Which Advent is it?

Question

^{I know, a bit late.}

Challenge

Your challenge is, given a date, output if it's not an Advent Sunday, or the Advent Sunday of the year.

Rules

The \$4^\text{th}\$ Advent is determined by the Sunday before Christmas day (the \$25^\text{th}\$ of December). The \$3^\text{rd}\$ Advent is the Sunday before the \$4^\text{th}\$ Advent, and so on until the \$1^\text{st}\$ Advent.
If Christmas Day falls on a Sunday, then the Sunday before is still the \$4^\text{th}\$ Advent.
Input will consist of day, month and year, as an iterable with any order, or a string, or 3 separate integers.
Input will be a valid date, numerically and also calendrically.
Output should be falsey (0 or Boolean False) if the date is not an Advent Sunday, or else an integer denoting which Advent Sunday it is.
This is code-golf, so the shortest answer wins.

Examples

[In]: [2, 12, 2021]
[Out]: 0

[In]: [6, 5, 2023]
[Out]: False

[In]: [12, 12, 2021]
[Out]: 3

[In]: [30, 11, 1997]
[Out]: 1

[In]: [25, 12, 2011]
[Out]: false

[In]: [21, 12, 3000]
[Out]: 4

Good Luck!

\$\begingroup\$ May we take input as 3 integers instead of a collection? \$\endgroup\$
– chunes
Dec 27, 2021 at 18:21 — chunes, Dec 27, 2021 at 18:21
\$\begingroup\$ @chunes yes, 'll add that in \$\endgroup\$
– math scat
Dec 27, 2021 at 18:32 — math scat, Dec 27, 2021 at 18:32

Engineer Toast · Accepted Answer · 2022-01-05 18:37:41Z

Excel, 71 66 Bytes

^{Saved 5 bytes thanks to Taylor Raine}

Aside: At the time of posting, this answer has the smallest byte count. That is baffling. Score one for Excel's date handling, I guess.

=LET(a,4-(INT(("12/23/"&YEAR(A1))/7)*7+1-A1)/7,(LEN(a)=1)*(a<5)*a)

Input is in cell A1. Output is wherever the formula is. Returns a number 1-4 for advent Sundays and a 0 for all other dates.

LET(a,4-(INT(("12/23/"&YEAR(A1))/7)*7+1-A1)/7

The first part calculates which advent Sunday an input is and does most of the heavy lifting.

("12/23/"&YEAR(A1)) finds Christmas Eve Eve (CEE) in the year of the input. We find Eve Eve hear because it saves us some extra bytes of adjustment later.
INT(~/7)*7 takes that date, divides by 7 (b/c 7 days per week), rounds it down, then multiplies it by 7. This finds the Saturday before CEE. If CEE is a Saturday, then it returns that date.
(INT(~/7)*7+1-A2)/7 adds one to get Sunday, subtracts the input, and divides by 7 to get the number of weeks between the input and the Sunday before Christmas. This may not be a whole number.
4-(~) subtracts that result from 4 to invert the result. I.E., the Sunday before Christmas needs to be the 4th Sunday, not the 1 week before.

LET(a,~,(LEN(a)=1)*(a<5)*a)

Now that the heavy lifting is out of the way, we can examine the results. Note that using TRUE and FALSE values in math will treat TRUE = 1 and FALSE = 0. By multiplying various statements that will evaluate to either TRUE or FALSE, we can filter out undesirable results.

(LEN(a)=1) filters out any non-integers as well as negative integers. This removes every non-Sunday in the advent calendar and all dates before the first Sunday. This does not filter out the 0th Sunday but the end result will still be 0 so it works.
(a<5) filters out Sundays after the 4th Sunday.
(~)*(~)*a returns the value if it has passed both those filters. Otherwise, it returns 0.

In this screenshot, I have added a column to return the value of a since that's so much of the calculation. I have also extended the data beyond the samples above to show some edge cases.

