14
\$\begingroup\$

Fairly obviously, you have to print the Greek alphabet. But I thought I'd make it slightly more interesting, so there's a twist: when your program is ran with an argument (anything), it should output the Greek alphabet in lowercase.

Information

  1. Greek alphabet (Capitals): ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
  2. Greek alphabet (Lowercase): αβγδεζηθικλμνξοπρστυφχψω

Rules/Requirements

  • Each submission must be a full program.
  • Builtins to print the Greek alphabet are not permitted
  • Standard loopholes apply
  • You must print exactly what is shown.

Scoring

Programs are scored according to bytes. If you are using a character set different to UTF-8 please specify. Try to get the least bytes out of everyone, this is !

Test cases

./program
==> ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ

./program 1
==> αβγδεζηθικλμνξοπρστυφχψω

./program 'lower'
==> αβγδεζηθικλμνξοπρστυφχψω

./program 123897883421
==> αβγδεζηθικλμνξοπρστυφχψω

./program ""
==> αβγδεζηθικλμνξοπρστυφχψω

greek();
==> ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ

greek("blah");
==> αβγδεζηθικλμνξοπρστυφχψω

greek(123);
==> αβγδεζηθικλμνξοπρστυφχψω

Submissions

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

Leaderboard

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

/* Configuration */

var QUESTION_ID = 97049; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 53406; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
  return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}

function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });  
}

getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    if (/<a/.test(lang)) lang = jQuery(lang).text();
    
    languages[lang] = languages[lang] || {lang: a.language, user: a.user, size: a.size, link: a.link};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang > b.lang) return 1;
    if (a.lang < b.lang) return -1;
    return 0;
  });

  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
                       .replace("{{NAME}}", lang.user)
                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }

}
body { text-align: left !important}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 290px;
  float: left;
}

table thead {
  font-weight: bold;
}

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>

\$\endgroup\$
  • 5
    \$\begingroup\$ Does an empty string count as having an argument or having no argument? Please use the Sandbox to get feedback before posting a challenge on the main site. \$\endgroup\$ – user45941 Oct 22 '16 at 6:37
  • 2
    \$\begingroup\$ Related. \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 8:36
  • 2
    \$\begingroup\$ Will the argument be always one argument, or can there be two or more arguments? \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 8:40
  • 1
    \$\begingroup\$ Kritixi approves the challenge. \$\endgroup\$ – user41805 Oct 22 '16 at 18:11
  • 4
    \$\begingroup\$ I'm a bit confused by your last three test cases; they look like function calls, yet the rules specifically ask for a full program. \$\endgroup\$ – Dennis Oct 31 '16 at 14:11

37 Answers 37

7
\$\begingroup\$

05AB1E, 16 bytes

Uses CP-1252 encoding.

24Ý17K913+çJDl¹s

Try it online!

Explanation

24Ý               # push range [0 ... 24]
   17K            # remove 17
      913+        # add 913 to each
          ç       # convert from code point
           J      # join to string
            Dl    # push a lowercase copy
              ¹   # push the input
               s  # swap the top 2 elements of the stack
                  # implicitly display the top of the stack
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Did you mean [0 ... 23] or [0, 24) instead of [0 ... 24] or [0, 24]? \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 10:34
  • 1
    \$\begingroup\$ @EriktheGolfer There is one other character between the range of greek alphabet code points. \$\endgroup\$ – jimmy23013 Oct 22 '16 at 10:42
  • \$\begingroup\$ @jimmy23013 Oh right, I was really confused. I am working it in another language, which I won't tell here. I'm commited to that right now. \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 10:43
8
\$\begingroup\$

Ruby, 56 bytes

Full program. I don't think a function/lambda answer will be shorter than this for this language.

