diff --git a/blogContent/headerImages/ConstantTimeComplexity.png b/blogContent/headerImages/ConstantTimeComplexity.png new file mode 100644 index 0000000..f5ae1bf Binary files /dev/null and b/blogContent/headerImages/ConstantTimeComplexity.png differ diff --git a/blogContent/posts/programming/everything-fibonacci.md b/blogContent/posts/programming/everything-fibonacci.md index 3ecb4a9..a34f14d 100644 --- a/blogContent/posts/programming/everything-fibonacci.md +++ b/blogContent/posts/programming/everything-fibonacci.md @@ -3,14 +3,16 @@ know what the fibonacci sequence is and how to calculate it. For those who don't know: [Fibonacci](https://en.wikipedia.org/wiki/Fibonacci) is a sequence of numbers starting with 0,1 whose next number is the sum of the two previous numbers. After having multiple of my CS classes -give lectures and multiple homeworks on the Fibonacci sequence; I decided -that it would be a good idea to write a blog post going over -the 4 main ways of calculating the nth term of the Fibonacci sequence -and proving their time complexities both mathematically and empirically. +gave lectures and multiple homework on the Fibonacci sequence; I decided +that it would be a great idea to write a blog post going over +the 4 main ways of calculating the nth term of the Fibonacci sequence. +In addition to providing python code for calculating the nth perm of the sequence, a proof for their validity +and analysis of their time complexities both mathematically and empirically will +be examined. # Slow Recursive Definition -By the definition of the Fibonacci sequence, it is natural to write it as +By the definition of the Fibonacci sequence, it is the most natural to write it as a recursive definition. ```Python @@ -20,29 +22,38 @@ def fib(n): return fib(n-1) + fib(n-2) ``` -#### Time Complexity +##Time Complexity Observing that each call has two recursive calls we can place an upper bound on this -function as O(2^n). However, if we solve this recurrence we can compute the exact value +function as $O(2^n)$. However, if we solve this recurrence we can compute the exact value and place a tight bound for time complexity. We can write a recurrence for the number of times fib is called: -```angular2html -T(1) = 1 -T(n) = T(n-1) + T(n-2) -a^n = a^{n-1} + a^{n-2} -a^2 = a + 1 -a = \frac{1 + sqrt(5)}{2} -T(n) = \frac{1 + sqrt(5)}{2}^n + \frac{1 1 sqrt(5}{2}^n +$$ + F(0) = 0\\ + F(1) = 1\\ + F(n) = F(n-1) + F(n-2)\\ +$$ -O(1.618^n) -``` +Next we replace each instance of F(n) with $a^n$ since we want to solve for the roots since that +will allow us to put a tight asymptotic limit on the growth. + +$$ + a^n = a^{n-1} + a^{n-2}\\ + \frac{a^n}{a^{n-2}} = \frac{a^{n-1} + a^{n-2}}{a^{n-2}}\\ + a^2 = a + 1\\ + a = \frac{1 + sqrt(5)}{2}\\ +$$ + +From this calculation we can conclude that F(n) $\in \Theta 1.681^n$ -#### Measured Performance -Here is a graph of the actual performance that I observed for this algorithm. +## Measured Performance + +Here is a graph of the actual performance that I observed from this recursive definition of Fibonacci. + ![Recursive Definition](media/fibonacci/RecursiveDefinition.png) @@ -73,20 +84,84 @@ In this code example fibHelper is a method which accumulates the previous two te The fibIterative is a wrapper method which sets the two initial terms equal to 0 and 1 representing the fibonacci sequence. -Proof in latex that fibHelper +## Proof for Fib Helper +**Lemma:** For any n $\epsilon$ N if n $>$ 1 then + $fibHelper(n, a, b) = fibHelper(n - 1, a, b) + fibHelper(n - 2, a, b)$. + +**Proof via Induction** + +Base Case: n = 2: +$$ + LHS = fibHelper(2, a, b)\\ + = fibHelper(1, b, a + b) = a + b\\ + RHS = fibHelper(2 -1, a, b) + fibHelper(2-2, a, b)\\ + = a + b\\ +$$ + +Inductive Step: + +Assume proposition is true for all n and show n+1 follows. + +$$ + RHS=fibHelper(n+1;a,b)\\ + = fibHelper(n;b,a+b)\\ + =fibHelper(n-1;b,a+b) + fibHelper(n-2;b,a+b)\\ + =fibHelper(n;a,b) + fibHelper(n-1;a,b)\\ + =LHS\\ +$$ + +$\Box$ + +## Proof That fibIterative = Fib -proof in latex that fib iterative = fib +**Lemma:** For any n $\in$ N, $fib(n)$ = $fibIterative(n, 0, 1)$ + +**Proof via Strong Induction** +Base Case: n = 0: +$$ + fibIterative(0, 0, 0) = 0\\ + = fib(0) +$$ -#### Time Complexity +Base Case: n = 1: +$$ + fibIterative(1, 0, 0) = 1\\ + = fib(1) +$$ -proof in latex for time complexity +Inductive Step: -#### Measured Performance +Assume proposition is true for all n and show n+1 follows. + +$$ + fib(n+1) = fib(n) + fib(n-1)\\ + = fibHelper(n, 0, 1) + fibHelper(n+1, 0 ,1) \quad\text{I.H}\\ + = fibHelper(n+1, 0, 1) \quad\text{from result in previous proof}\\ +$$ + +$\Box$ + + +## Time Complexity + +Suppose that we wish to solve for time complexity in terms of the number of additions needed to be +computed. By observing the algorithm for fibHelper we can see that we perform one addition every time +which we have a recursive call. We can now form a recurrence for time complexity and solve for it. + +$$ + T(0) = 0\\ + T(1) = 0\\ + T(n) = 1 + T(n-1)\\ + T(n) = n-1\\ +$$ + +From this recurrence we can say that fibHelper $\in \Theta(n)$. + +## Measured Performance Notice how much faster this solution is compared to the original recursive solution for -Fibonacci. Also, I only measured going out to 500 because beyond that I hit the maximum -number of recursive calls for my installation of Python. +Fibonacci. ![Iterative Performance](media/fibonacci/Iterative.png) @@ -95,13 +170,23 @@ number of recursive calls for my installation of Python. # Matrix Solution We can actually get better than linear time for performance while calculating -the Fibonacci sequence recursively. - -![Inductive Proof of Fibonacci Matrix](media/fibonacci/FibonacciMatrix.png) +the Fibonacci sequence recursively using this fact: + +$$ +\begin{bmatrix} +1 & 1\\ +1 & 0 +\end{bmatrix}^n = +\begin{bmatrix} +F_{n+1} & F_n\\ +F_n & F{n-1} +\end{bmatrix}^n +$$ Without any other tricks, raising a matrix to a power n times would not get us better than linear performance. However, if we use the [Exponentiation by Squaring](https://en.wikipedia.org/wiki/Exponentiation_by_squaring) -method, we can expect to see logarithmic time. +method, we can expect to see logarithmic time. Since two spots in the matrix are always equal, +I represented the matrix as an array with only three elements. ```Python @@ -128,17 +213,112 @@ def fibPower(n): return power(l, n)[1] ``` -#### Time Complexity - -latex proof for 9lg(n) performance +## Time Complexity + +For this algorythem lets solve for the time complexity as the number of additions and multiplications. + +Since we are always multiplying two 2x2 matrices, that is constant time. + +$$ + T_{multiply} = 9 +$$ + +Solving for the time complexity of fib power is slightly more complicated. +$$ + T_{power}(1) = 0\\ + T_{power}(n) = T(\left\lfloor\dfrac{n}{2}\right\rfloor) + T_{multiply}\\ + = T(\left\lfloor\dfrac{n}{2}\right\rfloor) + 9\\ + = T(\left\lfloor\dfrac{n}{2*2}\right\rfloor) + 9 + 9\\ + = T(\left\lfloor\dfrac{n}{2*2*2}\right\rfloor) + 9+ 9 + 9\\ + T_{power}(n) = T(\left\lfloor\dfrac{n}{2^k}\right\rfloor) + 9k\\ +$$ + +let $k=k_0$ such that $\left\lfloor\dfrac{n}{2^{k_0}}\right\rfloor = 1$ + +$$ + \left\lfloor\dfrac{n}{2^{k_0}}\right\rfloor = 1 \rightarrow 1 \leq \frac{n}{2^{k_0}} < 2\\ + \rightarrow 2^{k_0} \leq n < 2^{k_0 +1}\\ + \rightarrow k_0 \leq lg(n) < k_0+1\\ + \rightarrow k_0 = \left\lfloor lg(n)\right\rfloor\\ + T_{power}(n) = T(1) + 9*\left\lfloor lg(n)\right\rfloor\\ + T_{power}(n) = 9*\left\lfloor\ lg(n)\right\rfloor\\ + T_{fibPower}(n) = T_{power}(n)\\ +$$ + + + + +## Inductive Proof for Matrix Method + +**Lemma:** For any n $\epsilon$ N if n $>$ 0 then + $$ + \begin{bmatrix} + 1 & 1\\ + 1 & 0 + \end{bmatrix}^n = + \begin{bmatrix} + F_{n+1} & F_n\\ + F_n & F{n-1} + \end{bmatrix}^n + $$ + +Let + +$$ +A= +\begin{bmatrix} +1 & 1\\ +1 & 0 +\end{bmatrix}^n +$$ + +**Base Case:** n = 1 +$$ +A^1= +\begin{bmatrix} +1 & 1\\ +1 & 0 +\end{bmatrix}^n = +\begin{bmatrix} +F_{2} & F_2\\ +F_2 & F_{0} +\end{bmatrix}^n +$$ + +**Inductive Step:** Assume proposition is true for n, show n+1 follows +$$ +A^{n+1}= +\begin{bmatrix} +1 & 1\\ +1 & 0 +\end{bmatrix} +\begin{bmatrix} +F_{n+1} & F_n\\ +F_n & F{n-1} +\end{bmatrix}^n\\ += \begin{bmatrix} +F_{n+1} + F_n & F_n + F_{n-1}\\ +F_{n+1} & F_{n} +\end{bmatrix}\\ += \begin{bmatrix} +F_{n+2} & F_{n+1}\\ +F_{n+1} & F_{n} +\end{bmatrix}\\ +$$ + +$\Box$ + + +## Measured Performance -#### Measured Performance ![FibPower Performance](media/fibonacci/FibPower.png) + As expected by our mathmatical calcuations, the algorthem appears to be running in logarithmic time. -#### Measured Performance With Large Numbers +## Measured Performance With Large Numbers + ![FibPower Performance](media/fibonacci/FibPowerBigPicture.png) When calculating the fibonacci term for extremely large numbers dispite having a polynomial @@ -153,6 +333,61 @@ the 30th term of Fibonacci. # Closed Form Definition -It is actually possible to calculate Fibonacci in constant time using a closed form definition. +It is actually possible to calculate Fibonacci in constant time using Binet's Formula. + +$$ + F_n = \frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}} +$$ + +```Python +def fibClosedFormula(n): + p = ((1+ math.sqrt(5))/2)**n + v = ((1-math.sqrt(5))/2)**n + return (p-v)/math.sqrt(5) +``` + +## Derivation of Formula + +Similar to when we were calculating for the time complexity, we want to start by finding the +two roots of the equation. + +$$ + a^n = a^{n-1} + a^{n-2}\\ + \frac{a^n}{a^{n-2}} = \frac{a^{n-1} + a^{n-2}}{a^{n-2}}\\ + a^2 = a + 1\\ + 0 = a^2 - a - 1\\ + a = \frac{1 \pm sqrt(5)}{2}\\ +$$ + +Since there are two roots to the equation, the solution of $F_n$ is going to be +a linear combination of the two roots. + + +$$ + F_n = c_1(\frac{1 + \sqrt{5}}{2})^n + c_2(\frac{1 - \sqrt{5}}{2})^n +$$ + +Fact: $F_1$ = 1 + +$$ + F_1 = 1\\ + = c_1(\frac{1 + \sqrt{5}}{2}) + c_2(\frac{1 - \sqrt{5}}{2})\\ + = \frac{c_1}{2} + \frac{c_2}{2} + \frac{c_1\sqrt{5}}{2} - \frac{c_2\sqrt{5}}{2}\\ +$$ + +Let $c_1 = \frac{1}{\sqrt{5}}$, +Let $c_2 = \frac{-1}{\sqrt{5}}$ + +$$ + F_n = \frac{1}{\sqrt(5)}((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n)\\ + = \frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}} +$$ + +## Time Complexity + + +Since we managed to find the closed form of the fibonacci sequence we can expect to see constant performance. + +## Measured Performance -latex proof of closed form definition \ No newline at end of file +![FibPower Performance](media/fibonacci/ConstantTimeComplexity.png) \ No newline at end of file diff --git a/blogContent/posts/programming/media/fibonacci/ConstantTimeComplexity.png b/blogContent/posts/programming/media/fibonacci/ConstantTimeComplexity.png new file mode 100644 index 0000000..2c781f4 Binary files /dev/null and b/blogContent/posts/programming/media/fibonacci/ConstantTimeComplexity.png differ diff --git a/blogContent/posts/programming/media/fibonacci/FibonacciMatrix.png b/blogContent/posts/programming/media/fibonacci/FibonacciMatrix.png deleted file mode 100644 index f0d6014..0000000 Binary files a/blogContent/posts/programming/media/fibonacci/FibonacciMatrix.png and /dev/null differ diff --git a/posts/singlePost.js b/posts/singlePost.js index 04dc292..97ef5c7 100644 --- a/posts/singlePost.js +++ b/posts/singlePost.js @@ -1,36 +1,4 @@ -const utils = require('../utils/utils.js'); - -const sql = require('../utils/sql'); - -const Remarkable = require('remarkable'); -const hljs = require('highlight.js'); - -const pandoc = require('../utils/markdownToHTML.js'); - - -const md = new Remarkable( -{ - html: true, - highlight: function (str, lang) - { - if (lang && hljs.getLanguage(lang)) - { - try - { - return hljs.highlight(lang, str).value; - } - catch (err) {} - } - - try - { - return hljs.highlightAuto(str).value; - } - catch (err) {} - - return ''; // use external default escaping - } -}); +const postGenerator = require('../utils/renderBlogPost.js'); module.exports= @@ -43,69 +11,7 @@ module.exports= */ renderPreview: function(post) { - return new Promise(function(resolve, reject) - { - //var html = "
"; - - var html = "
"; - - //image - if(!(post.picture_url === "n/a")) - { - html +="\"\""; - } - - html += "
"; - //title - html += "