I'm just realising now that THIS IS SO UNDERRATED. Excel beat Mathematica!!! — math scat, Dec 30, 2021 at 7:55
You can get this down to 66 bytes if you use some implicit date conversion and Boolean multiplication as =LET(a,4-(INT(("12/23/"&YEAR(A1))/7)*7+1-A1)/7,(a<5)*(LEN(a)=1)*a) - but this will change the falsey output to be 0 rather than false — Taylor Alex Raine, Jan 5, 2022 at 17:54
@TaylorRaine This is great! I haven't golfed in a while so the refresher course on Excel tricks is fantastic. I welcome your comments. I don't think I even knew about implicit date conversion. — Engineer Toast, Jan 5, 2022 at 18:39

chunes · Accepted Answer · 2021-12-27 20:32:04Z

Factor + `calendar.holidays.us`, 85 bytes

[ <date> dup christmas-day 4 [ dup last-sunday ] times 0 5 narray reverse nip index ]

Takes input as three integers in <year> <month> <day> order. The last-sunday word postdates build 1525, the one TIO uses, so here's a screenshot of running the above code in build 2101's listener:

Explanation

                             ! 2021 12 12
<date>                       ! 2021-12-12
dup                          ! 2021-12-12 2021-12-12
christmas-day                ! 2021-12-12 2021-12-25
4 [ dup last-sunday ] times  ! 2021-12-12 2021-12-25 2021-12-19 2021-12-12 2021-12-5 2021-11-28
0 5 narray                   ! 2021-12-12 2021-12-25 { 2021-12-19 2021-12-12 2021-12-5 2021-11-28 0 }
reverse                      ! 2021-12-12 2021-12-25 { 0 2021-11-28 2021-12-5 2021-12-12 2021-12-19 }
nip                          ! 2021-12-12 { 0 2021-11-28 2021-12-5 2021-12-12 2021-12-19 }
index                        ! 3

tsh · Accepted Answer · 2021-12-28 08:37:29Z

4

JavaScript (Node.js), 78 bytes

(y,m,d)=>/Su/.test(c=new Date(y,m-1,d))*'01234'[(c-new Date(y,9,51))/6048e5|0]

Try it online!

Falsy may be 0 or NaN. Add 2 bytes (append |0) to always get 0. User must run this program with computer configured to UTC timezone or any other timezone that: 1. No DST shift during Nov 20 to Dec 31, 2. Name of timezone (in system language) does not contain substring "Su" (case sensitive).

new Date(y,9,51) is Nov 20 in JavaScript.

edited Dec 28, 2021 at 8:37

answered Dec 28, 2021 at 6:32

tsh

32.8k2 gold badges32 silver badges121 bronze badges

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Ajax1234 · Accepted Answer · 2021-12-27 16:23:12Z

3

Python3, 152 bytes:

from datetime import*
t=lambda x:0if x>=date(x.year,12,25)else 1+t(x+timedelta(days=7))
f=lambda x:(s:=date(*x)).weekday()==6and(0<(j:=(5-t(s)))<5)and j

Try it online!

edited Dec 27, 2021 at 16:23

answered Dec 27, 2021 at 16:15

Ajax1234

5,3931 gold badge13 silver badges26 bronze badges

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Neil · Accepted Answer · 2021-12-27 16:57:33Z

Retina 0.8.2, 116 bytes

3=`
;
\d+
$*_
((_*)\2\2\2_*);((_*)\4\4\4_*)
$4$3$2$1$1$1$1$1
\G_{7}

-_{12}-
____
-_{11}-_{26}

^(_{7}){1,4}$|.+
$#1

Try it online! Link includes test cases. Takes input in ISO date format. Explanation:

3=`
;

Separate the year from the century.

\d+
$*_

Convert to unary.

((_*)\2\2\2_*);((_*)\4\4\4_*)
$4$3$2$1$1$1$1$1
\G_{7}

Perform Zeller's congruence on the 25th of December of the input year.

-_{12}-
____
-_{11}-_{26}

Add this to the day of the month, adding an extra 4 for December or subtracting 26 for November 26 or later. Other months get ignored and fail to match the next stage at all.

^(_{7}){1,4}$|.+
$#1

Count the number of whole weeks, but the result must be an integer between 1 and 4 inclusive.

Noodle9 · Accepted Answer · 2021-12-28 11:03:35Z

3

C (gcc), ^{215 202} 192 bytes

#import<time.h>
i;*t;*l;f(d,m,y){5[t=calloc(9,8)]=y-1900;t[4]=m-1;t[3]=d;l=mktime(t);i=6[t=gmtime(&l)];y=t[7];t[4]=11;t[3]=25;l=mktime(t);d=7[t=gmtime(&l)]-y;m=4-d/7-!t[6];i=i|!d|m<1|m>4?0:m;}

Try it online!

