – caird coinheringaahing – 2017-04-02T01:09:40.143

There is no way around making a ZX81 BASIC program without using line numbers. You may take out the line number and run it in direct mode if you wish. – Shaun Bebbers – 2017-04-02T06:27:16.507

This is perhaps another reason why retrocomputing @ Stack Exchange should allow one-liners and code-golf therem, but apparently this would not constitute retro computing. Or something. – Shaun Bebbers – 2017-04-02T06:30:08.930

Correction on my last comment: there not therem. – Shaun Bebbers – 2017-04-02T07:24:22.993

2

# Lean Mean Bean Machine, 53 bytes

\\\
O))\ \
#)) o )
))u ))
))  ))
))  )u
))  ~
~~~


I like this.

2

# J, 21 bytes

,":,.$,:~}.,:,:'golf'  Try it online!  'golf' One dimensional array ,:,: Itemize twice (1x1x4 array) }. Drop first element (0x1x4) ,:~ Append to itself as distinct items (2x0x1x4)$                 Get dimensions (2 0 1 4)
,.                  Flatten items, essentially prints 2014 vertically.
(so there are no spaces)
,":                    To strings, flatten.


# 20 bytes

#.(#_),,~(,~,~#_),%_
#.(#_),,~(,~$,._),%_  # 15 bytes do'bbbc',~":_bd  # 11 bytes ,":_bk,:_be  2 ## ><>, 7 bytes aa+nen;  Try it online! aa + n e n ; ^^ ^ ^ ^ ^ ^ Push 10 to stack twice, add top two items in stack, print top item in stack as a number, push 14 to stack, print top item in stack as a number, stop.  26 bytes: eaa+n< – Jo King – 2018-04-12T00:43:43.133 2 # JavaScript, various solutions: 131, 44, 43, 30, 28, and 16 characters (16 bytes) Original answer: JS-fuck inspired JavaScript, 131 characters: (!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![]))-((!![]<<!![])<<!![]<<(!![]<<!![])<<!![])-!![]-!![]  Shifty truth edition: (true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true))-((true<<true)<<true<<(true<<true)<<true)-true-true  ## Addenda Edit: I'm having too much fun with this... combining type coercion with hex strings, 44 characters. a=+[];(a+'xade')-(a+'x'+(+!a+!a+!a)+a+a) a=+[];(a+'xade')-(a+'x'+(++a+a+a)+--a+a)  a is 0, 0xade is 2782. We need to subtract 768 to get 2014. 768 is 0x300. (a+'x'+(+!a+!a+!a)+a+a) and (a+'x'+(++a+a+a)+--a+a) are two ways of producing "0x300", so the final result is "0xade" - "0x300", which JavaScript coerces back to numbers, resulting in 2014. Here is a radix 36-based parseInt solution, 43 characters: (a=+[]);parseInt(++a+'jy',a+++a+++''+(a+a))  First, we initiate a variable a at 0 (clearly this requires non-strict mode), increment it to 1, concatenate with 'jy' - 1jy happens to be 2014 in radix 36. The quickest way call parseInt at this radix is to generate the string "36" and abuse type coercion again: a++ + a++ results in 3, with a set to 3 as well, which means (a+a) is 6, so a+++a+++''+(a+a) results in "36", resulting in 2014. After coming up with these two solutions I started looking at other JavaScript answers (thanks for the tip, Scrooble!), to see if combining ideas from other people gives interesting results. Zaq's approach can be shortened with the hex string trick to 28 characters: -~[]+!![]+[+[]]+ +(+[]+'xe')  How it works: -~[]+!![]+[+[]] results in "20", +[]+'xe' results in "0xe". +"0xe" is 14, so "20" + +"0xe" becomes "20" + 14, which becomes "2014". Now here's a fun fact: 20 in hexadecimal is 0x14! Oh, and 14 is 0xe. So +('0x' + 0xe) produces 20. Combining this with the above results in a 30 character solution: _=+[]+'x';+(_+ +(_+='e'))+[+_]"  Finally, initially inspired by Dan Prince's answer I came up with what might be the shortest possible JavaScript solution abusing hex-strings, at 16 characters: +[]+'xfbc'>>!![]  0xfbc is 4028. Shift right by one, and we have 2014. If anyone knows of a shorter way to generate 0 or 1, let me know. Edit2: having exhausted the hexadecimals, we should of course also look at binary and octal notation! // +'0b11111011110' +(a=+[],a+++'b'+(a+(((a+=[a]+a+a+(+[])))+a))) // +'0o3736' +(a=-~[],a=a+++a+++''+(a+a),+[]+'o'+((+a+1)+a))  2Welcome to the site, and nice first post! – Khuldraeseth na'Barya – 2018-03-08T20:55:48.233 Thanks. Do you know of any ways to show all answers on one page so I can quickly check for duplicates next time? ;) – Job – 2018-03-08T20:57:31.693 1@Job the stack snippet in the question body lists by shortest solutions and the shortest solutions by language. I've never been able to see them on mobile/in the app, though, so your mileage may vary. – Giuseppe – 2018-03-08T21:11:30.487 1 @Job You can do a search with inquestion. For just about all questions, there'll be just one page of results. The 17005 is this question's ID, found in the URL. – Khuldraeseth na'Barya – 2018-03-08T21:23:33.640 1Thanks for the tips! Added some more solutions :) – Job – 2018-03-09T00:28:49.897 2 # Pyt, 27 bytes ɳąḞḞ⬠⬠⬠π⎶⁻⦋ĐąžΠ²+ĐŚřƩ½*-⁻⁻Ɩ  Try it online! Not exactly a serious contender, just had some fun. ɳ push '0123456789' as string ą convert to array of digits [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] ḞḞ for each item in array, replace with xth fibonacci # (2x) [1, 1, 2, 3, 8, 34, 377, 17711, 9227465, 225851433717] ⬠⬠⬠ for each item in array, replace with xth pentagonal # (3x) [1, 1, 1820, 66045, 240027425, 29321506727800, 6947548864499411875070L, 165405818231059923692911546880492501L, 898044801648686628863443901192030771814779461710865094720L, 115670237695821250427139838385782853032222541808893547195455834936957002151009052998969975100L] π⎶⁻ push pi, round, and decrement (2) ⦋ get the 2th element of that list (1820) Đ Duplicate 1820 ąž make an array of digits and remove zeroes [1, 8, 2] Π multiply together (16) ²+ square and add (2076) ĐŚ Duplicate and sum digits (15) řƩ make a range from 1 to 15 and sum all (120) ½* multiply by 1/2 (60) - subtract (2016.0) ⁻⁻ decrement (2x) (2014.0) Ɩ cast to integer (2014) implicit output  2 ## JavaScript (ES6), 52 46 bytes t=-~'';console.log(${++t}${t-t}${t/t}${t*t})  Try it online! Will log through console.log the number 2014 as a string. Thanks to Jacob for multiple optimizations saving 6 bytes ### Explanation t=-~'';  In JavaScript, ~~ will convert the proceeding value to a number, in this case, true equates to 1. Set the value of t to 1 by using ~ on an empty string, which would equate to -1, then take the opposite of that number, 1. For more info about tilde in JavaScript, see this article. console.log( ... )  Logs the template string ... with ${} expressions available, where ... includes:

${++t}  Sets w to t+t, which would be 2, which would return the number 2. Added to string. Set t to itself + 1, and display the final result, 2 ${t-t}


Displays t-t, which would be 0, which would return the number 0.

${t/t}  Takes the value of t and divides by itself, returning 1. ${t*t}


Takes the value of w*w*w*w-w (or w ^ 4 - 1), where w (as previously set) is 2, and subtracts w from it, and returns the result. Added to string.

Takes the value of t*t (or t ^ 2), where t (as previously set) is 2.

The added expressions equate to 2014, which ... is in the log.

Technically you don't need the ~~ before the true – Jo King – 2018-06-19T00:46:50.713

t=-~'';console.log(\${++t}${t-t}${t/t}${t*t}) – Jakob – 2018-06-19T01:45:21.603

2

## Fortran: (43 27)

print*,z'FBC'/len('hi');end


Thanks to Hristo Iliev, the above is about 40% smaller! z'FBC' returns the decimal form of that hex value (which is 4028), len returns the length of hi (i.e.,2).

print*,ichar(',')*ichar(',')+ichar('N');end


Converts the string , and N to ASCII values: 44 & 78 respectively: 44**2 + 78 = 1936 + 78 = 2014.

Shorter version using hexadecimal literals: print*,z'FBC'/len('hi');end. – Hristo 'away' Iliev – 2014-01-08T12:31:53.937

@HristoIliev: Totally forgot about printing hex via z! Thanks a bunch! – Kyle Kanos – 2014-01-08T14:51:19.940

2

# PHP (21 chars)

<?=ord('').ord(''); //These are not empty strings ;)


If you don't believe it, see the proof.

1If it contains non-printables, you should provide a hex dump or list them. – mbomb007 – 2016-03-04T20:06:31.940

That looks like 19 characters to me. – Joe Z. – 2014-01-02T15:06:50.087

1(Oh wait, nonprintables.) – Joe Z. – 2014-01-02T15:07:17.337

2

# Clojure (177 characters)

In the true Lisp-ish spirit that "too many parentheses are never enough" I present:

(Integer. (clojure.string/join [(+ (second (range))  (second (range))) (first (range)) (second (range)) (+ (second (range)) (second (range)) (second (range)) (second (range)))]))


How it works:
The function range produces a lazy sequence of numbers. If no starting point and ending point are specified the range starts at zero and extends infinitely in the positive direction; however, because it's a lazy sequence the numbers are not produced until needed. Thus, applying the first function to the result of the range function without arguments produces the value 0, which is the first element in the sequence 0 to positive infinity. Applying the function second to such a range produces the value 1. From there it's a simple matter of producing enough 1's and summing them up to get 2 and 4, then converting them (implicitly) into strings to join then together, then converting the resulting string back to an integer. (I find it amusing that this is actually longer than some of the Brainf*ck answers - and to add to the horror, it's also legible :-).

