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To celebrate Rounded Pi Day, you must take advantage of today's date to make a program that takes in a circle's diameter and outputs its circumference by multiplying the diameter by 3.1416, where 3.1416 must be obtained using today's date.

Input 3
Output 9.4248
etc.

Does not take the date as input. You do not have to use all components of the date but the digits of pi must come from formatting a Date object or using a date object to obtain it. The answers so far look good.

Not sure what else to specify. All the answers so far meet what I was expecting.

My constraint was that you must use components of the date to come up with Pi. You can of course use the components, multiply by 0 then add 3.1416, but that's boring and wastes precious chars!

Shortest code wins!

share|improve this question
59  
What's so special about 14/3/16? – Neil Mar 14 at 13:01
5  
define using today's date. I could get the date as a number, divide by itself and multiply by a predefined constant for pi – Luis Mendo Mar 14 at 13:02
7  
@Neil: Because 3/14/16 and 31/4/16 are not valid actual dates in D/M/Y format. There aren't 14 months, and April only has 30 days. I personally think we should wait until 6/28/32 (M/D/Y) or maybe 6/2/83 (D/M/Y), but that's a whole other holy war. – Darrel Hoffman Mar 14 at 14:38
8  
@DarrelHoffman You're a Tau man I see. – jmasterx Mar 14 at 14:46
16  
Sadly, its only PI day in the US cultural area. Europeans, with their silly lexigraphcially-sensible date ordering, don't get to have yearly PI days. Spare a sad thought for them, as you're eating your pi(e) today. – T.E.D. Mar 14 at 18:03

37 Answers 37

up vote 8 down vote accepted

05AB1E, 16 13 bytes

žfžežg¦¦J*4°/

Try it online.

Unfortunately a bug with floats makes this a byte longer :/

Thanks to Adnan for golfing off 3 bytes.

Explanation

žfžežg¦¦J*4°/
žfžežg         push month day year
      ¦¦       slice off the first two chars from the year (2016 -> 16)
        J      join them together into a number
         *     multiply by input
          4°/  divide by 1e4
share|improve this answer
    
¦¦ instead of 2000- is three bytes shorter ;) – Adnan Mar 14 at 16:11
11  
+1 žfžežg sounds like a nice Czech word :) – yo' Mar 15 at 9:02
    
In UTF-8, this totals 19 bytes. – OldBunny2800 Mar 16 at 21:21
3  
@OldBunny2800 05AB1E uses CP1252 for encoding. – quartata Mar 16 at 21:21
    
Bug with floats? – CalculatorFeline Apr 7 at 5:18

C, 32 bytes

#define f(d)d*time(0)/464083315.

If losing a little more accuracy is fine, I can get it down to about 29 bytes being still accurate to 4 digits past the decimal (as of the time of this posting):

#define f(d)d*time(0)/46408e4
share|improve this answer
    
#include<time.h>? – cat Mar 14 at 15:08
    
Also, where's the executable code? this is just something for the preprocessor to copy/paste; a snippet. We like runnable implementations here. – cat Mar 14 at 15:09
3  
@tac most C compilers do not require you to #include standard library headers. Including them manually is almost always a good idea, but omitting them makes code golfing in C a lot easier. – Josh Mar 14 at 15:21
12  
@tac the answer behaves mostly the same as a function. Several answers on here are standalone functions as well. I don't see the point in providing a sample main function to demonstrate how to call this single argument macro. – Josh Mar 14 at 15:25
1  
@tac I think the standard [code-golf] rules is that unless forbidden functions are ok. – Mindwin Mar 14 at 19:32

AppleScript, 122 120 bytes

set a to current date
(display dialog""default answer"")'s text returned*(month of a+day of a*.01+year of a*1e-4 mod.01)

Variable a

Variable a is a date object. I call all of my date information off of it.

Month, day, and year

The month, day, and year calls actually return an object that normally returns a string. To properly use it as a number, I have surrounded it on both sides with mathematical operations to automatically cast it to a number.