Dangit, sigma ς. This is why we can't have nice things. And you too, (unrenderable character that serves as a placeholder for "uppercase" ς)

print *($*[0]?([*?α..?ω]-[?ς]):[*?Α..?Ρ,*?Σ..?Ω])
| improve this answer | |
\$\endgroup\$
  • 7
    \$\begingroup\$ I think this prints an extra lowercase sigma, ς \$\endgroup\$ – Angs Oct 22 '16 at 11:47
  • 1
    \$\begingroup\$ This now prints an extra U+03A2. \$\endgroup\$ – LegionMammal978 Oct 22 '16 at 18:02
5
\$\begingroup\$

JavaScript (ES6), 89 83 81 bytes

(...a)=>String.fromCharCode(...[...Array(24)].map((_,i)=>i*1.06+913+32*a.length))

If a character array is acceptable, then for 82 80 78 bytes:

(...a)=>[...Array(24)].map((_,i)=>String.fromCharCode(i*1.06+913+32*a.length))

Edit: Saved a bunch of bytes thanks to @ETHproductions.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Using String.fromCharCode(...array) is again shorter: (...a)=>String.fromCharCode(...[...Array(24)].map((_,i)=>i+(i>16)+913+32*!!a.length)) \$\endgroup\$ – ETHproductions Oct 22 '16 at 15:27
  • 1
    \$\begingroup\$ And since String.fromCharCode floors, you can save another two bytes like so: (...a)=>String.fromCharCode(...[...Array(24)].map((_,i)=>i*1.06+913+32*!!a.length)) \$\endgroup\$ – ETHproductions Oct 22 '16 at 15:29
  • \$\begingroup\$ @ETHproductions I can't believe I forgot to use String.fromCharCode(...) but I like that i*1.06 trick! Sadly comprehensions are a byte longer this time as far as I can tell. \$\endgroup\$ – Neil Oct 22 '16 at 16:48
  • \$\begingroup\$ I managed to get it down to (...a)=>String.fromCharCode(...[for(_ of Array(i=24))i++*42/41+889+32*!!a.length]) (82). \$\endgroup\$ – ETHproductions Oct 22 '16 at 17:16
  • \$\begingroup\$ You can remove the !!, as there will only be zero or one args. \$\endgroup\$ – ETHproductions Oct 22 '16 at 19:43
5
\$\begingroup\$

Haskell, 114 108 bytes

import System.Environment
q[a,b]=[a..b]
f[]=q"ΑΡ"++q"ΣΩ"
f _=q"αρ"++q"σω"
main=f<$>getArgs>>=putStr

Thanks to @xnor for saving 6 bytes

This might win in the category of longest imports if nothing else…

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ It looks like it's worth defining q[a,b]=[a..b] to use 4 times. \$\endgroup\$ – xnor Nov 6 '16 at 11:15
  • 1
    \$\begingroup\$ getArgs>>=putStr.f is shorter. \$\endgroup\$ – Ørjan Johansen Mar 5 '17 at 1:26
4
\$\begingroup\$

R, 104 99 92 bytes

g=function(x)for(i in sprintf("'\\u%04X'",913:937+32*!missing(x))[-18])cat(eval(parse(t=i)))

Golfing down on the second version I had previously. Work the same way as previous version.

Thanks to @JDL for shaving off 7 bytes!

Old versions at 104 bytes:

I have two different solutions with the same byte count:

f=function(x){a="ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ";cat(if(missing(x))a else tolower(a))}

Or:

g=function(x)for(i in sprintf("'\\u%04X'",if(missing(x))913:937 else 945:969)[-18])cat(eval(parse(t=i)))

Explanations:

#First one is trivial
f=function(x){
     a="ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ" #String to print
     #If argument missing, print uppercase else lowercase 
     cat(if(missing(x))a else tolower(a)) 
     }
#Second one more interesting:
g=function(x)
     #Create escaped unicode strings (i. e. "'\\u03B1'" etc.), loop over them...
     for(i in sprintf("'\\u%04X'",if(missing(x))913:937 else 945:969)[-18])
          #...eval the strings
          cat(eval(parse(t=i))) 

Usage:

> f()
ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
> g()
ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ
> f(1)
αβγδεζηθικλμνξοπρστυφχψω
> g(1)
αβγδεζηθικλμνξοπρστυφχψω
> f("a")
αβγδεζηθικλμνξοπρστυφχψω
> g("a")
αβγδεζηθικλμνξοπρστυφχψω

For some reasons, it doesn't work on R-Fiddle (it changes " to ' by default which makes the code throw an error) but you can try it out on Ideone.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Replace +'if'(missing(x),0,32) with +32*!missing(x)? \$\endgroup\$ – JDL Oct 27 '16 at 13:42
  • \$\begingroup\$ @JDL Thanks! Don't know why I didn't think of that one. \$\endgroup\$ – plannapus Oct 27 '16 at 13:50
4
\$\begingroup\$

CJam, 16 bytes

25,H-ea'α'Α?f+

Try it online!

