Take the 2-minute tour ×
Programming Puzzles & Code Golf Stack Exchange is a question and answer site for programming puzzle enthusiasts and code golfers. It's 100% free, no registration required.

Your challenge: Make a Christmas tree. The size must be choosable by some input method, but doesn't have to be directly related to any part of the tree; however, larger inputs should produce a larger tree.

How can you make it? You can make the tree any way you like, other than by printing the unicode character for the tree, such as ouputting an image, ascii art, with other aspects, etc. Whatever you do, remember that this is a , so be creative.

The answer with the most upvotes by the end of December wins, but I'll accept another if it gets higher

share|improve this question
    
Nice question ;) –  Timtech Dec 12 '13 at 22:34
    
A pity decorations weren't mandatory! –  Lohoris Dec 13 '13 at 17:10
    
@Lohoris I think it is good that they aren't mandatory. This way, people can do what they want, and those brilliant fractal trees are legitimate answers. –  Quincunx Dec 13 '13 at 17:12
    
@Quincunx you could add decorations even to fractal trees... –  Lohoris Dec 13 '13 at 17:13
    
@Lohoris yes, but they are plenty as is. The author of those trees could add decorations if they'd like to, but I'd rather not require it. –  Quincunx Dec 13 '13 at 17:14

26 Answers 26

up vote 72 down vote accepted

Python

A fractal Christmas tree using the turtle package:

enter image description here

n = input()*1.

from turtle import *
speed("fastest")
left(90)
forward(3*n)
color("orange", "yellow")
begin_fill()
left(126)
for i in range(5):
    forward(n/5)
    right(144)
    forward(n/5)
    left(72)
end_fill()
right(126)

color("dark green")
backward(n*4.8)
def tree(d, s):
    if d <= 0: return
    forward(s)
    tree(d-1, s*.8)
    right(120)
    tree(d-3, s*.5)
    right(120)
    tree(d-3, s*.5)
    right(120)
    backward(s)
tree(15, n)
backward(n/2)

import time
time.sleep(60)

n is the size parameter, the shown tree is for n=50. Takes a minute or two to draw.

share|improve this answer
7  
+1 for using the turtle as a 'stand'. –  marinus Dec 14 '13 at 20:44
    
This looks lovely :) –  Soner Gönül Dec 18 '13 at 12:00

JavaScript

enter image description here

Show the animated tree online.

var size = 400;

var canvas = document.createElement('canvas');
canvas.width = size;
canvas.height = size;
document.body.appendChild(canvas);

var ctx = canvas.getContext('2d');

var p3d = [];

var p = [Math.random(), Math.random(), Math.random(), 0];

for (var i = 0; i < 100000; i++) {
    p3d.push([p[0],p[1],p[2],p[3]]);
    var t = Math.random();
    if (t<0.4) {
        _y = 0.4 * p[1];
        _x = 0.1 * p[0];
        _z = 0.6 * p[2];
        var r = Math.floor(3*t/0.4)/3.0;
        var rc = Math.cos(Math.PI*2.0*r);
        var rs = Math.sin(Math.PI*2.0*r);
        p[1] = _x+0.1*r+0.5*_y*_y;
        p[0] = _y*rc+_z*rs;
        p[2] = _z*rc-_y*rs;
        p[3] = 0.2*t + 0.8*p[3];
    } else {
        p[1] = 0.2 + 0.8*p[1];
        p[0] = 0.8 * p[0];
        p[2] = 0.8 * p[2];
        p[3] = 0.2 + 0.8*p[3];
    }
}

var rot = 0.0;

function render() {
    rot = rot + 0.1;
    var rc = Math.cos(rot);
    var rs = Math.sin(rot);

    ctx.strokeStyle='#FF7F00';
    ctx.lineWidth=2;
    ctx.beginPath();
    ctx.moveTo(size/2,size/8);
    ctx.lineTo(size/2,size*15/16);
    ctx.stroke();

    var img = ctx.getImageData(0, 0, size, size);
    for (var j = 0; j < size*size; j++) {
        img.data[4*j+0] = 0.5*img.data[4*j+0];
        img.data[4*j+1] = 0.5*img.data[4*j+1];
        img.data[4*j+2] = 0.5*img.data[4*j+2];
        img.data[4*j+3] = 255;
    }

    for (var i = 0; i < p3d.length; i++) {
        var px = p3d[i][0];
        var py = 0.5 - p3d[i][1];
        var pz = p3d[i][2];
        var col = Math.floor(128.0*p3d[i][3]);

        var _x = rc*px + rs*pz;
        var _z = rc*pz - rs*px;

        var z = 3.0 * size / (_z + 4.0);
        var x = size / 2 + Math.round(_x * z);        
        var y = size / 2 + Math.round(py * z);

        if(x>=0&&y>=0&&x<size&&y<size) {
            img.data[4 * (y * size + x) + 0] = col;
            img.data[4 * (y * size + x) + 1] = 128+col;
            img.data[4 * (y * size + x) + 2] = col;
            img.data[4 * (y * size + x) + 3] = 255;
        }
    }

    ctx.putImageData(img, 0, 0);
}

setInterval(render, 1000 / 30);
share|improve this answer
    
I like it, but you should add a trunk. –  Fels Dec 13 '13 at 9:42
    
@Fels added trunk –  Howard Dec 13 '13 at 10:16
2  
This is very impressive, but what's up with those green pixels circling the tree? I didn't realize that Santa had little green elves who can fly. (they do not detract from the beauty of your answer, but I wonder where they came from) –  Quincunx Dec 15 '13 at 6:24
    
@Quincunx: They seem to be initial sample points from before the iteration converges. Skipping the first 100 points gets rid of them. –  Ilmari Karonen Dec 15 '13 at 14:25
    
Really cool, but it doesn't quite look like a Christmas tree. It looks more like a stick with a few fern leaves attached to it. Also, could you make the view slowly scroll along the y axis, so the 3D-ness is more apparent? –  The Guy with The Hat Dec 20 '13 at 17:04

Javascript

Output of given code sample

This is my first code golf!

var a=40,b=8,c=13,o="<div style='font-family:monospace;text-align:center;color:#094'>",w=1,x=0,y="|#|<br>";for(i=1;i<a;i++){for(j=0;j<w;j++){x%c==0?o+="<span style='color:#D00'>O</span>":o+="+";x++;}i%b==0?w-=4:w+=2;o+="<br>";}document.write(o+"<span style='color:#640'>"+y+y+y+"</span></div>");

It comes in at 295 Characters.

The size and decoration of the tree is set by the a,b,c variables:

  • a sets the amount of rows in the tree
  • b sets the amount of rows between decreases in width (set low for a skinny tree, high for a fat tree). Must be greater than or equal to 3.
  • c sets the amount of baubles (set zero for none, 1 for only baubles, higher numbers for less dense placement of baubles)

It looks best when a is a multiple of b, as in the example.