" + post.name + "

"; - //date - html += "
" + - post.published.toDateString() + "
"; - html +="
"; - - html += "
"; - - try - { - sql.getCategory(post.category_id).then(function(category) - { - var pathName = "blogContent/posts/" + category[0].url + "/" - + post.url + ".md"; - var markDown = utils.getFileContents(pathName).toString(); - - markDown = markDown.split("(media/").join("(" + "../blogContent/posts/" - + category[0].url + "/media/"); - var htmlPost = md.render(markDown).split("

"); - - for(var i = 0; i < 3; i++) - { - html+= "

" + htmlPost[i]; - } - - html = html.split("").join(""); - - html += "

\n" + - "

\n" + - "
\n"; - - html += "


"; - - resolve(html) - }).catch(function(error) - { - reject(error); - }); - } - catch(ex) - { - reject(ex); - } - }); + return postGenerator.generateBlogPost(post, 3); }, /** @@ -117,56 +23,6 @@ module.exports= */ renderPost: function(post) { - return new Promise(function (resolve, reject) - { - var htmlHead = "
"; - //image - if(!(post.picture_url === "n/a")) - { - htmlHead +="\"\""; - } - - htmlHead += "
"; - //title - htmlHead += "

" + post.name + "

"; - //date - htmlHead += "
" + - post.published.toDateString() + "
"; - htmlHead +="
"; - - var html = "
"; - try - { - sql.getCategory(post.category_id).then(function(category) - { - const pathName = "blogContent/posts/" + category[0].url + "/" - + post.url + ".md"; - var markDown = utils.getFileContents(pathName).toString(); - markDown = markDown.split("(media/").join("(" + "../blogContent/posts/" - + category[0].url + "/media/"); - //html += md.render(markDown); - - pandoc.convertToHTML(markDown).then(function(result) - { - html +=result; - - html = html.split("").join(""); - html += "


