How many countries are in the European Union?

Question

Tonight, 31 January 2020, Brexit will happen and the United Kingdom will leave the European Union, the first time that the number of EU member states will decrease.

Your job is to take a date and output the number of EU¹ members on that date, according to the following table:

| Start      | End        | Number of members 
| 0          | 1957-12-31 | 0
| 1958-01-01 | 1972-12-31 | 6
| 1973-01-01 | 1980-12-31 | 9
| 1981-01-01 | 1985-12-31 | 10
| 1986-01-01 | 1994-12-31 | 12
| 1995-01-01 | 2004-04-30 | 15
| 2004-05-01 | 2006-12-31 | 25
| 2007-01-01 | 2013-06-30 | 27
| 2013-07-01 | 2020-01-31 | 28
| 2020-02-01 | Inf        | 27

Input may take one of the following forms:

• your language's date format
• any ordering of integers for year, month and day with any separator (e.g. YYYYMMDD or DD/MM/YYYY)
• number of days lapsed since some epoch.

This is code-golf, so shortest code wins!

_{1. or EEC before 1993}

Answer 1

Javascript, 85 83 Bytes

a=d=>d<0?0:d<180?6:d<276?9:d<336?10:d<444?12:d<556?15:d<588?25:d<666?27:d<745?28:27

It takes an input as a float as the months / fractions of months passed since 1958-01-01
(0 for 1958-01-01T00:00:00 and a negative number for any previous date)

(Since number of days from some epoch is allowed, I assume that also a number of months is valid, as well)

Try it Online

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80 Bytes

(credits: Arnauld)

a=d=>d<0?0:d<180?6:d<276?9:d<336?10:d<444?12:d<556?15:d<588?25:27+(d>=666&d<745)

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_{Sad to see you go, UK}

Answer 2

05AB1E, 29 27 26 25 33 bytes

¨12βŽ₅b-‘´`<lp€¸‘Ç.¥@•¿“0p•12в<*O

Try it online!

Input is in the form [yyyy, mm, dd].

Answer 3

Python 2, 110 97 bytes

lambda d:ord('069:<?IKLK'[sum(d>i for i in(0,5479,8401,10227,13514,16922,17897,20270,22676))])-48

Try it online!

_{-13 bytes, thanks to Jonathan Allan}

Takes input as number of days since 1957-12-31 (so 1958-01-01 is day 1)

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Python 3, 93 bytes

lambda d:b'069:<?IKLK'[sum(d>i for i in(0,5479,8401,10227,13514,16922,17897,20270,22676))]-48

Try it online!

_{-4 bytes, thanks to mypetlion}

Answer 4

05AB1E, 42 36 bytes

•AÂʒë.š¡ε%ž·7í•ŽL}в.¥›O•=γ1sæΔ•₆вsè<

Input as number of days since 1957-12-31 (so 1957-12-30 is day -1; 1957-12-31 is day 0; 1958-01-01 is day 1; etc.)

Try it online or verify some more test cases.

Old 42 bytes answer taking input in the format yyyyMMdd:

¨¨•a3|}\§λ’Iœg½þ•ŽOΩв•32Ø•+@O•=γ1sæΔ•₆вsè<

Try it online or verify some more test cases.

•a3|}\§λ’Iœg½þ•ŽOΩв•32Ø•+ can alternatively be •Me1εä~.=ΔΩ»•Ž5ãв.¥•32Ù•+ for the same byte-count: Try it online or verify some more test cases.

Explanation:

•AÂʒë.š¡ε%ž·7í•    # Push compressed integer 813218926689775697373196902446
  ŽL}              # Push compressed integer 5480
     в             # Convert the larger integer to base-5480 as list:
                   #  [5479,2922,1826,3287,3408,975,2373,2406]
      .¥           # Undelta it with leading 0:
                   #  [0,5479,8401,10227,13514,16922,17897,20270,22676]
        ›          # Check for each if it's larger than the (implicit) input-integer
         O         # Take the sum to get the amount of truthy values
•=γ1sæΔ•           # Push compressed integer 122116126451824
 ₆в                # Convert it to base-36 as list:
                   #  [1,7,10,11,13,16,26,28,29,28]
   sè              # Swap to get the sum, and use it to index into this list
     <             # And decrease it by 1
                   # (since a compressed integer/list cannot contain a leading 0)
                   # (after which the result is output implicitly)

