Atbash Self Palindromes

Question

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

---

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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;

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• nicael: Related: Swap the Alphabet.
• Adám: AtBash is actually not a new thing. It is the Kabala (Jewish Mysticism) Hebrew letter transformation equivalent to this. Since Hebrew is written only with with wovels, any letter string can be read by inserting random wovels. A-T-B(a)SH is a mnemonic for transforming Alef (first Hebrew letter) to Tav (last), Beis (second) to SHin (next-to-last).
• kojiro: Consider giving -1000000 points if someone's solution code is itself an atbash self palindrome? :p
• Adám: @kojiro Non-trivial as opposed to code {Comment-symbol}{Atbash'ed Comment-symbol} Atbash'ed code...
• Conor O'Brien: @mbomb007 I said I may offer a bounty if such a nontrivial program is found

Answer 1

Pyth, (削除) 10 (削除ここまで) 9 bytes

qJrz0_XJG

Try this fiddle online or verify all test cases at once.

Explanation

qJrz0_XJG
 rz0 Lowercase input
 J Store a copy in J
 _XJG Translate J with the reverse alphabet and reverse
q Compare

• xnor: Since you're using rz0 twice, isn't it shorter to save it to a variable?
• PurkkaKoodari: Like @xnor suggests, q_Jrz0XJG is one byte shorter.

Answer 2

RX, (削除) 9 (削除ここまで) 8 bytes

Heavily inspired by Retina, I made this a few days ago. Code:

prR`w$rM

Explanation:

prR`w$rM
p # Start pattern
 r # Reversed lowercase alphabet
 R # Reversed uppercase alphabet
 ` # Next pattern
 w # Equivalent to a-zA-Z_0-9 (word pattern)
 $ # End pattern and compute regex
 r # Reverse input
 M # Change mode to Match mode, compares the atbash string with the reversed string.

Try it here!

• Martin Ender: So how does the language itself actually work? Is it some sort of stack-based string processing language? This is really impressive, but as far as I can tell there is no way to loop in the language yet which means it's highly unlikely that this meets our standards of a programming language at this stage.
• Adnan: @MartinBüttner This language is primarily based on the processing of the input using a stack model. It does not use integers (and probably never will). I do have implemented a loop, but that version has not been released yet.
• lirtosiast: @Martin Regexes are capable of primality testing on their own, so I'm pretty sure this is valid.
• Martin Ender: @ThomasKwa As far as I can see, the interpreter does not use any actual regular expressions.
• lirtosiast: @Martin Hmm, you're right.

Answer 3

Julia, 96 bytes

s->join([get(Dict(zip([u=map(Char,65:90);],reverse(u))),c,c)for c=(S=uppercase(s))])==reverse(S)

This is a lambda function that accepts a string and returns a string. To call it, assign it to a variable.

Ungolfed:

function f(s::AbstractString)
 # Get all of the uppercase letters A-Z
 u = map(Char, 65:90)
 # Create a dictionary for the transformation
 D = Dict(zip(u, reverse(u)))
 # Uppercase the input
 S = uppercase(s)
 return join([get(D, c, c) for c in S]) == reverse(S)
end

Answer 4

Bash + Linux utils, 56

tr a-z `printf %s {z..a}`<<<${1,,}|cmp - <(rev<<<${1,,})

Outputs the empty string for Truthy and something like - /dev/fd/63 differ: byte 1, line 1 for Falsey. If this is not acceptable then we can add -s for an extra 3 bytes and use the standard Unix return codes of 0 for Success (Truthy) and 1 for Failure (Falsey).

Answer 5

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!

This answer was largely inspired by DigitalTrauma's sed answer 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.

• Conor O'Brien: I guess it's better to have linefeeds be the stage separators and pilcrows be the literal than vice versa.
• Martin Ender: @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. ;)

Answer 6

GNU Sed, 105

s/.*/\l&/
h
y/abcdefghijklmnopqrstuvwxyz/zyxwvutsrqponmlkjihgfedcba/
G
:
s/\(.\)\n1円/\n/
t
/^.$/{c1
q}
c0

Outputs 1 for truthy and 0 for falsey.

I tried to do this in Retina, but couldn't figure out how to save the string before Atbash transliteration for reverse comparison with after. Perhaps there is a better way.

Sed's y transliteration command leaves a lot to be desired.