1e-4

1e-4 is a byte shorter than .0001.

mod.01

. acts as a separator to the AppleScript autocorrect. By using this, I can use modulo and still keep it a byte for myself.

No return statement/log

The last value calculated automatically is returned by the program. I output the number calculated via the return box.

And here's a gif of it running!

pi day gif

share|improve this answer
7  
Forget Java, this has got to be the most verbose language – Downgoat Mar 15 at 14:52
    
@Downgoat I guess that it's supposed to be "readable like English" or something... we use it a bit at one of my jobs and I really dislike whenever I have to make changes to such scripts. – Chris Cirefice Mar 15 at 15:49
    
@ChrisCirefice Yeah, no - this language has some annoying pieces to it. i.e. volume system volume. – VTCAKAVSMoACE Mar 15 at 16:00

Mathematica + coreutils, 20 bytes

<<"!date +%m.%d%y"#&

Luckily, Mathematica interprets the output of an external command as an expression. In this case the output is a valid float, so that multiplication with the function argument # is implied.

share|improve this answer
    
That's a combination I didn't know existed. – CalculatorFeline Apr 5 at 7:07

Lua, 30 27 Bytes

print(...*os.date"%m.%d%y")

Multiply the first command-line argument by the current date in format mm.ddyy, which is actually 03.1416.

share|improve this answer

Bash + bc, 25 20 bytes

date +%m.%d%y\*$1|bc

Thanks to manatwork for saving five bytes.

Usage:

$ ./pi-round.sh 3
9.4248
share|improve this answer
    
That here-string notation is so long: date +%m.%d%y\*$1|bc – manatwork Mar 14 at 14:28
    
@manatwork you are correct! I was playing around with the same myself but couldn't figure out how to combine it. Thanks! – andlrc Mar 14 at 14:31

PowerShell v2+, 46 28 25 bytes

$args[0]*(Date -F "M.dy")

Pretty straightforward. Takes input $args[0] and multiplies it by the date formatted as M.dy (the Get- is implied). Note that this may take a long time to run on v2 as it iterates possible substitutions for Date (e.g., checking your %PATH% environment variable, etc.) before settling on Get-Date.

share|improve this answer
    
Can you cut the space after -f? – briantist Mar 15 at 4:02
1  
@briantist Sadly, not here. We're abusing PowerShell's pattern recognition, since the actual flag is -Format. With flags like this, you just need to be unambiguous (for example, if you had -Debug and -Delimiter as potential options, you would need at least three letters to differentiate). Removing the space means PowerShell tries to parse the flag -F"M.dy" but can't find an argument that matches, and so throws an error. – TimmyD Mar 15 at 12:23
    
ahhh I see it now. That's what I get for reading this on a train after a long day. I was interpreting it as the -f operator. Makes total sense now. – briantist Mar 15 at 13:27

GNU Awk, 23 characters

$0*=strftime("%m.%d%y")

Sample run:

bash-4.3$ awk '$0*=strftime("%m.%d%y")' <<< 3
9.4248
share|improve this answer

R 3.2.4, 55 51 47 bytes

edit I realized I could use scan thanks @FryAmTheEggMan. Reduced 4 bytes thanks to @MickyT.

scan()*as.numeric(format(Sys.Date(),'%m.%d%y'))

First attempt at a golf. Happy pi day!

share|improve this answer
1  
Welcome to PPCG :) I'm no R programmer, but you don't need to supply a named function, just an expression that evaluates to a function. So I'm pretty sure you can drop the f=. – FryAmTheEggman Mar 14 at 16:11
2  
You can go even further and do scan()*as.numeric(format(Sys.Date(),'%m.%d%y')) – MickyT Mar 14 at 16:17
2  
as.double instead of as.numeric saves an additional byte – Flounderer Mar 15 at 4:09

Pyth, 19 bytes

*Qvs[.d4\..d5>2`.d3

Try it here!
Only works at 2016-03-14 of course.

