Next Friday the 13th

Question

What is the shortest program that can predict the next time Friday will fall on the 13th day of the month?

• Must be an complete functioning program (not just a function/subroutine)
• Must print out date in the following format: YYYY-MM-DD
• Allow user to provide a start date either as a command line argument or through STDIN
• If user provides no start date, use today as the start date.
• If the start date is a Friday the 13th, the program should find the next Friday the 13th.

If I were to run the program today (Feb 16 2011) I should get the following output.

$ ./fr13th 2013-05-09
2013-09-13
$ ./fr13th 2007-06-29
2007-07-13
$ ./fr13th 2007-07-13
2008-06-13
$ ./fr13th
2011-05-13

Answer 1

Windows PowerShell, 74

for($d="date $args"|iex;($d+='1').day*$d.dayofweek-65){}'{0:yyy-MM-d}'-f$d

Fairly straightforward. One perhaps confusing bit is the use of "Get-Date $args" | Invoke-Expression to get either the current date (if $args is empty) or the date specified in $args without raising an error.

72-byte variant:

for($d="date $args"|iex;($d+=9).day*$d.dayofweek-65){}'{0:yyy-MM-d}'-f$d

Takes ages, though ... this doesn't increment the datetime by a whole day each iteration but instead only 900 nanoseconds. But two bytes shorter.

67-byte variant:

for($d="date $args"|iex;($d+='1').day*$d.dayofweek-65){}'{0:d}'-f$d

This is a bit locale-sensitive; if it fails on your machine, try setting your date format to ISO-8601 beforehand. :-)

Oh, and it can be made into 65 bytes just like the 72-byte version.

History:

• 2011-02-17 00:33 (92) First attempt.
• 2011-02-17 00:35 (85) Improved getting an initial date.
• 2011-02-17 00:37 (79) Compared the product instead of day and day of week individually. Admittedly stolen from Ventero.
• 2011-02-17 00:40 (76) Pulled the first line into the for. Comparison just as subtraction instead of -eq which saves another two bytes.
• 2011-02-17 00:53 (75) Unix date format string is a bit shorter.
• 2011-02-17 11:42 (74) Reverted to the default date pattern but yyy-MM-d suffices (since the year is always longer than three characters and the day is always 13. Thanks to Ty Auvil for this.

Answer 2

Ruby, 96 75 characters

require"date"
d=Date.parse(gets||"thu")+1
d+=1 while d.wday*d.day!=65
$><<d

Takes the date from stdin. To not specify a date press ctrl-d.

Thanks very much for Ventero's help.

Ungolfed:

require "date"
# Date.parse("thu") will return this week's thursday
date = Date.parse(gets || "thu")+1
date += 1 while d.wday * d.day != 5 * 13
$stdout << date

Sample IO:

$ ruby fr13th.rb
2013-05-09
2013-09-13
$ ruby fr13th.rb
2007-06-29
2007-07-13
$ ruby fr13th.rb
2007-07-13
2008-06-13
$ ruby fr13th.rb
2011-05-13

Answer 3

bash, 75

until set `date +%F -d$1+day`
date -d$1|grep -q '^F.* 13'
do :
done
echo $1

This is a bit locale-sensitive; if it fails on your machine, try exporting LC_ALL=C beforehand.

$ bash fri13th.sh 2013-05-09
2013-09-13                                             
$ bash fri13th.sh 2007-06-29
2007-07-13                                             
$ bash fri13th.sh 2007-07-13
2008-06-13
$ bash fri13th.sh
2011-05-13

Answer 4

C#, 185

Based on Andrew Koester's C# solution, but heavily modified along the way. I eventually arrived at a solution similar to my PowerShell solution:

using System;class
P{static void
Main(string[]a){var
n=a.Length>0?DateTime.Parse(a[0]):DateTime.Now;for(;(n=n.AddDays(1)).Day*(int)n.DayOfWeek!=65;);Console.Write("{0:yyy-MM-d}\n",n);}}

Answer 5

R, 113 characters

f="%Y-%m-%d";o=format;a=c(as.Date(scan(,""),f),Sys.Date())[1];repeat{a=a+1;if(o(a,"%w%d")==513)break};cat(o(a,f))

Example runs:

> f="%Y-%m-%d";o=format;a=c(as.Date(scan(,""),f),Sys.Date())[1];repeat{a=a+1;if(o(a,"%w%d")==513)break};cat(o(a,f))
1: 2007-06-29
2: 
Read 1 item
2007-07-13