Share and enjoy.

:-)

Do you need all that whitespace? – cat – 2016-04-18T02:38:18.210

I suppose that you don't need to convert back to integer, instead add an output function. – Paŭlo Ebermann – 2014-01-05T18:15:34.983

2

# Lost, 5723 22 bytes

<^/*(
)/@+
">>v
^?%<<


Byte-count more than halved (-35 bytes) thanks to @JoKing.

Let me start with an explanation of Lost itself. Lost is a 2D path-walking language. Most 2D path-walking languages start at the top-left position and travel towards the right by default. Lost is unique however, in that both the start position AND starting direction it travels in is completely random. So making the program deterministic, meaning it will have the same output regardless of where it starts or travels, can be quite tricky.

A Lost program of 2 rows and 5 characters per row can have 40 possible program flows. It can start on any one of the 10 characters in the program, and it can start traveling up, down, left, or right.

In Lost you therefore want to lead everything to a starting position so it follows the designed path you want it to. In addition, you'll usually have to clean up the stack when it starts somewhere in the middle.

Program explanation:

The 22-bytes program is similar as the previous 23-bytes program below, but with a smarter path to save that byte:

v<<<<<>>>>>
>%?"^ <"*+@


The "^ <" will push the character-codepoints for the three characters in the string, being 94 32 60 respectively. The * multiplies the top two, and + adds the top two of the stack, so it becomes 94+(32*60), which results in 2014.

The @ will terminate the program, but only if the safety is 'off'. When the program starts the safety is always 'on', otherwise the program starting at the exit character immediately terminates without doing anything.
The % will turn the safety 'off'. So as soon as the % is encountered and the safety is 'off', the program can be terminated with an @.

The ? is to clean up the stack if it started somewhere in the middle.

And finally the v<<<<<>>>>>, > and use of ^ < in the string are to lead the program path towards the correct starting position for it to correctly print 2014. Note that the top line could have been v<<<<<<<<<<, but that the reversed part >>>>> will wrap-around to the other side, making the path shorter and therefore the performance slightly better. The byte-count remains the same anyway, so why not.

Now for the 22-bytes solution, and how it actually is the same as the 23-bytes solution, but with a different path.

The arrows are still used to lead the path into the given direction. The / are used as a mirror. So if we go from right to left and encounter the /, it will continue downwards; if we go from the top to the bottom and encounter the /, it will continue towards the left; etc.

The ( will pop the top value on the stack and push to to the scope, and the ) will do the reversed: it pops from the scope, and pushes it back to the stack.

So regardless of where we start and in which direction we travel, the path leads towards the first < of the bottom row. From there, the program flow travels in this order:

%?^        Direction changed upwards
" <^" <    Direction changed towards the left
(*/        Direction changed downwards
/          Direction changed towards the left
) +@


So it will:

• Turn the safety 'off' with %;
• Clean the stack with ?;
• Push the character-codepoints for " <^", which are 32 60 94 respectively;
• Pop the 94 and store it in the scope with (;
• Multiply the 32 60 with *, resulting in 1920;
• Push the 94 from the scope back onto the stack with );
• Add the 1920 94 together with +, resulting in 2014;
• And then terminates the program with @, implicitly outputting the top of the stack.

1

Why not calculate the 2014 code point and skip the -A flag? 23 bytes

– Jo King – 2018-11-01T09:46:25.533

@JoKing Ah, didn't thought about that.. Thanks a lot for halving the byte-count! :) – Kevin Cruijssen – 2018-11-01T10:18:02.807

122 bytes. How it actually works is a bit weird... – Jo King – 2018-11-01T10:26:42.087

@JoKing Oh, smart! That 23-byter was pretty obvious and I can't believe I missed it now. But that 22-byter is very smart and not something I would have thought of myself. Well done! It's only my first Lost answer, so hopefully I will get better at it. Btw, out of curiosity, is there a reason the language is lacking a divide and modulo operator? – Kevin Cruijssen – 2018-11-01T10:43:20.903

shrugs Minimalism? – Jo King – 2018-11-01T12:53:48.153

2

# Python 51

Using true = 1 and false = 0

t=True
print str(t+t)+str(t-t)+str(+t)+str(t+t+t+t)


clever. 40 chars in PHP: $t=true;echo$t+$t.$t-$t.$t.$t+$t+$t+$t; – zamnuts – 2014-01-05T10:08:56.750

Damn just wrote that while reading the answers well done, – Noelkd – 2014-01-05T10:37:55.350

2

## C, 31 bytes -- without a multi-character literal

main(){printf("%o",'\xe'*'J');}


Save 3 bytes by shortening main to f, since we don't require main to be used. – MD XF – 2017-05-12T19:52:12.727

2

# Python, 30 chars

s=int('RZ',ord('$'));print s+s  2014 => 2 * 1007 => RZ in base 36 => ascii code for$ character

In interpreted mode, without the print statement it is 24 chars:

s=int('RZ',ord('$'));s+s  2 ### Julia, 13 characters ('x'-'e')*'j'  In Julia, most arithmetic operations, when applied to a single character, convert this character to its ASCII integer value. x, e and j are respectively 120, 101 and 106, therefore (120-101)*106 is 19*106=2014. julia> ('x'-'e')*'j' 2014  Edit: 11 characters, thanks to Glen O A different choice of characters allows us to skip parentheses: '.'*'.'-'f'  Just thought I'd point out that a different sequence can save you a few characters. For instance, '.'*'.'-'f' is only 11 characters. – Glen O – 2014-06-06T03:36:23.017 @GlenO thanks! I added it as an edit. – plannapus – 2014-06-06T07:15:46.683 2 # Bash, 29 bytes Bash without using external programs: echo$((x=++y+y))$?$y$((x+x))  Reduce to 25 bytes by using: echo$[y=++x+x]$?$x$[y+y]. – Isaac – 2019-08-30T14:50:14.410 2 bc, 7 bytes. Try it Online!! K*ZZ+Y bc, 8 bytes. Try it Online!! K*A*A+E Which needs 14 bytes to run in bash: bc<<<"K*A*A+E" In bc the upper (single) letters maintan their meaning as a number in 10-36 range in any input base. A previous approach changed the input base: echo ibase=D\;BBC|bc Make numeric base 13 (D) and print BBC in that base --> 2014. 2 # Flobnar, 17 14 bytes <+\@:! +<>.!..  Try it online! ### Explanation:  @ Start, going left \ Push to the stack the value from the bottom row > .!.. Several print statements we will get back to < :! Add the not of the top of the stack to itself + This is !0+!0 = 2 . Print the 2 . Print the result of the print (0) .! Print the result of the not of the print (1) \ Continue forward after pushing the zero to the stack <+ :! Add the same 2 from the beginning to itself +< (!0 + !0)+(!0 + !0) = 4 And print implicitly as the last value  2 # 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. However, in Python 2, len('ߞ') returns 2. And since ord only takes a string of length 1 (or a char, but Python doesn't have chars), Python 2 doesn't really like that: TypeError: ord() expected a character, but string of length 2 found 2 # C#, 25 bytes (24 characters) Console.Write((int)'ߞ');  Try it online! ### Explanation The decimal Unicode of ߞ is 2014, so you can just cast it to an int and 2014 is printed. 1Welcome! Though admittedly, the answer is simple, in general it is best to add an explanation or link to an online interpreter. Code-only answers are usually automatically flagged as low-quality, meaning that those answers have to be reviewed by someone. – mbomb007 – 2019-08-30T21:19:43.300 @mbomb007 yeah i was thinking about adding a link to tio.run, i'll do that – Sagittarius – 2019-08-31T09:09:09.340 2 # 05AB1E, 221087 6 bytes T·žvÍJ  Try it online! How it works: • T pushes 10, · doubles it, we get 20. • žv pushes 16, Í subtracts 2, we get 14. Lastly, we concatenate them using J! 1 Hi, welcome to PPCG! See here for an existing 6- and 5-byte 05AB1E answer. But to golf your approach a bit more: 1. you can remove all « and only have a single J (join) at the end. 2. You could save values in between with © and use it later on with ® (i.e. žh¦¦©нžhнžh¦н®¦¦нJ). 3. There are builtins for the numbers 0, 1, and 2 if you don't change them later on, which are ¾, X, and Y respectively, so an alternative 6-byter could be Y¾XY·J (push 2; push 0; push 1; push 2 and double it; join all values on the stack together). – Kevin Cruijssen – 2019-09-02T08:12:57.003 (this last one doesn't really look like your initial approach anymore, but just stating it as example). And if you haven't seen it yet, tips for golfing in 05AB1E might be interesting to read through. :) – Kevin Cruijssen – 2019-09-02T08:13:55.680 Digits aren't allowed though. Otherwise just Ž7æ (compressed 2014) or 2014 itself would be enough. :) – Kevin Cruijssen – 2019-09-04T10:42:56.160 @KevinCruijssen Oh, how silly. I forgot. – mekb – 2019-09-04T10:43:40.397 I've found a 10 byte solution now. I don't think there might be any room for improvement. – mekb – 2019-09-04T11:08:52.633 +1 from me for effort. :) PS: žz could be ₁ (256), ₂ (26), or ₅ (255). – Kevin Cruijssen – 2019-09-04T11:21:24.207 Actually, I found a 8 byte solution. I changed 0 and 1 to 10 and reversed it. – mekb – 2019-09-05T07:36:15.440 2 # ~-~! (No Comment), 41 Pretty basic solution. '=~~~~~:''=~~,','@'':@''-~~:@''-~:@''+~~:  Pretty good for just 8 unique characters, eh? xD So this could theoretically be stored in 123 bits, or ~15.4 bytes. 2 # k [16 chars] (*/"i"$".,")-@""
2014


### Explanation

Get the ASCII value of ",.".

