Doomsday rule
You are encouraged to solve this task according to the task description, using any language you may know.
- 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 on 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
11l
F isleap(year)
R year % 4 == 0 & (year % 100 != 0 | year % 400 == 0)
F weekday(year, month, day)
V days = [‘Sunday’, ‘Monday’, ‘Tuesday’,
‘Wednesday’, ‘Thursday’, ‘Friday’, ‘Saturday’]
V 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]]
V c = year I/ 100
V r = year % 100
V s = r I/ 12
V t = r % 12
V c_anchor = (5 * (c % 4) + 2) % 7
V doomsday = (s + t + (t I/ 4) + c_anchor) % 7
V anchorday = dooms[isleap(year)][month - 1]
V weekday = (doomsday + day - anchorday + 7) % 7
R days[weekday]
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)]
print(‘#.-#02-#02 -> #.’.format(year, month, day, weekday(year, month, day)))- 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
ALGOL 68
BEGIN # find the day of the week of dates using John Conway's Doomsday rule #
# returns the day of the week (Sunday = 0, Monday = 1,...) for the date #
# specified by ccyy, mm and dd #
PROC dow = ( INT ccyy, mm, dd )INT:
BEGIN
INT doomsday = ( # Tuesday # 2
+ ( 5 * ( ccyy MOD 4 ) )
+ ( 4 * ( ccyy MOD 100 ) )
+ ( 6 * ( ccyy MOD 400 ) )
)
MOD 7;
BOOL is leap year = ccyy MOD 4 = 0
AND ( ccyy MOD 100 /= 0 OR ccyy MOD 400 = 0 );
INT anchor day = CASE mm
IN IF is leap year THEN 4 ELSE 3 FI
, IF is leap year THEN 1 ELSE 7 FI
, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5
ESAC;
( doomsday + ( dd - anchor day ) ) MOD 7
END # dow # ;
BEGIN # task test cases #
# prints a test date and its day of the week #
PROC test dow = ( INT ccyy, mm, dd )VOID:
BEGIN
[]CHAR digit = "0123456789"[ AT 0 ];
[]STRING day name =
[]STRING( "Sunday", "Monday", "Tuesday", "Wednesday"
, "Thursday", "Friday", "Saturday"
)[ AT 0 ];
print( ( whole( ccyy, 0 )
, "-", digit[ mm OVER 10 ], digit[ mm MOD 10 ]
, "-", digit[ dd OVER 10 ], digit[ dd MOD 10 ]
, ": ", day name[ dow( ccyy, mm, dd ) ]
, newline
)
)
END # test dow # ;
test dow( 1800, 1, 6 );
test dow( 1875, 3, 29 );
test dow( 1915, 12, 7 );
test dow( 1970, 12, 23 );
test dow( 2043, 5, 14 );
test dow( 2077, 2, 12 );
test dow( 2101, 4, 2 );
test dow( 2022, 6, 19 )
END
END- 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 2022-06-19: Sunday
ALGOL W
begin % find the day of the week of dates using John Conway's Doomsday rule %
% returns the day of the week (Sunday = 0, Monday = 1,...) for the date %
% specified by ccyy, mm and dd %
integer procedure dow ( integer value ccyy, mm, dd ) ;
begin
integer doomsday, anchorDay;
logical isLeapYear;
doomsday := ( % Tuesday % 2
+ ( 5 * ( ccyy rem 4 ) )
+ ( 4 * ( ccyy rem 100 ) )
+ ( 6 * ( ccyy rem 400 ) )
)
rem 7;
isLeapYear := ccyy rem 4 = 0 and ( ccyy rem 100 not = 0 or ccyy rem 400 = 0 );
anchorDay := case mm of ( if isLeapYear then 4 else 3
, if isLeapYear then 1 else 7
, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5
);
( doomsday + ( dd - anchorDay ) ) rem 7
end dow ;
begin % task test cases %
% prints a test date and its day of the week %
procedure testDow ( integer value ccyy, mm, dd ) ;
write( i_w := 1, s_w := 0
, ccyy
, "-", mm div 10, mm rem 10
, "-", dd div 10, dd rem 10
, ": ", dayName( dow( ccyy, mm, dd ) )
);
string(9) array dayName ( 0 :: 6 );
dayName( 0 ) := "Sunday"; dayName( 1 ) := "Monday"; dayName( 2 ) := "Tuesday";
dayName( 3 ) := "Wednesday"; dayName( 4 ) := "Thursday"; dayName( 5 ) := "Friday";
dayName( 6 ) := "Saturday";
testDow( 1800, 1, 6 );
testDow( 1875, 3, 29 );
testDow( 1915, 12, 7 );
testDow( 1970, 12, 23 );
testDow( 2043, 5, 14 );
testDow( 2077, 2, 12 );
testDow( 2101, 4, 2 );
testDow( 2022, 6, 19 )
end
end.- 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 2022-06-19: Sunday
