Doomsday rule

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Revision as of 07:04, 3 June 2021 by Wherrera (talk | contribs) (sp)
Doomsday rule is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
About the task

John Conway (1937-2020), was a mathematician who also invented several mathematically oriented computer pastimes, such as the famous Game of Life cellular automaton program. Dr. Conway invented a simple algorithm for finding the day of the week, given any date. The algorithm was based opon calculating the distance of a given date from certain "anchor days" which follow a pattern for the day of the week upon which they fall.

Algorithm

The formula is calculated assuming that Sunday is 0, Monday 1, and so forth with Saturday 7, and 

   doomsday = (Tuesday(or 2) + 5(y mod 4) + 4(y mod 100) + 6(y mod 400)) % 7

which, for 2021, is 0 (Sunday).

To calculate the day of the week, we then count days from a close doomsday, with these as charted here by month, then add the doomsday for the year, then get the remainder after dividing by 7. This should give us the number corresponding to the day of the week for that date. 

   Month	        Doomsday Dates for Month
   --------------------------------------------
   January (common years)   3, 10, 17, 24, 31
   January (leap years)     4, 11, 18, 25
   February (common years)  7, 14, 21, 28
   February (leap years)    1, 8, 15, 22, 29
   March                    7, 14, 21, 28
   April                    4, 11, 18, 25
   May                      2, 9, 16, 23, 30
   June                     6, 13, 20, 27
   July                     4, 11, 18, 25
   August                   1, 8, 15, 22, 29
   September                5, 12, 19, 26
   October                  3, 10, 17, 24, 31
   November                 7, 14, 21, 28
   December                 5, 12, 19, 26
Task

Given the following dates:

  •   1800-01-06 (January 6, 1800)
  •   1875-03-29 (March 29, 1875)
  •   1915-12-07 (December 7, 1915)
  •   1970-12-23 (December 23, 1970)
  •   2043-05-14 (May 14, 2043)
  •   2077-02-12 (February 12, 2077)
  •   2101-04-02 (April 2, 2101)


Use Conway's Doomsday rule to calculate the day of the week for each date.

see also



APL

Works with: Dyalog APL

<lang apl>weekday←{⎕IO←1 

   days←'Sunday' 'Monday' 'Tuesday' 'Wednesday' 'Thursday' 'Friday' 'Saturday'
   leap←4 7∊⍨2⊥0=4 100 400∘|
   ld←4 7 1 4 2 6 4 1 5 3 7 5
   nd←3 7 7 4 2 6 4 1 5 3 7 5
   y m d←⍵
   c←⌊y÷100 ⋄ r←100|y
   s←⌊r÷12  ⋄ t←12|r
   can←7|2+5×4|c
   doom←7|s+t+can+⌊t÷4
   anchor←m⊃(1+leap y)⊃nd ld
   (1+7|7+doom+d-anchor)⊃days

}</lang>

Output:
      weekday 1800 1 6
Monday
      weekday 1875 3 29
Monday
      weekday 1915 12 7
Tuesday
      weekday 1970 12 23
Wednesday
      weekday 2043 5 14
Thursday
      weekday 2077 2 12
Friday
      weekday 2101 4 2
Saturday

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.

<lang BASIC>10 DIM D$(7): FOR I=1 TO 7: READ D$(I): NEXT I 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 40 PRINT USING "##/##/####: ";M;D;Y; 50 C=Y\100: R=Y MOD 100 60 S=R\12: T=R MOD 12 70 A=(5*(C AND 3)+2) MOD 7 80 B=(S+T+(T\4)+A) MOD 7 90 PRINT D$((B+D-D(M,-(Y MOD 4=0 AND (Y MOD 100<>0 OR Y MOD 400=0)))+7) MOD 7+1) 100 GOTO 30 110 DATA Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday 120 DATA 3,7,7,4,2,6,4,1,5,3,7,5 130 DATA 4,1,7,4,2,6,4,1,5,3,7,5 140 DATA 1800,1,6, 1875,3,29, 1915,12,7, 1970,12,23 150 DATA 2043,5,14, 2077,2,12, 2101,4,2, 0</lang>

Output:
 1/ 6/1800: Monday
 3/29/1875: Monday
12/ 7/1915: Tuesday
12/23/1970: Wednesday
 5/14/2043: Thursday
 2/12/2077: Friday
 4/ 2/2101: Saturday

