Doomsday rule: Difference between revisions

{{trans|Python}}

 

R year % 4 == 0 & (year % 100 != 0 | year % 400 == 0)

 

L(year, month, day) [(1800, 1, 6), (1875, 3, 29), (1915, 12, 7),

(1970, 12, 23), (2043, 5, 14), (2077, 2, 12), (2101, 4, 2)]

 

{{out}}

=={{header|ALGOL 68}}==

{{Trans|ALGOL W}}

# returns the day of the week (Sunday = 0, Monday = 1,...) for the date #

# specified by ccyy, mm and dd #

test dow( 2022, 6, 19 )

END

{{out}}

<pre>

 

=={{header|ALGOL W}}==

% returns the day of the week (Sunday = 0, Monday = 1,...) for the date %

% specified by ccyy, mm and dd %

testDow( 2022, 6, 19 )

end

{{out}}

<pre>

=={{header|APL}}==

{{works with|Dyalog APL}}

days←'Sunday' 'Monday' 'Tuesday' 'Wednesday' 'Thursday' 'Friday' 'Saturday'

leap←4 7∊⍨2⊥0=4 100 400∘|

anchor←m⊃(1+leap y)⊃nd ld

(1+7|7+doom+d-anchor)⊃days

{{out}}

<pre> weekday 1800 1 6

 

=={{header|AppleScript}}==

tell yyyymmdd to set {y, m, d} to {word 1 as integer, word 2 as integer, word 3 as integer}

set doomsdayForYear to (y + y div 4 - y div 100 + y div 400 + 2) -- (mod 7 further down anyway)

end task

 

 

{{output}}

1875-03-29 --> Monday

1915-12-07 --> Tuesday

2043-05-14 --> Thursday

2077-02-12 --> Friday

 

=={{header|BASIC}}==

Note that 1/6/1800 is actually a Monday, not a Sunday. As far as I can tell this is actually the correct day.

 

20 DIM D(12,1): FOR I=0 TO 1: FOR J=1 TO 12: READ D(J,I): NEXT J,I

30 READ Y: IF Y=0 THEN END ELSE READ M,D

130 DATA 4,1,7,4,2,6,4,1,5,3,7,5

140 DATA 1800,1,6, 1875,3,29, 1915,12,7, 1970,12,23

{{out}}

<pre> 1/ 6/1800: Monday

Note that 1/6/1800 is actually a Monday, not a Sunday. As far as I can tell this is actually the correct day.

 

 

let dayname(n) =

writef("February 12, 2077 will be on a %S.*N", dayname(weekday(2077, 2, 12)))

writef("April 2, 2101 will be on a %S.*N", dayname(weekday(2101, 4, 2)))

{{out}}

<pre>January 6, 1800 was on a Monday.

 

=={{header|C}}==

#include <stdint.h>

#include <stdbool.h>

return 0;

{{out}}

<pre>January 6, 1800 was on a Monday.

 

=={{header|C++}}==

#include <cstdint>

 

return 0;

{{out}}

<pre>January 6, 1800 was on a Monday

 

=={{header|CLU}}==

return(year//4=0 & (year//100=0 | year//400=0))

end leap_year

stream$putl(po, weekday(d))

end

{{out}}

<pre>1 June 1800 was on a Sunday

 

=={{header|Cowgol}}==

 

record Date is

print_date(&dates[i]);

i := i + 1;

{{out}}

<pre>1/6/1800: Monday

=={{header|Factor}}==

{{works with|Factor|0.99 2021-02-05}}

generalizations kernel math math.order math.vectors sequences ;

 

2077 2 12

2101 4 2

{{out}}

<pre>

 

=={{header|FOCAL}}==

01.15 S X(7)=4;S X(8)=1;S X(9)=5;S X(10)=3;S X(11)=7;S X(12)=5

01.20 S Y=1800;S M= 1;S D= 6;D 2;D 3

04.50 I (E-2)4.6;T "TUESDAY";R

04.60 I (E-1)4.7;T "MONDAY";R

{{out}}

<pre>M= 1 D= 6 Y= 1800 DAY=MONDAY

 

=={{header|FreeBASIC}}==

{ 3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5 }, _

{ 4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5 } } 'the first doomsday in

print get_day( 2043, 05, 14 )

print get_day( 2077, 02, 12 )

{{out}}

<pre>Monday

=={{header|Go}}==

{{trans|Wren}}

 

import (

fmt.Printf("%s -> %s\n", date, days[dow])

}

 

{{out}}

 

=={{header|Haskell}}==

 

data Date = Date {year :: Int, month :: Int, day :: Int}

 

main :: IO ()

{{out}}

<pre>1800-01-06: Monday

 

To find the doomsday for a month, find the number of days for each month in that year, and compute the running sum from right to left. Then add 1 to each of those numbers, find the mod 7 remainder and add 1 again

3 7 7 4 2 6 4 1 5 3 7 5

1+7|1++/\.31 29 31 30 31 30 31 31 30 31 30 31

 

Also note that adding 7 is like adding 0 in modulo 7 arithmetic.

