Draw a regular polygon

Question

The goal of this code golf is to draw a regular polygon (one with equal side lengths) given the number of sides and radius (distance from center to vertex).

• The number of sides and the radius can be entered via a file, STDIN, or just a plain old variable. Use whatever is shorter in your language.
• -25% of total characters/bytes if the image is actually drawn instead of ASCII art.

Answer 1

LOGO 37 - 25% = 27.75 (with variables)

REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

LOGO 49 - 25% = 36.75 (as a function)

TO P:R:S REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]END

Triangle

Called with variables

Make "R 100
Make "S 3
REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

Used as a function P 100 3

Square

Called with variables

Make "R 100
Make "S 4
REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

Used as a function P 100 4

Pentagon

Called with variables

Make "R 100
Make "S 5
REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

Used as a function P 100 5

Decagon

Called with variables

Make "R 100
Make "S 10
REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

Used as a function P 100 10

Circle

Called with variables

Make "R 100
Make "S 360
REPEAT:S[FD:R*2*sin(180/:S)RT 360/:S]

Used as a function P 100 360

Answer 2

Mathematica, 40 - 25% = 30

ListPolarPlot[r&~Array~n]/.PointPolygon

Answer 3

Java 8 : 533 322 - 25% = 241.5

Well, it's Java :/ Just draws lines, point to point. Should work for any arbitrarily sized polygon. Cut it down quite a bit from original size. Huge credit to Vulcan (in comments) for the golf lesson.

import java.awt.*;class D{public static void main(String[]v){new Frame(){public void paint(Graphics g){int i=0,r=Short.valueOf(v[0]),s=Short.valueOf(v[1]),o=r+30,x[]=new int[s],y[]=x.clone();for(setSize(o*2,o*2);i<s;x[i]=(int)(Math.cos(6.28*i/s)*r+o),y[i]=(int)(Math.sin(6.28*i++/s)*r+o));g.drawPolygon(x,y,s);}}.show();}}

Line Breaks:

import java.awt.*;
class D{
    public static void main(String[]v){
        new Frame(){
            public void paint(Graphics g){
                int i=0,r=Short.valueOf(v[0]),s=Short.valueOf(v[1]),o=r+30,x[]=new int[s],y[]=x.clone();
                for(setSize(o*2,o*2);i<s;x[i]=(int)(Math.cos(6.28*i/s)*r+o),y[i]=(int)(Math.sin(6.28*i++/s)*r+o));
                g.drawPolygon(x,y,s);
            }
        }.show();
    }
}

Input is arguments [radius] [sides]:

java D 300 7

Output:

Answer 4

TeX/TikZ (60 – 80.25)

File polygon.tex:

\input tikz \tikz\draw(0:\r)\foreach\!in{1,...,\n}{--(\!*360/\n:\r)}--cycle;\bye

(80 bytes)

The radius and number of sides are provided as variables/macros \r and \n. Any TeX unit can be given for the radius. Without unit, the default unit cm is used. Examples:

\def\r{1}\def\n{5}    % pentagon with radius 1cm
\def\r{10mm}\def\n{8} % octagon with radius 10mm

(16 bytes without values)

If the page number should be suppressed, then it can be done by

\footline{}

(11 bytes)

Examples for generating PDF files:

pdftex "\def\r{1}\def\n{3}\input polygon"

pdftex "\def\r{1}\def\n{5}\input polygon"

pdftex "\def\r{1}\def\n{8}\input polygon"

pdftex "\def\r{1}\def\n{12}\input polygon"

Score:

It is not clear to, what needs counting. The range for the score would be:

• The base code is 80 bytes minus 25% = 60

• Or all inclusive (input variable definitions, no page number): (80 + 16 + 11) minus 25% = 80.25

• If the connection between the first and last point do not need to be smooth, then --cycle could be removed, saving 7 bytes.

Answer 5

Geogebra, 42 – 25% = 31.5 bytes

If you count in characters instead of bytes, this would be 41 – 25% = 30.75 characters.

(That is, if you consider Geogebra a language...)

Assumes the radius is stored in the variable r and the number of sides stored in the variable s.

Polygon[(0,0),(sqrt(2-2cos(2π/s))r,0),s]

This uses the cosine theorem c² = a² + b² – 2 a b cos C to calculate the side length from the given radius.

