Pi Day, Pi Minute, or Pi Second?

Question

In this challenge you will need to determine whether it's Pi Day, Pi Minute, or Pi Second.

Because Pi is irrational, it wants your code to be as short as possible.

Examples

No input is provided, your program should use the system time. I've just added it for clarity

March 14, 2016 0:00:00
Pi Day
December 25, 2015 3:14:45
Pi Minute
December 29, 2015 0:03:14
Pi Second
January 1, 2016 0:00:00
<No Output>

What is Pi Day / Minute / Second

• Pi Day is when the month is March, and the date is the 14th
• Pi Minute is when the hour is 3, and the minute is 14
• Pi Second is when the minute is 3, and the second is 14
• Pi Day should be preferred instead of Pi Minute or Pi Second, and Pi Minute should be preferred instead of Pi Second.
• For this challenge you should use 12-hour time (15:14 == 3:14). The date/time used to determine the Pi Day/Minute/Second should be based on system time.

Scoring & Bonus

-15 byte Bonus: If you print "No Pi Time" when it's not Pi time.

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As always, standard loopholes are disallowed. This is code-golf shortest code in bytes wins!

Answer 1

CJam, 41 bytes

et[3E]#"
Pi Day

Pi Minute
Pi Second
"N/=

Test it here. Alternatively use this link to stub the result of et for easier testing.

Explanation

et   e# Get the current datetime as an array with the following elements:
     e#   - Year
     e#   - Month
     e#   - Day
     e#   - Hour
     e#   - Minute
     e#   - Second
     e#   - Millisecond
     e#   - Weekday
     e#   - Tickrate or something.
[3E] e# Push the array [3 14].
#    e# Find the position of this subarray in the current datetime array. Let's see
     e# what results we can get:
     e#   - Year 3 is in the past and there is no 14th month, so we can't get 0.
     e#   - Pi day will give result 1.
     e#   - Day 3, hour 14 would give 2.
     e#   - Pi minute will give result 3.
     e#   - Pi second will give result 4.
     e#   - Second 3, millisecond 14 would give 5.
     e#   - Weekday and tickrate won't be 14, so we'll never get 6 or 7.
     e#   - If [3 14] isn't found at all we get -1.
"\Pi Day\\Pi Minute\Pi Second\"
     e# Push this string (with linefeeds instead of backslashes.
N/   e# Split into lines.
=    e# Select the corresponding element. The non-existent "pi hour" and "pi millisecond"
     e# would map to empty strings as well as the -1.

Answer 2

Javascript (ES6), 114 112 - 15 = 97 bytes

x=>['Pi Day','Pi Minute','Pi Second'].find((x,i)=>[/ar 14/,/(03|15):14:/,/03:14/][i].test(Date()))||'No Pi Time'

Ungolfed:

x=>
['Pi Day', 'Pi Minute', 'Pi Second']  // array of outputs
.find(                                // find first element in the array
    (x, i)=>                          // which returns truthy for this function
    [/ar 14/, /(03|15):14:/, /03:14/] // array of regex patterns
    [i]                               // get corresponding regex based on index
    .test(Date())                     // test it against current date, date is automatically cast to string
) || 'No Pi Time'                     // if no result, then return "No Pi Time"

Thanks for -2 bytes @edc65

Answer 3

Ruby, 125 124 chars

i=[*[(t=Time.new).month,t.day,t.hour,t.min,t.sec].each_cons(2)].index [3,14];i&&$><<['Pi Day','','Pi Minute','Pi Second'][i]

Alas, the cleverer %i[month day hour min sec].map{|x|Time.new.send x} is longer.

The key here is the use of each_cons to avoid repetition (see the last few lines of the explanation below).

i=                          # send i (index) to...
[*                          # convert to array (splat)...
 [
  (t=Time.new).month,       # the current month...
  t.day,t.hour,t.min,t.sec  # etc... (duh)
 ]
 .each_cons(2)              # each consecutive two elements
]                           # [[month, day], [day, hour], [hour, min], etc]
.index [3,14];              # first occurrence of [3, 14]
i&&                         # shorthand for "if i"...
$><<                        # output...
[
 'Pi Day',                  # [month=3, day=14] is Pi Day
 '',                        # [day=3, hour=14] isn't anything
 'Pi Minute',               # [hour=3, min=14] is Pi Minute
 'Pi Second'                # [min=3, sec=14] is Pi Second
][i]                        # index by index (obviously)

Answer 4

Python 2, 219 186 183 Bytes (198-15)

_{I tried}

Ungolfed:

from datetime import datetime

now = datetime.now()
output = ['Pi Day', 'Pi Minute', 'Pi Second', 'No Pi Time']

if now.month == 3 and now.day == 14:
    print output[0]
elif now.hour == 2 and now.minute == 13:
    print output[1]
elif now.minute = 2 and now.second == 13:
    print output[2]
else:
    print output[3]

Golfed:

from datetime import *
n=datetime.now()
a=n.minute
if n.month==3and n.day==14:print'Pi Day'
elif n.hour==2and a==13:print'Pi Minute'
elif a==2and n.second==13:print'Pi Second'
else:print'No Pi Time'

Answer 5

Japt, 78 - 15 = 63 bytes

D=Ð)g ¥3©Df ¥E?"Pi Day":Dd %C¥3©Dc ¥E?`Pi M©e`:Dc ¥3©Db ¥E?`Pi SeÖ:No Pi Te

Pretty straightforward - just checks the date for every case.