"i"$".," 46 44  Find the product */"i"$".,"
2024


Get the data type of char.

@""
10h


On running the complete code (2024-10)

(*/"i"$".,")-@"" 2014  212 chars: +/&" ~~~~h'"; 6 chars, 7 bytes, unicodey: \i$"ߞ" – zgrep – 2017-04-13T13:00:35.033

2

# Python, 52 49 chars

from math import*
print(int((e*pi+e)**pi+e/e+e))


Works in Python 2 and Python 3.

2

# 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!

1Ayy, somebody actually uses my language :D – Dion – 2020-08-06T05:22:17.537

I tried using _ in the online interpreter, but for some reason it didn't work. I made a Gtihub issue on it. I'd like to solve some more puzzles, but that operator is causing a bit of a problem. – Razetime – 2020-08-06T05:24:27.157

Ok, will check and fix. Thanks for notifying me! – Dion – 2020-08-06T05:25:34.183

2

# Labyrinth, 12 bytes

))!!)!))))!@


Try it online!

))))_#!!#!!@


Try it online!

))_#"
) !!@


Try it online!

))!"
)
)#!!@


Try it online!

I tried both linear and complex layouts, but I can't figure out how to remove a single byte from any of these programs.

111 bytes by reusing the prints. And an alternative – Jo King – 2020-09-30T08:09:49.477

2

# 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!

2

# Husk, 4 bytes

c'ߞ


Try it online!

2

# Python 3, 49 bytes

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


# Python 3, 15 bytes

print(ord('ߞ'))


2

Welcome to the site, and creative first answer! You can improve your score by remove the spaces before the [, before the print and before the end: Try it online!

– caird coinheringaahing – 2020-11-17T21:54:27.990

@cairdcoinheringaahing thanks for notice me! – Amir reza Riahi – 2020-11-17T21:57:28.097

2

## ><> (9 bytes ASCII)

In pure ASCII,

'd!:'*+n;


This pushes d, !, and : to the stack, then multiplies the numerical values of top two entries, and adds the value of the last entry before outputting the value on top of the stack as a number and ending.

Using Unicode this can be reduced to 6 bytes:

'ߞ'n;


Simply outputs the numerical value of ߞ and ends.

1You could shorten 'ߞ'n; to 'n;ߞ, I believe. – Addison Crump – 2015-11-01T12:03:30.323

2

# C#, 4 characters, 5 bytes

+'ߞ'


Note: you need LINQPad to run it, not Visual Studio. LinqPad is good for CodeGolfing in C#.

1It's 4 characters, yes, but 5 bytes. – Joe Z. – 2014-09-20T17:37:56.573

@JoeZ. ok, updated to reflect the number of bytes. Still way better than previous 63 and 64 bytes solutions. – Cœur – 2014-09-21T17:45:24.563

2

# J (13)

#.a.i.'_!!! '


Interprets the ASCII value of _!!!  (95 33 33 33 32) as a binary number (it's weird that this is possible, I agree). This produces 2014.

# J (15)

This one doesn't use any character strings. It's based on the weird coincidence that the sum of the first 46 primes is 4028: double 2014.

-:+/p:i.<:+:_bn


If anyone knows of a shorter way than <:+:_bn to represent 45 (preferably without strings), please let me know.

2

# Insomnia, 7

Each line is one program doing the same thing: print 2014 to output stream.

e}u#Hi-
e}u#Hs-
e}u#H}-
e}g#*i-
e}g#*s-
e}g#*}-
e}gKHi-
e}gKH}-
e}gKxi-
e}gKxs-
e}gKx}-
e}u#dK-
e}u#eK-
e}u#fK-
e}gKdK-
e}gKeK-
e}gKfK-


2

# JavaScript, 24 bytes

A bit long, but no idea how this way got left out...

alert("ߞ".charCodeAt())

Explanation

The character ߞ is obtained by doing String.fromCharCode(2014) . Thus the code is actually just converting that character back to its character code and alerting it.

Thanks to hsl for this shorter version

That code doesn't work. Did you mean alert("ߞ".charCodeAt())? – NinjaBearMonkey – 2014-12-27T21:12:41.190

@hsl String.charCodeAt is present only in Firefox, it seems. But I'll use charCodeAt since its multi browser and shorter . Thanks! – Optimizer – 2014-12-27T21:25:49.687

2

## Python 2 (19 bytes, ASCII only, CPython-specific)

print hash("w_'qe")


Tested only on 64-bit, but I assume/hope that since 2014 is small and positive the results would be the same on 32-bit? Originally tested on Python 3, but ProgramFOX confirms it also works on Python 2.

## Python 3 (31 bytes, ASCII only)

print(ord("\N{NKO LETTER KA}"))


Quite fond of this one, even though better solutions exist. The equivalent Python 2 code is no shorter, as it required a u string prefix.

1I tested on Python 2.7, and it works fine there; so you can save one character. – ProgramFOX – 2015-01-01T16:40:28.800

I found the same python 3 version, but shorter (16 bytes) as I didn’t restrict myself to ASCII :print(ord('ߞ')) – Frédéric Grosshans – 2015-11-04T14:37:06.237

2

# CMD - 42 bytes

set/aa=f
set/a%a%xAAA-%a%xFF-%a%xFF-%a%xCE


The 'trick' is that when using the /a switch on the set command, letters (and other invalid characters) are evaluated as 0. I then just use hexadecimal to evaluate 2014. The 0 is needed because in CMD hexadecimal must be expressed with the leading 0x. There is almost definitely a shorter way to get to 2014...

%a%xAAA - %a%xFF - %a%xFF - %a%xCE = 2730 - 255 - 255 - 206 = 2014


2

# Python (30 chars)

(10 + 9) * 106 = 2014

(ord('\t')+ord('\n'))*ord('j')


2

# Hassium, 77 Bytes

Really excited about this one. It gets Math.pi and divides it by itself in variable a (1), then uses increment and basic math operators to get it to 2014.

use Math;func main(){a=Math.pi;a/=a;print(++a)print(a-a)print(a/a)print(a*a)}


Run online and see expanded here

2

# Milky Way 1.0.0, 22 bytes

<^a:::+;:l+:>h<::++-<-


### Explanation

<          <     <   # rotate the stack leftward
^                   # pop the TOS without outputting
a                  # logical not on the TOS
:::  :   ::       # duplicate the TOS
+       ++     # push the sum the top two stack elements
;             # swap the top two stack elements
>           # rotate the stack rightward
h          # push the TOS to the power of the second stack element
- -  # push the difference of the top two stack elements


The stack defaults to ["", 0].

### Stack Visualization

["", 0]                # default stack

[0, ""]                # <
[0]                    # ^
[1]                    # a
[1, 1, 1, 1]           # :::
[1, 1, 2]              # +
[1, 2, 1]              # ;
[1, 2, 1, 1]           # :
[1, 2, 1, 10]          # l
[1, 2, 11]             # +
[1, 2, 11, 11]         # :
[11, 1, 2, 11]         # >
[11, 1, 2048]          # h
[1, 2048, 11]          # <
[1, 2048, 11, 11, 11]  # ::
[1, 2048, 33]          # ++
[1, 2015]              # -
[2015, 1]              # <
[2014]                 # -


By default, if nothing has been output manually, the bottom stack item is output on termination of the program.

# Milky Way (current version), 8 bytes

XZ*W+U+!


### Explanation

X         # push 20 to the stack
Z        # push 100 to the stack
*       # push the product of the TOS and STOS
W      # push 10 to the stack
+ +   # push the sum of the TOS and STOS
U    # push 4 to the stack
!  # output the TOS


2

# Jolf, 3 bytes

(It's almost 2016. Language obviously postdates question.)

@ߞ
@  Get charcode of next character
ߞ charcode 2014


2

# JavaScript, 9 bytes

btoaÛMx


This Base-64 encodes ÛMx to make 2014.

2

## Lua, 32 bytes

print(#'XX'..#''..#'X'..#'XXXX')


This makes use of the length operator # used on string literals to get numbers, and the lengths are concatenated (with ..) to print 2014

2

# Fourier, 12 bytes

Non-competing: Fourier is newer than the challenge

I know I'm two years too late, but it seemed like a fun challenge to do in Fourier.

^^ovvo^o^^^o


The command ^ increments the value of the accumulator (which starts at zero) and v decrements the value of the accumulator.

Try it online!

2

## Straw, 13 bytes

(…………………σ)«$>  « sum the codepoint of all characters in a string, $ convert from unary to decimal and > is the print operator.

Try it online

1

# Vim 8.0, 15 bytes

:h u
ggf:wywZZp


I didn't see a vim answer yet, so I figured I'd add one. This opens up a helpfile, so it is specifically vim 8.0, since it might not work with a future version that updates that file.

