Generate a British numberplate

Question

Modern British number plates (a.k.a. registration marks or license plates) use the format AANNXXX where AA is a "DVLA memory tag" (basically a location code), NN is the age identifier, and XXX is three random letters.

Your task is to, given a date, generate a random valid number plate with a date code matching the supplied date. Simplified from the actual implementation, this is how to generate them:

• The memory tag will just be generated randomly:

  • First letter: random from ABCDEFGHKLMNOPRSVWY (the alphabet, minus IJQTUXZ)
  • Second letter: random from ABCDEFGHJKLMNOPRSTUVWXY (the alphabet, minus IQZ)

• The age identifier follows this pattern:

  • 1st September 2001 - 28th February 2002: 51
  • 1st March 2002 - 31st August 2002: 02
  • September 2002 - February 2003: 52
  • March 2003 - August 2003: 03
  • ...
  • March 2010 - August 2010: 10
  • September 2010 - February 2011: 60
  • March 2011 - August 2011: 11
  • September 2011 - 29th February 2012: 61
  • ...
  • September 2020 - February 2021: 70
  • ...
  • September 2049 - February 2050: 99
  • March 2050 - August 2050: 00
  • September 2050 - February 2051: 50
  • March 2051 - August 2051: 01

• The 3 random letters are picked from ABCDEFGHJKLMNOPRSTUVWXYZ (the alphabet, minus IQ)

Rules

• You must output in uppercase with no spaces or dashes between letters. However, trailing newlines, or any other output from your language that is not easily suppressible, are allowed
• For the purposes of this challenge, "random" means picked pseudo-randomly with any distribution as long as all possibilities have a non-zero chance of occurring
• You may assume the date will always be in the range 1st September 2001 to 31st August 2051
• The date can be in any reasonable input format (for example, your language's native date object, april 28 2021, (28,4,21), 1619615864 (a Unix epoch timestamp))
• You may ignore leap seconds, daylight savings time, and timezone issues (just assume the date is always in UTC)
• You may use any reasonable I/O method
• Standard loopholes are forbidden
• This is code-golf, so the shortest code in bytes wins

Test cases

These are to test the date conversion alone.

23rd September 2001     51
1st March 2006          06
29th February 2012      61
1st December 2021       71
11th October 2049       99
19th January 2051       50

Answer 1

Jelly, 48 bytes

Ṗḅ12_3:6d2Ṛ%50ḅ50ṾŻṫ-“IQ“““Z“JTUX”;\ØAḟⱮ$X€ŒH¤jṚ

Try it online!

A full program taking the date as a list of integers [y,m,d] and printing the numberplate.

Thanks to @JonathanAllan for pointing out a couple of errors!

Answer 2

JavaScript (ES6),  121  116 bytes

Expects (month)(year). The day is ignored. Returns an array of characters.

m=>y=>[...'9512444'].map(n=>((y-(m<3))%50-50*~-~(m<3|m>8)+n)[n]||'ABCDEFGHKLMNOPRSVWYJTUXZ'[Math.random()*(28-n)|0])

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Commented

m =>                         // m = 1-indexed month
y =>                         // y = full year
[...'9512444']               // split 9,5,1,2,4,4,4 as characters
.map(n =>                    // for each value n above:
  (                          //   build a string:
    ( y -                    //     subtract 1 from y if ...
      (m < 3)                //     ... the month is January of February
    ) % 50                   //     reduce modulo 50
    - 50 *                   //     add 100 to the result for March to August
    ~-~(m < 3 | m > 8)       //     or 150 otherwise
    + n                      //     coerce to a string
  )[n]                       //   attempt to extract the n-th character
  ||                         //   if n is greater than or equal to 4:
  'ABCDEFGHKLMNOPRSVWYJTUXZ' //     use this string instead
  [ Math.random()            //     and extract a random character
    * (28 - n)               //     in [0 ... 18] (if n = '9')
    | 0                      //     or [0 ... 22] (if n = '5')
  ]                          //     or [0 ... 23] (if n = '4')
)                            // end of map()

Answer 3

Charcoal, 51 bytes

≔⁻⪪α¹⪪IQ¹ζ‽⁻ζ⪪JTUXZ¹‽⁻ζZ﹪%02d⁺﹪⁻θ›³η⁵⁰∧÷⁻⁸η⁶¦⁵⁰Ｆ³‽ζ

Try it online! Link is to verbose version of code. Takes input as [Y, M, D] (well the D is optional and unused). Explanation:

≔⁻⪪α¹⪪IQ¹θ

Make a copy of the uppercase alphabet with the letters I and Q removed.