25,H-         e# 0 to 24 excluding 17.
ea'α'Α?       e# If the argument list is truthy (non-empty), alpha, otherwise Alpha.
f+            e# Return characters with each number added to the character code.
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Can you add a explanation please? \$\endgroup\$ – XiKuuKy Oct 23 '16 at 9:47
  • \$\begingroup\$ @XiKuuKy I don't think so :/ \$\endgroup\$ – RGS May 2 at 21:45
  • 1
    \$\begingroup\$ @RGS Edited. :/ \$\endgroup\$ – jimmy23013 May 3 at 7:36
  • \$\begingroup\$ @jimmy23013 nice one :-P I simply scrolled through this answer and thought I'd comment if for posterity. Didn't think you would come and edit your answer! +1 \$\endgroup\$ – RGS May 3 at 8:01
3
\$\begingroup\$

Pyke, 55 21 20 bytes

24Fi16>913s.C)sz!hAl

Try it here!

24F          )       -   for i in range(24):
   i16>              -       i > 16
       913s          -      sum(i, ^, 913)
           .C        -     chr(^)
              s      -  sum(^) - created uppercase alphabet
                  Al - [len, lower, upper][V](^)
                 h   -  V + 1
               z!    -   not input()
| improve this answer | |
\$\endgroup\$
3
\$\begingroup\$

Python 3, 73 bytes

import sys
x=25
while x:print(end=chr(906-x+32*len(sys.argv))[:x^8]);x-=1

Try it online!

This still may not be as low as it can go!

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

Actually, 62 bytes

 "$ù"E"ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ"@ƒ.X

Try it online!

Because some of the uppercase Greek alphabet is not present in CP437, this solution is encoded in UTF-8 and scored accordingly. Here is a hexdump (reversible with xxd -r):

00000000: 2022 24c3 b922 4522 ce91 ce92 ce93 ce94   "$.."E"........
00000010: ce95 ce96 ce97 ce98 ce99 ce9a ce9b ce9c  ................
00000020: ce9d ce9e ce9f cea0 cea1 cea3 cea4 cea5  ................
00000030: cea6 cea7 cea8 cea9 2240 c692 2e58 0a    ........"@...X.

Explanation:

 "$ù"E"ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ"@ƒ.X
<space>                               push the number of items present on the stack
 `$ù`E                                select `ù` (lowercase) if the number of items is not 0, else `$` (stringify - does nothing to a string)
      "ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ"      push the uppercase Greek alphabet
                                @ƒ    call the selected function
                                  .   print
                                   X  discard
| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

Python 3, 80 77 76 bytes

import sys;a='ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ';print([a,a.lower()][-len(sys.argv)])

Old version:

import sys;a='ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ';print([a,a.lower()][len(sys.argv)-1])

Older version:

import sys;a='ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ';print(a.lower()if len(sys.argv)>1else a)
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ [a,a.lower()][len(sys.argv)-1] because the comments state that there will only ever be one argument. \$\endgroup\$ – Value Ink Oct 22 '16 at 18:33
  • \$\begingroup\$ You can shave one more byte with [a.lower(),a][-len(sys.argv)]. \$\endgroup\$ – toriningen Oct 22 '16 at 22:17
  • 2
    \$\begingroup\$ What encoding are you using that this is 76 bytes? \$\endgroup\$ – AdmBorkBork Oct 24 '16 at 13:43
  • \$\begingroup\$ I count 100 bytes. \$\endgroup\$ – Oliver Ni Oct 24 '16 at 18:01
2
\$\begingroup\$

Japt, 21 19 bytes

;Ck"rz" c+816+32*NÊ

Test it (no input)
Test it (empty string input)


Explanation

      :Implicit input.
;C    :The lowercase alphabet.
k"rz" :Remove r & z.
c+    :Map over the charcodes of the string and add...
816+  :    816 plus...
32*NÊ :    32 multiplied by the length of the array of input variables; 0 if no input is supplied, 1 if a single input is supplied.
      :Implicit output of resulting string
| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

R, 96 53 51 58 bytes

cat(intToUtf8((913:937)[-18]+32*!!length(commandArgs(T))))

Try it online!

