Day of the week
A company decides that whenever Xmas falls on a Sunday they will give their workers all extra paid holidays so that, together with any public holidays, workers will not have to work the following week (between the 25th of December and the first of January).
You are encouraged to solve this task according to the task description, using any language you may know.
In what years between 2008 and 2121 will the 25th of December be a Sunday?
Using any standard date handling libraries of your programming language; compare the dates calculated with the output of other languages to discover any anomalies in the handling of dates which may be due to, for example, overflow in types used to represent dates/times similar to y2k type problems.
ABAP
<lang ABAP>report zday_of_week data: lv_start type i value 2007,
lv_n type i value 114,
lv_date type sy-datum,
lv_weekday type string,
lv_day type c,
lv_year type n length 4.
write 'December 25 is a Sunday in: '. do lv_n times.
lv_year = lv_start + sy-index. concatenate lv_year '12' '25' into lv_date. call function 'DATE_COMPUTE_DAY' exporting date = lv_date importing day = lv_day.
select single langt from t246 into lv_weekday
where sprsl = sy-langu and
wotnr = lv_day.
if lv_weekday eq 'Sunday'.
write / lv_year.
endif.
enddo. </lang> Output:
December 25 is a Sunday in: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Ada
<lang ada>with Ada.Calendar.Formatting; use Ada.Calendar.Formatting; with Ada.Text_IO; use Ada.Text_IO;
procedure Yuletide is begin
for Year in Ada.Calendar.Year_Number loop -- 1901..2399
if Day_Of_Week (Time_Of (Year, 12, 25)) = Sunday then
Put_Line (Image (Time_Of (Year, 12, 25)));
end if;
end loop;
end Yuletide;</lang> Sample output:
1904-12-25 00:00:00 1910-12-25 00:00:00 1921-12-25 00:00:00 1927-12-25 00:00:00 1932-12-25 00:00:00 1938-12-25 00:00:00 1949-12-25 00:00:00 1955-12-25 00:00:00 1960-12-25 00:00:00 1966-12-25 00:00:00 1977-12-25 00:00:00 1983-12-25 00:00:00 1988-12-25 00:00:00 1994-12-25 00:00:00 2005-12-25 00:00:00 2011-12-25 00:00:00 2016-12-25 00:00:00 2022-12-25 00:00:00 2033-12-25 00:00:00 2039-12-25 00:00:00 2044-12-25 00:00:00 2050-12-25 00:00:00 2061-12-25 00:00:00 2067-12-25 00:00:00 2072-12-25 00:00:00 2078-12-25 00:00:00 2089-12-25 00:00:00 2095-12-25 00:00:00 2101-12-25 00:00:00 2107-12-25 00:00:00 2112-12-25 00:00:00 2118-12-25 00:00:00 2129-12-25 00:00:00 2135-12-25 00:00:00 2140-12-25 00:00:00 2146-12-25 00:00:00 2157-12-25 00:00:00 2163-12-25 00:00:00 2168-12-25 00:00:00 2174-12-25 00:00:00 2185-12-25 00:00:00 2191-12-25 00:00:00 2196-12-25 00:00:00 2203-12-25 00:00:00 2208-12-25 00:00:00 2214-12-25 00:00:00 2225-12-25 00:00:00 2231-12-25 00:00:00 2236-12-25 00:00:00 2242-12-25 00:00:00 2253-12-25 00:00:00 2259-12-25 00:00:00 2264-12-25 00:00:00 2270-12-25 00:00:00 2281-12-25 00:00:00 2287-12-25 00:00:00 2292-12-25 00:00:00 2298-12-25 00:00:00 2304-12-25 00:00:00 2310-12-25 00:00:00 2321-12-25 00:00:00 2327-12-25 00:00:00 2332-12-25 00:00:00 2338-12-25 00:00:00 2349-12-25 00:00:00 2355-12-25 00:00:00 2360-12-25 00:00:00 2366-12-25 00:00:00 2377-12-25 00:00:00 2383-12-25 00:00:00 2388-12-25 00:00:00 2394-12-25 00:00:00
ALGOL 68
<lang algol68># example from: http://www.xs4all.nl/~jmvdveer/algol.html - GPL # INT sun=0 # , mon=1, tue=2, wed=3, thu=4, fri=5, sat=6 #;
PROC day of week = (INT year, month, day) INT: (
# Day of the week by Zeller’s Congruence algorithm from 1887 # INT y := year, m := month, d := day, c; IF m <= 2 THEN m +:= 12; y -:= 1 FI; c := y OVER 100; y %*:= 100; (d - 1 + ((m + 1) * 26) OVER 10 + y + y OVER 4 + c OVER 4 - 2 * c) MOD 7
);
test:(
print("December 25th is a Sunday in:");
FOR year FROM 2008 TO 2121 DO
INT wd = day of week(year, 12, 25);
IF wd = sun THEN print(whole(year,-5)) FI
OD;
new line(stand out)
) </lang> Output:
December 25th is a Sunday in: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
AppleScript
<lang applescript>set ChristmasSundays to {} set Christmas to (current date) set month of Christmas to December set day of Christmas to 25 repeat with year from 2008 to 2121 set year of Christmas to year if weekday of Christmas is Sunday then set end of ChristmasSundays to year end repeat ChristmasSundays</lang>
AutoHotkey
<lang autohotkey>year = 2008 stop = 2121
While year <= stop {
FormatTime, day,% year 1225, dddd If day = Sunday out .= year "`n" year++
} MsgBox,% out</lang>
AutoIt
<lang AutoIt>#include <date.au3> Const $iSunday = 1 For $iYear = 2008 To 2121 Step 1
If $iSunday = _DateToDayOfWeek($iYear, 12, 25) Then
ConsoleWrite(StringFormat($iYear & "\n"))
EndIf
Next</lang>
AWK
<lang AWK>
- syntax: GAWK -f DAY_OF_THE_WEEK.AWK
- runtime does not support years > 2037 on my 32-bit Windows XP O/S
BEGIN {
for (i=2008; i<=2121; i++) {
x = strftime("%Y/%m/%d %a",mktime(sprintf("%d 12 25 0 0 0",i)))
if (x ~ /Sun/) { print(x) }
}
} </lang>
BBC BASIC
<lang bbcbasic> INSTALL @lib$+"DATELIB"
FOR year% = 2008 TO 2121
IF FN_dow(FN_mjd(25, 12, year%)) = 0 THEN
PRINT "Christmas Day is a Sunday in "; year%
ENDIF
NEXT</lang>
bc
Because bc has no date library, this program uses Zeller's rule, also known as Zeller's congruence, to calculate day of week.
<lang bc>scale = 0
/*
* Returns day of week (0 to 6) for year, month m, day d in proleptic * Gregorian calendar. Sunday is 0. Assumes y >= 1, scale = 0. */
define w(y, m, d) { auto a
/* Calculate Zeller's congruence. */ a = (14 - m) / 12 m += 12 * a y -= a return ((d + (13 * m + 8) / 5 + \ y + y / 4 - y / 100 + y / 400) % 7) }
for (y = 2008; y <= 2121; y++) { /* If December 25 is a Sunday, print year. */ if (w(y, 12, 25) == 0) y } quit</lang>
C
Because of problems with various C libraries (such as time_t overflowing during 2038, or strptime() or mktime() not filling in tm_wday), this program uses Zeller's Rule to calculate day of week.
