Print all Sundays in 2017

Question

Print the dates of all the Sundays in 2017 in the following format : dd.mm.yyyy.
Expected Output:

01.01.2017
08.01.2017
15.01.2017
22.01.2017
29.01.2017
05.02.2017
12.02.2017
19.02.2017
26.02.2017
05.03.2017
12.03.2017
19.03.2017
26.03.2017
02.04.2017
09.04.2017
16.04.2017
23.04.2017
30.04.2017
07.05.2017
14.05.2017
21.05.2017
28.05.2017
04.06.2017
11.06.2017
18.06.2017
25.06.2017
02.07.2017
09.07.2017
16.07.2017
23.07.2017
30.07.2017
06.08.2017
13.08.2017
20.08.2017
27.08.2017
03.09.2017
10.09.2017
17.09.2017
24.09.2017
01.10.2017
08.10.2017
15.10.2017
22.10.2017
29.10.2017
05.11.2017
12.11.2017
19.11.2017
26.11.2017
03.12.2017
10.12.2017
17.12.2017
24.12.2017
31.12.2017

Answer 1

Python 2, 81 bytes

x=0
exec"print'%05.2f.2017'%(x%30.99+1.01);x+=7+'0009ANW'.count(chr(x/7+40));"*53

Try it online!

No date libraries, computes the dates directly. The main trick is to treat the dd.mm as a decimal value. For example, 16.04.2017 (April 16) corresponds to the number 16.04. The number is printed formatted as xx.xx with .2017 appended.

The day and month are computed arithmetically. Each week adds 7 days done as x+=7. Taking x modulo 30.99 handles rollover by subtracting 30.99 whenever the day number gets too big. This combines -31 to reset the days with +0.01 to increment the month.

The rollover assumes each month has 31 days. Months with fewer days are adjusted for by nudging x upwards on certain week numbers with +[8,8,8,17,25,38,47].count(x/7). These list is of the week numbers ending these short months, with 8 tripled because February is 3 days short of 31.

This list could is compressed into a string by taking ASCII values plus 40. The shift of +40 could be avoided by using unprintable chars, and could be accessed shorter as a bytes object in Python 3.

Answer 2

PHP, 48 bytes

while($t<53)echo gmdate("d.m.2017
",605e3*$t++);

PHP, 46 bytes (for non-negative UTC offsets)

while($t<53)echo date("d.m.2017
",605e3*$t++);

Answer 3

Python 2, 90 79 bytes

-5 bytes with the help of xnor (avoid counting the weeks themselves)
-1 byte thanks to xnor (add back in e for 605000 as 605e3)

from time import*
i=0
exec"print strftime('%d.%m.2017',gmtime(i));i+=605e3;"*53

Try it online!

0 seconds since epoch is 00:00:00 on the 1st of January 1970, which, like 2017 was not a leap year. 605000 seconds is 1 week, 3 minutes, 20 seconds. Adding 52 of these "weeks" does not take us beyond midnight.

Answer 4

Bash + coreutils, 44 bytes

seq -f@%f 0 605e3 32e6|date -uf- +%d.%m.2017

may save 2 bytes -u if GMT is assumed

• Thanks Digital Trauma point out -f paramter for date which saves 10 bytes;
• And use 2017 in format string saves more bytes which idea is from the answer given by user63956

---

• @0 is 1970-1-1
• 605000 is one week (604800) plus 200 sec
  • 200 sec. should just work since there are only 52 weeks in a year
• @32000000 is just a little more than a year

Answer 5

PowerShell, 51 47

0..52|%{date((date 2017-1-1)+7.*$_)-u %d.%m.%Y}

Fairly straightforward. 2017-01-01 is a Sunday, so is every following seven days. We can save two bytes if we only need the script to be working in my lifetime:

0..52|%{date((date 17-1-1)+7.*$_)-u %d.%m.%Y}

Answer 6

Excel VBA 106 91 79 bytes

Sub p()
For i = #1/1/2017# To #12/31/2017#
If Weekday(i) = 1 Then MsgBox i
Next
End Sub

saved 15 bytes thanks to @Radhato

Assuming 1/1/2017 is Sunday it will save 12 more bytes.