Clever! In my case :h followed by 9wywZZp works. So down to 10 symbols/bytes – defhlt – 2016-12-05T17:22:50.943

@defhlt OK, good to know! Although technically that doesn't work because it uses the digit 9. You could do f:w in place of 9w for only one byte more. Which version of vim do you have? – James – 2016-12-05T17:25:04.843

Also, !!date +\%Y is 12 symbols (assuming you are reading this from 2014) – defhlt – 2016-12-05T17:26:02.593

You are totally right! I use nvim 0.1.6-dev. – defhlt – 2016-12-05T17:28:43.340

1

## JavaScript, 81 76 bytes

l="length";alert(("hi"[l]<<"javascript"[l])-"wow"[l]*"hello death"[l]-true);

1

# C, 24

(Using GCC-4.9.2)

f(){printf("%i",'Þ');}         (Properly formatted version on Pastebin)

IMPORTANT NOTE: There is a U+0007 in-front of the Þ character but stackexchange removes it for some reason. Make sure you edit it back in before compiling my code. Here is a pastebin of the code that does include the unicode characters needed.

1

# SmileBASIC, 10 bytes

?&HFBC>>!.


&HFBC is hexadecimal for 2014*2, which is right shifted by not(0.0)

?ASC("ߞ") looks shorter, but it's actually the same length when saved in UTF-8, and about 100000x more boring.

1

# √ å ı ¥ ® Ï Ø ¿, 8 bytes

TTX''_o


Try it online!

The rest of the code in the TIO link is the Python interpreter (because I can't be bothered to ask Dennis to add √ å ı ¥ ® Ï Ø ¿

1

# Braingolf, 3 bytes

#ߞ


Try it online!

The ordinal of ߞ is 2014, # pushes the ordinal of the next character to the stack, and Braingolf implicitly outputs the last item on the stack.

Braingolf is your language, right? – MD XF – 2017-06-06T19:11:25.403

@MDXF Indeed it is – Skidsdev – 2017-06-06T19:46:55.463

1

# Charcoal, 5 bytes

Ｉ²⁰¹⁴


Try it online!

Language was created after January 1, 2014, but as Charcoal uses the superindices ⁰¹²³⁴⁵⁶⁷⁸⁹ to represent the numbers, the answer is valid. :-)

1

# Lua, 29 bytes

b=("").byte print(b"?"..b"?")


NB: the two question marks are substitutes for characters that are not appearing properly in the post. See the tio link below for proof.

Try it online!

As an interesting point, although this is not the case with Lua, a language with an implementation of pi to at least 3137 digits would be able to print pi and find '2014' at digits 3133-3136!

1

# Python, 41 bytes

print(int(ord("j")/len("aa") * ord("&")))


Try it online!

2Welcome to the site! The aim of this question (and code golf in general) is to get the shortest functional code. You can remove the spaces around the * to start with. – caird coinheringaahing – 2018-01-07T20:51:51.207

1

# Pyt, 9 bytes

π!⬡⁻⁻⁻△⁻⁻


Try it online!

1

# LibreLogo (Paper Format: Tabloid), 55 bytes

The document must be in Tabloid Format for this method to work.

Code:

print ''.join(set(str(pagesize.pop()))).replace('.','')


Result:

Explanation:

pagesize                                           ; Returns [792.0, 1224.0] (Tabloid Format)
pagesize.pop()                                     ; Returns 1224.0
str(pagesize.pop())                                ; Returns "1224.0"
set(str(pagesize.pop()))                           ; Returns {u'2', u'0', u'1', u'.', u'4'}
''.join(set(str(pagesize.pop())))                  ; Returns "201.4"
''.join(set(str(pagesize.pop()))).replace('.','')  ; Returns "2014"


1

## Canvas, 7 bytes

ＡＡ＋ｑ⁷┤ｔ


Try it online!

ＡＡ       | Push 10 to stack twice
＋      | Add top two items in stack
ｑ     | Print top item in stack on current line
⁷    | Push 16 to stack
┤   | Decrement top item in stack twice
ｔ | Print top item in stack on current line, disable implicit output


1

# Perl 5 + Perligata, 25 bytes

Byte count uses the new rule that command-line switches for choosing appropriate modules don't count.

perl -MLingua::Romana::Perligata -e'MMXIVimum egresso scribe.'


Tested in Strawberry 5.26.0 with Perligata 0.601 (with " instead of ' in the above code).

1

# jshell, 5 bytes

How to make Java an useful code golf language? Simple, just remove the need to declare classes, methods, make things automatically print and so on.

+'ߞ'


1

# Stax, 4 bytes

ü◘♥┐


Run and debug (ha) it online!

Obligatory Stax answer. This unpacks to 2014. Implicit print.

I haven't yet found a shorter version.

1

# C# (.NET Core), 42 bytes

System.Console.Write((int)'ϯ'+(int)'ϯ');

// Btw. it's now 2018 ^^ so
System.Console.Write((int)'ϱ' + (int)'ϱ');


Try it online!

1

# CJam, 2 bytes

KE


K is predefined as 20, E is predefined as 14. The stack gets automatically printed after the program.

Try it online!

1

## ANSI C - 9547 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.

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.

You don't actually need to include stdio.h, most compilers will give a warning but include it for you. – nyuszika7h – 2014-01-01T21:58:50.720

This code yields 7 for me with tcc and 508830 with gcc. clang gives an error: character too large for enclosing character literal type. – nyuszika7h – 2014-01-01T22:02:40.410

@nyuszika7h ... 7 is equivalent with \a - this could be a problem with passing to the printf function, it would be interesting for me to see if int a = '\aÞ' is 2014 using tcc. 508830 looks very strange to me - i would guess some endian thing but n * 256 + 7 can never be that number. So this illuminates how string this code depends on the system. – Johannes – 2014-01-01T22:58:20.533

1

# Racket: 20 (19 chars)

(char->integer #\ߞ)


ߞ is a unicode character that has 2014 as it's code.

This abuses the fact that every top level form gets its evaluation printed to stdout. This is quite unique amongst LISPs which usualy only have this behaviour in the REPL and not when running programs.

# Scheme: 29 bytes (28 chars)

(display(char->integer #\ߞ))


So 29 bytes, then? – Joe Z. – 2014-01-01T18:47:54.733

1@JoeZ. Yes. I was just about to update it :) – Sylwester – 2014-01-01T18:49:15.593

At least in Racket, you don’t need to call display — the REPL will print the result of char->integer automatically — which would save you another 9 chars. – Matthew Butterick – 2014-10-19T17:19:16.387

@MatthewButterick Unlike Scheme and CL #!racket actually display every top level result to stdout even when running a compiled program. It's quite annoying but I can abuse that. thanks :) – Sylwester – 2014-10-19T18:20:24.210

Still beat it, I’m afraid – Matthew Butterick – 2014-10-19T20:39:53.373

1

maybe not the shortest, but one of the more readable ones in Smalltalk:

Transcript show: Date today year.


This doesn't work.. (lol) – Albert Renshaw – 2017-01-30T03:36:38.997

6

see the comments. You can't use current Year.

– Wasi – 2014-01-01T19:27:44.493

1

# 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));


# 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 -->


That would be 2013 newlines followed by <?=__LINE__ for 2024 chars, not 9. – Peter Taylor – 2014-01-02T09:28:01.420

@PeterTaylor I didn't write 9 chars to win the challenge (since there are lots of shortest answers!), but because the actual code is that <?=__LINE__;?>, which I thought would be funny :) Nvm, I'll edit that. – Vereos – 2014-01-02T09:31:52.200

If it doesn't work without the newlines, they're part of the "actual code". – Peter Taylor – 2014-01-02T09:32:21.890

I guess you're right, edited. – Vereos – 2014-01-02T09:37:10.877

1In PHP, you can skip ?> at end of the program. But interesting idea with __LINE__, even if it's ridiculous for such huge number. – Konrad Borowski – 2014-01-02T12:15:14.013

1

## Game Maker Language, 22

show_message(ord("ߞ"))


1

# 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"));}


1

## Python, 14 characters

print ord('ߞ')


Short way ;)

Does it work without the u before the string? – Joe Z. – 2014-01-07T02:16:27.130

TypeError: ord() expected a character, but string of length 2 found – boothby – 2014-01-07T05:41:04.490

1@JoeZ. this is valid Python 3 – Peter Gibson – 2014-01-07T06:21:59.677

2In Python 3: SyntaxError: invalid syntax (Python 3 needs brackets around stuff being printed). – Joe Z. – 2014-01-07T07:41:28.197

1

# C/C++ 39

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


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

1

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'.

1

C#, 28 bytes

You don't need to cast to an int as Hille does, so it becomes

Console.WriteLine('ϱ'+'ϱ')

1

# 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|"")


1

# Symbolic Python, 31 bytes

_=-~-~-~(_==_)
_=""[_::_]


Try it online!

Symbolic Python bans numbers anyway.

### Explanation:

_=-~-~-~(_==_)   # Set _ to 4
_=               # Set _ to
""         # From the representation of some unprintables
# Which is '\x12\x10\x11\x14'
[_::_]   # From the 4th character, take every 4th character
# Output the contents of _ implicitly


A more interesting solution at 34 bytes:

_=-~(_==_)
_=_+_-_+_/_+_*_


Try it online!

