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Jet Hughes
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@ -1,8 +1,14 @@
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---
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title: "04-requirements"
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sr-due: 2022-04-06
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sr-interval: 15
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sr-ease: 232
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tags:
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- info201
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- lecture
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sr-due: 2022-04-10
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sr-interval: 3
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sr-ease: 250
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---
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[requirements](notes/requirements.md)
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@ -1,17 +1,24 @@
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---
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title: "07-mergesort-1"
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sr-due: 2022-04-26
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sr-interval: 23
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sr-ease: 250
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tags:
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- cosc201
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- lecture
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---
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# Divide and conquer
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[mergeosrt](notes/mergeosrt.md)
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#unfinished
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# 1 Divide and conquer
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1. pre ⇒ break apartinto two or more smaller problems whose size add up to at most n
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2. Rec ⇒ solve those problems recursively
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3. post ⇒ combine solutions into a solution of the original problem
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## 1 quicksort
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## 1.1 quicksort
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pre ⇒ select pivot and split the array
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@ -27,78 +34,3 @@ works best of primitive types as they can be stored in the fastest memory locati
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- those advantages are limited when sorting objects of reference type
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- i that case each element of the array is just a reference to where the object really is
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- so there are no local access advantages
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# Mergesort
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a variant of a divide and conquer sorting array
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pre ⇒ split array into two pieces of nearly equal size,
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rec ⇒ sort the pieces,
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post ⇒ merge the pieces
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## 2 Merge
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take the two lowest values
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place the lowest of the two in the next place in the sorted array
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## 3 Implementation
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- given: a and b are sorted arrays. m in an array whose sixe is the sum fo their sizes
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- desired outcome: the elements of a and b have been copoed into m in sorted order
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- maiain indices, ai, bi, and mi of the active location in a b and m
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- if both ai and bi represent actual indices of a and b, find the one which points to the lesser value (break ties in favour of a) copy that vale into m at mi and increment mi and whichever of ai or bi was used for the copy.
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- once one of ai and bi is out of range, copy the rest of the other array into the remainder of m
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```java
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public static int[] merge (int[] a int[] b){
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int[] m = new int[a.length + b.length]
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int ai = 0, bi = 0, mi = 0;
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while(ai < a.length && bi < b.length) {
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if(a[ai] <= b[bi]) m[mi++] = a[ai++];
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else m[mi++] = b[bi++]
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}
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while (ai < a.length) m[mi++] = a[ai++];
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while (bi < b.length) m[mi++] = a[bi++];
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return m;
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}
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```
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```java
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public static void mergeSort(int[] a){
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mergeSort(a, 0, a.length);
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}
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public static void mergeSort(int[] a, int lo, int hi){
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if(hi - lo <= 1) return;
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int mid = (hi + lo)/2;
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mergeSort(a, lo, mid);
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mergeSort(a, mid, hi);
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merge(a, lo, mid, hi);
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}
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public static void merge(int[] a, int lo, int mid, int hi){
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int[] t = new int [hi-lo];
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//adjust code from 'merge' here so that the part of a from lo to mid, and the part of a from mid to hi are merged into t
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System.arraycopy(t, 0, a, lo, hi-lo) //copy back into a
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}
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```
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## 4 Complexity
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- n is the length of a plus the length of b
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- no obvious counter controlled loop
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- key ⇒ in each of the three loops mi in incremented by one.
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- ∴ the total number of loop bodies executed is always n
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- since each loop has a constant amount of work
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- ∴ so total cost is **ϴ(n)**
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@ -1,110 +1,11 @@
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---
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title: "08-mergesort-2"
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sr-due: 2022-04-06
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sr-interval: 8
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sr-ease: 270
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tags:
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- cosc201
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- lecture
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---
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recall definition of merge sort
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- pre ⇒ split
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- rec ⇒ sort pieces
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- post ⇒ merge
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## 1 Complexity
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no counters
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pre and post pahses are constant and ϴ(n)
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so M(n) = ϴ(n) + 2 * M(n/2)
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does this even help. what if n is odd
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pretend ϴ(n) is $C \times n$
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$$
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\begin{align*}
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M(n) &= C \times n+2 \times M(n/2) \\
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&= C \times n+2 \times (C \times (n/2) + 2 \times M(n/4))\\
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&= C \times (2n) + 4 \times M(n/4) \\
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&= C \times (2n) + 4 \times (C \times (n/4)) + 2 \times M(n/8))\\
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&= C \times (3n) + 8 \times M(n/8)\\ \\
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&= C \times (kn) + 2^k \times M(n/2^k)
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\end{align*}
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$$
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ends when we find base case
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when we get to $n/2^k = 1$
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we could split earlier.
