##### Lecture 0 • 2017/01/05
• Office Hours Tues/Thu 1-2pm
• 40% for 5 assignments, 15% for midterm, 45% for final exam
• Midterm March 9 (one crib sheet permitted), during class time
• End of class April 11
• Final exam TBD
##### Lecture 1 • 2017/01/10
• *** A significant portion of the lecture overlaps with comp 250, so I did not add much about it here ***
• f(n) is O(g(n)) iff there exists a point n0 beyond which f(n) is less than some fixed constant times g(n) → for all n ≥ n0, f(n) ≤ c * g(n) (for some c > 0)
• Let T1(n) = O(f(n)) & T2(n) = O(f(n))
• T1(n) + T2(n) = O(f(n))
• T1(n) / T2(n) is not necessarily O(1); big O is not necessarily the tightest upper bound.
T1(n) = 3n & T2(n) = 2n is an example.
• Heap – binary tree such that
• For any node n other than the root, key(n) ≥ key(parent(n))
• Let h be the height of the heap
• First h – 1 levels are full
• At depth h, leaves are packed on the left side of the tree
##### Lecture 2 • 2017/01/12
• Table S with n records of x
• X is key, key[x] → satellite data
• Insert(S, x): S ← S ∪ {x}
• Delete(S, x): S ← S \ {x}
• Search (S, k)
• Direct Address Table – each slot/position corresponds to key in U
• If x has key k, T[k] points to x
• If T[k] is empty, represented by NIL
• All operations O(1), but if #keys < #slots, lots of wasted space
• Hash Table – reduce storage, resolve conflicts by chaining, ave O(1) search time, not worse case
• Hashing with Chaining
• Insertion – O(1) – insert at beginning
• Deletion – search time + O(1) if we use doubly linked list
• Search
• Worst case – O(n)
• Time to computer has function + time to search the list
• Assuming time to compute is O(1)
• When all keys go on same slot
• Average – depends how keys are distributed
• Load factor α = n/m n = #keys, m = #slots
• Theorem – expected time of search is Θ(1 + α)
• O(1) if α < 1, O(n) if α is O(n)
• Proof – unsuccessful vs successful
• Successful search
• 1/m probability of collision – after finding x have been inserted in hash table before x (ie we insert at head)
• 1 + α/2 + α/(2n)
• Hash functions
• A good hash function should uniformly distribute the keys into slots, and should not be affected by patterns in keys
• Division method – h(k) = kmod d
• D must be d chosen carefully, usually 2r where r is a prime not too close to a power of 2 or 10
• Easy to implement, but division is slow
• Multiplication method – h(k) = (A kmod2w) >> (w – r)
• Extracted bits are in the middle of the binary key
• No chaining; if slot is not empty, try another has function
• Collisions exist; resolved by adding another slot
• We can delete elements in open address tables
• Cannot store more records than total number of slots in table
• Deletion is difficult
• Goal is uniform hashing – each key is equally likely to have any permutation as its probe sequence
• Theorems
• Expected # of probes in unsuccessful search is at most 1/(1 - α)
• Expected # of probes in successful search is at most 1/α * log(1/(1 - α))
• Probing
• Linear – h(k, i) = (h’(k) + i)mod m
• If slot is full, check next slot; tendency to create clusters
• Quadratic probing – h(k, i) = (h’(k) + c1i + c2i2)mod m
• Must ensure we have full permutation of <0, …, m – 1>
• Secondary clustering – 2 distinct keys have same h’ value if they have same probe sequence
• Double hashing – h(k, i) = (h1(k) + i * h2(k)) mod m
• h2(k) should be “relatively” prime to m to guarantee full permutation
##### Lecture 3 • 2017/01/17
• Max heap – largest element stored at root; all children are smaller
• Min heap – smallest element stored at root; all children are bigger
• Heaps as array – root = A, left[i] = A[2i], right[i] = A[2i + 1], parent[i] = A[i/2]
• Height - # of edges on longest simple path from node to leaf
• Height of heap = height of root = Θ(log n)
• Basic operations are O(log n)
• Heap Pseudocode (Comp 250)
• Maintaining heap property
• Fix offending node by exchanging value at node with larger of the values at its children
• Recursively fix until there are no more offenses
• Running time of buildMaxHeap is O(n)
• maxHeapify = O(log n); heap height = log n;
$$O \left( n \sum_{h=0}^{\lfloor log n \rfloor}\dfrac{h}{2^h} \right)$$
• HeapSort is O(n logn)
##### Lecture 4 • 2017/01/19
• BST search, insert, delete are Θ(h); h = height of BST
• Height(x) = 1 + max(height(x.left), height(x.right))
• A good BST is balanced, height = Θ(log n)
• A bad BST is unbalanced, height = Θ(n)
• AVL – self balancing BST – Adelson-Velsky & Landis
• Height of one child is at most one greater than the other
• Each node contains data to indicate which child is bigger, or if they have the same height
• Rotations are used to maintain balanced properties
• Rotations
• Remove zigzags
• If a.left = b and b.right = c, rotate b leftwards
• Result: a.left = c and c.left = b
• Reattach children properly
• If a.left = b, b.left = c, b.right = d, a.right = e, rotate b rightwards
• Result: b.right = a, a.right = e, b.left = c
• Reattach middle child (d) to right child or local root → a.left = d
##### Lecture 5 • 2017/01/24
• Recursive equation for best case running time of function heapify on heap size of n? Ω(1)
• Red Black Trees
• Always balanced height is O(logn) worst case operations are O(logn)
• +1 bit per node for attribute color: red or black
• Empty trees are black and will be referenced as nil
• Properties
• Every node is red or black
• The root and every leaf is black
• If a node is red, its children are black
• For each node, all paths from the node to descendant leaves contain same number of black nodes (same black height)
• Let
• h(x) = # of edges in longest path to leaf
• bh(x) = # of black nodes from x to leaf, not counting x and counting the leaf
• Black height of RBT = bh(root)
• Note
• h(x)/2 ≤ bh(x) ≤ h(x) ≤ 2bh(x)
• A subtree rooted at any node x has ≥ 2bh(x) – 1 internal nodes
• A RBT with n internal nodes has height ≤ 2lg(n+1) (proof by previous point)
• Pseudocode
/*
* Red Black Tree Implementation
*/
public class RedBlackTree {