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
}
- 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
AppleScript
on dayOfWeek(yyyymmdd)
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)
if ((m < 3) and ((y mod 4 = 0) and (y mod 100 > 0) or (y mod 400 = 0))) then set m to m + 12
set doomsdayInMonth to item m of {3, 28, 7, 4, 2, 6, 4, 8, 5, 10, 7, 12, 4, 29}
return item ((doomsdayForYear + d - doomsdayInMonth + 28) mod 7 + 1) of ¬
{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}
end dayOfWeek
on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join
on task()
set dateStrings to {"1800-01-06", "1875-03-29", "1915-12-07", "1970-12-23", "2043-05-14", "2077-02-12", "2101-04-02"}
set output to {}
repeat with yyyymmdd in dateStrings
set end of output to yyyymmdd & " --> " & dayOfWeek(yyyymmdd)
end repeat
return join(output, linefeed)
end task
task()
- 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"
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.
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
- 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.
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)))
$)- 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
#include <stdio.h>
#include <stdint.h>
#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;
}
- 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++
#include <iostream>
#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;
}
- 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
CLU
leap_year = proc (year: int) returns (bool)
return(year//4=0 & (year//100=0 | year//400=0))
end leap_year
weekday = proc (d: date) returns (string)
own leapdoom: array[int] := array[int]$[4,1,7,2,4,6,4,1,5,3,7,5]
own normdoom: array[int] := array[int]$[3,7,7,4,2,6,4,1,5,3,7,5]
own days: array[string] := array[string]$[0:
"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
"Friday", "Saturday"
]
c: int := d.year/100
r: int := d.year//100
s: int := r/12
t: int := r//12
c_anchor: int := (5 * (c//4) + 2) // 7
doom: int := (s + t + t/4 + c_anchor) // 7
anchor: int
if leap_year(d.year)
then anchor := leapdoom[d.month]
else anchor := normdoom[d.month]
end
return(days[(doom+d.day-anchor+7)//7])
end weekday
start_up = proc ()
po: stream := stream$primary_output()
dates: array[date] := array[date]$
[date$create( 1, 6,1800,0,0,0),
date$create(29, 3,1875,0,0,0),
date$create( 7,12,1915,0,0,0),
date$create(23,12,1970,0,0,0),
date$create(14, 5,2043,0,0,0),
date$create(12, 2,2077,0,0,0),
date$create( 2, 4,2101,0,0,0)]
for d: date in array[date]$elements(dates) do
stream$puts(po, date$unparse_date(d))
if d<now()
then stream$puts(po, " was on a ")
else stream$puts(po, " will be on a ")
end
stream$putl(po, weekday(d))
end
end start_up- Output:
1 June 1800 was on a Sunday 29 March 1875 was on a Monday 7 December 1915 was on a Tuesday 23 December 1970 was on a Wednesday 14 May 2043 will be on a Thursday 12 February 2077 will be on a Friday 2 April 2101 will be on a Saturday
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;- 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
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
- 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
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"- 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
FreeBASIC
dim shared as ubyte fdoom(0 to 1, 1 to 12) = {_
{ 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
'each month for common
'and leap years
dim shared as string*10 days(0 to 6) = {"Sunday", "Monday", "Tuesday", "Wednesday",_
"Thursday", "Friday", "Saturday"}
function doomsday(y as uinteger) as ubyte
' John Conway's doomsday formula
return (2 + 5*(y mod 4) + 4*(y mod 100) + 6*(y mod 400)) mod 7
end function
function leap(y as uinteger) as ubyte
'is it a leap year?