<lang c>#include <stdio.h>
/* Calculate day of week in proleptic Gregorian calendar. Sunday == 0. */ int wday(int year, int month, int day) { int adjustment, mm, yy;
adjustment = (14 - month) / 12; mm = month + 12 * adjustment - 2; yy = year - adjustment; return (day + (13 * mm - 1) / 5 + yy + yy / 4 - yy / 100 + yy / 400) % 7; }
int main() { int y;
for (y = 2008; y <= 2121; y++) { if (wday(y, 12, 25) == 0) printf("%04d-12-25\n", y); }
return 0; }</lang>
C++
<lang cpp>#include <boost/date_time/gregorian/gregorian.hpp>
- include <iostream>
int main( ) {
using namespace boost::gregorian ;
std::cout
<< "Yuletide holidays must be allowed in the following years:\n" ;
for ( int i = 2008 ; i < 2121 ; i++ ) {
greg_year gy = i ;
date d ( gy, Dec , 25 ) ;
if ( d.day_of_week( ) == Sunday ) {
std::cout << i << std::endl ;
} } std::cout << "\n" ; return 0 ;
}</lang> This produces the following output:
Yuletide holidays must be allowed in the following years: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
C#
<lang csharp>using System;
class Program {
static void Main(string[] args)
{
for (int i = 2008; i <= 2121; i++)
{
DateTime date = new DateTime(i, 12, 25);
if (date.DayOfWeek == DayOfWeek.Sunday)
{
Console.WriteLine(date.ToString("dd MMM yyyy"));
}
}
}
}</lang>
Using LINQ:
<lang csharp>using System; using System.Linq;
class Program {
static void Main(string[] args)
{
string[] days = (from day in
(from year in Enumerable.Range(2008, 2121 - 2007)
select new DateTime(year, 12, 25))
where day.DayOfWeek == DayOfWeek.Sunday
select day.ToString("dd MMM yyyy")).ToArray();
foreach (string day in days) Console.WriteLine(day); }
}</lang>
This looks better:
<lang csharp>using System; using System.Linq;
class Program {
static void Main(string[] args)
{
string[] days = Enumerable.Range(2008, 2121 - 2007)
.Select(year => new DateTime(year, 12, 25))
.Where(day => day.DayOfWeek == DayOfWeek.Sunday)
.Select(day => day.ToString("dd MMM yyyy")).ToArray();
foreach (string day in days) Console.WriteLine(day); }
}</lang>
Lambda expressions FTW:
<lang csharp>using System; using System.Linq;
class Program {
static void Main(string[] args)
{
Enumerable.Range(2008, 113).ToList()
.ConvertAll(ent => new DateTime(ent, 12, 25))
.Where(ent => ent.DayOfWeek.Equals(DayOfWeek.Sunday)).ToList()
.ForEach(ent => Console.WriteLine(ent.ToString("dd MMM yyyy")));
}
}</lang>
25 Dec 2011 25 Dec 2016 25 Dec 2022 25 Dec 2033 25 Dec 2039 25 Dec 2044 25 Dec 2050 25 Dec 2061 25 Dec 2067 25 Dec 2072 25 Dec 2078 25 Dec 2089 25 Dec 2095 25 Dec 2101 25 Dec 2107 25 Dec 2112 25 Dec 2118
Clojure
Utilizing Java interop
<lang clojure> (import '(java.util GregorianCalendar)) (defn yuletide [start end] (filter #(= (. (new GregorianCalendar % (. GregorianCalendar DECEMBER) 25) get (. GregorianCalendar DAY_OF_WEEK)) (. GregorianCalendar SUNDAY)) (range start (inc end))))
(yuletide 2008 2121) </lang>
(2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118)
CoffeeScript
<lang coffeescript> december = 11 # gotta love Date APIs :) sunday = 0 for year in [2008..2121]
xmas = new Date year, december, 25 console.log year if xmas.getDay() is sunday
</lang>
one-liner:
<lang coffeescript>console.log year for year in [2008...2121] when new Date(year, 11, 25).getDay() is 0</lang>
output
<lang> 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118 </lang>
ColdFusion
<lang ColdFusion> <cfloop from = "2008" to = "2121" index = "i">
<cfset myDate = createDate(i, 12, 25) />
<cfif dayOfWeek(myDate) eq 1>
December 25th falls on a Sunday in <cfoutput>#i#</cfoutput>
</cfif>
</cfloop> </lang>
Common Lisp
<lang lisp>(loop for year from 2008 upto 2121
when (= 6 (multiple-value-bind
(second minute hour date month year day-of-week dst-p tz)
(decode-universal-time (encode-universal-time 0 0 0 25 12 year))
day-of-week))
collect year)</lang>
D
<lang d>import std.stdio, std.datetime, std.conv, std.algorithm, std.range;
void main() {
writeln("Christmas comes on a Sunday in the years:");
foreach (year; 2008 .. 2122) {
immutable d = text(year) ~ "-Dec-25";
if (Date.fromSimpleString(d).dayOfWeek() == DayOfWeek.sun)
writeln(year);
}
}</lang>
- Output:
Christmas comes on a Sunday in the years: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Delphi
<lang delphi>procedure IsXmasSunday(fromyear, toyear: integer); var i: integer; TestDate: TDateTime; outputyears: string; begin outputyears := ;
for i:= fromyear to toyear do
begin
TestDate := EncodeDate(i,12,25);
if dayofweek(TestDate) = 1 then
begin
outputyears := outputyears + inttostr(i) + ' ';
end;
end;
form1.label1.caption := outputyears;
end;</lang>
Procedure called with year range to test and outputs a space-delimited array of years to a label. There is no error check that fromyear < toyear, but this is easily added.
output: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Erlang
<lang erlang>% Implemented by bengt kleberg -module(yuletide). -export([main/0, sunday_years/2]).
main() -> [io:fwrite("25 December ~p is Sunday~n", [X]) || X <- sunday_years(2008, 2121)].
sunday_years( Start, Stop ) -> [X || X <- lists:seq(Start, Stop), is_sunday(calendar:day_of_the_week({X, 12, 25}))].
is_sunday( 7 ) -> true; is_sunday( _ ) -> false.
</lang>
- Output:
25 December 2011 is Sunday 25 December 2016 is Sunday 25 December 2022 is Sunday 25 December 2033 is Sunday 25 December 2039 is Sunday 25 December 2044 is Sunday 25 December 2050 is Sunday 25 December 2061 is Sunday 25 December 2067 is Sunday 25 December 2072 is Sunday 25 December 2078 is Sunday 25 December 2089 is Sunday 25 December 2095 is Sunday 25 December 2101 is Sunday 25 December 2107 is Sunday 25 December 2112 is Sunday 25 December 2118 is Sunday
Euphoria
<lang Euphoria> --Day of the week task from Rosetta Code wiki --User:Lnettnay
--In what years between 2008 and 2121 will the 25th of December be a Sunday
include std/datetime.e
datetime dt
for year = 2008 to 2121 do
dt = new(year, 12, 25)
if weeks_day(dt) = 1 then -- Sunday = 1
? year
end if
end for </lang> Output
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Factor
<lang factor>USING: calendar math.ranges prettyprint sequences ; 2008 2121 [a,b] [ 12 25 <date> sunday? ] filter .</lang>
FBSL
<lang qbasic>#APPTYPE CONSOLE
'In what years between 2008 and 2121 will the 25th of December be a Sunday? dim date as integer, dayname as string for dim year = 2008 to 2121 date = year * 10000 + 1225 dayname = dateconv(date,"dddd") if dayname = "Sunday" then print "Christmas Day is on a Sunday in ", year end if next PAUSE </lang>
Forth
Forth has only TIME&DATE, which does not give day of week. Many public Forth Julian date calculators had year-2100 problems, but this algorithm works well. <lang forth> \ Zeller's Congruence
- weekday ( d m y -- wd) \ 1 mon..7 sun
over 3 < if 1- swap 12 + swap then 100 /mod dup 4 / swap 2* - swap dup 4 / + + swap 1+ 13 5 */ + + ( in zeller 0=sat, so -2 to 0= mon, then mod, then 1+ for 1=mon) 2- 7 mod 1+ ;
- yuletide
." December 25 is Sunday in " 2122 2008 do 25 12 i weekday 7 = if i . then loop cr ;
</lang>
<lang forth> cr yuletide December 25 is Sunday in 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
ok
</lang>
To show year-2100 problems with SwiftForth's provided Modified Julian Day support:
<lang forth>: yuletide
." December 25 is Sunday in " 2122 2008 do 25 12 i d/m/y 7 mod 0= if i . then loop cr ;
cr yuletide December 25 is Sunday in 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2100 2106 2117</lang>
In 4tH a library is available which provides the right answer:
<lang forth>include lib/time.4th
- yuletide
." December 25 is Sunday in " 2122 2008 do 25 12 i weekday 6 = if i . then loop cr ;
cr yuletide</lang>
The code is derived from "Collected Algorithms from ACM", Volume 1 Algorithms 1-220.