BCPL

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

<lang bcpl>get "libhdr"

let dayname(n) = 

   n = 0 -> "Sunday",
   n = 1 -> "Monday",
   n = 2 -> "Tuesday",
   n = 3 -> "Wednesday",
   n = 4 -> "Thursday",
   n = 5 -> "Friday",
   n = 6 -> "Saturday",
   dayname(n rem 7)

let leap(year) = year rem 4 = 0 & (year rem 100 ~= 0 | year rem 400 = 0)

let weekday(y, m, d) = valof $( let leapdoom = table 4,1,7,4,2,6,4,1,5,3,7,5 

   let normdoom = table 3,7,7,4,2,6,4,1,5,3,7,5

   let c = y / 100 and r = y rem 100
   let s = r / 12 and t = r rem 12
   
   let an = (5 * (c rem 4) + 2) rem 7
   let doom = (s + t + (t/4) + an) rem 7
   
   let anchor = (leap(y) -> leapdoom, normdoom)!(m-1)
   resultis (doom + d - anchor + 7) rem 7

$)

let start() be $( writef("January 6, 1800 was on a %S.*N", dayname(weekday(1800, 1, 6))) 

   writef("March 29, 1875 was on a %S.*N", dayname(weekday(1875, 3, 29)))
   writef("December 7, 1915 was on a %S.*N", dayname(weekday(1915, 12, 7)))
   writef("December 23, 1970 was on a %S.*N", dayname(weekday(1970, 12, 23)))
   writef("May 14, 2043 will be on a %S.*N", dayname(weekday(2043, 5, 14)))
   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)))

$)</lang>

Output:
January 6, 1800 was on a Monday.
March 29, 1875 was on a Monday.
December 7, 1915 was on a Tuesday.
December 23, 1970 was on a Wednesday.
May 14, 2043 will be on a Thursday.
February 12, 2077 will be on a Friday.
April 2, 2101 will be on a Saturday.


C

<lang c>#include <stdio.h>

  1. include <stdint.h>
  2. include <stdbool.h>

typedef struct { 

   uint16_t year;
   uint8_t month;
   uint8_t day;

} Date;

bool leap(uint16_t year) { 

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

}

const char *weekday(Date date) { 

   static const uint8_t leapdoom[] = {4,1,7,2,4,6,4,1,5,3,7,5};
   static const uint8_t normdoom[] = {3,7,7,4,2,6,4,1,5,3,7,5};
   static const char *days[] = {
       "Sunday", "Monday", "Tuesday", "Wednesday",
       "Thursday", "Friday", "Saturday"
   };
   
   unsigned c = date.year/100, r = date.year%100;
   unsigned s = r/12, t = r%12;
   
   unsigned c_anchor = (5 * (c%4) + 2) % 7;
   unsigned doom = (s + t + (t/4) + c_anchor) % 7;
   unsigned anchor = (leap(date.year) ? leapdoom : normdoom)[date.month-1];
   return days[(doom+date.day-anchor+7)%7];

}

int main(void) { 

   const char *past = "was", *future = "will be";
   const char *months[] = { "",
       "January", "February", "March", "April", "May", "June",
       "July", "August", "September", "October", "November", "December"
   };
   
   const Date dates[] = {
       {1800,1,6}, {1875,3,29}, {1915,12,7}, {1970,12,23}, {2043,5,14},
       {2077,2,12}, {2101,4,2}
   };
   
   int i;
   for (i=0; i < sizeof(dates)/sizeof(Date); i++) {
       printf("%s %d, %d %s on a %s.\n",
           months[dates[i].month], dates[i].day, dates[i].year,
           dates[i].year > 2021 ? future : past,
           weekday(dates[i]));
   }
   
   return 0;

}</lang>

Output:
January 6, 1800 was on a Monday.
March 29, 1875 was on a Monday.
December 7, 1915 was on a Tuesday.
December 23, 1970 was on a Wednesday.
May 14, 2043 will be on a Thursday.
February 12, 2077 will be on a Friday.
April 2, 2101 will be on a Saturday.

C++

<lang cpp>#include <iostream>

  1. include <cstdint>

struct Date { 

   std::uint16_t year;
   std::uint8_t month;
   std::uint8_t day;

};

constexpr bool leap(int year) { 

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

}

const std::string& weekday(const Date& date) { 

   static const std::uint8_t leapdoom[] = {4,1,7,2,4,6,4,1,5,3,7,5};
   static const std::uint8_t normdoom[] = {3,7,7,4,2,6,4,1,5,3,7,5};
   
   static const std::string days[] = {
       "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
       "Friday", "Saturday"
   };
   
   unsigned const c = date.year/100, r = date.year%100;
   unsigned const s = r/12, t = r%12;
   
   unsigned const c_anchor = (5 * (c%4) + 2) % 7;
   unsigned const doom = (s + t + t/4 + c_anchor) % 7;
   unsigned const anchor = (leap(date.year) ? leapdoom : normdoom)[date.month-1];
   return days[(doom+date.day-anchor+7)%7];