Paste into the console to create a tree. Looks better from far away!

share|improve this answer
1  
reduced to 264 a=40,b=8,c=13,o="<p style='font:monospace;color:#094' align='center'>",w=1,x=0,y="|#|<br>";for(i=1;i<a;i++){for(j=0;j<w;j++){x%c==0?o+‌​="<b style='color:red'>O</b>":o+="+";x++;}i%b==0?w-=4:w+=2;o+="<br>";}document.write(‌​o+"<b style='color:#640'>"+y+y+y+"</b></p>"); –  user1886419 Dec 13 '13 at 16:45
2  
You are really golfing on a popularity contest? Wow, I would have written nice readable code (easier to code in). +1 –  Quincunx Dec 13 '13 at 17:06
3  
ungh. You're using doc.write? No +1 from me. –  Jan Dvorak Dec 13 '13 at 20:14

C++

Let's make it in the spirit of IOCCC and have the code in the shape of a tree as well! :D

#include <iostream>
using namespace std;

               int
             main(){
              int a
                ; 
               cin
             >>a;int
           w=a*2+5;for
             (int x=
           0;x<a;++x){
         for(int y=2;y>=
       0;--y){for(int z=0;
           z<a+y-x;++z
        ){cout<<" ";}for(
     int z=0;z<x*2-y*2+5;++z
        ){cout<<".";}cout
      <<endl;}}for(int x=0;
    x<w/5+1;++x){for(int z=0;
  z<w/3+1;++z){cout<<" ";}for(
int z=0;z<w-(w/3+1)*2;z+=1){cout
           <<"#";}cout
           <<endl;;;}}

Takes an integer as input, and returns a Christmas tree with that many "stack levels". For example, an input of

5

Returns:

       .
      ...
     .....
      ...
     .....
    .......
     .....
    .......
   .........
    .......
   .........
  ...........
   .........
  ...........
 .............
      ###
      ###
      ###
      ###
share|improve this answer
    
output example? –  Jan Dvorak Dec 14 '13 at 9:48
    
I put one up now. –  Joe Z. Dec 14 '13 at 9:52
19  
The code looks more like a chrismas tree than the output... –  dystroy Dec 16 '13 at 8:57
1  
@dystroy for the record, the instructions do not say it's the output that must containt the tree. –  Pierre Arlaud Dec 16 '13 at 9:20
3  
@ArlaudPierre: But, the size must be adjustable, which is impossible with just the source code. I could make a program that outputs program code in the shape of a Christmas tree that outputs an actual Christmas tree, but I imagine that would be terribly difficult without resorting to just inserting stray semicolons everywhere. –  Joe Z. Dec 16 '13 at 11:02

Javascript

Quasirealistic fully 3D procedural fir tree generator.

Featuring: extensive configuration with even more configuration options present in code; a zigzagy trunk; branching branches; growth animation; rotation of a fully grown tree.

Not featuring: jQuery, Underscore.js or any other library; hardware dependency - only canvas support is required; messy code (at least that was the intention)

Live page: http://fiddle.jshell.net/honnza/NMva7/show/

Edit page: http://jsfiddle.net/honnza/NMva7/

screenshot:

fir tree

HTML:

<canvas id=c width=200 height=300 style="display:none"></canvas>
<div id=config></div>

Javascript:

var TAU = 2*Math.PI,
    deg = TAU/360,

    TRUNK_VIEW      = {lineWidth:3, strokeStyle: "brown", zIndex: 1},
    BRANCH_VIEW     = {lineWidth:1, strokeStyle: "green", zIndex: 2},
    TRUNK_SPACING   = 1.5,
    TRUNK_BIAS_STR  = -0.5,
    TRUNK_SLOPE     = 0.25,
    BRANCH_LEN      = 1,
    BRANCH_P        = 0.01,
    MIN_SLOPE       = -5*deg,
    MAX_SLOPE       = 20*deg,
    INIT_SLOPE      = 10*deg,
    MAX_D_SLOPE     =  5*deg,
    DIR_KEEP_BIAS   = 10,
    GROWTH_MSPF     = 10, //ms per frame
    GROWTH_TPF      = 10, //ticks per frame
    ROTATE_MSPF     = 10,
    ROTATE_RPF      = 1*deg; //radians per frame

var configurables = [
//    {key: "TRUNK_SPACING", name: "Branch spacing", widget: "number",
//     description: "Distance between main branches on the trunk, in pixels"},
    {key: "TRUNK_BIAS_STR", name: "Branch distribution", widget: "number",
     description: "Angular distribution between nearby branches. High values tend towards one-sided trees, highly negative values tend towards planar trees. Zero means that branches grow in independent directions."},
    {key: "TRUNK_SLOPE", name: "Trunk slope", widget: "number",
     description: "Amount of horizontal movement of the trunk while growing"},
    {key: "BRANCH_P", name: "Branch frequency", widget: "number",
     description: "Branch frequency - if =1/100, a single twig will branch off approximately once per 100 px"},
    {key: "MIN_SLOPE", name: "Minimum slope", widget: "number", scale: deg,
     description: "Minimum slope of a branch, in degrees"},
    {key: "MAX_SLOPE", name: "Maximum slope", widget: "number", scale: deg,
     description: "Maximum slope of a branch, in degrees"},
    {key: "INIT_SLOPE", name: "Initial slope", widget: "number", scale: deg,
     description: "Angle at which branches leave the trunk"},
    {key: "DIR_KEEP_BIAS", name: "Directional inertia", widget: "number",
     description: "Tendency of twigs to keep their horizontal direction"},
    {get: function(){return maxY}, set: setCanvasSize, name: "Tree height",
     widget:"number"}
    ];

var config = document.getElementById("config"),
    canvas = document.getElementById("c"),
    maxX   = canvas.width/2,
    maxY   = canvas.height,
    canvasRatio = canvas.width / canvas.height,
    c;

function setCanvasSize(height){
    if(height === 'undefined') return maxY;
    maxY = canvas.height = height;
    canvas.width = height * canvasRatio;
    maxX = canvas.width/2;
x}

var nodes = [{x:0, y:0, z:0, dir:'up', isEnd:true}], buds = [nodes[0]],
    branches = [],
    branch,
    trunkDirBias = {x:0, z:0};

////////////////////////////////////////////////////////////////////////////////

configurables.forEach(function(el){
    var widget;
    switch(el.widget){
        case 'number':
            widget = document.createElement("input");
            if(el.key){
                widget.value = window[el.key] / (el.scale||1);
                el.set = function(value){window[el.key]=value * (el.scale||1)};
            }else{
                widget.value = el.get();
            }
            widget.onblur = function(){
                el.set(+widget.value);
            };
            break;
        default: throw "unknown widget type";
    }
    var p = document.createElement("p");
    p.textContent = el.name + ": ";
    p.appendChild(widget);
    p.title = el.description;
    config.appendChild(p);
});
var button = document.createElement("input");
button.type = "button";
button.value = "grow";
button.onclick = function(){
    button.value = "stop";
    button.onclick = function(){clearInterval(interval)};
    config.style.display="none";
    canvas.style.display="";
    c=canvas.getContext("2d");
    c.translate(maxX, maxY);                
    c.scale(1, -1);
    interval = setInterval(grow, GROWTH_MSPF);
}
document.body.appendChild(button);
function grow(){
    for(var tick=0; tick<GROWTH_TPF; tick++){
        var budId = 0 | Math.random() * buds.length,
            nodeFrom = buds[budId], nodeTo, branch,
            dir, slope, bias