Fortran
Based on Forth example <lang fortran>PROGRAM YULETIDE
IMPLICIT NONE
INTEGER :: day, year
WRITE(*, "(A)", ADVANCE="NO") "25th of December is a Sunday in" DO year = 2008, 2121
day = Day_of_week(25, 12, year) IF (day == 1) WRITE(*, "(I5)", ADVANCE="NO") year
END DO
CONTAINS
FUNCTION Day_of_week(d, m, y)
INTEGER :: Day_of_week, j, k, mm, yy
INTEGER, INTENT(IN) :: d, m, y
mm=m
yy=y
IF(mm.le.2) THEN
mm=mm+12
yy=yy-1
END IF
j = yy / 100
k = MOD(yy, 100)
Day_of_week = MOD(d + ((mm+1)*26)/10 + k + k/4 + j/4 + 5*j, 7)
END FUNCTION Day_of_week
END PROGRAM YULETIDE</lang> Output
25th of December is a Sunday in 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
GAP
<lang gap>Filtered([2008 .. 2121], y -> WeekDay([25, 12, y]) = "Sun");
- [ 2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118 ]
- A possible implementation of WeekDayAlt
WeekDayAlt := function(args)
local d, m, y, k;
d := args[1];
m := args[2];
y := args[3];
if m < 3 then
m := m + 12;
y := y - 1;
fi;
k := 1 + RemInt(d + QuoInt((m + 1)*26, 10) + y + QuoInt(y, 4)
+ 6*QuoInt(y, 100) + QuoInt(y, 400) + 5, 7);
return days[k];
end;
Filtered([2008 .. 2121], y -> WeekDayAlt([25, 12, y]) = "Sun");
- [ 2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118 ]</lang>
Go
<lang go>package main
import "fmt" import "time"
func main() {
for year := 2008; year <= 2121; year++ {
if time.Date(year, 12, 25, 0, 0, 0, 0, time.UTC).Weekday() ==
time.Sunday {
fmt.Printf("25 December %d is Sunday\n", year)
}
}
}</lang> Output:
25 December 2011 is Sunday 25 December 2016 is Sunday 25 December 2022 is Sunday 25 December 2033 is Sunday 25 December 2039 is Sunday 25 December 2044 is Sunday 25 December 2050 is Sunday 25 December 2061 is Sunday 25 December 2067 is Sunday 25 December 2072 is Sunday 25 December 2078 is Sunday 25 December 2089 is Sunday 25 December 2095 is Sunday 25 December 2101 is Sunday 25 December 2107 is Sunday 25 December 2112 is Sunday 25 December 2118 is Sunday
Groovy
Solution: <lang groovy>def yuletide = { start, stop -> (start..stop).findAll { Date.parse("yyyy-MM-dd", "${it}-12-25").format("EEE") == "Sun" } }</lang>
Test program: <lang groovy>println yuletide(2008, 2121)</lang>
Output:
[2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118]
Haskell
Using the time library: <lang haskell>import Data.Time import Data.Time.Calendar.WeekDate
isXmasSunday year = wday == 7
where (_,_,wday) = toWeekDate $ fromGregorian year 12 25
main = mapM_ putStrLn ["25 December " ++ show year ++ " is Sunday"
| year <- [2008..2121], isXmasSunday year]</lang>
Output:
25 December 2011 is Sunday 25 December 2016 is Sunday 25 December 2022 is Sunday 25 December 2033 is Sunday 25 December 2039 is Sunday 25 December 2044 is Sunday 25 December 2050 is Sunday 25 December 2061 is Sunday 25 December 2067 is Sunday 25 December 2072 is Sunday 25 December 2078 is Sunday 25 December 2089 is Sunday 25 December 2095 is Sunday 25 December 2101 is Sunday 25 December 2107 is Sunday 25 December 2112 is Sunday 25 December 2118 is Sunday
The built-in System.Time module overflows at the Unix epoch in 2038: <lang haskell>import System.Time
isXmasSunday year = ctWDay cal == Sunday
where cal = toUTCTime $ toClockTime cal'
cal' = CalendarTime {
ctYear = year,
ctMonth = December,
ctDay = 25,
ctHour = 0,
ctMin = 0,
ctSec = 0,
ctPicosec = 0,
ctWDay = Friday,
ctYDay = 0,
ctTZName = "",
ctTZ = 0,
ctIsDST = False
}
main = mapM_ putStrLn ["25 December " ++ show year ++ " is Sunday"
| year <- [2008..2121], isXmasSunday year]</lang>
Output on 32-bit machine:
25 December 2011 is Sunday 25 December 2016 is Sunday 25 December 2022 is Sunday 25 December 2033 is Sunday *** Exception: user error (Time.toClockTime: invalid input)
HicEst
<lang HicEst>DO year = 1, 1000000
TIME(Year=year, MOnth=12, Day=25, TO, WeekDay=weekday) IF( weekday == 7) WRITE(StatusBar) year
ENDDO
END</lang>
No anomalies detected for the first million years :-)
Dec 25 = Sunday in
5 ... 2011 2016 2022 2033 2039 2044 2050 2061 2067
2072 2078 2089 2095 2101 2107 2112 2118 ... 999994
Icon and Unicon
<lang Icon>link datetime
procedure main() writes("December 25th is a Sunday in: ") every writes((dayoweek(25,12,y := 2008 to 2122)=="Sunday",y)," ") end</lang>
<lang Icon>procedure dayoweek(day, month, year) #: day of the week
static d_code, c_code, m_code, ml_code, y, C, M, Y
initial {
d_code := ["Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday"]
c_code := table()
c_code[16] := c_code[20] := 0
c_code[17] := c_code[21] := 6
c_code[18] := c_code[22] := 4
c_code[19] := c_code[23] := 2
m_code := table()
m_code[1] := m_code["January"] := 1
m_code[2] := m_code["February"] := 4
m_code[3] := m_code["March"] := 4
m_code[4] := m_code["April"] := 0
m_code[5] := m_code["May"] := 2
m_code[6] := m_code["June"] := 5
m_code[7] := m_code["July"] := 0
m_code[8] := m_code["August"] := 3
m_code[9] := m_code["September"] := 6
m_code[10] := m_code["October"] := 1
m_code[11] := m_code["November"] := 4
m_code[12] := m_code["December"] := 6
ml_code := copy(m_code)
ml_code[1] := ml_code["January"] := 0
ml_code[2] := ml_code["February"] := 3
}
if year < 1600 then stop("*** can't compute day of week that far back")
if year > 2299 then stop("*** can't compute day of week that far ahead")
C := c_code[(year / 100) + 1] y := year % 100 Y := (y / 12) + (y % 12) + ((y % 12) / 4) month := integer(month) M := if (year % 4) = 0 then ml_code[month] else m_code[month]
return d_code[(C + Y + M + day) % 7 + 1]
end</lang>
Sample Output:
December 25th is a Sunday in: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
J
<lang j> load 'dates' NB. provides verb 'weekday'
xmasSunday=: #~ 0 = [: weekday 12 25 ,~"1 0 ] NB. returns years where 25 Dec is a Sunday xmasSunday 2008 + i.114 NB. check years from 2008 to 2121
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118</lang>
Java
<lang java>import java.util.Calendar; import java.util.Date; import java.util.GregorianCalendar;
public class Yuletide{ public static void main(String[] args) { for(int i = 2008;i<=2121;i++){ Calendar cal = new GregorianCalendar(i, Calendar.DECEMBER, 25); if(cal.get(Calendar.DAY_OF_WEEK)==Calendar.SUNDAY){ System.out.println(cal.getTime()); } } } }</lang> Output:
Sun Dec 25 00:00:00 CST 2011 Sun Dec 25 00:00:00 CST 2016 Sun Dec 25 00:00:00 CST 2022 Sun Dec 25 00:00:00 CST 2033 Sun Dec 25 00:00:00 CST 2039 Sun Dec 25 00:00:00 CST 2044 Sun Dec 25 00:00:00 CST 2050 Sun Dec 25 00:00:00 CST 2061 Sun Dec 25 00:00:00 CST 2067 Sun Dec 25 00:00:00 CST 2072 Sun Dec 25 00:00:00 CST 2078 Sun Dec 25 00:00:00 CST 2089 Sun Dec 25 00:00:00 CST 2095 Sun Dec 25 00:00:00 CST 2101 Sun Dec 25 00:00:00 CST 2107 Sun Dec 25 00:00:00 CST 2112 Sun Dec 25 00:00:00 CST 2118
JavaScript
<lang javascript>for (var year = 2008; year <= 2121; year++){
var xmas = new Date(year, 11, 25)
if ( xmas.getDay() === 0 )
console.log(year)
}</lang> output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
K
<lang K> wd:{(__jd x)!7} / Julian day count, Sun=6
y@&6={wd 1225+x*10000}'y:2008+!114
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118 </lang>
Liberty BASIC
<lang lb> count = 0
for year = 2008 to 2121
