Is it an alphadrome?

Question

An alphadrome is a word in which each letter in the first half of the word "reflects" its alphabetical "opposite" in the second half of the word. Write a program or function that returns truthy if the given word is an alphadrome and falsey otherwise.

For example, BEEB is a palindrome. Its first letter B "reflects" its last letter B, and its second letter E reflects its second-to-last letter E.

In contrast, BEVY is an alphadrome. Its first letter B is the second letter of the alphabet, and it reflects the word's last letter Y, which is the second-to-last letter of the alphabet. Likewise, its second letter E is the 5th letter of the alphabet, and it reflects V, the 5th-from-the-end letter of the alphabet.

Like palindromes, an alphadrome can have an odd number of letters, too. WORLD is an alphadrome. W is the 4th-from-the-end letter of the alphabet and D is the 4th letter; O is the 12th-from-the-end letter of the alphabet and L is the 12th letter. The center R reflects itself.

Any single-letter word is an alphadrome.

Rules

• Input words will include only the characters a-z or A-Z, whichever is convenient.

• Output will be a truthy or falsey value.

• Default I/O rules apply, standard rules apply, and standard loopholes are forbidden.

• This is code golf. Shortest answer in bytes wins.

Test cases

A        true
ABA      false
ABZ      true
ABBA     false
BEVY     true
GOLF     false
ZOLA     true
WIZARD   true
BIGOTRY  true
RACECAR  false
ABCDEFGHIJKLMNOPQRSTUVWXYZ  true

Answer 1

><>, 17 15 bytes

Saved 2 bytes thanks to Jo King

l<*=""+{$n?(3

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Explanation

l<                # initialize as truthy (length of the input) and reverse direction
l        n?(3     # output when the stack holds at most 1 letter (if odd)
      +{$         # add the first and unhandeled letter
   =""            # compare with 155
  *               # multiply with current truthy/false value

Answer 2

Knight (v2), 52 bytes

;=tT;=n/L=pP2;W+1=n-nT=t&?155+A Gp nT A Gp--LpTnTtOt

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Answer 3

Desmos, 59 bytes

k=l.length
f(l)=∏_{n=1}^{k/2-.5}0^{(155-l[n]-l[k-n+1])^2}

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Try It On Desmos! - Prettified

Takes in a list of codepoints because Desmos doesn't support strings.

Answer 4

jq, 54 bytes

Using recursion (54 bytes)

def f:length<2or(.[1:-1]|f)and.[0]+last==155;explode|f

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Using mappings (55 bytes)

explode|[.,reverse]|[transpose[]|add==155,true]|sort[1]

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Answer 5

Pushy, 16 bytes

L2/:{+v;O155K=P#

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L2/:                 \ floor(length/2) times do:
    {                \    Cyclic shift the stack left
     +               \    Add the top two items
      v;             \    More the result to the second stack.
        O            \ Go to the second stack
         155         \ Push 155
            K=       \ Check equality in place, replacing every stack element with 0 or 1
              P#     \ Print the product of these

Answer 6

C (101 bytes)

a;main(z,s)char**s;{int l=strlen(s[1])-1;while(l/2>a)if(s[1][a]-65^90-s[1][l-a++])exit(0);puts("1");}

Answer 7

Thunno 2, 6 bytes

Ḳ+ṇ6oḣ

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Output with inverted booleans. Add the ! flag to take the logical NOT.

Port of Kevin Cruijssen's 05AB1E answer.

Explanation

Ḳ+ṇ6oḣ  # Implicit input
Ḳ+      # Bifurcate and add
  ṇ6o   # Remove 155s
     ḣ  # Remove the first item
        # Implicit output