Output with the same number of digits of pi as the length of the code

Question

Your challenge is to print x digits of pi where x is your code length.

Examples:

Source code (length) -> output
foo!        (4)         3.141
foobar      (6)         3.14159
kinda long  (10)        3.141592653
+++++       (5)         3.1415
123456789   (9)         3.14159265

You can use floor(π/10 * 10code_length) / 10code_length - 1 to determine the number of digits you need to print. Note that the decimal point does not count as a digit of pi - so code of length 3 should have output 3.14, of length 4.

Rules:

• Your code length must be larger than three characters.
• You may not use any standard loopholes.
• You may use any standard allowed output methods.
• You may not read your source to determine code length.
• You may not use a builtin pi constant.
• Pi must be completely accurate and not approximated.
• The decimal point in the output is necessary. If you choose to output via return value, you must return a floating point integer.
• The goal of this challenge is to find the shortest solution in each language, not the shortest language for the solution. Don't be afraid to post a solution in a language when a shorter solution is posted in the same language as long as your solution uses a different method.

Answer 1

Mathematica, 18 bytes

using zeta function

Sqrt[6Zeta@2]~N~18

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Answer 2

Jelly, 9 bytes

“Œı⁸Ç’÷ȷ8

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Outputs 3.14159265

How it Works

“Œı⁸Ç’     - The integer 314159265
      ÷    - divide
       ȷ8  - The integer 100000000

Answer 3

Jelly, 6 bytes

3;⁽4ƒṾ

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Prints 3,14159 with the allowed "European" decimal separator, ,.

How?

3;⁽4ƒṾ - Main link: no arguments
3      - literal 3                3
  ⁽4ƒ  - base 250 literal 14159   14159
 ;     - concatenate              [3,14159]
     Ṿ - unevaluate               ['3', ',', '1', '4', '1', '5', '9']
       - implicit print           3,14159

Answer 4

VBA (Excel), 15 bytes

MsgBox 4*Atn(1)

Output: 3.14159265358979

I think this is not acceptable as I golfed it just to fit the length of the PI. The un-golfed length of it is 17 bytes. D:

Answer 5

Python 3, 64 bytes

print("3.%d"%int("SIXW57LUPVBUA10HBQJOL57QLF54UT0U00KIN32O",36))

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Answer 6

SOGL V0.12, 10 9 7 bytes

4ΘΞ“6⌡υ

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Explanation:

4ΘΞ“     push 31415
    6⌡   6 times do
      υ    divide by 10

Answer 7

CJam, 5 bytes

3.14F

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How it works

3.14   e# Push 3.14.
    F  e# Push 15.

Answer 8

Desmos, 130 bytes

f\left(\right)=\frac{\operatorname{floor}\left(10^{128}\cdot 4\sum_{k=1}^{9^9}\frac{\left(-1\right)^{k+1}}{2k-1}\right)}{10^{128}}

The source code for this (Which can be accessed by copy and pasting inside and outside of Desmos) isn't optimal when generated using the Desmos editor, so a few bytes of whitespace were golfed down where possible.

This defines a function f which takes no arguments returns pi, calculated using the Gregory Sequence to k=9^9 (I can't confirm this is accurate enough, however I am of the believe that it is, it can be made more accurate with a greater value of k) it then floors the result to 128 decimal places, which, alongside the 3., is the length of the source code.

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Answer 9

Japt, 9 4 bytes

As short as it can allowably get :)

Includes an unprintable (reverse line feed, charcode 141) after the #.

3.#

Test it

In Japt, any character following # is converted to its character code and appended to any digits or decimal points which may precede it, in this case the 3.

Answer 10

Charcoal, 8 bytes

”i‴Ａχ:&”

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Just a compressed string.

Answer 11

Javascript, 16 bytes

_=>Math.acos(-1)

f=
_=>Math.acos(-1)
console.log(f())

Answer 12

Neim, 5 bytes

3FBρσ

Explanation:

3      Push 3
 FB    F pushes 45, B converts it into a character code, '.'
   ρ   Push 14
    σ  Push 15
Implicitly concatenate and print stack

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Answer 13

APL, 6 bytes

10○⍟-1

Takes the imaginary part of ln(-1).

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Answer 14

J, 16 bytes

17j15":4*_3 o.1:

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Similar to the VBA answer. Outputs: 3.141592653589793.

Answer 15

Math.JS, 17 bytes.

     log(-1,e).im

That's 5 leading spaces

This calculates to 15 decimal places, and implicitly prints it.

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Answer 16

05AB1E, 11 bytes

3•∊&SK•)'.ý

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COMMAND     # CURRENT STACK         | EXPLANATION
------------#-----------------------+-----------------------------------
3           # [3]                   | Push 3.
 •∊&SK•     # [3, 1415926535]       | Push compressed base-255 number 1415926535.
       )    # [[3,1415926535]]      | Wrap the 2 to an array.
        '.  # [[3,1415926535],'.']  | Push a decimal point.
          ý # ['3.1415926535']      | Join 3 and 1415926535 with a decimal point.