‽⁻θ⪪JTUXZ¹

Print a random letter excluding the letters J, T, U, X and Z.

‽⁻θZ

Print a random letter excluding Z.

﹪%02d⁺﹪⁻θ›³η⁵⁰∧÷⁻⁸η⁶¦⁵⁰

Decrement the year if the month is January or February, reduce modulo 50, then add 50 if appropriate, and format the result as two digits.

Ｆ³‽θ

Print three more random letters.

Answer 4

Excel, 142 bytes

=REPLACE(CONCAT(MID("ABCDEFGHKLMNOPRSVWYJTUXZ",INT(RANDARRAY(7)*{19;23;1;1;24;24;24}+1),1)),3,2,TEXT(MOD(B2,50)+IFS(A2<3,49,A2>8,50,1,),"00"))

Link to Spreadsheet

Assumes Month in A2 and Year is in B2

Excel date as input, 161 bytes

=LET(x,EOMONTH(A1,-2),REPLACE(CONCAT(MID("ABCDEFGHKLMNOPRSVWYJTUXZ",INT(RANDARRAY(7)*{19;23;1;1;24;24;24}+1),1)),3,2,TEXT(MOD(YEAR(x),50)+(MONTH(x)>6)*50,"00")))

Answer 5

Retina, 118 bytes

T`d`9d`.0?(?=/0[12])
T`5-9`d`^.
T`d`5-9`^.(?!./0[3-8])
L$`^..
==$&===
=
<26*=>
Y`=`L
I|Q

,4`[JTUXZ]

0`Z

<(\w)+>
$?1

Try it online! Takes input in the format YY/MM but link includes test suite that trims off any trailing /DD. Explanation:

T`d`9d`.0?(?=/0[12])

Decrement the year if the month is January or February.

T`5-9`d`^.

Reduce the year modulo 50.

T`d`5-9`^.(?!./0[3-8])

Add 50 to the year unless the month is in the range March to August.

L$`^..
==$&===

Remove the month and add placeholders for the letters.

=
<26*=>
Y`=`L

Expand each placeholder to a run of letters <A..Z>.

I|Q

,4`[JTUXZ]

0`Z

Delete the Is, Qs, the first J, T, U, X and Z, and the (originally second) Z.

<(\w)+>
$?1

Pick a random remaining letter from each run.

Answer 6

Vyxal, 60 bytes

«ÞTȮa⟩Ȯ2¬~Zyė∆₂ċ«⇧D3ȯ℅₴7ȯ℅₴6ɾ2+?ḟu=50*?50%+:₀<[\0p]₴2ȯDẊẊ℅ṅ₴

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It's... not terrible, but still bad.

Answer 7

Python 3.8 (pre-release), 167... 164 bytes

lambda m,y,p=lambda e=0:choice('ZJTUXABCDEFGHKLMNOPRSVWY'[e:]):p(5)+p(1)+[str((z:=y%50)+49+(m>2)+50*(z<1)),'0'*(z<10)+str(z)][2<m<9]+p()+p()+p()
from random import*

Try it online!

Similar technique to Arnauld: this was written by analyzing the age identifiers. If the month is between 3 and 8 (bounds included) add the y%50 string with a leading zero if y%50<10. Otherwise, add either 50 or 49 to y%50 and return that.

Takes two inputs: month and year (in that order).

I know that from random import* definitely saves bytes, but I'm yet to find a way to integrate that without breaking TIO and adding f= (which actually inflates the bytecount.)

- 1 3 bytes thanks to ovs.

+9 bytes to fix a bug.

Answer 8

Julia, 100 bytes

!n=rand(setdiff('A':'Z',"IQZJTUX"[1:n]))
m$y=!7*!3lpad((a=(m+12y-3)÷6)÷2%50+a%2*50,2,'0')*!2*!2*!2

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