/**
* Inserts a node at its proper position and makes it red
*
* @param tree the red black tree
* @param z    the node to insert
*/
void insert(tree, z) {
y = nil[tree] //initialize reference
x = tree.root
while (x != nil[tree]) {
y = x
if (z.key < x.key) x = x.left
else x = x.right
}
//we've found the proper location; add z as child
z.parent = y
if (y == nil[tree]) tree.root = z
else if (z.key < y.key) y.leftChild = z
else y.rightChild = z

/*
* Proceed to check the three cases below
* and continue "fixing" until the root is reached
* Grandparent of x is x.parent.parent
* Uncle of x is sibling of x.parent (other child of grandparent)
*/
}

/**
* Uncle is red;
* we will swap the colors of the parent, uncle, and grandparent,
* which brings the conflict one level higher
* we will assume that the grandparent exists
*
* @param z the node to insert
* @return new potential conflict node
*/
Node case1(tree, z) {
//by properties, the grandparent must be black -> swap
z.parent.parent.color = red
//swap colors for red father and uncle
z.parent.color = black
z.uncle.color = black
//Only conflict now is from grandparent and its parent
return z.parent.parent
}

/**
* Uncle is black, so we cannot swap it to red
* We will also assume that z, z.parent, and z.parent.parent are not inline (one is left child, one is right)
* We will now align them through rotation and proceed to case 3)
* In this case, we shall assume that the
* z.parent.parent.leftChild z.parent
* And z.parent.rightChild = z
*
* @param z the node to insert
* @return new potential conflict node
*/
Node case2(z) {
//this is simply a generic rotation
p = z.parent
z.parent = p.parent
p.parent = z
p.rightChild = z.leftChild
z.leftChild = p
return case3(z) //now that we have the proper arrangement
//we will proceed to case3
}

/**
* Uncle is black, z is inline with parent and grandparent
* We shall assume that z and its parent are both left children to their respective parents
*
* @param z the node to insert
* @return new potential conflict node
*/
Node case3(z) {
//swap colors for parent and grandparent
z.parent = black
z.parent.parent = red
rotateRight(z, z.parent) //z will now be the parent
z.color = black //z the local root is now black
z.parent.color = red //z.parent the child is now red
return z
}
}