Spanish ID card control character calculator

Question

This is a very very simple algorithm, that I am sure can be solved in many many different languages. In Spain ID cards (known as DNI) consist of 8 numbers and a control character. The control character is calculated with the following algorithm: divide the number by 23, take the remainder of the operation and replace it with a character according to this table:

0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22  
T  R  W  A  G  M  Y  F  P  D  X  B  N  J  Z  S  Q  V  H  L  C  K  E

If the DNI belongs to a foreign person living in Spain, the first digit is changed to X, Y or Z and it is called an NIE. In this case, the following substitutions are made before calculating the control character:

X Y Z
0 1 2

There are a lot of calculators online that help you get the control character, but, how short can you write that code? Write an algorithm (program or function) that receives a string with the DNI number (that will always consist of 8 alphanumeric characters) and returns just the single control character calculated and nothing more (a trailing newline is accepted).

Notes:

• The DNI is always written in uppercase, but in your algorithm you can choose the input and output to be upper- or lowercase, just be consistent.
• In real life, some NIEs issued before 2008 have 8 digits after the X, Y or Z, but for the purposes of this game, you can consider they have 7 digits as they have nowadays.
• You can consider that the input string will always have 8 characters, but if they are not in the "8 digits" format nor the "[XYZ] plus 7 digits" format, you must return an error (of your choice) or just throw an exception.

Test cases:

00000010 -> X (HRM Juan Carlos I's DNI number)
01234567 -> L
98765432 -> M
69696969 -> T
42424242 -> Y
Z5555555 -> W (Z=2)
Y0000369 -> S (Y=1)
A1234567 -> <Error code or exception>
1231XX12 -> <Error code or exception>

This is code-golf, so may the shortest code for each language win!

Answer 1

Python 3, 83 bytes

lambda n:'TRWAGMYFPDXBNJZSQVHLCKE'[int([n,str(ord(n[0])%4)+n[1:]][n[0]in'XYZ'])%23]

Try it online!

-5 thanks to AlixEinsenhardt (from 99 to 94). -1 thanks to JonathanAllan.

Answer 2

Haskell, 107 93 92 bytes

c(x:y)="TRWAGMYFPDXBNJZSQVHLCKE"!!mod(read(("X0Y1Z2"!x):y))23
(a:b:c)!x|x==a=b|2>1=c!x
_!x=x

Try it online!

Answer 3

Pyth, 35 34 bytes

The code contains some unprintable characters, so here is a reversible xxd hexdump.

00000000: 402e 5043 22fc eeff 1ffc adc7 e614 9451  @.PC"..........Q
00000010: 2247 2573 7358 637a 5d31 3e33 4755 3320  "G%ssXcz]1>3GU3
00000020: 3233                                     23

Uses lowercase characters.

Try it online. Test suite.

Printable version

@.P305777935990456506899534929G%ssXcz]1>3GU3 23

Explanation

• cz]1 splits the input at position 1, e.g. "y0000369" to ["y", "0000369"].
• >3G gets the last 3 characters of the alphabet, "xyz".
• U3 gets the range [0, 3[, [0, 1, 2].
• X maps xyz to [0, 1, 2] in the split array, e.g. ["y", "0000369"] to [1, "0000369"]. This replaces the first character if it is one of xyz, while leaving the tail of 7 characters untouched since any 7 character string can't be equal to a single character.
• s joins the array with the empty string, e.g. [1, "0000369"] to "10000369".
• s casts this string to integer, e.g. "10000369" to 10000369. This throws an error if any extra non-digit characters are left in the string.
• %…23 gets the value modulo 23, e.g. 10000369 to 15.
• C"…" converts the binary string from base 256 to integer (about 3.06 × 10²⁶).
• .P…G gets the permutation of the alphabet with that index.
• @ gets the correct character from the permutation.

Answer 4

MATL, 62 59 bytes

'RWAGMYFPDXBNJZSQVHLCKET'j'[\dXYZ]\d{7}'XXg'XYZ'I:47+XEU1))

The error for not valid input is A(I): index out of bounds (compiler running in Octave) or Index exceeds matrix dimensions (compiler running in Matlab).