I tried to make an R answer with command line args, but it's a bit longer(now a lot, lot shorter) than the loop based approach by plannapus. I used some of the tricks in their answer.

A crazy -43 bytes from Dominic Van Essen.

-2 more bytes from Dominic Van Essen.

+7 bytes after removing ς

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Cool to see command-line args in action in R: we don't use them often (so I'm not very good with them...)! There are still a couple of golfing tricks you can use: 1. you can skip the {} in the function definition if there's only 1 command (like here)... \$\endgroup\$ – Dominic van Essen Oct 1 at 6:42
  • \$\begingroup\$ 2. you could actually use the function argument, instead of leaving it out & wasting bytes by calling missing (like this)... \$\endgroup\$ – Dominic van Essen Oct 1 at 6:43
  • 1
    \$\begingroup\$ and 3. once you've done that, you'll probably want to remove the function altogether & incorporate it into the main expression (like this)... \$\endgroup\$ – Dominic van Essen Oct 1 at 6:44
  • \$\begingroup\$ 51 bytes with !! instead of (>0) \$\endgroup\$ – Dominic van Essen Oct 1 at 10:46
  • \$\begingroup\$ I'm afraid I've just noticed (from a comment in the answer above) that the Greek alphabet doesn't seem to be contiguous in Utf8-encoding! You need to remove the 'ς' character. This unfortunately brings it back up to 58 bytes... \$\endgroup\$ – Dominic van Essen Oct 1 at 11:23
1
\$\begingroup\$

Scala, 82 bytes

print((('Α'to'Ρ')++('Σ'to'Ω')).mkString.map(& =>if(args.size<1)&else&toLower))

Ungolfed:

print((('Α'to'Ρ')++('Σ'to'Ω')).mkString.map(s=>if(args.size==0)s else s.toLower))

Sadly, (('Α'to'Ρ')++('Σ'to'Ω')) is a Vector[Char], which would be printed as Vector('Α', 'Β', ..., so it has to be converted to a string with mkString.The argument to map is called & to save spaces between &else, else& and & toLower.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Jelly, 21 bytes

,⁸Ea32µ24R%18T+912+µỌ

A full program

TryItOnline! - note once run with an argument a refresh will be required to run with no argument.

How?

,⁸Ea32µ24R%18T+912+µỌ - Main link
,⁸                    - pair (,) with left argument (defaults: left=0, ⁸=[])
                            no argument: [0,[]]
                               argument: [argument,argument]
  E                   - all items equal? -> no argument: 0; argument: 1
   a32                - and 32           -> no argument: 0; argument: 32
      µ               - monadic chain separation
       24R            - range(25)        -> [  1,  2,  3,..., 17, 18, 19,..., 25]
          %18         - mod 18           -> [  1,  1,  1,...,  1,  0,  1,...,  1]
             T        - truthy indices   -> [  1,  2,  3,..., 17,     19,..., 25]
              +912    - add 912          -> [913,914,915,...,929,    931,...,937]
                  +   - add the 0 or 32 (lowercase letters are 32 code points higher)
                   µ  - monadic chain separation
                    Ọ - cast to ordinals
| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

PowerShell v2+, 68 bytes

(($a=-join([char[]](913..929+931..937))).toLower(),$a)[!$args.count]

ASCII-only, but outputs in UTF-16. Constructs a char-array of the appropriate characters, -joins it together into a string, stores it into $a. Then, uses !$args.count as an index into the tuple, to output either $a if no arguments are present, or $a.ToLower() if there is at least one argument.

PS C:\Tools\Scripts\golfing> .\print-the-greek-alphabet.ps1
ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ

PS C:\Tools\Scripts\golfing> .\print-the-greek-alphabet.ps1 ''
αβγδεζηθικλμνξοπρστυφχψω
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Python 3, 80 bytes

import sys;print(''.join(chr(x+881+len(sys.argv)*32)for x in range(25)if x!=17))
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C#, 174 bytes

Lazy implementation, can probably golf it a lot

class P{static void main(string[] a){System.Console.Write(a.Length>0?"αβγδεζηθικλμνξοπρστυφχψω":"ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ");}}
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tcl, 73

 puts [expr $argc>0?"αβγδεζηθικλμνξοπρστυφχψω":"ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ"]

demo

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PHP, 84 Bytes

Online Version

<?=($argc<2?trim:mb_strtolower)("ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ");