        if(nodeFrom.dir === 'up' && nodeFrom.isEnd){
            nodeFrom.isEnd = false; 
            rndArg = Math.random()*TAU;
            nodeTo = {
                x:nodeFrom.x + TRUNK_SPACING * TRUNK_SLOPE * Math.sin(rndArg),
                y:nodeFrom.y + TRUNK_SPACING,
                z:nodeFrom.z + TRUNK_SPACING * TRUNK_SLOPE * Math.cos(rndArg), 
                dir:'up', isEnd:true}
            if(nodeTo.y > maxY){
                console.log("end");
                clearInterval(interval);
                rotateInit();
                return;
            }
            nodes.push(nodeTo);
            buds.push(nodeTo);
            branch = {from: nodeFrom, to: nodeTo, view: TRUNK_VIEW};
            branches.push(branch);
            renderBranch(branch);
        }else{ //bud is not a trunk top
            if(!(nodeFrom.dir !== 'up' && Math.random() < BRANCH_P)){
                buds.splice(buds.indexOf(nodeFrom), 1)
            }
            nodeFrom.isEnd = false; 
            if(nodeFrom.dir === 'up'){
                bias = {x:trunkDirBias.x * TRUNK_BIAS_STR,
                        z:trunkDirBias.z * TRUNK_BIAS_STR};
                slope = INIT_SLOPE;
            }else{
                bias = {x:nodeFrom.dir.x * DIR_KEEP_BIAS,
                        z:nodeFrom.dir.z * DIR_KEEP_BIAS};
                slope = nodeFrom.slope;
            }
            var rndLen = Math.random(),
                rndArg = Math.random()*TAU;
            dir = {x: rndLen * Math.sin(rndArg) + bias.x,
                   z: rndLen * Math.cos(rndArg) + bias.z};
            var uvFix = 1/Math.sqrt(dir.x*dir.x + dir.z*dir.z);
            dir = {x:dir.x*uvFix, z:dir.z*uvFix};
            if(nodeFrom.dir === "up") trunkDirBias = dir;
            slope += MAX_D_SLOPE * (2*Math.random() - 1);
            if(slope > MAX_SLOPE) slope = MAX_SLOPE;
            if(slope < MIN_SLOPE) slope = MIN_SLOPE;
            var length = BRANCH_LEN * Math.random();
            nodeTo = {
                x: nodeFrom.x + length * Math.cos(slope) * dir.x,
                y: nodeFrom.y + length * Math.sin(slope),
                z: nodeFrom.z + length * Math.cos(slope) * dir.z,
                dir: dir, slope: slope, isEnd: true
            }
            //if(Math.abs(nodeTo.x)/maxX + nodeTo.y/maxY > 1) return;
            nodes.push(nodeTo);
            buds.push(nodeTo);
            branch = {from: nodeFrom, to: nodeTo, view: BRANCH_VIEW};
            branches.push(branch);
            renderBranch(branch);
        }// end if-is-trunk
    }// end for-tick
}//end func-grow

function rotateInit(){
    branches.sort(function(a,b){
        return (a.view.zIndex-b.view.zIndex);
    });
    interval = setInterval(rotate, ROTATE_MSPF);
}

var time = 0;
var view = {x:1, z:0}
function rotate(){
    time++;
    view = {x: Math.cos(time * ROTATE_RPF),
            z: Math.sin(time * ROTATE_RPF)};
    c.fillStyle = "white"
    c.fillRect(-maxX, 0, 2*maxX, maxY);
    branches.forEach(renderBranch);
    c.stroke();
    c.beginPath();
}

var prevView = null;
function renderBranch(branch){
    if(branch.view !== prevView){
        c.stroke();
        for(k in branch.view) c[k] = branch.view[k];
        c.beginPath();
        prevView = branch.view;
    }
    c.moveTo(view.x * branch.from.x + view.z * branch.from.z,
             branch.from.y);
    c.lineTo(view.x * branch.to.x + view.z * branch.to.z,
             branch.to.y);
}
share|improve this answer

Yet another Mathematica / Wolfram Language tree based on Vitaliy's answer:

PD = .5;
s[t_, f_] := t^.6 - f
dt[cl_, ps_, sg_, hf_, dp_, f_, flag_] :=
    Module[{sv, basePt},
           {PointSize[ps],
            sv = s[t, f];
            Hue[cl (1 + Sin[.02 t])/2, 1, .3 + sg .3 Sin[hf sv]],
            basePt = {-sg s[t, f] Sin[sv], -sg s[t, f] Cos[sv], dp + sv};
            Point[basePt],
           If[flag,
              {Hue[cl (1 + Sin[.1 t])/2, 1, .6 + sg .4 Sin[hf sv]], PointSize[RandomReal[.01]],
               Point[basePt + 1/2 RotationTransform[20 sv, {-Cos[sv], Sin[sv], 0}][{Sin[sv], Cos[sv], 0}]]},
              {}]
          }]

frames = ParallelTable[
                       Graphics3D[Table[{
                                         dt[1, .01, -1, 1, 0, f, True], dt[.45, .01, 1, 1, 0, f, True],
                                         dt[1, .005, -1, 4, .2, f, False], dt[.45, .005, 1, 4, .2, f, False]},
                                        {t, 0, 200, PD}],
                                  ViewPoint -> Left, BoxRatios -> {1, 1, 1.3}, 
                                  ViewVertical -> {0, 0, -1},
                                  ViewCenter -> {{0.5, 0.5, 0.5}, {0.5, 0.55}}, Boxed -> False,
                                  PlotRange -> {{-20, 20}, {-20, 20}, {0, 20}}, Background -> Black],
                       {f, 0, 1, .01}];

Export["tree.gif", frames]

Christmas tree

share|improve this answer
3  
This is the best one!! –  Murta Dec 28 '13 at 16:53

Bash

Sample output:

ASCII Xmas tree

The tree size (i.e. number of lines) is passed on the command line, and is limited to values of 5 or greater. The image above was produced from the command ./xmastree.sh 12. Here's the source code:

#!/bin/bash
declare -a a=('.' '~' "'" 'O' "'" '~' '.' '*')
[[ $# = 0 ]] && s=9 || s=$1
[[ $s -gt 5 ]] || s=5
for (( w=1, r=7, n=1; n<=$s; n++ )) ; do
  for (( i=$s-n; i>0; i-- )) ;  do
    echo -n " "
  done
  for (( i=1; i<=w; i++ )) ; do
    echo -n "${a[r]}"
    [[ $r -gt 5 ]] && r=0 || r=$r+1
  done
  w=$w+2
  echo " "
done;
echo " "
share|improve this answer
    
That is very pretty. –  pandubear Dec 28 '13 at 7:29

Disclaimer: this is based on my LaTeX christmas tree, first posted here: http://tex.stackexchange.com/a/87921/8463

The following code will generate a chrismtas tree, with random decorations. You can change both the size of the tree, and the random seed to generate a different tree.

To change the seed modify the value inside \pgfmathsetseed{\year * 6} to any other numerical value (the default one will generate a new tree every year)

To change the size of the tree modify the order=10 to be either larger, or smaller depending on the size of the tree you want.

