6
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What are Fibonacci networks?

In 1202, an Italian mathematician known as Fibonacci posed the following problem:

"A male and a female rabbit are born at the beginning of the year. We assume the following conditions:

1.After reaching the age of two months, each pair produces a mixed pair, (one male, one female), and then another mixed pair each month thereafter.

2.No deaths occur during the year."

How many rabbits will there be at the end of the year?

Fibonacci rabbits

Sources for graph and problem statement

Challenge

Your challenge is to write a program/function to draw an ASCII art to output or a text file, or a graphical picture to a file/screen of a graph with following settings:

  1. At 0th tick, there is one newly-born node.

  2. Every newly born node creates a normal node after one tick after being born and gets deactivated.

  3. A normal node creates a normal node and a newly-born node one tick after its birth and gets deactivated.

Your program will get the non-negative value n as input, and will draw the graph at the nth tick. Note that your program can draw newly-born nodes and normal nodes differently, but this is not necessary.

Rules

  • You must not use any built-in Fibonacci sequence function or mathematical constants or similar.

  • If your program outputs to a file, it must be either a text file in a suitable and viewable format and encoding, or an image file in a suitable and viewable format with recognizable colors (for example, an image format that can be uploaded to SE).

  • Your program can draw graph in any size, any shape and any alignment but the nodes and connections must be recognizable. Tree-shaped graphs (like the graph in "What are Fibonacci networks?" section) are recommended, but not necessary.

  • Common code golf rules apply to this challenge.

Examples

Input: 2
Possible Output: 
*
|
*
| \
* *

Input: 0
Possible Output: 
* 

Input: 5

Possible Graphical Output:Diagram in PNG

This is a . Shortest code wins. Good luck!

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  • 4
    \$\begingroup\$ Can you have a larger testcase? \$\endgroup\$ – Leaky Nun Jul 30 '16 at 13:12
  • \$\begingroup\$ @LeakyNun Well, every time i draw , i notice that i've drawn something wrong ... I'm trying hard to draw a larger test case ... \$\endgroup\$ – user55673 Jul 30 '16 at 13:15
  • 2
    \$\begingroup\$ Then draw a larger testcase before uploading. \$\endgroup\$ – Leaky Nun Jul 30 '16 at 13:15
  • \$\begingroup\$ @LeakyNun meta.codegolf.stackexchange.com/a/9681/17602 \$\endgroup\$ – Neil Jul 30 '16 at 13:24
  • \$\begingroup\$ @GLASSIC To draw the large test case, why not just write an ungolfed version of your challenge and run it? \$\endgroup\$ – Socratic Phoenix Jul 30 '16 at 15:20
6
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Pyth, 31 bytes

Ms+0m+Wd+J+G"| "+G1+\-gJtd_UHgb

Try it online

How it works

M                                  def g(G, H):
    m                     _UH        map for d in [H - 1, …, 0]:
         J+G"| "                       assign J = G + "| "
        +       +G1                    J + G + 1
     +Wd                               conditional +, if d is truthy (nonzero):
                   +\-gJtd               "-" + g(J, d - 1)
  +0                                 prepend 0
 s                                   concatenate
                             gbQ   g("\n", input)

Output for 6

0
| 
1-0
| | 
| 1-0
| | | 
| | 1-0
| | | 
| | 1
| | 
| 1-0
| | | 
| | 1
| | 
| 1-0
| | 
| 1
| 
1-0
| | 
| 1-0
| | | 
| | 1
| | 
| 1-0
| | 
| 1
| 
1-0
| | 
| 1-0
| | 
| 1
| 
1-0
| | 
| 1
| 
1-0
| 
1

Pyth, 43 bytes

Mc+WH`G++WG\-+j"
| "g0tH"
`-"j"
  "g1tHbjg0

Try it online

Output for 6

0-1-0-1-0-1-0
  |   |   `-1
  |   `-1-0-1
  |     `-1-0
  |       `-1
  `-1-0-1-0-1
    |   `-1-0
    |     `-1
    `-1-0-1-0
      |   `-1
      `-1-0-1
        `-1-0
          `-1
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5
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HTML5/JavaScript, 1262 bytes

This is my code which displays a Fibonacci network. You can just paste it into a file with a name like index.html and run it in your favourite browser (I only tested Chrome and Firefox though). When you launch it, it asks you for the desired number of generations. It uses the HTML5 <canvas> tag to display the image.