Try it online!

Explanation

'RWAGMYFPDXBNJZSQVHLCKET' % Push this string (output letters circularly shifted by 1)
j                         % Unevaluated input
'[\dXYZ]\d{7}'            % Push this string (regexp pattern)
XX                        % Regexp. Returns cell arary with matching string, or empty
g                         % Convert to standard array. Will be empty if non-valid input
'XYZ'                     % Push this string
I:47+                     % Push [47 48 49] (ASCII codes of '012')
XE                        % Transliterate
U                         % Convert to number
1)                        % Get first entry. Gives an error if empty
)                         % Index (modular, 1-based) into initial string
                          % Implicitly display

Answer 5

ES6, 83 82 81 bytes

i=>'TRWAGMYFPDXBNJZSQVHLCKE'[(/^[XYZ]/.test(i)?i.charCodeAt()%4+i.slice(1):i)%23]

In action!

Uppercase only, the error code for invalid numbers is undefined.

One byte saved thanks to Jonathan Allan.
Another byte saved thanks to Shaggy.

Answer 6

Java 8, 154 145 104 bytes

s->{s[0]-=s[0]<88|s[0]>90?0:40;return"TRWAGMYFPDXBNJZSQVHLCKE".charA‌​t(new Integer(new String(s))%23);}

-9 bytes thanks to @OliverGrégoire.
-41 bytes thanks to @OliverGrégoire again, by taking the input as a char-array (char[]).

If the input is invalid, it will either fail with a java.lang.NumberFormatException or java.lang.StringIndexOutOfBoundsException.

Explanation:

Try it here. (Invalid test cases are surrounded by try-catch so it doesn't stop at the first error.)

s->{                      // Method with char[] parameter and char return-type
  s[0]-=s[0]<88|s[0]>90?  // If the first character is not XYZ:
    0                     //  Leave the first character as is
   :                      // Else:
    40;                   //  Subtract 40 to convert it to 012
  return"TRWAGMYFPDXBNJZSQVHLCKE".charAt(
                          //    Get the char from the String
    new Integer(          //    by converting the following String to an integer:
      new String(s)       //     by converting the char-array to a String
    )%23);                //    And take modulo-23 of that integer
}                         // End of method

Answer 7

PHP, 88 bytes

prints 1 for an error

$a[0]=strtr(($a=$argn)[0],ZYX,210);echo!ctype_digit($a)?:TRWAGMYFPDXBNJZSQVHLCKE[$a%23];

Try it online!

Answer 8

Jelly, 42 bytes

“X0Y1Z2”⁸y@1¦RṪ€V%23ị“Ñ×ⱮEɼiʋ}'uƒẹsø’ṃØA$¤

Try it online!

Too long, Jelly! Dennis is disappointed of you!^{[citation-needed]}

Answer 9

q/kdb+, 68 bytes

Solution:

{"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}

Examples:

q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"00000010"
"X"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"01234567"
"L"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"98765432"
"M"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"69696969"
"T"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"42424242"
"Y"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"Z5555555"
"W"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"Y0000369"
"S"
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"A1234567"
" "
q){"TRWAGMYFPDXBNJZSQVHLCKE"mod["J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x;23]}"1231XX12"
" "

Explanation:

If the first character, x 0, is in the string "XYZ" then a will be 0, 1 or 2. If the first character is not in the string, then a will be 3. If a is less than 3, we switch out the first character for the string of a (0, 1 or 2), otherwise we switch out for the first character (thus effectively doing nothing). This string is cast to a long ("J"$), which is then mod'd with 23 to give the remainder. This remainder is used to index into the lookup table.

