How many friday the 13th in a year?

Question

Your challenge is to write a program which, given a year, outputs the number of "Friday 13ths" in it.

Rules & Details:

• You can take input via STDIN, or as an argument passed to your program.
• You should output the result to STDOUT.
• You may assume that input will be a valid year, and does not pre-date the Gregorian calendar (undefined behaviour is permitted in these cases).
• Calendar/Date libraries are allowed.

This is a code-golf, so the shortest code (in bytes) wins.

(Related challenge link)

Answer 1

APL (Dyalog APL) with cal from dfns, 29 bytes

+/{13∊⍎,⍉3↑¯5↑⍉2↓cal⍵}¨⎕,¨⍳12

Try it online!

⍳ 12 the integers one through twelve

⎕ ,¨ take numeric input and prepend to each of the twelve numbers

{…}¨ on each of the pairs, apply the function…

 cal⍵ get a calendar for that year-month

 2 ↓ drop two rows (caption and days)

 ⍉ transpose (so we can address columns instead of rows)

 ¯5 ↑ take the last five (two digits for each of Friday and Saturday plus one space)

 3 ↑ take the first two (two digits for Friday plus a space)

 ⍉ transpose (so we get reading order)

 , ravel

 ⍎ execute as APL expression (gives list of Fridays' dates)

 13 ∊ is thirteen a member of that list?

+/ sum the 12 Booleans

---

Using @Wrzlprmft's algorithm, we can do it without libraries for 53 bytes:

'21232211321211'⊃⍨14|2 3 ¯1+.×⊢,0≠.=400 100 4∘.|-∘0 1

-∘0 1 subtract zero and one

400 100 4 ∘.| division remainder table for the two years (across) divided by these numbers (down)

0 ≠.= inner "product" with 0, but using ≠ and = instead of +.×

⊢ , prepend the unmodified argument year

2 3 ¯1 +.× inner product with these numbers

14 | division remainder when divided by fourteen

'21232211321211' ⌷⍨ index into this string

Answer 2

Mathematica 49 46 45 44 42

As a pure function: 42 chars

DayName@{#,m,6}~Table~{m,12}~Count~Friday&

Example

DayName@{#,m,6}~Table~{m,12}~Count~Friday&[2013]

2

---

As a named function: 44 chars

f=DayName@{#,m,6}~Table~{m,12}~Count~Friday&

Examples

f[1776]
f[2012]
f[2013]
f[2014]

2
3
2
1

Answer 3

Ruby, 49 48 47 46

f=->m{(1..12).count{|i|Time.gm(m,i,6).friday?}}

Edit: Shaved a character by going back a week, thanks to Jan, and another by switching from Time.new to Time.gm

Edit: At the expense of obfuscating it a bit more, I can get to 46 with

f=->m{(1..12).count{|i|Time.gm(m,i,8).wday<1}}

Answer 4

Powershell, 68 63 58 52 50

Thanks Iszi for the tip.

$n=$args;(1..12|?{!+(date $n-$_).DayOfWeek}).Count

Using the fact that if the 1st day in the month is Sunday, the 13th will be Friday.

I've also tried:

(1..12|?{!+(date $args-$_).DayOfWeek}).Count

but it's not the same $args inside the script block.

Answer 5

R 76 72 57

sum(format(as.Date(paste(scan(),1:12,1,sep="-")),"%w")<1)

Answer 6

Python2.7 90 86

from datetime import*
s=c=0
exec's+=1;c+=date(%d,s,9).weekday()<1;'%input()*12
print c

Monday the 9th may not have quite the same ring to it but works just as well.

