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Atbash Self Palindromes

Consider the Atbash transformation:

A|B|C|D|E|F|G|H|I|J|K|L|M
Z|Y|X|W|V|U|T|S|R|Q|P|O|N

Where A ⇔ Z and L ⇔ O, e.g. There is an interesting property that some words share. When some strings are translated to their atbash-equivalent, said translation is the original word reversed. I call these Atbash Self Palindromes.

As an example, let's translate WIZARD:

W → D
I → R
Z → A
A → Z
R → I
D → W

The result is DRAZIW, which is WIZARD reversed. Thus, WIZARD is an atbash self palindrome.

Objective Given a string of printable ASCII characters, output or return a truthy value if that string is an atbash self palindrome, and a falsey value otherwise. (This is done through STDIN, closest equivalent, functional input, etc. If your language cannot do any of these, ~~(削除) consider choosing a different language (削除ここまで)~~ you may hardcode the input.) You should do this case-insensitively. If the input is a palindrome and is unaffected by the atbash seqeunce, you should still output true, since a palindrome + itself is a palindrome. This is a code-golf, so the shortest program in bytes wins.

Test cases

"Input" => true, false
"WIZARD" => true
"Wizard" => true // case doesn't matter
"wIzArD" => true 
"W I Z A R D" => true
"W IZ ARD" => false // the atbash of this is D RA ZIW, which is not a palindrome of W IZ ARD
"ABCXYZ" => true // ZYXCBA
"345 09%" => false // is not a palindrome
"ev" => true // ve
"AZGDFSSF IJHSDFIU HFIA" => false
"Zyba" => true
"-AZ" => false // -ZA is not a reverse of -AZ
"Tree vvig" => true // Givv eert 
"$%%$" => true // palindrome
"A$&$z" => true // z$&$A

Leaderboard

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

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 snippet:

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

var QUESTION_ID=68757,OVERRIDE_USER=44713;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>

Answer*

## [Retina](https://github.com/mbuettner/retina), 44 bytes

 $
 ¶$_
 T`lL`Ro`.+$
 +`(¶.*)(.)
 2ドル1ドル
 i`^(.+)1円$

Prints `1` or `0`. The byte count assumes that the file is encoded as ISO 8859-1.

[Try it online!][1] 

This answer was largely inspired [by DigitalTrauma's sed answer](https://codegolf.stackexchange.com/a/68776/8478) but then again I guess there aren't that many approaches to this challenge in the first place.

### Explanation

Whenever you see a `¶`, the first thing Retina does after splitting the code into lines is to replace all of those pilcrows with linefeeds. This allows the inclusion of linefeeds for a single byte even though linefeeds are Retina's stage separator.

 $
 ¶$_

We start by duplicating the input. We match the end of the input with `$` and insert a linefeed along with the input itself (using `$_`).

 T`lL`Ro`.+$

A transliteration stage. Let's start with the regex: `.+$`. It matches the second copy of the input (by ensuring the match goes until the end of the string). So only characters in the second copy will be transliterated. The transliteration itself makes use of some very recent features. `l` and `L` are character classes for lower and upper case letters, respectively. `o` refers to the other character set of the transliteration and `R` reverses it. So the two character sets expand to:

 abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
 ZYXWVUTSRQPONMLKJIHGFEDCBAzyxwvutsrqponmlkjihgfedcba

You'll notice that this swaps the case while doing the Atbash cypher, but we'll do the final comparison case-insensitively anyway.

 +`(¶.*)(.)
 2ドル1ドル

Now we reverse the second copy. Unfortunately, Retina doesn't have a convenient way for doing that yet, so we'll have to move one character from the end to the front at a time. This is done by repurposing the linefeed separator as a marker of which part hasn't been reversed yet. We match that part but capture the last character separately. That character goes in front, and the remainder is unchanged. The `+` tells Retina to do this repeatedly until it's no longer possible (because `¶` is at the end of the string).

 i`^(.+)1円$

Finally, we check whether the two strings are the same. The `i` makes the pattern case-insensitive - conveniently, in .NET, this means that backreferences are also case-insensitive. You might notice that we no longer have a separator between the original input and the modified copy. We don't need one though, because they are the same length, and if the string now consists exactly of the same string twice (up to case), then those must be the original and the modified string. If you're wondering what happened to the trailing linefeed that we used as a marker, it's still there, but in many regex flavours `$` also matches *before* the last character of the string if that character is a linefeed.

Since this stage only consists of a single line, it is taken to be a match stage, which counts the number of matches. If the input is an Atbash palindrome, we'll get exactly one match and the output is `1`. If not, then this regex won't match and the output will be `0`.

 [1]: http://retina.tryitonline.net/#code=JArCtiRfClRgbExgUm9gLiskCitgKMK2LiopKC4pCiQyJDEKaWBeKC4rKVwxJA&input=QUJDIFhZeg

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one. Explanations of your answer make it more interesting to read and are very much encouraged.
…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).

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\$\begingroup\$ I guess it's better to have linefeeds be the stage separators and pilcrows be the literal than vice versa. \$\endgroup\$

Conor O'Brien
– Conor O'Brien

2016年01月07日 16:54:33 +00:00
Commented Jan 7, 2016 at 16:54
\$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ For convenience, you can also insert linefeeds by escape sequences, \n in regex and $n in substitution, but that's wasted bytes for golfing. ;) \$\endgroup\$

Martin Ender
– Martin Ender

2016年01月07日 16:55:42 +00:00
Commented Jan 7, 2016 at 16:55

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