}

int main(void) { 

   const std::string months[] = {"",
       "January", "February", "March", "April", "May", "June",
       "July", "August", "September", "October", "November", "December"
   };
   
   const Date dates[] = {
       {1800,1,6}, {1875,3,29}, {1915,12,7}, {1970,12,23}, {2043,5,14},
       {2077,2,12}, {2101,4,2}
   };
   
   for (const Date& d : dates) {
       std::cout << months[d.month] << " " << (int)d.day << ", " << d.year;
       std::cout << (d.year > 2021 ? " will be " : " was ");
       std::cout << "on a " << weekday(d) << std::endl;
   }
   
   return 0;

}</lang>

Output:
January 6, 1800 was on a Monday
March 29, 1875 was on a Monday
December 7, 1915 was on a Tuesday
December 23, 1970 was on a Wednesday
May 14, 2043 will be on a Thursday
February 12, 2077 will be on a Friday
April 2, 2101 will be on a Saturday

Cowgol

<lang cowgol>include "cowgol.coh";

record Date is 

   year: uint16;
   month: uint8;
   day: uint8;

end record;

sub weekday(d: [Date]): (dayname: [uint8]) is 

   var daynames: [uint8][] := {
       "Sunday", "Monday", "Tuesday", "Wednesday", 
       "Thursday", "Friday", "Saturday"
   };
   
   var leapdoom: uint8[] := {4,7,1,4,2,6,4,1,5,3,7,5};
   var normdoom: uint8[] := {3,7,7,4,2,6,4,1,5,3,7,5};
   
   var c := d.year / 100;
   var r := d.year % 100;
   var s := r / 12;
   var t := r % 12;
   
   var can := (5 * (c%4) + 2) % 7;
   var doom := (s + t + (t/4) + can) % 7;
   var anchor: int16;
   if d.year%4 == 0 and (d.year%100 != 0 or d.year%400 == 0) then
       anchor := leapdoom[d.month-1] as int16;
   else
       anchor := normdoom[d.month-1] as int16;
   end if;
   
   var dayno := (doom as int16 + d.day as int16 - anchor + 7) % 7;
   dayname := daynames[dayno as @indexof daynames];

end sub;

sub print_date(d: [Date]) is 

   print_i8(d.month); print_char('/');
   print_i8(d.day); print_char('/');
   print_i16(d.year); print(": ");
   print(weekday(d));
   print_nl();

end sub;

var dates: Date[] := { 

   {1800,1,6}, {1875,3,29}, {1915,12,7}, {1970,12,23}, {2043,5,14},
   {2077,2,12}, {2101,4,2}

};

var i: @indexof dates := 0; while i < @sizeof dates loop 

   print_date(&dates[i]);
   i := i + 1;

end loop;</lang>

Output:
1/6/1800: Monday
3/29/1875: Monday
12/7/1915: Tuesday
12/23/1970: Wednesday
5/14/2043: Thursday
2/12/2077: Friday
4/2/2101: Saturday

Factor

Works with: Factor version 0.99 2021-02-05

<lang factor>USING: accessors calendar calendar.english formatting generalizations kernel math math.order math.vectors sequences ;

check-year ( year -- )  ! months and days checked by <date>
   1582 9999 between?
   [ "Year must be between 1582 and 9999." throw ] unless ;
doomsday ( year -- n )
   { 4 100 400 } [ mod ] with map { 5 4 6 } vdot 2 + 7 mod ;
anchorday ( year month -- m )
   1 - swap leap-year? { 4 1 } { 3 7 } ?
   { 7 4 2 6 4 1 5 3 7 5 } append nth ;
weekday ( date -- str )
   [ year>> dup check-year doomsday ] [ day>> + ]
   [ dup year>> swap month>> anchorday - 7 + 7 mod ] tri
   day-names nth ;
test ( date -- )
   [ "%B %d, %Y" strftime ]
   [ now before? "was" "will be" ? ]
   [ weekday ] tri
   "%s %s on a %s.\n" printf ;

1800 1 6 1875 3 29 1915 12 7 1970 12 23 2043 5 14 2077 2 12 2101 4 2 [ <date> test ] 3 7 mnapply</lang>

Output:
January 06, 1800 was on a Monday.
March 29, 1875 was on a Monday.
December 07, 1915 was on a Tuesday.
December 23, 1970 was on a Wednesday.
May 14, 2043 will be on a Thursday.
February 12, 2077 will be on a Friday.
April 02, 2101 will be on a Saturday.