Examples. order=11:

enter image description here

order=8

enter image description here

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc, lindenmayersystems,shapes,decorations,decorations.shapes}
\begin{document}

\def\pointlistleft{}
\def\pointlistright{}
\pgfmathsetseed{\year * 6}

\makeatletter
\pgfdeclarelindenmayersystem{Christmas tree}{
    \symbol{C}{\pgfgettransform{\t} \expandafter\g@addto@macro\expandafter\pointlistleft\expandafter{\expandafter{\t}}}
    \symbol{c}{\pgfgettransform{\t} \expandafter\g@addto@macro\expandafter\pointlistright\expandafter{\expandafter{\t}}}
    \rule{S -> [+++G][---g]TS}
    \rule{G -> +H[-G]CL}
    \rule{H -> -G[+H]CL}
    \rule{g -> +h[-g]cL}
    \rule{h -> -g[+h]cL}
    \rule{T -> TL}
    \rule{L -> [-FFF][+FFF]F}
}
\makeatother

\begin{tikzpicture}[rotate=90]
\draw [color=green!50!black,l-system={Christmas tree,step=4pt,angle=16,axiom=LLLLLLSLFFF,order=10,randomize angle percent=20}] lindenmayer system -- cycle;

\pgfmathdeclarerandomlist{pointsleft}{\pointlistleft}
\pgfmathdeclarerandomlist{pointsright}{\pointlistright}
\pgfmathdeclarerandomlist{colors}{{red}{blue}{yellow}}

\foreach \i in {0,1,...,5}
{
    \pgfmathrandomitem{\c}{pointsleft}
    \pgfsettransform{\c}
    \pgfgettransformentries{\a}{\b}{\c}{\d}{\xx}{\yy}
    \pgfmathrandomitem{\c}{pointsright}
    \pgfsettransform{\c}
    \pgfgettransformentries{\a}{\b}{\c}{\d}{\XX}{\YY}
    \pgftransformreset

    \pgfmathsetmacro\calcy{min(\yy,\YY)-max((abs(\yy-\YY))/3,25pt)}

    \draw[draw=orange!50!black, fill=orange!50, decorate, decoration={shape backgrounds, shape=star, shape sep=3pt, shape size=4pt}, star points=5] (\xx,\yy) .. controls (\xx,\calcy pt) and (\XX,\calcy pt) .. (\XX,\YY);
}

\foreach \i in {0,1,...,15}
{
    \pgfmathrandomitem{\c}{pointsleft}
    \pgfsettransform{\c}
    \pgftransformresetnontranslations
    \draw[color=black] (0,0) -- (0,-4pt);
    \pgfmathrandomitem{\c}{colors}
    \shadedraw[ball color=\c] (0,-8pt) circle [radius=4pt];
}

\foreach \i in {0,1,...,15}
{
    \pgfmathrandomitem{\c}{pointsright}
    \pgfsettransform{\c}
    \pgftransformresetnontranslations
    \draw[color=black] (0,0) -- (0,-4pt);
    \pgfmathrandomitem{\c}{colors}
    \shadedraw[ball color=\c] (0,-8pt) circle [radius=4pt];
}

\end{tikzpicture}
\end{document}
share|improve this answer
    
I would say that this is based on Stefan Kottwitz's tree, first posted here. Saying that it is based on your tree makes it seem like you wrote all the code for it, which isn't true. Someone else wrote most the code for the actual tree. –  The Guy with The Hat Dec 21 '13 at 13:37
    
@RyanCarlson the idea to use the lindenmayer system as a basis for putting on decorations was my idea, and if you check carefully both of the code you can see, that the lindenmayer part of the code was redesigned to also have additional parts (like a root for the tree, which the original lacks). Besides that I didn't forget to reference the original, where anyone can check for the references (including the one to Peitgen and Saupe, who might've been the first to draft the equation of the tree) –  SztupY Dec 21 '13 at 20:29

Befunge 93

This is an undecorated tree:

&::00p20pv
vp010    <
v              p00<
>:0`!#v_" ",1-
v,"/"$<
>10g:2+v
vp01   <
>:0`!#v_" ",1-
v,"\"$<
>91+,00g1-::0`!#v_^
v        *2+1g02<
>:0`!#v_"-",1-
v g02$<
>91+,v
v    <
>:0`!#v_" ",1-
"||"$ <@,,

Sample output, input is 10:

          /\
         /  \
        /    \
       /      \
      /        \
     /          \
    /            \
   /              \
  /                \
 /                  \
----------------------
          ||

Let's add some decorations:

&::00p20pv
vp010    <
v              p00<
>:0`!#v_" ",1-
v,"/"$<
>10g:2+v             >"O" v
vp01   <            >?v
>:0`!#v_           >?>>" ">,1-
v,"\"$<             >?<
>91+,00g1-::0`!#v_^  >"*" ^
v        *2+1g02<
>:0`!#v_"-",1-
v g02$<
>91+,v
v    <
>:0`!#v_" ",1-
"||"$ <@,,

Sample output:

          /\
         /O \
        /    \
       /   ** \
      /     *  \
     /    **   O\
    /* O* *  *   \
   / O      O     \
  / *  **   O*     \
 /   *OO   *  OOO * \
----------------------
          ||
share|improve this answer

Java

import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.Toolkit;
import java.awt.event.MouseEvent;
import java.awt.event.MouseListener;
import java.awt.geom.AffineTransform;
import java.awt.image.BufferedImage;
import java.io.IOException;
import java.net.URL;
import javax.imageio.ImageIO;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;

/**
 *
 * @author Quincunx
 */
public class ChristmasTree {
    public static double scale = 1.2;

    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            @Override
            public void run() {
                new ChristmasTree();
            }
        });
    }

    public ChristmasTree() {
        try {
            URL url = new URL("http://imgs.xkcd.com/comics/tree.png");
            BufferedImage img = ImageIO.read(url);

            JFrame frame = new JFrame();
            frame.setUndecorated(true);
            frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
            Dimension d = Toolkit.getDefaultToolkit().getScreenSize();

            BufferedImage result = 
                    new BufferedImage((int)(img.getWidth() * scale), 
                            (int) (img.getHeight() * scale), 
                            BufferedImage.TYPE_INT_RGB);
            Graphics2D g = (Graphics2D) result.getGraphics();
            g.drawRenderedImage(img, AffineTransform.getScaleInstance(scale, scale));

            JImage image = new JImage(result);
            image.setToolTipText("Not only is that terrible in general, but you "
                    + "just KNOW Billy's going to open the root present first, "
                    + "and then everyone will have to wait while the heap is "
                    + "rebuilt.");
            frame.add(image);
            frame.pack();
            frame.setVisible(true);

        } catch (IOException ex) {

        }
    }

    class JImage extends JPanel implements MouseListener {

        BufferedImage img;

        public JImage(){
            this(null);
        }
        public JImage(BufferedImage image){
            img = image;
            setPreferredSize(new Dimension(image.getWidth(), image.getHeight()));
            addMouseListener(this);
        }

        @Override
        public void paintComponent(Graphics g) {
            super.paintComponent(g);
            g.drawImage(img, WIDTH, WIDTH, this);
        }

        public void setImage(BufferedImage image) {
            img = image;
            repaint();
        }

        @Override
        public void mouseClicked(MouseEvent me) {
            System.exit(0);
        }

        @Override
        public void mousePressed(MouseEvent me) {

        }

        @Override
        public void mouseReleased(MouseEvent me) {

        }

        @Override
        public void mouseEntered(MouseEvent me) {

        }

        @Override
        public void mouseExited(MouseEvent me) {

        }
    }
}

To alter the size, change scale to some other double value (keep it around 1 if you want to see anything).

