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I have a weakness for 3d nets which when cut out and folded allow you to make 3d shapes out of paper or card. The task is simple, written the shortest program you can that draws nets for the 13 Archimedean solids. The output should be an image file in any sensible format (png, jpg).

All thirteen shapes are described at http://en.wikipedia.org/wiki/Archimedean_solid and in the following table taken from there.

enter image description here

Input: An integer from 1 to 13. Assume the shapes are numbered exactly as in the table above so that the "truncated tetrahedron" is number 1 and the "snub dodecahedron" is number 13.

Output: An image file containing the net for that shape. Just the outline including the internal lines is OK. There is no need to fill it in with colors

You can use any programming language you like as well as any library that was not made especially for this competition. Both should be available freely however (in both senses) online.

I will accept the answer with the smallest number of characters in exactly one week's time. Answers will be accepted whenever they come.

(No) Winner yet. Sadly no valid entrants. Maybe it is too hard?

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Maybe do away with the time-limit altogether? What if somebody finds this a year from now. Do you want them not to try?...Might've been better to do the Platonic first, wait, then the hard one. You may have divided the interest. For me personally (all of this is extrapolation), when I saw two similar ones, I kind of withdrew from it, feeling I didn't have the time to really look at both and plan out how to solve both. And I wouldn't want to do it any other way....On the other hand, others here have had difficulty with Part-2 challenges. See the Minsky Register Machine ones. Maybe it's not you. –  luser droog Feb 18 '13 at 0:15
    
@luserdroog Thanks. Question edited. I should maybe add that I have emailed the answer to the related question around lots of people who love it! FWIW. –  felipa Feb 20 '13 at 16:16
    
I don't think it's hard to do, but to golf it requires several hours of thinking and experimenting because there are many possible nets for each polyhedron and they won't compress equally well. –  Peter Taylor Mar 2 '13 at 12:58

2 Answers 2

Mathematica

Out of competition, not a free language

f[n_] := PolyhedronData[Sort[PolyhedronData["Archimedean", 
                                             {"FaceCount", "StandardName"}]][[n, 2]],  "NetImage"]

Usage:

f /@ Range@13

Mathematica graphics

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Java, not golfed

import java.awt.Graphics2D;
import java.awt.geom.AffineTransform;
import java.awt.image.BufferedImage;
import java.io.File;

import javax.imageio.ImageIO;

public class A {
    static int f = 1;
    static Graphics2D g;

    static void fixtrans() {
        double[] m = new double[6];
        g.getTransform().getMatrix(m);
        for (int i = 0; i < 6; ++i) {
            if (Math.abs(m[i] - Math.round(m[i])) < 1e-5) {
                m[i] = Math.round(m[i]);
            }
        }
        g.setTransform(new AffineTransform(m));
    }

    static void d(String s) {
        for (int i = 0; i < s.length();) {
            int n = s.charAt(i++) - '0';
            int t = s.charAt(i++) - '0';
            for (int j = 0; j < t; ++j) {
                g.drawLine(0, 0, 20, 0);
                g.translate(20, 0);
                g.rotate(f * Math.PI * 2 / n);
                fixtrans(); // optional, straightens some lines
            }
            f = -f;
        }
    }

    public static void main(String[] args) throws Exception {
        String[] a = {
                "33623368356:356;66",
                "33413341334535463547354735473444",
                "33823382338:3586338>358>358>358?88",
                "66456:466:466845684668466766",
                "334144453546354635474746464646464647354634463446344744",
                "88456:466:466:4668458<468<468<468:456846684668466788",
                "33343535353231333535313133353447353434353534313133353447353545343535313133353447353545343444",
                "33513351335233593554335433593554335935543359355433593559355835593559355935593455",
                "33:233:233:433:B35:833:833:B35:833:B35:833:B35:833:B35:B35:833:B35:B35:B35:B35:C::",
                "66566:576:57696869576969586969586:586969576969586857685868586766",
                "334155453546354635463547594658465846584658473546354634463546344635463446354634463547584657465746574657473546344634463446344755",
                "::456:466:466:466:466845:@46:@46:@46:@46:>4568466:4668466:4668466:4668466:4668466845:>46:>46:>46:>46:<45684668466846684667::",
// bad          "333531333434343132333434353531313335343434323232323334343435353231333133343556343434313233323335345935353532313331313132333233353535343557343133343556343434355934353531333459343434355935323335345935323335345935323335345935323335345935353455"
                "333531333434343132333434353531313335343434323232323334343435353231333133343556343434313233323335345935353532313331313132333233353535343557343133343556343434355934353535593432333234355935323335345935323335345935323335345935343459313334353455"};

        BufferedImage img = new BufferedImage(1300, 1300, BufferedImage.TYPE_INT_RGB);
        g = img.createGraphics();
        g.translate(500, 500);
        d(a[Integer.parseInt(args[0]) - 1]);
        String f = args[0] + ".png";
        ImageIO.write(img, "png", new File(f));
    }
}

Results (trimmed, negated, joined and scaled):

results

The shapes are quite unusual :) but correct as far as I can tell (let me know if you find any errors). They were generated (in a separate program) by constructing the face graph and cutting cycles in a DFS.

I'm sure this can be golfed quite a lot using e.g. python and turtle.

Edit: oops, the last case was self-intersecting a bit. I fixed it now (by hand), here's the updated image:

13 corrected

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Does the fix by hand mean the code still outputs a self intersection? Is that the only thing separating this from being a valid answer? –  githubphagocyte Apr 17 at 10:31
    
@githubphagocyte If it still outputted a self-intersection, it wouldn't be a fix. This is a valid answer. –  aditsu Apr 18 at 2:56

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