dateString$="12/25/";year
dayNumber=date$(dateString$)
if dayNumber mod 7 = 5 then
count = count + 1
print dateString$
end if
next year
print count; " years when Christmas Day falls on a Sunday" end</lang>
Lua
Library: LuaDate
<lang Lua>require("date")
for year=2008,2121 do
if date(year, 12, 25):getweekday() == 1 then
print(year)
end
end</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
M4
<lang M4>divert(-1)
define(`for',
`ifelse($#,0,``$0, `ifelse(eval($2<=$3),1, `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
dnl julian day number corresponding to December 25th of given year define(`julianxmas',
`define(`yrssince0',eval($1+4712))`'define(`noOfLpYrs',
eval((yrssince0+3)/4))`'define(`jd',
eval(365*yrssince0+noOfLpYrs-10-($1-1501)/100+($1-1201)/400+334+25-1))`'
ifelse(eval($1%4==0 && ($1%100!=0 || $1%400==0)),1,
`define(`jd',incr(jd))')`'jd')
divert
for(`yr',2008,2121,
`ifelse(eval(julianxmas(yr)%7==6),1,`yr ')')</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Mathematica
<lang Mathematica>Reap[If[DateString[{#,12,25},"DayName"]=="Sunday",Sow[#]]&/@Range[2008,2121]]2,1</lang> gives back: <lang Mathematica>{2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118}</lang>
MATLAB / Octave
<lang Matlab> t = datenum([[2008:2121]',repmat([12,25,0,0,0], 2121-2007, 1)]);
t = t(strmatch('Sunday', datestr(t,'dddd')), :);
datestr(t,'yyyy')
</lang>
Output:
ans = 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Maxima
<lang maxima>weekday(year, month, day) := block([m: month, y: year, k],
if m < 3 then (m: m + 12, y: y - 1),
k: 1 + remainder(day + quotient((m + 1)*26, 10) + y + quotient(y, 4)
+ 6*quotient(y, 100) + quotient(y, 400) + 5, 7),
['monday, 'tuesday, 'wednesday, 'thurdsday, 'friday, 'saturday, 'sunday][k]
)$
sublist(makelist(i, i, 2008, 2121),
lambda([y], weekday(y, 12, 25) = 'sunday));
/* [2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118] */</lang>
Modula-3
Modula-3 represents time using a (safe) wrapper around the C time interface. Consequently, it suffers from the same problem as C.
<lang modula3>MODULE Yule EXPORTS Main;
IMPORT IO, Fmt, Date, Time;
VAR date: Date.T;
time: Time.T;
BEGIN
FOR year := 2008 TO 2121 DO date.day := 25; date.month := Date.Month.Dec; date.year := year;
TRY
time := Date.ToTime(date);
EXCEPT
| Date.Error =>
IO.Put(Fmt.Int(year) & " is the last year we can specify\n");
EXIT;
END;
date := Date.FromTime(time);
IF date.weekDay = Date.WeekDay.Sun THEN
IO.Put("25th of December " & Fmt.Int(year) & " is Sunday\n");
END;
END;
END Yule.</lang>
Output:
25th of December 2011 is Sunday 25th of December 2016 is Sunday 25th of December 2022 is Sunday 25th of December 2033 is Sunday 2038 is the last year we can specify
МК-61/52
<lang>П9 7 П7 1 П8 НОП ИП8 2 2 - 1 0 / [x] П6 ИП9 + 1 8 9 9 - 3 6 5 , 2 5 * [x] ИП8 ИП6 1 2 * - 1 4 - 3 0 , 5 9 * [x] + 2 9 + ИП7 + П4 ИП4 7 / [x] 7 * - x=0 64 ИП9 С/П ИП9 1 + П9 БП 06</lang>
Input: РX: starting year.
Output: the year in which Christmas falls on a Sunday. For example, enter 2008, the first result: 2018 (January 7, 2018 is Sunday).
MUMPS
<lang MUMPS> DOWHOLIDAY
;In what years between 2008 and 2121 will December 25 be a Sunday? ;Uses the VA's public domain routine %DTC (Part of the Kernel) named here DIDTC NEW BDT,EDT,CHECK,CHKFOR,LIST,I,X,Y ;BDT - the beginning year to check ;EDT - the end year to check ;BDT and EDT are year offsets from the epoch date 1/1/1700 ;CHECK - the month and day to look at ;CHKFOR - what day of the week to look for ;LIST - list of years in which the condition is true ;I - the year currently being checked ;X - the date in an "internal" format, for input to DOW^DIDTC ;Y - the output from DOW^DIDTC SET BDT=308,EDT=421,CHECK="1225",CHKFOR=0,LIST="" FOR I=BDT:1:EDT SET X=I_CHECK D DOW^DIDTC SET:(Y=0) LIST=$SELECT($LENGTH(LIST):LIST_", ",1:"")_(I+1700) IF $LENGTH(LIST)=0 WRITE !,"There are no years that have Christmas on a Sunday in the given range." IF $LENGTH(LIST) WRITE !,"The following years have Christmas on a Sunday: ",LIST KILL BDT,EDT,CHECK,CHKFOR,LIST,I,X,Y QUIT
</lang>Usage:
USER>D ^DOW The following years have Christmas on a Sunday: 2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref savelog symbols nobinary
yearRanges = [int 2008, 2121] searchday = cal = Calendar
loop year = yearRanges[0] to yearRanges[1]
cal = GregorianCalendar(year, Calendar.DECEMBER, 25) dayIndex = cal.get(Calendar.DAY_OF_WEEK) if dayIndex = Calendar.SUNDAY then searchday = searchday year end year
say 'Between' yearRanges[0] 'and' yearRanges[1]', Christmas day falls on a Sunday on the following years:' searchday = searchday.strip.changestr(' ', ',') say ' 'searchday
return </lang>
- Output
Between 2008 and 2121, Christmas day falls on a Sunday on the following years: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Comparison of Some Common Day-of-Week Algorithms
The following program exercises some common "Day-0f-Week" algorithms to confirm they all arrive at the same result. <lang NetRexx>/* NetRexx */ options replace format comments java crossref savelog symbols nobinary
days = 'Monday Tuesday Wednesday Thursday Friday Saturday Sunday' yearRanges = [int 2008, 2121]
searchday = searchday['index'] = days.wordpos('Sunday') searchday[0] = 0
algorithmName = ['Java Calendar', 'Zeller[1]', 'Zeller[2]', 'Sakamoto', 'Gauss', 'Keith', 'Babwani']
loop alg = 0 to algorithmName.length - 1
sd = searchday[0] + 1
searchday[0] = sd
searchday['agorithm', sd] = algorithmName[alg]
loop year = yearRanges[0] to yearRanges[1]
select case alg
when 0 then dayIndex = getDaynumJavaLibrary(year, 12, 25)
when 1 then dayIndex = getDaynumZellersCongruenceMethod1(year, 12, 25)
when 2 then dayIndex = getDaynumZellersCongruenceMethod2(year, 12, 25)
when 3 then dayIndex = getDaynumSakamoto(year, 12, 25)
when 4 then dayIndex = getDaynumGauss(year, 12, 25)
when 5 then dayIndex = getDaynumKeith(year, 12, 25)
when 6 then dayIndex = getDaynumBabwani(year, 12, 25)
otherwise nop
end
if dayIndex = searchday['index'] then
searchday[sd] = searchday[sd] year
end year
end alg
-- display results say 'Between' yearRanges[0] 'and' yearRanges[1]', Christmas day falls on a Sunday in the following years:' loop r_ = 1 to searchday[0]
searchday[r_] = searchday[r_].strip.changestr(' ', ',')
say searchday['agorithm', r_].right(20)':' searchday[r_]
end r_
return
-- ----------------------------------------------------------------------------- method getDaynumJavaLibrary(Year = int, Month = int, Day = int, iso = Rexx 'Y') public static binary returns int
-- The day-of-week is an integer value where 1 is Sunday, 2 is Monday, ..., and 7 is Saturday -- For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use d = ((h - 1 + 6) mod 7) + 1
cal = Calendar
jmNumber = [ -
Calendar.JANUARY, Calendar.FEBRUARY, Calendar.MARCH, Calendar.APRIL -
, Calendar.MAY, Calendar.JUNE, Calendar.JULY, Calendar.AUGUST -
, Calendar.SEPTEMBER, Calendar.OCTOBER, Calendar.NOVEMBER, Calendar.DECEMBER -
]
mon = jmNumber[Month - 1] cal = GregorianCalendar(Year, mon, Day) h = cal.get(Calendar.DAY_OF_WEEK)
if 'YES'.abbrev(iso.upper, 1) then w = ((h - 1 + 6) // 7) + 1
else w = h
return w