Sample output (for 1.0 as scale, too lazy to take a screenshot, so just posted what it does):

enter image description here

The program takes this image from the internet, resizes it according to scale, then puts it in an undecorated window where it is displayed. Clicking on the window closes the program. Also, the tool tip text is there, but the link is not.

share|improve this answer
1  
This isn't "drawing" anything. –  Mrchief Dec 17 '13 at 18:28
    
@Mrchief The challenge isn't to draw something; it is to make a tree. This is making a tree, by taking the image –  Quincunx Dec 17 '13 at 18:35
1  
Did you generate the image yourself? No. You're just downloading it from somewhere. The question isn't to download an image and resize it. Check other answers and you'll understand what the OP actually asked for. –  Mrchief Dec 18 '13 at 5:23
5  
@Mrchief Did you notice who wrote the OP? This is one of the many possible solutions I was looking for when I asked this question. –  Quincunx Dec 18 '13 at 5:44
1  
Ha! Smack on the head! Since its your own post, I don't see much reason to argue, other than expressing my opinion that this is not same as your code "making" something (like rest of the answers do). Just like downloading a movie is not the same as making a movie. –  Mrchief Dec 18 '13 at 19:24

HTML and CSS

<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Scalable christmas tree</title>
<style>
body {
  background-color: skyblue;
  background-size: 11px 11px, 21px 21px, 31px 31px;
  background-image: radial-gradient(circle at 5px 5px,white,rgba(255,255,255,0) 1px), radial-gradient(circle at 5px 5px,white,rgba(255,255,255,0) 2px), radial-gradient(circle at 5px 5px,white 1px,skyblue 1px);
}
div {
  float: left;
  border-style: solid;
  border-color: transparent;
  border-bottom-color: green;
  border-width: 20px;
  border-top-width: 0;
  border-bottom-width: 30px;
  border-bottom-left-radius: 35%;
  border-bottom-right-radius: 35%;
  box-shadow: red 0 15px 5px -13px;
  animation-name: light;
  animation-duration: 1s;
  animation-direction: alternate;
  animation-iteration-count: infinite;
  -webkit-animation-name: light;
  -webkit-animation-duration: 1s;
  -webkit-animation-direction: alternate;
  -webkit-animation-iteration-count: infinite;
}
section {
  float: left;
}
header {
  color: yellow;
  font-size: 30px;
  text-align: center;
  text-shadow: red 0 0 10px;
  line-height: .5;
  margin-top: 10px;
  animation-name: star;
  animation-duration: 1.5s;
  animation-direction: alternate;
  animation-iteration-count: infinite;
  -webkit-animation-name: star;
  -webkit-animation-duration: 1.5s;
  -webkit-animation-direction: alternate;
  -webkit-animation-iteration-count: infinite;
}
footer {
  float: left;
  width: 100%;
  height: 20px;
  background-image: linear-gradient(to right, transparent 45%, brown 45%, #600 48%, #600 52%, brown 55%, transparent 55%);
}
:target {
  display: none;
}
@keyframes star {
  from { text-shadow: red 0 0 3px; }
  to { text-shadow: red 0 0 30px; }
}
@-webkit-keyframes star {
  from { text-shadow: red 0 0 3px; }
  to { text-shadow: red 0 0 30px; }
}
@keyframes light {
  from { box-shadow: red 0 15px 5px -13px; }
  to { box-shadow: blue 0 15px 5px -13px; }
}
@-webkit-keyframes light {
  from { box-shadow: red 0 15px 5px -13px; }
  to { box-shadow: blue 0 15px 5px -13px; }
}
</style>
</head>
<body>
<section>
<header>&#x2605;</header>
<div id="0"><div id="1"><div id="2"><div id="3"><div id="4"><div id="5"><div id="6"><div id="7"><div id="8"><div id="9"><div id="10"><div id="11"><div id="12"><div id="13"><div id="14"><div id="15"><div id="16"><div id="17"><div id="18"><div id="19"><div id="20"><div id="21"><div id="22"><div id="23"><div id="24">
</div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div>
<footer></footer>
</section>
</body>
</html>

Sizes between 1 and 25 are supported – just add the size to the URL as fragment identifier.

Works in Chrome, Explorer and Firefox. In Opera is ugly, but the scaling part works.

Sample access:

http://localhost/xmas.html#5

christmas tree of size 5

Sample access:

http://localhost/xmas.html#15

christmas tree of size 15

Live view:

http://dabblet.com/gist/8026898

(The live view contains no vendor prefixed CSS and includes links to change the size.)

share|improve this answer
    
+1 I like the idea of using :target to scale the output :-) –  squeamish ossifrage Dec 14 '13 at 23:03

Bash with Bc and ImageMagick

#!/bin/bash

size="${1:-10}"

width=$(( size*25 ))
height=$(( size*size+size*10 ))

cos=( $( bc -l <<< "for (i=0;i<3.14*2;i+=.05) c(i)*100" ) )
sin=( $( bc -l <<< "for (i=0;i<3.14*2;i+=.05) s(i)*100" ) )
cos=( "${cos[@]%.*}" )
sin=( "${sin[@]%.*}" )

cos=( "${cos[@]/%-/0}" )
sin=( "${sin[@]/%-/0}" )

snow=()
needle=()
decor=()
for ((i=2;i<size+2;i++)); do
  for ((j=3;j<=31;j++)); do
    (( x=width/2+i*10+cos[62-j]*i*10/100 ))
    (( y=i*i+sin[j]*i*5/100 ))

    for ((e=0;e<i;e++)); do
      needle+=(
        -draw "line $x,$y $(( x+RANDOM%i-i/2 )),$(( y+RANDOM%i-i/2 ))"
        -draw "line $(( width-x )),$y $(( width-x+RANDOM%i-i/2 )),$(( y+RANDOM%i-i/2 ))"
      )
    done

    (( RANDOM%2 )) && (( x=width-x ))
    snow+=(
      -draw "circle $x,$(( y-i/2 )) $(( x+i/3 )),$(( y-i/2+i/3 ))"
    )

    (( RANDOM%10 )) || decor+=(
      -fill "rgb($(( RANDOM%5*20+50 )),$(( RANDOM%5*20+50 )),$(( RANDOM%5*20+50 )))"
      -draw "circle $x,$(( y+i )) $(( x+i/2 )),$(( y+i+i/2 ))"
    )
  done
done

flake=()
for ((i=0;i<width*height/100;i++)); do
  flake+=(
    -draw "point $(( RANDOM%width )),$(( RANDOM%height ))"
  )
done

convert \
  -size "${width}x$height" \
  xc:skyblue \
  -stroke white \
  -fill white \
  "${snow[@]}" \
  -blur 5x5 \
  "${flake[@]}" \
  -stroke brown \
  -fill brown \
  -draw "polygon $(( width/2 )),0 $(( width/2-size )),$height, $(( width/2+size )),$height" \
  -stroke green \
  -fill none \
  "${needle[@]}" \
  -stroke none \
  -fill red \
  "${decor[@]}" \
  x:

Sample run:

bash-4.1$ ./xmas.sh 5

Sample output:

christmas tree of size 5

Sample run:

bash-4.1$ ./xmas.sh 15

Sample output:

christmas tree of size 15

share|improve this answer

TI Basic

Just because I wanted to see a Christmas Tree on my calculator. I shall type it here as it appears on my calculator.