'return 0 for common years, 1 for leap years
if y mod 4 > 0 then return 0
if y mod 100 = 0 and y mod 400 > 0 then return 0
return 1
end function
function get_day(y as uinteger, m as ubyte, d as ubyte) as string
dim as ubyte c = doomsday(y), diff
diff = (7 + d - fdoom( leap(y), m )) mod 7
return days( (c+diff) mod 7 )
end function
print get_day( 1800, 01, 06 )
print get_day( 1875, 03, 29 )
print get_day( 1915, 12, 07 )
print get_day( 1970, 12, 23 )
print get_day( 2043, 05, 14 )
print get_day( 2077, 02, 12 )
print get_day( 2101, 04, 02 )- Output:
Monday Monday Tuesday Wednesday Thursday Friday Saturday
FutureBasic
local fn IsLeapYear( year as NSUInteger ) as BOOL
BOOL result = YES
if ( year mod 4 > 0 ) then result = NO : exit fn
if ( year mod 100 = 0 ) and ( year mod 400 > 0 ) then result = NO : exit fn
end fn = result
local fn DoomsdayWeekdayForDate( month as NSUInteger, day as NSUInteger, year as NSUInteger ) as CFStringRef
CFArrayRef weekdayNames = @[@"Sunday", @"Monday", @"Tuesday", @"Wednesday", @"Thursday", @"Friday", @"Saturday"]
CFArrayRef normalYearDays = @[@4, @1, @7, @4, @2, @6, @4, @1, @5, @3, @7, @5]
CFArrayRef leapYearDays = @[@3, @7, @7, @4, @2, @6, @4, @1, @5, @3, @7, @5]
CFStringRef dayOfWeek = @""
NSUInteger anchorday
NSUInteger doomsday = ( 2 + 5 * ( year mod 4 ) + 4 * ( year mod 100 ) + 6 * ( year mod 400) ) mod 7
if ( fn IsLeapYear( year ) ) then anchorday = intVal( normalYearDays[month-1] ) else anchorday = intVal( leapYearDays[month-1] )
NSUInteger weekday = ( doomsday + day - anchorday + 7 ) mod 7
if ( weekday == 0 ) then dayOfWeek = @"Sunday" else dayOfWeek = weekdayNames[weekday]
end fn = dayOfWeek
printf @"01-06-1800 : %@", fn DoomsdayWeekdayForDate( 1, 6, 1800 )
printf @"03-29-1875 : %@", fn DoomsdayWeekdayForDate( 3, 29, 1875 )
printf @"12-07-1915 : %@", fn DoomsdayWeekdayForDate( 12, 7, 1915 )
printf @"12-23-1970 : %@", fn DoomsdayWeekdayForDate( 12, 23, 1970 )
printf @"05-14-2043 : %@", fn DoomsdayWeekdayForDate( 5, 14, 2043 )
printf @"02-12-2077 : %@", fn DoomsdayWeekdayForDate( 2, 12, 2077 )
printf @"04-02-2101 : %@", fn DoomsdayWeekdayForDate( 4, 2, 2101 )
HandleEvents- 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
Go
package main
import (
"fmt"
"strconv"
)
var days = []string{"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}
func anchorDay(y int) int {
return (2 + 5*(y%4) + 4*(y%100) + 6*(y%400)) % 7
}
func isLeapYear(y int) bool { return y%4 == 0 && (y%100 != 0 || y%400 == 0) }
var firstDaysCommon = []int{3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5}
var firstDaysLeap = []int{4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5}
func main() {
dates := []string{
"1800-01-06",
"1875-03-29",
"1915-12-07",
"1970-12-23",
"2043-05-14",
"2077-02-12",
"2101-04-02",
}
fmt.Println("Days of week given by Doomsday rule:")
for _, date := range dates {
y, _ := strconv.Atoi(date[0:4])
m, _ := strconv.Atoi(date[5:7])
m--
d, _ := strconv.Atoi(date[8:10])
a := anchorDay(y)
f := firstDaysCommon[m]
if isLeapYear(y) {
f = firstDaysLeap[m]
}
w := d - f
if w < 0 {
w = 7 + w
}
dow := (a + w) % 7
fmt.Printf("%s -> %s\n", date, days[dow])
}
}
- 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
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 "%4d-%02d-%02d" y m d
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} =
let 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
in (doom + d - anchor + 7) `mod` 7
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
- 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
J
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
1+7|1++/\.31 28 31 30 31 30 31 31 30 31 30 31
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
4 1 7 4 2 6 4 1 5 3 7 5
Also note that adding 7 is like adding 0 in modulo 7 arithmetic.
