Lecture 9 CSE 331 Sep 15, 2014.

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Presentation transcript:

Lecture 9 CSE 331 Sep 15, 2014

Run time of an algorithm (Worst-case) run time T(n) for input size n Maximum number of steps taken by the algorithm for any input of size n

g(n) is O(f(n)) c*f(n) for some c>0 g(n) n n0

g(n) is Ω(f(n)) g(n) ε*f(n) for some ε>0 n n1

Reading Assignments Sections 1.1, 1.2, 2.1, 2.2 and 2.4 in [KT]

Questions?

Today’s agenda Asymptotic run time Analyzing the run time of the GS algo

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

Implementation Steps How do we represent the input? How do we find a free woman w? How would w pick her best unproposed man m? How do we know who m is engaged to? How do we decide if m prefers w’ to w?

Arrays and Linked Lists Front Last n numbers a1,a2,…,an a1 a2 a3 an 1 2 3 n Array Linked List Access ith element O(1) O(i) Is e present? O(n) (O(log n) if sorted) O(n) Insert an element O(n) O(1) given pointer Delete an element O(n) O(1) given pointer Static vs Dynamic Static Dynamic

Today’s agenda O(n2) implementation of the Gale-Shapley algorithm More practice with run time analysis

Gale-Shapley Algorithm At most n2 iterations 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 O(1) time implementation 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