"; - - resolve(htmlHead + html); - }).catch(function(error) - { - reject(error); - }) - - }); - } - catch(ex) - { - reject(ex); - } - }); + return postGenerator.generateBlogPost(post, -1); } }; \ No newline at end of file diff --git a/utils/markdownToHTML.js b/utils/markdownToHTML.js deleted file mode 100644 index f2c9567..0000000 --- a/utils/markdownToHTML.js +++ /dev/null @@ -1,26 +0,0 @@ -const pandoc = require('node-pandoc'); -// const args = '-t html5'; - -const args = '-S --base-header-level=1 --toc --toc-depth=6 -N --normalize -s --mathjax -t html5'; - -console.log(""); -module.exports= - { - convertToHTML: function(markdownContents) - { - return new Promise(function(resolve, reject) - { - // Set your callback function - callback = function (err, result) - { - if (err) - { - reject(err); - } - resolve(result); - }; - console.log(markdownContents); - pandoc(markdownContents, args, callback); - }); - }, - } diff --git a/utils/renderBlogPost.js b/utils/renderBlogPost.js new file mode 100644 index 0000000..45d3898 --- /dev/null +++ b/utils/renderBlogPost.js @@ -0,0 +1,167 @@ +const pandoc = require('node-pandoc'); + +const utils = require('../utils/utils.js'); + +const sql = require('../utils/sql'); + +const argsFull = '-S --base-header-level=1 --toc --toc-depth=3 -N --normalize -s --mathjax -t html5'; +const argsPreview = '-S --normalize -s --mathjax -t html5'; + + +module.exports= + { + + /** + * Renders the entire blog post based on the sql data pulled + * from the database. + * + * @param post sql data which has title, date, and header img location + * @param blocks number of blocks to display for a preview or -1 for + * all the blocks + * @returns {Promise} async call which renders the entire blog post. + */ + generateBlogPost: function(post, blocks) + { + return new Promise(function(resolve, reject) + { + Promise.all([module.exports.generateBlogPostHeader(post), + module.exports.generateBlogPostBody(post, blocks), + module.exports.generateBlogPostFooter()]).then(function(content) + { + resolve(content.join('')); + }).catch(function(error) + { + reject(error); + }) + }); + }, + + /** + * Renders the header of the blog post which contains the header image, and date + * published. + * + * @param post sql data + * @returns {string} + */ + generateBlogPostHeader: function(post) + { + var htmlHead = "
"; + //image + if(!(post.picture_url === "n/a")) + { + htmlHead +="\"\""; + } + + htmlHead += "
"; + //title + htmlHead += "

" + post.name + "

"; + //date + htmlHead += "
" + + post.published.toDateString() + "
"; + htmlHead +="
" + "
"; + + return htmlHead; + }, + + + /** + * Method which renders the body of the blog post. This is responsible for getting + * the contents of the markdown/latex file and rendering it into beautiful html. + * + * @param post + * @param blocks + * @returns {Promise} + */ + generateBlogPostBody: function(post, blocks) + { + return new Promise(function(resolve, reject) + { + sql.getCategory(post.category_id).then(function(category) + { + const pathName = "blogContent/posts/" + category[0].url + "/" + + post.url + ".md"; + var markDown = utils.getFileContents(pathName).toString(); + markDown = markDown.split("(media/").join("(" + "../blogContent/posts/" + + category[0].url + "/media/"); + + module.exports.convertToHTML(markDown, blocks).then(function(result) + { + + result = result.split("
").join("
"); + + if(blocks == -1) + resolve(result); + + const htmlBlocks = result.split("

"); + var html = ""; + for(var i = 0; i < blocks; i++) + { + html += "

" + htmlBlocks[i]; + } + + html += "

\n" + + "

\n" + + "
\n"; + + resolve(html); + + }).catch(function(error) + { + reject(error); + }) + + }); + }) + }, + + + /** Method to return the footer of the html blog post. + * + * @returns {string} + */ + generateBlogPostFooter: function() + { + return "


"; + }, + + /** + * Converts markdown into html. + * + * @param markdownContents + * @param type + * @returns {Promise} + */ + convertToHTML: function(markdownContents, type) + { + return new Promise(function(resolve, reject) + { + // Set your callback function + callback = function (err, html) + { + if (err) + { + reject(err); + } + + + html = html.split("").join(""); + + + resolve(html); + }; + if(type == -1) + { + pandoc(markdownContents, argsFull, callback); + } + else + { + pandoc(markdownContents, argsPreview, callback); + } + }); + }, + }