Calculate the area of a regular polygon

Question

Task

Given an integer, n, where 3 <= n < 2^32, calculate the area of a regular n-gon, with an apothem of 1, using the formula A = n * tan(π / n).

The apothem of a regular polygon is a line segment from the center to the midpoint of one of its sides.

Output the area of this regular polygon as a floating point value showing no fewer than eight decimal places (i.e. 1.24217000, 3.14159265).

Tests

In: 3
Out: 5.1961524227

In: 6
Out: 3.4641016151

In: 10
Out: 3.2491969623

In: 20
Out: 3.1676888065

In: 99
Out: 3.1426476062

In: 1697
Out: 3.1415962425

In: 15000
Out: 3.1415926995

Note: The output values in the sample cases above each show ten decimal places -- two more than required.

Community · Accepted Answer · 2020-06-17 09:04:33Z

9

Mathematica, 16 bytes

N[Tan[Pi/#]#,9]&

Try it online!

of course mathematica has built-ins for this

Area@*RegularPolygon

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:12

ZaMoC

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\$\begingroup\$ Area@RegularPolygon should be Area@*RegularPolygon; as it is now, it cannot be captured in a variable. That is, f = Area@RegularPolygon; f[3] does not work. Relevant meta discussion \$\endgroup\$
– JungHwan Min
Sep 5, 2017 at 0:30
\$\begingroup\$ @JungHwanMin ok, I fixed it.(although I didn't post it as an answer.I was just showing the built-ins for fun) \$\endgroup\$
– ZaMoC
Sep 5, 2017 at 8:10

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Roman Gräf · Accepted Answer · 2017-09-04 08:30:47Z

7

Java (OpenJDK 9), 24 bytes

i->i*Math.tan(Math.PI/i)

Try it online!

answered Sep 4, 2017 at 8:30

Roman Gräf

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

Actually, 5 bytes

╦/Tß*

Try it online!

How?

╦/Tß*   Full program.

╦       Push Pi.
 /      Divide ^ by the input.
  T     Tangent.
   ß*   Multiply by the input.
        Output implicitly.

Alternative: ß╦/T*. _{o_O Actually actually beats Jelly!!!}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:23

Mr. Xcoder

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\$\begingroup\$ 2-byte builtin names... \$\endgroup\$
– Erik the Outgolfer
Sep 4, 2017 at 8:25
\$\begingroup\$ yes, I know... @EriktheOutgolfer 3 byte built-ins in Pyth though >.< \$\endgroup\$
– Mr. Xcoder
Sep 4, 2017 at 8:25
3

\$\begingroup\$ +1 for "Actually actually beats Jelly!!!" That pun never gets old. ;) \$\endgroup\$
– Kevin Cruijssen
Sep 4, 2017 at 13:59

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Cody Gray - on strike · Accepted Answer · 2017-09-05 06:51:30Z

x87 Machine Code, 11 bytes

D9 EB
DA 31
D9 F2
DD D8
DA 09
C3

The above bytes of code define a function that calculates the area of a regular n-gon with an apothem of 1. It uses x87 FPU instructions (the classic floating-point unit on the x86 processors) to do this computation.

Following a standard x86 register-based calling convention (in this case, __fastcall), the function's argument is a pointer to the integer, passed in the ECX register. The function's result is a floating-point value, returned at the top of the x87 floating-point stack (register ST0).

Try it online!

Ungolfed assembly mnemonics:

D9 EB  fldpi                  ; load constant PI at top of FPU stack
DA 31  fidiv DWORD PTR [ecx]  ; divide PI by integer input (loaded from pointer
                              ;   in ECX), leaving result at top of FPU stack
D9 F2  fptan                  ; compute tangent of value at top of FPU stack
DD D8  fstp  st0              ; pop junk value (FPTAN pushes 1.0 onto stack)
DA 09  fimul DWORD PTR [ecx]  ; multiply by integer input (again, loaded via ECX)
C3     ret                    ; return control to caller

As you can see, this is basically just a straightforward computation of the given formula,
result = n * tan(π / n)
Only a couple of interesting things bear pointing out:

The x87 FPU has a dedicated instruction for loading the constant value PI (FLDPI). This was rarely used, even back in the day (and obviously much less so now), but it's shorter in size than embedding a constant into your binary and loading that.
The x87 FPU instruction to calculate tangent, FPTAN, replaces the value of the input register (the top of the FPU stack) with the result, but also pushes a constant 1.0 onto the top of the FPU stack. This is done for backwards compatibility with the 8087 (I have no idea why this was done on the 8087; probably a bug). That means we need to pop this unneeded value off of the stack. The fastest and shortest way to do that is a simple FSTP st0, like we use here. We could have also done a multiply-and-pop, since multiplying by 1.0 won't change the result, but this is also 2 bytes (so no win in code size), will probably execute more slowly, and may introduce unnecessary indeterminacy into the result.

Although a modern programmer or compiler would use the SSE (and later) instruction set, rather than the aging x87, this would require more code to implement, as there's no single instruction to compute a tangent in these newer ISAs.

Erik the Outgolfer · Accepted Answer · 2017-09-04 08:11:23Z

3

Jelly, 6 bytes

ØP÷ÆT×

Try it online!

Jelly's π builtin has >8 decimal places.

answered Sep 4, 2017 at 8:11

Erik the Outgolfer

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\$\begingroup\$ Nice. I was trying to figure this out (and Jelly in its entirety), right now. :-) \$\endgroup\$
– Zach Gates
Sep 4, 2017 at 8:12

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Erik the Outgolfer · Accepted Answer · 2017-09-04 08:25:06Z

3

Brachylog, 9 bytes

;π/₍*₄;?×

Try it online!

answered Sep 4, 2017 at 8:25

Erik the Outgolfer

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TuxCrafting · Accepted Answer · 2017-09-04 13:33:20Z

3

Sakura, 4 bytes

*ĳ/π

This is expanded to *ĳ/π⓪⓪, which is

*              *
 ĳ     tan(   )
  /         /
   π       π
    ⓪        n
     ⓪          n

edited Sep 4, 2017 at 13:33

answered Sep 4, 2017 at 13:06

TuxCrafting

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1

\$\begingroup\$ Non-competing isn't a thing any more. \$\endgroup\$
– Shaggy
Sep 4, 2017 at 13:20
\$\begingroup\$ @Shaggy what do you mean? Since when? \$\endgroup\$
– shooqie
Sep 4, 2017 at 20:32
\$\begingroup\$ See this Meta, Tux. \$\endgroup\$
– Shaggy
Sep 4, 2017 at 20:36

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Giuseppe · Accepted Answer · 2017-09-04 15:16:37Z

3

R, 25 bytes

cat((n=scan())*tan(pi/n))

Input from stdin, output to stdout.

Try it online!

answered Sep 4, 2017 at 15:16

Giuseppe

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1

\$\begingroup\$ Works without cat(). 5 bytes less. \$\endgroup\$
– Rui Barradas
Sep 4, 2017 at 19:21

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

JavaScript (ES6), 24 bytes

x=>x*Math.tan(Math.PI/x)

Try it

o.innerText=(f=
x=>x*Math.tan(Math.PI/x)
)(+i.value);oninput=_=>o.innerText=f(+i.value)

<input id=i min=3 type=number value=3><pre id=o>

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:29

Shaggy

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Cinaski · Accepted Answer · 2017-09-04 08:14:14Z

2

MATL, 7 bytes

YPy/Z,*

Try it online!

answered Sep 4, 2017 at 8:14

Cinaski

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 7 bytes

*MtMP/U

Test it

Explanation

Just implements the forumla, where Mt is tan, MP is pi and U is the input.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:14

Shaggy

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Ohm v2, 7 bytes

απ/ÆT³*

Try it online!

How?

