是否有任何程序/网站可以显示递归Fibonacci计算的图形。 我想说明需要多少递归步骤。
答案 0 :(得分:1)
我发现这是一个很小的挑战,让我分享一下我的实施:
var canvas = document.getElementById("canvas");
var width = canvas.width;
var height = canvas.height;
var ctx = canvas.getContext("2d");
var FN = 8;
var FONT_SIZE = 11; // in points
var SHOW_BOXES = false;
var SHOW_DISCS = true;
var SHOW_FIB_N = true;
function Tree(fn) {
var pdata = {};
pdata.lhs = null;
pdata.rhs = null;
pdata.fn = fn;
this.getLeft = function() { return pdata.lhs; };
this.setLeft = function(node) { pdata.lhs = node; };
this.getRight = function() { return pdata.rhs; };
this.setRight = function(node) { pdata.rhs = node; };
this.getFn = function() { return pdata.fn; };
}
function fib(n) {
if(n == 0)
return new Tree(0);
if(n == 1)
return new Tree(1);
else {
var lhs = fib(n-1);
var rhs = fib(n-2);
var root = new Tree(lhs.getFn() + rhs.getFn());
root.setLeft(lhs);
root.setRight(rhs);
return root;
}
}
var root = fib(FN);
function Box(x0, y0, x1, y1) {
if(arguments.length < 4) {
x0 = 1;
y0 = 1;
x1 = -1;
y1 = -1;
}
this.x0 = x0;
this.y0 = y0;
this.x1 = x1;
this.y1 = y1;
this.width = function() { return this.x1 - this.x0; };
this.height = function() { return this.y1 - this.y0; };
this.offset = function(x, y) {
this.x0 += x;
this.y0 += y;
this.x1 += x;
this.y1 += y;
};
this.extend = function(x, y) {
if(this.x1 < this.x0 || this.y1 < this.y0) {
this.x0 = x;
this.x1 = x;
this.y0 = y;
this.y1 = y;
} else {
this.x0 = this.x0 < x ? this.x0 : x;
this.y0 = this.y0 < y ? this.y0 : y;
this.x1 = this.x1 > x ? this.x1 : x;
this.y1 = this.y1 > y ? this.y1 : y;
}
}
};
(function () {
// assume spheres of radius 0.5
function setBounds(node, offX, offY) {
var bbox = new Box(offX, offY, offX + 1, offY + 1);
if(node.getLeft() != null || node.getRight() != null) {
var lhs = node.getLeft(), rhs = node.getRight();
if(lhs != null) {
setBounds(lhs, offX + 0, offY + 1.1);
bbox.extend(lhs.bbox.x0, lhs.bbox.y0);
bbox.extend(lhs.bbox.x1, lhs.bbox.y1);
}
if(rhs != null) {
setBounds(rhs, offX + (lhs != null ? lhs.bbox.width() : 0), offY + 1.1);
bbox.extend(rhs.bbox.x0, rhs.bbox.y0);
bbox.extend(rhs.bbox.x1, rhs.bbox.y1);
}
}
node.bbox = bbox;
}
setBounds(root, 0, 0);
})();
var transf = (function() {
var b = 2;
var sx = (width - 2 * b) / root.bbox.width();
var sy = (height - 2 * b) / root.bbox.height();
return {
ox: b / sx - root.bbox.x0,
oy: b / sx - root.bbox.y0,
sx: sx,
sy: sy,
};
})();
transf.smin = Math.min(transf.sx, transf.sy);
ctx.clearRect(0, 0, width, height);
(function(g) {
g.font = FONT_SIZE + "pt Arial";
g.textAlign = "center";
g.strokeStyle = "#000000";
function draw(node, pX, pY) {
if(node == null) return;
var cX = (node.bbox.x0 + node.bbox.x1) / 2;
var cY = (node.bbox.y0 + 0.5);
var radius = 0.475;
cX = transf.sx * (cX + transf.ox);
cY = transf.sy * (cY + transf.oy);
radius *= transf.smin;
draw(node.getLeft(), cX, cY);
draw(node.getRight(), cX, cY);
if(SHOW_BOXES) {
g.fillStyle = "#ff0000";
g.beginPath();
g.moveTo(transf.sx * (node.bbox.x0 + transf.ox), transf.sy * (node.bbox.y0 + transf.oy));
g.lineTo(transf.sx * (node.bbox.x1 + transf.ox), transf.sy * (node.bbox.y0 + transf.oy));
g.lineTo(transf.sx * (node.bbox.x1 + transf.ox), transf.sy * (node.bbox.y1 + transf.oy));
g.lineTo(transf.sx * (node.bbox.x0 + transf.ox), transf.sy * (node.bbox.y1 + transf.oy));
g.closePath();
g.stroke();
}
if(SHOW_DISCS) {
if(arguments.length >= 3) {
g.beginPath();
g.moveTo(pX, pY);
g.lineTo(cX, cY);
g.stroke();
}
g.fillStyle = "#ff0000";
g.beginPath();
g.arc(cX, cY, radius, 0, 2 * Math.PI);
g.fill();
g.stroke();
}
if(SHOW_FIB_N) {
g.fillStyle = "#0000ff";
g.fillText(node.getFn(), cX, cY + FONT_SIZE / 2);
}
}
draw(root);
})(ctx);
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<canvas id="canvas" width="800" height="480" />
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