Efficiency of Algorithms Csci 107 Lecture 8 Last

Efficiency of Algorithms Csci 107 Lecture 8

• Last time – Data cleanup algorithms and analysis – (1), (n 2) • Today – Binary search and analysis – Order of magnitude (lg n) – Sorting • Selection sort

Searching • Problem: find a target in a list of values • Sequential search – Best-case : (1) comparison • target is found immediately – Worst-case: (n) comparisons • Target is not found, or is the last element in the list – Average-case: (n) comparisons • Target is found in the middle • Can we do better? – No…unless we have the input list in sorted order

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, A 1, A 2, …. 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 – 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? – Found in the middle • What is the worst case? – Initially the size of the list is n – After the first iteration through the repeat loop, if not found, then either start = middle or end = middle ==> 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?

log 2 x • log 2 x – The number of times you can half a (positive) number x before it goes below 1 – Examples: • log 2 16 = 4 [16/2=8, 8/2=4, 4/2=2, 2/2=1] • log 2 n = m <==> 2 m = n • log 2 8 = 3 <==> 23=8

log 2 x Increases very slowly • log 2 8 = 3 • log 2 32 = 5 • log 2 128 = 7 • log 2 1024 = 10 • log 2 1000000 = 20 • log 2 100000 = 30 • …

Orders of magnitude • Order of magnitude ( lg n) – Worst-case efficiency of binary search: ( lg n) • Comparing order of magnitudes (1) << (lg n) << (n 2)

Comparing (lg n) and (n) Does efficiency matter? • Say n = 109 (1 billion elements) • 10 MHz computer ==> 1 instr takes 10 -7 sec – Seq search would take • (n) = 109 x 10 -7 sec = 100 sec – Binary search would take • (lg n) = lg 109 x 10 -7 sec = 30 x 10 -7 sec = 3 microsec

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. 5 n 2 -. 5 n (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: (n 2) irrespective of the input • Other sorting algorithms? Can we find more efficient sorting algorithms?

Exam 1 • Wednesday: Lab 4 (more efficiency, binary search, etc) – Due next Wednesday, but… – Strongly encouraged to finish lab before Exam 1 • Exam 1 (Monday febr 21 st) – – Material: Algorithms and efficiency Open books, notes, labs Practice problems handout Study group: this time only, Sunday night (instead of Monday) • Tom will email