_=-~(_==_)               # Set _ to 2
_=_+_-_+_/_+_*_
# '2'+'2-2'+'2/2'+'2*2' = '2'+'0'+'1'+'4' = '2014'


Crazy that someone else is actually using this stupid language I made, haha – FlipTack – 2018-12-17T21:01:50.457

1

## 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'));}


You can use int instead of toascii for all of those to save a lot of characters. – Joe Z. – 2014-01-07T02:58:53.543

@JoeZ. OK, I didn't think of that. How can I calculate the bytes? – None – 2014-01-07T02:59:48.053

Save your program as a text file, and then view how many bytes it is in the file manager. – Joe Z. – 2014-01-07T03:00:11.220

2 is not allowed. – Danko Durbić – 2014-01-07T07:46:59.197

Oops, I'll fix that. – None – 2014-01-07T16:08:48.250

@DankoDurbić Fixed. – None – 2014-01-07T16:10:16.863

I only count 62 bytes (using http://www.charactercountonline.com/). Also, can \t be replaced with the literal tab character?

– Justin – 2014-03-05T06:03:38.150

@Quincunx Literal tab character? – None – 2014-03-05T06:07:11.050

@hosch250 is that not what \t stands for? (tab. #9 in ascii: http://www.asciitable.com/)

– Justin – 2014-03-05T06:13:31.173

@Quincunx Yes, I can use plain 9. However, I can have no numbers in my source code, so I cannot use 9 instead. TAB does not work. – None – 2014-03-05T06:15:18.157

1

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\@)))))


1

## vba (immediate window), 3826 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

How do you write that in the Immediate window? (I assume that is what you mean by “direct window”.) If I copy-paste it I get “?year((cdbl(asc("e^"))*cdbl(asc("^(2)"))))”. (Copy-pasting “ê” and “²” from charmap.exe results “?” both.) And of course, that way the calculation not gives 2014. – manatwork – 2014-01-07T16:08:07.817

I used ?chr(234),chr(178) to get the characters, or you can hold down the ALT key and type 234 (and 178) and release the ALT to get each character – SeanC – 2014-01-07T16:31:23.603

With Alt+234 I get “r”, with Alt+178 I get “¦”. Of course, it works with the chr() function. Anyway, nice trick to use year() this way. – manatwork – 2014-01-07T16:41:29.913

ok.. I was thinking back to DOS days - now it's 0234 and 0178, but I found another shorter way now – SeanC – 2014-01-07T16:44:53.607

Thanks, it works this way. Although here appears “ȩ” and “¸”, the calculation is correct. – manatwork – 2014-01-07T16:59:31.070

1

## 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)))

1

## Ruby 1.9, 10 bytes

p 'ߞ'.ord


These are 8 bytes in UTF8. – schmijos – 2017-07-24T14:42:36.130

nice! 7 chars if you are in the irb command prompt 'ߞ'.ord – Eduard Florinescu – 2014-01-08T16:54:48.760

@EduardFlorinescu Thanks, I know. – Timtech – 2014-01-08T21:38:43.083

1

# Python, 23

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


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

1

## 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.

1

### Edited: bash 81 chars!

Just for fun:

wc -c < <(echo {,}{,,}{,}{,,,}{,}{,,,}film dbugjkqstvxz{,}{,,} 'Happy New Year!')


there is no numbers, all letters are used and this print exactly:

2014


( This method could reasonably be used until 2016: by just adding one or two exclamation point after the wish:

   wc -c < <(echo {,}{,,}{,}{,,,}{,}{,,,}film dbugjkqstvxz{,}{,,} 'Happy New Year!!')
2015


;-)

### bash 27 chars

.;v=$?;echo$v$?${#v}$[v+v]  This will output: bash: .: filename argument required .: usage: . filename [arguments] 2014  Ok, this will generate some unwanted output, but 2014 is printed and is a valid token! The two following sample are error free (a little longer but near golfed) v=$(echo {V..v});echo $[${#v}#vu]
2014


or

printf -vv "%d" $?xfbc;echo$[v>>${#?}] 2014  or even: echo$[$[$[${#?}$?-${#?}]$?>>${#?}]#Iy] 2014  Inspired by comment from GammaFunction: echo$[$[a-a]xfbd>>${#?}]


You don't need to set a var in the 0xfbc solution: echo $[$?xfbc>>++j] works just fine (19 bytes). – GammaFunction – 2019-03-28T07:15:55.807

Nice! Late to post but you could! Unfortunely you have to ensure $? to be 0 and this could not be reused... But impressive! – F. Hauri – 2019-03-28T15:54:59.963 1@GammaFunction Post edited ;-) – F. Hauri – 2019-03-28T16:46:01.117 1 ## @, 9 chars -Σ*{~~}82  Explanation (syntactically invalid) - 2 Subtract the result by 2(2014) Σ Summation of all ASCII codes in the string(2016) * 8 Duplicate the string 8 times {~~} Define the string "~~"  There is no Try It Online for @. If the current year is 2016, it would be perfect(7 chars): Σ*{~~}8  1 # Javascript, 45 characters _=> new Date().getFullYear()-new Date().getDate()  Very Temporary!! Another Day, another very Temporary answer :-) f=new Date().getFullYear()-new Date().getDay()  Welcome to Code Golf! Submissions should be either a full program or a function, so I would recommend prepending _=> to your solution to turn it into an anonymous function. – Stephen – 2019-09-05T20:00:21.470 You could possibly remove the space between => and new to save a byte – Benji – 2020-03-19T15:26:39.367 1 # ><>, 7 bytes aa+nen;  You can try it on The Online ><> interpreter. ## Explanation a # push 10 a # push 10 + # push sum of top two values (10 + 10 = 20) n # output top value as number e # push 14 n # output top value as number ; # halt program  This doesn't print the following: 20 14  because n outputs the top value as a number without a newline. ## Alternate Solution (9 bytes) This one actually puts 2014 on the stack as a single number and outputs it. cbde**+n;  I basically just tried random values for this one :P 1 # Triangular, 12 bytes tE**%Cdd.>/  Try it online! Y'know, I was just about to post a 14-byte solution; luckily I went back to double-check and had a hunch. Ungolfed:  t E * * % C d d .  > / -- Actual Execution Order -- t*C The first multiplication does nothing. 12 is pushed twice, then direction changes and both are multiplied to get 144. E*d>/d% Push 14, then multiply 14*144=2016. Decrement twice and print.  The 14-byte I was going to post: t*CE.$$$$*...>dd%  1 # Wren, 27 bytes Just like most other answers, convert to code points and print. However this uses control characters in the source code. System.printAll("".bytes)  Try it online! ## Explanation  "" // An unprintable chain containing 0x14 (20 in decimal) and 0x0e (14 in decimal) .bytes // Convert them to a list of their decimal codes [20, 14] System.printAll( ) // Print them all without a separator (2014)  1 # ABC, 11 bytes aacncacaaac  ## Explanation a increments the accumulator, c outputs it and n sets it to 0. 1 ## Wj, 4 bytes ë‘"C  ## Explanation ë‘" % Push a list C % Convert to list of ord codes [20,14] flag:j % Join without a separator $$$$  1 # MathGolf, 2 bytes ID  Try it online! ## Explanation I # push 20 D # push 14 # implicit output  1 ## Spice, 145 97 bytes Improvement using multiplication to get the 9 from 2*2*2+1, as we already had those values. ;t;z;o;n;i@LEN i z;ADD i i i;LEN i o;PUT i i i;LEN i t;MUL t t i;MUL t i i;ADD o i n;OUT t z o n;  ## Un-golfed explanation ;t;z;o;n;i@ - Declare vars LEN i z; - Get the length of [], 0, and store in z ADD i i i; - Adding implicitly uses the first element, or 0 if there is none, so we insert 0 into i LEN i o; - Store length of [0], 1, in o PUT i i i; - Insert 0th element of i into i at position i[0] (we're increasing the array size) LEN i t; - Store length of [0, 0], 2, into t MUL t t i; - Multiply t, 2, by itself and store in i MUL t i i; - Multiply t, 2, by i, 4, and store in i ADD o i n; - Add o, 1, to i, 8 and store in n OUT t z o n;- Write to console - "[2] [0] [1] [9]"  ## Original ;t;z;o;n;i@LEN i z;ADD i i i;LEN i o;PUT i i i;LEN i t;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;PUT i i i;LEN i n;OUT t z o n;  ## Un-golfed explanation In Spice, all variables are double arrays. Importantly, variables that have no assigned value are either treated as an empty list [] or 0 depending on the operation. The built-in LEN will give the length of an array and we can therefore produce numbers: ;t;z;o;n;i@ - Declare vars LEN i z; - Get the length of [], 0, and store in z ADD i i i; - Adding implicitly uses the first element, or 0 if there is none, so we insert 0 into i LEN i o; - Store length of [0], 1, in o PUT i i i; - Insert 0th element of i into i at position i[0] (we're increasing the array size) LEN i t; - Store length of [0, 0], 2, into t PUT i i i; - Now repeat until there are 9 elements... PUT i i i; PUT i i i; PUT i i i; PUT i i i; PUT i i i; PUT i i i; LEN i n; - ... and store in n OUT t z o n; - Write to console - "[2] [0] [1] [9]"  For the original 2014 version, you save bytes for less PUTs - 95 bytes. So this solution will improve next year! 1 # FEU, 69 bytes a/abcdeghij m/a/bb/b/cc/c/dd/d/ee/e/ff/f/gg/g/hh/h/ii/i/jj/j/kk/g U/k  Try it online! 1 # Excel, 12 =UNICODE("ߞ  Second best I could do was =ARABIC("MMXIV at 14 bytes. 1 # MAWP, 30 28 bytes !!+!!!!++!*+/!+!!+!+!!++++*:  Try it! This will be fun to golf. This is longer than @Lyxal 's answer, but outputs only one time as one number. 15 bytes – Lyxal – 2020-09-28T09:10:59.877 1 # Poetic, 112 bytes twos a bad thing using a two?o,hardly!i am using a poem a numeric poem?oh,clearly,but a limited one for certain  Try it online! 1 # DROL, 13 bytes ziill<ukl<dfo I wanted to add a language that wasn't already included... DROL is a limited instruction assembler-like language with only two registers as storage. The language does include numbers, so it think it qualifies for this question. But numbers are only used for loop length indicators, as well as being used as the name of some instructions. It's described on the Esolang Wiki DROL page. Here's a rundown of what the code does: z set R1=0 ii increment R1 by 1 twice (2) ll square R1 three two (16) < shift left R1 (32) u set R2=R1 (32) k increment R2 (33) l square R1 (1024) < shift left R1 (2048) d decrement R1 (2047) f subtract R1 = R1 - R2 (2014) o print R1 as an integer  1 # Kotlin, 42 bytes fun main()=print("ް.".map{it.code}.sum())  I used U+1968 which is ް and a . which is U+46. 1 # Phooey, 16 bytes =@+$i>$i=$i<@+$i  Try it online!  # Stk Tape =@+$i>$i=$i<@+$i # (0) >0 0 (initial state) = # (0) >1 0 tape = stack == tape to get 1 @ # 1 >1 0 push to stack + # (0) >2 0 pop; add stack to tape$i            # (0) >2  0   print tape as integer (2)
>$i # (0) 2 >0 move tape ptr right, print (0) =$i      # (0)  2 >1   same as above to get 1 again, print (1)
<@    #  2  >2  1   move back, push
+$i # (0) >4 1 add stack to tape, print (4)  Thank goodness for Phooey's = operator. This would be impossible in Foo. # Phooey, 23 19 bytes This version actually generates the number 2014 instead of printing 2,0,1, and 4. =@+@@@+@**@@*@+--$i


Try it online!

