Doomsday rule: Difference between revisions
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{{
[[Category:Date and time]]
; About the task
John Conway (1937-2020), was a
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
"anchor days" which follow a pattern for the day of the week upon which they fall.
Line 60 ⟶ 61:
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="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)))</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|ALGOL 68}}==
{{Trans|ALGOL W}}
<syntaxhighlight lang="algol68">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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|ALGOL W}}==
<syntaxhighlight lang="algolw">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.</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|APL}}==
{{works with|Dyalog APL}}
<
days←'Sunday' 'Monday' 'Tuesday' 'Wednesday' 'Thursday' 'Friday' 'Saturday'
leap←4 7∊⍨2⊥0=4 100 400∘|
Line 75 ⟶ 230:
anchor←m⊃(1+leap y)⊃nd ld
(1+7|7+doom+d-anchor)⊃days
}</
{{out}}
<pre> weekday 1800 1 6
Line 91 ⟶ 246:
weekday 2101 4 2
Saturday</pre>
=={{header|AppleScript}}==
<syntaxhighlight lang="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()</syntaxhighlight>
{{output}}
<syntaxhighlight lang="applescript">"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"</syntaxhighlight>
=={{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
Line 109 ⟶ 303:
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</
{{out}}
<pre> 1/ 6/1800: Monday
Line 122 ⟶ 316:
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) =
Line 158 ⟶ 352:
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)))
$)</
{{out}}
<pre>January 6, 1800 was on a Monday.
Line 171 ⟶ 365:
=={{header|C}}==
<
#include <stdint.h>
#include <stdbool.h>
Line 223 ⟶ 417:
return 0;
}</
{{out}}
<pre>January 6, 1800 was on a Monday.
Line 232 ⟶ 426:
February 12, 2077 will be on a Friday.
April 2, 2101 will be on a Saturday.</pre>
=={{header|C#}}==
{{trans|Java}}
<syntaxhighlight lang="C#">
using System;
class Doom {
public static void Main(string[] args) {
Date[] dates = {
new Date(1800,1,6),
new Date(1875,3,29),
new Date(1915,12,7),
new Date(1970,12,23),
new Date(2043,5,14),
new Date(2077,2,12),
new Date(2101,4,2)
};
foreach (Date d in dates)
Console.WriteLine($"{d.Format()}: {d.Weekday()}");
}
}
class Date {
private int year, month, day;
private static readonly int[] leapDoom = {4,1,7,4,2,6,4,1,5,3,7,5};
private static readonly int[] normDoom = {3,7,7,4,2,6,4,1,5,3,7,5};
public static readonly 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 bool IsLeapYear() {
return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
}
public string Format() {
return $"{month:00}/{day:00}/{year:0000}";
}
public string Weekday() {
int c = year / 100;
int r = year % 100;
int s = r / 12;
int t = r % 12;
int cAnchor = (5 * (c % 4) + 2) % 7;
int doom = (s + t + t / 4 + cAnchor) % 7;
int anchor =
IsLeapYear() ? leapDoom[month - 1] : normDoom[month - 1];
return weekdays[(doom + day - anchor + 7) % 7];
}
}
</syntaxhighlight>
{{out}}
<pre>
01/06/1800: Monday
03/29/1875: Monday
12/07/1915: Tuesday
12/23/1970: Wednesday
05/14/2043: Thursday
02/12/2077: Friday
04/02/2101: Saturday
</pre>
=={{header|C++}}==
<
#include <cstdint>
Line 283 ⟶ 551:
return 0;
}</
{{out}}
<pre>January 6, 1800 was on a Monday
Line 292 ⟶ 560:
February 12, 2077 will be on a Friday
April 2, 2101 will be on a Saturday</pre>
=={{header|CLU}}==
<syntaxhighlight lang="clu">leap_year = proc (year: int) returns (bool)
return(year//4=0 & (year//100=0 | year//400=0))
end leap_year
weekday = proc (d: date) returns (string)
own leapdoom: array[int] := array[int]$[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$putl(po, weekday(d))
end
end start_up</syntaxhighlight>
{{out}}
<pre>1 June 1800 was on a Sunday
29 March 1875 was on a Monday
7 December 1915 was on a Tuesday
23 December 1970 was on a Wednesday
14 May 2043 will be on a Thursday
12 February 2077 will be on a Friday
2 April 2101 will be on a Saturday</pre>
=={{header|Cowgol}}==
<
record Date is
Line 346 ⟶ 671:
print_date(&dates[i]);
i := i + 1;
end loop;</
{{out}}
<pre>1/6/1800: Monday
Line 355 ⟶ 680:
2/12/2077: Friday
4/2/2101: Saturday</pre>
=={{header|EasyLang}}==
{{trans|C}}
<syntaxhighlight>
func leap year .