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USING BUILT-INS:

05AB1E, 4 bytes

žq5£

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---

3žs doesn't work for some reason...

Answer 17

Jelly, 6 bytes

-ÆAær5

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Uses a different approach than the other Jelly answers.

How?

-ÆA is inverse cosine of -1 (which is pi), ær5 retrieves the needed section. It worked out so rounding is equivalent to truncation in this case

Answer 18

ForceLang, 269 bytes

set S 0x5F258E476FC1B3B5571012206C637089460E41E814CB071E61431B5F0671F551D18D2C5D2D3A2565E408C6DE8D753F595B6E9979C3866D1C9965008DCFB02E3BD11D21DFFAF17978F05C8BBACF55A5ED5E90B1D8CAD8736AA4B728EB342B453F86353DB371D322B6A98613BC5CCB00AC2.mult 1e-270
io.write S.toString 268

Answer 19

MATL, 9 8 bytes

-1X;V9:)

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Explanation:

-1      % number literal
X;      % inverse cosine (radians), giving pi
V       % convert to string
9       % specify precision
:)      % keep first 9 characters - 8 digits of precision

Saved one byte thanks to Luis Mendo!

Answer 20

RPL (HP48), 86.5 bytes, 87 digits [Does this count?]

« GROB 8 43
14159265358979323846264338327950288419716939937510582097494459230781640628620899862803
→STR 9 OVER SIZE SUB 2 "." REPL »

GROB is an image keyword. The 2 numbers that follow are width and height. The hexadecimal digits that follow are bitmap. That's 1 byte of storage for every 2 digits plus the image metadata.

Empty program « » takes 10 bytes. A command takes 2.5 bytes. A float takes 10.5 bytes, but if it's equal to a single digit integer, it can take 2.5 bytes.

HP48 stores 4 rather than 8 bits at each memory location, so a byte occupies 2 consecutive locations (low endian). Often 20-bit chunks are used, and they do take 2.5 bytes (not 3).

Answer 21

TI-BASIC, 7 bytes

round(4tan⁻¹(1),6

round(2sin⁻¹(1),6

round(cos⁻¹(-1),6

These three statements will print 3.141593.

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Note: TI-BASIC is a tokenized language. Character count does not equal byte count.

Answer 22

Julia 1.0, 16 bytes

show(acos(-1))##

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This is the default precision printed, I sadly have to add 2 dummy characters but I don't think it's possible to make it shorter

Answer 23

Haskell, 15 bytes

 acos(-1)-3e-15

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Using @JungHwanMin formula.

Answer 24

Perl 5, 40 bytes

}{$\=$\/~(2*$_)*~$_+2for-130..-2

Requires the command line option -pMbignum, counted as 8.

Sample Useage:

$ perl -pMbignum test.pl < /dev/null
3.141592653589793238462643383279502884197

I apologize for no TIO link, it doesn't seem to support bignum (or an unmatched opening brace...).

Answer 25

MY, 7 bytes

’2ō"8↑↵

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How?

• ’, decrement the top of the stack (0 is popped if empty, making the stack [-1]). (Same symbol as Jelly)
• 2, push 2.
• ō, since the top of the stack is two, discard the 2, pop n, then push acos(n) (in radians, this gives pi, this symbol is o with a negative sign on top of it. o (for normal trig) comes from APL).
• ", pop the top element of the stack and push it as a string.
• 8, push 8.
• ↑, pop a then b, push b[:a] (This is another symbol taken from APL).
• ↵, output with a new line (OUTPUT <- STACK).

Answer 26

><>, 16 bytes

"]a:5;."3no>n<

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There's a couple of unprintables in there, with the first having value 26 and the second 141.

Answer 27

C (gcc), 99 bytes

i;main(){for(;i<46;)printf(i++?"%02d":"3.",i[L"0;5:a]î.+& O2TÅ©']KiR J^-\N¤>cV0ý*uO"]);}

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Hardcode 99 bottle of digits

Answer 28

Pyth, 9 bytes

<`.ttZ4 T

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Works with the fact that .

Answer 29

Ruby, 16 bytes

->{Math.acos -1}

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Answer 30

Excel VBA, 16 Bytes

Anonymous VBE immediate window functions that takes no input and output to any of Excel VBA's STDOUTs

_{Each line represents a separate function that returns Pi of length 16}

?Mid(4*Atn(1),1)

?Str([Acos(-1)])

Debug.?4*Atn(1);

MsgBox[Acos(-1)]

[A1]="=ACOS(-1)"