FOCAL

<lang focal>01.10 S X(1)=3;S X(2)=7;S X(3)=7;S X(4)=4;S X(5)=2;S X(6)=6 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 01.25 S Y=1875;S M= 3;S D=29;D 2;D 3 01.30 S Y=1915;S M=12;S D= 7;D 2;D 3 01.35 S Y=1970;S M=12;S D=23;D 2;D 3 01.40 S Y=2043;S M= 5;S D=14;D 2;D 3 01.45 S Y=2077;S M= 2;S D=12;D 2;D 3 01.50 S Y=2101;S M= 4;S D= 2;D 2;D 3 01.55 Q

02.10 S C=FITR(Y/100);S R=Y-C*100 02.15 S S=FITR(R/12);S T=R-S*12 02.20 S A=5*(C-FITR(C/4)*4)+2;S A=A-FITR(A/7)*7 02.25 S B=S+T+FITR(T/4)+A;S B=B-FITR(B/7)*7 02.30 S E=X(M) 02.35 I (Y-FITR(Y/400)*400)2.4,2.5,2.4 02.40 I (Y-FITR(Y/100)*100)2.45,2.7,2.45 02.45 I (Y-FITR(Y/4)*4)2.7,2.5,2.7 02.50 I (M-1)2.55,2.6,2.55 02.55 I (M-2)2.7,2.65,2.7 02.60 S E=E+7 02.65 S E=E-6 02.70 S E=B+D-E+7 02.75 S E=E-FITR(E/7)*7

03.10 T "M",%2,M," " 03.20 T "D",%2,D," " 03.30 T "Y",%4,Y," " 03.40 T "DAY=";D 4;T !

04.10 I (E-6)4.2;T "SATURDAY";R 04.20 I (E-5)4.3;T "FRIDAY";R 04.30 I (E-4)4.4;T "THURSDAY";R 04.40 I (E-3)4.5;T "WEDNESDAY";R 04.50 I (E-2)4.6;T "TUESDAY";R 04.60 I (E-1)4.7;T "MONDAY";R 04.70 T "SUNDAY"</lang>

Output:
M=  1 D=  6 Y= 1800 DAY=MONDAY
M=  3 D= 29 Y= 1875 DAY=MONDAY
M= 12 D=  7 Y= 1915 DAY=TUESDAY
M= 12 D= 23 Y= 1970 DAY=WEDNESDAY
M=  5 D= 14 Y= 2043 DAY=THURSDAY
M=  2 D= 12 Y= 2077 DAY=FRIDAY
M=  4 D=  2 Y= 2101 DAY=SATURDAY

Haskell

<lang haskell>import Text.Printf

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

instance Show Date where 

   show (Date {year = y, month = m, day = d}) = printf "%2d/%2d/%4d" m d y

leap :: Int -> Bool leap year = year`mod`4==0 && (year`mod`100/=0 || year`mod`400==0)

weekday :: Date -> Int weekday (Date {year = y, month = m, day = d}) = (doom + d - anchor + 7) `mod` 7 

   where doom     = (s + t + (t `div` 4) + c_anchor) `mod` 7
         anchor   = dooms !! pred m
         c_anchor = (5 * mod c 4 + 2) `mod` 7 
         dooms    = (if leap y then [4,1] else [3,7]) ++ [7,4,2,6,4,1,5,3,7,5]
         (c, r)   = y `divMod` 100
         (s, t)   = r `divMod` 12

days :: [String] days = words "Sunday Monday Tuesday Wednesday Thursday Friday Saturday"

dates :: [Date] dates = [ Date { year = 1800, month = 1, day = 6}, 

         Date { year = 1875, month =  3, day = 29},
         Date { year = 1915, month = 12, day =  7},
         Date { year = 1970, month = 12, day = 23},
         Date { year = 2043, month =  5, day = 14},
         Date { year = 2077, month =  2, day = 12},
         Date { year = 2101, month =  4, day =  2} ]

dateAndDay :: Date -> String dateAndDay d = printf "%s: %s" (show d) (days !! weekday d)

main :: IO () main = putStr $ unlines $ map dateAndDay dates </lang>

Output:
 1/ 6/1800: Monday
 3/29/1875: Monday
12/ 7/1915: Tuesday
12/23/1970: Wednesday
 5/14/2043: Thursday
 2/12/2077: Friday
 4/ 2/2101: Saturday

Java

<lang java>class Doom { 

   public static void main(String[] args) {
       final Date[] dates = {
           new Date(1800,1,6),
           new Date(1875,3,29),
           new Date(1915,12,7),
           new Date(1970,12,23),
           new Date(2043,5,14),
           new Date(2077,2,12),
           new Date(2101,4,2)
       };
       
       for (Date d : dates)
           System.out.println(
               String.format("%s: %s", d.format(), d.weekday()));
   }