-- ----------------------------------------------------------------------------- method getDaynumZellersCongruenceMethod1(Year = int, Month = int, Day = int, iso = Rexx 'Y') public static returns int
-- DayNum results in an integer in the range 0-6 where 0 represents Monday etc. -- For an ISO week date add 1
if Month = 1 | Month = 2 then do Month = Month + 12 Year = Year - 1 end
MonthFactor = 2 * Month + 3 * (Month + 1) % 5 YearFactor = Year + Year % 4 - Year % 100 + Year % 400 DayNum = (Day + MonthFactor + YearFactor) // 7
if 'YES'.abbrev(iso.upper, 1) then d = DayNum + 1
else d = DayNum
return d
-- ----------------------------------------------------------------------------- method getDaynumZellersCongruenceMethod2(Year = int, Month = int, Day = int, iso = Rexx 'Y') public static binary returns int
-- h is the day of the week (0 = Saturday, 1 = Sunday, 2 = Monday, ...) -- For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use d = ((h + 5) mod 7) + 1
if Month < 3 then do Month = Month + 12 Year = Year - 1 end q = Day m = Month Y = Year
h = (q + ((m + 1) * 26 % 10) + Y + (Y % 4) + 6 * (Y % 100) + (Y % 400)) // 7
if 'YES'.abbrev(iso.upper, 1) then d = ((h + 5) // 7) + 1
else d = h
return d
-- ----------------------------------------------------------------------------- method getDaynumSakamoto(y = int, m = int, d = int, iso = Rexx 'Y') public static binary returns int
-- h is the day of the week (0 = Sunday, 1 = Monday, 2 = Tuesday...) -- For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use d = ((h + 6) mod 7) + 1
t = [int 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4] y = y - (m < 3) h = (y + y % 4 - y % 100 + y % 400 + t[m - 1] + d) // 7
if 'YES'.abbrev(iso.upper, 1) then d = ((h + 6) // 7) + 1
else d = h
return d
-- ----------------------------------------------------------------------------- method getDaynumGauss(Year = int, Month = int, Day = int, iso = Rexx 'Y') public static binary returns int
-- W is week day (0 = Sunday, ..., 6 = Saturday) -- For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use d = ((h + 6) mod 7) + 1
Year = Year - (Month < 3) k = double Day C = double Year % 100 Y = double Year // 100 m = double ((Month + 9) // 12) + 1
W = modulo(int (k + Math.floor(2.6 * m - 0.2) + y + Math.floor(y / 4) + Math.floor(c / 4) - 2 * c), 7)
if 'YES'.abbrev(iso.upper, 1) then h = ((W + 6) // 7) + 1
else h = W
return h
-- ----------------------------------------------------------------------------- method getDaynumKeith(y = int, m = int, d = int, iso = Rexx 'Y') public constant binary returns int
-- W is week day (0 = Sunday, ..., 6 = Saturday) -- For an ISO week date Day-of-Week d (1 = Monday to 7 = Sunday), use d = ((h + 6) mod 7) + 1
if m < 3 then do d = d + y y = y - 1 end else do d = d + y - 2 end
h = (23 * m % 9 + d + 4 + y % 4 - y % 100 + y % 400) // 7
if 'YES'.abbrev(iso.upper, 1) then W = ((h + 6) // 7) + 1
else W = h
return W
-- ----------------------------------------------------------------------------- method getDaynumBabwani(Year = int, Month = int, Day = int, iso = Rexx 'Y') public constant binary returns int
-- return dow = Day of week: 0 = Saturday, 1 = Sunday, ... 6 = Friday -- For an ISO week date Day-of-Week W (1 = Monday to 7 = Sunday), use W = ((dow + 5) mod 7) + 1
y = Year m = Month d = Day
dow = int -- dow stands for day of week dowfg = double fmonth = int leap = int
if ((y // 100 == 0) & (y // 400 \= 0)) then -- leap function 1 for leap & 0 for non-leap leap = 0 else if (y // 4 == 0) then leap = 1 else leap = 0
fmonth = 3 + (2 - leap) * ((m + 2) % (2 * m)) + (5 * m + m % 9) % 2 -- f(m) formula fmonth = fmonth // 7 -- f(m) is brought in range of 0 to 6
century = y % 100 lastdigits = y // 100
dowfg = 1.25 * lastdigits + fmonth + d - 2 * (century // 4) -- function of weekday for Gregorian dow = int dowfg // 7 -- remainder on division by 7
if 'YES'.abbrev(iso.upper, 1) then W = ((dow + 5) // 7) + 1
else W = dow
return W
-- ----------------------------------------------------------------------------- method modulo(N = int, D = int) inheritable static binary returns int
return (D + (N // D)) // D
</lang>
- Output
Between 2008 and 2121, Christmas day falls on a Sunday in the following years:
Java Calendar: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Zeller[1]: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Zeller[2]: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Sakamoto: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Gauss: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Keith: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Babwani: 2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118
Objective-C
It should works also with Cocoa and OpenStep in general, but I can't test these.
<lang objc>#import <Foundation/Foundation.h>
int main() {
NSAutoreleasePool *pool = [[NSAutoreleasePool alloc] init];
NSUInteger i;
for(i=2008; i<2121; i++)
{
NSCalendarDate *d = [[NSCalendarDate alloc]
initWithYear: i
month: 12
day: 25
hour: 0 minute: 0 second:0
timeZone: [NSTimeZone timeZoneWithAbbreviation:@"CET"] ];
if ( [d dayOfWeek] == 0 )
{
printf("25 Dec %u is Sunday\n", i);
}
[d release];
}
[pool release];
return 0;
}</lang>
Output:
25 Dec 2011 is Sunday 25 Dec 2016 is Sunday 25 Dec 2022 is Sunday 25 Dec 2033 is Sunday 25 Dec 2039 is Sunday 25 Dec 2044 is Sunday 25 Dec 2050 is Sunday 25 Dec 2061 is Sunday 25 Dec 2067 is Sunday 25 Dec 2072 is Sunday 25 Dec 2078 is Sunday 25 Dec 2089 is Sunday 25 Dec 2095 is Sunday 25 Dec 2101 is Sunday 25 Dec 2107 is Sunday 25 Dec 2112 is Sunday 25 Dec 2118 is Sunday
OCaml
<lang ocaml>#load "unix.cma" open Unix
try
for i = 2008 to 2121 do
(* I'm lazy so we'll just borrow the current time
instead of having to set all the fields explicitly *)
let mytime = { (localtime (time ())) with
tm_year = i - 1900;
tm_mon = 11;
tm_mday = 25 } in
try
let _, mytime = mktime mytime in
if mytime.tm_wday = 0 then
Printf.printf "25 December %d is Sunday\n" i
with e ->
Printf.printf "%d is the last year we can specify\n" (i-1);
raise e
done
with _ -> ()</lang>
The output of a run on a 32 bit machine is
25 December 2011 is Sunday 25 December 2016 is Sunday 25 December 2022 is Sunday 25 December 2033 is Sunday 2037 is the last year we can specify
With a dedicated library
Unlike the previous example which only uses the OCaml standard library, here with the OCaml Calendar Library we can go until the year 2121:
<lang ocaml>open CalendarLib
let list_make_seq first last =
let rec aux i acc = if i < first then acc else aux (pred i) (i::acc) in aux last []
let print_date (year, month, day) =
Printf.printf "%d-%02d-%02d\n" year month day
let () =
let years = list_make_seq 2008 2121 in let years = List.filter (fun year -> Date.day_of_week (Date.make year 12 25) = Date.Sun) years in print_endline "December 25 is a Sunday in:"; List.iter (Printf.printf "%d\n") years</lang>
Output:
$ ocaml unix.cma str.cma -I +calendar calendarLib.cma xmas_sundays.ml December 25 is a Sunday in: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
ooRexx
<lang ooRexx> date = .datetime~new(2008, 12, 25) lastdate = .datetime~new(2121, 12, 25)
