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Efficiency of Algorithms. Csci 107 Lecture 8. Last time Data cleanup algorithms and analysis (1), (n), (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
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Efficiency of Algorithms Csci 107 Lecture 8
Last time • Data cleanup algorithms and analysis • (1), (n), (n2) • 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, 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 • 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?
log2 x • log2 x • The number of times you can half a (positive) number x before it goes below 1 • Examples: • log2 16 = 4 [16/2=8, 8/2=4, 4/2=2, 2/2=1] • log2 n = m <==> 2m = n • log2 8 = 3 <==> 23=8
log2 x Increases very slowly • log2 8 = 3 • log2 32 = 5 • log2 128 = 7 • log2 1024 = 10 • log2 1000000 = 20 • log2 1000000000 = 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) << (n2)
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 x10-7 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 5 7 2 1 6 4 8 3 | 5 7 2 1 6 4 3 |8 5 2 1 6 4 3 | 7 8 5 2 1 3 4 | 6 7 8 … |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?
Exam 1 • Wednesday: Lab 4 (more efficiency, binary search, etc) • Due next Wednesday, but… • Strongly encouraged to finish lab before Exam1 • Exam 1 (Monday febr 21st) • Material: Algorithms and efficiency • Open books, notes, labs • Practice problems handout • Study group: this time only, Sunday night (instead of Monday) • Tom will email