¨¨                 # Remove the last two digits from the (implicit) input (the "dd")
  •a3|}\§λ’Iœg½þ•  # Push compressed integer 2722385715080006519908031109868
   ŽOΩ             # Push compressed integer 6203
      в            # Convert the larger integer to base-6203 as list:
                   #  [1,1501,2301,2801,3701,4605,4901,5507,6202]
       •32Ø•       # Push compressed integer 195800
            +      # Add it to each value in the list:
                   #  [195801,197301,198101,198601,199501,200405,200701,201307,202002]
             @     # Check for each if it's larger than or equal to the input minus "dd"
O•=γ1sæΔ•₆вsè<     # Same as above

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer lists?) to understand how the compression works.

Answer 5

Turing Machine Code, 972 1009 bytes

My input is the date in the format specified in the test cases YYYY-MM-DD.

0 0 0 * halt
* 1 _ r €
€ 9 _ r %
* 2 _ r $
$ 0 _ r £
* I 2 r ¢
% 5 _ r ^
% 6 6 r c
% 7 _ r x
% 8 _ r B
% 9 _ r i
% * _ r c
^ * 0 r c
^ 8 6 r c
^ 9 6 r c
B 0 9 r c
B 1 1 r r
B 2 1 r r
B 3 1 r r
B 4 1 r r
B 5 1 r r
B * 1 r e
r * 0 r c
e * 2 r c
x 0 6 r c
x 1 6 r c
x 2 6 r c
x * 9 r c
i 0 1 r 🇪🇺
i 1 1 r 🇪🇺
i 2 1 r 🇪🇺
i 3 1 r 🇪🇺
i 4 1 r 🇪🇺
i * 1 r t
t * 5 r c
🇪🇺 * 2 r c
£ 0 _ r 🇬🇧
£ 1 _ r ★
£ 2 _ r 👑
👑 0 _ r !
👑 * 2 r ¢
! - _ r ~
~ 0 _ r 1
1 1 2 r ☆
1 * 2 r ¢
★ 0 2 r ¢ 
★ 1 2 r ¢
★ 2 2 r ¢
★ 3 _ r &
★ * 2 r ☆
& - _ r @
@ 0 _ r 6
6 0 2 r ¢
6 1 2 r ¢
6 2 2 r ¢
6 3 2 r ¢
6 4 2 r ¢
6 5 2 r ¢
6 6 2 r ¢
6 * 2 r ☆
☆ * 8 r c
🇺🇰 * 5 r c
🇬🇧 1 1 r 🇺🇰
🇬🇧 2 1 r 🇺🇰
🇬🇧 3 1 r 🇺🇰
🇬🇧 4 _ r +
🇬🇧 5 2 r t
🇬🇧 6 2 r t
🇬🇧 * 2 r ¢
¢ * 7 r c
+ - _ r ¬
¬ 0 _ r 4
¬ * 2 r t
4 0 1 r t
4 1 1 r t
4 2 1 r t
4 3 1 r t
4 4 1 r t
4 * 2 r t
6 _ 6 l c
c * _ r c
c _ _ * halt


 

Try it online!

Added a few bytes thanks to Grimmy being a bit more thorough in testing my code than I was.

Answer 6

Charcoal, 48 bytes

Ｉ⌕γ§ &)*,/9;<;ＬΦ⪪”)¶↶⌕βγ⦄J≦σν{:Xδp⁴E⊙≕⍘Ｈ⊙βg”⁶‹ιθ

Try it online! Link is to verbose version of code. Takes input as YYYYMMDD. Explanation:

                 ”...”      Compressed string of YYYYMM values
                ⪪     ⁶     Split into substrings of length 6
              ＬΦ       ‹ιθ  Count those that appear before the input
   § &)*,/9;<;              Look the count up in a translation table
Ｉ⌕γ                         Subtract 32 from the ASCII code

Answer 7

Python 2, 158 150 bytes

i=int(input()[:6]);print((((((((27,28)[202002>i>201306],25)[i<200701],15)[i<200405],12)[i<199501],10)[i<198601],9)[i<198101],6)[i<197301],0)[i<195801]

Try it online!

Takes input as a string in YYYYMMDD format

Alternative Python 2 approach without imports. Simply uses nested list indexing to create the equivalent of a big if/elif structure.