}

class Date { 

   private int year, month, day;
   
   private static final int[] leapdoom = {4,1,7,4,2,6,4,1,5,3,7,5};
   private static final int[] normdoom = {3,7,7,4,2,6,4,1,5,3,7,5};
   public static final String[] weekdays = {
       "Sunday", "Monday", "Tuesday", "Wednesday",
       "Thursday", "Friday", "Saturday"
   };
   
   public Date(int year, int month, int day) {
       this.year = year;
       this.month = month;
       this.day = day;
   }
   
   public boolean isLeapYear() {
       return year%4 == 0 && (year%100 != 0 || year%400 == 0);
   }
   
   public String format() {
       return String.format("%02d/%02d/%04d", month, day, year);
   }
   
   public String weekday() {
       final int c = year/100;
       final int r = year%100;
       final int s = r/12;
       final int t = r%12;
       
       final int c_anchor = (5 * (c%4) + 2) % 7;
       final int doom = (s + t + t/4 + c_anchor) % 7;
       final int anchor = 
           isLeapYear() ? leapdoom[month-1] : normdoom[month-1];
       
       return weekdays[(doom + day - anchor + 7) % 7];
   }

}</lang>

Output:
01/06/1800: Monday
03/29/1875: Monday
12/07/1915: Tuesday
12/23/1970: Wednesday
05/14/2043: Thursday
02/12/2077: Friday
04/02/2101: Saturday

Julia

<lang julia>module DoomsdayRule export get_weekday

const weekdaynames = ["Sunday", "Monday","Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]

  1. For months 1 through 12, the date of the first doomsday that month.

const leapyear_firstdoomsdays = [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5] const nonleapyear_firstdoomsdays = [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]

""" 

   get_weekday(year::Int, month::Int, day::Int)::String

Return the weekday of a given date in past or future. Uses Conway's doomsday rule (see also https://en.wikipedia.org/wiki/Doomsday_rule) """ function get_weekday(year::Int, month::Int, day::Int)::String 

   # sanity checks
   @assert 1582 <= year <= 9999 "Invalid year (should be after 1581 and 4 digits)"
   @assert 1 <= month <= 12 "Invalid month, should be between 1 and 12"
   @assert 1 <= day <= 31 "Invalid day, should be between 1 and 31"

   # Conway's doomsday algorithm
   doomsday = (2 + 5 * (year % 4) + 4 * (year % 100) + 6 * (year % 400)) % 7
   anchorday = (year % 4 != 0) || (year % 100 == 0 && year % 400 != 0) ?  # leap year determination
                nonleapyear_firstdoomsdays[month] : leapyear_firstdoomsdays[month]
   weekday = (doomsday + day - anchorday + 7) % 7 + 1
   return weekdaynames[weekday]

end

end # module

using .DoomsdayRule

println("January 6, 1800 was on a ", get_weekday(1800, 1, 6)) println("March 29, 1875 was on a ", get_weekday(1875, 3, 29)) println("December 7, 1915 was on a ", get_weekday(1915, 12, 7)) println("December 23, 1970 was on a ", get_weekday(1970, 12, 23)) println("May 14, 2043 will be on a ", get_weekday(2043, 5, 14)) 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))

</lang>

Output:
January 6, 1800 was on a Monday
March 29, 1875 was on a Monday
December 7, 1915 was on a Tuesday
December 23, 1970 was on a Wednesday
May 14, 2043 will be on a Thursday
February 12, 2077 will be on a Friday
April 2, 2101 will be on a Saturday

Perl

Translation of: Raku

<lang perl># 20210602 Perl programming solution

use strict; use warnings;

sub dow ($) { 

  my ($year, $month, $day) = split /-/;
  my @D = $year%($year%25?4:16) ? (3,7,7,4,2,6,4,1,5,3,7,5) : (4,1,7,2,4,6,4,1,5,3,7,5);
  my $c = int($year / 100);
  my $s = ($year % 100) / 12;
  my $t = ($year % 100) % 12;
  my $a = ( 5 * ($c % 4) + 2 ) % 7;
  my $b = ( $s + $t + int($t / 4) + $a ) % 7;

  qw ( Sunday Monday Tuesday Wednesday Thursday Friday Saturday ) [
     ($b + $day - $D[$month - 1] + 7) % 7 ]

}

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";

}</lang>

Output:
1800-01-06 is a : Monday
1875-03-29 is a : Monday
1915-12-07 is a : Tuesday
1970-12-23 is a : Wednesday
2043-05-14 is a : Thursday
2077-02-12 is a : Friday
2101-04-02 is a : Saturday