Thus:
get_weekday=: {{ 'Y M D'=. y
Y0=. todayno Y,1 1
Y1=. todayno 1+Y,0 0
aday=. M{_,1+7|1++/\.>.//./}.|:todate Y0+i.Y1-Y0
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).
get_weekday=: {{ 'Y M D'=. y
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:
get_weekday 1800 1 6
Monday
get_weekday 1875 3 29
Monday
get_weekday 1915 12 7
Tuesday
get_weekday 1970 12 23
Wednesday
get_weekday 2043 5 14
Thursday
get_weekday 2077 2 12
Friday
get_weekday 2101 4 2
Saturday
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];
}
}
- 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
jq
Works with gojq, the Go implementation of jq
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"];
# For months 1 through 12, the date of the first doomsday that month.
def leapyear_firstdoomsdays: [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5];
def 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 the past or future subject to 1582<=$year<=9999.
# Uses Conway's doomsday rule (see also https://en.wikipedia.org/wiki/Doomsday_rule)
def assert(exp; msg):
if env.JQ_ASSERT then
(exp as $e | if $e then . else . as $in | "assertion violation: \(msg) => \($e)" | error end)
else . end;
def get_weekday($year; $month; $day):
# sanity checks
assert(1582 <= $year and $year <= 9999; "Invalid year (\($year) - should be after 1581 and 4 digits)")
| assert( 1 <= $month and $month <= 12; "Invalid month, should be between 1 and 12")
| assert( 1 <= $day and $day <= 31; "Invalid day, should be between 1 and 31")
# Conway's doomsday algorithm
| ((2 + 5 * ($year % 4) + 4 * ($year % 100) + 6 * ($year % 400)) % 7) as $doomsday
# leap year determination:
| (if (($year % 4 != 0) or ($year % 100 == 0 and $year % 400 != 0))
then nonleapyear_firstdoomsdays[$month - 1]
else leapyear_firstdoomsdays[$month - 1]
end) as $anchorday
| (($doomsday + $day - $anchorday + 7) % 7) as $weekday # IO==0
| weekdaynames[$weekday] ;
("January 6, 1800 was on a " + get_weekday(1800; 1; 6)),
("March 29, 1875 was on a " + get_weekday(1875; 3; 29)),
("December 7, 1915 was on a " + get_weekday(1915; 12; 7)),
("December 23, 1970 was on a " + get_weekday(1970; 12; 23)),
("May 14, 2043 will be on a " + get_weekday(2043; 5; 14)),
("February 12, 2077 will be on a "+ get_weekday(2077; 2; 12)),
("April 2, 2101 will be on a " + get_weekday(2101; 4; 2)),
("April 2, 21011 will be on a " + get_weekday(21011; 4; 2))- 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 April 2, 21011 will be on a Tuesday
Julia
module DoomsdayRule
export get_weekday
const weekdaynames = ["Sunday", "Monday","Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
# 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))
- 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
Nim
We use the “times” standard module to parse the dates, check for leap years and compare results with those obtained with the doomsday rule (useful only to trap bugs in our implementation). We use also the “WeekDay” type of this module which is an enumeration starting with “dMon” for Monday and ending with “dSun” for Sunday.