Explanation

*Qvs[.d4\..d5>2`.d3   # Q = input

     .d4              # current month
        \.            # dot between month and day
          .d5         # current day of the month
             >2`.d3   # last 2 digits of the year
   s[                 # concat everything into a string
  v                   # eval -> convert to float
*Q                    # multiply with input to get the circumference
share|improve this answer

PHP, 45 26 25 24 bytes

Uses Windows-1252 encoding

<?=$argv[1]*date(~‘Ñ•†);

Run like this:

echo '<?=$argv[1]*date(~‘Ñ•†);' | php -- 3
  • Saved 19 bytes by using date() instead of DateTime::format
  • Saved a byte by using the echo tag
  • Saved a byte by using ISO-8859 encoding and negating the format string, so no need for double quotes. Might mess up your terminal a bit when running it from CLI, but works.
share|improve this answer
    
If you assume register_globals to true, you can use $n via /?n=3 in the url. – Martijn Mar 15 at 9:23
    
@Martijn That requires PHP 4.1 or setting it using the php.ini file. aross said before that wants to steer away from such method. – Ismael Miguel Mar 15 at 9:28

Python 2, 58 chars

import time
print float(time.strftime("%m.%d%y"))*input()

Try it!

Explanation:

import time                                   # Import time time module
print                                         # Print the following
      float(                                  # Cast to float
            time.strftime("%m.%d%y"))         # Format the time as MM.DDYY
                                     *input() # Multiply by input
share|improve this answer

JavaScript, 41 39 characters

This uses a proprietary Firefox-only method.

d=>new Date().toLocaleFormat`%m.%d%y`*d

Thanks to:

  • Ismael Miguel for the template string suggestion (-2 characters).

Sample run (Firefox Web Console):

> (d=>new Date().toLocaleFormat`%m.%d%y`*d)(3)
9.4248

Note: this will fail in Firebug Console. Seems that Firebug performs some expansion on the template string, transforming `%m.%d%y` into `%__fb_scopedVars(m).d%y` before passing it to the code.

share|improve this answer
1  
-2 bytes: d=>new Date().toLocaleFormat`%m.%d%y`*d – Ismael Miguel Mar 15 at 9:29
    
Oh. [facepalm] No idea why, I usually forget that trick. Although I use the template strings regularly. – manatwork Mar 15 at 9:33
    
I know. I really don't get either. What I'm trying to get is a way to remove that new from there. – Ismael Miguel Mar 15 at 9:35
    
I tried it on Firefox and it worked fine. I copy-pasted from the console. Since I've replaced ('...') with 2 backticks, that's 2 bytes. – Ismael Miguel Mar 15 at 9:41

Ruby, 40 bytes

->n{n*Time.new.strftime('%m.%d%y').to_f}
share|improve this answer

Mathematica, 54

.0001FromDigits[Now@{"Month","Day","YearShort"},100]#&
share|improve this answer

Vitsy + coreutils, 19 bytes

'y%d%.m%+ etad',W*N

Explanation:

'y%d%.m%+ etad',W*N
'y%d%.m%+ etad'       Push 'date +%m.%d%y' to the stack.
               ,      Execute as shell.
                W     Grab input from STDIN and eval it.
                 *    Multiply the top two items
                  N   Output as a number.

Cannot be run in safe mode, as this uses the Runtime's exec method.

share|improve this answer

SpecBAS, 39 bytes

1 INPUT n: ?n*VAL DATE$(TIME,"mm.ddyy")

Nothing out of the ordinary. Formats date as a string then multiplies the input by the VAL (value) of that string.