return if year mod 4 = 0 and (year mod 100 <> 0 or year mod 400 = 0)
.
func weekday year month day .
normdoom[] = [ 3 7 7 4 2 6 4 1 5 3 7 5 ]
c = year div 100
r = year mod 100
s = r div 12
t = r mod 12
c_anchor = (5 * (c mod 4) + 2) mod 7
doom = (s + t + (t div 4) + c_anchor) mod 7
anchor = normdoom[month]
if leap year = 1 and month <= 2
anchor = (anchor + 1) mod1 7
.
return (doom + day - anchor + 7) mod 7 + 1
.
wkdays$[] = [ "Sunday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Saturday" ]
dates$[] = [ "1800-01-06" "1875-03-29" "1915-12-07" "1970-12-23" "2043-05-14" "2077-02-12" "2101-04-02" ]
for d$ in dates$[]
write d$ & " -> "
a[] = number strsplit d$ "-"
print wkdays$[weekday a[1] a[2] a[3]]
.
</syntaxhighlight>
=={{header|Factor}}==
{{works with|Factor|0.99 2021-02-05}}
<
generalizations kernel math math.order math.vectors sequences ;
Line 390 ⟶ 744:
2077 2 12
2101 4 2
[ <date> test ] 3 7 mnapply</
{{out}}
<pre>
Line 403 ⟶ 757:
=={{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
Line 440 ⟶ 794:
04.50 I (E-2)4.6;T "TUESDAY";R
04.60 I (E-1)4.7;T "MONDAY";R
04.70 T "SUNDAY"</
{{out}}
<pre>M= 1 D= 6 Y= 1800 DAY=MONDAY
Line 449 ⟶ 803:
M= 2 D= 12 Y= 2077 DAY=FRIDAY
M= 4 D= 2 Y= 2101 DAY=SATURDAY</pre>
=={{header|Fortran}}==
<syntaxhighlight lang="FORTRAN">
program someday ! Calculates the day that a week falls on
use datestuff
implicit none
!
! PARAMETER definitions
!
integer , parameter :: numdates = 7
!
! Local variable declarations
!
character(30) , dimension(numdates) , save :: date_string
integer , dimension(numdates) , save :: day
character(9) , dimension(numdates) , save :: days
integer :: j
integer :: k
integer , dimension(numdates) , save :: month
integer , dimension(numdates) , save :: year
!
data days/'Monday ' , 'Tuesday ' , 'Wednesday' , 'Thursday ' , &
&'Friday ' , 'Saturday ' , 'Sunday '/
data month/01 , 03 , 12 , 12 , 05 , 02 , 04/
data day/06 , 29 , 07 , 23 , 14 , 12 , 02/
data year/1800 , 1875 , 1915 , 1970 , 2043 , 2077 , 2101/
data date_string/'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)'/ !
!*Code
print *
do j = 1 , numdates
k = get_the_day(day(j) , month(j) , year(j))
print '(a, t31,1a1,1x,a)' , date_string(j) , ':' , days(k)
end do
end program someday
module datestuff
implicit none
public :: get_the_day
private :: get_anchor_day , gregorian_leap_year
!
enum , bind(c) ! Use enums to get greater clarity in the code
enumerator :: january = 1 , february , march , april , may , &
& june , july , august , september , october , november ,&
& december
end enum
enum , bind(c)
enumerator :: sunday = 0 , monday , tuesday , wednesday , &
& thursday , friday , saturday
end enum
contains
!
pure function gregorian_leap_year(year) result(answer) ! Tests if the year is a leap year
!
! Function and Dummy argument declarations
!
logical :: answer
integer , intent(in) :: year
!
answer = .false. ! Set default to not a leap year
if( mod(year , 4)==0 )answer = .true. ! Year divisible by 4 = leap year
if( mod(year , 100)==0 )then
if( mod(year , 400)==0 )then ! Year divisible by 400 = century year that is leap year
answer = .true.
else
answer = .false. ! Century years are not leap years
end if
end if
end function gregorian_leap_year
!
pure function get_anchor_day(year) result(answer) ! Returns Anchor Days in doomsday calculation
!Note: The days start as Monday = 1, Tuesday =2, etc until Sunday = 7
!The Doomsday rule, Doomsday algorithm or Doomsday method is an algorithm of determination of the day of the week for a given date.
!It provides a perpetual calendar because the Gregorian calendar moves in cycles of 400 years.