> f="%Y-%m-%d";o=format;a=c(as.Date(scan(,""),f),Sys.Date())[1];repeat{a=a+1;if(o(a,"%w%d")==513)break};cat(o(a,f))
1:
Read 0 items
2013-12-13

> f="%Y-%m-%d";o=format;a=c(as.Date(scan(,""),f),Sys.Date())[1];repeat{a=a+1;if(o(a,"%w%d")==513)break};cat(o(a,f))
1: 2013-12-13
2: 
Read 1 item
2014-06-13

Answer 6

Perl 6, 69 bytes

$_=Date.new(@*ARGS[0]//Date.today);.++while .day*.day-of-week-65;.say

Try it online!

Thanks to @ASCII-only for -5

Answer 7

Perl (and others), 114

for(($y,$m,$d)=(shift//`date +%F`)=~/\d+/g,$d>12&&$m++;$m
>12&&($y++,$m=1),`cal $m $y`!~/14$/m;$m++){}say"$y-$m-13"

Perl 5.10 or later, run with -E 'code here' or -M5.010 file. Needs date (from coreutils for Linux) and cal (from util-linux)

Sample run:

$ perl -M5.010 fr13.pl 2013-05-09
2013-9-13
$ perl -M5.010 fr13.pl 2007-06-29
2007-07-13
$ perl -M5.010 fr13.pl 2007-07-13
2008-6-13
$ perl -M5.010 fr13.pl
2011-5-13

I'm not sure when the leading zero for months before October is kept. It's obviously lost when the year rolls over; it seems to be kept when the answer is just next month. Let's call that undefined behavior--hey, this is golfing!

Answer 8

PHP - 103

(brute force)

<?for($d=date_create(@$argv[1]);$d->modify('next fri')&&$d->format(@d)-13;);die($d->format("Y-m-d\n"));

Ungolfed:

<?
$d = new DateTime(@$argv[1]);
while ($d->modify('next fri')) {
        if ($d->format('d') == 13) {
                die($d->format("Y-m-d\n"));
        }
}

Test:

$ php 979.php 2013-05-09
2013-09-13
$ php 979.php 2007-06-29
2007-07-13
$ php 979.php 2007-07-13
2008-06-13
$ php 979.php 
2011-05-13

Answer 9

BASH

#!/bin/bash
from=$1
if [ "$from" = "" ]; then
from=`date +%Y-%m-%d`
fi
i=1
while [ "$isFri" = "" ] || [ "$is13" = "" ]
do
isFri=`date -d "${from} ${i} days" | grep Fri`
is13=`date -d "${from} ${i} days" +%Y-%m-%d | grep "\-13"`
((i++))
done
((i--))
date -d "${from} ${i} days" +%Y-%m-%d

CONCEPT USED:

$ date -d "2011-02-16 2 days" +%Y-%m-%d
2011-02-18

SAMPLE I/O

:~/aman> ./fr13th.sh
2011-05-13
:~/aman> ./fr13th.sh 2013-05-09
2013-09-13
:~/aman> ./fr13th.sh 2007-06-29
2007-07-13
:~/aman> ./fr13th.sh 2007-07-13
2008-06-13

Answer 10

Python - 166 chars

Reads from stdin, so you need to feed a blank line if you want todays date

from datetime import*
D=datetime
e=timedelta(1)
I=raw_input()
d=e+(I and D.strptime(I,"%Y-%m-%d")or D.now())
while(d.weekday()+1)*d.day-65:d+=e
print d.strftime("%F")

Answer 11

C#

240 characters. C# needs a "run inside of a function only" mode!

using System;class P{static void Main(string[] a){var n=DateTime.Now;if(a.Length>0)DateTime.TryParse(args[0],out n);while(true){n=n.AddDays(1);if((n.Day==13)&&(n.DayOfWeek==(DayOfWeek)5))break;}Console.WriteLine(n.ToString("yyyy-MM-dd"));}}

Un-golfed:

using System;

class P
{
    static void Main(string[] a)
    {
        var n = DateTime.Now;
        if (a.Length > 0) DateTime.TryParse(args[0], out n);
        while (true)
        {
            n = n.AddDays(1);
            if ((n.Day == 13) && (n.DayOfWeek == (DayOfWeek)5)) break;
        }
        Console.WriteLine(n.ToString("yyyy-MM-dd"));
    }
}

