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@ -281,6 +281,8 @@ schedule:
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- date: 09/26/22
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topic: Recursive Analysis; Sorts
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dow: Mon
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section_b:
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slides: slide/11b-RecursionAnalysis.html
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- date: 09/28/22
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topic: Recursive Analysis (contd...)
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dow: Wed
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---
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template: templates/cse250_2022_slides.erb
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title: Inductive Proofs, Divide and Conquer
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date: Sept 21, 2022
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textbook: Ch. 15
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---
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<section>
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<h4 class="slide_title">Fibonacci</h4>
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<p>What's the complexity? (in terms of <tt>n</tt>)</p>
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<pre><code>
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def fibb(n: Int): Long =
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if(n < 2){ 1 }
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else { fibb(n-1) + fibb(n-2) }
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</code></pre>
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</section>
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<section>
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<h4 class="slide_title">Fibonacci</h4>
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$$T(n) = \begin{cases}
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\Theta(1) & \textbf{if } n < 2\\
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T(n-1) + T(n-2) + \Theta(1) & \textbf{otherwise}
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\end{cases}$$
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<p>Solve for $T(n)$</p>
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</section>
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<section>
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<h4 class="slide_title">Divide and Conquer</h4>
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<p>Remember the Towers of Hanoi...</p>
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<ol>
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<li class="fragment">You can move $n$ blocks, if you know how to move $n-1$ blocks</li>
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<li class="fragment">You can move $n-1$ blocks, if you know how to move $n-2$ blocks</li>
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<li class="fragment">You can move $n-2$ blocks, if you know how to move $n-3$ blocks</li>
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<li class="fragment">You can move $n-3$ blocks, if you know how to move $n-4$ blocks</li>
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</ol>
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<p class="fragment">...</p>
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<ul>
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<li class="fragment">You can always move $1$ block</li>
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</ul>
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</section>
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<section>
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<h4 class="slide_title">Divide and Conquer</h4>
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<p>To solve the problem at $n$:</p>
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<ul>
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<li class="fragment"><b>Divide</b> the problem into a problem at $n-1$</li>
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<li class="fragment"><b>Conquer</b> the $n-1$ problems</li>
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<li class="fragment"><b>Combine</b> the solutions from $n-1$</li>
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</ul>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<p><b>Input:</b> An array with elements in unknown order.</p>
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<p><b>Output:</b> An array with elements in sorted order.</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<p><b>Observation:</b> Merging two sorted arrays can be done in $O(1)$.</p>
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<p><b>Idea:</b> Split the input in half, sort each half, and merge.</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<svg data-src="graphics/11b/merge_sort_onestep.svg"/>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<svg data-src="graphics/11b/merge_sort_allsteps.svg"/>
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</section>
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<section>
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<h4 class="slide_title">Divide and Conquer</h4>
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<p>To solve the problem at $n$:</p>
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<ul>
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<li class="fragment"><b>Divide</b> the problem into two problems of size $\frac{n}{2}$</li>
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<li class="fragment"><b>Conquer</b> the $\frac{n}{2}$ problems</li>
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<li class="fragment"><b>Combine</b> the solutions from $\frac{n}{2}$</li>
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</ul>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<ol>
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<li>if the sequence has $1$ or $0$ values: Sorted!</li>
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<li>if the sequence has $n > 1$ values:
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<ol>
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<li>Sort values $1$ to $\left\lfloor\frac{n}{2}\right\rfloor-1$</li>
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<li>Sort values $\left\lfloor\frac{n}{2}\right\rfloor$ to $n$</li>
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<li>Merge sorted halves</li>
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</ol>
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</li>
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</ol>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<svg data-src="graphics/11b/merge_example.svg"/>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<pre><code class="scala">
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def merge[A: Ordering](left: Seq[A], right: Seq[A]): Seq[A] =
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{
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val output = ArrayBuffer[A]()
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// Buffered iterators let us "peek" at the head
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val leftItems = left.iterator.buffered
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val rightItems = right.iterator.buffered
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while(leftItems.hasNext || rightItems.hasNext)
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{
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if(!left.hasNext) { output.append(left.next) }
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else if(!right.hasNext){ output.append(right.next) }
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else if(Ordering[A].lt( left.head, right.head )){
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output.append(left.next)
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} else {
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output.append(right.next)
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}
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}
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output.toSeq
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}
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</code></pre>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<p>Each time though loop advances either left or right.</p>
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<p><b>Total Runtime: </b> $\Theta(|\texttt{left}| + |\texttt{right}|)$</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<pre><code class="scala">
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def sort[A: Ordering](data: Seq[A]): Seq[A] =
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{
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if(data.length <= 1) { return data }
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else {
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val (left, right) = data.splitAt(data.length / 2)
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return merge(
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sort(left),