import strformat, times
const
NormDoom = [1: 3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]
LeapDoom = [1: 4, 1, 7, 2, 4, 6, 4, 1, 5, 3, 7, 5]
proc weekday(year, month, day: int): WeekDay =
let doom = (2 + 5 * (year mod 4) + 4 * (year mod 100) + 6 * (year mod 400)) mod 7
let anchor = if year.isLeapYear: LeapDoom[month] else: NormDoom[month]
let wd = (doom + day - anchor + 7) mod 7
result = if wd == 0: dSun else: WeekDay(wd - 1)
const Dates = ["1800-01-06", "1875-03-29", "1915-12-07",
"1970-12-23", "2043-05-14", "2077-02-12", "2101-04-02"]
for date in Dates:
let dt = date.parse("yyyy-MM-dd")
let wday = weekday(dt.year, ord(dt.month), dt.monthday)
if wday != dt.weekday:
echo &"For {date}, expected {dt.weekday}, found {wday}."
else:
echo date, " → ", wday
- 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
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 $_, "\n";
}
- Output:
1800-01-06 is a : Monday 1875-03-29 is a : Monday 1915-12-07 is a : Tuesday 1970-12-23 is a : Wednesday 2043-05-14 is a : Thursday 2077-02-12 is a : Friday 2101-04-02 is a : Saturday
Phix
with javascript_semantics include timedate.e function doomsday_rule(integer y, m, d) integer yr4 = remainder(y,4), yr100 = remainder(y,100), yr400 = remainder(y,400), anchor_day = remainder(2 + 5*yr4 + 4*yr100 + 6*yr400,7) bool leap_year = yr4=0 and (yr100!=0 or yr400=0) sequence anchors = iff(leap_year?{4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5} :{3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5}) integer w = d - anchors[m] if w<0 then w += 7 end if integer dow = remainder(anchor_day+w,7) -- return dow return iff(dow=0?7:dow) -- to agree with ISO 8601, and Phix. end function constant dates = {"1800-01-06", "1875-03-29", "1915-12-07", "1970-12-23", "2043-05-14", "2077-02-12", "2101-04-02"} for i=1 to length(dates) do timedate dt = parse_date_string(dates[i],{"YYYY-MM-DD"}) integer {y,m,d} = dt, bid = day_of_week(y,m,d), drd = doomsday_rule(y,m,d) string ds = format_timedate(dt,"Dddd Mmmm D, YYYY"), bis = day_of_week(y, m, d, bAsText:=true), ok = iff(bid==drd?"(ok)":"*** ERROR ***") printf(1,"%s\n%-30s (%d = %d ? %s)\n",{ds,bis,bid,drd,ok}) end for
- Output:
Monday January 6, 1800 Monday (1 = 1 ? (ok)) Monday March 29, 1875 Monday (1 = 1 ? (ok)) Tuesday December 7, 1915 Tuesday (2 = 2 ? (ok)) Wednesday December 23, 1970 Wednesday (3 = 3 ? (ok)) Thursday May 14, 2043 Thursday (4 = 4 ? (ok)) Friday February 12, 2077 Friday (5 = 5 ? (ok)) Saturday April 2, 2101 Saturday (6 = 6 ? (ok))
PL/M
100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
PRINT$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- 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
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)))
- 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
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 '';
}
- 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
/*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)
- 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
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);
}
- 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
uBasic/4tH
Dim @f(24)
Push 3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5 ' the first doomsday in each
Push 4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5 ' month for common and leap years
For x = 23 to 0 Step -1 : @f(x) = Pop() : Next
Dim @d(7)
Push Dup("Sunday"), Dup("Monday"), Dup("Tuesday"), Dup("Wednesday")
Push Dup("Thursday"), Dup("Friday"), Dup("Saturday")
For x = 6 To 0 Step -1 : @d(x) = Pop() : Next
Print Show(FUNC(_get_day(1800, 01, 06)))
Print Show(FUNC(_get_day(1875, 03, 29)))
Print Show(FUNC(_get_day(1915, 12, 07)))
Print Show(FUNC(_get_day(1970, 12, 23)))
Print Show(FUNC(_get_day(2043, 05, 14)))
Print Show(FUNC(_get_day(2077, 02, 12)))
Print Show(FUNC(_get_day(2101, 04, 02)))
End
_doomsday ' John Conway's doomsday formula
Param (1) : Return ((2 + 5*(a@ % 4) + 4*(a@ % 100) + 6*(a@ % 400)) % 7)
_leap ' is it a leap year?