<script>let t,p,o,c=[],q=50,r=20,s=25;function d(o){return JSON.parse(JSON.stringify(o));}function e(a,x,y){o.beginPath();o.fillStyle=a?"red":"#000";o.arc(x*q+s,y*q+s,r,0,2*3.14159);o.fill();}function f(a,b,c,d){o.beginPath();o.strokeStyle="blue";o.lineWidth=3;o.moveTo(a*q+s,b*q+s);o.lineTo(c*q+s,d*q+s);o.stroke();}function g(i){return c[i].a+c[i].c;}function h(i){return c[i].b+c[i].d;}function k(j,e,f,b,n,x,y,z){f=j;b=0;while(1){if(!e){f.t=g(t);f.u=0;f.v=f.t;f.w=f.b;}else{n=t-b+1;x=t-b;z=e.b%2?(g(n)+g(x)+1):(-(g(n)+h(x)+1));f.t=e.v+z;f.u=e.w+1;f.v=f.t;f.w=f.u+f.b;}if(!f.a)break;b+=f.b+1;e=f;f=f.z;}return {a:f.b==0,x:f.v,y:f.w};}function l(j,c,b){b=k(j);f(b.x,b.y,c.x,c.y);e(b.a,b.x,b.y);return b;}function m(j){while(j.a)j=j.z;return j;}function n(j,o,p,q,r,s){if(o>t)return;q=l(j,p);r=d(j);m(r).b++;n(r,o+1,q);if(m(j).b>0){r=d(j);s=m(r);s.a=!0;s.z={a:!1,b:0};n(r,o+1,q);}}function b(i,z,y){t=parseInt(prompt());if(isNaN(t)||t<0)t=0;p=document.getElementById("o");c.push({a:0,b:0,c:0,d:0,e:1});for(i=1;i<t+1;i++){z=c[c.length-1];y={a:(i>1?i%2:0)+z.c,b:(i>1?1-i%2:0)+z.d,c:z.a+z.c,d:z.b+z.d};y.e=y.a+y.b+y.c+y.d;c.push(y);}p.width=q*(c[c.length-1].e+1);p.height=q*(t+1);o=p.getContext("2d");n({a:!1,b:0},0,0);}</script><body onload="b()"><canvas id="o">

An example for 6 generations would be this one: The Fibonacci network after 6 generations

As you can see, newly born nodes are colored in red, the others are black.

For reference, this is my ungolfed, development code that does exactly the same:

<html>
	<head>
		<meta charset="utf-8">
		<title>Fibonacci Graph</title>
		<script type="text/javascript">
			let totalGenerations;
			let canvas, ctx;
			let individualCount = [];
			let individualDrawSize = 50, individualDrawRadius = 20, individualDrawOffset = 25;
		
			function generateIndividualCount()
			{
				individualCount.push(
				{
					babiesL: 0,			//number of babies on the left side of the main individual
					babiesR: 0,
					adultsL: 0,
					adultsR: 0,
					total: 1			//total number of individuals (sum of all other members+1, because the main individual counts as well)
				});
				
				for(let i=1; i<totalGenerations+1; i++)
				{
					let last = individualCount[individualCount.length - 1];
					let next =
					{
						//make an adult alternate between having a baby at the left and the right side (right side first)
						babiesL: (i>1 ? i%2 : 0) + last.adultsL,
						babiesR: (i>1 ? 1-i%2 : 0) + last.adultsR,
						adultsL: last.babiesL + last.adultsL,
						adultsR: last.babiesR + last.adultsR
					};
					next.total = next.babiesL + next.babiesR + next.adultsL + next.adultsR;
					individualCount.push(next);
				}
			}
			
			function setupContext()
			{
				canvas.width = individualDrawSize * (individualCount[individualCount.length - 1].total + 1);
				canvas.height = individualDrawSize * (totalGenerations + 1);
				ctx = canvas.getContext("2d");
			}
			
			function deepCopy(obj)
			{
				return JSON.parse(JSON.stringify(obj));
			}

			function drawIndividual(isBaby, x, y)
			{
				ctx.beginPath();
				ctx.fillStyle = isBaby ? "red" : "black";
				ctx.arc(x*individualDrawSize + individualDrawOffset, y*individualDrawSize + individualDrawOffset, individualDrawRadius, 0, 2*Math.PI);
				ctx.fill();
			}
			
			function drawLine(x1, y1, x2, y2)
			{
				ctx.beginPath();
				ctx.strokeStyle = "blue";
				ctx.lineWidth = 3;
				ctx.moveTo(x1*individualDrawSize + individualDrawOffset, y1*individualDrawSize + individualDrawOffset);
				ctx.lineTo(x2*individualDrawSize + individualDrawOffset, y2*individualDrawSize + individualDrawOffset);
				ctx.stroke();
			}