απ/ÆT³*   Full Program

απ        Push Pi.
  /       Divided by the input.
   ÆT     Tangent.
     ³*   Multiplied by the input.
          Implicitly output.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:38

Mr. Xcoder

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user73921user73921 · Accepted Answer · 2017-09-04 17:05:36Z

2

Perl, 14 + 16 = 30

perl -MMath::Trig -ple'$_*=tan(pi/$_)'

14 bytes for the program proper, and 16 for the command line switches

answered Sep 4, 2017 at 17:05

user73921

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Zongor · Accepted Answer · 2017-09-04 18:46:09Z

2

var'aq, 51 bytes

'Ij latlh HeHmI' tam boqHa''egh qojmI' boq'egh cha'

Explanation

'Ij        - read from STDIN
latlh      - duplicate top of stack
HeHmI'     - push PI onto stack
tam        - swap first 2 elements on stack
boqHa''egh - divide
qojmI'     - take tangent
boq'egh    - multiply
cha'       - print

edited Sep 4, 2017 at 18:46

answered Sep 4, 2017 at 18:15

Zongor

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Renzo · Accepted Answer · 2017-09-04 19:07:18Z

2

Common Lisp, 29 bytes

(lambda(n)(* n(tan(/ pi n))))

Try it online!

answered Sep 4, 2017 at 19:07

Renzo

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Husnain Raza · Accepted Answer · 2017-09-04 08:06:15Z

1

Python 2, 45 bytes

from math import*
n=input()
print n*tan(pi/n)

Try it online!

answered Sep 4, 2017 at 8:06

Husnain Raza

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2

\$\begingroup\$ 38 bytes \$\endgroup\$
– Mr. Xcoder
Sep 4, 2017 at 8:08
\$\begingroup\$ In challanges like this I really dislike using lambda to save bytes by just putting the print in the footer... Therefore: 44 bytes \$\endgroup\$
– Simon
Sep 4, 2017 at 9:48
4

\$\begingroup\$ @Simon why? Functions are a valid submission - you can either output a value, or return it from a function. The print in the footer isn't required. \$\endgroup\$
– Stephen
Sep 4, 2017 at 13:48

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth, 9 bytes

*.tc.n0Q2

Test suite.

How?

*.tc.n0Q2    Full program. Q means input.

    .n0      Pi. 
   c         Divided by:
       Q     The input.
 .t     2    Tangent.
*        Q   Multiply by the input.
             Output implicitly.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:15

Mr. Xcoder

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Gaia, 5 bytes

₵P÷ṫ×

Try it online!

How?

₵P÷ṫ×   Full program.

₵P      Push Pi.
  ÷     Divided by the input.
   ṫ    Tangent.
    ×   Multiply by the input.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 8:47

Mr. Xcoder

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Swift, 35 bytes

With compiler warnings:

import Foundation
{tan(M_PI/$0)*$0}

Try it here!

Without compiler warnings, 40 bytes:

import Foundation
{tan(Double.pi/$0)*$0}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 4, 2017 at 11:06

Mr. Xcoder

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Wernisch · Accepted Answer · 2017-09-04 16:22:02Z

1

Excel, 16 bytes

=A1*TAN(PI()/A1)

answered Sep 4, 2017 at 16:22

Wernisch

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qwertxzy · Accepted Answer · 2017-09-05 06:47:37Z

0

Prolog (SWI), 25 bytes

f(X,Y):-X is Y*tan(pi/Y).

Try it online!

This is my first submission to codegolf, hope I'm doing alright. Written as a function.

answered Sep 5, 2017 at 6:47

qwertxzy

1014 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

IBM/Lotus Notes Formula Language, 13 bytes

a*@Tan(@Pi/a)

Input taken via a field named a on the same form as the field containing the formula. No TIO available so screenshot of all test cases shown below:

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 5, 2017 at 8:55

ElPedro

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AdmBorkBork · Accepted Answer · 2017-09-05 13:23:10Z

0

PowerShell, 38 bytes

param($n)$n*[math]::tan([math]::pi/$n)

Try it online!