Sub p()
For i = #1/1/2017# To #12/31/2017#
MsgBox i
i = i + 6
Next
End Sub

Thanks @Toothbrush 66 bytes

Sub p:For i=#1/1/2017# To #12/31/2017#:MsgBox i:i=i+6:Next:End Sub

Edit: (Sub and End Sub is not necessary) 52 bytes

For i=#1/1/2017# To #12/31/2017#:MsgBox i:i=i+6:Next

Answer 7

PHP, 67 bytes

Using the fact the PHP automatically assign value 1 to undeclared loop variables, and using Linux epoch times,

<?php for(;54>$t+=1;)echo date("d.m.Y\n",604800*($t)+1482624000);?>

Answer 8

k6, 32 bytes

`0:("."/|"."\)'$2017.01.01+7*!53

Brief explanation:

                2017.01.01+7*!53 /add 0, 7, 14, ..., 364 to January 1st
   ("."/|"."\)'$                 /convert to string, turn Y.m.d into d.m.Y
                                 /   split by ".", reverse, join by "."
`0:                              /output to stdout (or stderr), line by line

Alas, this seems to only work in the closed-source, on-request-only interpreter.

Answer 9

Pyke, 26 24 bytes

53 Fy17y"RVs6)c"%d.%m.%Y

Try it online!

53 F                     - for i in range(53):, printing a newline between each
    y17y"                -  Create a time object with the year 2017. (Month and days are initialised to 1.)
         RV  )           -  Repeat i times:
           s6            -   Add 1 week
              c"%d.%m.%Y -  Format in "dd.mm.yyyy" time

Or 11 bytes

If allowed to ignore output format

y17y"52VDs6

Try it online!

y17y"       - Create a time object with the year 2017. (Month and days are initialised to 1.)
     52V    - Repeat 52 times:
        D   -  Duplicate the old time
         s6 -  Add 1 week

Answer 10

R, 79 67 58 bytes

cat(format(seq(as.Date("2017/01/01"),,7,53),"\n%d.%m.%Y"))

First of January being a sunday, this snippet creates a sequence of days, every 7 days starting from the 01-01-2017 to the 31-12-2017, format them to the desired format and print them.

Answer 11

Befunge-98 (PyFunge), 99 95 93 85 bytes, Leaves trailing newline

All the optimizations were made by @JoKing many thanks to them

s :1g2/10g\%:d1p10g\`+:b`#@_:1\0d1g#;1+:a/'0+,a%'0+,'.,j;a"7102"4k,d1g7+
>8><><>><><>

Try it online!

I felt like we were missing out on some esostericity here so I made a solution in my favourite Esosteric language.

Explanation:
>8><><>><><> Encodes the length of the 12 months
s Store the old day in the blank space
:1g2/ Get an ASCII value from the bottom row and divide it by two this gives us the length of a given month Ex: 8 = 56 in ASCII => 56/2 = 28 => The month (February) has 28 days
10g\% Get the previously saved day and modulo it by the length of the month which allows us to transition the date into the next month
:d1p Save a copy of the new updated day
10g\`+ Test if old date > new date => we transitioned into the next month => add 1 to the month counter
:b` Test if month counter > 11 that means we reached the end of the year (using 0 indexing)
#@_ Based on the previous if terminate the program
:1\0d1g Reorder the stack so it looks like this: Month, 1, Month, 0, Day
# skip the next insctruction (duh)
1+:a/'0+,a%'0+,'., Convert the number to 1 indexing, print, add a . at the end
j; Use the 0 from the stack to not jump and use the ; to go to to the print schedule again then use the 1 to jump over the ; next time
a"7102"4k, Print 2017\n
d1g Get the day value again 7+ Add a week before repeating

Answer 12

JavaScript, 111 106 bytes

for(i=14833e8;i<1515e9;i+=605e6)console.log(new Date(i).toJSON().replace(/(....).(..).(..).*/,'$3.$2.$1'))

Note: Stack Exchange's console isn't long enough to display the entire list, so here's the first half as a separate snippet:

for(i=14833e8;i<15e11;i+=605e6)console.log(new Date(i).toJSON().replace(/(....).(..).(..).*/,'$3.$2.$1'))

The custom format costs me 40 bytes...

Answer 13

Perl 5, 64 bytes

use POSIX;print strftime"%d.%m.%Y\n",0,0,0,7*$_+1,0,117for 0..52

Try it online!