Edit: A year and a half to notice that date is shorter than datetime :)

Answer 7

Not using any libraries or built-in date functions:

Golfscript – 51

~..({4/.25/.4/--@}2*2*\3*+-
14%' [3/=RI[)a%:*.'\=5%

' [3/=RI[)a%:*.' could as well be 'feefefgeeffgfe'

Python – 82 79

Essentially the same algorithm.

l=lambda y:y/4-y/100+y/400
i=input()
print"21232211321211"[(2*i+3*l(i)-l(i-1))%14]

Using this trick, this can be golfed down further to:

l=lambda y:y/4-y/100+y/400
i=input()
print 94067430>>(4*i+6*l(i)-2*l(i-1))%28&3

---

This exploits the fact that, calender-wise, there are only 14 different years, which are distinguishable by their last day and whether they are leaping. l calculates the number of leap years up to its argument (if the Gregorian calendar extended backwards to the year 1). (2*i+3*l(i)-l(i-1))%14 is short for l(i)-l(i-1)+(i+l(i))%7*2, where l(i)-l(i-1) tells us whether the argument is a leap year and i+l(i) sums up the shifts of the last day (one in a normal year, two in a leap year).

Answer 8

C 301+ 287

main(int x,char**v){char p[400],*a[]={"abbababbacaacbac","bacabbb","baabbaca","abbb","aabbacaac","abbbbcaac","abbbbaabb"},*b="adcadcadcaebcadcadcafbcadcadcagbcadcadcadc";int c=0,i,y=atoi(v[0]);for(i=0;i<42;i++)strcpy(&p[c],a[b[i]-'a']),c+=strlen(a[b[i]-'a']);printf("%d",p[y%400]-'`');}

Not the shortest answer, but uses no libraries.

Answer 9

JavaScript, 70

f=function(a){b=0;for(c=12;c--;)b+=!new Date(a,c,1).getDay();return b}

Answer 10

bash 47 36

seq -f$1-%g-6 12|date -f-|grep -c ^F

Thanks @DigitalTrauma for saving 10 chars by using seq with default start to 1.

date -f<(printf "%s\n" $1-{1..12}-6)|grep -c ^F

(Previous version using echo present a bug because of the empty line when <(echo $1-{1..12}-6$'\n'). So this function worked fine until today is a Friday.

Lets see:

set -- 2013
seq -f$1-%g-6 1 12|date -f-|grep -c ^F
2

date -f<(printf "%s\n" $1-{1..12}-13)|grep -c ^F
2

Is is locale dependant, si if it don't work, you may have to

export LANG=C

or

LANG=C date -f<(printf "%s\n" $1-{1..12}-13)|grep -c ^F

Into a function; +7 -> 43

f(){ seq -f$1-%g-6 12|date -f-|grep -c ^F;}

f 2013
2

for i in {2010..2017};do echo $i $(f $i) ;done
2010 1
2011 1
2012 3
2013 2
2014 1
2015 3
2016 1
2017 2

Bonus: +78 -> 121

From there, if my function become:

f(){ o=();for t in $(seq -f$1-%g-6 12|date -f- +%a,%b);do [ "${t:0:1}" = "F" ]&&o+=(${t#*,});done;echo ${#o[@]} ${o[@]};}

or

f(){ o=();
     for t in $(seq -f$1-%g-6 1 12|date -f- +%a,%b);do
         [ "${t:0:1}" = "F" ]&&o+=(${t#*,})
       done
     echo ${#o[@]} ${o[@]}
}

for i in {2010..2017};do echo $i $(f $i) ;done
2010 1 Aug
2011 1 May
2012 3 Jan Apr Jul
2013 2 Sep Dec
2014 1 Jun
2015 3 Feb Mar Nov
2016 1 May
2017 2 Jan Oct

Answer 11

C (151 145 137 131 130 chars)

I am surprised to see that there is only one other solution that doesn't use built-in calendar tools. Here is a (highly obfuscated) mathematical approach, also in C:

f(x){return(x+(x+3)/4-(x+99)/100+!!x)%7;}main(int x,char**v){int y=atoi(v[1])%400,a=f(y+1);putchar('1'+((f(y)&3)==1)+(a>2&&a-5));}

(The above compiles in GCC with no errors)

Alternative solution: C (287->215 chars)

I rather enjoyed Williham Totland's solution and his use of compression. I fixed two small bugs and tweaked the code to shorten its length:

main(int x,char**v){char p[400],*a[]={"1221212213113","213122221122131","12213113","22213113","22221122","2131"},*b="abababafcbababafdbababafebababab";*p=0;for(;*b;b++)strcat(p,a[*b-97]);putchar(p[atoi(v[1])%400]);}

Answer 12

PHP, 82

<?for($i=1,$c=0;$i<13;$i++)$c+=(date("N",mktime(0,0,0,$i,1,$argv[1]))==7);echo $c;

Based on

"Any month that starts on a Sunday contains a Friday the 13th, and there is at least one Friday the 13th in every calendar year."