Sample output for s=7, r=5

Answer 6

C, 359 Chars

My first attempt at golfing. At least it beats the Java solution ;-)

int r,n,l,g,i,j,x,y;char* b;float a,c,u,z,p,q,s,t;main(int j,char**v){r=atoi(v[1]);b=malloc(g=(l=r*2+1)*r*2+1);memset(b,32,g);for(j=g-2;j>0;j-=l){b[j]='\n';}b[g-1]=0;a=2*3.14/(n=atoi(v[2]));for(;i<=n;i++,p=s,q=t){c=i*a;s=sin(c)*r+r;t=cos(c)*r+r;if(i>0){u=(s-p)/r,z=(t-q)/r;for(j=0;j<r;j++){x=p+u*j;y=q+z*j;if(x>=0&&y>=0&&y<r*2&&x<l-1)b[y*l+x]='#';}}}puts(b);}

ungolfed:

int r,n,l,g,i,j,x,y;
char* b;
float a,c,u,z,p,q,s,t;
main(int j,char**v){
    r=atoi(v[1]);
    b=malloc(g=(l=r*2+1)*r*2+1);
    memset(b,32,g);
    for(j=g-2;j>0;j-=l){b[j]='\n';} 
    b[g-1]=0;
    a=2*3.14/(n=atoi(v[2]));
    for(;i<=n;i++,p=s,q=t){
        c=i*a;s=sin(c)*r+r;t=cos(c)*r+r;
        if(i>0){
            u=(s-p)/r,z=(t-q)/r;
            for(j=0;j<r;j++){
                x=p+u*j;y=q+z*j;
                if(x>=0&&y>=0&&y<r*2&&x<l-1)b[y*l+x]='#';
            }
        }
    }
    puts(b);
}

And it's the only program that outputs the polygon in ASCII instead of drawing it. Because of this and some floating point rounding issues, the output doesn't look particularly pretty (ASCII Chars are not as high as wide).

                 ######
               ###    ###
            ####        ####
          ###              ###
        ###                  ####
     ###                        ###
     #                            #
     #                            ##
    #                              #
    #                              #
   ##                              ##
   #                                #
  ##                                ##
  #                                  #
  #                                  #
 ##                                  ##
 #                                    #
##                                    ##
#                                      #
#                                      #
#                                      #
#                                      #
##                                    ##
 #                                    #
 ##                                  ##
  #                                  #
  #                                  #
  ##                                ##
   #                                #
   ##                              ##
    #                              #
    #                              #
     #                            ##
     #                            #
     ###                        ###
        ###                  ####
          ###              ###
            ###         ####
               ###    ###
                 ######

Answer 7

C: 229 180

#include<stdio.h>
#include<math.h>
main(){float n=5,r=10,s=tan(1.57*(1.-(n-2.)/n))*r*2.,i=0,j,x,c,t;int u,v;for(;i<n;i++)for(j=0;j<s;j++)x=i*6.28/n,c=cos(x),t=sin(x),x=j-s/2.,u=c*r+t*x+r*2.,v=-t*r+c*x+r*2,printf("\e[%d;%dH*",v,u);}

(r is radius of incircle)

Please run in ANSI terminal

Edit:

• take ace's suggestion
• use old variables (or #define) as input
• use circumcircle radius now

u;main(v){float p=3.14,r=R*cos(p/n),s=tan(p/n)*r*2,i=0,j,x,c,t;for(;i++<n;)for(j=0;j<s;)x=i*p/n*2,c=cos(x),t=sin(x),x=j++-s/2,u=c*r+t*x+r*2,v=c*x-t*r+r*2,printf("\e[%d;%dH*",v,u);}

compile:

gcc -opoly poly.c -Dn=sides -DR=radius -lm

Answer 8

Mathematica, 54 * 75% = 40.5

Graphics@Polygon@Table[r{Cos@t,Sin@t},{t,0,2Pi,2Pi/n}]

I don't even think there's a point for an ungolfed version. It would only contain more whitespace.

Expects the radius in variable r and number of sides in variable n. The radius is a bit meaningless without displaying axes, because Mathematica scales all images to fit.

Example usage:

Answer 9

HTML/JavaScript : 215 - 25% = 161.25, 212 - 25% = 159

<canvas><script>R=100;i=S=10;c=document.currentScript.parentNode;c.width=c.height=R*2;M=Math;with(c.getContext("2d")){moveTo(R*2,R);for(;i-->0;){a=M.PI*2*(i/S);lineTo(R+M.cos(a)*R,R+M.sin(a)*R)}stroke()}</script>

Ungolfed version :

