CSE 331: Review

Main Steps in Algorithm Design Problem Statement Algorithm Real world problem Problem Definition Precise mathematical def “Implementation” Data Structures Analysis Correctness/Run time

Stable Matching Problem Gale-Shaply Algorithm

Stable Marriage problem Set of men M and women W Matching (no polygamy in M X W) Perfect Matching (everyone gets married) Instablity m m w w m’m’w’w’ Preferences (ranking of potential spouses) Stable matching = perfect matching+ no instablity

Gale-Shapley Algorithm Intially all men and women are free While there exists a free woman who can propose Let w be such a woman and m be the best man she has not proposed to w proposes to m If m is free (m,w) get engaged Else (m,w’) are engaged If m prefers w’ to w w remains free Else (m,w) get engaged and w’ is free Output the engaged pairs as the final output At most n 2 iterations O(1) time implementation

GS algorithm: Firefly Edition Mal Wash Simon Inara Zoe Kaylee

GS algo outputs a stable matching Lemma 1: GS outputs a perfect matching S Lemma 2: S has no instability

Proof technique de jour Source: 4simpsons.wordpress.com Proof by contradiction Assume the negation of what you want to prove After some reasoning

Two obervations Obs 1: Once m is engaged he keeps getting engaged to “better” women Obs 2: If w proposes to m’ first and then to m (or never proposes to m) then she prefers m’ to m

Proof of Lemma 2 By contradiction m m w w m’m’w’w’ Assume there is an instability (m,w’) m prefers w’ to w w’ prefers m to m’ w’ last proposed to m’

Contradiction by Case Analysis Depending on whether w’ had proposed to m or not Case 1: w’ never proposed to m w’ prefers m’ to m Assumed w’ prefers m to m’ Source: 4simpsons.wordpress.com By Obs 2

By Obs 1 Case 2: w’ had proposed to m Case 2.1: m had accepted w’ proposal m is finally engaged to w Thus, m prefers w to w’ 4simpsons.wordpress.com Case 2.2: m had rejected w’ proposal m was engaged to w’’ (prefers w’’ to w’) m is finally engaged to w (prefers w to w’’) m prefers w to w’ 4simpsons.wordpress.com By Obs 1

Overall structure of case analysis Did w’ propose to m? Did m accept w’ proposal? 4simpsons.wordpress.com

Graph Searching BFS/DFS

O(m+n) BFS Implementation BFS(s) CC[s] = T and CC[w] = F for every w≠ s Set i = 0 Set L 0 = {s} While L i is not empty L i+1 = Ø For every u in L i For every edge (u,w) If CC[w] = F then CC[w] = T Add w to L i+1 i++ Array Linked List Input graph as Adjacency list Version in KT also computes a BFS tree

An illustration

O(m+n) DFS implementation BFS(s) CC[s] = T and CC[w] = F for every w≠ s Intitialize Q= {s} While Q is not empty Delete the front element u in Q For every edge (u,w) If CC[w] = F then CC[w] = T Add w to the back of Q O(n) O(1) Repeated n u times O(n u ) Repeated at most once for each vertex u Σ u O(n u ) = O(Σ u n u ) = O(m) Σ u O(n u ) = O(Σ u n u ) = O(m) O(1)

A DFS run using an explicit stack

Topological Ordering

Run of TopOrd algorithm

Greedy Algorithms

Interval Scheduling: Maximum Number of Intervals Schedule by Finish Time

End of Semester blues MondayTuesdayWednesdayThursdayFriday Project 331 HW Exam study Party! Write up a term paper Can only do one thing at any day: what is the maximum number of tasks that you can do?

Schedule by Finish Time Set A to be the empty set While R is not empty Choose i in R with the earliest finish time Add i to A Remove all requests that conflict with i from R Return A*=A O(n log n) time sort intervals such that f(i) ≤ f(i+1) O(n) time build array s[1..n] s.t. s[i] = start time for i Do the removal on the fly

The final algorithm MondayTuesdayWednesdayThursdayFriday Project 331 HW Exam study Party! Write up a term paper Order tasks by their END time

Proof of correctness uses “greedy stays ahead”

Interval Scheduling: Maximum Intervals Schedule by Finish Time

Scheduling to minimize lateness MondayTuesdayWednesdayThursdayFriday Project 331 HW Exam study Party! Write up a term paper All the tasks have to be scheduled GOAL: minimize maximum lateness All the tasks have to be scheduled GOAL: minimize maximum lateness

The Greedy Algorithm (Assume jobs sorted by deadline: d 1 ≤ d 2 ≤ ….. ≤ d n ) f=s For every i in 1..n do Schedule job i from s(i)=f to f(i)=f+t i f=f+t i

Proof of Correctness uses “Exchange argument”

Proved the following Any two schedules with 0 idle time and 0 inversions have the same max lateness Greedy schedule has 0 idle time and 0 inversions There is an optimal schedule with 0 idle time and 0 inversions

Shortest Path in a Graph: non- negative edge weights Dijkstra’s Algorithm

Shortest Path problem Input: Directed graph G=(V,E) Edge lengths, l e for e in E “start” vertex s in V Output: All shortest paths from s to all nodes in V s u w 5 s u 5 s u w

Dijkstra’s shortest path algorithm Input: Directed G=(V,E), l e ≥ 0, s in V R = {s}, d(s) =0 While there is a x not in R with (u,x) in E, u in R d’(w) = min e=(u,w) in E, u in R d(u)+l e Pick w that minimizes d’(w) Add w to R d(w) = d’(w) s s w w u u z z x x y y d(s) = s s u u d(u) = w w d(w) = 2 5 x x d(x) = y y d(y) = 3 z z d(z) = 4 Shortest paths

Dijkstra’s shortest path algorithm (formal) Input: Directed G=(V,E), l e ≥ 0, s in V S = {s}, d(s) =0 While there is a v not in S with (u,v) in E, u in S Pick w that minimizes d’(w) Add w to S d(w) = d’(w) At most n iterations O(m) time O(mn) time bound is trivial O(m log n) time implementation is possible

Proved that d’(v) is best when v is added