From http://en.wikipedia.org/wiki/Friday_the_13th

Answer 13

k

64 characters

{+/6={x-7*x div 7}(.:')x,/:(".",'"0"^-2$'$:1+!:12),\:".13"}[0:0]

Reads from stdin

Answer 14

Common Lisp (CLISP), 149

(print 
    (loop for i from 1 to 12 count 
        (= 4 (nth-value 6 
            (decode-universal-time
                (encode-universal-time 0 0 0 13 i
                    (parse-integer (car *args*)) 0))))))

Answer 15

PHP, 55 bytes

for(;++$i<13;)$c+=!date(w,strtotime($argn.-$i));echo$c;

Run with echo <year> | php -nR '<code>'.

Basically the same that Oleg tried and Damir Kasipovic did, just with better golfing:
Every month that starts with a sunday, has a Friday the 13th.
So I loop through the months and count the first days that are sundays.

breakdown

for(;++$i<13;)          // loop $i from 1 to 12
    $c+=!                   // 4. if result is not truthy (weekday==0), increment $c
        date(w,             // 3. get weekday (0 stands for Sunday)
            strtotime(      // 2. convert to timestamp (midnight 1st day of the month)
                $argn.-$i   // 1. concatenate year, "-" and month
            )
        )
    ;
echo$c;                 // output

Answer 16

C# 110 101 93 92

int f(int y){int c=0;for(int i=1;i<13;i++)c+=new DateTime(y,i,8).DayOfWeek>0?0:1;return c;}

C# Linq 88

int g(int y){return Enumerable.Range(1,12).Count(m=>new DateTime(y,m,8).DayOfWeek==0);}

Thanks to Jeppe Stig Nielsen for linq and suggestion of checking for sunday on the 8th.

Thanks to Danko Durbić for suggesting > instead of ==.

Answer 17

Perl 6, 55 53

{sum (1..12).map: {Date.new($_,$^a,1).day-of-week>6}}

Old answer:

{sum (1..12).map: {Date.new($_,$^a,13).day-of-week==5}}

Answer 18

Python 195 / 204

Works only for previous years, because monthdatescalendar returns a calendar for the given year until now. I think there is a lot of optimizing potential left :).

import calendar, sys
c=calendar.Calendar()
f=0
for m in range(1,12):
 for w in c.monthdatescalendar(int(sys.argv[1]),m):
  for d in w:
   if d.weekday() == 4 and d.day == 13:
    f=f+1
print(f)

Another solution, works for every date but it isn't smaller:

import datetime,sys
y=int(sys.argv[1])
n=datetime.date
f=n(y,1,1)
l=n(y,12,31)
i=0
for o in range(f.toordinal(), l.toordinal()):
 d=f.fromordinal(o)
 if d.day == 13 and d.weekday() == 4:
  i=i+1
print(i)

Answer 19

Bash (52 47 characters)

for m in {1..12};do cal $m $Y;done|grep -c ^15

Answer 20

K, 42

{+/1={x-7*x div 7}"D"$"."/:'$+(x;1+!12;1)}

.

k){+/1={x-7*x div 7}"D"$"."/:'$+(x;1+!12;1)}'1776 2012 2013 2014
2 3 2 1

Answer 21

Rebol, 63

f: 0 repeat m 12[d: do ajoin["6-"m"-"y]if d/weekday = 5[++ f]]f

Usage example in Rebol console:

>> y: 2012
== 2012

>> f: 0 repeat m 12[d: do ajoin["6-"m"-"y]if d/weekday = 5[++ f]]f
== 3

Alternative solution which collects all the Friday 13th in given year is:

>> collect[repeat m 12[d: do ajoin["13-"m"-"y]if d/weekday = 5[keep d]]]
== [13-Jan-2012 13-Apr-2012 13-Jul-2012]

Answer 22

Bash and Sed, 39

ncal $1|sed '/F/s/13/\
/g'|grep -c ^\ 2

ncal prints a calendar for the given year with days of the week down the left.

sed with a /g flag subs out all 13s with newlines

grep -c counts the lines that start with " 2" (20 always follows 13)

Thanks to @DigitalTrauma for finding a bug in my old version and proposing a solution!