Param (1)
If (a@ % 4 > 0) Then Return (0) ' return 0 for common years
If (a@ % 100 = 0) * (a@ % 400 > 0) Then Return (0)
Return (1) ' 1 for leap years
_get_day
Param (3)
Local (2)
d@ = FUNC(_doomsday(a@))
e@ = (7 + c@ - @f(FUNC(_leap(a@)) * 2 + (b@-1))) % 7
Return (@d((d@+e@) % 7))- Output:
Monday Monday Tuesday Wednesday Thursday Friday Saturday 0 OK, 0:711
V (Vlang)
const
(
days = ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"]
first_days_common = [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]
first_days_leap = [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]
)
fn main() {
dates := [
"1800-01-06",
"1875-03-29",
"1915-12-07",
"1970-12-23",
"2043-05-14",
"2077-02-12",
"2101-04-02"
]
mut y, mut m, mut d, mut a, mut f, mut w, mut dow := 0,0,0,0,0,0,0
println("Days of week given by Doomsday rule:")
for date in dates {
y = date[0..4].int()
m = date[5..7].int()
m--
d = date[8..10].int()
a = anchor_day(y)
f = first_days_common[m]
if is_leap_year(y) {
f = first_days_leap[m]
}
w = d - f
if w < 0 {
w = 7 + w
}
dow = (a + w) % 7
println('$date -> ${days[dow]}')
}
}
fn anchor_day(y int) int {
return (2 + 5 *(y % 4) + 4 * (y % 100) + 6 *(y % 400)) % 7
}
fn is_leap_year(y int) bool {
return y % 4 == 0 && (y % 100 != 0 || y % 400 == 0)
}- 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
Wren
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.
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)")
}- 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
Yabasic
dim fdoom(1, 12)
for x = 0 to arraysize(fdoom(),1)
for y = 1 to arraysize(fdoom(),2)
read fdoom(x, y)
next y
next x
dim days$(6)
for x = 0 to arraysize(days$(),1)
read days$(x)
next x
sub doomsday(y)
// John Conway's doomsday formula
return mod((2 + 5*mod(y, 4) + 4*mod(y, 100) + 6*mod(y, 400)), 7)
end sub
sub leap(y)
//is it a leap year?
//return 0 for common years, 1 for leap years
if mod(y, 4) > 0 then return 0 : fi
if mod(y, 100) = 0 and mod(y, 400) > 0 then return 0 : fi
return 1
end sub
sub get_day$(d, m, y)
c = doomsday(y)
diff = mod((7 + d - fdoom(leap(y), m)), 7)
return days$(mod((c+diff), 7))
end sub
print get_day$(06, 01, 1800)
print get_day$(29, 03, 1875)
print get_day$(07, 12, 1915)
print get_day$(23, 12, 1970)
print get_day$(14, 05, 2043)
print get_day$(12, 02, 2077)
print get_day$(02, 04, 2101)
end
//the first doomsday in each month for common and leap years
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
data "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"- Output:
Igual que la entrada de FreeBASIC.