Test output

\Debug> f13.exe 2013-05-09
2013-09-13

\Debug> f13.exe 2007-06-29
2007-07-13

\Debug> f13.exe 2007-07-13
2008-06-13

\Debug> f13.exe
2011-05-13

Answer 12

D: 227 Characters

import std.datetime,std.stdio;void main(string[]a){auto d=a.length<2?cast(Date)(Clock.currTime()):Date.fromISOExtendedString(a[1]);for(;d.dayOfWeek!=DayOfWeek.fri||d.day!=13;d+=dur!"days"(1)){}writeln(d.toISOExtendedString());}

More Legibly:

import std.datetime, std.stdio;

void main(string[] a)
{
    auto d = a.length < 2 ? cast(Date)(Clock.currTime()) : Date.fromISOExtendedString(a[1]);

    for(; d.dayOfWeek != DayOfWeek.fri || d.day != 13; d += dur!"days"(1)) {}

    writeln(d.toISOExtendedString());
}

The fun part is that while D's std.datetime makes this sort of code very easy to write, it's also incredibly verbose - due primarily to the precise (and therefore long) function names. So, the usability and maintainability of the code is very high, but it's code golfability is rather low.

Answer 13

C#, 206 194 characters

Update

This is a slightly different take on the problem, so I left my other attempt here in full.

using System:class p{static void Main(string[]a){var n=a.Length>0?DateTime.Parse(a[0]):DateTime.Now;for(;(n=n.AddDays(5-(int)n.DayOfWeek).AddDays(7)).Day!=13;);Console.Write("{0:yyy-MM-d}",n);}}

Here, I'm finding the "current" week's Friday and then incrementing by 7 until I find one that's a 13. I also used Joey's for loop and output formatting to shave a few characters.

Ungolfed:

using System;
class p
{
    static void Main(string[] a)
    {
        var n = a.Length > 0 ? DateTime.Parse(a[0]) : DateTime.Now;

        for (; (n = n.AddDays(5 - (int)n.DayOfWeek).AddDays(7)).Day != 13; ) ;

        Console.Write("{0:yyy-MM-d}", n);
    }
}

Original:

This is similar to Andrew's above, but there were enough differences I decided to post a separate answer rather than comment and suggest edits.

using System;class p{static void Main(string[]a){var n=a.Length>0?DateTime.Parse(a[0]):DateTime.Now;do n=n.AddDays(1);while(!(n.Day==13&&n.DayOfWeek+""=="Friday"));Console.Write(n.ToString("yyyy-MM-dd"));}}

Ungolfed:

using System;
class p
{
    static void Main(string[] a)
    {
        var n = a.Length > 0 ? DateTime.Parse(a[0]) : DateTime.Now;
        do
        {
            n = n.AddDays(1);
        } while (!(n.Day == 13 && n.DayOfWeek + "" == "Friday"));

        Console.Write(n.ToString("yyyy-MM-dd"));
    }
}

Answer 14

SQLite, 374 characters

(Line breaks added here for readability; not included in the count.)

Requirement to "Allow user to provide a start date either as a command line argument or through STDIN" omitted due to technical limitations.

CREATE TABLE R(N UNIQUE);
INSERT INTO R VALUES(0);
INSERT INTO R VALUES(1);
REPLACE INTO R SELECT A.N*2048|B.N*1024|C.N*512|D.N*256|E.N*128|F.N*64|
G.N*32|H.N*16|I.N*8|J.N*4|K.N*2|L.N FROM
R A,R B,R C,R D,R E,R F,R G,R H,R I,R J,R K,R L;
CREATE TABLE F AS SELECT DATE('2000-01-13','+'||N||'months') AS D
FROM R WHERE STRFTIME('%w',D)='5';
SELECT MIN(D) FROM F WHERE D>DATE('now');

The first 4 statements create a table (R) with a single column containing all the integers from 0 to 4095.

The 5th statement builds a table (F) of all Friday the 13ths between 2000-10-13 and 2340-12-13.

The 6th statement simply returns the first Friday the 13th after the current (UTC) date.