<canvas><script>
    var RADIUS = 100;
    var SIDES_COUNT = 10;
    var canvas = document.currentScript.parentNode;
    canvas.width = canvas.height = RADIUS * 2;
    var context = canvas.getContext("2d");
    context.moveTo(RADIUS * 2, RADIUS);
    for(i = 1 ; i <= SIDES_COUNT ; i++) {
        var angle = Math.PI * 2 * (i / SIDES_COUNT);
        context.lineTo(
            RADIUS + Math.cos(angle) * RADIUS,
            RADIUS + Math.sin(angle) * RADIUS
        );
    }
    context.stroke();
</script>

Answer 10

Postscript 156 - 25% = 117

translate exch 1 exch dup dup scale div currentlinewidth mul setlinewidth
1 0 moveto dup{360 1 index div rotate 1 0 lineto}repeat closepath stroke showpage

Pass the radius, number of sides, and center point on the command line

gs -c "100 9 300 200" -- polyg.ps

or prepend to the source

echo 100 9 300 200 | cat - polyg.ps | gs -

Translate to the center, scale up to the radius, move to (1,0); then repeat n times: rotate by 360/n, draw line to (1,0); draw final line, stroke and emit the page.

Answer 11

Sage, 44 - 25% = 33

Assumes the number of sides is stored in the s variable and the radius is stored in the r variable.

polytopes.regular_polygon(s).show(figsize=r)

Sample output:

s=5, r=3

s=5, r=6

s=12, r=5

Answer 12

bc + ImageMagick + xview + bash, 104.25 (139 bytes - 25%)

This challenge would be incomplete without an ImageMagick answer...

convert -size $[$2*2]x$[$2*2] xc: -draw "polygon `bc -l<<Q
for(;i++<$1;){t=6.28*i/$1;print s(t)*$2+$2,",";c(t)*$2+$2}
Q`" png:-|xview stdin

For example, ./polygon.sh 8 100 produces this image:

Answer 13

JavaScript 584 (867 ungolfed)

This code uses N Complex Roots of unity and translates the angles to X,Y points. Then the origin is moved to centre of the canvas.

Golfed

function createPolygon(c,r,n){
c.width=3*r;
c.height=3*r;
var t=c.getContext("2d");
var m=c.width/2;
t.beginPath(); 
t.lineWidth="5";
t.strokeStyle="green";
var q=C(r, n);
var p=pts[0];
var a=p.X+m;
var b=p.Y+m;
t.moveTo(a,b);
for(var i=1;i<q.length;i++)
{
p=q[i];
t.lineTo(p.X+m,p.Y+m);
t.stroke();
}
t.lineTo(a,b);
t.stroke();
}
function P(x,y){
this.X=x;
this.Y=y;
}
function C(r,n){
var p=Math.PI;
var x,y,i;
var z=[];
var k=n;
var a;
for(i=0;i<k;i++)
{
a = 2*i*p/n;
x = r*Math.cos(a);
y = r*Math.sin(a);
z.push(new P(x,y));
}
return z;
}

Sample output:

Ungolfed

function createPolygon(c,r,n) {
c.width = 3*r;
c.height = 3*r;
var ctx=c.getContext("2d");
var mid = c.width/2;
ctx.beginPath(); 
ctx.lineWidth="5";
ctx.strokeStyle="green";
var pts = ComplexRootsN(r, n);
if(null===pts || pts.length===0)
{
alert("no roots!");
return;
}
var p=pts[0];
var x0 = p.X + mid;
var y0 = p.Y + mid;
ctx.moveTo(x0,y0);
for(var i=1;i<pts.length;i++)
{
p=pts[i];
console.log(p.X +"," + p.Y);
ctx.lineTo(p.X + mid, p.Y + mid);
ctx.stroke();
}
ctx.lineTo(x0,y0);
ctx.stroke();
}

function Point(x,y){
this.X=x;
this.Y=y;
}

function ComplexRootsN(r, n){
var pi = Math.PI;
var x,y,i;
var arr = [];
var k=n;
var theta;
for(i=0;i<k;i++)
{
theta = 2*i*pi/n;
console.log('theta: ' + theta);
x = r*Math.cos(theta);
y = r*Math.sin(theta);
console.log(x+","+y);
arr.push(new Point(x,y));
}
return arr;
}

This code requires HTML5 canvas element, c is canvas object, r is radius and n is # of sides.

Answer 14

PHP 140 - 25% = 105

<?
for(;$i++<$p;$a[]=$r-cos($x)*$r)$a[]=$r-sin($x+=2*M_PI/$p)*$r;
imagepolygon($m=imagecreatetruecolor($r*=2,$r),$a,$p,0xFFFFFF);
imagepng($m);

Assumes two predefined variables: $p the number of points, and $r the radius in pixels. Alternatively, one could prepend list(,$r,$p)=$argv; and use command line arguments instead. Output will be a png, which should be piped to a file.

