Prime Wednesdays

Question

Prime Wednesdays

Your task is to count the number of Wednesdays that fall on a prime day of the month in a particular year. For instance, 7-13-16 is a prime Wednesday. For consistency use the Gregorian calendar for all dates.

Input

The input to your program/function will be a year (eg 2016) and is flexible. The year will be an integer between 1912 and 2233 inclusive.

Output

The output is also flexible and should be the number of prime Wednesdays (eg 18).

Scoring

This is code-golf so shortest code in bytes wins!

Test Cases

input -> output
--------------------
1912 -> 19
1914 -> 16
1984 -> 17
1996 -> 19
2063 -> 19
2150 -> 16
2199 -> 18
2233 -> 18

Answer 1

MATL, 38 36 34 bytes

FT+"@llI$YO]q&:t8XO!s9\~)9#1$ZOZps

Try it online! Or verify all test cases (takes a few seconds).

Explanation

FT+     % Input year implicitly. Add [0 1] element-wise. Gives array with input year
        % and next year
"       % For each of those two years
  @     %   Push year
  ll    %   Push 1 twice. This indicates January 1.
  I$YO  %   Convert year, month, day to serial date number
]       % End for each. We now have the serial date number for January 1 of the input
        % year and that of the following year
q       % Subtract 1 to the latter, to yield December 31 of the input year
&:      % Inclusive range between those two numbers. This gives an array of serial date
        % numbers for the whole input year
t       % Push another copy of that array
8XO     % Convert to date string with format 8. This gives weekday as "Mon", "Tue" etc.
        % The result is a 3-column 2D char array, where each row is a day
!s      % Transpose, sum of each column. 'Wed' gives 288 (sum of ASCII codes)
9\~     % 288 gives 0 modulo 9, and is the only weekday to do so. So we compute modulo 9
        % and negate. This gives true for Wednesdays, false for the rest
)       % Apply as logical index into the array of serial date numbers
9#1$ZO  % Array of month numbers corresponding to those serial date numbers
Zp      % Array that contains true for prime numbers, false for the rest
s       % Sum of array. Display implicitly

Answer 2

Python 2, 95 93 68 67 bytes

lambda y:0x10ea2c8dbb06c5619/5**((y+((y-22)/99-y/2002)*16)%28)%5+16

Thanks to @Josay for golfing off 1 byte!

Test it on Ideone.

Answer 3

Brain-Flak, 6588, 2310, 2308, 2290 bytes

First things first, I did not write nearly 100% of this program, which is probably evidenced by the enormous size of the program. Most of this code was written by my own Brain-Flak golfing algorithm. Along with an additional python script I wrote to prompt it in the right direction.

Try it online!

({}<(((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((()()())){}{}){})[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])()())[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])())()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())[()()()])())())[()])[()])()()())[()()])()())[()()()()])()())())[()()])())()())[()()()()])())()()())[()()()])()())[()()()])()())[()]))())()())[()()()()])())()()())>[(((((((((()()()()())){}{}){}){}){}){}[()]){}){}){}]){({}[()]<{}>)}{}({}<{{}}>)

While this program is quite long for code golf, it really is decently short for Brain-Flak. Currently the world record for integer division is over 1000 bytes.

Explanation

The algorithm is quite simple. Because there is a limited number of years available (321) it simply pushes the answers in reverse order under the input and uses a lookup algorithm to find the correct answer. While hardcoding all 321 possibilities may seem rather inefficient with an task as complex as this one and a language as esoteric as brain-flak it may very well be the best solution. (I plan to find out in the coming week).

Since most of the 321 numbers are about 18 on average and they differ by very little from year to year instead of pushing all the numbers individually I push the first year (2233) normally and then just duplicate and change the value a bit for each year after. This way instead of paying to push ~18 for all 321 years I only pay to push ~2 for every year.

Once all the answers have been pushed it subtracts 1912 from the input ({}[(((((((((()()()()())){}{}){}){}){}){}[()]){}){}){}]) (This may be suboptimal, I rewrote the optimizer to skip certain values that I believed would not be optimal because hardcoding numbers is a super-exponential process and running it to completion may have taken a few days).

It then subtracts one from the first element and pops the second element until the result reaches zero, {({}[()]<{}>)}.

It pops the zero, {} and all the elements below the top element ({}<{{}}>).

