95
\$\begingroup\$

The Tabula Recta (sometimes called a 'Vigenere Table'), was created by Johannes Trithemius, and has been used in several ciphers, including all variants of Bellaso's Vigenere cipher and the Trithemius cipher. It looks like this:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
BCDEFGHIJKLMNOPQRSTUVWXYZA
CDEFGHIJKLMNOPQRSTUVWXYZAB
DEFGHIJKLMNOPQRSTUVWXYZABC
EFGHIJKLMNOPQRSTUVWXYZABCD
FGHIJKLMNOPQRSTUVWXYZABCDE
GHIJKLMNOPQRSTUVWXYZABCDEF
HIJKLMNOPQRSTUVWXYZABCDEFG
IJKLMNOPQRSTUVWXYZABCDEFGH
JKLMNOPQRSTUVWXYZABCDEFGHI
KLMNOPQRSTUVWXYZABCDEFGHIJ
LMNOPQRSTUVWXYZABCDEFGHIJK
MNOPQRSTUVWXYZABCDEFGHIJKL
NOPQRSTUVWXYZABCDEFGHIJKLM
OPQRSTUVWXYZABCDEFGHIJKLMN
PQRSTUVWXYZABCDEFGHIJKLMNO
QRSTUVWXYZABCDEFGHIJKLMNOP
RSTUVWXYZABCDEFGHIJKLMNOPQ
STUVWXYZABCDEFGHIJKLMNOPQR
TUVWXYZABCDEFGHIJKLMNOPQRS
UVWXYZABCDEFGHIJKLMNOPQRST
VWXYZABCDEFGHIJKLMNOPQRSTU
WXYZABCDEFGHIJKLMNOPQRSTUV
XYZABCDEFGHIJKLMNOPQRSTUVW
YZABCDEFGHIJKLMNOPQRSTUVWX
ZABCDEFGHIJKLMNOPQRSTUVWXY

I frequently need this, but can't find it anywhere on the internet to copy and paste from. Because the square table is so long, and takes frigging ages to type, your code must be as short as possible.

Rules/Requirements

  • Each submission should be either a full program or function. If it is a function, it must be runnable by only needing to add the function call to the bottom of the program. Anything else (e.g. headers in C), must be included.
  • If it is possible, provide a link to a site where your program can be tested.
  • Your program must not write anything to STDERR.
  • Standard Loopholes are forbidden.
  • Your program can output in any case, but it must be printed (not an array or similar).

Scoring

Programs are scored according to bytes, in UTF-8 by default or a different character set of your choice.

Eventually, the answer with the least bytes will win.

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 = 86986; // 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\$
14
  • \$\begingroup\$ Regarding rule 1: do we have to include each header for each function we use? \$\endgroup\$
    – owacoder
    Jul 30, 2016 at 15:59
  • \$\begingroup\$ I meant if we use printf, we need to include stdio.h, if isalpha is used, ctype.h is needed, etc. Is this correct? \$\endgroup\$
    – owacoder
    Jul 30, 2016 at 16:03
  • 3
    \$\begingroup\$ You should probably put "lowercase is allowed" in the challenge specification itself. Just so people are less likely to miss it if they don't see these comments. \$\endgroup\$
    – Sherlock9
    Jul 30, 2016 at 18:25
  • 2
    \$\begingroup\$ do i have to print it or can i return a string/char array \$\endgroup\$ Jul 30, 2016 at 19:57
  • 1
    \$\begingroup\$ Thanks for the question for making me stay awake all night. (+1) \$\endgroup\$ Aug 22, 2016 at 18:45

173 Answers 173

1
\$\begingroup\$

Cheddar, 79 bytes

(|>26).map(l->String.letters.slice(l)+String.letters.slice(0,l)).vfuse.slice(1)

Pretty ugly answer. I would of had a much nicer answer if I fixed cycle in time:

(|>26).map(String.letters.cycle).vfuse
\$\endgroup\$
1
\$\begingroup\$

Ruby, 75 42 bytes

a=('A'..'Z').to_a;l=a.length;l.times {l.times {|i|$><<a[i]};a.rotate!;puts}

a=*?A..?Z;26.times {puts a.join;a.rotate!}
\$\endgroup\$
1
  • \$\begingroup\$ Rotate! I was searching so long for this method, but I couldn't remember its name. \$\endgroup\$
    – IMP1
    Feb 17, 2017 at 10:41
1
\$\begingroup\$