resultList = .array~new -- our collector of years
-- date objects are directly comparable loop while date <= lastdate
if date~weekday == 7 then resultList~append(date~year) -- step to the next year date = date~addYears(1)
end
say "Christmas falls on Sunday in the years" resultList~toString("Line", ", ") </lang>
Pascal
See Delphi
Perl
<lang perl>#! /usr/bin/perl -w
use Time::Local; use strict;
foreach my $i (2008 .. 2121) {
my $time = timelocal(0,0,0,25,11,$i);
my ($s,$m,$h,$md,$mon,$y,$wd,$yd,$is) = localtime($time);
if ( $wd == 0 )
{
print "25 Dec $i is Sunday\n";
}
}
exit 0;</lang>
Output:
25 Dec 2011 is Sunday 25 Dec 2016 is Sunday 25 Dec 2022 is Sunday 25 Dec 2033 is Sunday Day too big - 25195 > 24855 Sec too small - 25195 < 78352 Sec too big - 25195 > 15247 Cannot handle date (0, 0, 0, 25, 11, 2038) at ./ydate.pl line 8
Using the DateTime module from CPAN: <lang perl>#! /usr/bin/perl -w
use DateTime; use strict;
foreach my $i (2008 .. 2121) {
my $dt = DateTime->new( year => $i,
month => 12,
day => 25
);
if ( $dt->day_of_week == 7 )
{
print "25 Dec $i is Sunday\n";
}
}
exit 0;</lang> or shorter: <lang perl>#! /usr/bin/perl -w
use DateTime; use strict;
for (2008 .. 2121) {
print "25 Dec $_ is Sunday\n" if DateTime->new(year => $_, month => 12, day => 25)->day_of_week == 7;
}
exit 0;</lang> Output:
25 Dec 2011 is Sunday 25 Dec 2016 is Sunday 25 Dec 2022 is Sunday 25 Dec 2033 is Sunday 25 Dec 2039 is Sunday 25 Dec 2044 is Sunday 25 Dec 2050 is Sunday 25 Dec 2061 is Sunday 25 Dec 2067 is Sunday 25 Dec 2072 is Sunday 25 Dec 2078 is Sunday 25 Dec 2089 is Sunday 25 Dec 2095 is Sunday 25 Dec 2101 is Sunday 25 Dec 2107 is Sunday 25 Dec 2112 is Sunday 25 Dec 2118 is Sunday
Alternatively in one line using grep (read from right to left): <lang perl>#! /usr/bin/perl -w
use DateTime; use strict;
print join " ", grep { DateTime->new(year => $_, month => 12, day => 25)->day_of_week == 7 } (2008 .. 2121);
0;</lang> Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Perl 6
As Perl 5, except DateTime is built-in, so you don't need to download a module of that name:
<lang perl6>say join ' ', grep { Date.new($_, 12, 25).day-of-week == 7 }, 2008 .. 2121;</lang>
PHP
<lang php><?php for($i=2008; $i<2121; $i++) {
$datetime = new DateTime("$i-12-25 00:00:00");
if ( $datetime->format("w") == 0 )
{
echo "25 Dec $i is Sunday\n";
}
} ?></lang>
Output:
25 Dec 2011 is Sunday 25 Dec 2016 is Sunday 25 Dec 2022 is Sunday 25 Dec 2033 is Sunday 25 Dec 2039 is Sunday 25 Dec 2044 is Sunday 25 Dec 2050 is Sunday 25 Dec 2061 is Sunday 25 Dec 2067 is Sunday 25 Dec 2072 is Sunday 25 Dec 2078 is Sunday 25 Dec 2089 is Sunday 25 Dec 2095 is Sunday 25 Dec 2101 is Sunday 25 Dec 2107 is Sunday 25 Dec 2112 is Sunday 25 Dec 2118 is Sunday
PicoLisp
<lang PicoLisp>(for (Y 2008 (>= 2121 Y) (inc Y))
(when (= "Sunday" (day (date Y 12 25)))
(printsp Y) ) )</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Pike
<lang Pike>filter(Calendar.Year(2008)->range(Calendar.Year(2121))->years()->month(12)->day(25), lambda(object day){ return day->week_day()==7; })->year()->format_nice();</lang> Output:
Result: ({ /* 17 elements */
"2011",
"2016",
"2022",
"2033",
"2039",
"2044",
"2050",
"2061",
"2067",
"2072",
"2078",
"2089",
"2095",
"2101",
"2107",
"2112",
"2118"
})
PL/I
<lang PL/I> declare i picture '9999'; do i = 2008 to 2121;
if weekday(days('25Dec' || i, 'DDMmmYYYY')) = 1 then
put skip list ('Christmas day ' || i || ' is a Sunday');
end; </lang>
PowerShell
<lang powershell>2008..2121 | Where-Object { (Get-Date $_-12-25).DayOfWeek -eq "Sunday" }</lang>
Protium
<lang html><@ SAI> <@ ITEFORLI3>2121|2008| <@ LETVARCAP>Christmas Day|25-Dec-<@ SAYVALFOR>...</@></@> <@ TSTDOWVARLIT>Christmas Day|1</@> <@ IFF> <@ SAYCAP>Christmas Day <@ SAYVALFOR>...</@> is a Sunday</@><@ SAYKEY>__Newline</@> </@> </@> </@></lang>
English dialect variable-length space-padded opcodes <lang html><# suppressimplicitoutput> <# iterate foriteration literalstring3>2121|2008| <# let variable capture>Christmas Day|25-Dec-<# say value foriteration>...</#></#> <# test dayofweek variable literal>Christmas Day|1</#> <# if> <# say capture>Christmas Day <# say value foriteration>...</#> is a Sunday</#><# say keyword>__Newline</#> </#> </#>
</#></lang>
Output
Christmas Day 2011 is a Sunday Christmas Day 2016 is a Sunday Christmas Day 2022 is a Sunday Christmas Day 2033 is a Sunday Christmas Day 2039 is a Sunday Christmas Day 2044 is a Sunday Christmas Day 2050 is a Sunday Christmas Day 2061 is a Sunday Christmas Day 2067 is a Sunday Christmas Day 2072 is a Sunday Christmas Day 2078 is a Sunday Christmas Day 2089 is a Sunday Christmas Day 2095 is a Sunday Christmas Day 2101 is a Sunday Christmas Day 2107 is a Sunday Christmas Day 2112 is a Sunday Christmas Day 2118 is a Sunday
PureBasic
PureBasic's internal Date() is limited between 1970-01-01 00:00:00 and 2038-01-19 03:14:07 <lang PureBasic>For i=2008 To 2037
If DayOfWeek(Date(i,12,25,0,0,0))=0 PrintN(Str(i)) EndIf
Next</lang>
Python
<lang python>import datetime
def yuletide():
sunday = 6
days = (day.strftime('%d %b %Y') for day in (datetime.date(year, 12, 25) for year in range(2008,2122)) if day.weekday() == sunday)
print '\n'.join(days)
yuletide()</lang>Output:
25 Dec 2011 25 Dec 2016 25 Dec 2022 25 Dec 2033 25 Dec 2039 25 Dec 2044 25 Dec 2050 25 Dec 2061 25 Dec 2067 25 Dec 2072 25 Dec 2078 25 Dec 2089 25 Dec 2095 25 Dec 2101 25 Dec 2107 25 Dec 2112 25 Dec 2118
R
<lang R>years <- 2008:2121 xmas <- as.POSIXlt(paste(years, '/12/25', sep="")) years[xmas$wday==0]</lang>
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
<lang R> Also:
xmas=seq(as.Date("2008/12/25"), as.Date("2121/12/25"), by="year") as.numeric(format(xmas[weekdays(xmas)== 'Sunday'], "%Y")) </lang>
<lang r>#Still another solution, using ISOdate and weekdays (2008:2121)[weekdays(ISOdate(2008:2121, 12, 25)) == "Sunday"]
- Simply replace "Sunday" with whatever it's named in your country,
- or set locale first, with
Sys.setlocale(cat="LC_ALL", "en")</lang>
Racket
<lang Racket>
- lang racket
(require racket/date)
(define (xmas-on-sunday? year)
(zero? (date-week-day (seconds->date (find-seconds 0 0 12 25 12 year)))))
(for ([y (in-range 2008 2121)] #:when (xmas-on-sunday? y))
(displayln y))
</lang>
REBOL
<lang REBOL>REBOL [ Title: "Yuletide Holiday" Author: oofoe Date: 2009-12-07 URL: http://rosettacode.org/wiki/Yuletide_Holiday ]
for y 2008 2121 1 [ d: to-date reduce [y 12 25] if 7 = d/weekday [prin [y ""]] ]</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
REXX
using DATE weekday
The extended DATE parameters (arguments 2 and 3) are only supported by the newer REXX interpreters. <lang rexx> do year=2008 to 2121
if date('w', year'1225', 's') == 'Sunday' then say year
end</lang>
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
using DATE base
The extended DATE parameters (arguments 2 and 3) are only supported by the newer REXX interpreters. <lang rexx> do year=2008 to 2121
if date('b', year'1225', 's') // 7 == 6 then say year
end</lang>
output is the same as above
using DATE iso
The extended DATE parameters (arguments 2 and 3) are only supported by the newer REXX interpreters.