!It takes advantage of each year having a certain day of the week upon which certain easy-to-remember dates,
!called the doomsdays, fall; for example, the last day of February, 4/4, 6/6, 8/8, 10/10, and 12/12 all occur
! on the same day of the week in any year. Applying the Doomsday algorithm involves three steps: Determination of the anchor day
! for the century, calculation of the anchor day for the year from the one for the century, and selection of the closest date
! out of those that always fall on the doomsday, e.g., 4/4 and 6/6, and count of the number of days (modulo 7) between that
! date and the date in question to arrive at the day of the week. The technique applies to both the Gregorian calendar and the
! Julian calendar, although their doomsdays are usually different days of the week.
!
! Function and Dummy argument declarations
!
integer :: answer
integer , intent(in) :: year
!
! Local variable declarations
!
integer :: diffyear
integer :: div12
integer :: numyears
integer :: temp1
!
! End of declarations
!*Code
numyears = mod(year , 100) ! Get number of years greater than century
temp1 = year - numyears ! Turn into a century year
temp1 = mod(temp1 , 400) ! Now mod 400 to get base year for anchor day
select case(temp1) ! Select the base day
case(0)
answer = tuesday
case(100)
answer = sunday
case(200)
answer = friday
case(300)
answer = wednesday
case default ! Anything else is an error
ERROR Stop 'Bad Anchor Day' ! Finish with error
end select
!
!Calculate the doomsday of any given year
!
div12 = int(numyears/12) ! Get number of times 12 goes into year
temp1 = mod(numyears , 12) ! Get the remainer
diffyear = int(temp1/4) ! Div 4 (magic algorithm)
answer = diffyear + div12 + answer + temp1
answer = mod(answer , 7)
end function get_anchor_day ! Note: The days start as Sunday = 0, Monday = 1, Tuesday =2, etc until Saturda
!
pure function get_the_day(day , month , year) result(answer)
! Note: The days start as Sunday = 0, Monday = 1, Tuesday =2, etc until Saturday = 6
!
! Function and Dummy argument declarations
integer :: answer
integer , intent(in) :: day
integer , intent(in) :: month
integer , intent(in) :: year
!
! Local variable declarations
integer :: closest
integer :: doomsday
integer :: temp1
integer :: temp2
integer :: up_or_down
!
! End of declarations
!
! There are doomsdays in every month, so we know what month it is ...
! We need to find the doomsday in the relevant month
select case(month) ! Scratch Variables
case(january)
closest = merge(4,3,gregorian_leap_year(year)) ! Use merge as a ternary
case(february)
closest = merge(29,28,gregorian_leap_year(year)) ! Use merge as a ternary
case(march)
closest = 7
case(april)
closest = 4
case(may)
closest = 9
case(june)
closest = 6
case(july)
temp1 = abs(4 - day)
temp2 = abs(11 - day)
closest = merge(4,11,temp1<temp2) ! Use merge as a ternary
case(august)
closest = 8
case(september)
closest = 5
case(october)
temp1 = abs(10 - day)
temp2 = abs(31 - day)
closest = merge(10,31,temp1<temp2) ! Use merge as a ternary
case(november)
closest = 7
case(december)
closest = 12
case default
ERROR Stop 'Error in get the day' ! Stop on error
end select
!
! Ok now we get the doomsday in question - i.e. Monday, Tuesday for this year
doomsday = get_anchor_day(year) ! Get this years doomsday
! If closest day is less we need to count down, if it is bigger we count up
if( closest>day )then
up_or_down = -7
else if( closest<day )then
up_or_down = 7
else
up_or_down = 0 ! The days are equal. Set to zero so no counting needed
end if
temp1 = closest ! Set temp var to closest doomsday
if( up_or_down>0 )then
do while ( temp1<=day )
temp2 = temp1
temp1 = temp1 + up_or_down ! Count in sevens to the final
end do
temp1 = day - temp2
temp1 = (doomsday + 7) + temp1
else if( up_or_down<0 )then
do while ( temp1>=day )
temp2 = temp1
temp1 = temp1 + up_or_down ! Count in sevens to the final
end do
temp1 = temp2 - day ! See how far away I am from this day
temp1 = (doomsday + 7) - temp1 ! Subtract the difference in days from the known doomsday
else
temp1 = doomsday ! It fell on the doomsday
end if
answer = mod(temp1 , 7) ! Turn Sundays into Zeros
end function get_the_day
!
end module datestuff
!