Saved 13 bytes thanks to ceilingcat!!!

Inputs the day, month and year as separate parameters.
Returns which Advent Sunday it is or \$0\$ otherwise.

edited Dec 28, 2021 at 11:03

answered Dec 28, 2021 at 1:44

Noodle9

19.4k3 gold badges18 silver badges45 bronze badges

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Ben Izd · Accepted Answer · 2021-12-28 11:19:09Z

3

Wolfram Language (Mathematica), 88 79 bytes

Input format should be {year, month, day}.

Ramp[4-DayCount[#,{#&@@#,12,24},s=Sunday]]Boole[DayName@#==s&&Rest@#!={12,25}]&

Try it online!

Output:

79 bytes - Thanks to @att

edited Dec 28, 2021 at 11:19

answered Dec 27, 2021 at 18:19

Ben Izd

2491 silver badge6 bronze badges

1

\$\begingroup\$ 79 bytes \$\endgroup\$
– att
Dec 28, 2021 at 2:56

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ing · Accepted Answer · 2021-12-28 11:40:03Z

3

Python 3.8 (pre-release), 114 bytes

lambda a,b,c:(y:=int((c+10.5)/7))%5+y//5 if date(a,b,c).weekday()==6 and[b,c]!=[12,25]else 0
from datetime import*

Try it online!

answered Dec 28, 2021 at 11:40

ing

312 bronze badges

\$\begingroup\$ Welcome to Code Golf, and nice first answer! \$\endgroup\$
– caird coinheringaahin g ♦
Dec 28, 2021 at 13:23
1

\$\begingroup\$ You can make it 112 bytes by removing the space in 5 if and 6 and[b,c] \$\endgroup\$
– Wondercricket
Dec 28, 2021 at 21:56

Add a comment |

Jonathan Allan · Accepted Answer · 2021-12-27 17:32:22Z

2

Python 3.8, 128 116 bytes

lambda q:max(x*(date(*q)==(M:=date(q[0],11,27))+timedelta(7*x+~M.weekday()))for x in(1,2,3,4))
from datetime import*

Try it online!

edited Dec 27, 2021 at 17:32

answered Dec 27, 2021 at 17:00

Jonathan Allan

97.9k7 gold badges61 silver badges259 bronze badges

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Kjetil S · Accepted Answer · 2021-12-31 13:25:16Z

2

Perl 5, 144 bytes

sub{$_=join$",map{@t=gmtime$_*604800+3e5;19000100+$t[5]*1e4+$t[4]*100+$t[3]}1..6e4;pop=~/..../;/ $&$'( $&\d+)+/;$_=6-$&=~y///c/9;/^[1-4]$/?$_:0}

Try it online!

answered Dec 31, 2021 at 13:25

Kjetil S

5,1079 silver badges17 bronze badges

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Stack Exchange Network

Which Advent is it?

Challenge

Rules

Examples

10 Answers 10

Excel, 71 66 Bytes

Factor + `calendar.holidays.us`, 85 bytes

Explanation

JavaScript (Node.js), 78 bytes

Python3, 152 bytes:

Retina 0.8.2, 116 bytes

C (gcc), ^{215 202} 192 bytes

Wolfram Language (Mathematica), 88 79 bytes

Python 3.8 (pre-release), 114 bytes

Python 3.8, 128 116 bytes

Perl 5, 144 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
date
or ask your own question.

Hot Network Questions

Which Advent is it?

Challenge

Rules

Examples

10 Answers 10

Excel, 71 66 Bytes

Factor + calendar.holidays.us, 85 bytes

Explanation

JavaScript (Node.js), 78 bytes

Python3, 152 bytes:

Retina 0.8.2, 116 bytes

C (gcc), 215 202 192 bytes

Wolfram Language (Mathematica), 88 79 bytes

Python 3.8 (pre-release), 114 bytes

Python 3.8, 128 116 bytes

Perl 5, 144 bytes

Your Answer

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Factor + `calendar.holidays.us`, 85 bytes

C (gcc), ^{215 202} 192 bytes

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