                    # stack       | tape
=@+@@@+@**@@*@+--$i # (0) | 0 initial state = # (0) | 1 tape = tape == pop() (to get 1) @+ # (0) | 2 double by adding to self @@ # 2 2 | 2 push two copies to the stack @+ # 2 2 | 4 double @* # 2 2 | 16 square by multiplying by self * # 2 | 32 multiply by 2 @ # 2 32 | 32 push 32 for later @* # 2 32 | 1024 square @+ # 2 32 | 2048 double - # 2 | 2016 pop and subtract - # (0) | 2014 pop and subtract$i #         (0) | 2014  print 2014


1

# ARM assembly, 94 85 bytes (28 bytes compiled)

Textual assembly for ARM mode.

s:subs sb,sb
lsl sl,sb,sb
rsb sb,sl,lsl sl
bx lr


A function which returns 2014 in sb (r9)

Clobbers sl (r10) and sb (r9).

Expanded version:

        // It feels so empty here...
.globl s
s:
// use the quirky way subs affects the flags
// to set r9 to 1
subs    r9, r9, r9
// double r9 by adding it to itself
// lsl works just as well
// r9 = 2
// r10 = r9 << r9
// r10 = 2 << 2
// r10 = 8
lsl     r10, r9, r9
// r9 = r9 + (r10 << r9)
// r9 = 2 + (8 << 2)
// r9 = 2 + 32
// r9 = 34
add     r9, r9, r10, lsl r9
// r9 = (r10 << r10) - r9
// r9 = (8 << 8) - 34
// r9 = 2048 - 34
// r9 = 2014
rsb     r9, r9, r10, lsl r10
// Return
bx      lr


This uses the same idea as my Phooey answer, of generating 2048, then subtracting 34. While I do have access to push and pop, ARM isn't a stack machine. It is a register machine. Additionally, we have lsl for shifting left which makes a few cases easier.

It is yet another rare case where the inverted carry flag on ARM is useful for something other than 64-bit subtraction, as it allows us to set a register to 1 when combined with adc.

Additionally, this only works in ARM mode: Thumb-2 did not show the return of shifting by register (which kinda was a dumb waste of encoding bits)

It uses the classic register names instead of the format which is r[0-15].

1

# Python 3, 16 bytes

print(ord('ߞ'))


Try it online!

1

# JavaScript 45

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


1

# 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.

1Normally entries that explore edge cases are popular. This is the only answer they obeys the letter of the rules that is down voted (all other down voted answers rely on year=2014). Anyone care to comment on what rubs them wrong about this answer? – gmatht – 2014-04-08T00:59:09.303

It follows the letter of the rules in a way that's almost universally considered boring and unoriginal. Additionally, it was posted almost three months after the original problem, meaning that it's way back on the 4th page where barely anybody will see it, let alone vote it up. – Joe Z. – 2014-04-17T00:45:09.343

For that matter, I've edited the question again to make my intentions for the problem even clearer. Is "independently of any external variables" good enough? Even then, this answer isn't the shortest in either category, so I still won't accept it. – Joe Z. – 2014-04-17T00:46:59.303

Thanks. For the record, when I commented above this answer had been voted down to -2, so the context was more "would someone like to comment on why they are voting it down?" rather than "why isn't it upvoted?". It also wasn't intended as a criticism of Joe Z's question specification. – gmatht – 2014-05-04T13:04:14.423

1

# 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.

Arne, True implicitly casts to 1, so you can start with a=True and save 8 characters. Also, please specify Python 2. – isaacg – 2014-07-17T05:52:58.900

1

# Marbelous 14

CB
CE
CF
CD
~~


How it works

the first 4 lines are language literals, in hexadecimal. Their values are 203, 206, 207 and 205. They will fall down by one cell on each tick. If you perform an 8-bit binary not on those values (which is exactly what ~~ does) you get the following values: 52, 49, 48 and 50. These values happen to be the ascii values of 4, 1, 0 and 2 respectively. The literals then fall off the board which causes their corresponding ascii character to be printed to STDOUT.

1

# C, 27 Bytes

main(){printf("%d",'\aÞ');}


Just a reminder that multi-character constants do exist :)

Alternatively, three bytes more:

main(){printf("%x",' i'-'U');}


main is not needed, you can shorten it to f and save 3 bytes, it'll still be a valid submission. – MD XF – 2017-05-12T19:53:01.907

>

• int is not needed. 2. return doesn't print anything.
• < – Dennis – 2014-09-16T15:07:59.180

start="3">

• '^GÞ' (character codes 7 and 222) would require only one multi-character constant.
• < – Dennis – 2014-09-16T15:16:07.850

My compiler does not agree :( – Gerwin – 2014-09-16T15:21:22.850

That's odd. 7 * 256 + 222 = 2014, so it should work. What do you get? – Dennis – 2014-09-16T15:23:01.037

6178782, it seems like my compiler thinks of ^ as a separate character. – Gerwin – 2014-09-16T15:26:14.280

No, ^G is caret notation for BEL character; it's a single character with char code 7. You have to type the unprintable character somehow, depending on your OS and editor. – Dennis – 2014-09-16T15:28:17.763

My compiler tells me it's '\aÞ'. Thanks anyway. – Gerwin – 2014-09-16T15:55:49.970

1

# Lua - 30 bytes

b=#" "print(b..b-b..b/b..b+b)

# is the length operator, so b = 2.

1Would b/b work? – Lynn – 2014-11-08T16:20:07.340

@nooodl Good suggestion; indeed b/b can replace b-#" " and save 3 characters. – Seeker14491 – 2014-11-10T00:56:59.370

1

(Java, 553 bytes as .class, 112 bytes as it stands, 84 bytes after renaming the class to 'm' and removing whitespace.)

This probably isn't the kind of answer you're looking for, but there are a bunch of strings that share a hashcode of 2014.

public class make2014 {
public static void main(String[] args){
System.out.println("={".hashCode());
}
}


1Consider adding the language name (Java, I assume) and byte count like the other answers do. – NinjaBearMonkey – 2014-09-20T16:31:20.777

Thanks, I forgot to do that. I didn't think my file size would matter much - the minimum file size of a .class is far greater than most of the attempts here. – John P – 2014-09-20T21:43:15.493

I got the .java down to 84 bytes without changing functionality. I see other people shaved off 9 bytes by printing special characters to produce 2014 instead of using hashcodes... I kind of thought it would be cheating to do it like that, since you're really printing 2014 as a number in a different format. Oh well. – John P – 2014-09-20T21:50:40.717

1

# Xojo, 27 chars (all ASCII)

MsgBox Str(&hFBC/(&hC-&hA))


1

## TinyMUSH, 16

We need more MUD language entries.