</syntaxhighlight>
{{out}}
<pre>
1800-01-06 (January 6, 1800) : Monday
1875-03-29 (March 29, 1875) : Monday
1915-12-07 (December 7, 1915) : Tuesday
1970-12-23 (December 23, 1970): Wednesday
2043-05-14 (May 14, 2043) : Thursday
2077-02-12 (February 12, 2077): Friday
2101-04-02 (April 2, 2101) : Saturday
</pre>
=={{header|FreeBASIC}}==
<syntaxhighlight lang="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 )</syntaxhighlight>
{{out}}
<pre>Monday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday</pre>
=={{header|FutureBasic}}==
<syntaxhighlight lang="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
</syntaxhighlight>
{{output}}
<pre>
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
</pre>
=={{header|Go}}==
{{trans|Wren}}
<syntaxhighlight lang="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])
}
}</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Haskell}}==
<
data Date = Date {year :: Int, month :: Int, day :: Int}
instance Show Date where
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
dooms =
(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 = 2101, month = 4, day = 2}
]
dateAndDay :: Date -> String
Line 486 ⟶ 1,222:
main :: IO ()
main = putStr $ unlines $ map dateAndDay dates
{{out}}
<pre>
1970-12
=={{header|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
<syntaxhighlight lang="j"> 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</syntaxhighlight>
Also note that adding 7 is like adding 0 in modulo 7 arithmetic.
Thus:
<syntaxhighlight lang="j">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'
}}</syntaxhighlight>
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).
<syntaxhighlight lang="j">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'
}}</syntaxhighlight>
These two implementations produce equivalent results.
Task examples:
<syntaxhighlight lang="j"> 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</syntaxhighlight>
=={{header|Java}}==
<
public static void main(String[] args) {
final Date[] dates = {
Line 552 ⟶ 1,336:
return weekdays[(doom + day - anchor + 7) % 7];
}
}</
{{out}}
<pre>01/06/1800: Monday
Line 561 ⟶ 1,345:
02/12/2077: Friday
04/02/2101: Saturday</pre>
=={{header|jq}}==
{{trans|Julia}}
{{works with|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.
<syntaxhighlight lang="jq">
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))</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Julia}}==
<
export get_weekday
Line 602 ⟶ 1,444:
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
Line 612 ⟶ 1,454:
April 2, 2101 will be on a Saturday
</pre>
=={{header|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.
<syntaxhighlight lang="nim">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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Perl}}==
{{trans|Raku}}
<
use strict;
Line 635 ⟶ 1,513:
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>
Line 648 ⟶ 1,526:
=={{header|Phix}}==
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
Line 686 ⟶ 1,564:
<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>
Line 706 ⟶ 1,584:
=={{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;
Line 787 ⟶ 1,665:
END;
CALL EXIT;
EOF</
{{out}}
<pre>1/6/1800: MONDAY
Line 798 ⟶ 1,676:
=={{header|Python}}==
<
from calendar import isleap
Line 826 ⟶ 1,704:
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)))</
{{out}}
<pre>January 06, 1800 was a Monday
Line 838 ⟶ 1,716:
=={{header|Raku}}==
<syntaxhighlight lang="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];
Line 862 ⟶ 1,740:
say "Builtin - $_ is a: ", @dow[Date.new($_).day-of-week];
say '';
}</
{{out}}
<pre>Conway - 1800-01-06 is a: Monday
Line 887 ⟶ 1,765:
=={{header|REXX}}==
<
parse arg $ /*obtain optional arguments from the CL*/
if $='' | $="," then $= , /*Not specified? Then use the default.*/
Line 904 ⟶ 1,782:
/*──────────────────────────────────────────────────────────────────────────────────────*/
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)</
{{out|output|text= when using the default inputs:}}
<pre>
Line 914 ⟶ 1,792:
04/02/2077 falls on Friday
04/02/2101 falls on Saturday
</pre>
=={{header|RPL}}==
{{works with|HP|48G}}
« IP LASTARG FP 100 * IP LASTARG FP 10000 *
→ d m y
« y DUP 100 MOD 4 400 IFTE MOD
{ 3 7 } { 4 1 } IFTE { 7 4 2 6 4 1 5 3 7 5 } + +
m GET d SWAP - 7 MOD
2 5 y 4 MOD * + 4 y 100 MOD * + 6 y 400 MOD * + 7 MOD
+ 7 MOD
{ "Sunday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Satursday" }
SWAP 1 + GET