Batch, 94 bytes

@set a=ABCDEFGHIJKLMNOPQRSTUVWXYZ
@for /l %%i in (0,1,25)do @call echo %%a:~%%i%%%%a:~0,%%i%%
\$\endgroup\$
1
\$\begingroup\$

Brachylog, 19 16 bytes

25yeN,@A:N:$(iw@Nw\
@A~c[A:B]rcw@Nw\

Try it online!

Explanation of the 16-byte solution

@A~c[A:B]rcw@Nw\
@A~c[A:B]            @A is in the form of A+B.  (choice point)
                     (@A == 'abcdefghijklmnopqrstuvwxyz')
         rcw         Write B+A to STDOUT.
            @Nw      Write '\n' to STDOUT.
               \     Backtrack to last choice point and
                     make the next choice until there is
                     no more choice left.

Explanation of the 19-byte solution

25yeN,@A:N:$(iw@Nw\
25yeN                N is a number between 0 and 25 inclusive. (choice point)
      @A:N:$(iw      @A left-rotated N times, print it to STDOUT.
                     (@A == 'abcdefghijklmnopqrstuvwxyz')
               @Nw   Write '\n' to STDOUT.
                  \  Backtrack to last choice point and
                     make the next choice until there is
                     no more choice left.
\$\endgroup\$
1
\$\begingroup\$

Pushy, 5 bytes

Non-competing as the language postdates the challenge, but I though I'd post it as it's the same length as the 05AB1E answer...

AL:"{

Try it online!

This takes advantages of cyclic shifting:

A      \ Push the uppercase alphabet, as a list of charcodes.
 L:    \ Length (yields 26) times do:
   "   \   Print the characters
    {  \   Cyclically shift left, once.
\$\endgroup\$
1
\$\begingroup\$

Perl 6, 40 bytes

say [~] (|("A".."Z")xx 2)[^26+$_]for ^26

Joshua's answer already shows the canonical way to do this in Perl 6, but I thought it might be interesting to show a solution that doesn't use the rotate built-in.

How it works

                                 for ^26  # For each index from 0 to 25:
           "A".."Z"                       #   Generate the alphabet.
         |("A".."Z")                      #   Slip it into the outer list.
                    xx 2                  #   Repeat it, yielding a flat list of 52 characters.
        (               )[      ]         #   Index this list using:
                          ^26             #      The range from 0 to 25,
                             +$_          #      with the loop counter added to both endpoints.
    [~]                                   #   Concatenate the result.
say                                       #   Print it, followed by a newline.
\$\endgroup\$
1
\$\begingroup\$

Racket 126 bytes

(let((d display)(g integer->char))(for((i 26))(for((j(range i 26)))(d(g(+ j 65))))(for((k(range 0 i)))(d(g(+ k 65))))(d"\n")))

Ungolfed:

(define(f)
  (let ((d display)
        (g integer->char))
    (for ((i 26))
      (for ((j (range i 26)))
        (d (g (+ j 65))))
      (for ((k (range 0 i)))
        (d (g (+ k 65))))
      (d"\n"))))

Testing:

(f)