\encrypt($"#&,.)  1 ## C++ 30 main(){cout<<('&')*(']'-'(');}  Why do you need the brackets around '&'? Does main(){cout<<'&'*(']'-'(');} not work? – Joe Z. – 2015-03-24T07:39:15.007 1 # x86 machine code, 19 bytes B8 3A 0E 2C 08 CD 10 2C 02 CD 10 04 01 CD 10 04 04 CD 10 Assembly code equivalent: mov ax, 0E3Ah; ah = 0Eh (bios teletype), al = 3Ah (ascii semicolon) sub al, 08h; ascii 2 int 10h sub al, 02h; ascii 0 int 10h add al, 01h; ascii 1 int 10h add al, 04h; ascii 5 int 10h  Yeah, I know: it logs 2015 rather than 2014. But seeing that this challenge is old and now the year is 2015, it seemed more appropriate to use the current year (it's my excuse for not "going home" :) ) Note: This was tested using DOSBOX 1You have numbers in your source. – Elliot A. – 2016-01-31T11:06:14.133 @ElliotA. Read the challenge: "without using any of the characters 0123456789"; numbers = characters representing numbers. – SirPython – 2016-01-31T21:49:13.510 1 ## JavaScript (19) Obvious cheating, but these expression ran in REPL print strings "2014" and "2015": ''+'ߞ'.charCodeAt() // 2014 ''+'ߟ'.charCodeAt() //2015  TIL: .charCodeAt implicitly converts it's first argument to 0. 1 # Vitsy, 5 4 Bytes, 3 Characters When in Rome... 'Nߞ Get the character with the value 2014 and then print it as a number. Simple. ### More Interesting Version (1210 9 Bytes): "ca-^b-N- My language supports hexadecimal, too. ;) " Capture the entire source as string by looping around the source. ca- Push 2 to the stack ^ 45^2 b- -11 N Output as a number. - Only here for character value 45.  Am I right that this language is newer than the question? Besides, I believe this question is already 'closed'(winner chosen, and even an edit at the start discouraging more replies) – Sanchises – 2015-11-01T11:42:26.327 @sanchises You're absolutely right - I'll pull my edit request. I'd still like to add it to the list, though, even if marked as a new language.. :D – Addison Crump – 2015-11-01T11:46:12.780 Just put it under 'invalid' I guess, there's a section for that at the end. – Sanchises – 2015-11-01T11:49:00.107 Changed the edit suggestion. :P Forgot about the 'newer than question' thing, but I should've considered that. Thanks, @sanchises – Addison Crump – 2015-11-01T11:50:08.600 1 # Perl 5, 8 28 bytes say 38*53 Seems to do it. Oh, without cheating ? $z=ord("!")*ord("=");say++$z  D'oh. I thought it was without any of those numbers. – Dale Johnson – 2015-12-02T04:24:58.003 1 T-SQL 27 bytes PRINT ASCII('')*ASCII('j')  Note that the character that isn't rendered here is the DC3 (CHAR(19)) in the first set of quote marks. It's unicode U+009F which, it would appear, doesn't copy and paste here too well but I can assure you it works in SQL Management Studio. 1 # Quetzalcoatl, 114 5 bytes ## Noncompeting because this language is from 2016. ::ord('ߞ')  The box should be replaced by Unicode character 2014. ### Edit This is for an old version of Quetzalcoatl. New version: 'ߞ'O  This language is new though, this year. – NoOneIsHere – 2016-03-02T22:53:25.030 There is no ASCII character 2014... it has to be Unicode to go that high. – mbomb007 – 2016-03-04T19:40:13.443 If Quetzalcoatl is newer than the challenge, you're answer is non-competing and should say so in its body. Also, I'm not aware of any encoding in which ߞ would be a single byte. – Dennis – 2016-04-08T17:21:13.733 1 # SMBF, 15 bytes \x00 is a literal NUL byte. This program adds 5 to each of /,+- and prints. <[+++++.<]\x00/,+-  Only my Python interpreter can accept non-printable ASCII. Change the data line to this, and substitute the code you want to run with the non-printable values escaped (on line 169): data = bytearray(b'the above code goes here')  1 # Retina, 24 bytes (newer than challenge) Note the trailing space on lines 2 and 3. Language is newer than the challenge.  xx x xxxx +(x)*$#+


Try it online

1

# Base64, 9 Bytes

MjAxNA==

(You can decode it with: echo MjAxNA== |base64 -d)

1

## Jelly, 10 bytes

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.


1

# Mathematica, 10 bytes

N[E,E^E^E]


Prints the decimal expansion of the number e to over 3.8 million decimal places. The first occurrence of 2014 in that decimal expansion starts at the 3180th decimal place.

1

# VBA, 21 characters

?cells(,"BYL").column


Write and run the above code in the Immediate Window. Basically, the code converts column name BYL to its column index (2014).

For future reference, this may be rewritten as ?[BYL:BYL].Column – Taylor Scott – 2017-09-03T21:04:26.123

1

# Actually, 15 bytes

This language was created long after this challenge was made, but I thought I'd still try my hand at it. This answer avoids all numerals, including Actually's ² for a*a. Golfing suggestions welcome. Try it online!

╜⌐u;*⌐úl¬¬τu;*-


Ungolfing

╜    Push register 0 (initialized to 0).
⌐u   Add 2 and increment. Returns 3.
;*   Duplicate and multiply. Equivalent to squaring. Returns 9.
⌐    Add 2 again. Returns 11.
úl   Pushes the lowercase alphabet and gets its length. Returns 26.
¬¬   Subtracts 2 twice. Returns 22.
τ    double(). Returns 44.
u    Increment. Returns 45.
;*   Square. Returns 2025.
-    Subtract. Returns 2025 - 11 == 2014.


0

# 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"


0

## Pyke, 3 bytes, noncompetitive

w߾


Try it here!

Where ߾ is 0xDFBE

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

How does it work? – Pavel – 2016-12-13T04:04:06.843

@Pavel explanation added – Blue – 2016-12-13T07:35:11.363

0

# tcl, 16

puts [scan ߞ %c]


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

0

## Groovy, casting to int, 8 bytes

(int)'ߞ'


0

# 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.

0

# Python 3, 16 Bytes :

print(ord('ߞ'))


# 2017 version, 16 Bytes :

print(ord('ߡ'))


0

## 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.

0

# Common Lisp, 15 characters

(char-code #\ߞ)


Try it online!

0

# Ruby (8 bytes, 7 chars)

'ߞ'.ord


The question mark is ߞ represented in two bytes UTF8.

Duplicate of O-I's Ruby solution posted 3.5 years earlier.

– manatwork – 2017-07-24T09:24:57.533

– Martin Ender – 2017-07-24T10:44:08.340

I didn't said they aren't. Neither downvoted or flagged. (Just for the record, I never agreed with that rule. And never will.) But I think would be more enjoyable to avoid duplicated solutions. Especially on a question that already has 263 answers. – manatwork – 2017-07-24T11:03:30.777

Hm... hard to find the old solutions. But actually his solution is 9 bytes because the codepoints representations need 2 bytes. – schmijos – 2017-07-24T12:58:02.617

Btw: Having seen the solution of @o-i : ?ߞ.ord would then be only 7 bytes, right? – schmijos – 2017-07-24T13:05:15.750

1Initially O-I's solution was also 'ߞ'.ord, but then be updated because the requirement says to “prints the number 2014”. The general rule (which can be overridden by each challenge) is that solutions must handle input and output either themselves explicitly or benefit the interpreter's service if it has such thing like ruby's -n or -p. Code that expects input to be readily available in the memory or just leaves the value they produce in the memory are called snippets and generally are not accepted as solutions. – manatwork – 2017-07-24T14:29:48.217

BTW, looks like neither Timtech found O-I's solution and posted his own after just 1 week. I assume there were considerably fewer solutions that time.

– manatwork – 2017-07-24T14:35:35.270

0

# Clojure - 35 characters ASCII only

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


0

# Lua, 27 bytes

Should work in Lua 5.1, Lua 5.2, and Lua 5.3.

x="ɅɅ"print(x:byte()..#x)


Try it online!

This is mean to be saved with the UTF-8 encoding. The first byte of the string is 201, and its length is four. Lua is mostly encoding agnostic, so as long as these things are true in whatever encoding, it works.

With only ASCII, 28 bytes:

x=""io.write(x:byte(y,#x))


Note: the string must contain ASCII 20 and ASCII 14 (which are not printable characters). y here is an undefined variable, so it is nil, which byte defaults to 1 in the first parameter.

0

# Aceto, 10 bytes

IIppIpPiIp


Try it online!

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

0

# Implicit, 3 bytes

ߞ


Try it online!

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


0

# Triangularity, 18 bytes

.. ..
."ߞ".
o


Try it online!

0

# 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!

0

# Brainf*ck

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


> Move pointer pos over
[ 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.

Welcome to PPC...Jk – RaviRavioli – 2018-01-12T03:11:19.593

0

# Brainfuck, 23 Bytes

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


Try it online!

0

# PHP, 25 bytes

<?=IntlChar::ord("ߞ");?>

This requires the IntlChar library; and that´s a lot younger than the question. Nice idea, though. – Titus – 2018-03-10T01:31:49.937

Thanks. Can be solved without using IntlChar library with preg_replace('#\D#','',json_encode('—')). Pick your number from a unicode table :-) – Agnius Vasiliauskas – 2018-03-10T12:08:29.453

0

# 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...

0

# Python 3, 16 bytes

lambda:ord('ߞ')


Try it online!

0

# T-SQL, 32 bytes

Based on the idea of Steve Matthews, but avoiding unprintable characters:

SELECT -~(ASCII('=')*ASCII('!'))


