Happy Rounded Pi Day

Question

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!

Answer 1

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

Answer 2

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!

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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!

Answer 9

GNU Awk, 23 characters

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

Sample run:

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

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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.

Answer 14

Ruby, 40 bytes

->n{n*Time.new.strftime('%m.%d%y').to_f}

Answer 15

Mathematica, 54

.0001FromDigits[Now@{"Month","Day","YearShort"},100]#&

Answer 16

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.

Answer 17

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.

Answer 18

Oracle 11g, 50 49 bytes

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

one less byte, thanks to @MickyT

Answer 19

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'))

Answer 20

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>.

Answer 21

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))

Answer 22

APL, 19 bytes

⎕×0.01⊥⌽⍎2⌽8↑2↓⍕⎕TS

⎕TS is 2016 3 14 12 34 56 789 i.e. March 14^rd, 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 14^rd, 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

Answer 23

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

Answer 24

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]}

Answer 25

Java 114 bytes

public double p(double d){
  return (d*Double.parseDouble(new SimpleDateFormat("MM.ddyy").format(new Date())));
}

Answer 26

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.

Answer 27

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

Answer 28

R 48 bytes

d*as.double(format(Sys.Date(),format="%m.%d%y"))

Answer 29

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 

Answer 30

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.