• Martin Ender: Yeah, "storing" things is still cumbersome in Retina. You'd have to duplicate the string and then transliterate and reverse only one copy. I want to add some sort of branching/forking feature in the future, but I'm not quite sure yet about the details.
• Digital Trauma: Ah, I think I see - I tried doing something similar by separating the before and after strings with a colon. I fell down with the regex at the end of the T - I was assuming it applied to each character in turn, but it if my understanding is right, it applies to the whole pattern space, which is much more useful
• Martin Ender: The regex in T is applied to the input string. The transliteration is the only performed within the matches of that regex and everything unmatched is left unchanged. The regex defaults to [\s\S]+ so by omitting it, you're transliterating everything.
• Martin Ender: As you wish. :)
• Toby Speight: Since it's GNU sed, you can save a byte by trading -r flag for the backslashes in \( and \). I agree with you on the y command!

Answer 7

ESMin, 15 chars / 30 bytes

ïþ)Ī(ᶐ,ᶐᴙ)ᴙ≔ïþ)

Try it here (Firefox only).

Explanation

ïþ)Ī(ᶐ,ᶐᴙ)ᴙ≔ïþ) // implicit: ï=input, ᶐ=A-Z
ïþ) // ï.toUpperCase()
 Ī(ᶐ,ᶐᴙ)ᴙ // transliterate input from A-Z to Z-A, then reverse
 ≔ïþ) // check if that's still equal to ï.toUpperCase()
 // implicit output

Answer 8

Parenthetic, 658 bytes

((()()())(()(((()))))((()()((())))))((()()())(()(((())))()()())((())()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()))((()()())(()(((())))())((()())((()(((())))()()))((()()()())((()(())(()))(()(((())))()())(()((()))))(()((())))((()((()))())((()(())(())())((()(()))(()(((())))()()())((()(()()))((())()()()()()()()()()()()()()()()()()()()()()()()()())((()(()()))((()(()())())((()((()))(()))(()(((())))()())))(()(((())))()()())))))((()(((())))())((()((()))()())(()(((())))()())))))))((()(())(()))((()()()(()))((()()()()())(()(((()))))))((()(((())))())((()()()()())(()(((())))))))

Only works for all caps without whitespace right now, using this modified version of the script so that it supports reading from stdin:

#!/usr/bin/env python
from collections import defaultdict
from itertools import izip
import copy
import operator
import os
import sys
# map from paren strings to english names
# for the predefined symbols (lambda, etc)
to_english = defaultdict(lambda:None,\
 {'()': 'lambda',
 '()()': 'define',
 '(())': 'plus',
 '(()())': 'minus',
 '()(())': 'mult',
 '(())()': 'div',
 '()()()': 'if',
 '((()))': 'empty',
 '()()()()': 'charsof',
 '()()(())': 'reverse',
 '()(())()': 'LE',
 '()(()())': 'not',
 '(()())()': 'intofchar',
 '()((()))': 'readline',
 '((()))()': 'cons',
 '(())(())': 'equal',
 '((()))(())': 'car',
 '((()))()()': 'cdr',
 '(())(())()': 'char',
 '(())()(())': 'string'})
# map from english to parenthetic
to_scheme = defaultdict(lambda:None)
for k,v in to_english.iteritems():
 to_scheme[v] = k
def Error(errorString = 'unmatched parens', debug_mode = True):
 if debug_mode:
 print "Error: " + errorString
 sys.exit()
 else:
 raise Exception('paren mismatch')
def bracketsMatch(chars):
 """Returns False if any parentheses in `chars` are not matched
 properly. Returns True otherwise.
 """
 level = 0
 for p in chars:
 if p == '(':
 level += 1
 elif p == ')':
 level -= 1
 if level < 0:
 return False 
 return level == 0
def get_exprs(chars):
 """Returns a list of character sequences such that for each sequence,
 the first and last parenthesis match.
 For example, "(())()()" would be split into ["(())", "()", "()"]
 """
 level = 0
 current = []
 for p in chars:
 if p == '(' or p == ')':
 current.append(p)
 if p == '(':
 level += 1
 elif p == ')':
 level -= 1
 if level == 0:
 yield current
 current = []
## built-in functions ##
def builtin_accumulate(init, accumulate, environment, params):
 """Helper function that handles common logic for builtin functions.
 Given an initial value, and a two-parameter function, the environment, and
 a list of params to reduce, this function will reduce [init] + params using
 the accumulate function and finally returns the resulting value.
 """
 result = init
 for param in params:
 value = interpret(param, environment)
 try: result = accumulate(result, value)
 except: Error(str(value) + ' is not the correct type')
 return result
def builtin_plus(environment, params):
 if len(params) >= 1:
 return builtin_accumulate(interpret(params[0], environment), operator.add, environment, params[1:])
 else:
 return 0.0
def builtin_minus(environment, params):
 if len(params) == 0:
 Error('subtraction requires at least 1 param')
 return builtin_accumulate(interpret(params[0], environment), operator.sub, environment, params[1:])
def builtin_mult(environment, params):
 return builtin_accumulate(1.0, operator.mul, environment, params)
def builtin_div(environment, params):
 if len(params) == 0:
 Error('division requires at least 1 param')
 return builtin_accumulate(interpret(params[0], environment), operator.div, environment, params[1:])
def builtin_LE(environment, params):
 return interpret(params[0], environment) <= interpret(params[1], environment)
def builtin_lambda(environment, params):
 bodies = [body for body in params[1:]]
 params = params[0][1]
 if len(bodies) == 0:
 Error("a function had no body")
 for kind, name in params:
 if kind != 'symbol':
 Error('lambda must have only symbols as arguments')
 def ret(old_environment, arguments):
 #print bodies
 try:
 # create new environment based on args
 environment = copy.copy(old_environment)
 for param, arg in izip(params, arguments):
 environment[param[1]] = interpret(arg, old_environment)
 # evaluate the function bodies using the new environment
 return interpret_trees(bodies, environment, False)
 except:
 Error("Error evaluating a function")
 return ret
def builtin_equal(environment, params):
 for param1, param2 in izip(params[:-1], params[1:]):
 if interpret(param1, environment) != interpret(param2, environment):
 return False
 return True
def builtin_if(environment, params):
 if len(params) != 3:
 Error("'if' takes in exactly 3 params") 
 if interpret(params[0], environment):
 return interpret(params[1], environment)
 return interpret(params[2], environment)
def builtin_not(environment, params):
 return False if interpret(params[0], environment) else True
def builtin_cons(environment, params):
 return (interpret(params[0], environment), interpret(params[1], environment))
def builtin_car(environment, params):
 result = interpret(params[0], environment)
 if not isinstance(result, tuple):
 Error("car must only be called on tuples")
 return result[0]
def builtin_cdr(environment, params):
 result = interpret(params[0], environment)
 if not isinstance(result, tuple):
 Error("cdr must only be called on tuples")
 return result[1]
def builtin_char(environment, params):
 result = interpret(params[0], environment)
 if result != int(result):
 Error("char must only be called on integers")
 return chr(int(result))
def builtin_intofchar(environment, params):
 result = interpret(params[0], environment)
 result = ord(result)
 return result
def builtin_string(environment, params):
 result = ''
 cur = interpret(params[0], environment)
 while cur != ():
 if not isinstance(cur, tuple) or not isinstance(cur[1], tuple):
 Error("string only works on linked lists")
 result += cur[0]
 cur = cur[1]
 return result
def unmakelinked(llist):
 result = ()
 while llist != ():
 if not isinstance(llist, tuple) or not isinstance(llist[1], tuple):
 Error("only works on linked lists")
 result += (llist[0],)
 llist = llist[1]
 return result
def makelinked(tup):
 result = ()
 while tup != ():
 result = (tup[-1],result)
 tup = tup[:-1]
 return result
def builtin_reverse(environment, params):
 result = interpret(params[0], environment)
 result = makelinked(unmakelinked(result)[::-1])
 return result
def builtin_charsof(environment, params):
 result = interpret(params[0], environment)
 result = makelinked(tuple(result))
 return result
def builtin_readline(environment, params):
 result = raw_input()
 return result
# define the default (top-level) scope
default_environment = \
 {to_scheme['plus']: builtin_plus,
 to_scheme['minus']: builtin_minus,
 to_scheme['mult']: builtin_mult,
 to_scheme['div']: builtin_div,
 to_scheme['lambda']: builtin_lambda,
 to_scheme['if']: builtin_if,
 to_scheme['equal']: builtin_equal,
 to_scheme['LE']: builtin_LE,
 to_scheme['not']: builtin_not,
 to_scheme['empty']: (),
 to_scheme['car']: builtin_car,
 to_scheme['cdr']: builtin_cdr,
 to_scheme['cons']: builtin_cons,
 to_scheme['char']: builtin_char,
 to_scheme['string']: builtin_string,
 to_scheme['readline']: builtin_readline,
 to_scheme['charsof']: builtin_charsof,
 to_scheme['reverse']: builtin_reverse,
 to_scheme['intofchar']: builtin_intofchar}
# parse the tokens into an AST
def parse(tokens):
 """Accepts a list of parentheses and returns a list of ASTs.
 Each AST is a pair (type, value).
 If type is 'symbol', value will be the paren sequence corresponding
 to the symbol.
 If type is 'int', value will be a float that is equal to an int.
 If type is expr, value will be a list of ASTs.
 """
 # check for errors
 if not bracketsMatch(tokens):
 Error('paren mismatch')
 # to return - a list of exprs
 exprs = []
 for expr in get_exprs(tokens):
 # check for errors
 if len(expr) < 2:
 Error('too few tokens in: ' + ''.join(expr))
 elif expr[0] != '(' or expr[-1] != ')':
 Error('expression found without () as wrapper')
 # pop off starting and ending ()s
 expr = expr[1:-1]
 # symbol
 if expr[:2] == ['(', ')'] and len(expr) > 2:
 exprs.append(('symbol', ''.join(expr[2:])))
 # integer
 elif expr[:4] == ['(', '(', ')', ')'] and len(expr) >= 4:
 exprs.append(('num', expr[4:].count('(')))
 # expr
 else:
 exprs.append(('expr', parse(expr)))
 return exprs
def interpret(tree, environment):
 """Interpret a single tree (may not be a define) and return the result"""
 kind, value = tree
 if kind == 'num':
 return float(value)
 elif kind == 'symbol':
 if value in environment:
 return environment[value]
 else:
 Error('Unresolved symbol - ' + value)
 elif kind == 'expr':
 function = interpret(value[0], environment)
 if not hasattr(function, '__call__'):
 Error('Symbol "'+value[0]+'" is not a function.')
 return function(environment, value[1:])
 else:
 Error("Unknown tree kind")
def interpret_trees(trees, environment, doprint = True):
 """Interpret a sequence of trees (may contain defines)
 and output the result.
 The trees passed in should be ASTs as returned by parse().
 If doprint is true, the post-interpretation value of each tree is printed.
 """
 environment = copy.copy(environment)
 # hoist define statements (note: trees.sort is stable)
 #trees.sort(key = lambda x: 0 if x[0] == 'expr' and x[1][0][1] == to_scheme['define'] else 1)
 ret = None
 for tree in trees:
 if tree[0] == 'expr' and tree[1][0][0] == 'symbol' and tree[1][0][1] == to_scheme['define']:
 try:
 symbol = tree[1][1]
 if symbol[0] != 'symbol':
 Error('first argument to define must be a symbol')
 symbol = symbol[1]
 value = tree[1][2]
 environment[symbol] = interpret(value, environment)
 except:
 Error('error evaluating define statement')
 else:
 ret = interpret(tree, environment)
 if doprint:
 print ret,
 return ret
# read in the code ignoring all characters but '(' and ')' 
f = open(sys.argv[1],'r')
code = []
for line in f.readlines():
 code += [c for c in line if c in '()']
# parse and interpret the code. print 'Parenthesis Mismatch'
# if an error occured.
#try:
syntax_trees = parse(code)
interpret_trees(syntax_trees, default_environment)
#except:
# print 'Parenthesis Mismatch'

