Generate a 5-pointed star [closed]

Question

Your challenge is generate a 5 pointed star like below. The star doesn't need to be filled. Shortest code wins.

Answer 1

PHP, 1 character

Although very short it will probably not become the accepted answer.

☆

(this is U+2606). May work also in other languages (untested).

Answer 2

LOGO, 32 29 bytes

REPEAT 5[FW 9LT 144FW 9RT 72]

Run at logo.twentygototen.org

Answer 3

VBA PowerPoint, 92 bytes

Sub s()
ActivePresentation.Slides(1).Shapes.AddShape msoShape5pointStar, 0, 0, 99, 99
End Sub

Because everything looks better in PowerPoint:

Answer 4

Mathematica, 19 18 bytes

This is easy, using CirculantGraph.

I drew a pentagram as defined in the link the OP used.

5~CirculantGraph~2

---

A slightly longer way, but equally straightforward, is to use the curated "Lamina" entity. This returns a filled pentagram.

 "FilledPentagram"~LaminaData~"Image"

or

 EntityValue[Entity["Lamina","FilledPentagram"],"Image"]

---

Answer 5

Python 3, 137 bytes

Polar plot.

import pylab as P,numpy as N
S=N.sin
p=N.pi
a=.9*p
d=8e-4*p
A=lambda t:N.arange(0,t,d)
P.polar(A(4*p),list(S(a)/S(a-A(.8*p)))*5)
P.show()

Answer 6

Small Basic, 50 bytes

For i=1To 5
Turtle.Move(9)
Turtle.Turn(144)
EndFor

A verbose language, but it has built in turtle functionality. Slowly draws a very, very tiny pentagram.

Answer 7

Scratch, 4-6 blocks (52-79 scratchblocks code)

This produces this

If anyone's interested, here's the project.
EDIT: To golf it, remove the pen down. You could remove the then flag clicked, too but it would only run if you clicked it. To golf the scratch blocks version, change "100" to "1", although you get a microscopic pentagram.

Nongolfed scratchblocks:

when gf clicked
clear
pen down
repeat (5
move (100) steps
turn cw (144) degrees

Answer 8

HTML & CSS: 270 269 bytes

<p id=a/><p id=b></p><p id=c></p><p id=d></p><p id=e></p><style>p{position:absolute;width:300px;border-top:solid #000;margin:100px;transform-origin:50% 1666%}#b{transform:rotate(72deg)}#c{transform:rotate(144deg)}#d{transform:rotate(216deg)}#e{transform:rotate(288deg)}

Sample run:

• http://jsfiddle.net/fm6Hu/2/ (Explorer, Firefox, Opera)
• http://jsfiddle.net/fm6Hu/3/ (Chrome, Opera)

HTML & CSS: 139 bytes

Impressing rewrite by Neil.

div{height:20px;width:300px;border-bottom:solid #000;margin:100px;transform-origin:33% 5%;transform:rotate(72deg)}

<div><div><div><div><div>

(Since first posted, all desktope browsers dropped vendor prefixes, so now one CSS code fits all.)

Answer 9

HTML5, 37 bytes

<svg><path d="M4,0 7,8 0,3 8,3 1,8"/>

Answer 10

Tcl/Tk, 57 bytes

pack [canvas .c]
.c cr po 72 0 114 130 3 49 140 49 29 130

Answer 11

TurtleGolf, 70 24 22 bytes

Note: This is non-competing, as this language was made after this challenge was posted (though specifically for this challenge).

0;
aa*^aa*4b*+>1+:5=?;

Explaination:

0; - Initialize the loop counter and go to the next line.
aa* - Get the value 100
^ - Move forward by 100px
aa*4b*+ - Get the value 144
> - Rotate 144 degrees to the right
1+ - Increment the loop counter by 1
: - Duplicate it cause we need a copy of it.
5= - Check if the loop counter equals 5, push 1 if it is.
?; - If the top of the stack equals 1, terminate the line.

Old, loopless version.

aa*^aa*4a*+4+>aa*^aa*4a*+4+>aa*^aa*4a*+4+>aa*^aa*4a*+4+>aa*^aa*4a*+4+>

I didn't add loops yet, so the code is kind of repetitive right now.

Explaination

aa*^              Move aa*(100) pixels forward
aa*4a*+a+         Get the value 144
>                 Turn 144 pixels to the right

EDIT: Shaved off lots of bytes.
EDIT2: Shaved off two more bytes.