Output:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
BCDEFGHIJKLMNOPQRSTUVWXYZA
CDEFGHIJKLMNOPQRSTUVWXYZAB
DEFGHIJKLMNOPQRSTUVWXYZABC
EFGHIJKLMNOPQRSTUVWXYZABCD
FGHIJKLMNOPQRSTUVWXYZABCDE
GHIJKLMNOPQRSTUVWXYZABCDEF
HIJKLMNOPQRSTUVWXYZABCDEFG
IJKLMNOPQRSTUVWXYZABCDEFGH
JKLMNOPQRSTUVWXYZABCDEFGHI
KLMNOPQRSTUVWXYZABCDEFGHIJ
LMNOPQRSTUVWXYZABCDEFGHIJK
MNOPQRSTUVWXYZABCDEFGHIJKL
NOPQRSTUVWXYZABCDEFGHIJKLM
OPQRSTUVWXYZABCDEFGHIJKLMN
PQRSTUVWXYZABCDEFGHIJKLMNO
QRSTUVWXYZABCDEFGHIJKLMNOP
RSTUVWXYZABCDEFGHIJKLMNOPQ
STUVWXYZABCDEFGHIJKLMNOPQR
TUVWXYZABCDEFGHIJKLMNOPQRS
UVWXYZABCDEFGHIJKLMNOPQRST
VWXYZABCDEFGHIJKLMNOPQRSTU
WXYZABCDEFGHIJKLMNOPQRSTUV
XYZABCDEFGHIJKLMNOPQRSTUVW
YZABCDEFGHIJKLMNOPQRSTUVWX
ZABCDEFGHIJKLMNOPQRSTUVWXY
\$\endgroup\$
1
\$\begingroup\$

Ruby, 43 bytes

26.times{|i|puts(([*?A..?Z]*2)[i,26].join)}

Explanation

[*?A..?Z]
This creates a list from 'A' to 'Z', using ruby's literal character syntax ?A and the splat operator on a range.

It then doubles this list, and cycles through it 26 times, printing out 26 characters with an increasing offset.

\$\endgroup\$
1
\$\begingroup\$

tcl, 75

time {puts -nonewline [format %c [expr ([incr i]%27)?($i-1)%26+65:10]]} 702

Can be seen running on: http://rextester.com/live/BNZ46923

\$\endgroup\$
1
\$\begingroup\$

Python 2, 58 bytes

a=bytearray(range(65,91))*2
exec'print a[:26];a=a[1:];'*26

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Python 2, 60 bytes

s = string.uppercase
for i in range(0,26):print s[i:]+s[:i]

Needs the string library to run.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Dark Lord, Sir - I have some good news and some bad - bad news: you need to include import string if you are using the external string library. Good news: you can drop 2 bytes by condensing s = string... to s=string... AND I have prepared a TIO Link for you, Dark Lord, Sir. \$\endgroup\$ Jul 22, 2017 at 18:09
1
\$\begingroup\$

Recursiva, 13 bytes

{("P+}J''_Q(}

Try it online!

Explanation

{("P+}J''_Q(}
{             - For each
 (            - Upper-case alphabet yield
  "           - Iteration command begin
   P          - Print
    +         - Concatenate 'BCDE...XYZA'
     }        - Current element say 'B' (2nd iteration) 
      J''     - join with nothing 'CDE...XYZA'
         _    - Reverse; ['CDEF.... XYZ', 'A']
          Q(} - Split ( at }'i.e. B'; ['A','CDEF....XYZ']
\$\endgroup\$
1
\$\begingroup\$

Kotlin, 59 bytes

(0..25).map{i->(0..25).map{print('A'+(it+i)%26)}
println()}

Beautified

    (0..25).map{i->
        (0..25).map { print('A' + (it+i)%26) }
        println()
    }

Test

fun x() =
(0..25).map{i->(0..25).map{print('A'+(it+i)%26)}
println()}

fun main(args: Array<String>) {
    x()
}

TryItOnline

\$\endgroup\$
1
\$\begingroup\$

Pyth, 11 10 7 bytes

-1 based off Jakube's idea.

-3 from Jakube

I'd like to see how far down this could be golfed. I'm guessing it'll have mapping somewhere in there.

Also, I'm taking a stab in the dark assuming that non-fixed-input s need not support input. If there is a meta page, could somebody link it?

VG=+t
GN

Explanation:

         (G=lowercase alphabet)
VG       Print the alphabet; For in the all but the last character of the alphabet:
  =+t    (New line = print next arg + input which I assume is empty)
     GN  Append the first character of G (the current index) to the end of G. This is pretty neat, while returning the value, it does two things: print the tabula, and modify the G variable.

Try it online!