Explanation

(
 define
 (() ()())
 input [[[[]]]]
 (() (((()))))
 exec readline
 ( (() ()((()))) )
)
(
 define
 (() ()())
 value of 'A' [[[[]]]] [][][]
 (() (((())))()()())
 65
 ((()) ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()())
)
(
 define
 (() ()())
 atbash [[[[]]]] []
 (() (((())))())
 (
 lambda
 (() ())
 (
 list [[[[]]]] [][]
 (() (((())))()())
 )
 (
 if
 (() ()()())
 (
 equal
 (() (())(()))
 list
 (() (((())))()())
 empty
 (() ((())))
 )
 then return empty
 (() ((())))
 else
 (
 cons
 (() ((()))())
 (
 char
 (() (())(())())
 (
 plus
 (() (()))
 value of 'A' 65
 (() (((())))()()())
 (
 minus
 (() (()()))
 25
 ((()) ()()()()()()()()()()
 ()()()()()()()()()()
 ()()()()())
 (
 minus
 (() (()()))
 (
 intofchar
 (() (()())())
 (
 car
 (() ((()))(()))
 list
 (() (((())))()())
 )
 )
 value of 'A' 65
 (() (((())))()()())
 )
 )
 )
 )
 (
 atbash
 (() (((())))())
 (
 cdr
 (() ((()))()())
 list
 (() (((())))()())
 )
 )
 )
 )
 )
)
(
 equals
 (() (())(()))
 (
 reverse
 (() ()()(()))
 (
 charsof
 (() ()()()())
 input
 (() (((()))))
 )
 )
 (
 atbash
 (() (((())))())
 (
 charsof
 (() ()()()())
 input
 (() (((()))))
 )
 )
)