Answer 4

Bash + common utilities, 39

ncal $1|grep W|factor|egrep -c ': \S+$'

Takes the input year as a command-line parameter. Typically outputs messages like this to STDERR - I think this is legal as per this meta-answer:

factor: ‘We’ is not a valid positive integer

If you want to explicitly suppress the STDERR output then you can do this instead for a score of 43:

ncal $1|grep W|factor 2>-|egrep -c ': \S+$'

Answer 5

Octave, 86 bytes

This is not fast, by any stretch. But that's not really the goal of a code golf, is it?

function r=p(y)r=0;for(i=698346:7:815953)d=datevec(i);r+=d(1)==y*isprime(d(3));end;end

Octave can track dates by "date number" - number of days elapsed where Jan 1, 0 is day 1. By this measure, Jan 3, 1912 (the first Wednesday in our set) is day 698,346. Start there, and iterate through every 7th day (all Wednesdays) until the end of 2233, and add 1 if the year is the target year AND the month-day is prime.

Answer 6

Python 2.7, 166, 165, 150 bytes

from datetime import*
y=input()
d,c=date(y,1,1),0
while d.year==y:n=d.day;c+=n>1<2==d.weekday()>0<all(n%x for x in range(2,n));d+=timedelta(1)
print c

There is certainly room for improvement here. I am rather new to golfing in python. This uses the datetime module. It loops through all the days in the year adding one to an accumulator if it fits the criterion. It then prints the result. Most of the heavy lifting is in the module so the code can be quite slim.

One byte saved thanks to Morgan Thrapp and 15 bytes saved by Pietu1998.

Answer 7

J, 44 bytes

+/@(valdate*3=weekday)@,.&(,/(>:,"0/p:)i.12)

I just discovered that J has builtins for date manipulation.

Usage

Extra commands are used for formatting multiple input/output.

   f =: +/@(valdate*3=weekday)@,.&(,/(>:,"0/p:)i.12)
   (,.f"0) 1912 1914 1984 1996 2063 2150 2199 2233
1912 19
1914 16
1984 17
1996 19
2063 19
2150 16
2199 18
2233 18

Explanation

+/@(valdate*3=weekday)@,.&(,/(>:,"0/p:)i.12)  Input: year
                                       i.12   The range [0, ..., 11]
                              >:              Increment each to get the months [1, ..., 12]
                                    p:        Get the first 12 primes [2, ..., 37]
                                ,"0/          Make a table between each month and prime
                           ,/                 Join the rows
                       ,.&                    Prepend the year to each
                                              The date format is YYYY MM DD
            3=weekday                         Check if each date occurs on Wednesday
    valdate*                                  and is a valid date
+/@                                           Count the number of true values and return

Answer 8

PowerShell v3+, 99 95 bytes

Brute force approach --

param($y)(1..12|%{$m=$_;2,3,5,7,11,13,17,19,23,29,31|?{(date "$m-$_-$y").DayofWeek-eq3}}).Count

Takes input $y, loops from 1 to 12, stores the month temporarily into $m, then loops over every prime from 2 to 31. For each of those, we construct a Get-Date of that particular day, then select only those with DayOfWeek -equal to 3 (i.e., Wednesday). Encapsulates that all in a parens to formulate an array, and takes the .Count thereof.

---

Alternatively, mathematical approach --

PowerShell v3+, 105 bytes

param($y)(16,19,18,20,16,18,19)[($a=(date "1-1-$y").DayOfWeek)]+(1,-3,0,1,2)[$y%5]*($a-in0,2,3,4)*!($y%4)

Winds up being just a hair longer than the brute force approach, but I'm including it here since it may be beneficial to others.

Again takes input $y as the year. This time we're performing strictly math operations based on the first day of the year. We first calculate what day of the week that is, and store that into $a for use later. That indexes into the first array, which gets us the number that is usually correct. We have to add to that a second index based on whether it's a potential leap year, whether it's a Sunday, Tuesday, Wednesday, or Thursday, and based on what the year is.

This is based on the following observation. The first column is what day of week January 1st is, the second is the usual output. Unless the year is one of the middle numbers, then it's the number in parens instead. The final column describes how the %5 indexing works.

Jan-1 -> #  ... Except if $y=       (then it's this number) | $y % 5 =
Sun   -> 16 ... 1928 1956 1984 etc. (17)                    |    3
Mon   -> 19
Tue   -> 18 ... 1924 1952 1980 etc. (20)                    |    4
Wed   -> 20 ... 1936 1964 1992 etc. (17)                    |    1
Thur  -> 16 ... 1920 1948 1976 etc. (17)                    |    0
Fri   -> 18
Sat   -> 19

---

Note: Both of these assume en-us is the current PowerShell setting for culture/date information. The date formatting and DayOfWeek number may need to be adjusted accordingly for other culture variants.

