Average Case Analysis of Quicksort: Probability of Adjustments and Comparisons - Prof. Mic, Study notes of Algorithms and Programming

Download Average Case Analysis of Quicksort: Probability of Adjustments and Comparisons - Prof. Mic and more Study notes Algorithms and Programming in PDF only on Docsity! Algorithm Design and Analysis 2/23/2008 Lecture 18 - Average case analysis of Quicksort Instructor: Mike Kowalczyk 1 Analysis of the random usher, and the connection to Quicksort (This lecture is a continuation of lecture 16) Suppose we have k people seated already in a row of n people, and we seat the next person. What’s the probability that some fixed person who’s already seated has to adjust? The probability will be 1n−k if that person has “contiguous seating to one end”, and 2 n−k otherwise. How many adjustments do we expect to see overall? One upper bound is to say on each seating, there can be at most k adjustments, when there’s k people already seated. This leads us to ∑n−1 k=0 k = Θ(n 2) adjustments in the worst case. More precisely, if there’s k people already seated, and 2n−k chance of each getting adjusted (using the worst case here), then total number of expected adjustments is 2kn−k . To find the total adjustments overall, we can just use a summation. Note that the second equality below holds by reversing the series (write out both series longhand to convince yourself that the second step is correct). n−1∑ k=0 2k n− k = 2 n−1∑ k=1 k n− k = 2 n−1∑ k=1 n− k k = 2 n−1∑ k=1 (n k − 1 ) = −2(n− 1) + 2 n−1∑ k=1 n k = −2(n− 1) + 2n n−1∑ k=1 1 k ∼ −2(n− 1) + 2n ∫ n−1 k=1 1 k ∼ −2n+ 2n ln(n) = θ(n lg n) So Θ(n2) steps in the worst case, and θ(n lg n) steps on average. Isn’t Quicksort supposed to have a similar behavior? Let’s take another look at Quicksort: 1