• Zorgatone: Do you want your code to be the longest? :P

Answer 9

Python 3, (削除) 90 (削除ここまで) 85 bytes

s=input().upper()
print(s[::-1]==''.join(chr([o,155-o][64<o<91])for o in map(ord,s)))

We convert the input to uppercase, and then calculate the Atbashed string by subtracting all ordinals from 155 if they're in the uppercase alphabet range.

Answer 10

Kerf, 73 bytes

Kerf is a proprietary language in the same general family as APL, J and K. It's possible to write cryptic, compact oneliners and avoid the use of explicit loops:

{[s]s:_ s;n:?isnull i:s search\<a:char 97+^26;f:(/a)[i];s match/f[n]:s[n]}

However, using the spelled-out aliases for commands instead of the shorthand symbols and using meaningful identifiers makes the program much clearer, and fairly easy to follow even if you aren't familiar with Kerf:

def atbash_palindrome(str) {
 str: tolower str
 alpha: char range(97, 123)
 indices: str search mapleft alpha
 encoded: (reverse alpha)[indices]
 notfound: which isnull indices
 return str match reverse encoded[notfound]:str[notfound]
}

In action:

KeRF> p: {[s]s:_ s;n:?isnull i:s search\<a:char 97+^26;f:(/a)[i];s match/f[n]:s[n]};
KeRF> p mapdown ["WIZARD","Wizard","W I Z A R D","ABCXYZ","345 09%","ev","Zyba","$%%$","-AZ"]
 [1, 1, 1, 1, 0, 1, 1, 1, 0]

Kerf probably isn't going to win a ton of codegolf competitions, especially against purpose-built languages, but it might be worth tinkering with if you like the idea of APL-family languages but find the syntax too weird. (Disclaimer: I'm the author of the reference manual for Kerf.)

Answer 11

Prolog, 121 bytes

a(W):-upcase_atom(W,X),atom_codes(X,C),b(C,Z),!,reverse(Z,C).
b([A|T],[R|S]):-(A>64,A<91,R is 77-A+78;R=A),(b(T,S);S=[]).

This is called with an atom as input, e.g. a('WIZARD')..

Answer 12

JavaScript (ES6), 91

x=>(x=[...x.toLowerCase()]).every(c=>((v=parseInt(c,36))>9?(45-v).toString(36):c)==x.pop())

TEST

F=x=>(x=[...x.toLowerCase()]).every(c=>((v=parseInt(c,36))>9?(45-v).toString(36):c)==x.pop())
console.log=x=>O.textContent+=x+'\n'
;[
 ["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
,["$%ZA%$", true]
].forEach(t=>{var i=t[0],x=t[1],r=F(i);
 console.log(i+' -> '+r+(x==r?' OK':' Fail (expected:'+x+')'))})

<pre id=O></pre>

Answer 13

C, (削除) 101 (削除ここまで) 97 bytes

As the question specified ASCII characters, this doesn't handle any other encodings.

f(char*s){char*p=s+strlen(s);while(*s&&!(isalpha(*--p)?*s<64||*s+*p-27&31:*s-*p))++s;return s>p;}

Explanation

int f(char*s)
{
 char *p = s + strlen(s);
 while (*s && !(isalpha(*--p) ? *s<64||*s+*p-27&31 : *s-*p))
 ++s;
 return s > p;
}

We make a pointer p that begins at the end of the string. We then loop, moving both s and p towards each other s reaches the end. This means that every pair of characters will be checked twice, but this saves a couple of bytes compared to stopping as soon as the pointers cross over.

At each iteration, we check whether *p is a letter. If so, check that *s is in the range of letters (ASCII 64 upward), and that *p and *s add up to 27 (mod 32). Non-letters over 64 will fail that test, so we don't need to check isalpha(*s).