Phix

with javascript_semantics

include timedate.e

function doomsday_rule(integer y, m, d)
    integer yr4 = remainder(y,4),
            yr100 = remainder(y,100),
            yr400 = remainder(y,400),
            anchor_day = remainder(2 + 5*yr4 + 4*yr100 + 6*yr400,7)
    bool leap_year = yr4=0 and (yr100!=0 or yr400=0)
    sequence anchors = iff(leap_year?{4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5}
                                    :{3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5})
    integer w = d - anchors[m]
    if w<0 then w += 7 end if
    integer dow = remainder(anchor_day+w,7)
--  return dow
    return iff(dow=0?7:dow)     -- to agree with ISO 8601, and Phix.
end function

constant dates = {"1800-01-06",
                  "1875-03-29",
                  "1915-12-07",
                  "1970-12-23",
                  "2043-05-14",
                  "2077-02-12",
                  "2101-04-02"}

for i=1 to length(dates) do
    timedate dt = parse_date_string(dates[i],{"YYYY-MM-DD"})
    integer {y,m,d} = dt,
            bid = day_of_week(y,m,d),
            drd = doomsday_rule(y,m,d)
    string ds = format_timedate(dt,"Dddd Mmmm D, YYYY"),
           bis = day_of_week(y, m, d, bAsText:=true),
           ok = iff(bid==drd?"(ok)":"*** ERROR ***")
    printf(1,"%s\n%-30s (%d = %d ? %s)\n",{ds,bis,bid,drd,ok})
end for
Output:
Monday January 6, 1800
Monday                         (1 = 1 ? (ok))
Monday March 29, 1875
Monday                         (1 = 1 ? (ok))
Tuesday December 7, 1915
Tuesday                        (2 = 2 ? (ok))
Wednesday December 23, 1970
Wednesday                      (3 = 3 ? (ok))
Thursday May 14, 2043
Thursday                       (4 = 4 ? (ok))
Friday February 12, 2077
Friday                         (5 = 5 ? (ok))
Saturday April 2, 2101
Saturday                       (6 = 6 ? (ok))

PL/M

<lang plm>100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT; PRINT$CHAR: PROCEDURE (C); DECLARE C BYTE; CALL BDOS(2,C); END PRINT$CHAR;

PRINT$NUMBER: PROCEDURE (N); 

   DECLARE S (6) BYTE INITIAL ('.....$');
   DECLARE (N, P) ADDRESS, C BASED P BYTE;
   P = .S(5);

DIGIT: 

   P = P-1;
   C = N MOD 10 + '0';
   N = N / 10;
   IF N >0 THEN GO TO DIGIT;
   CALL PRINT(P);

END PRINT$NUMBER;

PRINT$DATE: PROCEDURE (YEAR, MONTH, DAY); 

   DECLARE YEAR ADDRESS, (MONTH, DAY) BYTE;
   CALL PRINT$NUMBER(MONTH);
   CALL PRINT$CHAR('/');
   CALL PRINT$NUMBER(DAY);
   CALL PRINT$CHAR('/');
   CALL PRINT$NUMBER(YEAR);

END PRINT$DATE;

DAY$NAME: PROCEDURE (N) ADDRESS; 

   DECLARE N BYTE;
   DO CASE N;
       RETURN .'SUNDAY$';
       RETURN .'MONDAY$';
       RETURN .'TUESDAY$';
       RETURN .'WEDNESDAY$';
       RETURN .'THURSDAY$';
       RETURN .'FRIDAY$';
       RETURN .'SATURDAY$';
  END;

END DAY$NAME;

LEAP$YEAR: PROCEDURE (YEAR) BYTE; 

   DECLARE YEAR ADDRESS;
   RETURN (YEAR MOD 4 = 0) AND ((YEAR MOD 100 <> 0) OR (YEAR MOD 400 = 0));

END LEAP$YEAR;

WEEK$DAY: PROCEDURE (YEAR, MONTH, DAY) BYTE; 

   DECLARE LEAP$DOOM DATA (4,1,7,4,2,6,4,1,5,3,7,5);
   DECLARE NORM$DOOM DATA (3,7,7,4,2,6,4,1,5,3,7,5);
   DECLARE YEAR ADDRESS, (MONTH, DAY) BYTE;
   DECLARE (C, R, S, T, C$ANCHOR, DOOM$DAY, ANCHOR$DAY) BYTE;
   
   C = YEAR / 100;
   R = YEAR MOD 100;
   S = R / 12;
   T = R MOD 12;
   C$ANCHOR = (5 * (C MOD 4) + 2) MOD 7;
   DOOM$DAY = (S + T + T/4 + C$ANCHOR) MOD 7;
   IF LEAP$YEAR(YEAR)
       THEN ANCHOR$DAY = LEAP$DOOM(MONTH-1);
       ELSE ANCHOR$DAY = NORM$DOOM(MONTH-1);
   
   RETURN (DOOM$DAY + DAY - ANCHOR$DAY + 7) MOD 7;

END WEEK$DAY;

FORMAT$OUT: PROCEDURE (YEAR, MONTH, DAY); 