Answer 23

Scala, 76 68 characters

In 78 chars:

def f(y:Int)=0 to 11 count(new java.util.GregorianCalendar(y,_,6).get(7)==6)

Nothing out of ordinary, except for using magical numbers for DAY_OF_WEEK = 7 and FRIDAY = 6.

68 character version:

def f(y:Int)=0 to 11 count(new java.util.Date(y-1900,_,6).getDay==5)

Yes, Java changed values of day of the week constants between API's.

Answer 24

Clojure, 207 187 bytes

-20 bytes by getting rid of the import, and some whitespace I missed.

(import '[java.time LocalDate DayOfWeek])#(loop[d(LocalDate/parse(str %"-01-01"))c 0](if(=(.getYear d)%)(recur(.plusDays d 1)(if(and(=(.getDayOfMonth d)13)(= (.getDayOfWeek d) DayOfWeek/FRIDAY))(inc c)c))c))

Starting on Janauary 1st of the given year, it loops over each day. If the day is Friday the 13th, it increments the count. It continues to loop until it reaches the next year.

(import '[java.time LocalDate DayOfWeek])

(defn count-friday-13ths [year]
  (loop [d (LocalDate/parse (str year "-01-01")) ; Starting date
         c 0] ; The count
    (if (= (.getYear d) year) ; If we haven't moved onto the next year...
      (recur (.plusDays d 1) ; Add a day...
             (if (and (= (.getDayOfMonth d) 13) ; And increment the count if it's Friday the 13th
                      (= (.getDayOfWeek d) DayOfWeek/FRIDAY))
               (inc c) c))
      c))) ; Return the count once we've started the next year.

Answer 25

PHP, no builtins, 81 bytes

echo 0x5da5aa76d7699a>>(($y=$argn%400)+($y>102?$y>198?$y>299?48:32:16:0))%28*2&3;

Run with echo <year> | php -nR '<code>'.

breakdown

Weekdays repeat every 400 years.
In the results for 1600 to 1999 (for example), there is a 28-length period with just three gaps:

  0:2212122131132131222211221311
 28:2212122131132131222211221311
 56:2212122131132131222211221311
 84:2212122131132131122
103:       131132131222211221311
124:2212122131132131222211221311
152:2212122131132131222211221311
180:2212122131132131222
199:       131132131222211221311
220:2212122131132131222211221311
248:2212122131132131222211221311
276:221212213113213122221122
300:            2131222211221311
316:2212122131132131222211221311
344:2212122131132131222211221311
372:2212122131132131222211221311

After adjusting the year for these gaps, we can get the result with a simple hash:

$y=$argn%400;foreach([300,199,103]as$k)$y<$k?:$y+=16;
echo"2212122131132131222211221311"[$y%28];

Not short (95 bytes) but pretty. And we can golf

• 4 bytes by using a ternary chain for the offset,
• 8 bytes by converting the hash map from a base4 string to integer,
• one more by using the hex representation,
• and one by merging the expressions.

Answer 26

Python(v2) 120

import datetime as d,sys
s=0
for m in range(1, 13):
    if d.datetime(int(sys.argv[1]),m,6).weekday()==4: s=s+1
print s

Answer 27

Perl + lib POSIX 55

With the idea of not looking for 13th but first, and as sunday is 0 this let save 3 chars! Thanks @ Iszi and Danko Durbić!