Answer 15

Another Javascript, 153

I post another javascript answer because I can't comment the first...

a=new Date(process.argv[2]||Date.now());for(b=1;b;b=a.getDate()!=13||!/^F/.test(a))a.setTime(a.getTime()+864e5);console.log(a.toISOString().substr(0,10))

Execute with nodeJS :

$ node fr13th
2013-12-13

$ node fr13th 2007-06-29
2007-07-13

$ node fr13th 2013-05-09
2013-09-13

Answer 16

Python 3.3, 166 characters

import datetime as d
t=input()
t=d.date(*map(int,t.split()))if t!=""else d.date.today()
while 1:
 t+=d.timedelta(1)
 if t.day==13and t.weekday()==4:
  print(t);break

Input is in the format 2013 1 1,

>>> ================================ RESTART ================================
>>> 
2013 1 1
2013-09-13

or just press enter to use today's date (would be 2013/12/11 for this output)

>>> ================================ RESTART ================================
>>> 

2013-12-13

(there is actually a mix of my solution and @gnibbler 's that counts 144 chars)

import datetime as d
t=input()
t=d.date(*map(int,t.split()))if t!=""else d.date.today()
while t.day*t.weekday()-65:
 t+=d.timedelta(1)
print(t)

The very nice line while t.day*t.weekday()-65: is from @gnibbler's solution.

Answer 17

Japt, 35 bytes

ÐUªKs3 ¯A
@e ¶5©D¶Uf}a@f1Uf Ä
s3 ¯A

-8 bytes thanks to @ASCIIOnly!

Try it!

Answer 18

Swift 4, 310 bytes

import Foundation
let (u,f)=(Calendar.current,DateFormatter())
f.dateFormat="yyyy-MM-dd"
var t={(x:Date)->Int in let c=u.dateComponents([.weekday,.day],from:x);return c.weekday!*c.day!},d=readLine().map{f.date(from:$0)!} ?? Date()
while t(d) != 65{d=u.date(byAdding:.day,value:1,to:d)!}
print(f.string(from:d))

Try it online!

-4 thanks to TagTaco.

Alas...:

• needs Foundation for Date/Calendar.
• Swift doesn't allow Ints to be used as Bool.
• The shorthand enum syntax is helpful, though not that much.
• Space is required around ??.
• != also needs spacing, lest it's interpreted as an unwrapping.

Answer 19

Javascript

F13=function(x){
    z=function(x){return (''+x).replace(/^(.)$/,'0$1')}
    D=x?new Date(x):new Date(),Z=864e5,X=D.getDay()%7,X+=+D+(X?Z*(5-X):0);
    do{
        X+=Z*7;
        D=new Date(X);
    }while(D.getDate()!=13)
    return D.getFullYear()+"-"+z(D.getMonth()+1)+"-"+z(D.getDate());
}

ps: yeah, I know, I broke the first rule (couldn't be just a function)

Tests on javascript shell

F13("2013-05-09") // 2013-09-13
2013-09-13
F13("2007-06-29") // 2007-07-13
2007-07-13
F13("2007-07-13") // 2008-06-13
2008-06-13
F13() //2011-05-13
2011-05-13

Answer 20

T-SQL 359 285 253 Characters

CREATE PROCEDURE f13(@d DateTime=null)AS
SET @d=ISNULL(@d,GETDATE())
;WITH d AS
(SELECT @d+1 d
UNION ALL SELECT d+1 FROM d
WHERE DATEPART(dw,d)<>6 OR DAY(d)<>13)SELECT CAST(d AS DATE) FROM d
WHERE DATEPART(dw,d)=6 AND DAY(d)=13
OPTION (MAXRECURSION 999)

I just wanted to lay the smack down on the SQLite solution with my non-clever verbose date function using T-SQL procedure.

Update: My original fear that doing a day increment would take more space than a month increment was very incorrect.

Test results (SSMS):

f13 '2013-05-09';
GO

f13 '2007-06-29';
GO

f13 '2007-07-13';
GO

f13;
GO
--

d
----------
2013-09-13

d
----------
2007-07-13

d
----------
2008-06-13

d
----------
2011-05-13

Answer 21

Java 8, 200 197 bytes

interface M{static void main(String[]a){java.time.LocalDate z=null,r=a.length<1?z.now():z.parse(a[0]);for(;(r=r.plusDays(1)).getDayOfWeek().getValue()*r.getDayOfMonth()!=65;);System.out.print(r);}}

Explanation:

Try it here (remove the argument to use the current date).