			function getIndividualCountLeft(i)		//left like in "the opposite of right"
			{
				return individualCount[i].babiesL + individualCount[i].adultsL;
			}
			
			function getIndividualCountRight(i)
			{
				return individualCount[i].babiesR + individualCount[i].adultsR;
			}
			
			function getPosition(ancestorInformation)
			{
				let prevInfo;
				let tempInfo = ancestorInformation;
				let currDepth = 0;
				
				while(1)
				{
					if(prevInfo == undefined)
					{
						tempInfo.birthX = getIndividualCountLeft(totalGenerations);
						tempInfo.birthY = 0;
						
						tempInfo.givingBirthX = tempInfo.birthX;
						tempInfo.givingBirthY = tempInfo.ageAtGivingBirth;
					}
					else
					{
						let generationsLeftForParent = totalGenerations - (currDepth-1);
						let generationsLeftForChild = totalGenerations - currDepth;				//left like in "to go" or "There are 42 cookies left"
						let isLeftBranch = (prevInfo.ageAtGivingBirth % 2 == 0);
						let distanceToParent;
						
						
						if(isLeftBranch)
							distanceToParent = -(getIndividualCountLeft(generationsLeftForParent) + getIndividualCountRight(generationsLeftForChild) + 1);
						else
							//some of these directions may seem odd, but because the parent subtree begins with a left branch (and individualCount only contains data for trees beginning with a right branch), it needs to be inversed
							//if individualCount would have been generated for trees beginning with a left node, all of the getIndividualCountLeft/getIndividualCountRight function would need to be inversed
							//or, when individualCount would have been generated for both cases, functions like getIndividualCountRight_LeftBegin and getIndividualCountLeft_RightBegin could be used
							distanceToParent = getIndividualCountLeft(generationsLeftForParent) + getIndividualCountLeft(generationsLeftForChild) + 1;		//+1, because the previous value of distanceToParent is just the number of empty fields (on the x axis) between the parent and the child
						
						tempInfo.birthX = prevInfo.givingBirthX + distanceToParent;
						tempInfo.birthY = prevInfo.givingBirthY + 1;
						
						tempInfo.givingBirthX = tempInfo.birthX;
						tempInfo.givingBirthY = tempInfo.birthY + tempInfo.ageAtGivingBirth;
					}
					if(!tempInfo.isParent)
						break;
					currDepth += tempInfo.ageAtGivingBirth + 1;
					prevInfo = tempInfo;
					tempInfo = tempInfo.child;
				}
				
				return { isBaby: tempInfo.ageAtGivingBirth==0, x: tempInfo.givingBirthX, y: tempInfo.givingBirthY };
			}
			
			function draw(ancestorInformation, parentPos)
			{
				let pos = getPosition(ancestorInformation);
				drawLine(pos.x, pos.y, parentPos.x, parentPos.y);
				drawIndividual(pos.isBaby, pos.x, pos.y);
				return pos;
			}
			
			function getDeepestChild(ancestorInformation)
			{
				while(ancestorInformation.isParent)
					ancestorInformation = ancestorInformation.child;
				return ancestorInformation;
			}
			
			function drawAll(ancestorInformation, depth, parentPos)
			{
				if(depth > totalGenerations)
					return;
				
				let pos = draw(ancestorInformation, parentPos);
				let newAncestorInformation = deepCopy(ancestorInformation);
				getDeepestChild(newAncestorInformation).ageAtGivingBirth++;
				drawAll(newAncestorInformation, depth + 1, pos);
				
				if(getDeepestChild(ancestorInformation).ageAtGivingBirth > 0)
				{
					newAncestorInformation = deepCopy(ancestorInformation);
					let deepestChild = getDeepestChild(newAncestorInformation);
					deepestChild.isParent = true;
					deepestChild.child = { isParent: false, ageAtGivingBirth: 0 };
					drawAll(newAncestorInformation, depth + 1, pos);
				}
			}
		
			function load()
			{
				totalGenerations = parseInt(prompt("Number of generations?"));
				if(isNaN(totalGenerations)  ||  totalGenerations < 0)
					totalGenerations = 0;
				canvas = document.getElementById("output");
				generateIndividualCount();
				setupContext();
				drawAll({ isParent: false, ageAtGivingBirth: 0 }, 0, 0);					//although this individual does not really give birth after it has been born, it still has an attribute called "ageAtGivingBirth" (like its parents) to achieve a similar effect
			}
		</script>
	</head>
	<body onload="load()">
		<canvas id="output" style="position:absolute; top:10px; left:10px" width="1200" height="600">Canvas not supported...</canvas>
	</body>
</html>

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