   DECLARE YEAR ADDRESS, (MONTH, DAY) BYTE;
   CALL PRINT$DATE(YEAR, MONTH, DAY);
   CALL PRINT(.': $');
   CALL PRINT(DAY$NAME(WEEK$DAY(YEAR, MONTH, DAY)));
   CALL PRINT(.(13,10,'$'));

END FORMAT$OUT;

DECLARE YEARS (7) ADDRESS INITIAL (1800, 1875, 1915, 1970, 2043, 2077, 2101); DECLARE MONTHS (7) BYTE INITIAL ( 1, 3, 12, 12, 5, 2, 4); DECLARE DAYS (7) BYTE INITIAL ( 6, 29, 7, 23, 14, 12, 2);

DECLARE I BYTE; DO I=0 TO LAST(DAYS); 

   CALL FORMAT$OUT(YEARS(I), MONTHS(I), DAYS(I));

END; CALL EXIT; EOF</lang>

Output:
1/6/1800: MONDAY
3/29/1875: MONDAY
12/7/1915: TUESDAY
12/23/1970: WEDNESDAY
5/14/2043: THURSDAY
2/12/2077: FRIDAY
4/2/2101: SATURDAY

Python

<lang python>from datetime import date from calendar import isleap

def weekday(d): 

   days = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
           "Friday", "Saturday"]
   dooms = [
       [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5],
       [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]
   ]
   
   c = d.year // 100
   r = d.year % 100
   s = r // 12
   t = r % 12
   c_anchor = (5 * (c % 4) + 2) % 7
   doomsday = (s + t + (t // 4) + c_anchor) % 7
   anchorday = dooms[isleap(d.year)][d.month - 1]
   weekday = (doomsday + d.day - anchorday + 7) % 7
   return days[weekday]

dates = [date(*x) for x in 

   [(1800, 1, 6), (1875, 3, 29), (1915, 12, 7), (1970, 12, 23),
    (2043, 5, 14), (2077, 2, 12), (2101, 4, 2)]

]

for d in dates: 

   tense = "was" if d < date.today() else "is" if d == date.today() else "will be"
   print("{} {} a {}".format(d.strftime("%B %d, %Y"), tense, weekday(d)))</lang>
Output:
January 06, 1800 was a Monday
March 29, 1875 was a Monday
December 07, 1915 was a Tuesday
December 23, 1970 was a Wednesday
May 14, 2043 will be a Thursday
February 12, 2077 will be a Friday
April 02, 2101 will be a Saturday

Raku

<lang perl6>my @dow = < Sunday Monday Tuesday Wednesday Thursday Friday Saturday >;

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];

sub conway ($date) { 

   my ($year, $month, $day) = $date.comb(/\d+/)».Int;
   my $is-leap = ($year %% 4) && (($year % 100) || ($year %% 400));

   my $c = $year div 100;
   my $s = ($year % 100) div 12;
   my $t = ($year % 100)   % 12;
   my $a = ( 5 * ($c % 4) + 2 ) % 7;
   my $b = ( $s + $t + ($t div 4) + $a ) % 7;

   ($b + $day - %doomsday{$is-leap}[$month - 1] + 7) % 7

}


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

   say "Conway  - $_ is a: ", @dow[.&conway];
   # Or, we could use the method built into the compiler...
   say "Builtin - $_ is a: ", @dow[Date.new($_).day-of-week];
   say ;

}</lang>

Output:
Conway  - 1800-01-06 is a: Monday
Builtin - 1800-01-06 is a: Monday

Conway  - 1875-03-29 is a: Monday
Builtin - 1875-03-29 is a: Monday

Conway  - 1915-12-07 is a: Tuesday
Builtin - 1915-12-07 is a: Tuesday

Conway  - 1970-12-23 is a: Wednesday
Builtin - 1970-12-23 is a: Wednesday

Conway  - 2043-05-14 is a: Thursday
Builtin - 2043-05-14 is a: Thursday

Conway  - 2077-02-12 is a: Friday
Builtin - 2077-02-12 is a: Friday

Conway  - 2101-04-02 is a: Saturday
Builtin - 2101-04-02 is a: Saturday

REXX

<lang rexx>/*REXX program finds the day─of─week for a specified date using Conway's Doomsday rule. */ parse arg $ /*obtain optional arguments from the CL*/ if $= | $="," then $= , /*Not specified? Then use the default.*/ 