Answer 12

Ruby with Shoes, 36 bytes

Shoes.app{star(points:5).move 99,99}

Answer 13

[C (msvc)], 88 bytes

#include<Windows.h>
main(){int p[]={50,1,20,91,97,35,2,35,79,91};Polygon(GetDC(0),p,5);}

Answer 14

Desmos, 46 characters

r=\sec\left(.4\arccos\left(\cos2.5\theta\right)\right)

Here is a longer version that allows you to choose some n:

r=\sec\left(\frac{2}{n}\arccos\left(\cos\frac{n\theta}{2}\right)\right)

Answer 15

PostScript, 27 bytes

$ hexdump -C pentagram_binary.ps 
00000000  39 20 39 92 ad 30 20 39  92 6b 36 7b 30 20 39 92  |9 9..0 9.k6{0 9.|
00000010  63 31 34 34 92 88 7d 92  83 92 a7                 |c144..}....|
0000001b

Download to try.

Output:

Un-golfed ASCII version:

9 9 translate
0 9 moveto
6{
  0 9 lineto
  144 rotate
}repeat
stroke

Answer 16

Python, 47 bytes

from turtle import*
for i in[144]*5:fd(i);rt(i)

Try it online

The following (42 bytes) also works (Python 2 only), but Trinket doesn't allow exec, so you can't run it online. Also, the result is much smaller.

from turtle import*
exec'fd(9);rt(144);'*5

Answer 17

Google Blockly, 3 blocks

In response to the Scratch answer

Blockly turtle doesn't need "so much" boilerplate code.

This "code" can be ran here and produces this pentagram

Five pointed star, 5 blocks

The five pointed star can be tested here

Answer 18

Shell, 60 bytes

(Coordinates shamelessly stolen from user1455003's SVG answer. For artistic merits upvote his answer.)

convert -size 9x9 xc: -draw 'polygon 4,0 7,8 0,3 8,3 1,8' x:

Sample output:

Answer 19

R, 50 54 bytes

plot(1:9,t="n",as=1);polygon(c(5,8,1,9,2),c(1,9,4,4,9))

Output:

Answer 20

R, 40 bytes

symbols(1,st=cbind(8,3,8,3,8,3,8,3,8,3))

Answer 21

Swift 3, 394 370 bytes

import UIKit
UIGraphicsBeginImageContextWithOptions(CGSize(width:2,height:2),false,100)
let c1=0.3,c2=0.8,s1=1.0,s2=0.6,o=(0.0,1.0),p=[o,(s2,-c2),(-s1,c1),(s1,c1),(-s2,-c2),o].map{CGPoint(    x:$0.0,y:$0.1)},c=UIGraphicsGetCurrentContext()!
c.translateBy(x:1,y:1)
c.setLineWidth(1/100)
c.move(to:p[0])
p.forEach{c.addLine(to:$0)}
c.fillPath()
UIImage(cgImage:c.makeImage()!)

Answer 22

Octave, 24 bytes

polar(t=[1:.8:5]*pi,~~t)

Output:

Or, for 36 bytes you can have just the outline:

polar(t=[0:.2:2]*pi,1-mod(t,pi/2.5))

Answer 23

Tikz, 126 120 125 bytes

5 bytes saved thanks to Julian Wolf

\documentclass[tikz]{standalone}\usetikzlibrary{shapes}\begin{document}\tikz{\node[draw=red,star]{};}\end{document}

This draws a red star shaped node at (0,0).

Answer 24

Excel VBA, ₃₇ 35 Bytes

Anonymous VBE immediates window function that takes no input and outputs a five pointed star to the top left of the Sheet1 (default sheet) object

Sheet1.Shapes.AddShape 92,0,0,99,99

Sample Output

Answer 25

Python 2, 52 bytes

from turtle import*;exec('fd(99);rt(144);'*5);done()

Answer 26

JavaFX / FXML, 138 bytes

<?import javafx.scene.layout.*?><?import javafx.scene.shape.*?><Pane><children><SVGPath content="M4,0 7,8 0,3 8,3 1,8"/></children></Pane>

I haven't tested it yet, though

<?import javafx.scene.layout.*?>
<?import javafx.scene.shape.*?>

<Pane>
   <children>
      <SVGPath content="M4,0 7,8 0,3 8,3 1,8" stroke="BLACK"/>
   </children>
</Pane>

Answer 27

Python + Pylab, 48-byte copout answer

from pylab import*
plot(0,0,marker=(5,1))
show()

Plots a single point using a five-pointed star as a marker.

Answer 28

TI-83/84 Basic, 32 bytes

Line(0,1,1,1:Line(0,0,.5,1.5:Line(1,0,.5,1.5

Note I haven't actually tested this out but I'm assuming it's going to show up as a low resolution, distorted star-shaped object.

• Line( command is one byte