\$\endgroup\$
5
  • \$\begingroup\$ There are already shorter Pyth solutions here, but this was a short golf I immediately found for your solution: VG=+t<newline>GhG \$\endgroup\$
    – Jakube
    Sep 18, 2017 at 7:03
  • \$\begingroup\$ Actually, this is even shorter: VG=+t<newline>GN \$\endgroup\$
    – Jakube
    Sep 18, 2017 at 7:05
  • \$\begingroup\$ @Jakube Neither of these work, as they start at "bcde..." and not "abcd...", nice try though! \$\endgroup\$
    – Stan Strum
    Sep 18, 2017 at 10:23
  • \$\begingroup\$ No, they don't: TryItOnline \$\endgroup\$
    – Jakube
    Sep 18, 2017 at 12:33
  • \$\begingroup\$ @Jakube that's something I didn't know worked \$\endgroup\$
    – Stan Strum
    Sep 18, 2017 at 13:00
1
\$\begingroup\$

Pyth, 8 bytes

j.u.<N1G

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Mathematica, 56 bytes

StringRiffle[NestList[RotateLeft,Alphabet[],25],"\n",""]

Pretty simple. Just gets a list of all the rotations of the alphabet, then joins them with newlines and spaces within the lines.

\$\endgroup\$
1
\$\begingroup\$

Husk, 8 bytes

U¡ṙ1…"AZ

Try it online!

Explanation

     "AZ  -- string: "AZ"
    …     -- fill the gaps: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
 ¡        -- iterate the following infinitely times & accumulate results in list
  ṙ1      -- rotate by 1 (eg. first time: "BCDEFGHIJKLMNOPQRSTUVWXYZA")
U         -- only keep the longest prefix with unique elements
\$\endgroup\$
1
\$\begingroup\$

uBASIC, 68 bytes

That pesky Chr$() and adding extra spaces forces the use of Left$() for +7 Bytes

0ForI=0To25:ForJ=0To25:?Left$(Chr$(65+(I+J)Mod 26),1);:NextJ:?:NextI

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Pyth, 9 bytes

VlG.<rG1N

Try it online!

If lowercase is allowed, the code can be shortened to 7 bytes:

VlG.<GN

which ties Leaky Nun's (also lowercase) answer.

\$\endgroup\$
1
  • \$\begingroup\$ The accepted answer is lowercase, which implies that it is allowed. \$\endgroup\$
    – Jo King
    Mar 1, 2018 at 6:05
1
\$\begingroup\$

brainfuck, 137 135 128 bytes

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

Try it online!

How it Works:

++[[+<]>+>++]<  Generates the number 27
[[-<+>>+<]>-]   Copies and decrements the 27 until it is 0
                Tape looks like 27 26 25 24 23 21 ... 3 2 1 0 0'
<----[<[+<]>[>]<----] Adds 63 to each number, making them correspond to the uppercase alphabet and @
<, Pop the excess @
There are going to be two strings separated by a 0
The second is intitially blank
<[ While the first string exists
   [[.<]<] Print the two strings of characters
   >>[[>]>]<< Go to the start of the first string
   Copy the first character of the first string to the end of the second string
   [[<]<[<]<+>>[>]>[>]<-]<[<]<[<]<[->+<] 
   ++++++++++., Print a newline
   >[>]>[>]< Go to the start of the first string
] Ends if the first string is empty
\$\endgroup\$
1
\$\begingroup\$

Red, 66 65 bytes

repeat n 26[repeat m 26[prin to-char n + m - 2 % 26 + 65]print""]

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Small Basic, 117 bytes

Script that takes no input and outputs to the TextWindow object

For I=0To 25
s=""
For J=0To 25
s=s+Text.GetCharacter(65+Math.Remainder(I+J,26))
EndFor
TextWindow.WriteLine(s)
EndFor

Try it at SmallBasic.com! Requires Silverlight and thus must be opened with IE

\$\endgroup\$
1
\$\begingroup\$

Pepe, 79 bytes

rEeEeeeeeEREeEeEEeEeREERrEEEEErEEEeReerEEEEREEreeereeErRrEEEEEEEREEEEEEEReereee

Try it online!

\$\endgroup\$
1
\$\begingroup\$

T-SQL, 155 154 bytes

~Removed semicolon to save 1 byte.

It isn't as short as the other T-SQL answer, but I wanted to create a recursive SQL sample, and only have SQL server installed.