NB: Posted by a British guy who is sorry to see us leave. I'm still all for the Community and working together (whether EU or here).

Answer 8

Jelly, 38 37 bytes

“€ɓ⁴5O/ṖṪOṁṪḋg’ḃ⁽×ỵÄŻ>⁸Sị“÷ñıЀ½µ©¡ñ‘

Try it online!

A monadic link taking the zero-indexed count of days since 1958-01-01 as a integer argument and returning an integer.

Thanks to @JonathanAllan for saving a byte!

Answer 9

AWK, 98 bytes

{m=$1*12+$2-23496}1,$0=m<1?0:m<181?6:m<277?9:m<337?10:m<445?12:m<557?15:m<589?25:27+(m>666&&m<746)

Try it online!

Input format: YYYY MM DD

Answer 10

JavaScript (ES6), 95 bytes

Expects a number of days since 1957-12-31.

n=>[x=0,5479,2922,1826,3287,3408,975,2373,2406].map((d,i)=>x-=(n-=d)>0&&~('52012910'[i]||~1))|x

Try it online!

This is however a bit longer than @FabrizioCalderan's answer, even if it is fixed to use days instead of months.

Answer 11

x86 (32-bit) assembly, 43 bytes (13 bytes code + 30 bytes data)

This routine expects input in ax as a signed short giving number of days since 1950-01-01, and returns output in al (clobbering edi along the way).

    .text
    .globl eu_members
eu_members: 
    mov $.Ltbl-1, %edi
.Lloop:
    inc %edi
    scas (%edi), %ax
    jg .Lloop
    mov (%edi), %al
    ret
    .section .rodata
.Ltbl:
    .short 2921
    .byte 0
    .short 8400
    .byte 6
    .short 11322
    .byte 9
    .short 13148
    .byte 10
    .short 16435
    .byte 12
    .short 19843
    .byte 15
    .short 20818
    .byte 25
    .short 23191
    .byte 27
    .short 25597
    .byte 28
    .short 32767
    .byte 27

Exact opcodes of the code part (in a linked context so that the relocation of .Ltbl doesn't confuse things):

0804930b <eu_members>:
 804930b:       bf 49 a0 04 08          mov    $0x804a049,%edi
 8049310:       47                      inc    %edi
 8049311:       66 af                   scas   %es:(%edi),%ax
 8049313:       7f fb                   jg     8049310 <eu_members+0x5>
 8049315:       8a 07                   mov    (%edi),%al
 8049317:       c3                      ret    

And here's the test harness I used (note that the assembly code is not PIC safe, so on Debian or Ubuntu systems you will need to pass -no-pie to gcc):

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <string.h>

unsigned char eu_members_wrap(short days_since_1950_01_01) {
    unsigned char res;
    __asm__("call eu_members" : "=a"(res) : "0"(days_since_1950_01_01) : "edi", "cc");
    return res;
}

int main(int argc, char* argv[]) {
  struct tm tm;
  time_t epoch, req;
  short days_since_epoch;
  memset(&tm, 0, sizeof(tm));
  tm.tm_year = 50;
  tm.tm_mon = 0;
  tm.tm_mday = 1;
  epoch = mktime(&tm);
  memset(&tm, 0, sizeof(tm));
  tm.tm_year = atoi(argv[1]) - 1900;
  tm.tm_mon = atoi(argv[2]) - 1;
  tm.tm_mday = atoi(argv[3]);
  req = mktime(&tm);
  days_since_epoch = (((unsigned int) (req - epoch)) / (24 * 60 * 60));
  /* printf("epoch = %ld, req = %ld\n", epoch, req);
  printf("Raw number of days: %hd\n", days_since_epoch); */
  printf("EU members: %hhu\n", eu_members_wrap(days_since_epoch));
  return 0;
}

Answer 12

Perl 5, 46 bytes

($q,%d)=split;$q--while!exists$d{$q};say$d{$q}

Try it online!

...or try like this on some decent command line:

query=19730101
data="0 0 19580101 6 19730101 9 19810101 10 19860101 12 19950101 15 20040501 25 20070101 27 20130701 28 20200201 27"
echo "$query $data" | perl -nE '($q,%d)=split;$q--while!exists$d{$q};say$d{$q}'