{ "TRWAGMYFPDXBNJZSQVHLCKE" mod["J"$$[3>a:"XYZ"?x 0;string a;x 0],1_x;23] } / ungolfed solution
{                                                                         } / lambda function
                            mod[                                     ;23]   / performds mod 23 of the stuff in the gap
                                                                  1_x       / 1 drop input, drops the first character
                                                                 ,          / concatenation
                                    $[             ;        ;   ]           / if COND then TRUE else FALSE - $[COND;TRUE;FALSE]
                                        a:"XYZ"?x 0                         / "XYZ" find x[0], save result in a
                                      3>                                    / is this result smaller than 3
                                                    string a                / if so, then string a, e.g. 0 -> "0"
                                                             x 0            / if not, just return first character x[0]
                                "J"$                                        / cast to long
  "TRWAGMYFPDXBNJZSQVHLCKE"                                                 / the lookup table

Notes:

" " is returned in the error scenarios, this is because the cast returns a null, and indexing into a string at index null is an empty char. I could add 4 bytes at the beginning ("!"^) to make it more obvious that an error had occurred:

q){"!"^"TRWAGMYFPDXBNJZSQVHLCKE"("J"$$[3>a:"XYZ"?x 0;($)a;x 0],1_x)mod 23}"1231XX12"
"!"

Answer 10

JavaScript(ES6), 121 byte

f=i=>{c=+i[0];a=3;while(a--){i[0]=="XYZ"[a]&&(c=a)}b=7;while(b--){c= +i[7-b]+c*10}return "TRWAGMYFPDXBNJZSQVHLCKE"[c%23]}

console.log([f("00000010"),f("01234567"),f("98765432"),f("69696969"),f("42424242"),f("Z5555555"),f("Y0000369"),f("A1234567"),f("1231XX12")])

Answer 11

Japt, 50 bytes

Similar to most of the other approaches.

Input and output is lowercase, outputs undefined for invalid input.

`tr°gmyfpdxbnjzsqvhlcke`g("xyz"øUg)?Uc %4+UÅ:U %23

Test it
Test all valid test cases

Answer 12

Rust, 206 bytes

I don't think rust is well suited for code golfing -_-

let b=|s:&str|{s.chars().enumerate().map(|(i,c)|match i{0=>match c{'X'=>'0','Y'=>'1','Z'=>'2',_=>c},_=>c}).collect::<String>().parse::<usize>().ok().and_then(|x|"TRWAGMYFPDXBNJZSQVHLCKE".chars().nth(x%23))};

Answer 13

05AB1E, 41 40 39 bytes

ć…xyz2ÝJ‡ìDd_i.ǝ}23%.•Xk¦fΣT(:ˆ.Îðv5•sè

Takes the input in lowercase (to save 1 byte yay)

Try it online!

Prints the input to STDERR if it is malformed

Explanation

ć…xyz2ÝJ‡ìDd_i.ǝ}23%.•Xk¦fΣT(:ˆ.Îðv5•sè
ć                                       # Get head of input and put the rest of the input under it on the stack
 …xyz                                   # Push xyz
     2ÝJ                                # Push 012
        ‡                               # Transliterate
         ì                              # Prepend to the rest of the input
          Dd_                           # Does the result contain something other than numbers?
             i.ǝ}                       # If so print input to STDERR
                 23%                    # Modulo 23
                    .•Xk¦fΣT(:ˆ.Îðv5•   # Pushes the character list
                                     sè # Get the char at the index of the modulo

Answer 14

Dyalog APL, 95 bytes

{'TRWAGMYFPDXBNJZSQVHLCKE'[1+23|(10⊥¯1+'0123456789'⍳{(⍕{('XYZ'⍳⍵)<4:('XYZ'⍳⍵)-1⋄⍵} ⊃⍵),1↓⍵}⍵)]}

This is a monadic operator that accepts a character string as its operand and returns its result.

FIXME it doesn't check its input. It's not properly golfed.

Usage:

    OP ← {'TRWAGMYFPDXBNJZSQVHLCKE'[1+23|(10⊥¯1+'0123456789'⍳{(⍕{('XYZ'⍳⍵)<4:('XYZ'⍳⍵)-1⋄⍵} ⊃⍵),1↓⍵}⍵)]}

      OP '01234567'
L

      OP '00000010'
X