» » '<span style="color:blue">WKDAY</span>' STO
{ 6.011800 29.031875 7.121915 23.12197 14.052043 12.022077 2.042101 } 1 « <span style="color:blue">WKDAY</span> » DOLIST
{{out}}
<pre>
1: { "Monday" "Monday" "Tuesday" "Wednesday" "Thursday" "Friday" "Satursday" }
</pre>
=={{header|Rust}}==
<syntaxhighlight 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);
}</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Scala}}==
{{trans|Java}}
<syntaxhighlight lang="Scala">
object Doom extends App {
val dates = Array(
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)
)
dates.foreach(d => println(s"${d.format}: ${d.weekday}"))
}
class Date(val year: Int, val month: Int, val day: Int) {
import Date._
def isLeapYear: Boolean = year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)
def format: String = f"$month%02d/$day%02d/$year%04d"
def weekday: String = {
val c = year / 100
val r = year % 100
val s = r / 12
val t = r % 12
val cAnchor = (5 * (c % 4) + 2) % 7
val doom = (s + t + t / 4 + cAnchor) % 7
val anchor = if (isLeapYear) leapdoom(month - 1) else normdoom(month - 1)
weekdays((doom + day - anchor + 7) % 7)
}
}
object Date {
private val leapdoom = Array(4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5)
private val normdoom = Array(3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5)
val weekdays = Array("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday")
}
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|uBasic/4tH}}==
{{trans|FreeBASIC}}
<syntaxhighlight lang="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 "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "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))</syntaxhighlight>
{{Out}}
<pre>Monday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
0 OK, 0:711</pre>
=={{header|UNIX Shell}}==
{{trans|raku}}
{{works with|Bash}}
<syntaxhighlight lang=bash>#!/usr/bin/env bash
day-of-the-week()
if [[ "$1" =~ ([0-9]{4})-([0-9]{2})-([0-9]{2}) ]]
then
local -ra names=({Sun,Mon,Tues,Wednes,Thurs,Fri,Satur}day) doomsday=({37,41}7426415375)
local -i i c s t a b
local -i {year,month,day}=${BASH_REMATCH[++i]}
echo ${names[
c=year/100,
s=(year%100)/12,
t=(year % 100) % 12,
a=(5*(c%4)+2) % 7,
b=(s + t + (t / 4) + a ) % 7,
(b + day - ${doomsday[(year%4 == 0) && ((year%100) || (year%400 == 0))]:month-1:1} + 7) % 7
]}
else return 1
fi
for date in 1800-01-06 1875-03-29 1915-12-07 1970-12-23 2043-05-14 2077-02-12 2101-04-02
do day-of-the-week "$date"
done</syntaxhighlight>
{{out}}
<pre>Monday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
</pre>
=={{header|V (Vlang)}}==
{{trans|Go}}
<syntaxhighlight lang="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)
}
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
Line 919 ⟶ 2,075:
{{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"]
Line 956 ⟶ 2,112:
var d = Date.parse(date, Date.isoDate)
System.print("%(date) -> %(d.weekDay)")
}</
{{out}}
Line 977 ⟶ 2,133:
2077-02-12 -> Friday
2101-04-02 -> Saturday
</pre>
=={{header|XPL0}}==
{{trans|ALGOL W}}
<syntaxhighlight lang "XPL0">
\Finds the day of the week of a date 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.
function DOW ( CCYY, MM, DD );
integer CCYY, MM, DD;
integer Doomsday, AnchorDay, LeapYear, Dooms;
begin
Doomsday := rem(( \Tuesday \2
+ 5 * rem( CCYY/4 )
+ 4 * rem( CCYY/100 )
+ 6 * rem( CCYY/400 )
) / 7);
LeapYear := rem(CCYY/4) = 0 and ( rem(CCYY/100) # 0 or rem(CCYY/ 400) = 0 );
Dooms := [[4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5],
[3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]];
AnchorDay := Dooms(if LeapYear then 0 else 1, MM-1);
return rem( ( Doomsday + ( DD - AnchorDay ) + 7 ) / 7);
end; \DOW
\Prints a test date and its day of the week
procedure TestDOW ( CCYY, MM, DD );
integer CCYY, MM, DD;
integer DayName;
begin
DayName := ["Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday",
"Saturday"];
IntOut(0, CCYY);
ChOut(0, ^-); IntOut(0, MM/10); IntOut(0, rem(0));
ChOut(0, ^-); IntOut(0, DD/10); IntOut(0, rem(0));
Text(0, ": "); Text(0, DayName( DOW( CCYY, MM, DD ) ));
CrLf(0);
end;
begin \task test cases
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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Yabasic}}==
{{trans|FreeBASIC}}
<syntaxhighlight lang="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"</syntaxhighlight>
{{out}}
<pre>
Igual que la entrada de FreeBASIC.
</pre>
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