Language note: the DATE built-in function always returns the day-of-week in English, no matter what the native language is in effect.
<lang rexx>/*REXX program displays which years December 25th falls on a Sunday. */
parse arg start finish . /*get the start and finish years.*/
if start== then start=2008 /*None specified? Assume default*/
if finish== then finish=2121 /*None specified? Assume default*/
do y=start to finish /*process the years specified. */
if date('Weekday',y"-12-25",'ISO')\=='Sunday' then iterate
/* if date('w' ,y"-12-25",'i' )\== ... (same as above). */
/* option yyyy-mm-dd fmt */
say 'December 25th,' y "falls on a Sunday."
end /*y*/
/*stick a fork in it, we're done.*/</lang>
output when using the default input
December 25th, 2011 falls on a Sunday. December 25th, 2016 falls on a Sunday. December 25th, 2022 falls on a Sunday. December 25th, 2033 falls on a Sunday. December 25th, 2039 falls on a Sunday. December 25th, 2044 falls on a Sunday. December 25th, 2050 falls on a Sunday. December 25th, 2061 falls on a Sunday. December 25th, 2067 falls on a Sunday. December 25th, 2072 falls on a Sunday. December 25th, 2078 falls on a Sunday. December 25th, 2089 falls on a Sunday. December 25th, 2095 falls on a Sunday. December 25th, 2101 falls on a Sunday. December 25th, 2107 falls on a Sunday. December 25th, 2112 falls on a Sunday. December 25th, 2118 falls on a Sunday.
old school DOW
This DOW (day-of-week) version will work with any version of a REXX interpreter. <lang rexx>/*REXX program (old school) displays which years 12/25 falls on a Sunday*//
parse arg start finish . /*get the start and finish years.*/ if start== then start=2008 /*None specified? Assume default*/ if finish== then finish=2121 /*None specified? Assume default*/
do y=start to finish /*process the years specified. */
if dow(12,25,y)==1 then say 'December 25th,' y "falls on a Sunday."
end /*y*/
exit /*stick a fork in it, we're done.*/ /*─────────────────────────────────────DOW (day of week) subroutine─────*/ dow: procedure; arg m,d,y; if m<3 then do; m=m+12; y=y-1; end yL=left(y,2); yr=right(y,2); w=(d + (m+1)*26%10+yr+yr%4+yL%4+5*yL) // 7 if w==0 then w=7; return w /*Sunday=1, Monday=2, ... Saturday=7*/</lang> output using the default input
December 25th, 2011 falls on a Sunday. December 25th, 2016 falls on a Sunday. December 25th, 2022 falls on a Sunday. December 25th, 2033 falls on a Sunday. December 25th, 2039 falls on a Sunday. December 25th, 2044 falls on a Sunday. December 25th, 2050 falls on a Sunday. December 25th, 2061 falls on a Sunday. December 25th, 2067 falls on a Sunday. December 25th, 2072 falls on a Sunday. December 25th, 2078 falls on a Sunday. December 25th, 2089 falls on a Sunday. December 25th, 2095 falls on a Sunday. December 25th, 2101 falls on a Sunday. December 25th, 2107 falls on a Sunday. December 25th, 2112 falls on a Sunday. December 25th, 2118 falls on a Sunday.
Ruby
<lang ruby>require 'date'
(2008..2121).each do |year|
puts "25 Dec #{year}" if Date.new(year, 12, 25).wday == 0 # Ruby 1.9: if Date.new(year, 12, 25).sunday?
end</lang> Output:
25 Dec 2011 25 Dec 2016 25 Dec 2022 25 Dec 2033 25 Dec 2039 25 Dec 2044 25 Dec 2050 25 Dec 2061 25 Dec 2067 25 Dec 2072 25 Dec 2078 25 Dec 2089 25 Dec 2095 25 Dec 2101 25 Dec 2107 25 Dec 2112 25 Dec 2118
The Time class overflows at the Unix epoch in 2038: <lang ruby>SUNDAY = 0
(2008..2121).each do |year|
begin
day = Time.local(year, 12, 25)
puts "25 Dec #{year}" if day.wday == SUNDAY # Ruby 1.9: if day.sunday?
rescue ArgumentError
puts '%d is the last year we can specify' % (year-1)
break
end
end</lang> Output on 32-bit machine:
25 Dec 2011 25 Dec 2016 25 Dec 2022 25 Dec 2033 2037 is the last year we can specify
Run BASIC
<lang runbasic>for year = 2008 to 2121
if val(date$("12-25-";year)) mod 7 = 5 then print "For ";year;"xmas is Sunday"
next year</lang>
For 2011 xmas is Sunday For 2016 xmas is Sunday For 2022 xmas is Sunday For 2033 xmas is Sunday For 2039 xmas is Sunday For 2044 xmas is Sunday For 2050 xmas is Sunday For 2061 xmas is Sunday For 2067 xmas is Sunday For 2072 xmas is Sunday For 2078 xmas is Sunday For 2089 xmas is Sunday For 2095 xmas is Sunday For 2101 xmas is Sunday For 2107 xmas is Sunday For 2112 xmas is Sunday For 2118 xmas is Sunday
SAS
<lang sas>data _null_; do y=2008 to 2121; a=mdy(12,25,y); if weekday(a)=1 then put y; end; run;
/* 2011 2016 2022 2033 2039 2044 2050 2061 2067
2072 2078 2089 2095 2101 2107 2112 2118 */</lang>
Scala
<lang scala> import java.util.{Calendar, GregorianCalendar} import Calendar._
object DayOfTheWeek {
def main(args:Array[String]) {
for (year <- 2008 to 2121;
date <- Some(new GregorianCalendar(year, DECEMBER, 25));
if date.get(DAY_OF_WEEK) == SUNDAY) {
println(year)
}
}
} </lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Scheme
<lang scheme>(define (day-of-week year month day) (if (< month 3)
(begin (set! month (+ month 12)) (set! year (- year 1))))
(+ 1
(remainder (+ 5 day (quotient (* (+ 1 month) 13) 5)
year (quotient year 4) (* (quotient year 100) 6) (quotient year 400))
7)))
(define (task) (let loop ((y 2121) (v '())) (if (< y 2008)
v
(loop (- y 1)
(if (= 7 (day-of-week y 12 25))
(cons y v)
v)))))
(task)
- (2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118)</lang>
Seed7
The library time.s7i defines the function dayOfWeek, which returns 1 for monday, 2 for tuesday, and so on up to 7 for sunday. <lang seed7>$ include "seed7_05.s7i";
include "time.s7i";
const proc: main is func
local
var integer: year is 0;
begin
for year range 2008 to 2122 do
if dayOfWeek(date(year, 12, 25)) = 7 then
writeln("Christmas comes on a sunday in " <& year);
end if;
end for;
end func;</lang>
Output:
Christmas comes on a sunday in 2011 Christmas comes on a sunday in 2016 Christmas comes on a sunday in 2022 Christmas comes on a sunday in 2033 Christmas comes on a sunday in 2039 Christmas comes on a sunday in 2044 Christmas comes on a sunday in 2050 Christmas comes on a sunday in 2061 Christmas comes on a sunday in 2067 Christmas comes on a sunday in 2072 Christmas comes on a sunday in 2078 Christmas comes on a sunday in 2089 Christmas comes on a sunday in 2095 Christmas comes on a sunday in 2101 Christmas comes on a sunday in 2107 Christmas comes on a sunday in 2112 Christmas comes on a sunday in 2118
Smalltalk
<lang smalltalk>2008 to: 2121 do: [ :year | |date|
date := Date newDay: 25 monthIndex: 12 year: year.