If *p is not a letter, then we simply test whether it is equal to *s. In either case, we terminate the loop before s and p cross over.

If s and p have crossed, then every pair of letters matched correctly, so we return true; otherwise we return false.

Test program

Pass the strings to be tested as command-line arguments. This produces correct output for all the test cases. There is no supplied requirement for the empty string; my implementation returns false for that input.

#include <stdio.h>
int main(int argc, char **argv)
{
 while (*++argv)
 printf("\"%s\" => %s\n", *argv, f(*argv)?"true":"false");
 return 0;
}

• cat: you can drop f's type declaration for a K&R style prototype: f(char*s)

Answer 14

Perl 5, 70 bytes

A subroutine:

{$"='';reverse(map/[A-Z]/?chr(155-ord):$_,(@_=split'',uc$_[0]))eq"@_"}

See it in use:

print sub{...}->("W i z a r d")

Answer 15

Vyxal, 7 bytes

⇩ḂkaḂL·=

Try it Online!

⇩ # Lowercase input
 Ḃ # Duplicate and reverse
 L· # Transliterate by...
 kaḂ # lowercase alphabet and backwards lowercase alphabet
 = # is it equal to the original?

• u-ndefined: Still new in using this lang but using Ḃ allows to save 2 bytes: ⇩ḂkaḂL·=
• emanresu A: @u-ndefined Thanks!

Answer 16

MATL, 23 bytes

Uses current release.

jkt"@@2Y2m?_219+]h]tP=A

Examples

>> matl
 > jkt"@@2Y2m?_219+]h]tP=A
 > 
> Tree vvig
1
>> matl
 > jkt"@@2Y2m?_219+]h]tP=A
 > 
> W IZ ARD
0

Answer 17

CJam, 18 bytes

qeu_'[,65>_W%erW%=

Try it online

Works by converting input to uppercase, performing the translation of letters, flips the string, and checks for equality.

Answer 18

Japt, (削除) 30 (削除ここまで) 27 bytes

U=Uv)w \Ur"[a-z]"_c +4^31 d

Try it online!

How it works

This is based largely on my Japt answer on Swap the Alphabet.

U=Uv)w \Ur"[a-z]"_c +4^31 d
U=Uv) // Set U to U.toLowerCase().
w \ // Reverse it, and check if it is equal to:
Ur"[a-z]" // Take the input and replace each letter with:
 _c +4 // Take its char code and add 4. This results in
 // the string "abc...xyz"
 // becoming "efg...|}~".
 ^31 // XOR the result by 31. This flips its last five 5 bits.
 // We now have "zyx...cba".
 d // Convert back from a char code.
 // Implicit: output last expression

Answer 19

Ruby, 56 bytes

->s{s.upcase!;s==s.tr(x=[*?A..?Z]*'',x.reverse).reverse}

It's an anonymous function which takes a string and returns true or false. It's rather clumsy: In order to save some bytes, it uses the destructive variant of upcase (with a ! after it). upcase! unfortunately returns nil if nothing is changed (like all numeric input), so some bytes are lost trying to deal with that. Still works tho :)

Answer 20

Python 3.8 (pre-release), 73 bytes / 76 bytes

Lambda version: (73 bytes)

lambda x:(a:=[*map(ord,x.upper())])==[[c,155-c][64<c<91]for c in a][::-1]

Try it online!

Code version: (76 bytes)

a=[*map(ord,input().upper())];print(a==[[c,155-c][64<c<91]for c in a][::-1])

Try it online!

How it works:

• .upper() Upper all the char of the input
• a=[*map(ord,input().upper())] In both program we store in a the list containing the ascii value of each character.
• [... for c in a] We iterate through this list
• [c,155-c][64<c<91] If our ascii value is beetween 64 and 91 (if the char is uppercase) we "flip" it.
• a==[...][::-1] We compare our list with the new formed list reversed

Note:

If we used Python2, we could save 1 byte on the second program because :

• -3 bytes: map() returns a list so we don't need the [*map()] syntax anymore
• -2 bytes (and a little adjustment): print doesn't need parenthesis
• but +4 bytes because we have to use raw_input()instead of input

Answer 21

Fig, \8ドル\log_{256}(96)\approx\$ 6.585 bytes

s$Pcacza

Try it online!