$==$_;$_=grep{!strftime"%w",0,0,0,1,$_,$=-1900}(0..11)

Could compute 2010 to 2017 (for sample) in this way:

perl -MPOSIX -pE '$==$_;$_=grep{!strftime"%w",0,0,0,1,$_,$=-1900}(0..11)' <(
    printf "%s\n" {2010..2017})
11321312

(Ok, there is no newline, but that was not asked;)

Old post: 63

$==grep{5==strftime"%w",0,0,0,13,$_,$ARGV[0]-1900}(0..11);say$=

In action:

for i in {2010..2017};do
    echo $i $(
        perl -MPOSIX -E '
            $==grep{5==strftime"%w",0,0,0,13,$_,$ARGV[0]-1900}(0..11);say$=
            ' $i );
  done
2010 1
2011 1
2012 3
2013 2
2014 1
2015 3
2016 1
2017 2

Answer 28

In Smalltalk (Squeak/Pharo flavour), implement this method in Integer (86 chars)

countFriday13^(1to:12)count:[:m|(Date year:self month:m day:13)dayOfWeekName='Friday']

Then use it like this: 2014countFriday13.

Of course, we could use a shorter name, but then it would not be Smalltalk

Answer 29

Japt -x, 10 bytes

CÆ5¥ÐUXD e

Try it

CÆ5¥ÐUXD e     :Implicit input of integer U
C              :12
 Æ             :  Map each X in the range [0,C) (months are 0-indexed in JavaScript)
  5¥           :  Check if 5 is equal to
    ÐUXD       :  new Date(U,X,13)
         e     :  0-based index of day of the week
               :Implicitly reduce by addition and output

Answer 30

JavaScript, 56 bytes

y=>(g=m=>m?g(--m)-(new Date(y,m,13).getDay()!=5):12)(12)

Try It Online!

Answer 31

C++ - Too many bytes :(

I tried a solution which doesn't make use of any date libraries.

I found a pretty cool (if I may say so myself) solution. Unfortunately I can't get it shorter than this, which really bugs me because it feels like there should be a better way.

The solution hinges on this algorithm which is only 44 bytes in itself. Unfortunately I need another 100 bytes to wrap it nicely...

#include<stdlib.h>
main(int,char**v){int f=0,d,m,y;for(m=1;m<13;++m)d=13,y=atoi(v[1]),(d+=m<3?y--:y-2,23*m/9+d+4+y/4-y/100+y/400)%7-5||++f;return f;}

Output through the return code (in C++, using cout or printf or anything like that requires another #include, which would blow up the solution even more).

Driver / test program:

# Make sure we're not running an old version
rm doomsday.exe

gcc doomsday.cpp -o doomsday.exe

./doomsday.exe 1776
echo 1766: $?

./doomsday.exe 2012
echo 2013: $?

./doomsday.exe 2013
echo 2013: $?

./doomsday.exe 2014
echo 2014: $?

echo `wc doomsday.cpp -c` characters

Output of the driver program:

$ ./test_doomsday 
1766: 2
2013: 3
2013: 2
2014: 1
150 doomsday.cpp characters

Answer 32

VB.net

Not a CodeGolf attempt

Iterator Function SpecificDayOfTheWeekInYear(Year As Integer, DayOfTheWeek As DayOfWeek ) As IEnumerable(of Date)
  Dim FirstDayOfTheYear = New Date(Year,1,1)
  Dim CurrentDay = FirstDayOfTheYear.AddDays( DayOfTheWeek - FirstDayOfTheYear.DayOfWeek )
  If CurrentDay.Year < Year Then CurrentDay = CurrentDay.AddDays(7)
  While CurrentDay.Year = Year
    Yield CurrentDay 
    CurrentDay=CurrentDay.AddDays(7)
  End While
End Function

Which makes looking for Friday the 13th's simple.

Dim FridayThe13ths =  From d in SpecificDayOfTheWeekInYear(2013,DayOfWeek.Friday )
                     Where d.Day=13

or a much simpler

Iterator Function YieldDayInMonth(Year As Integer, Day As Integer) As IEnumerable(of date)
  For M=1 To 12
   Yield New Date(Year,M,Day )
  Next
End Function

This makes the coding thus.

Dim f13 = YieldDayInMonth( 2013, 13).Where( Function(d) d.DayOfWeek = DayOfWeek.Friday)