WITH f AS(SELECT 1n,CAST('abcdefghijklmnopqrstuvwxyz'AS VARCHAR(MAX))v UNION ALL SELECT n+1,RIGHT(v,LEN(v)-1)+LEFT(v,1)v FROM f WHERE n<26)SELECT v FROM f

~Added link to test environment~

https://rextester.com/CVII57986

\$\endgroup\$
2
  • \$\begingroup\$ Hello and welcome to PPCG. I currently cannot test your solution and therefore simply trust you that this answer is valid. It would be helpful if you could link to an online testing environment, though you do not have to. \$\endgroup\$ Mar 1, 2019 at 17:29
  • 1
    \$\begingroup\$ Good point, added a permalink. \$\endgroup\$
    – Aesais
    Mar 1, 2019 at 17:37
1
\$\begingroup\$

Gol><>, 21 bytes

`Z9sF:M|lFlKasRo}ao|;

2 bytes knocked off courtesy of JoKing, who also designed a version that has a leading newline (click here for it)

Try it online!

Gol><>, 23 bytes

`Z9sF:M|asFasKasRo}ao|;

Golfed off a ton of bytes by pushing the letters of the alphabet in! I'm going to try to make the output code smaller.

Try it online!

Old version, 44 bytes

"ABCDEFGHIJKLMNOPQRSTUVWXYZ"rasFasKasRo}ao|;

This is a brute force method of doing this, but I will golf it down some more.

Try it online!

\$\endgroup\$
0
1
\$\begingroup\$

Japt-R, 14 8 bytes

;26ÆBéYn

-6 bytes thanks to @Shaggy!

Try it Online!

\$\endgroup\$
3
  • \$\begingroup\$ 10 bytes (or 8 with the -R flag). Sorry I haven't helped you golf your other solutions so far; I've been spending the time I normally spend golfing working on my new interpreter the past few days. I'll try to have a look at them tomorrow for you, though. \$\endgroup\$
    – Shaggy
    Mar 6, 2019 at 23:14
  • \$\begingroup\$ @Shaggy Thanks, there are so many different functions in Japt, and the docs aren't always clear on what they do plus the fact I don't usually program in Javascript makes learning Japt pretty hard. Thank you for your aid! \$\endgroup\$
    – Gymhgy
    Mar 7, 2019 at 3:00
  • \$\begingroup\$ I read through the string methods section at least 20 times, how did I miss the one that does exactly what I need... \$\endgroup\$ May 27, 2021 at 21:43
1
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Malbolge, 4829 bytes

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Try it online!

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1
  • \$\begingroup\$ It works! (15chrs) \$\endgroup\$ Aug 17, 2020 at 5:53
1
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Poetic, 510 bytes

THE BOOGEYMAN WAS HERE,i read
a story i saw,a story i saw
it reads:WE SAW IN STARS AN END TO EARTH
i saw magic spells,oracles of chaos
i also saw hexes
o,i know a curse is a legend/a lie
i think i know a curse i saw
a curse is hogwash
now i browse a book
o,i laugh
it is trash
o,i giggle
in a bang,i heard a POW
a flash,a puff o flames
IT IS SATAN,said an unholy devil
go satan
i saw a story,i said,a story i saw
a coming doom,i said
a demon i saw
i scream:o,do beware of looming evils
o no

Try it online!

Poetic is an esolang I made in 2018 for a class project. It's basically brainfuck with word-lengths instead of symbols.

The point of the language is to allow for programs to be written in free-verse poetry. This specific poem was based on the brainfuck solution by Dorian (I had one, but it took longer to run and it was more bytes).

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1
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Python 3, 76 bytes

a="".join(list(map(chr,range(65,91))))
for i in range(26):print(a[i:]+a[:i])

Try it online!

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1
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Pip, 17 15 bytes

Fi,26Pz@>i.z@<i

Loops 26 times over Pip's built-in a-z variable and prints the table shifted each time.

-2 bytes from Dlosc because he actually read the question

Try it online!

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1
  • \$\begingroup\$ yep it workss :D \$\endgroup\$
    – Razetime
    Aug 31, 2020 at 1:09

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