share|improve this answer
    
Interpreter / compiler link? – cat Mar 14 at 15:06
    
sites.google.com/site/pauldunn – Brian Mar 14 at 16:06
    
I've edited the link into your answer – cat Mar 14 at 16:07

Oracle 11g, 50 49 bytes

SELECT &n*to_char(sysdate,'MM.DDYY')FROM dual;

one less byte, thanks to @MickyT

share|improve this answer
    
you can drop the space before the FROM – MickyT Mar 14 at 18:05

Python 3, 74 54 bytes

using the time module instead of datetime like Loovjo's Answer

import time
lambda n:n*float(time.strftime('%m.%d%y'))

previous solution

from datetime import*
lambda n:n*float(datetime.now().strftime('%m.%d%y'))
share|improve this answer
2  
@Zenadix The meta consensus is that unnamed functions are OK. – FryAmTheEggman Mar 14 at 16:09

Google Sheets, 13 bytes

Bytes are calculated with one byte per character, zero bytes per cell except for the designated input cell, two bytes. The equals sign to start a formula is not counted. (I don't know if this is standard, please correct me if I am wrong.)

Run snippet to see the code.

table {
  empty-cells: show;
}
table,
th,
td {
  border: 1px solid black;
}
<table border="1">
  <tr>
    <th></th>
    <th>1</th>
    <th>2</th>
    <th>3</th>
  </tr>
  <tr>
    <td>A</td>
    <td>&lt;input></td>
    <td style="text-align:right">3.1416</td>
    <td style="font-family:'Ubuntu Mono',monospace">=<span style="color:rgb(257,142,29)">A1</span>*<span style="color:rgb(126,55,148)">A2</span>
    </td>
  </tr>
</table>

This works because you can define your own date formats in Google Sheets. I made it so that it is <month>.<day><year>.

share|improve this answer
    
You might want to ask on meta about this scoring system. ;P – VTCAKAVSMoACE Mar 16 at 21:27
    
Will do when I have time (maybe tonight in Maryland) – OldBunny2800 Mar 16 at 21:29

Pyke, 16 15 bytes, ASCII encoding

C867wä%J"bT4^/*

Explanation:

C867            - b,c,d = year, day, month
    wä          - a = base_96("ä") - 100
      %         - a = a%b
       J"       - a = "".join([a,c,d])
         b      - b = base(a, 10)
          T4^   - a = 10^4
             /  - a = a/b
              * - imp_print(a*eval_or_not(input))

or 11 bytes (non-competitive), (adds string singles, 2 digit year, string sumables)

C856\.R3sb*

Explanation:

C856        - a,b,c = 2d_year, day, month
    \.      - d = "."
      R3    - a,b,c = b,c,a
        s   - a = sum(a,b,c)
         b  - a = base(a, 10)
          * - imp_print(a*eval_or_not(input))
share|improve this answer
    
I'm guessing this is a different Pyke than pyke.sourceforge.net ? Can you post a link to the language spec? – Robert Fraser Mar 15 at 4:08
    
    
Just wondering, what encoding does Pyke use? If it's UTF-8, this is 16 bytes. – OldBunny2800 Mar 16 at 21:30
1  
@OldBunny2800 It doesn't matter; in this case it uses good old ascii as all code points are below 255. According to meta.codegolf.stackexchange.com/a/5879/32686, the answer can define it's own encoding at no cost – muddyfish Mar 16 at 21:42

JavaScript ES6, 68 66 bytes

Saved 2 bytes thanks to dev-null!

x=>x*((a=new Date).getMonth()+1+"."+a.getDate()+(a.getYear()-100))

Anonymous function. Some uses:

f(1)    == 3.1416
f(5)    == 15.708
f(f(2)) == 19.73930112
share|improve this answer
1  
You should try using with – Ismael Miguel Mar 15 at 9:36

Tcl/Tk, 58 bytes

puts [expr {[gets stdin]*[clock f [clock se] -f %N.%d%g]}]

(That's a complete program. If you want to cheat and remove the explicit puts statement then it's only 51 bytes -- you'd have to start tclsh and type/paste the following in directly, though:)

expr {[gets stdin]*[clock f [clock se] -f %N.%d%g]}
share|improve this answer

Java 114 bytes

public double p(double d){
  return (d*Double.parseDouble(new SimpleDateFormat("MM.ddyy").format(new Date())));
}
share|improve this answer
    
