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completes dummy assignment #15

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193 changes: 174 additions & 19 deletions Fibonacci.ipynb
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Original file line number Diff line number Diff line change
Expand Up @@ -2,14 +2,20 @@
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"# Fibonacci Numbers [[1]](https://en.wikipedia.org/wiki/Fibonacci_number)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"In mathematics, the Fibonacci numbers are the numbers in the following integer sequence, called the Fibonacci sequence, and characterized by the fact that every number after the first two is the sum of the two preceding ones:\n",
"\n",
Expand All @@ -18,7 +24,10 @@
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"In mathematical terms, the sequence $F_n$ of Fibonacci numbers is defined by the recurrence relation:\n",
"\n",
Expand All @@ -31,7 +40,10 @@
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"In matrix notation this definition is equivalent to:\n",
"\n",
Expand Down Expand Up @@ -63,7 +75,10 @@
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"If we want to compute only the $n^{th}$ Fibonacci number, then the following identity is useful:\n",
"\n",
Expand All @@ -87,71 +102,203 @@
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"**Follow the instructions in the next sections. Feel free to create extra cells (for instance, you can try different values for $F_1$ and $F_0$).**"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"### 1. Fast Fibonacci Transform Implementation"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"Implement a function that returns $F_n$ as we described above (for this assignment we are not concerned about the efficiency of your implementation, i.e. you can use $M^n$ assuming octave does matrix exponentiation for you):"
]
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 4,
"metadata": {
"collapsed": true,
"deletable": true,
"editable": true
},
"outputs": [],
"source": [
"function fn = fastFibonacci (n)\n",
" fn = 0;\n",
" if (n == 0)\n",
" return;\n",
" elseif(n == 1)\n",
" fn = 1;\n",
" return;\n",
" else\n",
" multiplier=[1,1;1,0];\n",
" baseVector=[1;0];\n",
" fn = ((multiplier**(n-1))*baseVector)(1,1);\n",
" return;\n",
" endif\n",
"endfunction"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
"source": [
"function fn = calculateRatio (n)\n",
" fn = 0;\n",
" fn = fastFibonacci(n+1)/fastFibonacci(n);\n",
" return;\n",
"endfunction"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"### 2. Plot $F_{n+1} / F_n$ ratio"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"Initialize $F_0 = 0$ and $F_1 = 1$, then plot the $\\dfrac{F_{n+1}}{F_{n}}$ values for $ 1 \\leq n \\leq 100$. As $n \\to \\infty$, we expect $\\dfrac{F_{n+1}}{F_{n}} \\to \\dfrac{\\sqrt{5}+1}{2}$. "
]
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 8,
"metadata": {
"collapsed": true
"collapsed": false,
"deletable": true,
"editable": true
},
"outputs": [],
"source": []
"outputs": [
{
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"text/plain": [
"<IPython.core.display.Image object>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"x = 1:1:100;\n",
"plot(x,arrayfun(@calculateRatio,x));"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"### 3. Plot $F_{n+1} / F_n$ ratio starting with $F_0 = 2$ and $F_1 = 1 - \\sqrt{5}$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"metadata": {
"deletable": true,
"editable": true
},
"source": [
"Initialize $F_0 = 2$ and $F_1 = F_1 = 1 - \\sqrt{5}$, then plot the $\\dfrac{F_{n+1}}{F_{n}}$ values for $ 1 \\leq n \\leq 100$. If we would represent $\\sqrt{5}$ exactly in our floating point arithmetic, then as $n \\to \\infty$, we expect $\\dfrac{F_{n+1}}{F_{n}} \\to \\dfrac{1 - \\sqrt{5}}{2}$, but for the very large values of $n$, this ratio unexpectedly converges to $\\dfrac{\\sqrt{5} + 1}{2}$."
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"function fn = fastFibonacciSecondVersion (n)\n",
" fn = 2;\n",
" if (n == 0)\n",
" return;\n",
" elseif(n == 1)\n",
" fn = (1-sqrt(5));\n",
" return;\n",
" else\n",
" multiplier=[1,1;1,0];\n",
" baseVector=[2;(1-sqrt(5))];\n",
" fn = ((multiplier**(n-1))*baseVector)(1,1);\n",
" return;\n",
" endif\n",
"endfunction"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"function fn = calculateRatioSecondVersion (n)\n",
" fn = 0;\n",
" fn = fastFibonacciSecondVersion(n+1)/fastFibonacciSecondVersion(n);\n",
" return;\n",
"endfunction"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false,
"deletable": true,
"editable": true
},
"outputs": [
{
"data": {
"image/png": 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QkkE7Qz9odV3v/mlv0M53PFbnj94ll+xyzvf391VVmdU+rW3bu7u7uq7btu267vPnz8no\nnaFpmlLKer3e/TVh0E5aLBb9HTS7rxi0c/Sv8fQ77aTjd/ccj9X5o3fhNzX0JbQ/xThGbwSDdo6D\ny28G7aTjITJo5xs9el4dBEAIbmoAIARBAiAEQQIgBEECIARBAiCE/wHrvcqQ4IYSQAAAAABJRU5E\nrkJggg==\n",
"text/plain": [
"<IPython.core.display.Image object>"
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"plot(x,arrayfun(@calculateRatioSecondVersion,x));"
]
},
{
"cell_type": "code",
"execution_count": null,
Expand All @@ -171,14 +318,22 @@
"language_info": {
"file_extension": ".m",
"help_links": [
{
"text": "GNU Octave",
"url": "https://www.gnu.org/software/octave/support.html"
},
{
"text": "Octave Kernel",
"url": "https://github.com/Calysto/octave_kernel"
},
{
"text": "MetaKernel Magics",
"url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
}
],
"mimetype": "text/x-octave",
"name": "octave",
"version": "0.16.1"
"version": "4.0.0"
}
},
"nbformat": 4,
Expand Down