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sort(right)
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)
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}
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}
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</code></pre>
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</section>
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<section>
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<h4 class="slide_title">Divide and Conquer</h4>
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If we solve a problem of size $n$ by:
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<ul>
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<li>Dividing into $a$ sub-problems</li>
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<li>...where each problem is of size $\frac{n}{b}$ (usually $b=a$)</li>
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<li>...and stop recurring at $n \leq c$</li>
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<li>...and the cost of dividing is $D(n)</li>
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<li>...and the cost of combining is $C(n)</li>
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</ul>
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The total cost will be:
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$$T(n) = \begin{cases}
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\Theta(1) & \textbf{if } $n \leq c \\
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a\cdot T(\frac{n}{b}) + D(n) + C(n) & \textbf{otherwise}
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\end{cases}$$
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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<dl>
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<dt>Divide: Split the sequence in half</dt>
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<dd>$D(n) = \Theta(n)$ (can do in $\Theta(1)$)</dd>
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<dt>Conquer: Sort left and right halves</dt>
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<dd>$a = 2$, $b = 2$, $c = 1$</dd>
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<dt>Combine: Merge halves together</dt>
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<dd>$C(n) = \Theta(n)$</dd>
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</dl>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort</h4>
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$$T(n) = \begin{cases}
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\Theta(1) & \textbf{if } $n \leq 1$ \\
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2\cdot T(\frac{n}{2}) + \Theta(1) + \Theta(n) & \textbf{otherwise}
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\end{cases}$$
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<p class="fragment">How can we find a closed-form hypothesis?</p>
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</section>
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<section>
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<p><b>Idea: </b> Draw out the cost of each level of recursion.</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Recursion Tree</h4>
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$$T(n) = \begin{cases}
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\Theta(1) & \textbf{if } $n \leq 1$ \\
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2\cdot T(\frac{n}{2}) + \Theta(1) + \Theta(n) & \textbf{otherwise}
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\end{cases}$$
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<svg data-src="graphics/11b/recursion_tree.svg"/>
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<p>Each node of the tree shows $D(n) + C(n)$ </p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Recursion Tree</h4>
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<svg data-src="graphics/11b/recursion_tree.svg"/>
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<p>At level $i$ there are $2^i$ tasks, each with runtime $\Theta(\frac{n}{2^i})$, and there are $\log(n)$ levels.</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Recursion Tree</h4>
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<p>At level $i$ there are <span class="fragment highlight-current-blue" data-fragment-index="2">$2^i$</span> tasks, each with <span class="fragment highlight-current-blue" data-fragment-index="3">runtime $\Theta(\frac{n}{2^i})$</span>, and there are <span class="fragment highlight-current-blue" data-fragment-index="1">$\log(n)$</span> levels.</p>
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<p>
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<span class="fragment highlight-current-blue" data-fragment-index="1">$\sum_{i=0}^{\log(n)}$</span>
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<span class="fragment highlight-current-blue" data-fragment-index="2">$\sum_{j=1}^{2^i}$</span>
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<span class="fragment highlight-current-blue" data-fragment-index="3">$\Theta(\frac{n}{2^i})$</span>
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</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Recursion Tree</h4>
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<p>$$\sum_{i=0}^{\log(n)} \sum_{j=1}^{2^i} \Theta(\frac{n}{2^i})$$</p>
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<p class="fragment">$$\sum_{i=0}^{\log(n)} (2^i+1-1)\Theta(\frac{n}{2^i})$$</p>
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<p class="fragment">$$\sum_{i=0}^{\log(n)} 2^i\Theta(\frac{n}{2^i})$$</p>
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<p class="fragment">$$\sum_{i=0}^{\log(n)} \Theta(n)$$</p>
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<p class="fragment">$$(\log(n) - 0 + 1) \Theta(n)$$</p>
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<p class="fragment">$$\Theta(n\log(n)) + \Theta(n)$$</p>
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<p class="fragment">$$\Theta(n\log(n))$$</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Proof By Induction</h4>
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<p>Now use induction to prove that there is a $c, n_0$ such that $T(n) \leq c \cdot n\log(n)$ for any $n > n_0$</p>
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$$T(n) = \begin{cases}
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c_0 & \textbf{if } $n \leq 1$ \\
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2\cdot T(\frac{n}{2}) + c_1 + c_2\cdot n & \textbf{otherwise}
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\end{cases}$$
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Proof By Induction</h4>
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<p><b>Base Case:</b> $T(1) \leq c \cdot 1$</p>
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<p class="fragment">$$c_0 \leq c$$</p>
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<p class="fragment">True for any $c > c_0$</p>
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</section>
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<section>
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<h4 class="slide_title">Merge Sort: Proof By Induction</h4>
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<p><b>Assume:</b> $T(\frac{n}{2}) \leq c \cdot \frac{n}{2} \log\left(\frac{n}{2}\right)$</p>
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<p><b>Show:</b> $T(n) \leq c \cdot n \log\left(n\right)$</p>
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<p class="fragment">$$2\cdot T(\frac{n}{2}) + c_1 + c_2\cdot n \leq c\cdot n \log(n)$$</p>
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<p class="fragment">By the assumption and transitivity, showing the following inequality suffices:<br>
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$$2\cdot c \cdot \frac{n}{2} \log\left(\frac{n}{2}\right) + c_1 + c_2\cdot n \leq c\cdot n \log(n)$$</p>
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<p class="fragment">$$c \cdot n \log(n) - c n \log(2) + c_1 + c_2\cdot n \leq c\cdot n \log(n)$$</p>
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<p class="fragment">$$c_1 + c_2\cdot n \leq c\cdot n \log(n) - c \cdot n \log(n) + c n \log(2)$$</p>
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<p class="fragment">$$c_1 + c_2\cdot n \leq c n \log(2)$$</p>
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<p class="fragment">$$\frac{c_1}{n \log(2)} + \frac{c_2}{\log(2)} \leq c$$</p>
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<p class="fragment">True for any $n_0 \geq \frac{\log(2)}{c_1}$ and $c > \frac{c_2}{\log(2)}+1$</p>
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</section>
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<section>
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<h4 class="slide_title">Next time...</h4>
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<ul>
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<b>Quick Sort</b>
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<b>Average Runtime</b>
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</ul>
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</section>
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After Width: | Height: | Size: 13 KiB |
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Reference in New Issue