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the work done base case is (bounded by) some constatn D
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so if $k$ is large enough that $n/2^k$ is a base case, we get
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$$
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M(n) = C \times (kn) + 2^k \times D
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$$
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how big is $k$
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$k <=lg(n)$
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so:
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$$
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M(n) ≤ C \times (n lg(n)) + D(n) = ϴ(n lg(n))
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$$
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which is true
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> In a divide and consiwer algo wher pre and pst processign work are Ο(n) and the division is into parts of size at least n for some contatn c > 0 tge total time complexity is Ο(n lg n) and generally ϴ(n log n)
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## 2 Variations of mergesort
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unite and conquer
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5 | 8 | 2 | 3 | 4 | 1 | 7 | 6
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5 8 | 2 3 | 1 4 | 6 7
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2 3 5 8 | 1 4 6 7
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1 2 3 4 5 6 7 8
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```java
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public static void mergeSort(int[] a) {
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int blockSize = 1;
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while(blockSize < a.length) {
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int lo = 0;
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while (lo + blockSize < a.length) {
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int hi = lo + 2*blockSize;
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if (hi > a.length) hi = a.length;
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merge(a, lo, lo + blockSize, hi);
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lo = hi;
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}
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blockSize *=2;
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}
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}
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```
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outer loop is executed lg n times, where n is the length of a
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inner loop proceeds until we find a block that "runs out of elements"
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inner loop is having 2 x blocksize added each time, to runs most n/2 x blocksize
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inside inner is call to merge which is ϴ(blocksize)
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### 2.1 complexity from bottom up
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- $n$ is the numbe of elemetns in a
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- outer loop is executed
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![[Pasted image 20220329114859.png#invert]]
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### 2.2 improvments
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some arrays have sections that are already sorted
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you canm
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### 2.3 timsort
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used by python java rust etc
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[mergeosrt](notes/mergeosrt.md)
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@ -6,7 +6,7 @@ tags:
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---
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links: [[notes/cosc-201]]
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- [[notes/07-mergesort-1]]
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- [[notes/08-mergesort-2]]
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- [[notes/09-stacks-and-queues]]
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- [[notes/10-heaps-and-heapsort]]
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- [07-mergesort-1](notes/07-mergesort-1.md)
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- [08-mergesort-2](notes/08-mergesort-2.md)
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- [09-stacks-and-queues](notes/09-stacks-and-queues.md)
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- [10-heaps-and-heapsort](notes/10-heaps-and-heapsort.md)
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156
content/notes/mergeosrt.md
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156
content/notes/mergeosrt.md
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---
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title: "mergesort"
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tags:
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- cosc201
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- algorithm
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---
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Mergesort is a [divide-and-conquer](notes/divide-and-conquer.md) algorithm. It works by recursively splitting the array in half then merging the two (sorted) halfs together . It has three main steps. These are:
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- pre-processing: split the array into two pieces
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- recurive step: sort each of the pieces
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- post-processing: merge the two pieces
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e.g.,
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7 5 3 9 1 8 2 5 4 0
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7 5 3 9 1 | 8 2 5 4 0
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7 5 | 3 9 1 | 8 2 5 | 4 0
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5 7 | 1 9 3 | 2 8 5 | 0 4
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1 3 5 7 9 | 0 2 4 5 8
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0 1 2 3 4 4 6 7 8 9
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# 1 Implementation
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## 1.1 Merge
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Given: a and b are sorted arrays. m is an array whose size is the sum of their sizes
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Get: a sorted array containing the elements of a and b
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Keep track of indices, ai, bi, and mi of the active location of a, b, and m.
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Find which of ai or bi is lesser (break ties is favour of a), and copy that value into m at mi, and increment mi and whichever of ai or bi was used.