:tree()
:Prgm
:
:Local size
:Prompt size
:
:Local d,x,y,str,orn
:size→x
:0→y
:{" "," "," "," "," "," "," "," ","°","∫","θ","O","©"}→orn
:size→d
:
:While d≥0
: d-1→d
: ""→str
: 
: While x≥0
:  str&" "→str
:  x-1→x
: EndWhile
: 
: str&"/"→str
: 2*y→x
:
: While x>0
:  str&elementAt(orn,rand(colDim(list▶mat(orn))))→str
:  x-1→x
: EndWhile
: 
: str&"\"→str
: Disp str
: y+1→y
: d→x
:EndWhile
:
:""→str
:
:For x,1,2*(size+2)
: str&"-"→str
:EndFor
:Disp str
:
:""→str
:For x,1,size+1
: str&" "→str
:EndFor
:str&"||"→str
:Disp str
:
:EndPrgm


:elementAt(l,i)
:Func
:Local m
:list▶mat(l)→m
:
:Local cd
:colDim(m)→cd
:
:If i>cd or i≤0 Then
: 1/0
:Else
: If i=cd Then
:  Return sum(mat▶list(subMat(m,1,i))) - sum(mat▶list(subMat(m,1,i+1)))
: Else
:  Return sum(mat▶list(subMat(m,1,i))) - sum(mat▶list(subMat(m,1,i+1)))
: EndIf
:EndIf
:EndFunc

This works the same way as my Befunge answer, but I use different ornaments. Yes, my elementAt function's repeated uses slows the program down because of many conversions between lists and matrices, but as I wrote it before, I decided against editing it. Also, I learned while typing this answer that © makes a comment (I thought it looked like @, but that is another character). Never knew what it was before now.

Sample output:

size?
7
        /\
       / O\
      /∫   \
     / ©  ∫°\
    / θ ∫   ∫\
   /°  °  ©  ©\
  /O  ∫  O °   \
 / θ  °©  ∫ O  θ\
------------------
        ||

I love those 's; they look like candy canes.

share|improve this answer

Wolfram Language ( Mathematica )

As discussed at famed Reddit thread: t*sin (t) ≈ Christmas tree

PD = .5; s[t_, f_] := t^.6 - f;
 dt[cl_, ps_, sg_, hf_, dp_, f_] := 
    {PointSize[ps], Hue[cl, 1, .6 + sg .4 Sin[hf s[t, f]]], 
     Point[{-sg s[t, f] Sin[s[t, f]], -sg s[t, f] Cos[s[t, f]], dp + s[t, f]}]};

 frames = ParallelTable[

    Graphics3D[Table[{dt[1, .01, -1, 1, 0, f], dt[.45, .01, 1, 1, 0, f], 
                      dt[1, .005, -1, 4, .2, f], dt[.45, .005, 1, 4, .2, f]}, 
 {t, 0, 200, PD}],

     ViewPoint -> Left, BoxRatios -> {1, 1, 1.3}, ViewVertical -> {0, 0, -1}, 
    ViewCenter -> {{0.5, 0.5, 0.5}, {0.5, 0.55}}, Boxed -> False, Lighting -> "Neutral", 
    PlotRange -> {{-20, 20}, {-20, 20}, {0, 20}}, Background -> Black],

   {f, 0, 1, .01}];

Export["tree.gif", frames]

enter image description here

share|improve this answer
    
Where you set the size and how it affects the generated tree? –  manatwork Dec 28 '13 at 15:24
1  
@manatwork this is basically graph t*sin(t) as i said in the post. So plotting for larger t's will just make a bigger tree: Table[..., {t, 0, 200, PD}] –  Vitaliy Kaurov Dec 28 '13 at 19:41

Processing

The original fractal Christmas tree. Mouse Y position determines size, use up and down arrow keys to change the number of generations.

int size = 500;

int depth = 10;

void setup() {
  frameRate(30);
  size(size, size, P2D);
}

void draw() {
  background(255);
  tree(size/2.0, size, 0.0, radians(0), radians(108), (size - mouseY)/3, depth);
  tree(size/2.0, size, 0.0, radians(108), radians(0), (size - mouseY)/3, depth);
}

void keyPressed() {
  if(keyCode == UP) depth++;
  if(keyCode == DOWN) depth--;
  depth = max(depth, 1);
}

void tree(float posX, float posY, float angle, float forkRight, float forkLeft, float length, int generation) {
  if (generation > 0) {
    float nextX = posX + length * sin(angle);
    float nextY = posY - length * cos(angle);

    line(posX, posY, nextX, nextY);

    tree(nextX, nextY, angle + forkRight, forkRight, forkLeft, length*0.6, generation - 1);
    tree(nextX, nextY, angle - forkLeft,  forkRight, forkLeft, length*0.6, generation - 1);
  }
}

What a lovely Christmas tree, don't you think, ma?

Or, if you prefer a fuller tree:

int size = 500;

int depth = 10;

void setup() {
  frameRate(30);
  size(size, size, P2D);
}

void draw() {
  background(255);
  tree(size/2.0 - 5, size, 0.0, 0.0, (size - mouseY)/3, depth);
}

void keyPressed() {
  if(keyCode == UP) depth++;
  if(keyCode == DOWN) depth--;
  depth = max(depth, 1);
}

void tree(float posX, float posY, float angle, float fork, float length, int generation) {
  if (generation > 0) {
    float nextX = posX + length * sin(angle);
    float nextY = posY - length * cos(angle);

    line(posX, posY, nextX, nextY);

    tree(nextX, nextY, angle + fork + radians(108), fork, length*0.6, generation - 1);
    tree(nextX, nextY, angle - fork,                fork, length*0.6, generation - 1);
    tree(nextX, nextY, angle + fork - radians(108), fork, length*0.6, generation - 1);
  }
}

I'm super bummed that somebody beat me to posting a fractal tree even though I was was ahead of them in the idea and in actually starting work on it :/

share|improve this answer
    
+1 for using Processing –  The Guy with The Hat Dec 21 '13 at 13:28

Ruby

((1..20).to_a+[6]*4).map{|i|puts ('*'*i*2).center(80)}

Happy New Year

You can customize output by changing *.
For a green tree: ((1..20).to_a+[6]*4).map{|i|puts "\e[32m"+('*'*i*2).center(80)}

Approach 2 (Xmas tree that doesn't look like an arrow)

((1..6).to_a+(3..9).to_a+(6..12).to_a+[3]*4).map{|i|puts "\e[32m"+('*'*i*4).center(80)}

Happy New Year

Approach 3

((1..20).to_a+[6]*4).map{|i|puts' '*(20-i)+'*'*(2*i-1)}
share|improve this answer
2  
This is more an arrow, then a christmas tree! –  klingt.net Dec 14 '13 at 13:40
3  
@klingt.net *than. –  NARKOZ Dec 14 '13 at 16:43
    
Time to light up the christmas tree. –  NARKOZ Dec 14 '13 at 17:08
    
And where you specify the size? –  manatwork Dec 14 '13 at 17:31
    
@manatwork in ranges. 1..20, 1..6, etc. –  NARKOZ Dec 14 '13 at 17:56

I made this for a challenge on /r/dailyprogrammer, (not sure if reusing code is against the spirit/rules of this) but:

Brainfuck. Takes as input a number (length of bottom row of leaves) and two characters. One space between each.