s # Are the following equal, ignoring case:
 # Implicit input
 $ # Reverse of... 
 a # Lowercased (implicit) input
 P # Transliterated from
 ca # Alphabet
 cz # To reversed alphabet

Answer 22

Python, (削除) 156 (削除ここまで) 112 bytes

a=map(chr,range(65,91))
s=raw_input().upper()
print ''.join([dict(zip(a,a[::-1])).get(i,i) for i in s])==s[::-1]

Basically, it makes a dictionary of the translation with uppercase letters and the input is capitalized (if everything were lowercase instead, that would add 5 bytes). Then, for each character in the capitalized input, do the translation and append to a list unless the character is not in the alphabet, in which case append the character as is. Join the entire list and compare to the reversed list.

Shout-out to @Artyer for posting almost exactly as I was going to post before me. But I need to confirm, this is my work and I did this independently.

Based on the Julia answer by Alex A. Try it here

• Yytsi: There is an unnecessary whitespace after .get(i,i). +1.

Answer 23

Factor, (削除) 118 (削除ここまで) 113 bytes

This is an anonymous function.

[ >upper dup >array [ 1string 65 90 [a,b] [ 1string ] map dup reverse zip >hashtable at ] map "" join reverse = ]

I don't know of a shorter way to generate an associative array of the alphabet :c

Answer 24

Clojure, 100 bytes

(defn x[b](let[a(.toUpperCase b)c(reverse a)](=(map #(char(if(<= 65(int %)90)(- 155(int %))%))a)c)))

It should be possible to cut it down to a single anonymous function, cutting around 10 more bytes (of declarations) but I did not find a way yet.

Answer 25

Ruby, (削除) 79 (削除ここまで) 77 bytes

s=$*[0].upcase
exit(s==s.reverse.tr('A-Z','ZYXWVUTSRQPONMLKJIHGFEDCBA'))?0:1

Accepts the word to test as a command-line argument. Exits with code 0 (which is truthy to the shell) if the argument is an atbash self palindrome, or with code 1 otherwise.

• cat: Wouldn't putsing the result be shorter than exit with a ternary?
• Jordan: FYI $* is an alias for ARGV.

Answer 26

MATLAB, 61 bytes

Not the shortest solution, but still interesting

f=@(a)~any(changem(upper(a),90:-1:65,65:90)-fliplr(upper(a)))

Answer 27

Japt v2.0a0, 13 bytes

Note the unprintable (charcode 155) at the end of the second line.

u
ÔøUr\l_cn#

Try it

Answer 28

Factor + spelling, 60 bytes

[ >lower dup ALPHABET dup reverse zip substitute reverse = ]

Try it online!

Explanation:

• >lower Convert to lowercase.

  Stack: (e.g.) "wizard"

• dup Duplicate.

  Stack: "wizard" "wizard"

• ALPHABET Place the lowercase alphabet on the stack.

  Stack: "wizard" "wizard" "abcdefghijklmnopqrstuvwxyz"

• dup Duplicate.

  Stack: "wizard" "wizard" "abcdefghijklmnopqrstuvwxyz" "abcdefghijklmnopqrstuvwxyz"

• reverse Reverse.

  Stack: "wizard" "wizard" "abcdefghijklmnopqrstuvwxyz" "zyxwvutsrqponmlkjihgfedcba"

• zip Make an associative array from two sequences.

  Stack: "wizard" "wizard" { { 97 122 } { 98 121 } { 99 120 } { 100 119 } { 101 118 } { 102 117 } { 103 116 } { 104 115 } { 105 114 } { 106 113 } { 107 112 } { 108 111 } { 109 110 } { 110 109 } { 111 108 } { 112 107 } { 113 106 } { 114 105 } { 115 104 } { 116 103 } { 117 102 } { 118 101 } { 119 100 } { 120 99 } { 121 98 } { 122 97 } }

• substitute Substitute elements in a sequence that have keys in an assoc with their values.

  Stack: "wizard" "draziw"

• reverse Reverse.

  Stack: "wizard" "wizard"

• = Are they equal?

  Stack: t

Answer 29

05AB1E, 8 bytes

Code:

lDAAR‡RQ

Explanation:

l # Lowercase the implicit input
 D # Duplicate top of the stack
 AAR # Push the lowercase alphabet (A) and the lowercase alphabet reversed (AR)
 ‡ # Transliterate a -> b
 R # Reverse this string
 Q # Compare with the input string

Try it online!