Explanation:

• D=Ð)g get the current date (Ð), store it in the variable D and get the month (g). Why store it in the variable, if it's already one-letter? Because since then you can dress any part of date with Da, where a is the function, returning year, month, date, etc. But otherwise you'd have to do Ð a, see the space.

• ¥3 is ==3, checking if the month is March.

• © is &&, i.e. "and".
• Df is the day of the month.
• E is 14
• ?...:... - typical sets of ternary operators
• Dd %C the reminder of dividing the hour (Dd) by 12 (C)
• Dc is the minutes
• Db are seconds

Try it online!

---

To emulate Pi Day:

D=Ð"3/14/11 11:11:11";
Dg ¥2©Df ¥E?"Pi Day":Dd %C¥3©Dc ¥E?`Pi M©e`:Dc ¥3©Db ¥E?`Pi SeÖ:No Pi Te

To emulate Pi Minute:

D=Ð"11/11/11 3:14:11";
Dg ¥2©Df ¥E?"Pi Day":Dd %C¥3©Dc ¥E?`Pi M©e`:Dc ¥3©Db ¥E?`Pi SeÖ:No Pi Te

To emulate Pi Second:

D=Ð"11/11/11 00:3:14";
Dg ¥2©Df ¥E?"Pi Day":Dd %C¥3©Dc ¥E?`Pi M©e`:Dc ¥3©Db ¥E?`Pi SeÖ:No Pi Te

Answer 6

TI-BASIC, 124 bytes

Thanks to FlagAsSpam for shaving a few bytes.

"→Str1
getTime
If min({3,14}={Ans(2),Ans(3
"Pi Second→Str1
getTime
If Ans(2)=14 and max(Ans(1)={3,14
"Pi Minute→Str1
getDate
If min({3,14}={Ans(2),Ans(3)
"Pi Day→Str1
Str1

Answer 7

Perl, 80 - 15 = 65 bytes

print'No 'x localtime!~/(ar | 15:|03:)14/,'Pi ',(Time,Day,Minute,Second)["@-"/4]

Take 2, parsing the string representation of localtime. At present, this looks something like this:

Sun Jan  3 15:14:15 2016

The position of the matched string is used to determine the correct Pi Time.

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Perl, 100 bytes

@t=localtime;$t[2]%=12;3-/3/^{@t[$_,$_+1]}->{14}||exit!print'Pi ',(Second,Minute,_,Day)[$_]for 3,1,0

localtime returns the months zero indexed, hence the need for 3-/3/.

Answer 8

Python 3, 137 - 15 = 122 bytes

import time
T=list(time.localtime())
T[3]%=12
print({1:'Pi Day',3:'Pi Minute',4:'Pi Second'}.get(bytes(T[1:6]).find(b'\x03\x0e'),'No Pi Time'))

The 12-hour requirement was unfortunate, as this would've been 118-15=103 bytes without it:

import time
print({1:'Pi Day',3:'Pi Minute',4:'Pi Second'}.get(bytes(time.localtime()[1:6]).find(b'\x03\x0e'),'No Pi Time'))

Answer 9

AppleScript, 202 190 187 183 181 bytes

Hey, this isn't so bad after all.

set n to current date
set b to n's time string
if n's date string contains"March 14"
log"Pi Day"
else if b contains"3:14:"
log"Pi Minute"
else if b contains"3:14"
log"Pi Second"
end

I actually found a use for AppleScript's method calling. Go figure. Nope. Just turns out that I'm an idiot. Setting a variable is shorter.

(for those wondering, current date command returns a date-type with the contents "Saturday, January 2, 2016 at 2:46:01 PM" or the like)

Answer 10

PHP, 85 - 15 = 70 bytes

<?=['No Pi Time','Pi Day','Pi Minute','Pi Second'][strpos(@date(Ymdhi_is),'0314')/4];

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The main trick use here is the Ymdhi_is date format, at the time of writing, date('Ymdhi_is') returns 201501030258_5828.

• md, hi and is are the values who will be replaced by 0314 if it's Pi-something. Note that all those strings will be always be replaced by a 4-character long string.
• They are put in that specific order since strpos will stop searching at the first occurrence of the needle, so we put them in the order of priority.
• A separator between hi and is is necessary because we don't want strpos to match a value that would overlap both (thanks to primo for saving bytes here).
• The needle is 0314 because 314 would wrongly match 10:31:42 as Pi-Second.

The Y part is the trickiest. We need a prefix to to offset the first occurrence of Pi-something, allowing us to distinguish strpos's return values between false (not found, Pi-nothing) and 0 (found at index 0, Pi-day).

And we want this prefix to be either 4 or 5-character long, since we are planning to divide strpos's return value by 4.

Y is 4-character long, but:

• it will be 5-character long someday, and this will break the program (think about year 10314): the PHP documentation says that Y will be replaced by 4 digits, but it's not true.
• if you come back in time, at year 314, it will break the program.

Since PHP didn't exist in year 314, and will likely not exist anymore in year 10314, I guess these bugs can be safely ignored.

Note that 0314 can overlap Ymd since:

• Ymmd configuration: there's no 31st month.
• YYmm configuration: there's no 14th month.
• YYYm configuration: there are less than 40 months.

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Also, there's a version without the bugs related to the year, which is 86 - 15 = 71 bytes:

<?=['No Pi Time','Pi Day','Pi Minute','Pi Second'][strpos(@date(D_mdhi_is),'0314')/4];

Answer 11

Python 3, 179 bytes

import functools as F,datetime as D
T,G=D.datetime.now(),getattr
F.reduce(lambda i,j:print("Pi "+i.title())if G(T,i)/G(T,j)==3/14else j,"month day hour minute second".split(" "))