PHP, 87 Bytes

$s="ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ";echo$argc>1?mb_strtolower($s):$s;
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  • \$\begingroup\$ gotcha. You seem somehow absent. $argc>1 is 8 bytes shorter than isset($argv[1]) and doesn´t need a blank. \$\endgroup\$ – Titus Jan 26 '17 at 1:00
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Retina, 55 bytes

.+
αβγδεζηθικλμνξοπρστυφχψω
^$
ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ

Try it online

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APL (Dyalog Extended), 26 bytes

The normal way to pass multiple arguments to APL programs is through a lists. So this program prompts for such a list, which may have 0 or 1 arguments.

'ς΢'~⍨⍳'Ω'×~≢⎕

Try it online!

 prompt for argument list

 tally the number of arguments (0 or 1)

~ negate that (1 or 0)

'Ω'× "multiply" Omega by that (1 keeps uppercase, 0 folds to lowercase)

 All characters from ΑΩ or αω respectively

'ς΢'~⍨ remove lowercase and "uppercase" final sigmas

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Braingolf, 24 bytes

l?#α:#Α|# 9->[.!@ 1+];

Try it online!

Explanation:

l?#α:#Α|# 9->[.!@ 1+];  Implicit input of args to stack
l                       Push length of stack
 ?                      If last item (length) is != 0..
  #α                    ..Push lowercase alpha
    :                   else
     #Α                 ..Push uppercase alpha
       |                endif
        #<space>        Push 32 (ordinal of a space)
          9-            Subtract 9 (23)
            >           Move last item to start of stack
             [......]   Do-While loop, will run 24 times
              .         Duplicate last item
               !@       Print without popping
                  1     Push 1
                   +    Pop last 2 items (1 and duplicate) and push sum
                     ;  Suppress implicit output
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PowerShell, 81 bytes

'ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ'|%('*wer','*per')[!$args.Count]

Try it online!

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Python 3, 78 bytes

import sys;a=(len(sys.argv)-1)<<5;print(''.join(map(chr,range(a+913,a+938))))

Explanation

a is an offset, 32 when an argument is given to use the lowercase range. 913 to 937 is the range of unicode values for the uppercase greek alphabet.

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Python 3, 88 bytes

import sys
s=945 if sys.argv[1:] else 913
for i in range(s,s+25):
  print(chr(i),end='')
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Python 3, 74 bytes

This an improvement to @HackerBoss es' answer (I don't have enough reputation to comment).

import sys;a=len(sys.argv)-1<<5;print(*map(chr,range(a+913,a+938)),sep='')

Subtraction has higher precedence than bit shift, so no need for parentheses, we save 2 bytes. Also instead of using ''.join() we unpack the map() and set print()'s default seperator to an empty string to omit the spaces, that saves 1 byte.

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Python 2, 108 bytes

#coding=iso-8859-7
import sys;print'αβγδεζηθικλμνξοπρστυφχψω'if~-len(sys.argv)else'ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ'

Terribly golfed, maybe :(

And no, I can't use A.lower().

Thanks to 13285 (alexwlchan) for -11 bytes.

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  • \$\begingroup\$ This is the most golfed version I can do. \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 10:34
  • \$\begingroup\$ You can save 8 bytes by skipping the assignment to A,a, and just using those strings directly in the print. \$\endgroup\$ – alexwlchan Oct 22 '16 at 10:53
  • \$\begingroup\$ @alexwlchan Right, I just didn't think of it... still terribly golfed, though. \$\endgroup\$ – Erik the Outgolfer Oct 22 '16 at 10:57
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Mathematica, 91 bytes

Print@{#&,ToLowerCase}[[Length@$ScriptCommandLine]][""<>"Α"~CharacterRange~"Ω"~Drop~{18}]

Script. Just takes the character range from Α to Ω, removes U+03A2/ς, either converts to lowercase or doesn't, and prints.

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Perl, 39 + 3 (-C2 flag) = 44 42 bytes

 perl -C2 -E 'say map{chr$_+954-!@ARGV*32}-9..7,9..15'
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JavaScript, 95 bytes

95 bytes, but only 71 characters. Byte counter. Using @Neil s way to determine if arguments are passed.

(...a)=>'ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ'[`to${a.length?'Low':'Upp'}erCase`]()
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