Example input: 13 = +

Example output:

      +
     +++
    +++++
   +++++++
  +++++++++
 +++++++++++
+++++++++++++
     ===

Code:

                   >
                  ,--
                 -----
                -------
               ---------
              ---------[+
             +++++++++++++
            +++++++++++++++
           +++<[>>++++++++++
          <<-]>--------------
         ---------------------
        -------------[<+>-]>[<<
       +>>-]<,------------------
      --------------],>,,>>++++[<
     ++++++++>-]++++++++++>+<<<<<-
    [>>>>>>+>+<<<<<<<--]>>>>>>>[<<<
   <<<<++>>>>>>>-]<<<<<<<[>>>>>>[<<<
  .>>>>+<-]>[<+>-]<<[<<<.>>>>>+<<-]>>
 [<<+>>-]<<<.>>><-<++<<<<<--]>>...>>>-
--[>+<<<<..>>>--]<.>>[<<<.>>>>+<-]<<<<<
                ...>>>.
share|improve this answer
    
At last, a brainfuck answer! –  Pharap Aug 22 at 5:11

Game Maker Language

spr_tree

spr_tree

The tree's Create Event

image_speed=0
size=get_integer("How big do you want me to be?#Integer between 1 and 10, please!",10)
image_index=size-1

Room, 402 by 599

The tree is placed at (0,0)

Bonus! You can resize the Christmas tree after the initial input with the keys 0-9.

share|improve this answer

Processing

I made this tree generator using an L-System and a Turtle.
In my extremely biased opinion, it is the most realistic tree here so far.

A Tree

code:

//My code, made from scratch:
final int THE_ITERATIONS = 7;
final float SCALE = 1;
final int ANGLE = 130;
final int SIZE = 4;

final int ITERATIONS = THE_ITERATIONS - 1;

int lineLength;

String lSystem;
ArrayList<Turtle> turtles;

int turtleIndex;
int lSystemIndex;

void setup()
{
  size(320, 420);
  background(255);
  translate(width / 2, height - 70);

  lineLength = ITERATIONS * 2 + 2;

  lSystem = "[-F][+F]F";

  turtles = new ArrayList<Turtle>(ITERATIONS + 1);
  lSystemIndex = 0;

  calculateLSystem();
  println(lSystem);

  turtles.add(new Turtle(0, 0));

  doTurtles();
  save("Tree.png");
}

void doTurtles()
{
  while(lSystemIndex < lSystem.length())
  {
    print("\"" + lSystem.charAt(lSystemIndex) + "\": ");
    if(lSystem.charAt(lSystemIndex) == 'F')
    {
      turtleForward();
    }
    else if(lSystem.charAt(lSystemIndex) == '[')
    {
      lineLength -= 2;
      addTurtle();
      println(turtles.size());
    }
    else if(lSystem.charAt(lSystemIndex) == ']')
    {
      lineLength += 2;
      removeTurtle();
      println(turtles.size());
    }
    else if(lSystem.charAt(lSystemIndex) == '+')
    {
      turtleRight();
    }
    else if(lSystem.charAt(lSystemIndex) == '-')
    {
      turtleLeft();
    }
    lSystemIndex++;
  }
}

void addTurtle()
{
  turtles.add(new Turtle(turtles.get(turtles.size() - 1)));
}

void removeTurtle()
{
  turtles.remove(turtles.size() - 1);
}

void turtleLeft()
{
  turtles.get(turtles.size() - 1).left(ANGLE + random(-5, 5));
}

void turtleRight()
{
  turtles.get(turtles.size() - 1).right(ANGLE + random(-5, 5));
}

void turtleForward()
{
  print(turtles.get(turtles.size() - 1) + ": ");
  strokeWeight(min(lineLength / SIZE, 1));
  stroke(5 + random(16), 90 + random(16), 15 + random(16));
  if(turtles.size() == 1)
  {
    strokeWeight(lineLength / 2);
    stroke(100, 75, 25);
  }
  turtles.get(turtles.size() - 1).right(random(-3, 3));
  turtles.get(turtles.size() - 1).forward(lineLength);
}

void calculateLSystem()
{
  for(int i = 0; i < ITERATIONS; i++)
  {
    lSystem = lSystem.replaceAll("F", "F[-F][+F][F]");
  }
  lSystem = "FF" + lSystem;
}

void draw()
{

}

//——————————————————————————————————————————————————————
// Turtle code, heavily based on code by Jamie Matthews
// http://j4mie.org/blog/simple-turtle-for-processing/
//——————————————————————————————————————————————————————

class Turtle {
  float x, y; // Current position of the turtle
  float angle = -90; // Current heading of the turtle
  boolean penDown = true; // Is pen down?
  int lineLength;

  // Set up initial position
  Turtle (float xin, float yin) {
    x = xin;
    y = yin;
    //lineLength = lineLengthin;
  }

  Turtle (Turtle turtle) {
    x = turtle.x;
    y = turtle.y;
    angle = turtle.angle;
    //lineLength = turtle.lineLength - 1;
  }

  // Move forward by the specified distance
  void forward (float distance) {
    distance = distance * SIZE * random(0.9, 1.1);
    // Calculate the new position
    float xtarget = x+cos(radians(angle)) * distance;
    float ytarget = y+sin(radians(angle)) * distance;

    // If the pen is down, draw a line to the new position
    if (penDown) line(x, y, xtarget, ytarget);
    println(x + ", " + y + ", " + xtarget + ", " + ytarget);
    // Update position
    x = xtarget;
    y = ytarget;
  }

  // Turn left by given angle
  void left (float turnangle) {
    angle -= turnangle;
    println(angle);
  }

  // Turn right by given angle
  void right (float turnangle) {
    angle += turnangle;
    println(angle);
  }
}
share|improve this answer
2  
Yes! An L-System! –  luser droog Dec 18 '13 at 8:47
1  
I agree about it being the most realistic :) –  Timtech Dec 20 '13 at 22:31
    
@Timtech I disagree. I think this is the second most realistic, after Jan Dvorak's Javascript answer. I mean, look at those needles (on this one)! They are all clustered together in patterns that I have not seen on an evergreen tree. However, Jan's tree looks a lot like a fir sapling to me. –  Quincunx Dec 21 '13 at 0:20
1  
@Quincunx I must have missed that one. Still, this one looks pretty good. –  Timtech Dec 21 '13 at 0:38
    
@Quincunx Jan's tree—and Howard's, too—is really cool, but not necessarily completely realistic. Jan even admits that the tree is only "quasirealistic". – RyanCarlson 10 secs ago –  The Guy with The Hat Dec 21 '13 at 13:51

Mathematica ASCII

I really like ASCII art, so I add another very different answer - especially so it's so short in Mathematica:

Column[Table[Row[RandomChoice[{"+", ".", "*", "~", "^", "o"}, k]], {k, 1, 35, 2}], 
Alignment -> Center]

enter image description here

And now with a bit more sophistication a scalable dynamic ASCII tree. Watch closely - the tree is also changing - snow sticks to branches then falls ;-)