---

Output

$r=100; $p=5;

$r=100; $p=6;

$r=100; $p=7;

$r=100; $p=50;

Answer 15

HTML + JavaScript, 12 + 127 = 139 bytes

<canvas id=c

s=>r=>(g=n=>n?g(n-1,d.lineTo(r+r*M.cos(x=M.PI*2*n/s),r+r*M.sin(x))):d.fill())(s,c.width=c.height=r+r,d=c.getContext`2d`,M=Math)

Try it on Codepen

HTML (SVG) + JavaScript, 20 + 140 = 160 bytes

<svg id=c><path id=p

s=>r=>p.setAttribute(`d`,`M${M=Math,v=c.style,v.width=v.height=r+r},`+r+(g=n=>n?`L${r+r*M.cos(x=M.PI*2*n/s)},`+(r+r*M.sin(x))+g(n-1):``)(s))

Try it on Codepen

Answer 16

R, 42 bytes -25% = 31.5

function(n,r,p=r*1i^(4/n*0:n))plot(p,,"l")

Try it at rdrr.io

Calculates coordinates of vertices in the complex plane as r times a power-series of the (4/n)th-root of i.

Example output of regular_polygon(7,3):

Answer 17

TI-80 BASIC, 25 bytes - 25% = 18.75

PARAM
2π/ANS->TSTEP
"COS T->X1ᴛ
"SIN T->Y1ᴛ
DISPGRAPH

Assumes all settings are set to the default values. Run the program like 5:PRGM_POLYGON (for a pentagon)

It works by drawing a circle with a very low number of steps. For example, a pentagon would have steps of 2π/5 radians.

The window settings are good enough by default, and TMIN and TMAX are set to 0 and 2π, so all we need to change is TSTEP.

Answer 18

SmileBASIC 3, 183 159 - 25% = 119.25 bytes

Takes the sides and radius from INPUT, calculates and stores the points, and then draws them by using GLINE. I feel like this could be shorter but it's like 1 AM, whatever. Assumes a clean and default display env, so you might need to ACLS when running it from DIRECT.

INPUT S,R
DIM P[S,2]FOR I=0TO S-1
A=RAD(I/S*360)P[I,0]=COS(A)*R+200P[I,1]=SIN(A)*R+120NEXT
FOR I=0TO S-1GLINE P[I,0],P[I,1],P[(I+1)MOD S,0],P[(I+1)MOD S,1]NEXT

Answer 19

OpenSCAD: 31 less 25% = 23.25

module p(n,r){circle(r,$fn=n);}

First post here! I know I'm late to the party, but this seemed like as good a question as any to start with. Call using p(n,r).

Answer 20

APL (dzaima/APL), 50 - 12.5 = 37.5 37 - 9.25 = 27.75 bytes

P5.draw←{P5.G.ln ↑r+r×2 1○ᑈs÷⍨2×○0…s}

Shortened heavily with the help of The APL Orchard.

Explanation:

P5.draw←{P5.G.ln ↑r+r×2 1○ᑈs÷⍨2×○0…s}
P5.draw←{                           } Draw the following on the canvas every frame:
         P5.G.ln                      Draw a line using array(x1,y1,x2,y2...)
                                 0…s  Range from 0 to number of sides
                              2×○     times 2π
                           s÷⍨        divided by number of sides
                      2 1○ᑈ           Take cos and sin of each of the values from the left
                 ↑r+r×                Multiply the list by r and add r to center it, 
                                      then mix the values of the list together

Previous version:

p←(2×○⍳s+1)÷s
P5.draw←{P5.G.ln r+r×⍉(2○p)↑⍤⍮(1○p)}

Takes input as variables r and s, draws polygon with center at \$(r,r)\$.

Made with the help of the formula used here.

Outputs:

100 3

100 5

200 7

To execute the program, use the setup here, for a full view of the drawn polygon.

Answer 21

ActionScript 1, Flash Player 6: 92 - 25% = 69

n=6
r=100
M=Math
P=M.PI*2
lineStyle(1)
moveTo(r,0)
while((t+=P/n)<=P)lineTo(M.cos(t)*r,M.sin(t)*r)

Answer 22

C# in LINQPAD

Credit for the math part goes to Geobits (I hope you don't mind!) with the Java answer. I'm hopeless at math :)

I did this in LINQPAD as it's got a built in output window. So essentially you can drag and drop the following in to it and it'll draw the polygon. Just switch it to 'C# Program', and import the System.Drawing lib into the query properties.