The task given was 2017, not any year, so I hardcoded in:

• 117 (which is perlish for year 2017, 1900+117)
• +1 because 1st of January is a sunday in 2017
• 0..52 because 2017 have 53 sundays

POSIX is a core module and is always installed with Perl5. Doing the same without using modules in 101 bytes, removing whitespace:

$$_[5]==117&&printf"%02d.%02d.%d\n",$$_[3],$$_[4]+1,$$_[5]+1900
  for map[gmtime(($_*7+3)*86400)],0..1e4

Answer 14

Ruby, 75 bytes

Straightforward solution to figure out the dates with Time.

t=Time.new 2017
365.times{puts t.strftime("%d.%m.%Y")if t.sunday?
t+=86400}

Answer 15

SAS, 52 50 bytes

Saved 2 bytes thanks to @user3490.

data;do i=20820to 21184 by 7;put i ddmmyyp10.;end;

Answer 16

Mathematica 90 84 bytes

Fairly wordy. numbermaniac and Scott Milner saved 5 and 1 bytes, respectively.

Column[#~DateString~{"Day",".","Month",".","Year"}&/@DayRange["2017","2018",Sunday]]

Answer 17

VBA, 81 bytes (maybe 64)

Sub p()
For i = 0 To 52
MsgBox format(42736 + i * 7, "dd.mm.yyyy")
Next i
End Sub

My first post. Building on newguy's solution by removing the check for weekdays and just specifying every 7th day. Removing the dates saves 12 bytes a piece. 42736 is 1/1/2017. Output date format depends on system setting. Is that allowed? If so, it's 64 bytes because you don't need the format method.

MsgBox #1/1/2017# + i * 7

Answer 18

AHK, 67 bytes

d=20170101
Loop,52{
FormatTime,p,%d%,dd.MM.yyyy
Send,%p%`n
d+=7,d
}

Nothing magical happens here. I tried to find a shorter means than FormatTime but I failed.

Answer 19

Java 8+, 104 100 99 bytes

()->{for(int t=0;t<53;)System.out.printf("%1$td.%1$tm.2017%n",new java.util.Date(t++*604800000L));}

Java 5+, 109 105 104 bytes

void f(){for(int t=0;t<53;)System.out.printf("%1$td.%1$tm.2017%n",new java.util.Date(t++*604800000L));}

Uses the date-capabilities of the printf format.

Test it yourself!

Savings

1. 104 -> 100: changed the loop values and the multiplicand.
2. 100 -> 99: golfed the loop

Answer 20

T-SQL, 94 Bytes

DECLARE @ INT=0,@_ DATETIME='2017'W:SET @+=1SET @_+=7PRINT FORMAT(@_,'dd.MM.yyy')IF @<52GOTO W

if you don't like SQL GOTO or WHILE, here is a 122 bytes CTE solution

WITH C AS(SELECT CAST('2017'AS DATETIME)x UNION ALL SELECT x+7FROM C WHERE X<'12-31-17')SELECT FORMAT(x,'dd.MM.yyy')FROM C

Answer 21

Ruby, 60+7 = 67 bytes

Uses the -rdate flag.

(d=Date.new 1).step(d+365,7){|d|puts d.strftime"%d.%m.2017"}

Answer 22

Groovy, 81 77 63 60 56 bytes

d=new Date(0);53.times{printf('%td.%<tm.2017%n',d);d+=7}

The above can be run as groovy script.

My first code golf entry. Fortunately, the year 1970 was not a leap year, thus can use it as a base.

Thanks to Dennis, here's a: Try it online!

Answer 23

C#, 138 111 102 bytes

Saved 9 more bytes thanks to Johan du Toit!

Saved 27 bytes thanks to Kevin Cruijssen's suggestions!

()=>{for(int i=0;i<53;)Console.Write(new DateTime(2017,1,1).AddDays(7*i++).ToString("dd.MM.yyyy\n"));}

Anonymous function which prints all Sundays in 2017.

Full program with ungolfed method:

using System;

class P
{
    static void Main()
    {
        Action f =
        ()=>
        {
            for (int i = 0; i < 53; )
                Console.Write(new DateTime(2017, 1, 1).AddDays(7 * i++).ToString("dd.MM.yyyy\n"));
        };

        
        
        f();
    }
}

Answer 24

C#, 110 109 bytes

using System;()=>{for(int i=42729;i<43100;Console.Write(DateTime.FromOADate(i+=7).ToString("dd.MM.yyy\n")));}

You can enjoy this program online here

In this soluion we :

• Use OLE Automation Date (OADate) to avoid "AddDay()" from datetime.
  FromOADate() seem big but it's equal to new DateTime(2017,1,1)

• Start the loop on the last sunday of 2016. to allow us to increment with operator += only. This operator return the value after the increment is done.

