Transcript Document
Efficiency of Algorithms February 19th Today • Binary search – Algorithm and analysis • Order-of-magnitude analysis of algorithm efficiency – Review • Sorting – Selection sort Searching a sorted list • Problem: find a target in a sorted list – How can we exploit that the list is sorted, and come up with an algorithm faster than sequential search in the worst case? – How do we search in a phone book? – Can we come up with an algorithm? • Check the middle value • If smaller than target, go right • Otherwise go left Binary search • • • • Get values for list, A1, A2, ….An, n , target Set start =1, set end = n Set found = NO Repeat until ?? – Set m = middle value between start and end – If target = m then • Print target found at position m • Set found = YES – Else if target < Am then end = m-1 Else start = m+1 • If found = NO then print “Not found” • End Efficiency of binary search • What is the best case? • What is the worst case? – Initially the size of the list in n – After the first iteration through the repeat loop, if not found, then either start = m or end = m ==> size of the list on which we search is n/2 – Every time in the repeat loop the size of the list is halved: n, n/2, n/4,…. – How many times can a number be halved before it reaches 1? Orders of magnitude • Order of magnitude ( lg n) – Worst-case efficiency of binary search: ( lg n) • Comparing order of magnitudes (1) << (lg n) << (n) << (n2) Sorting • Problem: sort a list of items into alphabetical or numerical order • Why sorting? – Sorting is ubiquitous (very common)!! – Examples: • Registrar: Sort students by name or by id or by department • Post Office: Sort mail by address • Bank: Sort transactions by time or customer name or accound number … • For simplicity, assume input is a list of n numbers • Ideas for sorting? Selection Sort • Idea: grow a sorted subsection of the list from the back to the front 57216483| 5721643|8 521643|78 52134|678 … |1 2 3 4 5 6 7 8 Selection Sort • Pseudocode (at a high level of abstraction) – Get values for n and the list of n items – Set marker for the unsorted section at the end of the list – Repeat until unsorted section is empty • Select the largest number in the unsorted section of the list • Exchange this number with the last number in unsorted section of list • Move the marker of the unsorted section forward one position – End Selection Sort • Level of abstraction – It is easier to start thinking of a problem at a high level of abstraction • Algorithms as building blocks – We can build an algorithm from “parts” consisting of previous algorithms – Selection sort: • Select largest number in the unsorted section of the list • We have seen an algorithm to do this last time • Exchange 2 values Selection Sort Analysis • Iteration 1: – Find largest value in a list of n numbers : n-1 comparisons – Exchange values and move marker • Iteration 2: – Find largest value in a list of n-1 numbers: n-2 comparisons – Exchange values and move marker • Iteration 3: – Find largest value in a list of n-2 numbers: n-3 comparisons – Exchange values and move marker • … • Iteration n: – Find largest value in a list of 1 numbers: 0 comparisons – Exchange values and move marker Total: (n-1) + (n-2) + …. + 2 + 1 Selection Sort • Total work (nb of comparisons): – (n-1) + (n-2) + …. + 2 + 1 – This sum is equal to .5n2 -.5n (proved by Gauss) => order of magnitude is ( ? ) • Questions – best-case, worst-case ? – we ignored constants, and counted only comparisons.. Does this make a difference? • Space efficiency – extra space ? Selection Sort • In conclusion: Selection sort – Space efficiency: • No extra space used (except for a few variables) – Time efficiency • There is no best-case and worst-case • the amount of work is the same: (n2) irrespective of the input • Other sorting algorithms? Can we find more efficient sorting algorithms? This week… • Wednesday: Lab 3 • Thursday: Exam 1 • Material: Algorithms • Chapter 1, 2 & 3 from textbook • For next time – Review chapter 1,2 & 3 – Look over practice exam