Does exactly what it says on the tin, just takes a little longer due to the lengthy [math]:: .NET calls.

answered Sep 5, 2017 at 13:23

AdmBorkBork

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Snack · Accepted Answer · 2017-09-05 17:44:59Z

0

Ruby, 27 bytes

->n{n*Math.tan(Math::PI/n)}

Try it online!

answered Sep 5, 2017 at 17:44

Snack

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alephalpha · Accepted Answer · 2017-09-06 12:00:09Z

0

Pari/GP, 14 bytes

n->tan(Pi/n)*n

Try it online!

answered Sep 6, 2017 at 12:00

alephalpha

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Amit Gawande · Accepted Answer · 2017-09-07 05:00:05Z

0

C# (Mono C# compiler), 24 bytes


=>n*Math.Tan(Math.PI/n)

Try it online!

edited Sep 7, 2017 at 5:00

answered Sep 5, 2017 at 18:26

Amit Gawande

12 bronze badges

3

\$\begingroup\$ Unfortunately, this is neither a complete program, not a function. Instead it is a snippet, which isn't allowed here. However, I think that you can add n=> to the start to make this into an arrow function (take this with a pinch of salt, I don't know C#) which is valid. \$\endgroup\$
– caird coinheringaahin g ♦
Sep 5, 2017 at 18:33
\$\begingroup\$ You can place the snippet into a System.Func<T, T>, which would take a float as input and another one as output. The declaration would look like this: System.Func<float, float> f = n=>n*Math.Tan(Math.PI/n);, where the bytecount would start at n=>. In my example i omitted two of your brackets to save 2 bytes ;) \$\endgroup\$
– Ian H.
Sep 6, 2017 at 12:12

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Community · Accepted Answer · 2020-06-17 09:04:33Z

RPNGolf 0.6/0.7, 12 bytes

tbp-1mBsdmcc

My first post using RPNGolf, my new stack-based language!

This is a full program that reads an integer from standard input and prints the output to standard output (without a trailing newline).

Explanation:

tb              # push user input from STDIN as int
  p             # duplicate top of stack
   -1           # push -1
     mB         # pop i, push inverse cosine of i
       s        # swap top two items on the stack
        d       # pop b, pop a, push a/b
         mc     # pop i, push tangent of i
           c    # pop b, pop a, push a*b
# RPNGolf implicity prints the stack upon normal exit

chunes · Accepted Answer · 2022-05-18 10:19:06Z

0

Factor, 19 bytes

[ pi over / tan * ]

Try it online!

answered May 18, 2022 at 10:19

chunes

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Kevin Cruijssen · Accepted Answer · 2022-05-18 10:54:32Z

0

05AB1E, 7 bytes

žqI/Å¼*

Try it online or verify all test cases.

Explanation:

žq       # Push PI (3.141592653589793)
  I/     # Divide it by the input
    Å¼   # Get the tangent of this
      *  # Multiply it by the (implicit) input
         # (after which the result is output implicitly)

answered May 18, 2022 at 10:54

Kevin Cruijssen

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Kip the Malamute · Accepted Answer · 2023-01-13 13:09:08Z

0

Pyt, 6 bytes

Đπ/⅟ť*

Try it online!

Đ            implicit input (n); duplicate on stack
 π           push pi
  /          divide n by pi
   ⅟         take the reciprocal
    ť        tan(pi/n)
     *       multiply by n; implicit print

answered Jan 13 at 13:09

Kip the Malamute

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Calculate the area of a regular polygon

Task

Tests

30 Answers 30

Mathematica, 16 bytes

Java (OpenJDK 9), 24 bytes

Actually, 5 bytes

How?

x87 Machine Code, 11 bytes

Jelly, 6 bytes

Brachylog, 9 bytes

Sakura, 4 bytes

R, 25 bytes

JavaScript (ES6), 24 bytes

Try it

MATL, 7 bytes

Japt, 7 bytes

Explanation

Ohm v2, 7 bytes

How?

Perl, 14 + 16 = 30

var'aq, 51 bytes

Explanation

Common Lisp, 29 bytes

Python 2, 45 bytes

Pyth, 9 bytes

How?

Gaia, 5 bytes

How?

Swift, 35 bytes

Excel, 16 bytes

Prolog (SWI), 25 bytes

IBM/Lotus Notes Formula Language, 13 bytes

PowerShell, 38 bytes

Ruby, 27 bytes

Pari/GP, 14 bytes

C# (Mono C# compiler), 24 bytes

RPNGolf 0.6/0.7, 12 bytes

Factor, 19 bytes

05AB1E, 7 bytes

Pyt, 6 bytes

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