DynamicModule[{atoms, tree, pos, snow, p = .8, sz = 15},

 atoms = {
   Style["+", White],
   Style["*", White],
   Style["o", White],
   Style[".", Green],
   Style["~", Green],
   Style["^", Green],
   Style["^", Green]
   };

 pos = Flatten[Table[{m, n}, {m, 18}, {n, 2 m - 1}], 1];

 tree = Table[RandomChoice[atoms, k], {k, 1, 35, 2}];

 snow = Table[
   RotateLeft[ArrayPad[{RandomChoice[atoms[[1 ;; 2]]]}, {0, sz}, " "],
     RandomInteger[sz]], {sz + 1}];

 Dynamic[Refresh[

   Overlay[{

     tree[[Sequence @@ RandomChoice[pos]]] = RandomChoice[atoms];
     Column[Row /@ tree, Alignment -> Center, Background -> Black],

     Grid[
      snow = 
       RotateRight[
        RotateLeft[#, 
           RandomChoice[{(1 - p)/2, p, (1 - p)/2} -> {-1, 0, 1}]] & /@
          snow
        , {1, 0}]]

     }, Alignment -> Center]

   , UpdateInterval -> 0, TrackedSymbols -> {}]
  ]
 ]

enter image description here

share|improve this answer
    
Nice one. Where you adjust the size? Is that 35 in Table's parameters? –  manatwork Jan 7 at 9:29
    
@manatwork yes the 35 in Table. Thanks ;) –  Vitaliy Kaurov Jan 7 at 11:16

JavaScript (run on any page in console)

I was golfing this but then decided not to, so as you can see there are TONS of magic numbers :P

// size
s = 300

document.write('<canvas id=c width=' + s + ' height=' + s + '>')
c = document.getElementById('c').getContext('2d')
c.fillStyle = '#0D0'
for (var i = s / 3; i <= s / 3 * 2; i += s / 6) {
  c.moveTo(s / 2, s / 10)
  c.lineTo(s / 3, i)
  c.lineTo(s / 3 * 2, i)
  c.fill()
}
c.fillStyle = '#DA0'
c.fillRect(s / 2 - s / 20, s / 3 * 2, s / 10, s / 6)

Result for s = 300:

screenshot

s = 600:

screenshot

share|improve this answer
2  
Are you going to decorate the tree? :-) –  Quincunx Dec 12 '13 at 23:53
    
@Quincunx Possibly, I may work on that later :) –  Doorknob Dec 12 '13 at 23:54
1  
GAH! Where are the semicolons?! –  Andrew Larsson Dec 17 '13 at 0:46
1  
@AndrewLarsson: Semicolons are not required in JavaScript –  ProgramFOX Dec 18 '13 at 14:08

Turtle Graphics

enter image description here

Based on properties of the Euler spiral.

Code:

enter image description here

The scale is determined by the step size (move forward by: 6). An interactive version is available here.

P.S. Inspired by this question.

share|improve this answer
    
Where is this "scalable"? Other than that, this is great! –  Quincunx Jan 11 at 4:40
    
@Quincunx You can change scale by changing move forward by 6. For example, 10 will produce a larger tree. Actually there is no input in this "language" (or the whole code can be treated as input ;) ). –  ybeltukov Jan 11 at 18:14

Decorated FORTRAN tree

    character(len=36) :: f='/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\'
    character(len=18) :: g = '                  '
    character(len=14) :: p = ".x$.+oO+oO.#,~"
    character(len=10) :: q

    n = iargc()
    if (n /= 1) then 
        k = len(g)
    else
        call getarg(1,q) 
        read(q,*) k
    end if
    l = modulo(k,len(g))+1
    do i=1,l
        j = mod(i,len(p)-1)+1
        print *,g(1:l-i),p(j:j),f(1:2*i),p(j+1:J+1)
        end do
    print *,g(1:l-1),f(1:4)
    end

The tree has a limited range of sizes, but it believe it accurately reflects the life of most Christmas trees.

From infant tree:

$./tree
 x/\$
 /\/\

To adolescent tree:

$./tree 6
       x/\$
      $/\/\.
     ./\/\/\+
    +/\/\/\/\o
   o/\/\/\/\/\O
  O/\/\/\/\/\/\+
 +/\/\/\/\/\/\/\o
       /\/\

To adult:

$./tree 17
                  x/\$
                 $/\/\.
                ./\/\/\+
               +/\/\/\/\o
              o/\/\/\/\/\O
             O/\/\/\/\/\/\+
            +/\/\/\/\/\/\/\o
           o/\/\/\/\/\/\/\/\O
          O/\/\/\/\/\/\/\/\/\.
         ./\/\/\/\/\/\/\/\/\/\#
        #/\/\/\/\/\/\/\/\/\/\/\,
       ,/\/\/\/\/\/\/\/\/\/\/\/\~
      ./\/\/\/\/\/\/\/\/\/\/\/\/\x
     x/\/\/\/\/\/\/\/\/\/\/\/\/\/\$
    $/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\.
   ./\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\+
  +/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\o
 o/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\O
                  /\/\    
share|improve this answer

Rebol

With a dialect to display the symbols. To change the tree size, just change the parameter of make-tree.

make-tree: func [int /local tr] [
  tr: copy []
  length: (int * 2) + 3
  repeat i int [
      repeat j 3 [
            ast: to-integer ((i * 2) - 1 + (j * 2) - 2)
            sp: to-integer (length - ast) / 2
            append/dup tr space sp 
            append/dup tr "*" ast 
            append tr lf
      ]
  ]
  append/dup tr space (length - 1) / 2
  append tr "|"
  append tr lf
  tr
]

print make-tree 3

a tree with 3 layers a tree with 5 layers

share|improve this answer
3  
Where you set the size and how it affects the generated tree? –  manatwork Dec 24 '13 at 16:13
    
@manatwork To change the tree block. You can resize it or change some of the symbols at your will. –  WayneTsui Dec 24 '13 at 16:19
    
@WayneTsui this doesn't conform to the OP's requirements, "size must be choosable by some input method" and "larger inputs should produce a larger tree." –  ulidtko Dec 25 '13 at 14:28
1  
a new make-tree function from @kealist –  WayneTsui Dec 25 '13 at 15:49

Ti-Basic 84

Asks for a size:

              :
            Input 
           S:Lbl 1
            S-1→S
              :
             "||
            "+""→
           Str1:" 
             "→Str2
         :Disp Str2+
        "   **":Disp 
       Str2+"  "+"*"+
      Str1+"*":Disp " 
     "+Str1+"*-"+"||-*
    ":Disp Str1+"*--||-
   -*":Disp "   *---||--
  -*":Disp "  *----||----
 *":Disp Str1+"   "+Str2+"
":If S>0:Goto 2:Goto 1:Lbl 
              2

Output (size 1):

       **
      *||*
     *-||-*
    *--||--*
   *---||---*
  *----||----*
       ||
share|improve this answer
1  
Where is the size choosable? This is a fixed-size tree –  Quincunx Dec 20 '13 at 23:00
1  
@Quincunx Fixed. –  Timtech Dec 20 '13 at 23:33

Python

import sys
w = sys.stdout.write
def t(n,s):
    for i in range(n):
        for a in range(n-i):
            w(" ")
        w("[")
        for l in range(i<<1):
            if i==n-1:
                w("_")
            else:
                w("~")
        w("]")
        print("")
    for o in range(s):
        for i in range(n):
            w(" ")
        print("[]")

t(10, 2)

enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.