Answer 9

Ruby, 83 + 15 (-rdate -rprime flags) = 98 bytes

Try it online! (Imported modules are inlined because idk if I can use flags in repl.it)

->y{k=0;Prime.each(31){|d|k+=(1..12).count{|m|Date.new(y,m,d).wday==3 rescue p}};k}

Answer 10

R, 149 147 bytes

y=function(x){s=strftime;b=ISOdate
a=seq(b(x,1,1),t=b(x,12,31),b='d')
length(a[s(a,'%u')==3&trimws(s(a,'%e'))%in%c(2,3,5,7,11,13,17,19,23,29,31)])}

Test it on Ideone.

Answer 11

Batch, 248 bytes

@set/ad=0,l=1,n=20
@for /l %%i in (1913,1,%1)do @set/ad=(d+l+1)%%7,l=!(%%i%%4)-!(%%i%%100)+!(%%i%%400)
@goto %l%%d%
:03
:06
@set/an-=1
:12
:13
:16
@set/an-=1
:01
:04
:14
@set/an-=1
:00
:05
:10
:15
@set/an-=1
:02
:11
@echo %n%

Explanation: d is the day of the week, with 0 for Monday, which is conveniently 1st Jan 1912. l is a flag for whether the year is a leap year, 1 for 1912. We then loop through from 1913 to the input year, updating the day of week and recalculating the leap year flag as we go. Finally we use the leap year flag and day of week to index into what is effectively a big switch statement to determine n, the number of prime Wednesdays. Setting n to 20 and decrementing it with fall though is cheaper than using flow control logic, but the upshot is that if the 1st of January of a non-leap year is Thursday or Sunday then there are 16 prime Wednesdays and so on for the other cases.

Answer 12

JavaScript ES6 206 203 199 197 195 183 182 179

Not the shortest, but best I can do for now... Golfing suggestions welcome...

p=n=>--d-1?n%d&&p(n):1;v=Date;D=(x,y)=>new v(x.setDate(x.getDate()-y));W=a=>eval('for(Z=0,z=D(w=new v(a,11,31),(w.getDay()+4)%7);z>new v(a,0,1);)Z+=~~p(d=z.getDate()),z=D(z,7);Z')

Changes:

1. altering ternary component from: 3>=x?3-x:10-x to 6-(x+10)%7, saving: 3 Changes to declaration locations;
2. merged x=w.getDay();z=D(w,6-(x+10)%7) to z=D(w,6-(w.getDay()+10)%7), saving: 4
3. shifted Z=0 from for loop to Date declaration and pushed z=D(w,6-(x+10)%7) into for loop to tidy up, saving: 2
4. shifted w=new Date(a,Z=0,1) declaration into for loop, merging with existing w declaration, saving: 2
5. rewriting the prime finding function to a prime testing function, saving: 12
6. changing +!! to ~~ to reduce and still convert p(d=1) from NaN to 0, allowing Prime Test function to still work, saving: 1
7. Moved all extra functions out of main calling function W, redefined for loop - going in reverse from 31st Dec, writing Date object out as a separate variable, then re-wrote for loop into eval call; saving 3.

@PandaCoder, I'm catching up to you, mate!

Answer 13

JavaScript ES6, 187 182 181 179 bytes

179 Swapped in a for-loop for the while-loop

z=y=>{D=new Date("1/3/1912");N=0;a=()=>D.getFullYear();b=()=>D.getDate();c=()=>D.setDate(b()+7);for(;a()<=y;c())N+=y-a()?0:-1<[2,3,5,7,11,13,17,19,23,29,31].indexOf(b());return N}

181 Compacted the ternary

z=y=>{D=new Date("1/3/1912");N=0;a=()=>D.getFullYear();b=()=>D.getDate();c=()=>D.setDate(b()+7);while(a()<=y){N+=y-a()?0:-1<[2,3,5,7,11,13,17,19,23,29,31].indexOf(b());c()}return N}

182 Combined the two loops

z=y=>{D=new Date("1/3/1912");N=0;a=()=>D.getFullYear();b=()=>D.getDate();c=()=>D.setDate(b()+7);while(a()<=y){N+=a()==y?-1<[2,3,5,7,11,13,17,19,23,29,31].indexOf(b()):0;c()}return N}

187

z=y=>{D=new Date("1/3/1912");N=0;a=()=>D.getFullYear();b=()=>D.getDate();c=()=>D.setDate(b()+7);while(a()<y)c();for(;a()==y;c())N+=-1<[2,3,5,7,11,13,17,19,23,29,31].indexOf(b());return N}

Answer 14

Groovy, 126

Groovy doesn't have prime number validation, had to build that too.

{n->p={x->x<3||(2..Math.sqrt(x)).every{x%it}};(new Date("1/1/$n")..new Date("12/31/$n")).collect{it[7]==4&&p(it[5])?it:0}-[0]}