interface M{                        // Class
  static void main(String[]a){      //  Mandatory main-method
    java.time.LocalDate z=null,     //   LocalDate to reduce bytes when using static calls
     r=a.length<1?                  //   If no argument is given:
        z.now()                     //    Start at the current date
       :                            //   Else:
        z.parse(a[0]);              //    Start at the date of the first argument
    for(;(r=r.plusDays(1))          //   Before every iteration, go to the next day
                                    //   Loop as long as:
         .getDayOfWeek().getValue()
                                    //    the 1-indexed day of the week (Friday = 5)
         *r.getDayOfMonth()         //    multiplied by the day of the month
        !=65;                       //    is not 65
                                    //    (5 and 13 are primes, so only 5*13 results in 65)
    );                              //   End of loop
    System.out.print(r);            //   Print the result Friday the 13th
  }                                 //  End of main-method
}                                   // End of class

NOTE: Java's default format when printing is already yyyy-MM-dd.

Answer 22

VB.net (96c*)

Entry

Function NextFridayThe13th(d As Date) As Date
  While d.DayOfWeek<>DayOfWeek.Friday
    d=d.AddDays(1)
  End While
  While d.Day<>13
    d=d.AddDays(7)
  End While
  Return d
End Function

• I think the CodeGolf count for vb.net shouldn't include the function signature, closing end function and the return. So it just makes it about the internal implementation.

So my scoring is broken down as so

While d.DayOfWeek<>DayOfWeek.Friday    '35c
d=d.AddDays(1)                         '13c  48c
End While                              ' 9c  57c
While d.Day<>13                        '15c  72c
d=d.AddDays(7)                         '14c  86c
End While                              ' 9c  95c
Return d                               ' 1c  96c

Answer 23

05AB1E, 141 bytes

gĀi'-¡ëžežfžg)}V[Y`2ô0Kθ4ÖUD2Qi\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝVY`UÐ3‹12*+>13*5÷s3‹Xα©т%D4÷®т÷©4÷®·()O7%6QYн13Q*#}YRεDgi0ì]'-ý

05AB1E doesn't have any builtins for Date objects or calculations. The only builtin regarding dates it has is today's year/month/day/hours/minutes/seconds/microseconds.

So because of that, almost all of the code you see are manual calculations to go to the next day, and calculating the day of the week.

Mostly derived from my 05AB1E answer in The Work Day Countdown challenge (which is why I edited that one about an hour ago when I came across a bug..)

Input is a string in the format dd-MM-yyyy (but output is in the format yyyy-MM-dd, since it's one of the rules of the challenge).

Try it online or verify all test cases.

Explanation:

gĀi        # If an input is given:
   '-¡    '#  Split it by "-"
  ë        # Else:
   že      #  Push today's day
     žf    #  Push today's month
       žg  #  Push today's year
         ) #  Wrap them into a single list
  }V       # After the if-else statement: pop and store it in variable `Y`
[          # Start an infinite loop
 Y`2ô0Kθ4ÖUD2Qi\28X+ë<7%É31α}‹iY¬>0ëY1¾ǝDÅsD12‹i>1ë\1Dǝ¤>2}}ǝV
           #  Go to the next day
           #  (see my linked The Work Day Countdown answer for an explanation)
 Y`UÐ3‹12*+>13*5÷s3‹Xα©т%D4÷®т÷©4÷®·()O7%
           #  Calculate the day of the week (0 = Saturday, 1 = Sunday, ..., 6 = Friday)
           #  (see my linked The Work Day Countdown answer for an explanation)
 6Q        #  Check if the day of the week is a Friday
 Yн        #  Push the days of the current date
   13Q     #  Check if it's the 13th
 *         #  And if both checks are truthy:
  #        #   Stop the infinite loop
}YR        # After the infinite loop: push the resulting date-list, and reverse it
ε          # Map each value to:
 Dgi       #  If it's only a single digit:
    0ì     #   Prepend a leading "0"
]          # Close both the if-statement and map
 '-ý      '# Join the result by "-"
           # (and output the result implicitly)

Answer 24

Rebol, 136

d: any[do system/script/args now]until[d: d + 1 all[d/day = 13 d/weekday = 5]]print format/pad[4"-"-2"-"-2]reduce[d/year d/month d/day]

Ungolfed:

d: any [do system/script/args  now]

until [
    d: d + 1 
    all [d/day = 13 d/weekday = 5]
]

print format/pad [4 "-" -2 "-" -2] reduce [d/year d/month d/day]

Example usage:

$ rebol friday13th.reb 2013-05-09
2013-09-13
$ rebol friday13th.reb
2014-06-13