           '01/06/1800 03/29/1875 12/07/1915 12/23/1970 05/14/2043 04/02/2077 04/02/2101'

d= 'Sun Mon Tues Wednes Thurs Fri Satur' /*list of days of the week, sans "day".*/ y.0= 3 7 7 4 2 6 4 1 5 3 7 5 /*doomsday dates for non-leapyear month*/ y.1= 4 1 7 4 2 6 4 1 5 3 7 5 /* " " " leapyear " */ 

     do j=1  for words($);        datum= word($, j)    /*process each of the dates.    */
     parse var  datum    mm  '/'  dd  "/"  yy          /*parse the date  ──►  mm dd yy */
     ly= leapyear(yy)                                  /*get indication of a leapyear. */
     wd= (doomsday(yy)+dd-word(y.ly, mm) + 7) // 7 + 1 /*obtain a code for the weekday.*/
     say datum    ' falls on '    word(d, wd)"day"     /*display day-of-week for date. */
     end   /*j*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ doomsday: parse arg ?; return (2 + 5 * (?//4) + 4 * (?//100) + 6 * (?//400) ) // 7 leapyear: arg #; ly= #//4==0; if ly==0 then return 0; return ((#//100\==0) | #//400==0)</lang>

output   when using the default inputs:
01/06/1800  falls on  Monday
03/29/1875  falls on  Monday
12/07/1915  falls on  Tuesday
12/23/1970  falls on  Wednesday
05/14/2043  falls on  Thursday
04/02/2077  falls on  Friday
04/02/2101  falls on  Saturday

Rust

<lang rust>fn day_of_week(year: u32, month: u32, day: u32) -> u32 { 

   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];
   assert!(year > 1581 && year < 10000);
   assert!(month >= 1 && month <= 12);
   assert!(day >= 1 && day <= 31);
   let doomsday = (2 + 5 * (year % 4) + 4 * (year % 100) + 6 * (year % 400)) % 7;
   let anchorday = if year % 4 != 0 || (year % 100 == 0 && year % 400 != 0) {
       NONLEAPYEAR_FIRSTDOOMSDAYS[month as usize - 1]
   } else {
       LEAPYEAR_FIRSTDOOMSDAYS[month as usize - 1]
   };
   (doomsday + day + 7 - anchorday) % 7

}

fn print_day_of_week(year: u32, month: u32, day: u32) { 

   const DAY_NAMES: [&str; 7] = [
       "Sunday",
       "Monday",
       "Tuesday",
       "Wednesday",
       "Thursday",
       "Friday",
       "Saturday",
   ];
   println!(
       "{:04}-{:02}-{:02}: {}",
       year,
       month,
       day,
       DAY_NAMES[day_of_week(year, month, day) as usize]
   );

}

fn main() { 

   print_day_of_week(1800, 1, 6);
   print_day_of_week(1875, 3, 29);
   print_day_of_week(1915, 12, 7);
   print_day_of_week(1970, 12, 23);
   print_day_of_week(2043, 5, 14);
   print_day_of_week(2077, 2, 12);
   print_day_of_week(2101, 4, 2);

}</lang>

Output:
1800-01-06: Monday
1875-03-29: Monday
1915-12-07: Tuesday
1970-12-23: Wednesday
2043-05-14: Thursday
2077-02-12: Friday
2101-04-02: Saturday

Wren

Library: 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. <lang ecmascript>import "/date" for Date

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

var anchorDay = Fn.new { |y| (2 + 5 * (y%4) + 4 *(y%100) + 6 * (y%400)) % 7 }

var isLeapYear = Fn.new { |y| y%4 == 0 && (y%100 != 0 || y%400 == 0) }

var firstDaysCommon = [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5] var firstDaysLeap = [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]

var dates = [ 

   "1800-01-06",
   "1875-03-29",
   "1915-12-07",
   "1970-12-23",
   "2043-05-14",
   "2077-02-12",
   "2101-04-02"

]

System.print("Days of week given by Doomsday rule:") for (date in dates) { 

   var y = Num.fromString(date[0..3])
   var m = Num.fromString(date[5..6]) - 1
   var d = Num.fromString(date[8..9])
   var a = anchorDay.call(y)
   var w = d - (isLeapYear.call(y) ? firstDaysLeap[m] : firstDaysCommon[m])
   if (w < 0) w = 7 + w
   var dow = (a + w) % 7
   System.print("%(date) -> %(days[dow])")

}

System.print("\nDays of week given by Date module:") for (date in dates) { 

   var d = Date.parse(date, Date.isoDate)
   System.print("%(date) -> %(d.weekDay)")

}</lang>

Output:
Days of week given by Doomsday rule:
1800-01-06 -> Monday
1875-03-29 -> Monday
1915-12-07 -> Tuesday
1970-12-23 -> Wednesday
2043-05-14 -> Thursday
2077-02-12 -> Friday
2101-04-02 -> Saturday

Days of week given by Date module:
1800-01-06 -> Monday
1875-03-29 -> Monday
1915-12-07 -> Tuesday
1970-12-23 -> Wednesday
2043-05-14 -> Thursday
2077-02-12 -> Friday
2101-04-02 -> Saturday
Retrieved from "https://rosettacode.org/wiki/Doomsday_rule?oldid=312617"