From 5a7fd4c8fee396c37e574ef10cc4a39a2f2c24b6 Mon Sep 17 00:00:00 2001
From: eggy <danielchen04@hotmail.ca>
Date: Fri, 3 Nov 2023 12:40:16 -0400
Subject: [PATCH] ece250: add counting and quick sort

---
 docs/2a/ece250.md | 71 +++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 71 insertions(+)

diff --git a/docs/2a/ece250.md b/docs/2a/ece250.md
index 739245e..cdc0b5a 100644
--- a/docs/2a/ece250.md
+++ b/docs/2a/ece250.md
@@ -78,3 +78,74 @@ fn insert(A: Vec, &n: usize, key: i32) {
 	A[i] = k;
 }
 ```
+
+## Sorting algorithms
+
+### Quicksort
+
+Quicksort operates by selecting a **pivot point** that ensures that everything to the left of the pivot is less than anything to the right of the pivot, which is what partitioning does.
+
+```rust
+fn partition(A: Vec, left_bound: usize, right_bound: usize) {
+	let i = left_bound;
+	let j = right_bound;
+	
+	while true {
+		while A[j] <= A[right_bound] { j -= 1; }
+		while A[i] >= A[left_bound] { i += 1; }
+		
+		if i < j { A.swap(i, j); }
+		else { return j }  // new bound!
+	}
+}
+```
+
+Sorting calls partitioning with smaller and smaller bounds until the collection is sorted.
+
+```rust
+fn sort(a: Vec, left: usize, right: usize) {
+	if left < right {
+		let pivot = partition(A, left, right);
+		sort(A, left, pivot);
+		sort(A, pivot+1, right);
+	}
+}
+```
+
+- In the best case, if partitioning is even, the time complexity is $T(n)=T(n/2)+\Theta(n)=\Theta(n\log n)$.
+- In the worst case, if one side only has one element, which occurs if the list is sorted, the time complexity is $\Theta(n^2)$.
+
+### Counting sort
+
+If items are or are linked to a number from $1..n$ (duplicates are allowed), counting sort counts the number of each number, then moves things to the correct position. Where $k$ is the size of the counter array, the time complexity is $O(n+k)$.
+
+First, construct a count prefix sum array:
+
+```rust
+fn count(A: Vec, K: usize) {
+	let counter = vec![0; K];
+	
+	for i in A {
+		counter[i] += 1;
+	}
+	
+	for (index, val) in counter.iter_mut().enumerate() {
+		counter[index + 1] += val;   // ignore bounds for cleanliness please :)
+	}
+	return counter
+}
+```
+
+Next, the prefix sum represents the correct position for each item.
+
+```rust
+fn sort(A: Vec) {
+	let counter = count(A, 100);
+	let sorted = vec![0; A.len()];
+	
+	for i in n..0 {
+		sorted[counter[A[i]]] = A[i];
+		counter[A[i]] -= 1;
+	}
+}
+```