• Increment by 7 days to jump from sunday to sunday before printing the date.

• We stop once the last sunday of 2017 has been hit.

• Use Debug instead of Console to save two characters

• Avoid having explicit variable declaration and assignment

Answer 25

TSQL, 112 105 bytes

SELECT CONVERT(VARCHAR,DATEADD(d,number*7,42734),104)FROM master..spt_values WHERE type='p' AND number<53

Demo

T-SQL Convert Syntax

Answer 26

JavaScript (ES6), 123 bytes

It's my first post here, hello!

a=x=>`0${x}.`.slice(-3);[].map.call('155274263153',(x,i)=>{for(j=0;j<4+(2633>>i&1);j++)console.log(a(+x+j*7)+a(i+1)+2017)})

This solution uses hardcoded data and is designed to work specifically for the year 2017. It relies on no date/time APIs.

As for the digits in the string 155274263153, each digit is a number of its own and denotes the first Sunday of each consecutive month. Output for the entire year can be generated by successively adding 7 to those.

What about the magic number, 2633, used in the loop?
Well... 2633 (decimal) is 101001001001 in binary. Now what could those 1s mean? Starting from the right, the 1st, 4th, 7th, 10th and 12th bit are set. This corresponds to months which happen to have five Sundays, as opposed to those that have only four. Golfed down to this neat expression, it initially looked like this: for(j=0;j<4+ +[0,3,6,9,11].includes(i);j++).

I guess the remaining parts are fairly self-explanatory.

Answer 27

T-SQL, 79 77 Bytes

After helping Salman A improve his answer. I decided to write my own using a loop and PRINT.

I ended with this 90 bytes solution.

DECLARE @d DATETIME=42734 WHILE @d<43100BEGIN PRINT CONVERT(VARCHAR,@d,104)SET @d=@d+7 END

Then I looked at the current leader in T-SQL which was 94 bytes from WORNG ALL with this answer. This guy had found very good tricks.

1. Name the variable only @
2. Loop with GOTO instead of actual LOOP
3. Save one character using FORMAT instead of CONVERT. (SSMS2012+ only)

Using these tricks, this solution was trimmed down to the solution below which is 79 bytes.

DECLARE @ DATETIME=42734G:PRINT FORMAT(@,'dd.MM.yyy')SET @+=7IF @<43100GOTO G

Answer 28

Octave, 37 bytes

A lot shorter than all other non-golfing languages, and it's even tied with Jelly! Way to go Octave! :)

disp(datestr(367:7:737,'DD.mm.2017'))

Try it online!

Luckily, year 2 AD looks exactly the same as year 2017 AD. Both starts and ends at a Sunday, and neither is a leap year. This saves a lot of bytes, since 367:7:737 is quite a bit shorter than 736696:7:737060.

This converts the number of days since 01.01.0001, to a string on the format DD.mm, with a trailing .2017.

Answer 29

Haskell, 133 130 bytes

import Data.Time.Calendar
((\[f,g,h,i,j]->i:j:'.':f:g:".2017\n").drop 5.show)=<<take 53(iterate(addDays 7)$fromGregorian 2017 1 1)

Try it online!

Without a calendar library, 148 144 140 bytes

(\l->last l!l++length l!l++"2017\n").fst.span(>0).(<$>scanl((+).(+28))0[3,0,3,2,3,2,3,3,2,3,2]).(-)=<<[1,8..365]
a!_=['0'|a<=9]++show a++"."

This is funny since using an operator for the padding function saves two bytes even though the second argument is unused, because less parentheses are needed – p(last l) is longer than last l!l. Works by calculating day/month pairs by subtracting the cumulative month start dates from the day of the year. The month start dates are compressed as (scanl((+).(+28))0[3,0,3,2,3,2,3,3,2,3,2]). Month number is the number of positive elements and day number is the last positive element.

Try it online!

Answer 30

C#, 116 114 113 bytes

for(long i=(long)636188256e9;i<636502857e9;i+=(long)605e10)Out.Write(‌​new DateTime(i).ToString("dd.MM.yyyy\n"));

Can be run in the interactive windows of Visual Studio (or any other C# REPL based on Roslyn)

Down to 113 bytes: thanks to Kevin Cruijssen.