Bi-directional translator

Question

Bi-directional translator

The challenge is to create a translator that can translate a [a-zA-Z ] string into another string according to the mappings below. The translator should also be able to translate translated text back into its normal form (hence bi-directional). Case doesn't have to be maintained and can be either lower/uppercase (or both lower and upper-cased) in the output, however your program should be able to handle either lower/uppercase strings.

Here is the specification of the mappings for this challenge:

-------------------
|Letter | Mapping |
-------------------
|   S   |    9    |
|   F   |    1    |
|   H   |    5    |
|   X   |    =    |
|   Z   |   {N}   |
|   M   |   --*   |
|   N   |   :)    |
|   G   |    ]    |
|   T   |    [    |
-------------------

Here are some example in/outputs: (The " or ... or ..." obviously shouldn't be in your actual output, its just to demonstrate the possible outputs)

> ABS
< AB9 or ab9 or Ab9 or aB9

> foo
< 1oo or 1OO or ...

> FIZZ BUZZ
> 1i{N}{N} bu{N}{N} or 1I{N}{N} BU{N}{N} or ...

> {n}[:) 
< ztn or ZTN or ...

> bar
< bar or BAR or ...

The inputs will always be in [a-zA-Z ] or a "translated" string containing the mapped symbol, other symbols or characters outside wouldn't be inputted.

Answer 1

Retina 0.8.2, 70 68 57 bytes

.[}*]|\)

T`l`L
T`-{=519[]:NGTSF\HXZM`Ro
{
{N}
-
--*
:
:)

Try it online! Link includes test cases. Uppercases the result. Explanation:

.[}*]|\)

Remove the suffixes from {N}, --* and :) leaving their first character.

T`l`L

Uppercase everything.

T`-{=519[]:NGTSF\HXZM`Ro

Translate the first characters. The string is mapped to its reverse, so that - maps to M etc. and vice versa. The H is a special character and needs to be quoted. The - is normally a special character but placing it at the beginning avoids this.

{
{N}
-
--*
:
:)

Expand the {, - and : to the full strings again.

Answer 2

05AB1E, 65 63 bytes

u"{N}"¬.:"--*"¬.:":)"¬.:"SFHXZMNGT[]:-{=519"‡"{-:N-)}*"3ιD€нs‡

Uses the same approach as Neil's answer. Probably not going to be too competitive among the golf answers, but there's definitely improvements to be made here (I'm still a novice with 05AB1E, especially with string manipulation).

Explanation:

u            uppercase the input

"{N}"        push '{N}'
     ¬       extract and push head ('{')
      .:     replace all {N} => {

"--*"        push '--*'
     ¬       extract and push head ('-')
      .:     replace all --* => -

":)"         push ':)'
    ¬        extract and push head (':')
     .:      replace all :) => :

"SFH...519"  push lookup string

            bifurcate (push a, reversed(a))
 ‡           transliterate input with lookup

             set up stack for reverse process
             to do { -> {N} ... etc

"{-:N-)}*"   push literal string
3ι           uninterleave by 3
  D          duplicate
   €         apply single function map 
    н        extract first element
     s       swap
      ‡      transliterate

The bi-directional bit is achieved by transliterating with a bifurcated lookup string; each character that should be swapped is in the opposite position with the one it should be swapped with in the lookup string. The main problem (and potential savings) is that I repeat a lot of logic for the "{N}" -> "{" ... part at the beginning as in the "{" -> "N" ... part at the end. Originally, I thought transliterate could go both ways, so I was going to set up the stack carefully at the beginning, but I don't think transliterate likes it when there's a multi-character string in the list (it probably goes by character, rather than doing a replace all?).

Answer 3

JavaScript (ES6),  118 113  109 bytes

s=>s.toUpperCase(A=[':)',...'NF1G]H5S9T[X=Z','{N}','M','--*']).replace(/[{-]..|:?./g,c=>A[A.indexOf(c)^1]||c)

Try it online!

How?

Because the input is guaranteed to contain exclusively [a-zA-Z] and/or the special patterns, we can use the following minimal regular expression to match them all:

/[{-]..|:?./g
 [{-]..        // either "{" or "-" followed by 2 other characters
               // -> to match "{N}" or "--*"
        :?.    // or any character, preceded by an optional ":"
               // -> to match ":)" or any single character

Working on the uppercase version of the input, we look for the position of each matched pattern in the lookup array A[] and replace it with its counterpart, or leave it unchanged if it is not found (which may only happen for ABCDEIJKLOPQRUVWY).

Answer 4

Perl 5 -p, 90, 88 bytes

$_=uc;s/.[}*]|://g;y/-{)SFHXGT915=][ZNM/MZN915=][SFHXGT{:-/;s/{/{N}/g;s/:/:)/g;s/-/--*/g

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using the same algorithm as Neil's answer

saved 2 bytes removing the backslash before the opening brace, and removing the colon instead of closing parenthesis in :)

Answer 5

Python 3, 189 147 127 135 bytes

s=input().upper()
b=[*'915=][',':)','--*','{N}']
for k,v in zip(*['SFHXGTNMZ',b][::0>max(map(s.find,b))or-1]):s=s.replace(k,v)
print(s)

Try it online! If this were not codegolf the conditional should obviously be outside of the for rather than repeated at each iteration. What wouldn't we do for a few bytes...

• saved 42 bytes by replacing the dict with two lists;
• saved 20 bytes thanks to ovs' [::True or-1] dark magic.
• added 8 bytes because of missing .upper()

Answer 6

Haskell, 186 bytes

import Data.Char
k="S 9 F 1 H 5 X = Z {N} M --* N :) G ] T [ "
v=words$k++map toLower k
w=v++reverse v
_!""=""
[]!(t:u)=t:w!u
(a:b:c)!s|n<-length a,take n s==a=b++w!drop n s|1<2=c!s
(w!)

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Answer 7

Python 3, ^{226 189} 181 bytes

Saved a whopping 37 bytes thanks to ovs!!!
Saved 6 bytes thanks to Stef!!!

from re import*
def f(s,m="9 1 5 = --* :\) ] \[ {N}".split()):
 for t,r in zip(*(m,"SFHXMNGTZ")[::2*bool(search(f"({'|'.join(m)})",s,I))-1]):s=sub(t,r,s,0,I)
 return sub(r'\\','',s)

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Will reverse the translation if any of the mappings are in the input.

Answer 8

C (gcc), 333 bytes I'--* e=ac[ly 5al1 o1 9a[a:) N

void f(char*i,char*o,int*a,int*e,char**c){int t=0;int s=0;char*p=0;while(i[0]){char x=e[(uint8_t)i[0]];if(x){int l;s=a[(s*23)+x-1];if((l=a[((s+1)*23)-1]))t=l;if(!p)p=i;if(s){i++;continue;}}s=0;if(t){if(t>9)o+=sprintf(o,c[t-10]);else o+=sprintf(o,c[t+8]);t=0;p=0;continue;}if(p){while(p<i){o[0]=p[0];o++;p++;}p=0;}o[0]=i[0];o++;i++;}}

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