date dayName = #Sunday
ifTrue: [ date displayNl ]
]</lang>
Output:
25-Dec-2011 25-Dec-2016 25-Dec-2022 25-Dec-2033 25-Dec-2039 25-Dec-2044 25-Dec-2050 25-Dec-2061 25-Dec-2067 25-Dec-2072 25-Dec-2078 25-Dec-2089 25-Dec-2095 25-Dec-2101 25-Dec-2107 25-Dec-2112 25-Dec-2118
Suneido
<lang Suneido>year = 2008 while (year <= 2121)
{
if Date('#' $ year $ '1225').WeekDay() is 0
Print(year)
++year
}</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Tcl
<lang tcl>package require Tcl 8.5
for {set y 2008} {$y <= 2121} {incr y} {
if {[clock format [clock scan "$y-12-25" -format {%Y-%m-%d}] -format %w] == 0} {
puts "xmas $y is a sunday"
}
}</lang> outputs
xmas 2011 is a sunday xmas 2016 is a sunday xmas 2022 is a sunday xmas 2033 is a sunday xmas 2039 is a sunday xmas 2044 is a sunday xmas 2050 is a sunday xmas 2061 is a sunday xmas 2067 is a sunday xmas 2072 is a sunday xmas 2078 is a sunday xmas 2089 is a sunday xmas 2095 is a sunday xmas 2101 is a sunday xmas 2107 is a sunday xmas 2112 is a sunday xmas 2118 is a sunday
TI-83 BASIC
Works with TI-84+/SE only <lang ti83b>
- For(A, 2008,2121
- dayofWk(A,12,25
- If Ans=1
- Disp A
- End
</lang> outputs
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118 Done
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT PRINT "25th of December will be a Sunday in the following years: " LOOP year=2008,2121 SET dayofweek = DATE (number,25,12,year,nummer) IF (dayofweek==7) PRINT year ENDLOOP </lang> Output:
25th of December will be a Sunday in the following years: 2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
UNIX Shell
Unix commands may use time_t to count seconds since the epoch. For systems with 32-bit time, the counter overflows during 19 January 2038. These scripts continue to 2121 and may need a system with 64-bit time, to prevent the overflow.
With GNU date
This solution uses date -d, which seems to be a GNU extension, so it only works with those systems.
<lang bash>#! /bin/bash
for (( i=2008; i<=2121; ++i )) do
date -d "$i-12-25"
done |grep Sun
exit 0</lang>
The first lines of output (from a GNU/Linux system with 32bit time_t, date version 6.9) are
<lang bash>Sun Dec 25 00:00:00 CET 2011 Sun Dec 25 00:00:00 CET 2016 Sun Dec 25 00:00:00 CET 2022 Sun Dec 25 00:00:00 CET 2033 date: invalid date `2038-12-25'</lang>
I.e., starting from year 2038, the date command (which uses the glibc library, at least on GNU systems), is not able to recognise the date as a valid one!
Different machine/OS version (64 bit time_t): This is the same command run on RedHat Linux. <lang bash>bash-3.00$ date --version date (coreutils) 5.2.1 Written by David MacKenzie.
Copyright (C) 2004 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. bash-3.00$ uname -a Linux brslln01 2.6.9-67.ELsmp #1 SMP Wed Nov 7 13:56:44 EST 2007 x86_64 x86_64 x86_64 GNU/Linux bash-3.00$ for((i=2009; i <= 2121; i++)); do date -d "$i-12-25" |egrep Sun; done Sun Dec 25 00:00:00 GMT 2011 Sun Dec 25 00:00:00 GMT 2016 Sun Dec 25 00:00:00 GMT 2022 Sun Dec 25 00:00:00 GMT 2033 Sun Dec 25 00:00:00 GMT 2039 Sun Dec 25 00:00:00 GMT 2044 Sun Dec 25 00:00:00 GMT 2050 Sun Dec 25 00:00:00 GMT 2061 Sun Dec 25 00:00:00 GMT 2067 Sun Dec 25 00:00:00 GMT 2072 Sun Dec 25 00:00:00 GMT 2078 Sun Dec 25 00:00:00 GMT 2089 Sun Dec 25 00:00:00 GMT 2095 Sun Dec 25 00:00:00 GMT 2101 Sun Dec 25 00:00:00 GMT 2107 Sun Dec 25 00:00:00 GMT 2112 Sun Dec 25 00:00:00 GMT 2118 bash-3.00$</lang>
With GNU date and GNU seq (UnixPipes)
Like the previous solution, this solution uses date -d, which seems to be a GNU extension. Output is same as previous solution.
<lang bash>seq 2008 2121 | xargs -IYEAR -n 1 date +%c -d 'Dec 25 YEAR' | grep Sun</lang>
With Unix cal
The cal command is a tradition since Version 6 AT&T UNIX. This solution assumes that cal will always output a calendar in this format.
$ cal 12 2011
December 2011
Su Mo Tu We Th Fr Sa
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30 31
This format always puts Sunday in columns 1 and 2. The solution uses tail to delete the first 2 lines (month, year, names of days), cut to extract Sunday's columns, and grep to check if "25" appears in those columns.
<lang bash>y=2008 while test $y -lt 2122; do cal 12 $y | tail +3 | cut -c1-2 | grep -Fq 25 && echo 25 Dec $y y=`expr $y + 1` done</lang>
Running this script with OpenBSD, the output is identical to the C# program. OpenBSD cal accepts any year from 1 to 9999, so 2008 to 2122 is well within range.
With zsh
<lang bash>zmodload zsh/datetime for (( year = 2010; year <= 2121; year++ ));
if $(strftime '%A' $(strftime -r '%F' $year-12-25)) == Sunday print $year</lang>
If the system has 32-bit time, this script will malfunction for years >= 2038; it will print no year from 2038 to 2121 (unless today is Sunday, then it prints every year from 2038 to 2121). This happens because strftime -r '%F' $year-12-25 yields -1 for an out-of-range date, and strftime '%A' -1 yields name of today.
Ursala
A standard library, stt, provides basic date manipulation functions,
and is imported in this example. Unix era times denominated in seconds since
1969 (excluding leap seconds) are represented as natural numbers with
unlimited precision. Results are valid for the arbitrarily distant
future assuming the Gregorian calendar remains in effect.
The algorithm relies on the string_to_time function converting a date
expressed as a character string to seconds without needing a weekday field in
the input, and the time_to_string function outputting the corresponding
date with the weekday included. The output is then filtered for Sundays.
<lang Ursala>#import std
- import nat
- import stt
christmases = time_to_string* string_to_time*TS 'Dec 25 0:0:0 '-*@hS %nP* nrange/2008 2121
- show+
sunday_years = ~&zS sep` * =]'Sun'*~ christmases</lang> output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
Vedit macro language
<lang vedit>Buf_Switch(Buf_Free) for (#3 = 2008; #3 < 2122; #3++) {
Reg_Set(10, "12/25/")
Num_Str(#3, 10, LEFT+APPEND)
if (JDate(@10) % 7 == 0) {
Num_Ins(#3, NOCR)
}
}</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118
XPL0
The original routine in the library only worked correctly between the years 1980 and 2099. It was upgraded with this new routine that handles all dates in the Gregorian calendar, from 1583 onward. It's based on Zeller's Congruence.
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations
proc WeekDay(Year, Month, Day); \Return day of week (0=Sat 1=Sun..6=Fri) int Year, Month, Day; [if Month<=2 then [Month:= Month+12; Year:= Year-1]; return rem((Day + (Month+1)*26/10 + Year + Year/4 + Year/100*6 + Year/400) / 7); ]; \WeekDay
int Year;
[for Year:= 2008 to 2121 do
if WeekDay(Year, 12, 25) = 1 then \25th of December is a Sunday
[IntOut(0, Year); CrLf(0)];
]</lang>
Output:
2011 2016 2022 2033 2039 2044 2050 2061 2067 2072 2078 2089 2095 2101 2107 2112 2118