Thus:

 

Y0=. todayno Y,1 1

Y1=. todayno 1+Y,0 0

dday=. 7|2 5 4 6+/ .*1,4 100 400|/Y

'day',~;(7|D+dday-aday){;:'Sun Mon Tues Wednes Thurs Fri Satur'

 

That said, it's more concise to look up these anchor days. We can use the nonleap year sequence and adjust january and february's values for leap years by adding 1 or 2 (under 1 offset 7 modulo arithmetic).

 

leap=. 0~:/ .=4 100 400|/Y

aday=. 1+7|(M*leap*3>M)+M{_ 2 6 6 3 1 5 3 0 4 2 6 4

dday=. 7|2 5 4 6+/ .*1,4 100 400|/Y

'day',~;(7|D+dday-aday){;:'Sun Mon Tues Wednes Thurs Fri Satur'

 

These two implementations produce equivalent results.

Task examples:

 

Monday

get_weekday 1875 3 29

Friday

get_weekday 2101 4 2

 

=={{header|Java}}==

public static void main(String[] args) {

final Date[] dates = {

return weekdays[(doom + day - anchor + 7) % 7];

}

{{out}}

<pre>01/06/1800: Monday

 

Note that the assertions as defined here are only checked if the environment variable JQ_ASSERT is set.

def weekdaynames: ["Sunday", "Monday","Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"];

 

("February 12, 2077 will be on a "+ get_weekday(2077; 2; 12)),

("April 2, 2101 will be on a " + get_weekday(2101; 4; 2)),

{{out}}

<pre>

 

=={{header|Julia}}==

export get_weekday

 

println("February 12, 2077 will be on a ", get_weekday(2077, 2, 12))

println("April 2, 2101 will be on a ", get_weekday(2101, 4, 2))

<pre>

January 6, 1800 was on a Monday

We use also the “WeekDay” type of this module which is an enumeration starting with “dMon” for Monday and ending with “dSun” for Sunday.

 

 

const

echo &"For {date}, expected {dt.weekday}, found {wday}."

else:

 

{{out}}

=={{header|Perl}}==

{{trans|Raku}}

 

use strict;

for (qw( 1800-01-06 1875-03-29 1915-12-07 1970-12-23 2043-05-14 2077-02-12 2101-04-02 )) {

print $_, " is a : ", dow $_, "\n";

{{out}}

<pre>

 

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n%-30s (%d = %d ? %s)\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">ds</span><span style="color: #0000FF;">,</span><span style="color: #000000;">bis</span><span style="color: #0000FF;">,</span><span style="color: #000000;">bid</span><span style="color: #0000FF;">,</span><span style="color: #000000;">drd</span><span style="color: #0000FF;">,</span><span style="color: #000000;">ok</span><span style="color: #0000FF;">})</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

{{out}}

<pre>

 

=={{header|PL/M}}==

BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;

EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;

END;

CALL EXIT;

{{out}}

<pre>1/6/1800: MONDAY

 

=={{header|Python}}==

from calendar import isleap

 

for d in dates:

tense = "was" if d < date.today() else "is" if d == date.today() else "will be"

{{out}}

<pre>January 06, 1800 was a Monday

=={{header|Raku}}==

 

 

my %doomsday = False => [3,7,7,4,2,6,4,1,5,3,7,5], True => [4,1,7,4,2,6,4,1,5,3,7,5];

say "Builtin - $_ is a: ", @dow[Date.new($_).day-of-week];

say '';

{{out}}

<pre>Conway - 1800-01-06 is a: Monday

 

=={{header|REXX}}==

parse arg $ /*obtain optional arguments from the CL*/

if $='' | $="," then $= , /*Not specified? Then use the default.*/

/*──────────────────────────────────────────────────────────────────────────────────────*/

doomsday: parse arg ?; return (2 + 5 * (?//4) + 4 * (?//100) + 6 * (?//400) ) // 7

{{out|output|text=&nbsp; when using the default inputs:}}

<pre>

 

=={{header|Rust}}==

const LEAPYEAR_FIRSTDOOMSDAYS: [u32; 12] = [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5];

const NONLEAPYEAR_FIRSTDOOMSDAYS: [u32; 12] = [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5];

print_day_of_week(2077, 2, 12);

print_day_of_week(2101, 4, 2);

 

{{out}}

{{libheader|Wren-date}}

We only use the above module to check the dates of the week given by Conway's method. The latter are worked out from scratch.

 

var days = ["Sunday", "Monday", "Tuesday", "Wednesday","Thursday", "Friday", "Saturday"]

var d = Date.parse(date, Date.isoDate)

System.print("%(date) -> %(d.weekDay)")

 

{{out}}

=={{header|Yabasic}}==

{{trans|FreeBASIC}}

for x = 0 to arraysize(fdoom(),1)

for y = 1 to arraysize(fdoom(),2)

data 3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5

data 4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5

{{out}}

<pre>