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Once ai or bi is out of range, copy the rest of the other array into the remainder of m
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```java
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public static int[] merge (int[] a int[] b){
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int[] m = new int[a.length + b.length]
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int ai = 0, bi = 0, mi = 0;
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while(ai < a.length && bi < b.length) {
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if(a[ai] <= b[bi]) m[mi++] = a[ai++];
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else m[mi++] = b[bi++]
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}
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while (ai < a.length) m[mi++] = a[ai++];
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while (bi < b.length) m[mi++] = a[bi++];
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return m;
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}
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```
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### 1.1.1 Complexity of merge
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$\theta(n)$
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$n$ is the sum of the lengths of $a$ and $b$
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Each time we loop the parameter $mi$ must increase by one, and this parameter runs from $0$ to $n-1$
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Since the total number of loop bodies executed is $n$ and each loop does a constant amount of work, the total time complexity is $\theta(n)$
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## 1.2 Sort
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```java
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public static void mergeSort(int[] a) {
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int blockSize = 1;
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while (blockSize < a.length) {
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int lo = 0;
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while (lo + blockSize < a.length) {
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int hi = lo + 2*blockSize;
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if (hi > a.length) hi = a.length;
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merge(a, lo, lo + blockSize, hi);
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lo = hi;
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}
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blockSize *= 2;
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}
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}
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```
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### 1.2.1 Complexity of Sort
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$\theta(n\ lg\ n)$
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> In a divide and conquer algorithm where pre and pst processing work are Ο(n) and the division is into parts of size at least n for some contatn c > 0 the total time complexity is Ο(n lg n) and generally ϴ(n log n)
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#### 1.2.1.1 Top down
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We can split into the three steps to analyse this.
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- Pre has constant time i.e., $\theta(1)$
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- Post has linear time i.e., $\theta(n)$
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Therefore: $M(n) = \theta(n) + 2 \times M(n/2))$
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Substitue $\theta(n)$ with $C \times n$ then
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$$
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\begin{align*}
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M(n) &= C \times n+2 \times M(n/2) \\
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&= C \times n+2 \times (C \times (n/2) + 2 \times M(n/4))\\
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&= C \times (2n) + 4 \times M(n/4) \\
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&= C \times (2n) + 4 \times (C \times (n/4)) + 2 \times M(n/8))\\
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&= C \times (3n) + 8 \times M(n/8)\\ \\
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&= C \times (kn) + 2^k \times M(n/2^k)
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\end{align*}
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$$
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This stops (at least) when we reach the base case of $n/2^k=1$ . We could stop earlier and
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If we do a constant amount of work $D$ when we reach the base case we get:
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$$
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M(n) = C \times (kn) + 2^k \times D
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$$
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where $k\leq lg(n)$ so:
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$$
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M(n) ≤ C \times (n lg(n)) + D(n) = ϴ(n\ lg(n))
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$$
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#### 1.2.1.2 Bottom Up
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- Let n be the number of elements in the array, $a$.
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- The outer while loop (controlled by blockSize, or $b$) is executed $lg(n)$ times since its upper bound is n and b is doubled each time.
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- In the inner loop, the update on lo is to add $2b$ so it is executed $n/(2b)$ times.
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- The inner loop merges two arrays of size b, so each instance does $\theta(b)$ work.
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- That gives an upper bound on the work done in one instance of the outer loop of the form:
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$$
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(n/(2b)) \times (A \times b) = (A/2) \times n
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$$
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and a matching lower bound.
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- Thus, the work done in one instance of the outer loop is $\theta(n)$
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- And so, the total complexity is $\theta(n\ lg\ n)$.
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The bottom-up version does exactly the same thing as the top-down version, just in an apparently different order, so this analysis applies to the top-down version as well.
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# 2 Variations of Mergesort
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[[unite and conquer]] #unfinished
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56
content/notes/unite-and-conquer.md
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56
content/notes/unite-and-conquer.md
Normal file
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---
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title: "unite-and-conquer"
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tags:
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- cosc201
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---
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unite and conquer
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5 | 8 | 2 | 3 | 4 | 1 | 7 | 6
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5 8 | 2 3 | 1 4 | 6 7
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2 3 5 8 | 1 4 6 7
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1 2 3 4 5 6 7 8
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```java
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public static void mergeSort(int[] a) {
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int blockSize = 1;
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while(blockSize < a.length) {
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int lo = 0;
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while (lo + blockSize < a.length) {
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int hi = lo + 2*blockSize;
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if (hi > a.length) hi = a.length;
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merge(a, lo, lo + blockSize, hi);
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lo = hi;
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}
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blockSize *=2;
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}
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}
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```
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outer loop is executed lg n times, where n is the length of a
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inner loop proceeds until we find a block that "runs out of elements"
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inner loop is having 2 x blocksize added each time, to runs most n/2 x blocksize
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inside inner is call to merge which is ϴ(blocksize)
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### 0.1.1 complexity from bottom up
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- $n$ is the numbe of elemetns in a
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- outer loop is executed
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![[Pasted image 20220329114859.png#invert]]
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### 0.1.2 improvments
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some arrays have sections that are already sorted
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you canm
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### 0.1.3 timsort
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used by python java rust etc
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