//using System.Drawing;

void Main()
{
// Usage: (sides, radius)
    DrawSomething(4, 50);
}

void DrawSomething(int sides, int radius)
{
    var points = new Point[sides];
    var bmpSize = radius*sides;
    var bmp = new Bitmap(bmpSize,bmpSize);
    using (Graphics g = Graphics.FromImage(bmp))
    {   
        var o = radius+30;
        for(var i=0; i < points.Length; i++)
        {
            // math thanks to Geobits
            double w = Math.PI*2*i/sides;
            points[i].X = (int)(Math.Cos(w)*radius+o);
            points[i].Y = (int)(Math.Sin(w)*radius+o);
        }
        g.DrawPolygon(new Pen(Color.Red), points);
    }
    Console.Write(bmp);
}

Answer 23

Matlab 58 bytes - 25% = 43.5

Saw no Matlab solution, so here is one which is pretty straightforward:

f=@(n,r) plot(r*cos(0:2*pi/n:2*pi),r*sin(0:2*pi/n:2*pi));

You can shave off some bytes if r and n are already in the workspace.

Example call:

f(7,8)

Answer 24

Python 2, 222 bytes

from math import*
def f(s,r):
 r*=cos(pi/s)
 v,R=2*pi/s,[(2*t)/98.-1for t in range(99)]
 print"P1 99 99 "+" ".join(["0","1"][all(w*(w-x)+z*(z-y)>0for w,z in[(r*cos(a*v),r*sin(a*v))for a in range(s)])]for x in R for y in R)

Checks if a pixel is on the inner side of all hyperplanes (lines) of the polygon. The radius is touched because actually the apothem is used.

Answer 25

Mathematica 27 ( = 36 - 25%)

Graphics[Polygon[CirclePoints[r, n]]]

When we submit Mathematica code we often forget about new functions that keep getting built into the language, with current language vocabulary amassing around 5000 core functions. Large and expanding language vocabulary is btw quite handy for code-golfing. CirclePoints were introduced in the current version 11.X. A specific example of 7-sided radius 5 is:

Also you just need to enter angle parameter to control orientation of your polygon:

Graphics[Polygon[CirclePoints[{1, 2}, 5]]]

Answer 26

Python 2, 74 bytes - 25% = 55.5

Input is in the variables r,n. If included in the count, it would be r,n=input(), for 12 bytes more.

import math,turtle as t
exec"t.fd(2*r*math.sin(180/n));t.rt(360/n);"*n

Try it online - (uses different code since exec is not implemented in the online interpreter)

Answer 27

SmileBASIC, 85 75 - 25% = 56.25 bytes

FOR I=0TO S
A=I/S*6.28N=X
M=Y
X=R+R*COS(A)Y=R+R*SIN(A)GLINE N,M,X,Y,-I
NEXT

Variables S and R are used for input.

Explained:

FOR I=0 TO Sides        'Draw n+1 sides (since 0 is skip)
 Angle=I/Sides*6.28     'Get angle in radians
 OldX=X:OldY=Y          'Store previous x/y values
 X=Radius+Radius*COS(A) 'Calculate x and y
 Y=Radius+Radius*SIN(A)
 GLINE OldX,OldY,X,Y,-I 'Draw line. Skip if I is 0 (since old x and y aren't set then)
NEXT

The sides are drawn using the color -I, which is usually close to -1 (&HFFFFFFFF white) (except when I is 0, when it's transparent).

You can also draw a filled polygon by using GTRI N,M,X,Y,R,R,-I instead of GLINE...

Answer 28

Tikz, 199 bytes

\documentclass[tikz]{standalone}\usetikzlibrary{shapes.geometric}\begin{document}\tikz{\def\p{regular polygo}\typein[\n]{}\typein[\r]{}\node[draw,minimum size=\r,\p n,\p n sides=\n]{}}\end{document}

This solution uses the tikz library shapes.geometric.

Here is what a polygon with 5 sides and radius 8in looks like when viewed in evince.

Answer 29

MATLAB/Octave, 26.25 bytes (35 bytes-25%)

@(n,r)polar((0:n)/n*2*pi,r+(0:n)*0)

Can be tested on octave-online.
@ptev already posted a MATLAB answer but mine uses far less bytes and isn't distorted.
We can save up another 6 bytes (or 4.5 including bonus) if we assume input can be given as variables r&n already existing in workspace - then just calling polar((0:n)/n*2*pi,r+(0:n)*0) is enough.

Also, example output (n=5,r=3):