Alternative (but rather obvious) solution, 23 bytes:

PRINT UNICODE(N'ߞ')


0

# Objective-C, 27

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


0

# 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. 0 # Ahead, 5 bytes ezpz 'ߞO@ ' push next cell to stack ߞ U+7DE (2014 decimal) O output as number @ die  Try it online! 0 # 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. 0 # 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 0 Objective C NSDateFormatter *formatter = [[NSDateFormatter alloc] init]; [formatter setDateFormat:@"yyyy"]; NSLog(@"%@",[formatter stringFromDate:[NSDate date]]);  NSLog(@"%i",'&'*('F'-'A')); – Albert Renshaw – 2017-02-16T18:57:18.297 1 From the comments: Gelatin: “Is it acceptable to use the current year?” Joe Z.: “No, it has to be 2014 exactly.” – manatwork – 2014-01-01T13:17:34.027 Because the question is a code-golf question, please add the character count. – ProgramFOX – 2014-01-01T13:31:16.810 1What happened to you, Smalltalk ? You look...different. – bug – 2014-01-07T01:17:56.403 @AlbertRenshaw that just prints 190. – Cœur – 2018-03-17T10:07:45.653 1@Cœur sorry meant this NSLog(@"%i",'&'*('V'-'!')); – Albert Renshaw – 2018-03-17T20:08:23.010 Should be noted that OP's source code just prints 2018 ;) – Albert Renshaw – 2018-03-17T20:10:01.140 0 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 Yes, but sadly your code is far from being the shortest. :P – Joe Z. – 2014-01-01T18:26:18.263 1@Joe Z. Right, you could include an HTML script src instead :P – Cilan – 2014-01-01T23:10:12.560 location.href.match(/-(\d+)-/).pop() – Alf Eaton – 2014-01-03T09:18:57.627 location='http://goo.gl/miVwHe' would be shorter in my opinion. – Konrad Borowski – 2014-01-04T11:23:18.060 @xfix Thanks for the tip, I even removed 'http:' :) – Cilan – 2014-02-08T19:24:21.323 You can take out those two slashes at the start too. ;) – cjfaure – 2014-02-08T19:33:15.343 @Trimsty I wish, but unfortunately not ;( – Cilan – 2014-02-08T19:33:56.223 @TheWobbuffet Hm. That's really weird, I really thought it'd work! Sorry. – cjfaure – 2014-02-08T19:39:28.973 0 # ;# - 203 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#  This language is invalid according to the community consensus on what counts as a programming language – MilkyWay90 – 2019-07-05T04:33:29.127 @MilkyWay90 Um, this is from just under a year ago... I don't think there was a consensus back then. Either way, it's not like i even got an upvote, so let's just leave it, eh? – seadoggie01 – 2019-07-05T16:46:42.040 The consensus was made in 2014. You can take it up with Meta, but until then this is invalid – MilkyWay90 – 2019-07-05T16:49:01.617 – pppery – 2020-09-30T02:42:18.187 0 ## 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 π is 2 bytes and → is 3, bringing the byte count to 49. – Joe Z. – 2014-01-07T02:18:28.000 @JoeZ. All characters at http://tibasicdev.wikidot.com/83lgfont are 1 byte. – Timtech – 2014-01-07T12:10:20.987 0 # K (ngn/k), 10 bytes "&"*-/"U "  Try it online! and 2018 just because this is so late # K (ngn/k), 11 bytes -"^"-*/"@!"  Try it online! 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 The non-competing rule has been abolished – MilkyWay90 – 2019-07-05T04:33:55.100 0 ### C# (56 characters) Class P{static void Main(){Console.Write(','*','+'N');}}  This doesn't print anything. – shamp00 – 2014-01-05T15:05:17.700 26 characters = 4 bytes?? I don't think so... – jub0bs – 2014-01-06T12:24:21.490 Sorry. I just included the logic only. Now I included the whole program. – Merin Nakarmi – 2014-01-07T03:34:18.220 You need System.Console to use Console. – shamp00 – 2014-01-08T14:24:47.447 using System; will be on the header. So.... – Merin Nakarmi – 2014-01-08T17:50:00.877 0 # Aheui (esotope), 8 chars (24 bytes) 반밝따바뱟해망어  Try it online! 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. 0 # 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. 0 # Perl, 24 bytes print-ord(A)+ord for U,O  Pure ascii, no nonprinting characters or utf8-only characters used. Uses the 21st and 15th letters of the alphabet to print 20, 14. 0 # 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  0 BAT (windows batch), 7 characters echo %*  save as a.bat and invoke as a 2014 is there any restriction on using command line parameters? – Einacio I don't think so, but echo$1 still has a number in it. – Joe Z.

not a so valid answer, but noone was posting an answer in this wonderful language

Does BAT have an equivalent to sh’s $@ for “all parameters”? – Zev Eisenberg – 2014-01-10T23:29:33.147 @ZevEisenberg that's what i used, altought it seems someone didn't like it – Einacio – 2014-01-13T15:14:43.497 0 # k 11 & 17 chars this 11-char one is from a colleague @:[:']*.(.)  this 17-char one is the best i could come up with on my own; it's a variant on ASCII abuse: .,/$-/"i"zzfl


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

.,/$#:'(;();;) / another approach  @David Carraher my latest improvements on my port of your algo--two implementations, both 39 chars: -_-(x*p*p:exp acos@-#)-f*x*x*x:f*f:# f+(x*_p*p:exp acos@-#)-f*x*x*x:f*f:#  (i don't have the rep to comment on your post yet) – Aaron Davies – 2014-01-10T02:02:03.027 0 PHP (27) - not shortest ASCII PHP but more readable <?=hexdec($e=hexdec(E)).$e;  0 # Bash, 15 bytes echo "^T "|od -x  (^T is one character.) How about: printf %x \'— 13 characters. – Isaac – 2019-08-30T22:26:22.403 0 # Bat (Windows batch), 13 bytes (characters) echo.^T |od -x  (^T is one character.) 0 # PYTHON print(str(len('Happy new year to me')) + str(len('Happy new year')))  This has a 1 in it. – Joe Z. – 2014-01-11T23:16:32.597 oh whoops let me fix it – Oliver Ni – 2014-01-11T23:17:52.873 1Must str('Happy new year') be wroted as str(len('Happy new year'))? – AMK – 2014-01-11T23:28:14.173 sorry. I fixed it. – Oliver Ni – 2014-01-13T04:53:30.753 0 PHP, 22 bytes echo ord('').ord('')  Edit SE is stripping the characters. Working version here: http://codepad.org/unzjXNY2. this produces "00" when I run it. Please explain. – Darren Stone – 2014-01-14T20:32:21.080 SE is stripping the characters. Added a codepad example. – MichaelRushton – 2014-01-14T22:40:51.837 You can swap out echo with <?=, to make it 2 characters shorter. – cjfaure – 2014-02-08T19:36:44.783 0 # Forth, 29 bytes hex fbc 'B' '!' / / decimal .  Try it online! 0 *nix shell (POSIX/bourne) 41 bytes a=.. c=. d=....;echo${#a}${#b}${#c}${#d} ### Other solutions printf %x \'— # 13 characters Try it online! echo$[x=++y+y]$?$y$[x+x] # 25 characters Try it online! 0 # C# .NET 75 bytes public class p{public static void Main(){System.Console.Write((int)'ߞ');}}  Just displaying the integer value of the unicode character ߞ Try online (also has 2015-2019) 0 # k4, 20 bytes */(-)."j"$("Ud";"/")


      "j"$("Ud";"/") / convert "Ud" and "/" to underlying integer representation -> (85 100;47) (-). / subtract 47 from 85 and 100 -> 38 53 */ / multiply over  0 ## Delphi (26bytes & 26 chars) ord('-')*ord('/')-ord('e')  Ascii values - : 45 / : 47 e : 101 45*47 = 2115 - 101 = 2014 0 # MAWP 2.1, 15 bytes !+!:!!-:!!$:!+:


Try it online lol.

0

# Vyxal, 10 3 bytes

»÷⌈


Try it Online!

Simply compress the number 2014

Can't be bothered to download the interpreter, but does !!!!!''.$... work for 11 bytes? – Jo King – 2020-10-09T04:29:02.837 0 # Neim, 4 2 bytes χρ  Try it online! Push 20, push 14 and print the entire stack 0 # BRASCA, 4 bytes 'ߞn  Try it online! ### Explanation 'ߞ - Push the character ߞ (Codepoint 2014) n - Print as number  ## Alternative solution, 9 bytes ll+'.E-Sn  Try it online! ### Explanation ll+ - 10+10 = 20 '.E- - code point 46-32 = 14 S - Concatenate 20 and 14 n - Print as number  1 According to Wikipedia, 1 is a number. I don't know how accurate it is, though. – EasyasPi – 2021-01-28T13:25:37.797 I'm too used to those dang numbers. – SjoerdPennings – 2021-01-28T13:36:41.833 It's now fixed. No more stray numbers. – SjoerdPennings – 2021-01-28T13:48:52.393 0 ## 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). 0 # 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 :) 0 ## 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'))  0 # 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. 0 # 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.  0 # 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));}  0 # q, 20 17 bytes {x+y*z}."j"$"$+."  0 # TeaScript, 7 bytes 'ߞ'c()  Takes the char code of the character with a char code of 2014 0 # Groovy, 48 bytes For 2015 change cabe to cabf... n={it.each{print(((int)it)-(int)'a')}} n('cabe')    0 # 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. 0 # JavaScript, 17 Bytes atob("MjAxNA==")  0 # 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. 0 # Swift 2.2, 28 bytes print("\(ENOTDIR)\(EFAULT)")  Falling back on Darwin/glibc error codes from errno.h. 0 # Mathematica, 16 bytes FromDigits@"JAE"  0 ## 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  -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
`

2This isn't a valid answer. You have to count the shell command also, since it's a non-standard configuration. Also, if you have unprintables in your code, you should include a hexdump, for extra clarity. – None – 2016-09-27T15:46:01.797

– MD XF – 2017-05-21T23:37:40.480

There is more than one way to ^Skin a cat, so these comments are incorrect as written; I clarified my post to address the confusion these folks (and others) may be having. It's still 29 if you use emacs as an editor. MD XF: how do you enter your [U+07DE] character into C-source; also, my compiler throws an error when replacing [main()] with [f()] (excluding the obvious gcc -Df=main hack), but if there is a compiler that accepts [f()], then all C source scores drop by 3. So you need to bump your score by 3, or reduce all other C solutions. – Brian Carcich – 2017-05-22T14:29:36.393