96 bytes: float p(float d){return d*Float.parseFloat(new SimpleDateFormat("MM.ddyy").format(new Date()));} – VTCAKAVSMoACE Mar 14 at 18:04

Racket, 112 characters

(define d(seconds->date(current-seconds)))(*(read)(+(date-month d)(*(date-day d).01)(*(-(date-year d)2e3)1e-4)))

Reads the number from input in standard reader syntax.

share|improve this answer

TI-84 Basic, 30 bytes

Works on TI-83/84 calculators; E is the scientific notation token and ~ is the negative token.

Prompt D:getDate:D(Ans(2)+E~4Ans(1)-.2+.01Ans(3

Test Case

D=?3
          9.4248
share|improve this answer

R 48 bytes

d*as.double(format(Sys.Date(),format="%m.%d%y"))
share|improve this answer

MATL, 17 bytes

Z'2$'mm.ddyy'XOU*

Try it online!

Z'              % get current date and time as float 
2$'mm.ddyy'XO   % format as a string with custom format 
U               % convert to number 
*               % multiply by implicit input 
share|improve this answer

TI-BASIC, 16 13 9 bytes

Xround(π,1+min(getDate

We round π to a number of decimal places equal to the minimum of {month,day,year}, and then multiply it by the input.

This is a function that takes input through X. Store it to one of the Y-variables, for example Y1, and then call like Y1([number]) on the homescreen.

share|improve this answer
    
There seems to be some disagreement that taking input via Ans is allowed. – TimmyD Mar 15 at 15:55
    
How is this only 13 bytes? Not counting Input A, I count 22 characters. – OldBunny2800 Mar 16 at 21:15
1  
@OldBunny2800 TI-BASIC is token based, each atom is represented as one or two bytes. – Adám Mar 16 at 21:22
    
@OldBunny2800 The code page is here. – lirtosiast Mar 16 at 22:10

APL, 19 bytes

⎕×0.01⊥⌽⍎2⌽8↑2↓⍕⎕TS

⎕TS is 2016 3 14 12 34 56 789 i.e. March 14rd, 2016 right before 12:35 pm
make into string, i.e. '2016 3 14 12 34 56 789'
8↑2↓ drop first two ('20') then take next eight ('16 3 14 ')
2⌽ rotate two characters, giving ' 3 14 16'
make into numbers (3 14 16)
reverse the list, giving 16 14 3
0.01⊥ evaluate in base ¹⁄₁₀₀, = 16 × 0.01² + 15 × 0.01¹ + 3 × 0.01⁰ = 0.0016 + 0.15 + 3 = 3.1416
⎕× multiply with input

or

⎕×1E¯4⊥⌽⍎2⌽7↑2↓⍕⎕TS

⎕TS is 2016 3 14 12 34 56 789, i.e. March 14rd, 2016 right before 12:35 pm
make into string, i.e. '2016 3 14 12 34 56 789'
7↑2↓ drop first two ('20') then take next seven ('16 3 14')
2⌽ rotate two characters, giving ' 3 1416'
make into numbers (3 1416)
reverse the list, giving 1416 3
1E¯4⊥ evaluate in base ¹⁄₁₀₀₀₀, = 1416 × 0.01¹ + 3 × 0.01⁰ = 0.1416 + 3 = 3.1416
⎕× multiply with input

share|improve this answer
    
How does this work? – OldBunny2800 Mar 16 at 20:49
1  
@OldBunny2800 Clear? – Adám Mar 16 at 21:09
    
Yes, thank you, but in UTF-8 this is 38 bytes. – OldBunny2800 Mar 16 at 21:14
    
@OldBunny2800 meta.codegolf.stackexchange.com/questions/5878/… – Adám Mar 16 at 21:21
1  
@OldBunny2800 APL uses its own code page -- each of these characters are one byte. (This was before ASCII mind you...) – quartata Mar 16 at 21:26

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